SOME GENERALIZATION IN GEOMETRY
P.D. Barry.
c
Copyright
April 2, 2015
ii
Contents
1 Geometric algebra for a Euclidean plane
1.1 Sensed-area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Sensed-area of a triple of points . . . . . . . . . . . . . . .
1.1.2 Menelaus’ theorem . . . . . . . . . . . . . . . . . . . . . .
1.2 Dual sensed-area . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Point of intersection of two lines . . . . . . . . . . . . . .
1.2.2 Properties of dual sensed-area . . . . . . . . . . . . . . . .
1.2.3 Further identification of dual sensed-area . . . . . . . . .
1.2.4 Linear functions . . . . . . . . . . . . . . . . . . . . . . .
1.3 More material related to Ceva’s theorem . . . . . . . . . . . . . .
1.3.1 Coordinates for the point of concurrency . . . . . . . . . .
1.3.2 Non-concurrent lines . . . . . . . . . . . . . . . . . . . . .
1.4 Projectivities for ranges of points . . . . . . . . . . . . . . . . . .
1.4.1 Central perspectivities . . . . . . . . . . . . . . . . . . . .
1.4.2 Parallel perspectivities . . . . . . . . . . . . . . . . . . . .
1.4.3 A property common to central and parallel perspectivities
1.4.4 Projectivities . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.5 Composition and decomposition of projectivities . . . . .
1.5 Quadrangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.2 Definition of sensed area of quadruple . . . . . . . . . . .
1.5.3 Linking sensed-areas of quadruples and triples . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
1
1
4
5
5
7
8
9
10
10
11
12
12
13
14
14
16
19
19
21
23
2 Pencils of lines
2.1 Pencils of lines . . . . . . . . . . . . . . .
2.1.1 A basic feature of dual sensed-area
2.1.2 A pencil of lines . . . . . . . . . .
2.1.3 Lines in two pencils . . . . . . . .
2.1.4 An example for §2.1.3 . . . . . . .
2.2 Quadrilateral . . . . . . . . . . . . . . . .
2.2.1
. . . . . . . . . . . . . . . . . . .
2.2.2 Pairs of opposite vertices . . . . .
2.3 Projectivities for pencils of pairs . . . . .
2.3.1
. . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
27
27
27
29
30
31
31
31
33
34
34
3 Areal coordinates
3.1 Areal point coordinates . . . . . . . . . . . . . .
3.1.1 Areal point coordinates . . . . . . . . . .
3.1.2 Sensed-area in terms of areal coordinates
3.2 Equation of a line in areal coordinates . . . . . .
3.2.1 Preliminary identities . . . . . . . . . . .
3.2.2 Equation of a line . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
37
37
37
38
38
38
39
iii
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
iv
CONTENTS
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
39
40
41
41
43
43
44
44
46
46
4 Projective and affine transformations
4.1 Special transformations . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Transformations depending on intersecting lines or parallel lines
4.1.2 Definition of projective and affine transformations . . . . . . . .
4.1.3 Transformations in Cartesian coordinates . . . . . . . . . . . . .
4.1.4 Images of lines . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Equipoised quotients; cross-ratio of ranges and pencils . . . . . . . . . .
4.2.1 Equipoised quotients . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 Values of permutations of Magnus quotient . . . . . . . . . . . .
4.2.3 Cross-ratio of a range of collinear points . . . . . . . . . . . . . .
4.2.4 Cross-ratio of a pencil of concurrent lines . . . . . . . . . . . . .
4.3 Constant-sized angle property of a circle . . . . . . . . . . . . . . . . . .
4.3.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Ptolemy’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Transformations in terms of equipoised quotients . . . . . . . . . . . . .
4.5.1 Affine transformations . . . . . . . . . . . . . . . . . . . . . . . .
4.5.2 Projective transformations . . . . . . . . . . . . . . . . . . . . . .
4.6 Perspectivities and projectivities as restrictions . . . . . . . . . . . . . .
4.6.1 Perspectivities . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.2 Projectivities . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 Quadrangle; diagonal triangle . . . . . . . . . . . . . . . . . . . . . . . .
4.7.1 Harmonic property of diagonal triangle . . . . . . . . . . . . . .
4.7.2 The diagonal triple . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8 Sensed-area and sensed-angles . . . . . . . . . . . . . . . . . . . . . . . .
4.8.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
49
49
49
51
52
52
53
55
55
56
57
58
59
59
60
61
61
62
62
63
63
63
64
66
66
66
67
67
5 Line equations of points; dual concepts
5.1 Areal pair-coordinates; line equation of a point . . . . . .
5.1.1 Line equation of a point . . . . . . . . . . . . . . .
5.1.1 Equation of a parallel pencil . . . . . . . . . . . . .
5.1.2 Linear expressions in areal pair-coordinates . . . .
5.1.3 Determinant form of equation . . . . . . . . . . . .
5.1.4 Line on two points . . . . . . . . . . . . . . . . . .
5.1.5 Parametric point equations . . . . . . . . . . . . .
5.1.6 Pair-coordinates under change of triple of reference
5.1.7 Illustration using multiples of line-coordinates . . .
5.2 Properties and identities for dual sensed-area . . . . . . .
5.2.1 Equipoised quotients of dual sensed-areas . . . . .
5.3 Parametric equations with pair-coordinates . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
73
73
73
74
76
77
78
79
79
79
80
80
80
3.3
3.4
3.2.3 Coefficient triples . . . . . . . . . . . . . . . .
3.2.4 Determinant form of equation . . . . . . . . .
3.2.5 Point of intersection of two lines . . . . . . .
3.2.6 Parallel lines . . . . . . . . . . . . . . . . . .
3.2.7 Identifying a line for drawing . . . . . . . . .
3.2.8 Parametric equations of a line . . . . . . . . .
Areal coordinates under change of triple of reference
3.3.1 Change of triple of reference . . . . . . . . . .
Reducible quadratic equations . . . . . . . . . . . . .
3.4.1
. . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
v
CONTENTS
5.3.1
5.3.2
5.3.3
Dual of parametric equations of a line . . . . . . . . . . . . . . . . . . . . .
The question rephrased . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The question answered . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Grassmann properties of rotors; geometrical applications
6.1 Rotors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Definition of rotors . . . . . . . . . . . . . . . . . . . . .
6.1.2 Multiplication of a rotor by a number . . . . . . . . . .
6.1.3 Associativity . . . . . . . . . . . . . . . . . . . . . . . .
6.1.4 Addition of rotors . . . . . . . . . . . . . . . . . . . . .
6.1.5 Commutativity of addition . . . . . . . . . . . . . . . .
6.1.6 Additive identity . . . . . . . . . . . . . . . . . . . . . .
6.1.7 Addition associative . . . . . . . . . . . . . . . . . . . .
6.1.8 Subtraction of rotors . . . . . . . . . . . . . . . . . . . .
6.1.9 First distributive law . . . . . . . . . . . . . . . . . . . .
6.1.10 Second distributive law . . . . . . . . . . . . . . . . . .
6.1.11 Moment of a force . . . . . . . . . . . . . . . . . . . . .
6.1.12 Carrier of a rotor . . . . . . . . . . . . . . . . . . . . . .
6.2 Dual sensed-area and rotors . . . . . . . . . . . . . . . . . . . .
6.2.1 Dual sensed-area and rotors . . . . . . . . . . . . . . . .
6.2.2 Areal pair-coordinates and rotors . . . . . . . . . . . . .
6.2.3 Dual sensed-area and product of a rotor by a scalar . .
6.2.4 Dual sensed-area and addition of rotors . . . . . . . . .
6.2.5 A basic feature of dual sensed-area extended . . . . . .
6.3 Pencils of rotors . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3.2 Amenable pairs of pairs of points . . . . . . . . . . . . .
6.3.3 Rotors common to two pencils . . . . . . . . . . . . . .
6.3.4 Dual-parallelism of pencils . . . . . . . . . . . . . . . . .
6.3.5 Formula for a rotor common to two pencils of rotors . .
6.4 Rotors as duals of points . . . . . . . . . . . . . . . . . . . . . .
6.4.1 Sensed distance and distance from one rotor to another
6.4.2 Division in a given ratio . . . . . . . . . . . . . . . . . .
6.4.3 Dual of line, half-line and segment . . . . . . . . . . . .
6.4.4 Sensed-angle in dual situation . . . . . . . . . . . . . . .
6.5 Geometrical applications of rotors . . . . . . . . . . . . . . . .
6.5.1 Analogue of isosceles triangle . . . . . . . . . . . . . . .
6.5.2 Dual of Pythagoras’ theorem . . . . . . . . . . . . . . .
6.5.3 Dual parallelograms . . . . . . . . . . . . . . . . . . . .
6.5.4 Mid-rotors of opposite pairs of vertex rotors . . . . . . .
6.5.5 Menelaus’ theorem for rotors and pencils . . . . . . . .
6.6 Dual of angle property of circles . . . . . . . . . . . . . . . . . .
6.6.1 Original property of circle . . . . . . . . . . . . . . . . .
6.6.2 Point of contact of tangents to a circle . . . . . . . . . .
6.6.3 Dual of angle property . . . . . . . . . . . . . . . . . . .
6.7 Double rotors . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.7.1 Equality of two couples of rotors . . . . . . . . . . . . .
6.7.2 Scalar multiplication of a double rotor . . . . . . . . . .
6.7.3 Affine line property for double rotors . . . . . . . . . . .
6.7.4 Affine line property for doubly dual sensed-area . . . . .
6.7.5 Menelaus’ theorem for double rotors and ranges . . . . .
6.7.6 Ceva’s theorem for rotors and pencils . . . . . . . . . .
6.8 Miscellaneous topics . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
80
81
82
89
89
89
89
90
90
92
92
92
93
96
96
96
97
97
97
97
97
97
98
98
98
98
100
101
104
105
105
106
106
106
109
109
111
112
114
115
117
117
118
119
120
120
120
121
123
123
124
125
vi
CONTENTS
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
125
126
127
127
127
7 Conic sections
7.1 Definition of a conic section . . . . . . . . . . . . . . . . . . . . . . . .
7.1.1 Preparatory examples . . . . . . . . . . . . . . . . . . . . . . .
7.1.2 Steiner’s definition of a conic . . . . . . . . . . . . . . . . . . .
7.1.3 Ambiguous cases . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.4 Conic through the points of the triple of reference . . . . . . .
7.1.5 Two-term form of equation . . . . . . . . . . . . . . . . . . . .
7.1.6 Parametric equations for a conic . . . . . . . . . . . . . . . . .
7.2 Intersection of a line and a conic . . . . . . . . . . . . . . . . . . . . .
7.2.1 Intersection of a line and a conic; interior and exterior regions .
7.2.2 Parametric equation for the line; equation of incidence . . . . .
7.2.3 Tangent to a conic . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.4 Some aspects of tangents . . . . . . . . . . . . . . . . . . . . .
7.2.5 Second parametric form for an incident line; polar of a point .
7.2.6 Centre of a conic . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.7 A symmetrical relation; conjugate points . . . . . . . . . . . . .
7.2.8 Centre not on a polar . . . . . . . . . . . . . . . . . . . . . . .
7.2.9 What lines are polars? Poles . . . . . . . . . . . . . . . . . . .
7.3 Lines of deficient incidence; asymptotes of hyperbola . . . . . . . . . .
7.3.1 No lines of deficient incidence for an ellipse/circle . . . . . . . .
7.3.2 Lines of deficient incidence for a parabola . . . . . . . . . . . .
7.3.3 Lines of deficient incidence for a hyperbola; asymptotes . . . .
7.3.4 Intersection of a line and a proper conic . . . . . . . . . . . . .
7.3.5 More on tangents . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.6 Intersection of a polar with a conic . . . . . . . . . . . . . . . .
7.3.7 Parallel polars . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.8 Self-conjugate triples or triangles . . . . . . . . . . . . . . . . .
7.4 Conjugate diametral lines . . . . . . . . . . . . . . . . . . . . . . . . .
7.4.1 Locus of mid-points of chords on parallel lines . . . . . . . . . .
7.4.2 Incidence of diametral line with central conic . . . . . . . . . .
7.4.3 Conjugate diametral lines . . . . . . . . . . . . . . . . . . . . .
7.5 Newton’s property of conics . . . . . . . . . . . . . . . . . . . . . . . .
7.5.1 Newton’s property . . . . . . . . . . . . . . . . . . . . . . . . .
7.6 Classical Greek property of conics . . . . . . . . . . . . . . . . . . . .
7.6.1 Classical Greek property for an ellipse or hyperbola . . . . . .
7.6.2 Classical Greek property for a parabola . . . . . . . . . . . . .
7.7 An affine characterisation of conics . . . . . . . . . . . . . . . . . . . .
7.7.1 A re-working of the Greek properties . . . . . . . . . . . . . . .
7.7.2 An affine characterization of conics . . . . . . . . . . . . . . . .
7.8 Axes; standard forms of equation . . . . . . . . . . . . . . . . . . . . .
7.8.1 Perpendicular conjugate diametral lines . . . . . . . . . . . . .
7.8.2 Standard forms of equation . . . . . . . . . . . . . . . . . . . .
7.8.3 Equation of a circle . . . . . . . . . . . . . . . . . . . . . . . .
7.8.4 Equations of ellipse and hyperbola . . . . . . . . . . . . . . . .
7.8.5 Equation of parabola . . . . . . . . . . . . . . . . . . . . . . . .
7.9 Images of conics under projective and affine transformations . . . . . .
7.9.1 Image of a conic under projective and affine transformations .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
129
129
129
132
133
135
136
137
141
142
143
145
147
148
149
151
151
151
153
153
153
155
157
160
161
164
164
165
165
167
167
170
170
171
171
172
174
174
175
176
176
177
179
180
181
182
182
6.9
6.8.1 Dual of transversals of a pencil . . . . .
6.8.2 Quadrilaterals . . . . . . . . . . . . . .
6.8.3 Sensed area associated with a quadruple
Identities . . . . . . . . . . . . . . . . . . . . .
6.9.1
. . . . . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
of points
. . . . . .
. . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
CONTENTS
vii
7.9.2
7.9.3
7.9.4
Conics through four or five points . . . . . . . . . . . . . . . . . . . . . . . 182
The Chasles-Steiner theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 184
Image of a circle under some projective transformations . . . . . . . . . . . 185
8 Dual conics
8.1 Point-pair and rotor conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.1 Dual of definition of a conic . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.2 Point-pair conic containing pairs of vertices of triple of reference . . . . . .
8.1.3 Two-term form of equation . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.4 Equation of point-pairs in a tangent line to a conic . . . . . . . . . . . . . .
8.1.5 Sketching point-pair conics . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Incidence of a pencil and a point-pair conic . . . . . . . . . . . . . . . . . . . . . .
8.2.1 Intersection of a pencil and a point-pair conic; interior and exterior regions
8.2.2 Equation of incidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.3 Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.4 Deficient incidence in dual situation . . . . . . . . . . . . . . . . . . . . . .
8.3 Duals of diametral lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.1 Analogue of diametral lines . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.2 Dual of Newton property . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4 Normalised areal line coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4.1 Definition of normalized point-pair coordinates; identities . . . . . . . . . .
8.4.2 Dependence of normalized pair-coordinates on lines . . . . . . . . . . . . . .
8.4.3 Equation of a line; coefficient triples . . . . . . . . . . . . . . . . . . . . . .
8.4.4 Equation of a line through two points . . . . . . . . . . . . . . . . . . . . .
8.4.5 Point of intersection of two lines . . . . . . . . . . . . . . . . . . . . . . . .
8.4.6 Parallel lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4.7 Equation of point-pairs in a concurrent or parallel pencil . . . . . . . . . . .
8.4.8 Pair of points common to two pencils . . . . . . . . . . . . . . . . . . . . .
8.4.9 Dual-parallelism of concurrent pencils . . . . . . . . . . . . . . . . . . . . .
8.4.10 Determinant form of equation of pencil . . . . . . . . . . . . . . . . . . . .
8.4.11 Parametric equations of point-pairs in a concurrent pencil . . . . . . . . . .
8.4.12 Correspondence of our two parametrisations . . . . . . . . . . . . . . . . . .
8.4.13 Parametric equations of pairs in a parallel pencil . . . . . . . . . . . . . . .
8.5 Further dual concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.5.1 Dual harmonic pencil and dual mid-pair . . . . . . . . . . . . . . . . . . . .
8.5.2 Dual mid-line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.5.3 Sensed-distance between point-pairs in a concurrent pencil . . . . . . . . . .
8.5.4 Parametric equations of dual-parallel concurrent pencils . . . . . . . . . . .
8.5.5 Sensed-distances between point-pairs in a parallel pencil . . . . . . . . . . .
8.5.6 Change of triple of reference . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6 Z0 -deleted point-pair conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6.1 Deleted point-pair conics . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6.2 Incidence equation; type of Z0 -deleted point-pair conic . . . . . . . . . . . .
8.6.3 Contact for a Z0 -deleted point-pair conic . . . . . . . . . . . . . . . . . . .
8.6.4 Pole of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6.5 Dual-centre of a point-pair conic . . . . . . . . . . . . . . . . . . . . . . . .
8.6.6 Locus of mid-pairs of pairs in a point-pair conic which are dual-parallel . .
8.6.7 Conjugate dual-diametral concurrent pencils . . . . . . . . . . . . . . . . .
8.6.8 Second dual of Newton’s property . . . . . . . . . . . . . . . . . . . . . . .
8.6.9 Dual of classical Greek property . . . . . . . . . . . . . . . . . . . . . . . .
189
189
189
190
190
192
194
195
195
196
197
199
200
200
201
202
202
204
205
205
205
206
206
207
207
208
208
210
211
212
212
213
214
215
218
218
218
218
219
219
220
220
221
222
224
228
viii
CONTENTS
9 Affine methods and results
9.1 Affine transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.1 Affine transformations recalled . . . . . . . . . . . . . . . . . . . .
9.2 Geometrical affine invariants . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.1 Lines, segments and half-lines . . . . . . . . . . . . . . . . . . . . .
9.2.2 Pairs of lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.3 Closed half-planes . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.4 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.5 Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3 Behaviour of distance and sensed-distance . . . . . . . . . . . . . . . . . .
9.3.1 Magnification ratio . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3.2 Distances on parallel lines . . . . . . . . . . . . . . . . . . . . . . .
9.3.3 Sensed distances . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3.4 Mid-points, centroids and harmonic ranges . . . . . . . . . . . . .
9.3.5 Ratio of sensed-areas, orientation of triples of non-collinear points
9.4 Deduction of properties of ellipses . . . . . . . . . . . . . . . . . . . . . .
9.4.1 Ellipse as image of a circle . . . . . . . . . . . . . . . . . . . . . . .
9.4.2 Chord, centre, diameter . . . . . . . . . . . . . . . . . . . . . . . .
9.4.3 Tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5 Results for ellipses, based mainly on distance and sensed distance . . . . .
9.5.1 Locus of mid-points . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5.2 Conjugate diameters . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5.3 Elliptic analogue of Grassmann supplement for a position vector .
9.5.4 Classical Greek property . . . . . . . . . . . . . . . . . . . . . . . .
9.5.5 Axes of ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5.6 Chord joining endpoints of conjugate diameters . . . . . . . . . . .
9.5.7 Inscribed triangle with centroid at centre . . . . . . . . . . . . . .
9.5.8 An elliptic Pythagoras property . . . . . . . . . . . . . . . . . . . .
9.5.9 Tangents from an exterior point . . . . . . . . . . . . . . . . . . .
9.5.10 Products of lengths of segments . . . . . . . . . . . . . . . . . . . .
9.5.11 A tangent cutting a pair of parallel tangents . . . . . . . . . . . .
9.6 Results on ellipses, based mainly on angle-measure . . . . . . . . . . . . .
9.6.1 Behaviour of magnitude of an angle . . . . . . . . . . . . . . . . .
9.6.2 Mid-line of an angle-support . . . . . . . . . . . . . . . . . . . . .
9.6.3 Analogue of orthocentre . . . . . . . . . . . . . . . . . . . . . . . .
9.6.4 Lines parallel to conjugate diametral lines . . . . . . . . . . . . . .
9.6.5 Analogue of Euler line . . . . . . . . . . . . . . . . . . . . . . . . .
9.6.6 Analogue of Wallace-Simson line . . . . . . . . . . . . . . . . . . .
9.6.7 Analogue of constant-sized angle property of a circle . . . . . . . .
9.6.8 Analogue of incentre . . . . . . . . . . . . . . . . . . . . . . . . . .
9.7 Results on ellipses, based mainly on sensed-areas . . . . . . . . . . . . . .
9.7.1 Elliptic sine and cosine . . . . . . . . . . . . . . . . . . . . . . . . .
9.7.2 Sum of squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.7.3 Elliptic addition of angles . . . . . . . . . . . . . . . . . . . . . . .
9.7.4 The duplication formula . . . . . . . . . . . . . . . . . . . . . . . .
9.7.5 Analogue of areal form of Pythagoras’ theorem . . . . . . . . . . .
9.8 Some results of Appolonius’ recalled . . . . . . . . . . . . . . . . . . . . .
9.8.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.8.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.8.3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.9 Analogous results for hyperbolas . . . . . . . . . . . . . . . . . . . . . . .
9.9.1 Triangle with centroid at centre of hyperbola . . . . . . . . . . . .
9.9.2 Hyperbolic theorem of Pythagoras . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
229
229
229
230
230
231
231
231
232
232
232
233
234
234
235
236
236
236
237
238
238
238
238
241
241
242
243
244
244
246
247
247
247
249
249
250
251
252
254
254
256
256
257
257
257
258
258
259
259
260
261
261
262
ix
CONTENTS
9.9.3 Analogue for hyperbola of constant-sized angle property of circle
9.9.4 Hyperbolic sine and cosine . . . . . . . . . . . . . . . . . . . . .
9.9.5 Hyperbolic analogue of Grassmann’s supplement . . . . . . . . .
9.10 Oblique analogues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.10.1 Oblique axial symmetries . . . . . . . . . . . . . . . . . . . . . .
9.10.2 Oblique analogue of rotation . . . . . . . . . . . . . . . . . . . .
9.10.3 A three-axes result . . . . . . . . . . . . . . . . . . . . . . . . . .
9.10.4 Oblique frame of reference . . . . . . . . . . . . . . . . . . . . . .
9.10.5 Conjugate directions . . . . . . . . . . . . . . . . . . . . . . . . .
9.11 Generalisation to ellipses not in normal position . . . . . . . . . . . . . .
9.11.1 Analogue of inner product and supplement . . . . . . . . . . . .
9.11.2 Grassmann supplement as conjugate semi-diameter . . . . . . . .
9.11.3 Components along a vector and its generalized supplement . . .
9.11.4 Analogue of similar triangles . . . . . . . . . . . . . . . . . . . .
10 Projective methods and results
10.1 Maclaurin property of conics . . . . . . . . . . . . . .
10.1.1
. . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Involution . . . . . . . . . . . . . . . . . . . . . . . . .
10.2.1 Definition and some basic occurrences . . . . .
10.2.2 Desargues’ involution theorem . . . . . . . . .
10.2.3 The lines through a point, intersecting a conic
10.2.4 Conjugate points lying on one line . . . . . . .
10.2.5 Conjugate diametral lines . . . . . . . . . . . .
10.3 Pascal’s and Carnot’s theorems . . . . . . . . . . . . .
10.3.1 Pascal’s theorem . . . . . . . . . . . . . . . . .
10.3.2 Carnot’s theorem (1803) . . . . . . . . . . . . .
10.4 Identification of coefficients in equations of conics . . .
10.4.1
. . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
264
266
267
268
268
268
269
269
270
271
271
272
273
273
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
275
275
275
276
276
277
279
280
280
280
280
281
283
283
11 Similarity methods and results
11.1 Similarity transformations . . . . . . . . . . . . . . . . . . .
11.1.1 Definition of similarity transformations . . . . . . .
11.1.2 Similarity invariants . . . . . . . . . . . . . . . . . .
11.2 Specialised similarity transformations . . . . . . . . . . . . .
11.2.1 Coefficients in a similarity transformation . . . . . .
11.2.2 Isometries . . . . . . . . . . . . . . . . . . . . . . . .
11.2.3 Similarity non-invariants which are metric invariants
11.2.4 Dilatations . . . . . . . . . . . . . . . . . . . . . . .
11.2.5 Enlargements . . . . . . . . . . . . . . . . . . . . . .
11.2.6 Translations . . . . . . . . . . . . . . . . . . . . . . .
11.3 Eccentricity, focus-directrix properties . . . . . . . . . . . .
11.3.1 Eccentricity, focus, directrix . . . . . . . . . . . . . .
11.3.2 Images of conics under similarity transformations . .
11.3.3 Reference to other text-books . . . . . . . . . . . . .
11.3.4 Sum or difference of focal distances . . . . . . . . . .
11.3.5 Another way of generating central conics . . . . . .
11.3.6 An other characterization of a focus . . . . . . . . .
11.4 Proto-foci . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.4.1 Proto-foci in the case of an ellipse . . . . . . . . . .
11.5 Similar and similarly situated conics . . . . . . . . . . . . .
11.5.1 Condition on coefficients . . . . . . . . . . . . . . . .
11.5.2 Images of two circles under one affine transformation
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
285
285
285
285
287
287
288
288
288
289
289
290
290
291
291
292
293
294
294
294
296
296
298
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
x
CONTENTS
11.6 Analogue of power property of a circle; radical axis . . . . . . . . . . .
11.6.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.7 Miscellaneous results . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.8 Two-conic analogue of Pascal’s theorem . . . . . . . . . . . . . . . . .
11.8.1 Locus with respect to four lines . . . . . . . . . . . . . . . . . .
11.8.2 Point-pairs in lines tangent to a conic . . . . . . . . . . . . . .
11.8.3 Dual result for pairs on tangent lines to a conic again . . . . .
11.9 Cartesian and Plücker duality . . . . . . . . . . . . . . . . . . . . . . .
11.9.1 Cartesian duality . . . . . . . . . . . . . . . . . . . . . . . . . .
11.9.2 Relationship between Cartesian and areal dual coordinates . .
11.9.3 Derivation of formulae . . . . . . . . . . . . . . . . . . . . . . .
11.9.4 Tangential equations for a circle and other conics . . . . . . . .
11.9.5 Plücker duality . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.9.6 Features of Plücker reciprocation . . . . . . . . . . . . . . . . .
11.9.7 Plücker reciprocal of focus-directrix property of ellipse . . . . .
11.9.8 Plücker reciprocal of sensed area . . . . . . . . . . . . . . . . .
11.9.9 Polar reciprocation . . . . . . . . . . . . . . . . . . . . . . . . .
11.9.10 Conics and conic envelopes as images under polar reciprocation
11.9.11 A tangent to a conic meeting the directrix . . . . . . . . . . . .
11.9.12 Analogue of constant-sized angle property of a circle . . . . . .
11.9.13 Five tangents to a conic . . . . . . . . . . . . . . . . . . . . . .
11.10Nine-point homothetic conic . . . . . . . . . . . . . . . . . . . . . . . .
11.10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.10.2 Conjugate directions . . . . . . . . . . . . . . . . . . . . . . . .
11.10.3 Nine-point homothetic conics . . . . . . . . . . . . . . . . . . .
11.10.4 Miquel’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . .
11.10.5 First Lemoine circle . . . . . . . . . . . . . . . . . . . . . . . .
11.10.6 Components along a vector and its supplement . . . . . . . . .
12 The Complex Euclidean Plane
12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.1.1 Complexification . . . . . . . . . . . . . . .
12.1.2 The trigonometric tangent function . . . . .
12.2 Initial geometry of the complex Euclidean plane . .
12.2.1 Complex sensed-area and areal coordinates
12.2.2 Complex lines . . . . . . . . . . . . . . . . .
12.2.3 Parallel lines . . . . . . . . . . . . . . . . .
12.2.4 Perpendicular lines . . . . . . . . . . . . . .
12.2.5 Complex axial symmetries . . . . . . . . . .
12.3 Hesse’s normal form of equation for a line . . . . .
12.3.1 Difficulties due to Square Roots . . . . . . .
12.3.2 Hesse’s normal form of equation for a line .
12.4 Insertion of supplementary material for RE2 . . . .
12.4.1 Use of polar coordinates in RE2 . . . . . . .
12.4.2 Duo-sectors in RE2 . . . . . . . . . . . . . .
12.5 Return to CE2 . . . . . . . . . . . . . . . . . . . .
12.5.1 Parametric equations of a line in CE2 . . .
12.5.2 Use of polar parametric equations in CE2 .
12.5.3 Duo-sectors in CE2 . . . . . . . . . . . . . .
12.5.4 Duo-angles in CE2 . . . . . . . . . . . . . .
12.5.5 Complex Rotations . . . . . . . . . . . . . .
12.5.6 Dropping a perpendicular . . . . . . . . . .
12.5.7 Mid-lines . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
298
298
300
300
303
305
306
309
309
312
313
313
314
317
322
324
324
326
328
328
329
329
329
330
330
331
332
334
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
337
337
337
339
345
345
346
347
349
350
351
351
352
353
353
354
355
355
358
358
359
359
360
360
xi
CONTENTS
12.5.8 Axial symmetries . . . . . . . . . . . .
12.5.9 Composition of two axial symmetries .
12.5.10 Some material on triangles . . . . . .
12.6 Square of distance, circles . . . . . . . . . . .
12.6.1 Square of distance . . . . . . . . . . .
12.6.2 Complex circle . . . . . . . . . . . . .
12.7 Independence of particular frame of reference
12.7.1 Concepts based on sensed-area . . . .
12.7.2 Duo-sectors . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
13 Extension of Euclidean to projective planes
13.1 Homogeneous coordinates, the Real Projective Plane . . . . . .
13.1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . .
13.1.2 Use of Plücker’s homogeneous coordinates for projective
13.1.3 Projective lines of p-points . . . . . . . . . . . . . . . .
13.1.4 Further properties of p-lines . . . . . . . . . . . . . . . .
13.1.5 p-Pencils of p-lines . . . . . . . . . . . . . . . . . . . . .
13.1.6 Overview . . . . . . . . . . . . . . . . . . . . . . . . . .
13.1.7 Cross-ratio of a range of points . . . . . . . . . . . . . .
13.1.8 p-Segments . . . . . . . . . . . . . . . . . . . . . . . . .
13.2 Selecting representatives of p-points . . . . . . . . . . . . . . .
13.2.1 Use of Plücker coordinates; first method . . . . . . . . .
13.2.2 Extended dual sensed-area . . . . . . . . . . . . . . . .
13.2.3 A property of dual sensed-area . . . . . . . . . . . . . .
13.2.4 Equation of a p-line in extended areal coordinates . . .
13.2.5 Use of Plücker coordinates; second method . . . . . . .
13.2.6 p-Duo-sectors . . . . . . . . . . . . . . . . . . . . . . . .
13.2.7 Extended duo-angles . . . . . . . . . . . . . . . . . . . .
13.2.8 Extended orientation . . . . . . . . . . . . . . . . . . . .
13.2.9 Extended duo-angles in standard position; addition . . .
13.2.10 Sine, cosine, tangent of extended duo-angles . . . . . . .
13.2.11 Sensed extended duo-angles . . . . . . . . . . . . . . . .
13.2.12 Extended duo-angles with vertex on the ideal line . . .
13.3 Projective coordinates . . . . . . . . . . . . . . . . . . . . . . .
13.3.1 Projective coordinates . . . . . . . . . . . . . . . . . . .
13.3.2 More detail in use of projective coordinates . . . . . . .
13.3.3 Conics and the ideal line . . . . . . . . . . . . . . . . . .
13.4 The theory independent of the frame of reference . . . . . . . .
13.5 THE COMPLEX PROJECTIVE PLANE . . . . . . . . . . . .
13.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . .
13.5.2 Development . . . . . . . . . . . . . . . . . . . . . . . .
13.5.3 Mid-lines . . . . . . . . . . . . . . . . . . . . . . . . . .
13.5.4 Further material on sensed duo-angles . . . . . . . . . .
13.5.5 Duo-angles . . . . . . . . . . . . . . . . . . . . . . . . .
13.5.6 Tangent of a duo-angle . . . . . . . . . . . . . . . . . . .
13.5.7 Sensed duo-angles . . . . . . . . . . . . . . . . . . . . .
13.5.8 Complex-valued distance, square of distance . . . . . . .
13.5.9 Another route to specialization . . . . . . . . . . . . . .
13.6 EXTENSION OF RATIOS OF SENSED-AREAS . . . . . . . .
13.6.1 Ratio of sensed-areas . . . . . . . . . . . . . . . . . . . .
13.6.2 Ratios of distances . . . . . . . . . . . . . . . . . . . . .
13.6.3 Complex extended conics . . . . . . . . . . . . . . . . .
13.6.4 Generalization of distance . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
362
363
364
365
365
366
366
366
368
. . . .
. . . .
points
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
371
371
371
372
374
377
379
379
379
381
382
382
382
385
386
387
387
388
389
390
391
393
393
394
394
396
397
398
400
400
401
402
402
403
405
406
406
407
407
407
408
408
409
xii
CONTENTS
13.6.5 Mapping an ellipse to a hyperbola . . . . . . . . .
13.6.6 A result on sensed-areas . . . . . . . . . . . . . . .
13.7 EXTENDED REAL CONICS . . . . . . . . . . . . . . . .
13.7.1 Equations in areal coordinates . . . . . . . . . . .
13.7.2 Type of conic given by intersections with ideal line
13.7.3 Centre of extended parabola . . . . . . . . . . . .
13.7.4 Asymptotes . . . . . . . . . . . . . . . . . . . . . .
13.7.5 Diametral lines as polars . . . . . . . . . . . . . . .
13.7.6 Identification of the foci . . . . . . . . . . . . . . .
13.7.7 Plücker reciprocation, areal line coordinates . . . .
13.7.8 Equation of a line . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
410
410
411
411
411
412
412
412
412
413
415
14 Further projective methods and results
14.1 EXTENDED CONICS, THE IDEAL CIRCULAR POINTS . . . . . . . . . . . . .
14.1.1 Projective transformations in terms of homogeneous coordinates . . . . . .
14.1.2 Duo-sectors in projective terms . . . . . . . . . . . . . . . . . . . . . . . . .
14.1.3 The ideal circular points . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.1.4 Ratio of sensed areas interpreted . . . . . . . . . . . . . . . . . . . . . . . .
14.1.5 Ratio of distances interpreted . . . . . . . . . . . . . . . . . . . . . . . . . .
14.1.6 Angle-measure interpreted . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.2 UTILIZATION OF ANALOGUES OF RATIOS OF SENSED-AREAS . . . . . . .
14.2.1 Basic expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.2.2 Sine and cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.2.3 Analogue of rotors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.3 ANALOGUES OF RATIOS OF COMPLEX-VALUED DISTANCES . . . . . . . .
14.3.1 Basic expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.3.2 Distance 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.3.3 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.3.4 Complex-valued ratio constant . . . . . . . . . . . . . . . . . . . . . . . . .
14.3.5 Ratio of square of distance . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4 UTILIZATION OF ANALOGUE OF ANGLE-MEASURE . . . . . . . . . . . . .
14.4.1 Basic expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.2 Angles of equal measures . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.3 Analogue of null angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.4 Analogue of perpendicularity . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.5 Alternate angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.6 Mid-lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.7
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.4.8 Analogue of isosceles triangle . . . . . . . . . . . . . . . . . . . . . . . . . .
14.5 ANALOGUES OF APPOLONIUS’, PYTHAGORAS’ AND PTOLEMY’S THEOREMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.5.1 Appolonius’ characterisation of a circle . . . . . . . . . . . . . . . . . . . . .
14.5.2 Analogue of Pythagoras’ theorem . . . . . . . . . . . . . . . . . . . . . . . .
14.5.3 Ptolemy’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.6 THE NEWTON AND CLASSICAL GREEK PROPERTIES . . . . . . . . . . . .
14.6.1 Newton’s property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.6.2 Classical Greek property . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.7 GENERALIZATIONS FROM THE HYPERBOLA . . . . . . . . . . . . . . . . . .
14.7.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.7.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.8 NEWTON’S ORGANIC METHOD OF GENERATION . . . . . . . . . . . . . . .
14.8.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.9 ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS . . . . . . . . .
417
417
417
418
418
419
419
420
420
420
421
424
425
425
425
426
426
428
428
428
429
429
429
430
430
430
431
432
432
433
434
436
436
438
440
440
441
442
442
443
xiii
CONTENTS
14.9.1 Analogue of centroid . . . . . . . . . . . . . . . .
14.9.2 Analogue of orthocentre . . . . . . . . . . . . . .
14.9.3 Analogue of the Euler line . . . . . . . . . . . . .
14.9.4 Analogue of the Wallace-Simson line . . . . . . .
14.10ANALOGUE OF POWER PROPERTY OF CIRCLES
14.10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.11ANALOGUE OF PROTO-FOCI . . . . . . . . . . . . .
14.11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.12ANALOGUE OF OBLIQUE AXIAL SYMMETRIES .
14.12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.13GENERALIZATIONS FROM THE HYPERBOLA . . .
14.13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.14NEWTON’S ORGANIC METHOD OF GENERATION
14.14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.15ANALOGUES OF SOME AFFINE AND SIMILARITY
14.15.1 Analogue of centroid . . . . . . . . . . . . . . . .
14.15.2 Analogue of orthocentre . . . . . . . . . . . . . .
14.15.3 Analogue of the Euler line . . . . . . . . . . . . .
14.15.4 Analogue of the Wallace-Simson line . . . . . . .
14.16ANALOGUE OF POWER PROPERTY OF CIRCLES
14.16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.17ANALOGUE OF PROTO-FOCI . . . . . . . . . . . . .
14.17.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.18ANALOGUE OF OBLIQUE AXIAL SYMMETRIES .
14.18.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
RESULTS
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
443
445
446
446
447
447
449
449
451
451
452
452
453
454
454
455
455
457
458
458
459
459
461
461
463
463
15 Inversive Methods and Results
15.1 GENERAL INVERSION . . . . . . . . . . . . . . . . .
15.1.1 Definition of inversion . . . . . . . . . . . . . . .
15.1.2 Lineo-loci . . . . . . . . . . . . . . . . . . . . . .
15.1.3 Parallel lineo-loci . . . . . . . . . . . . . . . . . .
15.2 PARAMETRIC EQUATIONS FOR A LINEO-LOCUS .
15.2.1 Parametric equations . . . . . . . . . . . . . . . .
15.2.2 Lineo-segments and lineo-half-loci . . . . . . . .
15.2.3 Analogue of mid-points . . . . . . . . . . . . . .
15.3 Duo-angles between tangents . . . . . . . . . . . . . . .
15.4 CIRCULAR INVERSION . . . . . . . . . . . . . . . . .
15.4.1 Circular inversion . . . . . . . . . . . . . . . . . .
15.4.2 Complex-valued distance . . . . . . . . . . . . .
15.4.3 Angle-measure . . . . . . . . . . . . . . . . . . .
15.4.4 Sensed-area . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
465
465
465
465
466
466
466
466
467
467
467
467
468
468
469
A More Complicated Geometric Algebra
A.1 Sensed-area, points given indirectly . . . . . . . . . . .
A.1.1 Sensed-area, two points given indirectly . . . .
A.2 Sensed-area, three points given indirectly . . . . . . .
A.3 Pappus’ theorem and Desargues’ perspective theorem
A.3.1 Pappus’ theorem . . . . . . . . . . . . . . . . .
A.3.2 Desargues’ perspective theorem . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
471
471
471
472
475
475
476
B Lines common to two pencils
.
.
.
.
.
.
479
xiv
CONTENTS
C Material on dual sensed-area
C.0.3 A determinant for dual sensed-area . . . . . . . . . . . . . . . . . . . . . . .
C.0.4 A mixed identity for sensed-area and dual sensed-area . . . . . . . . . . . .
C.0.5 An identity for dual sensed-area . . . . . . . . . . . . . . . . . . . . . . . .
483
483
484
484
D Rotors common to two pencils
487
E Note on a paper of W.K. Clifford’s
497
F More on trigonometric
F.1 . . . . . . . . . . . .
F.1.1
. . . . . . .
F.1.2
. . . . . . .
499
499
499
499
tangent function
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
CONTENTS
DRAFT PREFACE
• This book addresses the question of what the ordinary geometry of straight lines and circles
leads on to. Thus it is a second and not a first book on geometry. Many books contain
the material pre-requisite for starting this, but the one most closely aligned is my textbook
‘Geometry with Trigonometry’ [2]. Any concepts not defined anew will be found there, apart
from an elementary treatment of three-dimensional Euclidean geometry.
Down the centuries many adults have studied as a hobby further material on the geometry
they learned at school, and derived great pleasure from this. With the decline in study of
geometry over the last fifty years, this pool is probably greatly reduced but I have a strong
hope that persons of this type will notice this book and be encouraged to study it. All results
in Chapters 1 to 11 have quite faithful diagrams.
• A major objective of the approach is to motivate in the familiar context of Euclidean geometry
basic concepts of projective geometry. Since Felix Klein’s ‘Erlanger Program’ of 1870, it is
generally accepted by mathematicians that any type of geometry consists of the study of
concepts which are invariant under a specific group of transformations relevant to it. Thus
Euclidean geometry consists of the study of concepts which are invariant under the group
of isometries or rigid motions, while projective geometry is the study of concepts which are
invariant under the group of projective transformations.
Projective transformations are those of the form
x′ =
d1 x + e1 y + f1
,
d3 x + e3 y + f3
y′ =
d2 x + e2 y + f2
,
d3 x + e3 y + f3
which do not degenerate, while Euclidean transformations are those of the form
x′ = d1 x + e1 y + f1 ,
y ′ = d2 x + e2 y + f2 ,
where d1 e2 − e1 d2 = ±1. Thus the set of Euclidean transformations is a subset of the set
of projective transformations, but a very small subset. It follows that while every projective
invariant is also a Euclidean invariant the converse is by no means true. To put it rather
tautologically, the only Euclidean invariants which are also projectively invariant are the ones
which happen to be so.
Thus the concepts studied in projective geometry are generally unfamiliar to those (surely all)
who have first studied Euclidean geometry, and the intuition built up in Euclidean geometry
can be a hindrance as much as a help. There is a jump discontinuity in the study of geometry,
such as does not occur in such other mainstream mathematical topics as algebra and analysis,
and this jump must be a barrier to the study and understanding of advanced geometry.
Of course earlier authors have provided such motivation from familiar concepts; e.g. Seidenberg [27] gives over his Chapter 1 to it. But this I think is the first occasion so large a part
of a book is devoted to this purpose. Moreover results which are projective in nature are
proved in a Euclidean context, and not just mentioned.
While we work in Chapters 1 to 12 in the Euclidean plane we do not do so just in a selfsufficient introverted looking way, but we select concepts which lend themselves to generalization. In Chapter 13 we extend to the real and complex projective planes.
• Many of the basic early results in projective geometry (e.g. Pappus’ theorem, Desargues’
perspective or two-triangle theorem) depend on the collinearity of points or the concurrence
of lines. It seems profitable to have expressions which deal with collinearity and concurrency
and notation for them, preferably with geometrical interpretations. Such expressions should
be expected to have frequent use.
I think that this role is played effectively by the concept of sensed area δF (Z1 , Z2 , Z3 ), which
was introduced in 1827 by Möbius [19]. In the first place, the more precise an expression
2
CONTENTS
the more likely it is to produce formulae; for that reason we give preference to equations of
lines in the form δF (Z, Z2 , Z3 ) = 0 or 2δF (Z, Z2 , Z3 ) = 0. Secondly, the point of intersection
of two lines is expeditiously handled by using parametric equations and sensed areas. The
concurrency of lines can be further handled by the related concept of dual sensed-area
δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] which is introduced here.
Areal coordinates, also introduced by Möbius in 1827, have been commonly used down the
years since then and to great effect. But reducing all expressions to areal coordinates tends
to obscure patterns which are of considerable value, so here as well as areal coordinates
we consider sensed-areas in non-reduced form. This was done briefly by Möbius in 1827,
Magnus [17] in 1833 and Clifford [4, pages 80–109, 110–114] in 1865, and they were handled
by a few minor authors as well, but ours seems to be the first sustained effort to develop and
utilize them. For us they have a fundamental role.
In the study of vector spaces linear transformations are prominent. Now in vector spaces there
is a distinguished vector, as the null vector has a special role. By contrast in the Euclidean
plane there are no distinguished points. The geometric analogue of linear functionals is surely
sensed-area. We keep this in mind as we examine its properties in §1.1.
• According to Field and Gray [8, page 10], in 1639 Desargues, who was then essentially
initiating the subject of projective geometry, abandoned the many proofs based on area
contained in the Conics of Appolonius. With such a pioneering work, it would be harsh to
think of this as a lost opportunity but it should provide food for some thought.
In 1823, Poncelet [25, pages 23-25, sections 45-47] continuing from Carnot in pioneering the
revival of projective geometry, while stating that he based his work on ratios of distances and
of sines of angles, mentioned the invariance of ratios of areas but said that he would omit
consideration of the latter as the subject was already large enough. Thus Poncelet envisaged
a development such as ours. This does look like a missed opportunity.
Before the era of pure projective geometry some proofs based on area were used in deriving
projective results. However, references to area in modern treatments of projective geometry
are rare indeed. In fact some authors have stated that area is inappropriate for projective
geometry as it is metric in nature. But being metric in nature did not prevent distance
from being the basis of the concept of cross-ratio, which is almost ubiquitous in projective
geometry.
• A major emphasis in projective geometry as it was developed in the 19th century, was the
role of ranges of points on lines and the role of pencils of lines on points. This was combined
with cross-ratio of four collinear points generalizing from the role of distance, and dually the
cross-ratio of four concurrent lines generalizing from angle-measure.
By contrast we let our points and lines roam freely over the plane. However cross-ratio
of collinear points can readily be handled by our equipoised quotients, by adjoining a fifth
point not on the line, and equipoised quotients in which one particular vertex occurs in every
sensed-area, handle cross-ratio of pencils of lines. Thus our results are more general than
these traditional types of results, and underpin and straddle them.
• Throughout the 19th century sustained efforts were made to place projective geometry on
a self-contained basis and not have it dependent on metrical concepts. Major milestones on
the way were the noting of the circular points at infinity by Poncelet in 1823, the handling of
angle-measure by Laguerre in 1853, and the handling of distance by Cayley via an absolute
conic in 1859. The process culminated in success with the magnificent synthetic achievements
of von Staudt.
When Klein broadened out Cayley’s work in the 1870’s not only was projective geometry a
magnificent and elegant self-contained topic in its own right, but it contained as subcases Euclidean, hyperbolic and elliptic geometries. This top-down, descent-from-on-high, approach is
3
CONTENTS
presented, without undue dwelling on difficulties, in Meserve [18]. Our bottom-up approach
is in stark contrast to this top-down approach, and is intended to precede it as motivation.
• There was a considerable body of Euclidean geometry by the start of the 19th century. The
pioneering projective geometers aimed at keeping their Euclidean baggage as light as possible,
and many old results had no role or analogues in the trim, elegant treatment of projective
geometry. An example of this trend can be seen in Coxeter [?, pp. 118-119] where the only
appearance of cross-ratio is in exercises.
Affine geometry, which was started by Euler in 1748, lies in between Euclidean and projective
geometries. Affine transformations are those of the form
x′ = d1 x + e1 y + f1 ,
y ′ = d2 x + e2 y + f2 ,
with d1 e2 − e2 d2 6= 0, and so are intermediate between Euclidean and projective transformations. While there was a considerable body of Euclidean geometry, and projective geometry
was being vigorously pursued, it seems to me that there was a relative neglect of affine
geometry.
In our approach, we establish many new results and develop (rescue?) old ones. We organize
the whole material under the headings of affine, projective and similarity methods and results.
• Our projective results could be included in a conventional approach to projective geometry.
For, once homogeneous coordinates are introduced for the real or complex projective plane,
our equipoised quotients are well-defined provided we use the same homogeneous coordinates
for any point, at all its occurrences in a quotient. Our proofs would remain valid. Thus a
top-down approach could be based on this material. In fact there is extensive feedback and
to-ing and fro-ing in our Chapters 12 and 13.
• One of the irritations of Euclidean geometry, especially for those used to projective geometry,
is the proliferation of exceptional cases, which need to be dealt with separately, especially
those linked to the phenomenon of parallelism. We have drafted our material so that the
proofs cover the cases of three concurrent lines and three parallel lines. To deal with standard
material this led us into rather ponderous expressions, although with study the patterns can
be fairly readily observed. To avoid a barrier of difficulty, we have placed this material, which
falls naturally into Chapter 1, in Appendix A.
• There is in projective geometry a complete and perfect duality in which the roles of points
and lines can be interchanged in many results. There is a more diffuse and partial duality
in Euclidean geometry in which the dual of a point can be a line, a pair of points or a less
common concept named a rotor.
The last of these concepts was introduced in 1844 in the pioneering work of Grassmann [13],
whose own books are acknowledged to be very difficult to read. An old-style presentation of
this material in English was given in 1898 in Whitehead [31]. There was an effort to make
it accessible to a wider mathematical audience by Forder [9] in 1941 and a further one in
1980 by Fearnley-Sander [7]. In 1922 Neville [21, page 66] strongly urged the use of rotors in
mathematics and included some material on them.
We have been led to use rotors to buttress our material on duality and give considerable
geometrical application. Having an introduction as here to one of Grassmann’s basic concepts
via conventional material, may be a helpful preliminary to a direct study of his material.
• Areal coordinates were introduced by Möbius in 1827 and trilinear coordinates by Plücker c.
1840; the latter have been used much more frequently than the former ever since, although
as shown here the former are a much more convenient basis for generalization. I think that
it may have been the fact that workers in the field were able to define line-coordinates from
trilinear coordinates but not from areal coordinates. We have been able to introduce dual
coordinates from areal point coordinates; this is based on a key identity in 2.2.1.
4
CONTENTS
Chapter 1
Geometric algebra for a Euclidean
plane
1.1
1.1.1
Sensed-area
Sensed-area of a triple of points
Suppose that F = ([O, I , [O, J ) is a frame of reference for the Euclidean plane Π, where [O, I and
[O, J are half-lines on a pair OI, OJ of perpendicular lines and |O , I | = |O , J | = 1. When a point
Z has Cartesian coordinates relative to F we write Z ≡F (x, y), so that in particular I ≡F (1, 0)
and J ≡F (0, 1).
−−→
−−→
For points Z1 ≡F (x1 , y1 ), Z2 ≡F (x2 , y2 ), we recall for the position vectors OZ1 and OZ2 , the
sum, difference and scalar multiple
−−→ −−→ −−−→
OZ1 + OZ2 = OW1 ,
−−→ −−→ −−−→
OZ2 − OZ1 = OW2 ,
−−→ −−−→
k OZ1 = OW3 ,
(1.1.1)
respectively, where
W1 ≡F (x1 + x2 , y1 + y2 ),
We note that
W2 ≡F (x2 − x1 , y2 − y1 ),
−→ −−−→
(x2 − x1 )OI = OW4 ,
W3 ≡F (kx1 , ky1 ).
−→ −−−→
(y2 − y1 )OJ = OW5 ,
where W4 ≡F (x2 − x1 , 0), W5 ≡F (0, y2 − y1 ), and so
−→
−→ −−−→ −−−→
(x2 − x1 )OI + (y2 − y1 )OJ =OW4 + OW5
−−−→
=OW2
−−→ −−→
=OZ2 − OZ1 .
At the risk of some confusion we extend the convention of Barry [2, §11.4.4] and write Z1 + Z2
for W1 , Z2 − Z1 for W2 and kZ1 for W3 , for the points in (1.1.1).
For an ordered triple of points (Z1 , Z2 , Z3 ) of points with Z1 ≡F (x1 , y1 ), Z2 ≡F (x2 , y2 ) and
Z3 ≡F (x3 , y3 ), we recall from [2, §10.5] the sensed-area δF (Z1 , Z2 , Z3 ) defined by the formula
δF (Z1 , Z2 , Z3 ) =
=
=
1
2 [x1 (y2 − y3 ) − y1 (x2 − x3 ) + x2 y3 − x3 y2 ]
1
2 [(x2 − x1 )(y3 − y1 ) − (x3 − x1 )(y2 − y1 )]

x1
1
det  x2
2
x3
1
y1
y2
y3

1
1 .
1
2
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
We wish to note sufficient of its properties to characterize it and then derive other properties from
these.
We note first that
δF (Z1 , Z2 , Z3 ) =δF (Z2 , Z3 , Z1 ) = δF (Z3 , Z1 , Z2 )
(1.1.2)
δF (Z1 , Z3 , Z2 ) = − δF (Z1 , Z2 , Z3 ),
(1.1.3)
and
from which it follows that
δF (Z3 , Z2 , Z1 ) =δF (Z2 , Z1 , Z3 ) = −δF (Z2 , Z3 , Z1 ) = −δF (Z1 , Z2 , Z3 ),
and
δF (Z2 , Z1 , Z3 ) =δF (Z3 , Z2 , Z1 ) = −δF (Z3 , Z1 , Z2 ) = −δF (Z1 , Z2 , Z3 ).
From these it also follows that
δF (Z1 , Z2 , Z2 ) = δF (Z3 , Z2 , Z3 ) = δF (Z1 , Z1 , Z3 ) = 0.
To bring in position vectors, we note that
δF (Z1 , Z2 , Z3 ) = δF (O, Z2 − Z1 , Z3 − Z1 ),
(1.1.4)
and this is equal to
δF (O, Z4 , Z5 ),
where
Z2 − Z1 =Z4 − O,
Z3 − Z1 = Z5 − O,
and so
m.p.(Z1 , Z4 ) =m.p.(O, Z2 ),
m.p.(Z1 , Z5 ) = m.p.(O, Z3 ),
where we have used the notation m.p. to denote mid-point. This formula (1.1.4) shows that
δF (Z1 , Z2 , Z3 ) is the signed magnitude of the vector or cross product 21 (Z2 − Z1 ) × (Z3 − Z1 ).
To bring in a linear functional property we note that
δF (O, k1 Z6 + k2 Z7 , Z8 ) = k1 δF (O, Z6 , Z8 ) + k2 δF (O, Z7 , Z8 ),
(1.1.5)
in all cases, and so
δF O, k1 (Z4 − Z1 ) + k2 (Z5 − Z1 ), Z3 − Z1 =k1 δF (O, Z4 − Z1 , Z3 − Z1 ) + k2 δF (O, Z5 − Z1 , Z3 − Z1 ),
=k1 δF (Z1 , Z4 , Z3 ) + k2 δF (Z1 , Z5 , Z3 ).
From (1.1.2) to (1.1.5) we have that
δF (Z1 , Z2 , Z3 ) = δF (O, Z2 − Z1 , Z3 − Z1 )
=δF O, (x2 − x1 )I + (y2 − y1 )J, (x3 − x1 )I + (y3 − y1 )J
=(x2 − x1 )δF O, I, (x3 − x1 )I + (y3 − y1 )J + (y2 − y1 )δF O, J, (x3 − x1 )I + (y3 − y1 )J
=(x2 − x1 )(x3 − x1 )δF (O, I, I) + (x2 − x1 )(y3 − y1 )δF (O, I, J)
+ (y2 − y1 )(x3 − x1 )δF (O, J, I) + (y2 − y1 )(y3 − y1 )δF (O, J, J)
= (x2 − x1 )(y3 − y1 ) − (y2 − y1 )(x3 − x1 ) δF (O, I, J),
3
1.1. SENSED-AREA
and this yields the correct value provided that we add
δF (O, I, J) = 21 .
(1.1.6)
Thus the properties (1.1.2), (1.1.3), (1.1.4), (1.1.5) and (1.1.6) characterize sensed area.
To deduce further properties, we note that
δF Z1 , Z2 , Z4 + s(Z6 − Z5 ) = δF O, Z2 − Z1 , Z4 − Z1 + s[Z6 − Z1 − (Z5 − Z1 )]
=δF O, Z2 − Z1 , Z4 − Z1 + sδF O, Z2 − Z1 , Z6 − Z1 − (Z5 − Z1 )
=δF O, Z2 − Z1 , Z4 − Z1 + s[δF O, Z2 − Z1 , Z6 − Z1 − δF O, Z2 − Z1 , Z5 − Z1 ]
=δF Z1 , Z2 , Z4 + s[δF Z1 , Z2 , Z6 − δF Z1 , Z2 , Z5 ].
(1.1.7)
A particular case of this is
δF Z1 , (1 − s)Z4 + sZ5 , Z3 = (1 − s)δF (Z1 , Z4 , Z3 ) + sδF (Z1 , Z5 , Z3 ).
(1.1.8)
Anticipating somewhat we consider an affine transformation
f (x, y) = (x′ , y ′ ) = (ax + by + k1 , cx + dy + k2 )
where ad − bc 6= 0. Now note that
f (x4 , y4 ) + s[(x6 , y6 ) − (x5 , y5 )]
=f (x4 + s(x6 − x5 ), y4 + s(y6 − y5 ))
= a[x4 + s(x6 − x5 )] + b[y4 + s(y6 − y5 )] + k1 ,
=c[x4 + s(x6 − x5 )] + d[y4 + s(y6 − y5 )] + k2
= ax4 + by4 + k1 + s[ax6 + by6 + k1 − (ax5 + by5 + k1 )],
=cx4 + dy4 + k2 + s[cx6 + dy6 + k2 − (cx5 + dy5 + k2 )]
=f (x4 , y4 ) + s[f (x6 , y6 ) − f (x5 , y5 )].
Because of this we shall refer to (1.1.7) as sensed-area having the affine line property in each
vertex.
Next we note that
δF (W1 , Z1 , Z2 ) − δF (W2 , Z1 , Z2 )
=δF (W1 − Z1 , O, Z2 − Z1 ) − δF (W2 − Z1 , O, Z2 − Z1 )
=δF W1 − Z1 − (W2 − Z1 ), O, Z2 − Z1 )
=δF (W1 − W2 , O, Z2 − Z1 )
=δF (W1 − W2 , Z1 − Z2 , O).
(1.1.9)
This is then equal to
− δF (Z1 − Z2 , W1 − W2 , O)
= − δF (Z1 , W1 , W2 ) + δF (Z2 , W1 , W2 ).
Concerning (1.1.9) it is easily checked that lines Z1 Z2 and W1 W2 are parallel if and only if
δF (W1 − W2 , Z1 − Z2 , O) = 0.
(1.1.10)
As a final property of sensed-area we have the result
δF (Z4 , Z2 , Z3 ) + δF (Z4 , Z3 , Z1 ) + δF (Z4 , Z1 , Z2 ) = δF (Z1 , Z2 , Z3 ).
(1.1.11)
4
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
For
δF (Z4 , Z2 , Z3 ) + δF (Z4 , Z3 , Z1 ) + δF (Z4 , Z1 , Z2 )
=δF (Z2 , Z3 , Z4 ) − δF (Z1 , Z3 , Z4 ) + δF (Z4 − Z1 , O, Z2 − Z1 )
=δF (Z2 − Z1 , Z3 − Z4 , O) − δF (Z4 − Z1 , Z2 − Z1 , O)
=δF (Z4 − Z3 , Z2 − Z1 , O) − δF (Z4 − Z1 , Z2 − Z1 , O)
=δF Z4 − Z3 − (Z4 − Z1 ), Z2 − Z1 , O
=δF −(Z3 − Z1 ), Z2 − Z1 , O
= − δF (Z3 − Z1 , Z2 − Z1 , O)
= − δF (Z3 , Z2 , Z1 )
=δF (Z1 , Z2 , Z3 ).
For a point Z = (1 − s)Z2 + sZ3 on Z2 Z3 and a point Z1 6∈ Z2 Z3 , we have that
δF (Z1 , Z2 , Z) = δF (Z1 , Z2 , (1 − s)Z2 + sZ3 ) = sδF (Z1 , Z2 , Z3 ),
δF (Z1 , Z, Z3 )
= δF (Z1 , (1 − s)Z2 + sZ3 , Z3 ) = (1 − s)δF (Z1 , Z2 , Z3 ),
and so we have the expression for a sensed-ratio (see e.g. [2, 7.6.1])
Z2 Z
s
δF (Z1 , Z2 , Z)
=
=
.
1−s
δF (Z1 , Z, Z3 )
ZZ3
(1.1.12)
−→
−−→
We also recall the representation in [2, §11.4.4] of arbitrary position vectors OZ = pOZ1 +
−−→
−−→
q OZ2 + rOZ3 , with p + q + r = 1, where (Z1 , Z2 , Z3 ) is a triple of non-collinear points. This was
written as Z = pZ1 + qZ2 + rZ3 , and it follows from (1.1.8) as in §11.4.5 there, that
δF (Z1 , Z2 , pZ4 + qZ5 + rZ6 ) = pδF (Z1 , Z2 , Z4 ) + qδF (Z1 , Z2 , Z5 ) + rδF (Z1 , Z2 , Z6 ),
(1.1.13)
for any points Z1 , Z2 , Z4 , Z5 , Z6 when p, q, r are real numbers for which p + q + r = 1.
1.1.2
Menelaus’ theorem
We start with three lines W1 W2 , W3 W4 , W5 W6 and consider the points
W7 =
λ
1
µ
1
ν
1
W1 +
W2 , W8 =
W3 +
W4 , W9 =
W5 +
W6 ,
1+λ
1+λ
1+µ
1+µ
1+ν
1+ν
on them, respectively. Then
λ
1
µ
1
ν
1
W1 +
W2 ,
W3 +
W4 ,
W5 +
W6 .
δF (W7 , W8 , W9 ) = δF
1+λ
1+λ
1+µ
1+µ
1+ν
1+ν
On expanding this in terms of λ, µ and ν by repeated use of (1.1.8), we obtain that
δF (W7 , W8 , W9 )(1 + λ)(1 + µ)(1 + ν)
=λµνδF (W2 , W4 , W6 ) + µνδF (W1 , W4 , W6 ) + νλδF (W2 , W3 , W6 ) + λµδF (W2 , W4 , W5 )
+ λδF (W2 , W3 , W5 ) + µδF (W1 , W4 , W5 ) + νδF (W1 , W3 , W6 ) + δF (W1 , W3 , W5 ).
(1.1.14)
It follows that the points W7 , W8 and W9 are collinear if and only if this expression is equal to 0.
From this we can deduce the following.
Menelaus’ theorem (c. 100A.D.). For non-collinear points Z1 , Z2 , Z3 , let Z4 ∈ Z2 Z3 ,
Z5 ∈ Z3 Z1 and Z6 ∈ Z1 Z2 . Then Z4 , Z5 , Z6 are collinear if and only if
δF (Z1 , Z2 , Z4 ) δF (Z2 , Z3 , Z5 ) δF (Z3 , Z1 , Z6 )
= −1.
δF (Z1 , Z4 , Z3 ) δF (Z2 , Z5 , Z1 ) δF (Z3 , Z6 , Z2 )
5
1.2. DUAL SENSED-AREA
b
Z1
Z6
b
b
b
Z2
Figure 1.1.
Z5
b
b
Z4
Z3
Proof. We apply (1.1.14) with
W1 = Z2 , W2 = Z3 , W3 = Z3 , W4 = Z1 , W5 = Z1 , W6 = Z2 , W7 = Z4 , W8 = Z5 , W9 = Z6 .
From (1.1.14) we obtain λµν = −1, while from (1.1.8)
δF (Z2 , Z3 , Z5 )
δF (Z3 , Z1 , Z6 )
δF (Z1 , Z2 , Z4 )
= λ,
= µ,
= ν,
δF (Z1 , Z4 , Z3 )
δF (Z2 , Z5 , Z1 )
δF (Z3 , Z6 , Z2 )
and so the stated conclusion follows.
1.2
Dual sensed-area
1.2.1
Point of intersection of two lines
Suppose that W1 W2 and W3 W4 are non-parallel lines and denote by W their point of intersection.
Then we have W = (1 − r)W1 + rW2 for some r ∈ R. Thus
0 = δF (W, W3 , W4 ) = (1 − r)δF (W1 , W3 , W4 ) + rδF (W2 , W3 , W4 ),
so that
δF (W2 , W3 , W4 )
1−r
=−
,
r
δF (W1 , W3 , W4 )
1 δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
=
r
δF (W1 , W3 , W4 )
δF (W2 , W3 , W4 )
δF (W1 , W3 , W4 )
, 1−r =−
.
r=
δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
(1.2.1)
Thus
W =−
δF (W1 , W3 , W4 )
δF (W2 , W3 , W4 )
W1 +
W2 . (1.2.2)
δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
There is a similar formula to this in which (W1 , W2 ) and (W3 , W4 ) are interchanged. For the
denominators here we refer to (1.1.9).
For this W and arbitrary (W5 , W6 ) we note that
δF (W2 , W3 , W4 )
δF (W1 , W5 , W6 )
δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
δF (W1 , W3 , W4 )
δF (W2 , W5 , W6 ).
+
δF (W1 , W3 , W4 ) − δF (W2 , W3 , W4 )
δF (W, W5 , W6 ) = −
(1.2.3)
6
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
Independently of how it was derived, we introduce the expression in (1.2.3) without the denominators by the following
δF (W1 , W3 , W4 )δF (W2 , W5 , W6 ) − δF (W2 , W3 , W4 )δF (W1 , W5 , W6 ),
(1.2.4)
and denote it by
δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )].
We shall refer to δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )] in (1.2.5)
conveniently express it as a determinant
δF (W1 , W3 , W4 ) δF (W2 , W3 , W4 )
δF (W1 , W5 , W6 ) δF (W2 , W5 , W6 )
(1.2.5)
as dual sensed-area. We can
.
We deduce from (1.2.3) that if W1 W2 , W3 W4 are lines which are distinct and intersecting, then
these lines and W5 W6 are concurrent if and only if
δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )] = 0.
(1.2.6)
On the other hand, suppose that W1 W2 , W3 W4 , W5 W6 are lines such that W1 W2 and W3 W4
are distinct and parallel. Then δF (W1 , W3 , W4 ) = δF (W2 , W3 , W4 ) 6= 0, and
δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )]
=δF (W1 , W3 , W4 )δF (W2 , W5 , W6 ) − δF (W2 , W3 , W4 )δF (W1 , W5 , W6 )
=δF (W1 , W3 , W4 )[δF (W2 , W5 , W6 ) − δF (W1 , W5 , W6 )]
Thus (1.2.6) is satisfied again if and only if W1 W2 k W5 W6 . Hence if W1 W2 and W3 W4 are distinct
parallel lines, then all three of W1 W2 , W3 W4 and W5 W6 are parallel if and only if (1.2.6)holds.
A small result is that if Z1 Z2 = Z3 Z4 , then δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] = 0 as δF (Z1 , Z3 , Z4 ) =
δF (Z2 , Z3 , Z4 ) = 0 in the expansion.
We also note the property
δF (W1 , W2 ), (W3 , W4 ), W5 , (1 − s)W7 + sW8
=(1 − s)δF [(W1 , W2 ), (W3 , W4 ), (W5 , W7 )] + sδF [(W1 , W2 ), (W3 , W4 ), (W5 , W8 )].
Ceva’s theorem (1678). For non-collinear points Z1 , Z2 , Z3 , let Z4 ∈ Z2 Z3 , Z5 ∈ Z3 Z1 and
Z6 ∈ Z1 Z2 . Then Z1 Z4 , Z2 Z5 , Z3 Z6 are concurrent or all parallel if and only if
δF (Z1 , Z2 , Z4 ) δF (Z2 , Z3 , Z5 ) δF (Z3 , Z1 , Z6 )
= 1.
δF (Z1 , Z4 , Z3 ) δF (Z2 , Z5 , Z1 ) δF (Z3 , Z6 , Z2 )
For with the notation of Menelaus’ theorem the conclusion follows if and only if
δF [(Z1 , Z4 ), (Z2 , Z5 ), (Z3 , Z6 )] = 0. Now we have
1
λ
1
µ
1
ν
δF (Z1 ,
Z2 +
Z3 ), (Z2 ,
Z3 +
Z1 ), (Z3 ,
Z1 +
Z2 )
1+λ
1+λ
1+µ
1+µ
1+ν
1+ν
1
µ
λ
1
ν
1
Z3 +
Z 1 δF
Z2 +
Z3 , Z3 ,
Z1 +
Z2
=δF Z1 , Z2 ,
1+µ
1+µ
1+λ
1+λ
1+ν
1+ν
1
1
λ
1
µ
ν
− δF
Z2 +
Z3 , Z2 ,
Z3 +
Z 1 δF Z 1 , Z 3 ,
Z1 +
Z2
1+λ
1+λ
1+µ
1+µ
1+ν
1+ν
1
1
µ
1
λ
ν
=
δF (Z1 , Z2 , Z3 )
δF (Z2 , Z3 , Z1 ) −
δF (Z3 , Z2 , Z1 )
δF (Z1 , Z3 , Z2 )
1+µ
1+λ1+ν
1+λ1+µ
1+ν
1
1
1
=
(1 − λµν)δF (Z1 , Z2 , Z3 )2 ,
1+λ1+µ1+ν
and this is equal to 0 if and only if λµν = 1.
7
1.2. DUAL SENSED-AREA
1.2.2
Properties of dual sensed-area
To give us control of dual sensed-area we express it as the value of a determinant as follows. With
W1 ≡ (u1 , v1 ) and so on, we note that
4δF (W1 , W3 , W4 )δF (W2 , W5 , W6 )
u1 v1 1 u2 v2 1 = u3 v3 1 u5 v5 1 u4 v4 1 u6 v6 1 =[u1 (v3 − v4 ) + v1 (u4 − u3 ) + (u3 v4 − u4 v3 )][u2 (v5 − v6 ) + v2 (u6 − u5 ) + (u5 v6 − u6 v5 )]
and on interchanging subscripts 1 and 2 in this
4δF (W2 , W3 , W4 )δF (W1 , W5 , W6 )
u2 v2 1 u1 v1 1 = u3 v3 1 u5 v5 1 u4 v4 1 u6 v6 1 =[u2 (v3 − v4 ) + v2 (u4 − u3 ) + (u3 v4 − u4 v3 )][u1 (v5 − v6 ) + v1 (u6 − u5 ) + (u5 v6 − u6 v5 )].
On subtracting the second of these from the first we are left with the terms
4δF (W1 , W3 , W4 )δF (W2 , W5 , W6 ) − 4δF (W2 , W3 , W4 )δF (W1 , W5 , W6 )
=(v1 − v2 )[(u4 − u3 )(u5 v6 − u6 v5 ) − (u6 − u5 )(u3 v4 − u4 v3 )]
+ (u2 − u1 )[(v5 − v6 )(u3 v4 − u4 v3 ) − (v3 − v4 )(u5 v6 − u6 v5 )]
+ (u1 v2 − u2 v1 )[(v3 − v4 )(u6 − u5 ) − (v5 − v6 )(u4 − u3 )]
v1 − v2 u2 − u1 u1 v2 − u2 v1 = v3 − v4 u4 − u3 u3 v4 − u4 v3 .
v5 − v6 u6 − u5 u5 v6 − u6 v5 (1.2.7)
By (1.2.7) we see that the value of dual sensed-area is unaltered if we exchange the pairs
(W1 , W2 ), (W3 , W4 ), (W5 , W6 ) while retaining cyclic order, and its value is multiplied by −1 either
if we vary the cyclic order among the pairs or exchange the order of points within any pair such
as by replacing (W1 , W2 ) with (W2 , W1 ).
We can rewrite the determinant in (1.2.7) as
u2 − u1 v2 − v1 δF (O, W1 , W2 ) 2 u4 − u3 v4 − v3 δF (O, W3 , W4 ) ,
u6 − u5 v6 − v5 δF (O, W5 , W6 ) and on expanding this in terms of the elements of the third column we find the identity
δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )]
=δF (O, W1 , W2 )δF (O, W4 − W3 , W6 − W5 ) + δF (O, W3 , W4 )δF (O, W6 − W5 , W2 − W1 )
+ δF (O, W5 , W6 )δF (O, W2 − W1 , W4 − W3 ).
(1.2.8)
The identity (1.2.8) leads to a more general one, as follows. From it, for any point W , we note
that
δF (O, W1 − W, W2 − W )δF O, W4 − W − (W3 − W ), W6 − W − (W5 − W )
+ δF (O, W3 − W, W4 − W )δF O, W6 − W − (W5 − W ), W2 − W − (W1 − W )
+ δF (O, W5 − W, W6 − W )δF O, W2 − W − (W1 − W ), W4 − W − (W3 − W )
=δF [(W1 − W, W2 − W ), (W3 − W, W4 − W ), (W5 − W, W6 − W )].
8
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
On the left we have
δF (W, W1 , W2 )δF (O, W4 − W3 , W6 − W5 ) + δF (O, W3 , W4 )δF (W, W6 − W5 , W2 − W1 )
+ δF (W, W5 , W6 )δF (O, W2 − W1 , W4 − W3 ),
while on the right we get
δF (W1 − W, W3 − W, W4 − W )δF (W2 − W, W5 − W, W6 − W )
− δF (W2 − W, W3 − W, W4 − W )δF (W1 − W, W5 − W, W6 − W )
=δF O, W3 − W − (W1 − W ), W4 − W − (W1 − W ) δF O, W5 − W − (W2 − W ), W6 − W − (W2 − W )
− δF O, W3 − W − (W2 − W ), W4 − W − (W2 − W ) δF O, W5 − W − (W1 − W ), W6 − W − (W1 − W )
=δF (O, W3 − W1 , W4 − W1 )δF (O, W5 − W2 , W6 − W2 ) − δF (O, W3 − W2 , W4 − W2 )δF (O, W5 − W1 , W6 − W1 )
=δF (W1 , W3 , W4 )δF (W2 , W5 , W6 ) − δF (W2 , W3 , W4 )δF (W1 , W5 , W6 )
=δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )].
Thus we have the identity
δF (W, W1 , W2 )δF (O, W4 − W3 , W6 − W5 ) + δF (W, W3 , W4 )δF (O, W6 − W5 , W2 − W1 )
+ δF (W, W5 , W6 )δF (O, W2 − W1 , W4 − W3 ),
=δF [(W1 , W2 ), (W3 , W4 ), (W5 , W6 )].
1.2.3
(1.2.9)
Further identification of dual sensed-area
If we take
Z4 = (1 − r)Z2 + rZ3 ,
Z5 = (1 − s)Z2 + sZ3 ,
as points on Z2 Z3 , then
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )]
=δF (Z4 , Z6 , Z7 )δF (Z5 , Z8 , Z9 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z8 , Z9 )
= (1 − r)δF (Z2 , Z6 , Z7 ) + rδF (Z3 , Z6 , Z7 ) (1 − s)δF (Z2 , Z8 , Z9 ) + sδF (Z3 , Z8 , Z9 )
− (1 − s)δF (Z2 , Z6 , Z7 ) + sδF (Z3 , Z6 , Z7 ) (1 − r)δF (Z2 , Z8 , Z9 ) + rδF (Z3 , Z8 , Z9 )
=(s − r) δF (Z2 , Z6 , Z7 )δF (Z3 , Z8 , Z9 ) − δF (Z3 , Z6 , Z7 )δF (Z2 , Z8 , Z9 )
=(s − r)δF [(Z2 , Z3 ), (Z6 , Z7 ), (Z8 , Z9 )]
=
Z4 Z5
δF [(Z2 , Z3 ), (Z6 , Z7 ), (Z8 , Z9 )],
Z2 Z3
as Z5 − Z4 = (1 − s)Z2 + sZ3 − [(1 − r)Z2 + rZ3 ] = (s − r)(Z3 − Z2 ).
On applying this, we have that if the lines Z4 Z5 , Z6 Z7 and Z8 Z9 meet at non-collinear points
Z1 , Z2 , Z3 , with
Z4 , Z5 ∈ Z2 Z3 , Z6 , Z7 ∈ Z3 Z1 , Z8 , Z9 ∈ Z1 Z2 ,
then
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )] =
Z4 Z5 Z6 Z7 Z8 Z9
δF [(Z2 , Z3 ), (Z3 , Z1 ), (Z1 , Z2 )],
Z2 Z3 Z3 Z1 Z1 Z2
and the final term is equal to
δF (Z2 , Z3 , Z1 )δF (Z3 , Z1 , Z2 ) = δF (Z1 , Z2 , Z3 )2 .
(1.2.10)
9
1.2. DUAL SENSED-AREA
This identifies the dual sensed-area but we also note that it can be re-expressed as
δF (Z1 , Z2 , Z3 )2
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )]
=
.
Z4 Z5 Z6 Z7 Z8 Z9
Z2 Z3 Z3 Z1 Z1 Z2
This last was taken as a fundamental concept by Clifford [4, pp.90-91, 111], and referred to as
the projector of a triangle.
Z6
b
b
Z3
b
Z7
Z2
b
Z5
b
b
b
Z1
Z4
Z8
b
Z3
Z8
b
Figure 1.2. Dual sensed-area.
b
b
Z7
b
b
b
Z5
Z1
Z6
b
Z9
b
Z10
b
Z2
b
b
Z4
Z9
The dual sensed-area is equal to 0 if Z4 Z5 , Z6 Z7 , Z8 Z9 are all parallel, so a remaining case to
be identified is when Z3 Z4 and Z5 Z6 are distinct parallel lines and Z7 Z8 is a transversal. The
most basic case of this is when Z2 Z1 k Z3 Z4 and Z2 6= Z3 . We then have
δF [(Z1 , Z2 ), (Z2 , Z3 ), (Z3 , Z4 )] = δF (Z1 , Z2 , Z3 )δF (Z2 , Z3 , Z4 ),
and also
δF (Z1 , Z2 , Z3 ) = δF (Z1 , Z2 , Z4 ),
δF (Z3 , Z4 , Z2 ) = δF (Z3 , Z4 , Z1 ).
The more general case can be expressed in terms of this, as the earlier case was expressed in terms
of (Z1 , Z2 , Z3 ) and sensed-ratios.
1.2.4
Linear functions
Further light can be thrown on the identity (1.1.11) by the following argument. Let f be a linear
function, that is of the form
f (W ) = au + bv + c,
where W ≡ (u, v). We show that f is determined by its values at three non-collinear points. For
suppose that
au1 + bv1 + c =d1 ,
au2 + bv2 + c =d2 ,
au3 + bv3 + c =d3 .
10
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
Then by Cramer’s rule we have the unique solution
1
a=
2δF (W1 , W2 , W3 ) 1
b=
2δF (W1 , W2 , W3 ) 1
c=
2δF (W1 , W2 , W3 ) d1
d2
d3
v1
v2
v3
u1
u2
u3
d1
d2
d3
u1
u2
u3
v1
v2
v3
,
1 1 ,
1 d1 d2 .
d3 1
1
1
If now we suppose that f takes equal values at three non-collinear points we can deduce that
it must be a constant function. For then a = b = 0 as in each of the relevant determinants one
column is a multiple of another.
We can use this to give an easy proof of (1.1.11). For
δF (W, W2 , W3 ) + δF (W, W3 , W1 ) + δF (W, W1 , W2 )
(1.2.11)
has the given form and has the value δF (W1 , W2 , W3 ) at each of the points W1 , W2 and W3 so it
must be constant when these are non-collinear. For the remaining cases, first consider when W1 =
W2 ; then (1.2.11) is equal to δF (W, W1 , W3 ) + δF (W, W3 , W1 ) = 0 while also δF (W1 , W1 , W3 ) = 0.
Finally suppose that W3 = (1 − s)W1 + sW2 ; then we have that
δF W, W2 , (1 − s)W1 + sW2 + δF W, (1 − s)W1 + sW2 , W1 + δF (W, W1 , W2 )
=(1 − s)δF (W, W2 , W1 ) + sδF (W, W2 , W1 ) + δF (W, W1 , W2 )
=0,
while
δF W1 , W2 , (1 − s)W1 + sW2 = 0.
1.3
More material related to Ceva’s theorem
1.3.1
Coordinates for the point of concurrency
Consider a triangle [Z4 , Z5 , Z6 ] with points Z7 ∈ Z5 Z6 , Z8 ∈ Z6 Z4 , Z9 ∈ Z4 Z5 such that
Z4 Z7 , Z5 Z8 , Z6 Z9 are concurrent at a point which we denote by G. We wish to find coordinates of G.
We first suppose that G = (1 − r)Z5 + rZ8 and use the fact that G ∈ Z6 Z9 . Then by (1.2.2)
with W1 = Z5 , W2 = Z8 , W3 = Z6 , W4 = Z9 , W = G we have that
G=−
δF (Z5 , Z6 , Z9 )
δF (Z8 , Z6 , Z9 )
Z5 +
Z8 .
δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )
δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )
Now we let Z8 = (1 − s)Z6 + sZ4 and Z9 = (1 − t)Z4 + tZ5 . Then
δF (Z5 , Z6 , Z9 ) = δF (Z5 , Z6 , (1 − t)Z4 + tZ5 ) = (1 − t)δF (Z5 , Z6 , Z4 ).
Similarly
and
δF (Z8 , Z6 , Z9 ) = δF ((1 − s)Z6 + sZ4 , Z6 , Z9 ) = sδF (Z4 , Z6 , Z9 ),
δF (Z4 , Z6 , Z9 ) = δF (Z4 , Z6 , (1 − t)Z4 + tZ5 ) = −tδF (Z4 , Z5 , Z6 ).
On combining these we have that
δF (Z8 , Z6 , Z9 ) = sδF (Z4 , Z6 , Z9 ) = −stδF (Z4 , Z5 , Z6 ).
(1.3.1)
11
1.3. MORE MATERIAL RELATED TO CEVA’S THEOREM
Also
δF (Z5 , Z6 , Z9 ) = δF (Z5 , Z6 , (1 − t)Z4 + tZ5 ) = (1 − t)δF (Z5 , Z6 , Z4 ).
It follows that
stδF (Z4 , Z5 , Z6 )
Z5
(1 − t)δF (Z4 , Z5 , Z6 ) + stδF (Z4 , Z5 , Z6 )
(1 − t)δF (Z4 , Z5 , Z6 )
+
[(1 − s)Z6 + sZ4 ]
(1 − t)δF (Z4 , Z5 , Z6 ) + stδF (Z4 , Z5 , Z6 )
−st
1−t
=
Z5 +
[(1 − s)Z6 + sZ4 ]
1 − t + st
1 − t + st
1
=
[(1 − t)sZ4 + stZ5 + (1 − t)(1 − s)Z6 ].
1 − t + st
G=
This can be tested when G is the centroid of the triangle as s = t =
Z6 ].
1.3.2
1
2
yields that G = 31 [Z4 + Z5 +
Non-concurrent lines
Consider a triangle [Z4 , Z5 , Z6 ] with points Z7 , Z8 , Z9 such that Z5 Z8 meets Z6 Z9 at a point G,
Z6 Z9 meets Z4 Z7 at a point H, and Z4 Z7 meets Z5 Z8 at a point K. We wish to obtain δF (G, H, K).
By (1.2.2) we have that
G = aZ5 + (1 − a)Z8 =
−δF (Z8 , Z6 , Z9 )
δF (Z5 , Z6 , Z9 )
Z5 +
Z8 .
δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )
δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )
Similarly
H = b′ Z4 + (1 − b′ )Z7 =
−δF (Z7 , Z6 , Z9 )
δF (Z4 , Z6 , Z9 )
Z4 +
Z7 .
δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )
δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )
Further
K = cZ4 + (1 − c)Z7 =
δF (Z4 , Z5 , Z8 )
−δF (Z7 , Z5 , Z8 )
Z4 +
Z7 .
δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )
δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )
Then
=δF (G, H, K)
=δF [aZ5 + (1 − a)Z8 , b′ Z4 + (1 − b′ )Z7 , cZ4 + (1 − c)Z7 ]
=cδF [aZ5 + (1 − a)Z8 , b′ Z4 + (1 − b′ )Z7 , Z4 ] + (1 − c)δF [aZ5 + (1 − a)Z8 , b′ Z4 + (1 − b′ )Z7 , Z7 ]
=c(1 − b′ )δF [aZ5 + (1 − a)Z8 , Z7 , Z4 ] + (1 − c)b′ δF [aZ5 + (1 − a)Z8 , Z4 , Z7 ]
=[b′ (1 − c) − c(1 − b′ )]δF [aZ5 + (1 − a)Z8 , Z4 , Z7 ]
=(b′ − c)[aδF (Z5 , Z4 , Z7 ) + (1 − a)δF (Z8 , Z4 , Z7 )].
Now
−δF (Z7 , Z6 , Z9 )
−δF (Z7 , Z5 , Z8 )
−
δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 ) δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )
δF (Z7 , Z5 , Z8 )[δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )] − δF (Z7 , Z6 , Z9 )[δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )]
=
[δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )][δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )]
δF (Z4 , Z6 , Z9 )δF (Z7 , Z5 , Z8 ) − δF (Z7 , Z6 , Z9 )δF (Z4 , Z5 , Z8 )
=
[δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )][δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )]
δF [(Z4 , Z7 ), (Z6 , Z9 ), (Z5 , Z8 )]
.
=
[δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )][δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )]
b′ − c =
12
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
The multiple of b′ − c is
aδF (Z5 , Z4 , Z7 ) + (1 − a)δF (Z8 , Z4 , Z7 )
−δF (Z8 , Z6 , Z9 )δF (Z5 , Z4 , Z7 ) + δF (Z5 , Z6 , Z9 )δF (Z8 , Z4 , Z7 )
δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )
δF [(Z5 , Z8 ), (Z6 , Z9 ), (Z4 , Z7 )]
.
=
δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )
=
We conclude that
δF (G, H, K)
=−
δF [(Z4 , Z7 ), (Z5 , Z8 ), (Z6 , Z9 )]2
.
[δF (Z4 , Z6 , Z9 ) − δF (Z7 , Z6 , Z9 )][δF (Z4 , Z5 , Z8 ) − δF (Z7 , Z5 , Z8 )][δF (Z5 , Z6 , Z9 ) − δF (Z8 , Z6 , Z9 )]
More complicated material of the type used up to now is contained in Appendix A.
1.4
1.4.1
Projectivities for ranges of points
Central perspectivities
Z3
b
Z1
b
b
b
b
W1
Z′
b
Z Z2
b
b
Z4
b
b
Figure 1.7.
Consider distinct lines Z1 Z2 and Z3 Z4 and a point W1 not on either. Then points Z ≡ (x, y)
on Z1 Z2 have the form
x=
λ
1
λ
1
x1 +
x2 , y =
y1 +
y2 ,
1+λ
1+λ
1+λ
1+λ
and points Z ′ ≡ (x′ , y ′ ) on Z3 Z4 have the form
x′ =
1
µ
1
µ
x3 +
x4 , y ′ =
y3 +
y4 .
1+µ
1+µ
1+µ
1+µ
The points Z, Z ′ , W1 are collinear if and only if δF (W1 , Z, Z ′ ) = 0. We then say that we have a
central perspectivity from the line Z1 Z2 to the line Z3 Z4 , with centre the point W1 . This condition
expands as
δF (W1 , Z2 , Z4 )λµ + δF (W1 , Z2 , Z3 )λ + δF (W1 , Z1 , Z4 )µ + δF (W1 , Z1 , Z3 ) = 0.
(1.4.1)
λ
1
Z1 + 1+λ
Z2 has no point corresponding to λ = −1 and does not
The representation Z = 1+λ
contain Z2 . For the latter we note that
Z=
1
1+
1 Z2 +
λ
1
λ
1+
1
λ
Z1 ,
and so we can cater for Z2 by taking
Z=
λ′
1
Z
+
Z1
2
1 + λ′
1 + λ′
1.4. PROJECTIVITIES FOR RANGES OF POINTS
13
and putting λ′ = 0, in short taking 1/λ = 0. To be meticulous we should take the representation
Z=
ν′
ν
Z1 +
Z2
ν + ν′
ν + ν′
for Z1 Z2 , with ν +ν ′ 6= 0. Then for points other than Z1 and Z2 we can use either of the ratios ν/ν ′
or ν ′ /ν, but for Z1 we must revert to ν = 0 and for Z2 to ν ′ = 0. We make the above convention
for interpretation and use the simpler form; this is commonly done. We make a similar convention
for µ and Z4 .
These conventions fit in with the interpretation of (1.4.1). Suppose first that Z4 is not the
image of Z2 . Then
1
µ
δF W1 , Z2 ,
Z3 +
Z4 = 0,
1+µ
1+µ
so that δF (W1 , Z2 , Z3 ) + µδF (W1 , Z2 , Z4 ) = 0, and this is what we obtain on dividing across by λ
in (1.4.1) to get
1
1
δF (W1 , Z2 , Z4 )µ + δF (W1 , Z2 , Z3 ) + δF (W1 , Z1 , Z4 ) µ + δF (W1 , Z1 , Z3 ) = 0,
λ
λ
and then putting 1/λ = 0. Similarly if Z is the pre-image of Z4 we have
λ
1
Z1 +
Z2 , Z4 = 0,
δF W1 ,
1+λ
1+λ
so that δF (W1 , Z1 , Z4 ) + λδF (W1 , Z2 , Z4 ) = 0, and this is what we obtain on dividing across by µ
in (1.4.1) to get
δF (W1 , Z2 , Z4 )λ + δF (W1 , Z2 , Z3 )λ
1
1
+ δF (W1 , Z1 , Z4 ) + δF (W1 , Z1 , Z3 ) = 0,
µ
µ
and then putting 1/µ = 0.
Suppose next that Z2′ = Z4 . Then we have δF (W1 , Z2 , Z4 ) = 0 and this is what we obtain if
we divide across by 1/λµ in (1.5.1) to get
δF (W1 , Z2 , Z4 ) + δF (W1 , Z2 , Z3 )
1
1
11
+ δF (W1 , Z1 , Z4 ) + δF (W1 , Z1 , Z3 )
= 0,
µ
λ
λµ
and then putting 1/λ = 1/µ = 0.
If the lines Z1 Z2 and Z3 Z4 are parallel, then with the convention just mentioned, the correspondence Z → Z ′ is from the line Z1 Z2 onto Z3 Z4 . However if the lines Z1 Z2 and Z3 Z4 meet in
a unique point, then the point Z in which the line through W1 parallel to Z3 Z4 meets Z1 Z2 will
have no image, and the point Z ′ in which the line through W1 parallel to Z1 Z2 meets Z3 Z4 will
have no pre-image; thus the domain and range will each be a line less one of its points.
1.4.2
Parallel perspectivities
Let neither of the distinct lines Z1 Z2 and Z3 Z4 be parallel to Z7 Z8 . Now consider points Z =
µ
λ
1
1
′
′
1+λ Z1 + 1+λ Z2 on Z1 Z2 and Z = 1+µ Z3 + 1+µ Z4 on Z3 Z4 such that ZZ is always parallel to
Z7 Z8 . Then
δF Z 7 , Z 8 ,
δF (Z7 , Z8 , Z) =δF (Z7 , Z8 , Z ′ ),
1
1
λ
µ
Z1 +
Z2 =δF Z7 , Z8 ,
Z3 +
Z4 ,
1+λ
1+λ
1+µ
1+µ
(1 + µ)[δF (Z7 , Z8 , Z1 ) + λδF (Z7 , Z8 , Z2 )] =(1 + λ)[δF (Z7 , Z8 , Z3 ) + µδF (Z7 , Z8 , Z4 )].
14
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
This expands to
λµ[δF (Z7 , Z8 , Z2 ) − δF (Z7 , Z8 , Z4 )] + λ[δF (Z7 , Z8 , Z2 ) − δF (Z7 , Z8 , Z3 )]
+ µ[δF (Z7 , Z8 , Z1 ) − δF (Z7 , Z8 , Z4 )] + [δF (Z7 , Z8 , Z1 ) − δF (Z7 , Z8 , Z3 )] = 0.
(1.4.2)
We refer to this correspondence Z → Z ′ as a parallel perspectivity from Z1 Z2 to Z3 Z4 . We
need the same convention as to handling Z2 and Z4 . Now the domain and range are each a line.
1.4.3
A property common to central and parallel perspectivities
Consider a central or parallel perspectivity Z → Z ′ from Z1 Z2 to Z3 Z4 under which
Z1 → Z3 ,
Z2 → Z4
where Z1 6= Z3
and Z2 6= Z4 .
Then from (1.2.6) we have
δF [(Z, Z ′ ), (Z1 , Z3 ), (Z2 , Z4 )]
δF (Z, Z1 , Z3 )δF (Z ′ , Z2 , Z4 ) − δF (Z ′ , Z1 , Z3 )δF (Z, Z2 , Z4 )
and as
Z=
1
λ
Z1 +
Z2 ,
1+λ
1+λ
Z′ =
= 0,
= 0,
1
µ
Z3 +
Z4
1+µ
1+µ
this expands to
λδF (Z2 , Z1 , Z3 )δF (Z3 , Z2 , Z4 ) − µδF (Z4 , Z1 , Z3 )δF (Z1 , Z2 , Z4 ) = 0.
(1.4.3)
This determines Z ′ uniquely in terms of Z and shows that a central or parallel perspectivity is
determined by two points and their images, neither point being the point of intersection in the case
of intersecting lines.
We note that (1.4.3) is the simplest formula for perspectivities.
1.4.4
Projectivities
Arising from (1.4.1) and (1.4.2) we now consider a correspondence Z → Z ′ from a line Z1 Z2 to
µ
1
λ
1
a line Z3 Z4 (possibly the same) where Z = 1+λ
Z1 + 1+λ
Z2 , Z ′ = 1+µ
Z3 + 1+µ
Z4 and λ and µ
satisfy a bilinear relationship
aλµ + bλ + cµ + d = 0,
(1.4.4)
with the convention as to Z2 and Z4 noted in §1.4.1. When a 6= 0, we then have
(aλ + c)(aµ + b) = bc − ad,
(1.4.5)
and to prevent a degenerate situation we make the assumption that
ad − bc 6= 0.
(1.4.6)
We then have a function Z → Z ′ from Z1 Z2 to Z3 Z4 except that there is no image for the point
Z with λ = −c/a and there is no pre-image for the point Z ′ with µ = −b/a. When a = 0 we have
bλ + cµ + d = 0,
(1.4.7)
and the condition (1.4.6) implies that b 6= 0, c 6= 0 and so we again we have a function Z → Z ′
from Z1 Z2 to Z3 Z4 . This function, defined by the bilinear relation (1.4.4), subject to the condition
(1.4.6) is said to be a projectivity from Z1 Z2 to Z3 Z4 .
Projectivities are very prominent in modern treatments of projective geometry.
15
1.4. PROJECTIVITIES FOR RANGES OF POINTS
The central perspectivity in (1.4.1) is an example of this as it clearly has the required form and
the condition for non-degeneracy is
δF (W1 , Z2 , Z4 )δF (W1 , Z1 , Z3 ) − δF (W1 , Z2 , Z3 )δF (W1 , Z1 , Z4 )
= δF [(Z4 , Z3 ), (W1 , Z2 ), (W1 , Z1 )] 6= 0.
This simplifies to
δF (W1 , Z4 , Z3 )δF (Z1 , W1 , Z2 ) 6= 0,
and so it is satisfied.
The parallel perspectivity in (1.4.2) is also an example of a projectivity as it has the required
form and the condition for non-degeneracy is
−[δF (Z1 , Z5 , Z2 ) − δF (Z1 , Z5 , Z4 )]δF (Z1 , Z5 , Z3 ) + [δF (Z1 , Z5 , Z2 ) − δF (Z1 , Z5 , Z3 )]δF (Z1 , Z5 , Z4 )
= δF (Z1 , Z5 , Z2 )[δF (Z1 , Z5 , Z4 ) − δF (Z1 , Z5 , Z3 )]
and this is non-zero as we need Z5 6∈ Z1 Z2 and Z3 Z4 6k Z5 Z1 .
Suppose now that under (1.4.4) and (1.4.6) we have three points and their images,
Z1 → Z3 , Z2 → Z4 , Z7 → Z8 ,
where Z1 , Z2 and Z7 are distinct and collinear so that Z3 , Z4 and Z8 are distinct and collinear,
and none of them is the point of intersection in the case of intersecting lines, so that
Z1 6= Z3 ,
Z2 6= Z4 ,
Z7 6= Z8 .
With
1
λ
1
µ
Z1 +
Z2 , Z ′ =
Z3 +
Z4 ,
1+λ
1+λ
1+µ
1+µ
we then have that λ = 0 ⇒ µ = 0 and so we must have d = 0. Thus
Z=
a+b
1
1
+ c = 0,
µ
λ
and we then have that 1/λ = 0 ⇒ 1/µ = 0, so that a = 0. Then bλ + cµ = 0 so that bλ7 + cµ8 = 0
and so we have that
µ8 λ − λ7 µ = 0,
(1.4.8)
where
λ7 =
Z1 Z7
,
Z7 Z2
µ8 =
Z3 Z8
.
Z8 Z4
We can also write
δF (W1 , Z1 , Z7 )
δF (W2 , Z3 , Z8 )
, µ8 =
,
δF (W1 , Z7 , Z2 )
δF (W2 , Z8 , Z4 )
for any W1 6∈ Z1 Z2 and W2 6∈ Z3 Z4 .
From this we see that projectivities between different lines are determined uniquely by three
distinct points and their images, none of the points being the point of intersection in the case of
intersecting lines.
For a property which projectivities hold in common, when a 6= 0 we note by (1.4.4) that for
distinct λ1 , λ2 , λ3 , λ4 , and their images µ1 , µ2 , µ3 , µ4 , respectively,
λ7 =
λ1 − λ3
λ2 − λ4
λ1 − λ4
λ2 − λ3
µ1 − µ3
,
(aµ1 + b)(aµ3 + b)
µ2 − µ4
= (ad − bc)
,
(aµ2 + b)(aµ4 + b)
µ1 − µ4
= (ad − bc)
,
(aµ1 + b)(aµ4 + b)
µ2 − µ3
= (ad − bc)
.
(aµ2 + b)(aµ3 + b)
= (ad − bc)
16
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
On combining these we obtain
(λ1 − λ3 )(λ2 − λ4 )
(µ1 − µ3 )(µ2 − µ4 )
=
.
(λ1 − λ4 )(λ2 − λ3 )
(µ1 − µ4 )(µ2 − µ3 )
When a = 0 this is established even more readily from (1.4.7).
Thus cross-ratio of collinear points is invariant under each projectivity.
1.4.5
Composition and decomposition of projectivities
We note that a composition of two projectivities is also a projectivity. For if we have
a1 λµ + b1 λ + c1 µ + d1
a2 µν + b2 µ + c2 ν + d2
=
=
0,
0,
(a1 λ + c1 )µ + b1 λ + d1
= 0,
(a2 ν + b2 )µ + c2 ν + d2
= 0,
then
and so on eliminating µ we have
(a1 λ + c1 )(c2 ν + d2 ) − (b1 λ + d1 )(a2 ν + b2 ) = 0,
so that
a3 λν + b3 λ + c3 ν + d3 = 0,
where
a3 = a1 c2 − b1 a2 , b3 = a1 d2 − b1 b2 , c3 = c1 c2 − d1 a2 , d3 = c1 d2 − d1 b2 ,
(1.4.9)
and
a3 d3 −b3 c3 = (a1 c2 −b1 a2 )(c1 d2 −d1 b2 )−(a1 d2 −b1 b2 )(c1 c2 −d1 a2 ) = (a1 d1 −b1 c1 )(a2 d2 −b2 c2 ) 6= 0.
The composition of course can have a domain or range which fails to be a full line, but we
note in particular that when a1 6= 0 and a2 6= 0 so that λ = −c1 /a1 has no image under the first
projectivity and ν = −b2 /a2 has no pre-image under the second projectivity, we have
−c1
−b2
−c1 −b2
+ b3
+ c3
+ d3
a1 a2
a1
a2
1
[b2 c1 (a1 c2 − a2 b1 ) − a2 c1 (a1 d2 − b1 b2 ) − a1 b2 (c1 c2 − a2 d1 ) + a1 a2 (c1 d2 − b2 d1 )]
=
a1 a2
=0,
(1.4.10)
a3
and so in the composition this value of λ gives rise to this value of ν.
The composition of the central perspectivity (1.4.1) with another of the same form, Z ′ → Z ′′
where
1
ν
Z ′′ =
Z5 +
Z6 ,
1+ν
1+ν
and so
δF (W2 , Z4 , Z6 )µν + δF (W2 , Z4 , Z5 )µ + δF (W2 , Z3 , Z6 )ν + δF (W2 , Z3 , Z5 ) = 0.
will be found, from (1.4.9), to be
δF [(Z4 , Z3 ), (W1 , Z2 ), (Z6 , W2 )]λν + δF [(Z4 , Z3 ), (W1 , Z2 ), (Z5 , W2 )]λ
+ δF [(Z4 , Z3 ), (W1 , Z1 ), (Z6 , W2 )]ν + δF [(Z4 , Z3 ), (W1 , Z1 ), (Z5 , W2 )] = 0. (1.4.11)
17
1.4. PROJECTIVITIES FOR RANGES OF POINTS
Similarly the composition of the parallel perspectivity (1.4.2)
λµ[δF (Z7 , Z8 , Z2 ) − δF (Z7 , Z8 , Z4 )] + λ[δF (Z7 , Z8 , Z2 ) − δF (Z7 , Z8 , Z3 )]
+ µ[δF (Z7 , Z8 , Z1 ) − δF (Z7 , Z8 , Z4 )] + [δF (Z7 , Z8 , Z1 ) − δF (Z7 , Z8 , Z3 )] = 0.
with another of the same form with Z ′ → Z ′′ where
Z ′′ =
1
ν
Z5 +
Z6 ,
1+ν
1+ν
and Z ′ Z ′′ k Z9 Z10 ,
µν[δF (Z9 , Z10 , Z4 ) − δF (Z9 , Z10 , Z6 )] + µ[δF (Z9 , Z10 , Z4 ) − δF (Z9 , Z10 , Z5 )]
+ ν[δF (Z9 , Z10 , Z3 ) − δF (Z9 , Z10 , Z6 )] + [δF (Z9 , Z10 , Z3 ) − δF (Z9 , Z10 , Z5 )] = 0,
can be tackled from (1.4.2) and (1.4.9). We have for example
a3
=[δF (Z7 , Z8 , Z2 ) − δF (Z7 , Z8 , Z4 )][δF (Z9 , Z10 , Z3 ) − δF (Z9 , Z10 , Z6 )]
− [δF (Z7 , Z8 , Z2 )δF (Z7 , Z8 , Z3 )][δF (Z9 , Z10 , Z4 ) − δF (Z9 , Z10 , Z6 )].
Now
δF (Z2 , Z7 , Z8 ) − δF (Z4 , Z7 , Z8 ) =δF (Z2 − Z4 , Z7 − Z8 , O),
δF (Z3 , Z9 , Z10 ) − δF (Z6 , Z9 , Z10 ) =δF (Z3 − Z6 , Z9 − Z10 , O),
δF (Z2 , Z7 , Z8 ) − δF (Z3 , Z7 , Z8 ) =δF (Z2 − Z3 , Z7 − Z8 , O),
δF (Z4 , Z9 , Z10 ) − δF (Z6 , Z9 , Z10 ) =δF (Z4 − Z6 , Z9 − Z10 , O).
Thus we obtain
a3 = δF (Z2 −Z4 , Z7 −Z8 , O)δF (Z3 −Z6 , Z9 −Z10 , O)−δF (Z2 −Z3 , Z7 −Z8 , O)δF (Z4 −Z6 , Z9 −Z10 , O),
which is of a form which we have not encountered.
We also consider the composition of two central or parallel perspectivities of the form (1.4.3)
λδF (Z2 , Z1 , Z3 )δF (Z3 , Z2 , Z4 ) − µδF (Z4 , Z1 , Z3 )δF (Z1 , Z2 , Z4 ) =
µδF (Z4 , Z3 , Z5 )δF (Z5 , Z4 , Z6 ) − νδF (Z6 , Z3 , Z5 )δF (Z3 , Z4 , Z6 ) =
0,
0.
On eliminating µ we form the composition
ν=
δF (Z2 , Z1 , Z3 )δF (Z3 , Z2 , Z4 )δF (Z4 , Z3 , Z5 )δF (Z5 , Z4 , Z6 )
λ.
δF (Z4 , Z1 , Z3 )δF (Z1 , Z2 , Z4 )δF (Z6 , Z3 , Z5 )δF (Z3 , Z4 , Z6 )
(1.4.12)
As a central or parallel perspectivity is determined by its images at two points and a projectivity
by its images at three points, it seems likely that these concepts do not coincide. To pursue this
we note that under the composition Z → Z ′′ in (1.4.12) we have
Z1 → Z5 ,
Z2 → Z6 ,
and by (1.4.3) the central or parallel perspectivity Z → Z ′′ for which this holds has
λδF (Z2 , Z1 , Z5 )δF (Z5 , Z2 , Z6 ) − νδF (Z6 , Z1 , Z5 )δF (Z1 , Z2 , Z6 ).
This is the same as the formula (1.4.12) for the composition if and only if
δF (Z2 , Z1 , Z3 )δF (Z3 , Z2 , Z4 )δF (Z4 , Z3 , Z5 )δF (Z5 , Z4 , Z6 )δF (Z6 , Z1 , Z5 )δF (Z1 , Z2 , Z6 )
= 1.
δF (Z4 , Z1 , Z3 )δF (Z1 , Z2 , Z4 )δF (Z6 , Z3 , Z5 )δF (Z3 , Z4 , Z6 )δF (Z2 , Z1 , Z5 )δF (Z5 , Z2 , Z6 )
(1.4.13)
18
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
Thus the composition of two central or parallel perspectivities is a central or parallel perspectivity
if and only if (1.4.13) holds. This shows that the concept of projectivity is more general than that
of a central or parallel perspectivity.
Turning to a related problem, as in the lead up to (1.4.8)consider points Z1 , Z2 and Z7 which are
distinct and collinear, and points Z5 , Z6 and Z8 which are distinct and collinear, and a projectivity
Z → Z ′′ from Z1 Z2 to the different line Z5 Z6 , where with Z1 → Z5 Z2 → Z6 , Z7 → Z8 and
Z1 6= Z5 Z2 6= Z6 , Z7 6= Z8 , so that with
Z=
λ
1
Z1 +
Z2 ,
1+λ
1+λ
Z ′′ =
1
ν
Z5 +
Z6 ,
1+ν
1+ν
by (1.4.8) we have
µ8 λ = λ7 ν.
(1.4.14)
We ask if we can express this as a composition of two central perspectivities. We take two points
W1 , W2 on Z7 Z8 and let W1 Z1 meet W2 Z5 at Z3 , W1 Z2 meet W2 Z6 at Z4 . For this, suppose that
W1 =
1
ρ
Z7 +
Z8 ,
1+ρ
1+ρ
Then we have
Z3 =
W2 =
1
σ
Z7 +
Z8 .
1+σ
1+σ
τ1
1
Z1 +
W1 ,
1 + τ1
1 + τ1
and as it lies on Z5 W2 we have
δF Z 1 , Z 5 ,
δF (Z1 , Z5 , W2 ) + τ1 δF (W1 , Z5 , W2 ) =
1
σ
ρ
1
σ
1
=
Z7 +
Z8 + τ1 δF
Z7 +
Z8 , Z5 ,
Z7 +
Z8
1+σ
1+σ
1+ρ
1+ρ
1+σ
1+σ
0,
0,
from which it follows that
τ1 (σ − ρ)δF (Z5 , Z7 , Z8 ) = (1 + ρ) [δF (Z1 , Z5 , Z7 ) + σδF (Z1 , Z5 , Z8 )] .
(1.4.15)
By a similar argument, we have
Z4 =
1
τ2
Z2 +
W1 ,
1 + τ2
1 + τ2
and as it lies on Z5 W2 we have
δF Z 2 , Z 6 ,
σ
1
Z7 +
Z8 + τ2 δF
1+σ
1+σ
δF (Z2 , Z6 , W2 ) + τ2 δF (W1 , Z6 , W2 ) =
1
ρ
1
σ
=
Z7 +
Z8 , Z6 ,
Z7 +
Z8
1+ρ
1+ρ
1+σ
1+σ
0,
0,
from which it follows that
τ2 (σ − ρ)δF (Z6 , Z7 , Z8 ) = (1 + ρ) [δF (Z2 , Z6 , Z7 ) + σδF (Z2 , Z6 , Z8 )] .
(1.4.16)
We can choose τ1 in (1.4.15) as Z5 6∈ Z1 Z8 , Z1 6∈ Z5 Z8 and W1 6= W2 . We also need to avoid
τ1 = −1 and so must ensure that
−(σ − ρ)δF (Z5 , Z7 , Z8 ) 6= (1 + ρ) [δF (Z1 , Z5 , Z7 ) + σδF (Z1 , Z5 , Z8 )] .
(1.4.17)
Similarly we can choose τ2 in (1.4.16) as Z6 6∈ Z2 Z7 , Z2 6∈ Z6 Z8 and W1 6= W2 . We also need to
avoid τ2 = −1 and so must ensure that
−(σ − ρ)δF (Z6 , Z7 , Z8 ) 6= (1 + ρ) [δF (Z2 , Z6 , Z7 ) + σδF (Z2 , Z6 , Z8 )] .
(1.4.18)
19
1.5. QUADRANGLES
b
Z2
Z1
Z7
b
b
b
b
b
Z9
b
W2
Z4
b
Z3
Z5
b
b
Z8
Figure 1.8
Z6
b
W1
We can choose ρ and σ to satisfy (1.4.17) and (1.4.18), by taking ρ − σ to be sufficiently small,
and thus make our choice of W1 and W2 . With this choice, the perspectivity with centre W1 maps
Z1 and Z2 to Z3 and Z4 , and the perspectivity with centre W2 maps Z3 and Z4 to Z5 and Z6 , so
that their composition maps Z1 and Z2 to Z5 and Z6 , respectively. If Z3 Z4 is not parallel to Z7 Z8
it will meet it at a point Z9 and the above central perspectivities will map Z7 to Z9 and Z9 to Z8 ,
respectively. If instead we have Z3 Z4 k Z7 Z8 , then Z7 will have no image under the perspectivity
with centre W1 , and Z8 will have no pre-image under the perspectivity with centre W2 , and so as
in (1.4.10) Z7 will map to Z8 under the composition. This shows that any projectivity between two
distinct lines can be expressed as the composition of two central perspectivities.
Finally suppose that we have a projectivity Z → Z ′′ from a line Z1 Z2 to itself. We take a distinct
line Z10 Z11 and a central perspectivity Z → Z ′ from Z1 Z2 to Z10 Z11 . Then the correspondence
Z ′ → Z ′′ is a projectivity from one line to a different one so it can be expressed as a composition
of two central perspectivities. It follows that the original projectivity Z → Z ′′ , on one line, can be
expressed as a composition of three central perspectivities. Thus any projectivity on one line has
this property.
1.5
Quadrangles
1.5.1
D3
b
b
Z1
b
Z4
b
b
Z3
D2
b
Z2
b
D1
Figure 1.9.
20
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
For a quadruple Z1 , Z2 , Z3 , Z4 of distinct points, we use the notation for lines
l1 =Z2 Z3 , l2 = Z3 Z1 , l3 = Z1 Z2 ,
l4 =Z1 Z4 , l5 = Z2 Z4 , l6 = Z3 Z4 .
Suppose that {l1 , l4 } are non-parallel, with point of intersection denoted by D1 , that {l2 , l5 } are
non-parallel, with point of intersection denoted by D2 , and that {l3 , l6 } are non-parallel, with point
of intersection denoted by D3 . Then by (1.2.1) we have that
D1 = (1 − r)Z2 + rZ3 , D2 = (1 − s)Z3 + sZ1 , D3 = (1 − t)Z1 + tZ2 ,
where
δF (Z2 , Z1 , Z4 )
δF (Z3 , Z1 , Z4 )
, r=
δF (Z2 , Z1 , Z4 ) − δF (Z3 , Z1 , Z4 )
δF (Z2 , Z1 , Z4 ) − δF (Z3 , Z1 , Z4 )
δF (Z3 , Z2 , Z4 )
δF (Z1 , Z2 , Z4 )
, s=
,
1−s=−
δF (Z3 , Z2 , Z4 ) − δF (Z1 , Z2 , Z4 )
δF (Z3 , Z2 , Z4 ) − δF (Z1 , Z2 , Z4 )
δF (Z1 , Z3 , Z4 , )
δF (Z2 , Z3 , Z4 )
, t=
.
1−t=−
δF (Z1 , Z3 , Z4 ) − δF (Z2 , Z3 , Z4 )
δF (Z1 , Z3 , Z4 ) − δF (Z2 , Z3 , Z4 )
1−r =−
Then
δF (D1 , D2 , D3 ) =δF (1 − r)Z2 + rZ3 , (1 − s)Z3 + sZ1 , (1 − t)Z1 + tZ2
=[(1 − r)(1 − s)(1 − t) + rst]δF (Z1 , Z2 , Z3 ).
From this, with j1 , j2 , j3 standing for the denominators above, we obtain that
j1 j2 j3 δF (D1 , D2 , D3 ) = −δF (Z3 , Z1 , Z4 )δF (Z1 , Z4 , Z2 )δF (Z2 , Z4 , Z3 )
+ δF (Z2 , Z4 , Z1 )δF (Z3 , Z4 , Z2 )δF (Z1 , Z4 , Z3 ) δF (Z1 , Z2 , Z3 )
= − 2δF (Z3 , Z2 , Z4 )δF (Z1 , Z3 , Z4 )δF (Z2 , Z1 , Z4 )δF (Z1 , Z2 , Z3 ).
From this we see that D1 , D2 , D3 are non-collinear if and only if no three of the of the original
points Z1 , Z2 , Z3 and Z4 are collinear. In that case we call (Z1 , Z2 , Z3 , Z4 ) a quadrangle, and
(D1 , D2 , D3 ) its diagonal triple. We also write
d1 = D2 D3 ,
d2 = D3 D1 ,
d3 = D1 D2 .
If we advance the original points cyclically, taking
(Z1′ , Z2′ , Z3′ , Z4′ ) = (Z2 , Z3 , Z4 , Z1 ),
then
l1′ =Z2′ Z3′ = Z3 Z4 = l6 ,
l2′ = Z3′ Z1′ = Z4 Z2 = l5 ,
l3′ =Z1′ Z2′ = Z2 Z3 = l1 ,
l4′ = Z1′ Z4′ = Z2 Z1 = l3 ,
l5′
l6′ = Z3′ Z4′ = Z4 Z1 = l4 .
=Z2′ Z4′
= Z 3 Z 1 = l2 ,
It follows that
l1′ ∩ l4′ = l6 ∩ l3 ,
l2′ ∩ l5′ = l5 ∩ l2 ,
l3′ ∩ l6′ = l1 ∩ l4 ,
and so
D1′ = D3 ,
D2′ = D2 ,
D3′ = D1 .
If instead we interchange the first two points, taking
(Z1′′ , Z2′′ , Z3′′ , Z4′′ ) = (Z2 , Z1 , Z3 , Z4 ),
21
1.5. QUADRANGLES
then
l1′′ =Z2′′ Z3′′ = Z1 Z3 = l2 ,
l3′′
l5′′
=Z1′′ Z2′′
=Z2′′ Z4′′
= Z 2 Z 1 = l3 ,
= Z 1 Z 4 = l4 ,
l2′′ = Z3′′ Z1′′ = Z3 Z2 = l1 ,
l4′′ = Z1′′ Z4′′ = Z2 Z4 = l5 ,
l6′′ = Z3′′ Z4′′ = Z3 Z4 = l6 .
It follows that
l1′′ ∩ l4′′ = l2 ∩ l5 ,
l2′′ ∩ l5′′ = l1 ∩ l4 ,
l3′′ ∩ l6′′ = l3 ∩ l6 ,
and so
D1′′ = D2 ,
D2′′ = D1 ,
D3′ = D3 .
On combining these, we conclude that the diagonal triple depends just on the original set of
points {Z1 , Z2 , Z3 , Z4 } although the notation depends on their order.
1.5.2
Definition of sensed area of quadruple
For any ordered set (Z1 , Z2 , Z3 , Z4 ) of four points we define
δF (Z1 , Z2 , Z3 , Z4 ) = δF (Z1 , Z2 , Z3 ) + δF (Z1 , Z3 , Z4 )
x y1 1 1 x1 y1 1 1 1
= x2 y2 1 + x3 y3 1 2
2
x3 y3 1 x4 y4 1 x y1 1 1 x1 y1 1 1 1
= − x3 y3 1 + x3 y3 1 2
2
x2 y2 1 x4 y4 1 x1
y1
1 x1
y1
1
1
1 x3
y3
1 = x3 − x1 y3 − y1 0
= 2
2
x4 − x2 y4 − y2 0 x4 − x2 y4 − y2 0
1 x − x1 y3 − y1 .
= 3
2 x4 − x2 y4 − y2 We refer to δF (Z1 , Z2 , Z3 , Z4 ) as the sensed-area associated with this ordered quadruple of points.
On advancing cyclically through 1, 2, 3, 4 we have that
δF (Z2 , Z3 , Z4 , Z1 ) =δF (Z2 , Z3 , Z4 ) + δF (Z2 , Z4 , Z1 )
1 x4 − x2 y4 − y2 = 2 x1 − x3 y1 − y3 1 x3 − x1 y3 − y1 = 2 x4 − x2 y4 − y2 =δF (Z1 , Z2 , Z3 , Z4 ).
By repetition of this, δF (Z3 , Z4 , Z1 , Z2 ) and δF (Z4 , Z1 , Z2 , Z3 ) are both equal to the first two.
22
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
Z1
Z1
b
b
Z4
b
(i)
b
(ii)
Z4
b
b
b
b
b
Z4
b
Z2
(iii)
Z1
Z5
Z3
b
Z2
b
Z3
b
Z4
b
Z1
Z2
Z3
b
Z2
b
Z3
b
Figure 1.10. (iv)
If [Z1 , Z2 , Z3 , Z4 ] is a convex quadrilateral, then (Z1 , Z2 , Z3 ) and (Z1 , Z3 , Z4 ) are similarly
oriented; when both are anticlockwise we have
δF (Z1 , Z2 , Z3 , Z4 ) = |δF (Z1 , Z2 , Z3 )| + |δF (Z1 , Z3 , Z4 )| = ∆[Z1 , Z2 , Z3 , Z4 ];
when both are clockwise we have
δF (Z1 , Z2 , Z3 , Z4 ) = −|δF (Z1 , Z2 , Z3 )| − |δF (Z1 , Z3 , Z4 )| = −∆[Z1 , Z2 , Z3 , Z4 ].
When (Z1 , Z2 , Z3 ) is anticlockwise and (Z1 , Z3 , Z4 ) is clockwise, we have that
δF (Z1 , Z2 , Z3 , Z4 ) = |δF (Z1 , Z2 , Z3 )| − |δF (Z1 , Z3 , Z4 )|.
This equals ∆[Z1 , Z2 , Z4 ] + ∆[Z2 , Z3 , Z4 ] in (iii) and ∆[Z2 , Z3 , Z5 ] − ∆[Z1 , Z4 , Z5 ] in (iv).
Note that for (1.1.9) we have that
δF (W1 − W2 , Z1 − Z2 , O) =δF (W1 , Z1 , Z2 ) − δF (W2 , Z1 , Z2 )
=δF (Z1 , W2 , Z2 ) + δF (Z1 , Z2 , W1 )
=δF (Z1 , W2 , Z2 , W1 ).
(1.5.1)
Similar to the identity (1.1.11), given a set of four points Z1 , Z2 , Z3 , Z4 in a given cyclic order,
23
1.5. QUADRANGLES
we note that, for all points Z,
δF (Z, Z1 , Z2 ) + δF (Z, Z2 , Z3 ) + δF (Z, Z3 , Z4 ) + δF (Z, Z4 , Z1 )
=[δF (Z, Z1 , Z2 ) + δF (Z, Z2 , Z3 ) + δF (Z, Z3 , Z1 )]
+ [δF (Z, Z1 , Z3 ) + δF (Z, Z3 , Z4 ) + δF (Z, Z4 , Z1 )]
=δF (Z1 , Z2 , Z3 ) + δF (Z3 , Z4 , Z1 )
=δF (Z1 , Z2 , Z3 ) + δF (Z1 , Z3 , Z4 )
=δF (Z1 , Z2 , Z3 , Z4 ).
1.5.3
Linking sensed-areas of quadruples and triples
b
Z3
Z2
b
b
Z1
b
Z4
Figure 1.11.
b
Z0
From (1.1.11), if W3 ∈ W1 W2 then
δF (W, W1 , W3 ) + δF (W, W3 , W2 ) = δF (W, W1 , W2 ).
Now given Z0 , Z1 , Z2 , let Z3 ∈ Z2 Z0 , Z4 ∈ Z1 Z0 . Then we have that
δF (Z0 , Z1 , Z2 )
=δF (Z, Z1 , Z2 ) + δF (Z, Z2 , Z0 ) + δF (Z, Z0 , Z1 ).
But
δF (Z, Z2 , Z0 ) + δF (Z, Z0 , Z3 ) + δF (Z, Z3 , Z2 ) = δF (Z2 , Z0 , Z3 ) = 0,
and so
δF (Z, Z0 , Z1 ) = δF (Z, Z3 , Z0 ) + δF (Z, Z2 , Z3 ).
Similarly
δF (Z, Z0 , Z1 ) = δF (Z, Z1 , Z4 ) + δF (Z, Z4 , Z0 ) = δF (Z0 , Z1 , Z4 ) = 0,
and so
δF (Z, Z0 , Z1 ) = δF (Z, Z4 , Z1 ) + δF (Z, Z0 , Z4 )
Hence
δF (Z0 , Z1 , Z2 )
=δF (Z, Z1 , Z2 ) + δF (Z, Z2 , Z3 ) + δF (Z, Z3 , Z0 ) + δF (Z, Z0 , Z4 )
+ δF (Z, Z4 , Z1 ) + δF (Z, Z3 , Z4 ) − δF (Z, Z3 , Z4 )
=δF (Z1 , Z2 , Z3 , Z4 ) + δF (Z0 , Z4 , Z3 ).
Hence
δF (Z1 , Z2 , Z3 , Z4 ) = δF (Z0 , Z1 , Z2 ) − δF (Z0 , Z4 , Z3 ).
24
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
Exercises
1.1 Suppose that Z1 , Z2 , Z3 are non-collinear points. Show that the point Z ≡ (x, y) is on the
mid-line of |Z2 Z1 Z3 when
δF (Z, Z1 , Z2 ) δF (Z, Z1 , Z3 )
+
= 0.
|Z1 , Z2 |
|Z1 , Z3 |
1.2 Suppose that Z1 , Z2 , Z3 are non-collinear points. Show that the median of [Z1 , Z2 , Z3 ] which
passes through Z1 has the equation
δF (Z1 , Z2 , Z) − δF (Z, Z3 , Z1 ) = 0.
1.3 Suppose that Z1 , Z2 , Z3 are non-collinear points, and that
Z4 = (1 − r1 )Z2 + r1 Z3 , Z5 = (1 − r2 )Z2 + r2 Z3 , Z6 = (1 − s1 )Z3 + s1 Z1 ,
Z7 = (1 − s2 )Z3 + s2 Z1 , Z8 = (1 − t1 )Z1 + t1 Z2 , Z9 = (1 − t2 )Z1 + t2 Z2 .
Show that δF (Z4 , Z6 , Z8 ) = δF (Z5 , Z7 , Z9 ) if and only if
(1 − r1 )(1 − s1 )(1 − t1 ) + r1 s1 t1 = (1 − r2 )(1 − s2 )(1 − t2 ) + r2 s2 t2 ,
and that a sufficient condition for this is
r1 + r2 = 1, s1 + s2 = 1, t1 + t2 = 1.
1.4 (Eperson, 1996). For a triangle [A, B, C], let the centroid be G, the incentre be I, and the
points of contact of the excircles with the side-lines be P, Q and R so that AP, BR, CQ are
concurrent at a point E. Suppose that the incentre of the circle through the mid-points of
the sides is J. Prove that E, J, G and I are collinear.
1.5 (Klamkin and Liu, 1992). Suppose that Z1 , Z2 , Z3 are non-collinear points, and that
Z4
=
Z6
=
Z8
=
1
λ1
1
λ2
Z2 +
Z3 , Z5 =
Z3 +
Z2 ,
1 + λ1
1 + λ1
1 + λ2
1 + λ2
1
µ1
1
µ2
Z3 +
Z1 , Z7 =
Z1 +
Z3 ,
1 + µ1
1 + µ1
1 + µ2
1 + µ2
ν1
1
ν2
1
Z1 +
Z2 , Z9 =
Z2 +
Z1 .
1 + ν1
1 + ν1
1 + ν2
1 + ν2
Prove that the lines Z5 Z6 , Z7 Z8 , Z9 Z4 are either concurrent or all parallel if and only if
λ1 µ1 ν1 + λ2 µ2 ν2 + λ1 λ2 + µ1 µ2 + ν1 ν2 = 1.
1.6 (Abeles, 1972). For non-collinear points Z1 , Z2 , Z3 , let
Z4 ∈ Z2 Z3 ,
Z5 ∈ Z3 Z1 ,
Z6 ∈ Z1 Z2 ,
be such that Z1 Z4 , Z2 Z5 , Z3 Z6 are concurrent at Z0 . Prove that then
Z0 Z4
Z0 Z5
Z0 Z6
+
+
= −1.
Z4 Z1
Z5 Z2
Z6 Z3
[Hint. In the notation used in the proof of Ceva’s theorem in Barry [2, 11.5.2], show that
u
v
w
−
−
1−u 1−v
1−w
v
w
u
−
−
−
1−u 1−v
1−w
−
=
=
s
t
1−s
1
t
+
+
=
+
,
1−v 1−w
1−v
1−v
1−w
1−t
1−r
r
1−t
1
+
+
=
+
,
1−w 1−u 1−u
1−w 1−u
25
1.5. QUADRANGLES
whence by addition
−2
and so
u
v
w
+
+
1−u 1−v 1−w
=
1
1
1
+
+
,
1−u 1−v 1−w
v
w
u
+
+
= −1.]
1−u 1−v
1−w
1.7 For points Z1 , Z3 not on the line Z2 Z4 , suppose that Z1 Z3 meets Z2 Z4 at Z5 . Show that
Z1 Z5
δF (Z2 , Z4 , Z3 )
.
=−
δF (Z2 , Z4 , Z1 )
Z5 Z3
26
CHAPTER 1. GEOMETRIC ALGEBRA FOR A EUCLIDEAN PLANE
Chapter 2
Pencils of lines
2.1
Pencils of lines
2.1.1
A basic feature of dual sensed-area
We now consider when dual sensed-area has a property similar to the affine line property noted in
(1.1.7). If we take
x9
x10
=
=
(1 − s)x5 + sx7 , y9 = (1 − s)y5 + sy7 ,
(1 − s)x6 + sx8 , y10 = (1 − s)y6 + sy8 ,
we find that
δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z9 , Z10 )]
= (1 − s)δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] + sδF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )],
or in a more suggestive notation
δF [(Z1 , Z2 ), (Z3 , Z4 ), (1 − s)(Z5 , Z6 ) + s(Z7 , Z8 )]
= δF [(Z1 , Z2 ), (Z3 , Z4 ), (1 − s)Z5 + sZ7 , (1 − s)Z6 + sZ8 ]
= (1 − s)δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] + sδF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )],
(2.1.1)
for all s ∈ R, if and only if at least one of
Z1 = Z2 ; Z3 = Z4 ; Z1 Z2 k Z3 Z4 ;
Z5 = Z7 ; Z6 = Z8 ; Z5 Z7 k Z6 Z8 ,
(2.1.2)
holds.
To prove this note that
=
=
=
δF [ (1 − s)Z5 + sZ7 , (1 − s)Z6 + sZ8 , (Z1 , Z2 ), (Z3 , Z4 )]
δF (1 − s)Z5 + sZ7 , Z1 , Z2 δF (1 − s)Z6 + sZ8 , Z3 , Z4
− δF (1 − s)Z6 + sZ8 , Z1 , Z2 δF (1 − s)Z5 + sZ7 , Z3 , Z4
(1 − s)δF (Z5 , Z1 , Z2 ) + sδF (Z7 , Z1 , Z2 ) (1 − s)δF (Z6 , Z3 , Z4 ) + sδF (Z8 , Z3 , Z4 )
− (1 − s)δF (Z6 , Z1 , Z2 ) + sδF (Z8 , Z1 , Z2 ) (1 − s)δF (Z5 , Z3 , Z4 ) + sδF (Z7 , Z3 , Z4 )
(1 − s)2 δF (Z5 , Z1 , Z2 )δF (Z6 , Z3 , Z4 ) − δF (Z6 , Z1 , Z2 )δF (Z5 , Z3 , Z4 )
+ s2 δF (Z7 , Z1 , Z2 )δF (Z8 , Z3 , Z4 ) − δF (Z8 , Z1 , Z2 )δF (Z7 , Z3 , Z4 )
+ s(1 − s) δF (Z5 , Z1 , Z2 )δF (Z8 , Z3 , Z4 ) + δF (Z7 , Z1 , Z2 )δF (Z6 , Z3 , Z4 )
− δF (Z6 , Z1 , Z2 )δF (Z7 , Z3 , Z4 ) − δF (Z8 , Z1 , Z2 )δF (Z5 , Z3 , Z4 ) .
27
28
CHAPTER 2. PENCILS OF LINES
Our expression has the form
u(1 − s)2 + vs2 + ws(1 − s) = u + (w − 2u)s + (u + v − w)s2 ,
and to make this identically equal to
(1 − s)δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] + sδF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )],
we first need u + v − w = 0 and then have
u + (w − 2u)s = u + (v − u)s,
so we need
u = δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )],
v = δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )].
Now above we have
u
= δF (Z5 , Z1 , Z2 )δF (Z6 , Z3 , Z4 ) − δF (Z6 , Z1 , Z2 )δF (Z5 , Z3 , Z4 )
= δF [(Z5 , Z6 ), (Z1 , Z2 ), (Z3 , Z4 )] = δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )],
as required.
Secondly above we had
v
=
=
δF (Z7 , Z1 , Z2 )δF (Z8 , Z3 , Z4 ) − δF (Z8 , Z1 , Z2 )δF (Z7 , Z3 , Z4 )
δF [(Z7 , Z8 ), (Z1 , Z2 ), (Z3 , Z4 )] = δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )],
as required.
Finally we have that
=
=
=
=
=
=
=
−u − v + w
−δF (Z5 , Z1 , Z2 )δF (Z6 , Z3 , Z4 ) + δF (Z6 , Z1 , Z2 )δF (Z5 , Z3 , Z4 )
− δF (Z7 , Z1 , Z2 )δF (Z8 , Z3 , Z4 ) + δF (Z8 , Z1 , Z2 )δF (Z7 , Z3 , Z4 )
+ δF (Z5 , Z1 , Z2 )δF (Z8 , Z3 , Z4 ) + δF (Z7 , Z1 , Z2 )δF (Z6 , Z3 , Z4 )
− δF (Z6 , Z1 , Z2 )δF (Z7 , Z3 , Z4 ) − δF (Z8 , Z1 , Z2 )δF (Z5 , Z3 , Z4 )
δF (Z6 , Z3 , Z4 ) δF (Z7 , Z1 , Z2 ) − δF (Z5 , Z1 , Z2 )
− δF (Z5 , Z3 , Z4 ) δF (Z8 , Z1 , Z2 ) − δF (Z6 , Z1 , Z2 )
− δF (Z8 , Z3 , Z4 ) δF (Z7 , Z1 , Z2 ) − δF (Z5 , Z1 , Z2 )
+ δF (Z7 , Z3 , Z4 ) δF (Z8 , Z1 , Z2 ) − δF (Z6 , Z1 , Z2 )
− δF (Z8 , Z3 , Z4 ) − δF (Z6 , Z3 , Z4 ) δF (Z7 , Z1 , Z2 ) − δF (Z5 , Z1 , Z2 )
+ δF (Z7 , Z3 , Z4 ) − δF (Z5 , Z3 , Z4 ) δF (Z8 , Z1 , Z2 ) − δF (Z6 , Z1 , Z2 )
δF (Z8 − Z6 , Z1 − Z2 , O)δF (Z7 − Z5 , Z3 − Z4 , O)
− δF (Z7 − Z5 , Z1 − Z2 , O)δF (Z8 − Z6 , Z3 − Z4 , O)
δF [(Z7 − Z5 , Z8 − Z6 ), (Z3 − Z4 , O), (Z1 − Z2 , O)]
δF [(Z1 − Z2 , O), (Z7 − Z5 , Z8 − Z6 ), (Z3 − Z4 , O)]
−δF (O, Z7 − Z5 , Z8 − Z6 )δF (Z1 − Z2 , Z3 − Z4 , O).
Equating to 0 the second of these final two terms corresponds to the first three entries in
(2.1.2); equating to 0 the first term similarly gives the final three entries in (2.1.2). In practice
it is the condition on ((Z5 , Z6 ), (Z7 , Z8 )) that we utilize, as we do not wish to have the choice of
((Z1 , Z2 ), (Z3 , Z4 )) unduly restricted to go with it.
29
2.1. PENCILS OF LINES
2.1.2
A pencil of lines
Given any non-collinear points Z4 , Z5 , Z8 , any line which passes through Z4 and is not parallel to
Z5 Z8 will meet Z5 Z8 in a unique point Zr = (1 − r)Z5 + rZ8 for some r ∈ R. Then all such lines
are of the form Z4 Zr and are said to form a concurrent pencil of lines with vertex Z4 .
For distinct points Z1 , Z2 we have several times used the fact that points Z of the form
Z = (1 − r)Z1 + rZ2 ,
are the points on the line Z1 Z2 as
δF (Z1 , Z2 , Z) = (1 − r)δF (Z1 , Z2 , Z1 ) + rδF (Z1 , Z2 , Z2 ) = 0.
We now consider a dual of this so as to obtain a parametric form of the lines on a point.
b
b
b
b
b
b
b
Z5
Vr
Ur
Z7
Z4
b
Z8
b
Z8
b
b
Z4
b
b
Z7 Ur
b
b
b
b
b
Vr
b
b
b
Z4 = Z7 = Ur
Figure 2.1.
Z5
b
b
Z5
b
b
b
Z8 Vr
For pairs of points (Z4 , Z5 ), (Z7 , Z8 ), where Z4 6= Z5 , Z7 6= Z8 we consider the pairs
(Ur , Vr ) = (1 − r)(Z4 , Z5 ) + r(Z7 , Z8 ),
for a real number r. Then we have that
Ur = (1 − r)Z4 + rZ7 ,
Vr = (1 − r)Z5 + rZ8 ,
so that Ur ∈ Z4 Z7 or Ur = Z4 = Z7 , and Vr ∈ Z5 Z8 , and, unless Z4 = Z7 ,
Z4 Ur
r
Z5 Vr
=
.
=
1−r
Ur Z7
Vr Z8
By (2.1.1) and (2.1.2), if
δF (O, Z7 − Z4 , Z8 − Z5 ) = 0,
(2.1.3)
which is equivalent to Z4 Z7 k Z5 Z8 when Z4 6= Z7 and Z5 6= Z8 , then
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Ur , Vr )]
=(1 − r)δF [(Z4 , Z5 ), (Z7 , Z8 ), (Ur , Vr )] + rδF [(Z4 , Z5 ), (Z7 , Z8 ), (Ur , Vr )]
=0,
(2.1.4)
and so the lines Z4 Z5 , Z7 Z8 , Ur Vr are concurrent or all parallel. When these lines are concurrent,
we take Z4 to be the point of concurrence and Z7 = Ur = Z4 Thus we have a concurrent pencil
with vertex Z4 . When the lines Z4 Z5 , Z7 Z8 , Ur Vr are parallel we say that we have a parallel
pencil. We emphasize that when Z4 = Z7 points Vr do not occur on all lines through the point of
intersection Z4 ; a single line is an exception, the one through Z4 which is parallel to Z5 Z8 .
We refer to two pairs ((Z4 , Z5 ), (Z7 , Z8 )) satisfying (2.1.3) as an amenable pair of pairs of
points for the lines Z4 Z5 and Z7 Z8 .
30
CHAPTER 2. PENCILS OF LINES
2.1.3
Lines in two pencils
Consider two pencils P1 and P2 , based on amenable pairs of pairs ((Z4 , Z5 ), (Z7 , Z8 )) and
((Z10 , Z11 ), (Z13 , Z14 )). We first take a pair (Ur , Vr ) in P1 , so that we are considering
(Ur , Vr ) = (1 − r)(Z4 , Z5 ) + r(Z7 , Z8 ),
and ask when Z10 Z11 , Z13 Z14 , Ur Vr are concurrent or are all parallel. Now
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Ur , Vr )]
=(1 − r)δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] + rδF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )]
=δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] − rJ1 ,
(2.1.5)
where
J1 = δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )].
(2.1.6)
Then Z10 Z11 , Z13 Z14 , Ur Vr are concurrent or are all parallel if and only if (2.1.5) is equal to 0
There would be no such r or infinitely many such values of r if and only if J1 = 0, with the
former or the latter case according as δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] is or is not non-zero.
When J1 6= 0, there is a unique solution given by
r=
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )]
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )]
, 1−r =−
.
J1
J1
(2.1.7)
We also consider pairs (Ur′ ′ , Vr′′ ) in P2 ,
(Ur′ ′ , Vr′′ ) = (1 − r′ )(Z10 , Z11 ) + r′ (Z13 , Z14 ),
and similarly deal with when Z4 Z5 , Z7 Z8 , Ur′ ′ Vr′′ are concurrent or are all parallel. In this
connection we deal with
J2 = δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] − δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z13 , Z14 )].
It is a formidable task to elucidate all the cases when J1 = 0 or J2 = 0, or both. As a first
preliminary, we note that if W4 − W3 = k1 (W2 − W1 ), then we have that
δF (W, W3 , W4 ) =δF (W − W3 , O, W4 − W3 )
=k1 δF (W − W3 , W1 − W2 , O)
=k1 δF (W − W3 , O, W2 − W1 )
=k1 [δF (W, W1 , W2 ) − δF (W3 , W1 , W2 )].
(2.1.8)
As a second preliminary, we note that we can rewrite J1 as
δF (Z10 , Z13 , Z14 )δF (Z11 , Z4 , Z5 ) − δF (Z11 , Z13 , Z14 )δF (Z10 , Z4 , Z5 )
− [δF (Z10 , Z13 , Z14 )δF (Z11 , Z7 , Z8 ) − δF (Z11 , Z13 , Z14 )δF (Z10 , Z7 , Z8 )]
=δF (Z10 , Z13 , Z14 )[δF (Z11 , Z4 , Z5 ) − δF (Z11 , Z7 , Z8 )]
− δF (Z11 , Z13 , Z14 )[δF (Z10 , Z4 , Z5 ) − δF (Z10 , Z7 , Z8 )].
As expressions of the form δF (W, W1 , W2 ) − δF (W, W3 , W4 ) feature prominently in this we
analyze it as follows. It is equal to
u v 1 u v 1 1
1
u1 v1 1 − u3 v3 1 2 u v 1 2 u v 1 2
2
4
4
1
= 2 [v4 − v3 ) − (v2 − v1 )]u − [u4 − u3 ) − (u2 − u1 )]v + u1 v2 − u2 v1 − (u3 v4 − u4 v3 )
=δF O, W, W4 − W3 − (W2 − W1 ) + δF (O, W1 , W2 ) − δF (O, W3 , W4 ).
31
2.2. QUADRILATERAL
so that
δF (W, W1 , W2 ) − δF (W, W3 , W4 )
=δF O, W, W4 − W3 − (W2 − W1 ) + δF (O, W1 , W2 ) − δF (O, W3 , W4 ).
(2.1.9)
The expression in (2.1.9) is constant (as W varies) if and only if W4 − W3 = W2 − W1 . This
constant value is equal to 0 if in addition δF (O, W1 , W2 ) = δF (O, W3 , W4 ). By (2.1.8) with k1 = 1,
then
δF (O, W3 , W4 ) = δF (O, W1 , W2 ) − δF (W3 , W1 , W2 ),
and this is equal to δF (O, W1 , W2 ) if and only if W3 ∈ W1 W2 . We conclude that
δF (W, W1 , W2 ) − δF (W, W3 , W4 ) = 0 for all W ∈ Π
if and only if
W4 − W3 = W2 − W1
and W3 ∈ W1 W2 .
(2.1.10)
Details of the cases that occur are given in Appendix B.
2.1.4
An example for §2.1.3
Dually to §1.1.2 we start with three pencils based on amenable quadruples of points
λ
1
(W1 , W2 ) +
(W3 , W4 ),
1+λ
1+λ
µ
1
(W5 , W6 ) +
(W7 , W8 ),
(Z14 , W14 ) =
1+µ
1+µ
1
ν
(Z15 , W15 ) =
(W9 , W10 ) +
(W11 , W12 ).
1+ν
1+ν
(Z13 , W13 ) =
Then by repeated use of (2.1.1) we have that
δF [(Z13 , W13 ), (Z14 , W14 ), (Z15 , W15 )](1 + λ)(1 + µ)(1 + ν)
=δF [(W1 , W2 ), (W5 , W6 ), (W9 , W10 )] + νδF [(W1 , W2 ), (W5 , W6 ), (W11 , W12 )]
+µ δF [(W1 , W2 ), (W7 , W8 ), (W9 , W10 )] + νδF [(W1 , W2 ), (W7 , W8 ), (W11 , W12 )]
+λ δF [(W3 , W4 ), (W5 , W6 ), (W9 , W10 )] + νδF [(W3 , W4 ), (W5 , W6 ), (W11 , W12 )]
+λµ δF [(W3 , W4 ), (W7 , W8 ), (W9 , W10 )] + νδF [(W3 , W4 ), (W7 , W8 ), (W11 , W12 )] .
(2.1.11)
It follows that the lines Z13 W13 , Z14 , W14 and Z15 W15 are either concurrent or all parallel if and
only this expression is equal to 0.
2.2
Quadrilateral
2.2.1
Dually to §1.5 we consider a quadruple
(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 ), (Z7 , Z8 ),
of distinct pairs of distinct points. With these, we suppose first that the pair-pencils on the
amenable quadruples ((Z3 , Z4 ), (Z5 , Z6 )) and ((Z1 , Z2 ), (Z7 , Z8 )) have a unique pair in common
(D1 , D2 ) = (1 − r)(Z3 , Z4 ) + r(Z5 , Z6 ).
32
CHAPTER 2. PENCILS OF LINES
Then by §2.1.3 J1 6= 0, J2 6= 0, where
J1 = δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z3 , Z4 )] − δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z5 , Z6 )],
is as in (2.1.6) and J2 is given similarly. Moreover as in (2.1.7)
r=
1
1
δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z3 , Z4 )], 1 − r = − δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z5 , Z6 )].
J1
J1
(2.2.1)
We suppose secondly that the pair-pencils on the amenable quadruples ((Z5 , Z6 ), (Z1 , Z2 )) and
((Z3 , Z4 ), (Z7 , Z8 )) have a unique pair in common
(D3 , D4 ) = (1 − s)(Z5 , Z6 ) + s(Z1 , Z2 ),
so that J3 6= 0, J4 6= 0, where
J3 = δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z5 , Z6 )] − δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z1 , Z2 )],
and
s=
1
1
δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z5 , Z6 )], 1 − s = − δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z1 , Z2 )].
J3
J3
(2.2.2)
We suppose thirdly that the pair-pencils on the amenable quadruples ((Z1 , Z2 ), (Z3 , Z4 )) and
((Z5 , Z6 ), (Z7 , Z8 )) have a unique pair in common
(D5 , D6 ) = (1 − t)(Z1 , Z2 ) + t(Z3 , Z4 ),
so that J5 6= 0, J6 6= 0, where
J5 = δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z1 , Z2 )] − δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z3 , Z4 )],
and
t=
1
1
δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z1 , Z2 )], 1 − t = − δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z3 , Z4 )].
J3
J3
(2.2.3)
Then we have that
δF [(D1 , D2 ), (D3 , D4 ), (D5 , D6 )]
=δF [(1 − r)(Z3 , Z4 ) + r(Z5 , Z6 ), (1 − s)(Z5 , Z6 ) + s(Z1 , Z2 ), (1 − t)(Z1 , Z2 ) + t(Z3 , Z4 )]
=[(1 − r)(1 − s)(1 − t) + rst]δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )].
Now
[(1 − r)(1 − s)(1 − t) + rst]J1 J3 J5
= − δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z5 , Z6 )]δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z1 , Z2 )]δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z3 , Z4 )]
+δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z3 , Z4 )]δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z5 , Z6 )]δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z1 , Z2 )],
and so
δF [(D1 , D2 ), (D3 , D4 ), (D5 , D6 )]J1 J3 J5
=2δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )]δF [(Z1 , Z2 ), (Z5 , Z6 ), (Z7 , Z8 )]•
δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )]δF [(Z3 , Z4 ), (Z5 , Z6 ), (Z7 , Z8 )].
33
2.2. QUADRILATERAL
D4
D3
b
b
Z8
b
b
b
D5
Z6
K6
D2
b
Z5
b
b
Z7
b
b
D6
b
K2
D1
b
b
b
b
b
Z1
K4
Z2
K3
b
b
K1
b
Z3
K5
b
Z4
Figure 2.3.
From this we see that the lines D1 D2 , D3 D4 , D5 D6 are either non-concurrent or not all parallel if
and only if no three of the lines Z1 Z2 , Z3 Z4 , Z5 Z6 , Z7 Z8 are either concurrent or all parallel. In
this case we say that the lines Z1 Z2 , Z3 Z4 , Z5 Z6 , Z7 Z8 form a quadrilateral and we refer to
the lines D1 D2 , D3 D4 , D5 D6 as its diagonal lines. Note that this definition of a quadrilateral
is different from that in elementary geometry but it is quite usual.
2.2.2
Pairs of opposite vertices
In §2.2.1, suppose that Z3 Z4 , Z5 Z6 meet at K1 , Z5 Z6 , Z1 Z2 meet at K2 , Z1 Z2 , Z3 Z4 meet at K3 ,
Z1 Z2 , Z7 Z8 meet at K4 , Z3 Z4 , Z7 Z8 meet at K5 , Z5 Z6 , Z7 Z8 meet at K6 . Then {K1 , K4 }, {K2 , K5 },
{K3 , K6 } are called pairs of opposite vertices.
Let
K4 =
λ0
1
µ0
1
ν0
1
K2 +
K3 , K5 =
K3 +
K1 , K6 =
K1 +
K2 .
1 + λ0
1 + λ0
1 + µ0
1 + µ0
1 + ν0
1 + ν0
As K4 , K5 , K6 are collinear, all lying on Z7 Z8 , by Menelaus’ theorem we have λ0 µ0 ν0 = −1.
The mid-points of the pairs of opposite vertices are
1
λ0
1
µ0
1
1
M1 = [K1 +
K2 +
K3 ], M2 = [K2 +
K3 +
K1 ],
2
1 + λ0
1 + λ0
2
1 + µ0
1 + µ0
1
ν0
1
M3 = [K3 +
K1 +
K2 ].
2
1 + ν0
1 + ν0
34
CHAPTER 2. PENCILS OF LINES
The sensed area of these three mid-points simplifies to
δF (M1 , M2 , M3 )
1 1 1 1 1 1 1 ν0
=
−
2 2 2 2 2 1 + µ0 2 1 + ν 0
1 1 1 µ0 1
1 1 1 1 1 1
−
+
2 1 + λ0 2 1 + µ0 2 1 + ν0
2 1 + λ0 2 1 + µ0 2
1 λ0 1 µ0 1 ν0 1 λ0 1 1 1
+
δF (K1 , K2 , K3 )
−
2 1 + λ0 2 2 1 + ν0
2 1 + λ0 2 1 + µ0 2 1 + ν0
1 + λ0 µ0 ν0
1
δF (K1 , K2 , K3 ) = 0.
=
4 (1 + λ0 )(1 + µ0 )(1 + ν0 )
It follows that the mid-points of the pairs of opposite vertices of a quadrilateral are collinear.
2.3
Projectivities for pencils of pairs
2.3.1
Z4
b
b
Z8
b
W
b
W1
W′
b
b
b
b
Z1 = Z3 = Z
Z6
b
Z2
b
b
Z5 = Z7 = Z ′
b
b
W2
Figure 2.4.
Dually to the material in 1.4 we consider two amenable double pairs ((Z1 , Z2 ), (Z3 , Z4 )) and
((Z5 , Z6 ), (Z7 , Z8 )), and a further pair (W1 , W2 ) of distinct points. Then as in 2.1.2, on taking the
pencils
λ
1
(Z1 , Z2 ) +
(Z3 , Z4 ),
1+λ
1+λ
µ
1
(Z5 , Z6 ) +
(Z7 , Z8 ),
(Z ′ , W ′ ) =
1+µ
1+µ
(Z, W ) =
(2.3.1)
(2.3.2)
we suppose that W1 W2 , ZW, Z ′ W ′ are concurrent or all parallel. For this it is necessary and
sufficient that
δF [(W1 , W2 ), (Z, W ), (Z ′ , W ′ )] = 0,
from which we obtain
λµδF [(W1 , W2 ), (Z3 , Z4 ), (Z7 , Z8 )] + λδF [(W1 , W2 ), (Z3 , Z4 ), (Z5 , Z6 )]
+ µδF [(W1 , W2 ), (Z1 , Z2 ), (Z7 , Z8 )] + δF [(W1 , W2 ), (Z1 , Z2 ), (Z5 , Z6 )] = 0.
This has the form (1.4.4) and we note that
=
ad − bc
δF [(W1 , W2 ), (Z3 , Z4 ), (Z7 , Z8 )]δF [(W1 , W2 ), (Z1 , Z2 ), (Z5 , Z6 )]
− δF [(W1 , W2 ), (Z3 , Z4 ), (Z5 , Z6 )]δF [(W1 , W2 ), (Z1 , Z2 ), (Z7 , Z8 )],
(2.3.3)
2.3. PROJECTIVITIES FOR PENCILS OF PAIRS
35
and it can be verified that this is equal to
δF [(W1 , W2 ), (Z1 , Z2 ), (Z3 , Z4 )]δF [(W1 , W2 ), (Z5 , Z6 ), (Z7 , Z8 )] 6= 0.
(2.3.4)
We say that the correspondence (Z, W ) → (Z ′ , W ′ ) is a pencil-perspectivity from the first pencil
to the second.
Any relationship from the pencil (2.3.1) to the pencil (2.3.2) of the form (1.4.4) and satisfying
(1.4.6) is called a projectivity. If we need to distinguish this concept from that in §1.4 we can
call the latter a range projectivity and this a pencil projectivity.
We could now develop results similar and dual to those in §1.4, but the expressions become
more complicated. We do not develop the methods of this chapter as they will be replaced by more
efficient ones in Chapter 6.
36
CHAPTER 2. PENCILS OF LINES
Chapter 3
Areal coordinates
In this chapter we develop further the material introduced in Chapter 1. The traditional use made
of sensed areas is to define and apply areal coordinates which we now introduce. They were first
used by Möbius in 1827. We deferred this introduction from Chapter 1 as their use can obscure
patterns which we wish to be evident.
3.1
3.1.1
Areal point coordinates
Areal point coordinates
We recall from Barry [2, 11.4.2] that given non-collinear points Z1 , Z2 , Z3 , any point Z of the plane
can be expressed in the form Z = uZ1 + vZ2 + wZ3 , with u + v + w = 1. For we first seek v, w
such that
v(x2 − x1 ) + w(x3 − x1 ) =
v(y2 − y1 ) + w(y3 − y1 ) =
x − x1 ,
y − y1 .
These equations have the solution
v=
δF (Z, Z3 , Z1 )
δF (Z, Z1 , Z2 )
, w=
,
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
and now we take u = 1 − v − w so that
u=
δF (Z, Z2 , Z3 )
.
δF (Z1 , Z2 , Z3 )
By (1.1.13) we see that for any such u, v, w, for all Z4 , Z5
δF (Z, Z4 , Z5 ) = uδF (Z1 , Z4 , Z5 ) + vδF (Z2 , Z4 , Z5 ) + wδF (Z3 , Z4 , Z5 ).
On applying this by choosing Z4 , Z5 among the vertices, we must have
δ(Z, Z2 , Z3 ) = uδF (Z1 , Z2 , Z3 ), δ(Z, Z3 , Z1 ) = vδF (Z1 , Z2 , Z3 ),
δ(Z, Z1 , Z2 ) = wδF (Z1 , Z2 , Z3 ).
This confirms that this representation is unique.
For non-collinear points Z1 , Z2 , Z3 , for any Z we write
α=
δF (Z, Z2 , Z3 )
,
δF (Z1 , Z2 , Z3 )
β=
δF (Z, Z3 , Z1 )
,
δF (Z1 , Z2 , Z3 )
37
γ=
δF (Z, Z1 , Z2 )
,
δF (Z1 , Z2 , Z3 )
(3.1.1)
38
CHAPTER 3. AREAL COORDINATES
and call (α, β, γ) normalized areal point coordinates of Z with respect to (Z1 , Z2 , Z3 ). This notation
is different from that in Barry [2, 11.4.3], where we used this notation for un-normalized areal point
coordinates. Note that we now have
u = α, v = β, w = γ,
and α + β + γ = 1.
We note from Barry [2, 11.4.3] that
x = x1 α + x2 β + x3 γ,
3.1.2
y = y1 α + y2 β + y3 γ.
(3.1.2)
Sensed-area in terms of areal coordinates
To establish the relationship between areal coordinates and sensed area of any triangle, suppose
that the points Z4 , Z5 , Z6 have areal coordinates (α4 , β4 , γ4 ), (α5 , β5 , γ5 ), (α6 , β6 , γ6 ), respectively.
Then we have
=
=
=
δF (Z4 , Z5 , Z6 )
δF (α4 Z1 + β4 Z2 + γ4 Z3 , α5 Z1 + β5 Z2 + γ5 Z3 , α6 Z1 + β6 Z2 + γ6 Z3 )
[α4 (β5 γ6 − β6 γ5 ) − β4 (α5 γ6 − α6 γ5 ) + γ4 (α5 β6 − α6 β5 )]δF (Z1 , Z2 , Z3 )


α4 β4 γ4
δF (Z1 , Z2 , Z3 ) det  α5 β5 γ5  .
α6 β6 γ6
From this we find that

1
δF (Z1 , Z5 , Z6 ) = δF (Z1 , Z2 , Z3 ) det  α5
α6
and similarly
δF (Z2 , Z5 , Z6 ) = δ1 (γ5 α6 − α5 γ6 ),
0
β5
β6

0
γ5  = δ1 (β5 γ6 − γ5 β6 ),
γ6
δF (Z3 , Z5 , Z6 ) = δ1 (α5 β6 − β5 α5 ),
where δ1 = δF (Z1 , Z2 , Z3 ).
3.2
3.2.1
Equation of a line in areal coordinates
Preliminary identities
Any line U V has an equation δF (Z, U, V ) = 0. With the notation of §3.1, given any distinct points
U, V we ask if it is possible to have an identity of the form
δF (Z, U, V ) = uδF (Z, Z2 , Z3 ) + vδF (Z, Z3 , Z1 ) + wδF (Z, Z1 , Z2 ).
On inserting Z as Z1 , Z2 , Z3 , respectively, we find a necessary condition to be
δF (Z1 , U, V ) = uδF (Z1 , Z2 , Z3 ), δF (Z2 , U, V ) = vδF (Z1 , Z2 , Z3 ),
δF (Z3 , U, V ) = wδF (Z1 , Z2 , Z3 ).
This suggests the identity
δF (Z1 , Z2 , Z3 )δF (Z, U, V ) = δF (Z1 , U, V )δF (Z, Z2 , Z3 )
+ δF (Z2 , U, V )δF (Z, Z3 , Z1 ) + δF (Z3 , U, V )δF (Z, Z1 , Z2 ),
39
3.2. EQUATION OF A LINE IN AREAL COORDINATES
and this is indeed valid as
δF (Z1 , Z2 , Z3 )δF (Z, U, V ) − δF (Z1 , U, V )δF (Z, Z2 , Z3 ) = δF [(Z1 , Z), (Z2 , Z3 ), (U, V )]
and
δF (Z2 , U, V )δF (Z3 , Z1 , Z) − δF (Z3 , U, V )δF (Z2 , Z1 , Z) = δF [(Z2 , Z3 ), (U, V ), (Z1 , Z)].
This is a fundamental identity; we re-write it as
δF (Z, U, V ) = δF (Z1 , U, V )α + δF (Z2 , U, V )β + δF (Z3 , U, V )γ.
(3.2.1)
We also introduce the notation
l = δF (Z1 , U, V ),
3.2.2
m = δF (Z2 , U, V ),
n = δF (Z3 , U, V ).
(3.2.2)
Equation of a line
Given any distinct points Z4 and Z5 , the variable point Z will lie on the line Z4 Z5 if and only if
δF (Z, Z4 , Z5 ) = 0. With the notation of §3.1, let
l4,5 = δF (Z1 , Z4 , Z5 ),
Then Z ∈ Z4 Z5 if and only if
m4,5 = δF (Z2 , Z4 , Z5 ),
n4,5 = δF (Z3 , Z4 , Z5 ).
l4,5 α + m4,5 β + n4,5 γ = 0.
(3.2.3)
We call (3.2.3) an equation of the line Z4 Z5 in areal point coordinates. We shall refer to
(l4,5 , m4,5 , n4,5 ) as areal pair-coordinates of (Z4 , Z5 ).
3.2.3
Coefficient triples
Can all triples (u, v, w) occur as coefficients in equations such as (3.2.3)? If u = v = w we have
u
δF (Z, Z3 , Z1 )
δF (Z, Z1 , Z2 )
δF (Z, Z2 , Z3 )
+v
+w
= u.
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
When u = 0 the equation is then satisfied by all points in the plane, and when u 6= 0 it is satisfied
by no points. Thus there is one case which must be excluded, namely u = v = w.
Conversely, given any triple u, v, w, with u = v = w excluded, we seek Z4 , Z5 such that
δF (Z, Z4 , Z5 ) = 0 if and only if uα + vβ + wγ = 0. We show in fact that we can find points U and
V and a constant k 6= 0 so that
δF (Z, U, V ) = k[uα + vβ + wγ],
for all points Z.
We must have at least two of
w 6= u,
u 6= v,
v 6= w,
so without loss of generality we suppose that w 6= u and u 6= v.
With this we suppose first that u 6= 0. We then take
U=
from which we have that
w
u
Z3 −
Z1 ,
u−w
u−w
V =
v
u
Z1 −
Z2 ,
v−u
v−u
u
w
v
u
Z3 −
Z1 ,
Z1 −
Z2
u−w
u−w
v−u
v−u
u
[uδF (Z, Z2 , Z3 ) + vδF (Z, Z3 , Z1 ) + wδF (Z, Z1 , Z2 )],
=
(u − w)(v − u)
δF (Z, U, V ) =δF Z,
40
CHAPTER 3. AREAL COORDINATES
as required.
As a second case we continue with w 6= u and u 6= v but now take u = 0. We first take v 6= w
as well and with it
w
v
U = Z1 , V =
Z2 −
Z3 ,
w−v
w−v
with which we have
δF (Z, U, V ) =δF Z, Z1 ,
=
w
v
Z2 −
Z3
w−v
w−v
1
[vδF (Z, Z3 , Z1 ) + wδF (Z, Z1 , Z2 )],
w−v
as required.
As a final case we take w 6= u, v = w 6= 0, u = 0, and with it
U = Z1 ,
V = Z1 + Z2 − Z3 .
Then by (1.1.3)
δF (Z, U, V ) =δF (Z, Z1 , Z1 + Z2 − Z3 )
=δF (Z, Z1 , Z1 ) + δF (Z, Z1 , Z2 ) − δF (Z, Z1 , Z3 )
=δF (Z, Z3 , Z1 ) + δF (Z, Z1 , Z2 ),
as required.
We also ask when two pairs (U, V ), (U ′ , V ′ ) have the same areal pair-coordinates, i.e.
δF (Z1 , U, V ) = δF (Z1 , U ′ , V ′ ), δF (Z2 , U, V ) = δF (Z2 , U ′ , V ′ ), δF (Z3 , U, V ) = δF (Z3 , U ′ , V ′ ).
If this holds, then by (3.2.1) we must have
δF (Z, U, V ) = δF (Z, U ′ , V ′ ),
for all points Z, and so by (2.1.10) we must have
V − U = V ′ − U ′,
U ′ ∈ U V.
The converse also holds, as if we have
δF (Z, U, V ) = δF (Z, U ′ , V ′ ),
for all Z, then it must hold in particular for the cases Z1 , Z2 and Z3 and the above argument
applies.
3.2.4
Determinant form of equation
Given areal coordinates (α4 , β4 , γ4 ), (α5 , β5 , γ5 ) of
check that

α β
det  α4 β4
α5 β5
distinct points Z4 , Z5 , respectively, we wish to

γ
γ4  = 0,
γ5
is an equation for the line Z1 Z2 . This equation is clearly of the correct form and satisfied by the
coordinates of Z4 and Z5 . It remains to check that
β4 γ5 − β5 γ4 = γ4 α5 − γ5 α4 = α4 β5 − α5 β4 ,
cannot occur.
3.2. EQUATION OF A LINE IN AREAL COORDINATES
41
Now, as in (3.1.2) and (3.2.3),
δ1 (β4 γ5 − β5 γ4 ) = δF (Z1 , Z4 , Z5 ) = l4,5 , δ1 (γ4 α5 − γ5 α4 ) = δF (Z2 , Z4 , Z5 ) = m4,5 ,
δ1 (α4 β5 − α5 β4 ) = δF (Z3 , Z4 , Z5 ) = n4,5 .
Thus equality of the coefficients is equivalent to
δF (Z1 , Z4 , Z5 ) = δF (Z2 , Z4 , Z5 ) = δF (Z3 , Z4 , Z5 ).
By §3.2.3, this leads to the conclusion that
δF (Z4 , Z2 , Z3 ) = δF (Z5 , Z2 , Z3 ), δF (Z4 , Z3 , Z1 ) = δF (Z5 , Z3 , Z1 ), δF (Z4 , Z1 , Z2 ) = δF (Z5 , Z1 , Z2 ),
and thus that Z4 = Z5 , which gives a contradiction.
3.2.5
Point of intersection of two lines
Suppose that
l 6 α + m6 β + n 6 γ
l 9 α + m9 β + n 9 γ
=
=
0,
0,
are equations of distinct intersecting lines Z4 Z5 and Z7 Z8 . Then the coordinates of their point of
intersection is given by the equations
l 6 α + m6 β + n 6 γ
l 9 α + m9 β + n 9 γ
=
=
0,
0,
α+β+γ
=
1,
and so
α =
β
=
γ
=
m6 n 9 − n 6 m9
,
m6 n 9 − n 6 m9 + n 6 l 9 − l 6 n 9 + l 6 m9 − m6 l 9
n6 l 9 − l 6 n9
,
m6 n 9 − n 6 m9 + n 6 l 9 − l 6 n 9 + l 6 m9 − m6 l 9
l 6 m9 − m6 l 9
,
m6 n 9 − n 6 m9 + n 6 l 9 − l 6 n 9 + l 6 m9 − m6 l 9
We note that
m6 n 9 − n 6 m9
n6 l 9 − l 6 n9
l 6 m9 − m6 l 9
3.2.6
=
=
=
δF [(Z2 , Z3 ), (Z4 , Z5 ), (Z7 , Z8 )],
δF [(Z3 , Z1 ), (Z4 , Z5 ), (Z7 , Z8 )],
δF [(Z1 , Z2 ), (Z4 , Z5 ), (Z7 , Z8 )].
Parallel lines
Suppose that
l 6 α + m6 β + n 6 γ
=
0,
l 9 α + m9 β + n 9 γ
=
0,
are equations of distinct lines. Then these are parallel if and only if the equations
l 6 α + m6 β + n 6 γ
l 9 α + m9 β + n 9 γ
=
=
0,
0,
α+β+γ
=
1,
42
CHAPTER 3. AREAL COORDINATES
have no solution. A necessary and sufficient condition for this is that

l6
det  l9
1
m6
m9
1

n6
n9  = 0.
1
(3.2.4)
We remark that this is also the condition that
l 6 α + m6 β + n 6 γ
=
0,
l 9 α + m9 β + n 9 γ
α+β+γ
=
=
0,
0,
have a solution other than (α, β, γ) = (0, 0, 0).
When (3.2.4) is satisfied, the rows must be linearly dependent, so there are numbers (j1 , j2 , j3 ) 6=
(0, 0, 0) such that
j1 (l6 , m6 , n6 ) + j2 (l9 , m9 , n9 ) + j3 (1, 1, 1) = (0, 0, 0).
First we rule out j1 = j2 = 0 as that would imply 1 = 0. Next we note that if we had j1 = 0 we
clearly could not also have j2 6= 0 as then we would have l9 = m9 = n9 which is ruled out. Thus
j1 6= 0 and similarly j2 6= 0. Further we cannot have j3 = 0 as that would make (l9 , m9 , n9 ) a
multiple of (l6 , m6 , n6 ) and we would have only one line. Thus we must have
(l9 , m9 , n9 ) = k1 (l6 , m6 , n6 ) + k2 (1, 1, 1),
for some k1 , k2 , both non-zero. This is our conclusion.
If l6 α + m6 β + n6 γ = 0 is the equation of a line Z4 Z5 and (α7 , β7 , γ7 ) are areal coordinates of
some point Z7 not on it, we seek the equation of the line which passes through Z7 and is parallel
to Z4 Z5 . Now for points Z on Z4 Z5 we have δF (Z, Z4 , Z5 ) = 0 and so for Z on any line parallel
to Z4 Z5 we have δF (Z, Z4 , Z5 ) = k for some constant k. Thus such a parallel line has an equation
l 6 α + m6 β + n 6 γ = k
which we rewrite as
l6 α + m6 β + n6 γ = k(α + β + γ).
As this is to pass through Z7 we must have
l6 α7 + m6 β7 + n6 γ7 = k(α7 + β7 + γ7 ).
On substituting this value for k in the equation we obtain
α+β+γ
l 6 α + m6 β + n 6 γ
=
,
l6 α7 + m6 β7 + n6 γ7
α7 + β7 + γ7
as the equation of the line parallel to Z4 Z5 which passes through Z7 .
This equation can be checked as follows. It has the correct form and in it l6 α + m6 β + n6 γ = 0
would imply α + β + γ = 0, so it has no point in common with the given line. It is genuinely the
equation of a line as if the coefficients in it were equal to each other we would have
l6 (α7 + β7 + γ7 ) − l6 α7 − m6 β7 − n6 γ7 =m6 (α7 + β7 + γ7 ) − l6 α7 − m6 β7 − n6 γ7
=n6 (α7 + β7 + γ7 ) − l6 α7 − m6 β7 − n6 γ7 ,
which would imply that l6 = m6 = n6 .
43
3.2. EQUATION OF A LINE IN AREAL COORDINATES
3.2.7
Identifying a line for drawing
Suppose that we have an equation of a line
uα + vβ + wγ = 0,
where u = v = w is excluded, and that this is the equation of
l6 α + m6 β + n6 γ = 0.
Then by §3.2.3 we have that
u = kl6 ,
v = km6 ,
w = kn6 ,
for some k 6= 0. On inserting α = 1 − β − γ in the initial equation of the line, we obtain
u(1 − β − γ) + vβ + wγ = 0,
and so
(v − u)β + (w − u)γ + u = 0.
u
When u 6= v we then have that the point with (β, γ) = u−v
, 0 is where this line meets the
u
line Z1 Z2 , and when u 6= w the point with (β, γ) = 0, u−w
is where this line meets he line Z1 Z3 .
This would enable us to draw the line in question.
3.2.8
Parametric equations of a line
We recall from Barry [2] that if Z ≡ (x, y) lies on the line Z4 Z5 , then
x = (1 − t)x4 + tx5 ,
y = (1 − t)y4 + yx5 ,
for some real number t. Then if (α, β, γ) are areal coordinates of Z, we have that
δF (Z, Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF ((1 − t)Z4 + tZ5 , Z2 , Z3 )
=
δF (Z1 , Z2 , Z3 )
δF (Z4 , Z2 , Z3 )
δF (Z5 , Z2 , Z3 )
=(1 − t)
+t
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
=(1 − t)α4 + tα5 ,
α=
and by a similar argument
β = (1 − t)β4 + tβ5 ,
γ = (1 − t)γ4 + tγ5 .
Conversely suppose that for some t ∈ R,
α = (1 − t)α4 + tα5 ,
β = (1 − t)β4 + tβ5 ,
γ = (1 − t)γ4 + tγ5 .
By (3.1.2) we have that
x = x1 α + x2 β + x3 γ,
y = y1 α + y2 β + y3 γ,
and then
x =x1 [(1 − t)α4 + tα5 ] + x2 [(1 − t)β4 + tβ5 ] + +x3 [(1 − t)γ4 + tγ5 ]
=(1 − t)[x1 α4 + x2 β4 + x3 γ4 ] + t[x1 α5 + x2 β5 + x3 γ5 ]
=(1 − t)x4 + tx5 .
44
CHAPTER 3. AREAL COORDINATES
Similarly
y = (1 − t)y4 + ty5 .
It follows that Z ∈ Z4 Z5 if and only if
α = (1 − t)α4 + tα5 ,
β = (1 − t)β4 + tβ5 ,
γ = (1 − t)γ4 + tγ5 ,
for some t ∈ R.
Now, by §3.2.4
m4,5 − n4,5
=γ4 α5 − α4 γ5 − (α4 β5 − β4 α5 )
δ1
=α5 (γ4 + β4 ) − α4 (β5 + γ5 )
=α5 (1 − α4 ) − α4 (1 − α5 )
=α5 − α4 ,
and by a similar argument
n4,5 − l4,5
= β5 − β4 ,
δ1
l4,5 − m4,5
= γ5 − γ4 .
δ1
It follows that the line Z4 Z5 also has parametric equations
α = α4 + t
n4,5 − l4,5
l4,5 − m4,5
m4,5 − n4,5
, β = β4 + t
, γ = γ4 + t
.
δ1
δ1
δ1
The line through Z7 parallel to Z4 Z5 has parametric equations
α = α7 + t(α5 − α4 ), β = β7 + t(β5 − β4 ), γ = γ7 + t(γ5 − γ4 ),
(t ∈ R).
For then
l4,5 α + m4,5 β + n4,5 γ
=l4,5 α7 + m4,5 β7 + n4,5 γ7 + t l4,5 (α5 − α4 ) + m4,5 (β5 − β4 ) + n4,5 (γ5 − γ4 )
=l4,5 α7 + m4,5 β7 + n4,5 γ7 + t.0
=l4,5 α7 + m4,5 β7 + n4,5 γ7 ,
and this is equal to 0 if and only if Z7 ∈ Z4 Z5 .
It follows that this line also has parametric equations
α = α7 + t
3.3
3.3.1
n4,5 − l4,5
l4,5 − m4,5
m4,5 − n4,5
, β = β7 + t
, γ = γ7 + t
,
δ1
δ1
δ1
(t ∈ R).
Areal coordinates under change of triple of reference
Change of triple of reference
Given two triples of non-collinear points (Z1 , Z2 , Z3 ) and (Z4 , Z5 , Z6 ), we denote normalized areal
coordinates of any point Z of the plane with respect to the former as (α, β, γ) and with respect to
the latter as (α′ , β ′ , γ ′ ), so that
Z = αZ1 + βZ2 + γZ3 ,
Z = α′ Z4 + β ′ Z5 + γ ′ Z6 ,
and α + β + γ = 1, α′ + β ′ + γ ′ = 1. To note the correspondence between these, we take
Z1 = α′1 Z4 + β1′ Z5 + γ1′ Z6 , Z2 = α′2 Z4 + β2′ Z5 + γ2′ Z6 , Z3 = α′3 Z4 + β3′ Z5 + γ3′ Z6 ,
45
3.3. AREAL COORDINATES UNDER CHANGE OF TRIPLE OF REFERENCE
with the sum of the coefficients being 1 in each case. Then
Z
= [α′1 Z4 + β1′ Z5 + γ1′ Z6 ]α + [α′2 Z4 + β2′ Z5 + γ2′ Z6 ]β + [α′3 Z4 + β3′ Z5 + γ3′ Z6 ]γ,
= [α′1 α + α′2 β + α′3 γ]Z4 + [β1′ α + β2′ β + β3′ γ]Z5 + [γ1′ α + γ2′ β + γ3′ γ]Z6 ,
and so
α′ =α′1 α + α′2 β + α′3 γ,
β ′ =β1′ α + β2′ β + β3′ γ,
γ ′ =γ1′ α + γ2′ β + γ3′ γ.
(3.3.1)
We also have that
Z4 = α4 Z1 + β4 Z2 + γ4 Z3 , Z5 = α5 Z1 + β5 Z2 + γ5 Z3 , Z6 = α6 Z1 + β6 Z2 + γ6 Z3 ,
and so
Z
= [α4 Z1 + β4 Z2 + γ4 Z3 ]α′ + [α5 Z1 + β5 Z2 + γ5 Z3 ]β ′ + [α6 Z1 + β6 Z2 + γ6 Z3 ]γ ′ ,
= [α4 α′ + α5 β ′ + α6 γ ′ ]Z1 + [β4 α′ + β5 β ′ + β6 γ ′ ]Z2 + [γ4 α′ + γ5 β ′ + γ6 γ ′ ]Z3 .
We thus have
α =α4 α′ + α5 β ′ + α6 γ ′ ,
β =β4 α′ + β5 β ′ + β6 γ ′ ,
γ =γ4 α′ + γ5 β ′ + γ6 γ ′ .
(3.3.2)
These show how we move forwards and backwards between the two triples of reference.
On using matrix notation we can rewrite (3.3.1) as

  ′
α′
α1
 β ′  =  β1′
γ′
γ1′
α′2
β2′
γ2′


α′3
α
β3′   β  ,
γ3′
γ

α5
β5
γ5
 ′ 
α6
α
β6   β ′  .
γ6
γ′
and (3.3.2) as
 
α
α4
 β  =  β4
γ
γ4
From these we can see that
 ′   ′
α
α1
 β ′  =  β1′
γ′
γ1′
α′2
β2′
γ2′

α′3
α4
β3′   β4
γ3′
γ4
α5
β5
γ5
 ′ 
α6
α
β6   β ′  .
γ6
γ′
On multiplying the square matrices here and equating coefficients, we can deduce nine identities.
Similarly from

 
α
α4
 β  =  β4
γ
γ4
α5
β5
γ5
we can write down nine more identities.
 ′
α1
α6
β6   β1′
γ1′
γ6
α′2
β2′
γ2′


α′3
α
β3′   β  ,
γ3′
γ
46
CHAPTER 3. AREAL COORDINATES
3.4
Reducible quadratic equations
3.4.1
We consider when a homogeneous quadratic form in areal coordinates can be factorized. Take
S(α, β, γ) = aα2 + bβ 2 + cγ 2 + 2f βγ + 2gγα + 2hαβ.
We start with the case a 6= 0 and note that then
aS(α, β, γ) = (aα + hβ + gγ)2 − {(h2 − ab)β 2 + 2(gh − af )βγ + (g 2 − ac)γ 2 }.
We wish this to be the difference of two squares and for this note that on taking out h2 − ab as a
common factor in the second main term we obtain
β2 + 2
g 2 − ac 2
gh − af
βγ + 2
γ ,
2
h − ab
h − ab
and on completing the square in this we obtain
2
2
g 2 − ac
gh − af
gh − af
}γ 2 .
γ +{ 2
−
β+ 2
h − ab
h − ab
h2 − ab
The term in γ 2 here is equal to
(g 2 − ac)(h2 − ab) − (gh − af )2 2
γ ,
(h2 − ab)2
and this is equal to 0 if and only if
(g 2 − ac)(h2 − ab) = (gh − af )2 .
(3.4.1)
On cancelling across this by a which is non-zero we obtain
abc + 2f gh − af 2 − bg 2 − ch2 = 0
which has the determinant form

a
det(M ) = det  h
g
If (3.4.1) holds and h2 − ab > 0 we have
h
b
f

g
f  = 0.
c
2
p
gh − af
2
h − abβ + √
γ
aS(α, β, γ) =(aα + hβ + gγ) −
h2 − ab
p
gh − af
γ}•
={aα + (h − h2 − ab)β + g − √
h2 − ab
p
gh − af
2
√
• {aα + (h + h − ab)β + g +
γ}.
h2 − ab
(3.4.2)
2
This gives aS(α, β, β) as the product of two real linear equations. On equating these to 0 we may
have two alternative lines or one repeated line but are not assured of two lines as the coefficients
may be equal in either or both.
If instead h2 − ab < 0 we have the sum of two squares instead
gh − af 2
γ
aS(α, β, γ) =(aα + hβ + gγ)2 + (ab − h2 ) β −
ab − h2
p
gh − af 2
=(aα + hβ + gγ)2 +
ab − h2 β − √
γ ,
ab − h2
47
3.4. REDUCIBLE QUADRATIC EQUATIONS
and this is equal to 0 if and only if we have
aα + hβ + gγ = 0,
p
gh − af
ab − h2 β − √
γ = 0,
ab − h2
(3.4.3)
simultaneously. Whether either of these is a line depends on not having its coefficients all equal.
If instead we have h2 − ab = 0, then by (3.4.1) we also have gh − af = 0, and so
aS(α, β, γ) = (aα + hβ + gγ)2 − (g 2 − ac)γ 2 ,
and we have a difference of two squares or a sum of two squares according as g 2 − ac is positive or
negative. When also g 2 − ac = 0, we have aS(α, β, γ) = (aα + hβ + gγ)2 .
This completes the case when a 6= 0 and a similar argument obtains if b 6= 0 or c 6= 0. We now
move to the case when a = b = c = 0 so that
S(α, β, γ) = 2f βγ + 2gγα + 2hαβ,
and the case f = g = h = 0 is excluded. We suppose that f 6= 0 and note that
1
2 f S(α, β, γ)
=f 2 βγ + f gγα + f hαβ
=(f β + gα)f γ + f hαβ
=(f β + gα)f γ
if h = 0,
and also
1
2 f S(α, β, γ)
=f 2 βγ + f gγα + f hαβ
=(f β + hα)f γ + f gγα
=(f β + hα)f γ if g = 0.
Now it can be checked that in this case det(M ) = 2f gh = 0 and so at least one of g, h is equal to
0 when f 6= 0.
48
CHAPTER 3. AREAL COORDINATES
Chapter 4
Projective and affine
transformations
4.1
4.1.1
Special transformations
Transformations depending on intersecting lines or parallel lines
If the distinct lines Z4 Z5 , Z7 Z8 meet at Z1 , then as k1 varies,
δF (Z, Z4 , Z5 )
= k1
δF (Z, Z7 , Z8 )
gives the lines through Z1 . If Z2 ∈ Z4 Z5 , Z3 ∈ Z7 Z8 are both different from Z1 , these lines can be
represented more simply by
δF (Z, Z1 , Z2 )
= k.
δF (Z, Z3 , Z1 )
Similarly if Z4 Z5 k Z7 Z8 , the lines parallel to these have equations
δF (Z, Z4 , Z5 )
= k1 ,
δF (Z, Z7 , Z8 )
but these are more simply given by
δF (Z, Z4 , Z5 ) = k.
We make use of these simplifications in what follows.
(i) For triples (Z1 , Z2 , Z3 ), (W1 , W2 , W3 ) of non-collinear points if we take
δF (Z, Z3 , Z1 )
δF (W, W3 , W1 )
= k1
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
δF (Z, Z1 , Z2 )
δF (W, W1 , W2 )
= k2
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
for non-zero k1 and k2 , then lines parallel to Z3 Z1 are mapped to lines parallel to W3 W1 , and
lines parallel to Z1 Z2 are mapped to lines parallel to W1 W2 . Taking (α, β, γ) as normalized areal
coordinates of Z with respect to (Z1 , Z2 , Z3 ) and (α′ , β ′ , γ ′ ) as normalized areal coordinates of W
with respect to (W1 , W2 , W3 ), these are
β ′ = k1 β,
γ ′ = k2 γ.
Then
α′ = 1 − β ′ − γ ′ = α + β + γ − k1 β − k2 γ,
49
50
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
so that we have
α′
β′
=
=
α + (1 − k1 )β + (1 − k2 )γ,
k1 β,
γ′
=
k2 γ.
Thus the correspondence Z → W is a function. By working in the opposite direction we find that
it has an inverse function
1
1
α = α′ + 1 −
β′ + 1 −
γ ′,
k1
k2
1 ′
β,
β =
k1
1 ′
γ =
γ.
k2
(ii) If instead we take
β′
β
γ
γ′
=
k
= k2 ,
,
1
′
′
α
α
α
α
the lines through Z3 are mapped to the lines through W3 , and the lines through Z2 are mapped
to the lines through W2 . By addition
α + k1 β + k2 γ
α′ + β ′ + γ ′
=
α′
α
and so
α′
=
β′
=
γ′
=
α
,
α + k1 β + k2 γ
k1 β
,
α + k1 β + k2 γ
k2 γ
.
α + k1 β + k2 γ
Thus we again have a function, and its inverse is easily found by working the other way around as
α
=
β
=
γ
=
α′
,
α′ + β ′ /k1 + γ ′ /k2
β ′ /k1
,
α′ + β ′ /k1 + γ ′ /k2
γ ′ /k2
.
α′ + β ′ /k1 + γ ′ /k2
(iii) If instead we take the relation
α
α′ = k1 ,
β
1
β ′ = k2 ,
β
then lines through Z3 are mapped to lines through W3 , and lines parallel to Z3 Z1 are mapped to
lines parallel to W3 W1 . Then we have
γ′
=
=
k1 α + k2
β
−(k1 + k2 )α + (1 − k2 )β − k2 γ
,
β
1−
so that this relation too is a function. Its inverse can readily be found from
α=
k2 α′
1
, β = k2 ′ .
k1 β ′
β
51
4.1. SPECIAL TRANSFORMATIONS
4.1.2
Definition of projective and affine transformations
It is convenient now to take the transformations in §4.1.1 in the slightly different forms
(i)
δF (Z, Z6 , Z4 )
δF (Z, Z4 , Z5 )
δF (W, Z1 , Z2 )
δF (W, Z3 , Z1 )
= k1
,
= k2
,
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
(ii)
δF (W, Z3 , Z1 )
δF (Z, Z6 , Z4 )
= k1
,
δF (W, Z2 , Z3 )
δF (Z, Z5 , Z6 )
δF (Z, Z4 , Z5 )
δF (W, Z1 , Z2 )
= k2
,
δF (W, Z2 , Z3 )
δF (Z, Z5 , Z6 )
(iii)
δF (Z, Z5 , Z6 )
δF (W, Z2 , Z3 )
= k1
,
δF (Z1 , Z2 , Z3 )
δF (Z, Z6 , Z4 )
δF (Z1 , Z2 , Z3 )
δF (W, Z3 , Z1 )
= k2
,
δF (Z1 , Z2 , Z3 )
δF (Z, Z6 , Z4 )
where now all the areal coordinates are with respect to the triple of reference (Z1 , Z2 , Z3 ). Then
each is in the form
α′
=
β′
=
γ′
=
where
l 1 α + m1 β + n 1 γ
,
l 4 α + m4 β + n 4 γ
l 2 α + m2 β + n 2 γ
,
l 4 α + m4 β + n 4 γ
l 3 α + m3 β + n 3 γ
,
l 4 α + m4 β + n 4 γ

l1
M =  l2
l3
m1
m2
m3
(4.1.1)

n1
n2 
n3
(4.1.2)
is non-singular, and the case l4 = m4 = n4 = 0 is excluded. This is called a projective transformation of the plane Π. When l4 = m4 = n4 then l4 α + m4 β + n4 γ = l4 and it is called an affine
transformation.
By addition in (4.1.1)
1=
(l1 + l2 + l3 )α + (m1 + m2 + m3 )β + (n1 + n2 + n3 )γ
,
l 4 α + m4 β + n 4 γ
and so
l 4 = l 1 + l 2 + l 3 , m4 = m1 + m2 + m3 , n 4 = n 1 + n 2 + n 3 .
On writing the transformation in matrix form as
 
α′
l1
 β ′  =  l2
γ′
l3

m1
m2
m3
we have



α
l4 α+m4 β+n4 γ
β
l4 α+m4 β+n4 γ
γ
l4 α+m4 β+n4 γ


 =
=
where L1 is the cofactor of l1 in M , etc.

n1

n2  
n3
α
l4 α+m4 β+n4 γ
β
l4 α+m4 β+n4 γ
γ
l4 α+m4 β+n4 γ


,
−1  ′ 
α
n1
n2   β ′ 
γ′
n3
 ′ 
L
L2 L3
α
1  1
M1 M2 M3   β ′  ,
det M
γ′
N1 N2 N3

l1
 l2
l3
m1
m2
m3

52
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
By addition in this
1
1
=
[(L1 + L2 + L3 )α′ + (M1 + M2 + M3 )β ′ + (N1 + N2 + N3 )γ ′ ],
l 4 α + m4 β + n 4 γ
det M
and so
α =
β
=
γ
=
L1 α′ + L2 β ′ + L3 γ ′
,
(L1 + L2 + L3
+ (M1 + M2 + M3 )β ′ + (N1 + N2 + N3 )γ ′
M1 α′ + M2 β ′ + M3 γ ′
,
′
(L1 + L2 + L3 )α + (M1 + M2 + M3 )β ′ + (N1 + N2 + N3 )γ ′
N1 α′ + N2 β ′ + N3 γ ′
.
′
(L1 + L2 + L3 )α + (M1 + M2 + M3 )β ′ + (N1 + N2 + N3 )γ ′
)α′
(4.1.3)
This is the inverse of the projective transformation, and is also projective.
The domain and range of an affine transformation are both the plane Π, and the inverse
transformation is affine also. The domain of the above projective transformation is the plane less
the line with equation l4 α + m4 β + n4 γ = 0, and its range is the plane less the line with equation
(L1 + L2 + L3 )α′ + (M1 + M2 + M3 )β ′ + (N1 + N2 + N3 )γ ′ = 0.
4.1.3
Transformations in Cartesian coordinates
By the formulae of §3.1.1 the transformation of the last section can be expressed in terms of
Cartesian coordinates in the form
u=
where
d2 x + e2 y + f2
d1 x + e1 y + f1
, v=
,
d3 x + e3 y + f3
d3 x + e3 y + f3

d1
 d2
d3
e1
e2
e3

f1
f2 
f3
is non-singular, and conversely by (3.1.2) any such transformation corresponds to a projective
transformation.
The more special form
ax + py
bx + qy − ab
u=
, v=
,
x
x
was introduced by Isaac Newton in 1686, and the general form was given by Waring in 1762.
4.1.4
Images of lines
By §3.2.1 and §3.2.2, any line has an equation of the form
l6 α + m6 β + n6 γ = 0.
Now from the inverse of our general projective transformation in (4.1.3), for such points Z the
corresponding points W satisfy
l6 [L1 α′ + L2 β ′ + L3 γ ′ ] + m6 [M1 α′ + M2 β ′ + M3 γ ′ ] + n6 [N1 α′ + N2 β ′ + N3 γ ′ ] = 0,
that is
[l6 L1 + m6 M1 + n6 N1 ]α′ + [l6 L2 + m6 M2 + n6 N2 ]β ′ + [l6 L3 + m6 M3 + n6 N3 ]γ ′ = 0.
53
4.1. SPECIAL TRANSFORMATIONS
Thus the coordinates of the image points satisfy the equation
l9 α + m9 β + n9 γ = 0,
where
l 9 = l 6 L 1 + m6 M 1 + n 6 N 1 , m9 = l 6 L 2 + m6 M 2 + n 6 N 2 , n 9 = l 6 L 3 + m6 M 3 + n 6 N 3 .
(4.1.4)
Conversely if (α′ , β ′ , γ ′ ) satisfy this latter equation, then by using the projective transformation
(4.1.1) directly, the corresponding (α, β, γ) must satisfy
[l6 L1 + m6 M1 + n6 N1 ][l1 α + m1 β + n1 γ] + [l6 L2 + m6 M2 + n6 N2 ][l2 α + m2 β + n2 γ]
+ [l6 L3 + m6 M3 + n6 N3 ][l3 α + m3 β + n3 γ] = 0,
that is
{l6 [L1 l1 + L2 l2 + L3 l3 ] + m6 [M1 l1 + M2 l2 + M3 l3 ] + n6 [N1 l1 + N2 l2 + N3 l3 ]}α
+ {l6 [L1 m1 + L2 m2 + L3 m3 ] + m6 [M1 m1 + M2 m2 + M3 m3 ] + n6 [N1 m1 + N2 m2 + N3 m3 ]}β
+ {l6 [L1 n1 + L2 n2 + L3 n3 ] + m6 [M1 n1 + M2 n2 + M3 n3 ] + n6 [N1 n1 + N2 n2 + N3 n3 ]}γ = 0,
which is
det M [l6 α + m6 β + n6 γ] = 0,
and so is the equation of the original line.
Either the transformation is affine, when l4 = m4 = n4 , or else l4 α + m4 β + n4 γ = 0 is an
equation of a line. This latter line has no image, and we suppose that the line with equation
l6 α + m6 β + n6 γ = 0,
is not it when the transformation is not affine.
To show that we cannot have l9 = m9 = n9 , we note that by the above argument applied to
the inverse transformation, (l6 , m6 , n6 ) is a multiple of
(l9 l1 + m9 m1 + n9 n1 , l9 l2 + m9 m2 + n9 n2 , l9 l3 + m9 m3 + n9 n3 ).
If we had l9 = m9 = n9 , then (l6 , m6 , n6 ) would be a multiple of
(l1 + m1 + n1 , l2 + m2 + n2 , l3 + m3 + n3 ) = (l4 , m4 , n4 ).
When l4 = m4 = n4 , this would make l6 = m6 = n6 , which is ruled out. In the other case, it would
make l6 α + m6 β + n6 γ = 0 an equation of the line with equation l4 α + m4 β + n4 γ = 0, which we
also ruled out. Thus in each case, l9 α + m9 β + n9 γ = 0 is an equation of a line.
When the transformation is affine, the image of each a line is a line. When the transformation
is not affine, l4 α + m4 β + n4 γ = 0 is an equation of a line. This line itself has no image, but for
any line parallel to it but distinct from it the image is a line. For any line not parallel to it, the
point of intersection has no image and the image is a line less a point.
4.1.5
Suppose now that
l4 α + m4 β + n4 γ = 0, l6 α + m6 β + n6 γ = 0, l7 α + m7 β + n7 γ = 0,
are equations of concurrent lines, where l4 , m4 , n4 are as in the transformation (4.1.1) and so
l 4 = l 1 + l 2 + l 3 , m4 = m1 + m2 + m3 , n 4 = n 1 + n 2 + n 3 .
54
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
Then we have that
l6
l
7
l1 + l2 + l3
m6
m7
m1 + m2 + m3
n6
n7
n1 + n2 + n3
= 0.
(4.1.5)
Now by (4.1.4) the images under (4.1.1) of the first two of our concurrent lines have the equations
(l6 L1 + m6 M1 + n6 N1 )α′ + (l6 L2 + m6 M2 + n6 N2 )β ′ + (l6 L3 + m6 M3 + n6 N3 )γ ′ =0,
(l7 L1 + m7 M1 + n7 N1 )α′ + (l7 L2 + m7 M2 + n7 N2 )β ′ + (l7 L3 + m7 M3 + n7 N3 )γ ′ =0.
We wish to show that these two image lines are parallel, and so consider
l 6 L 1 + m6 M 1 + n 6 N 1 l 6 L 2 + m6 M 2 + n 6 N 2 l 6 L 3 + m6 M 3 + n 6 N 3
l 7 L 1 + m7 M 1 + n 7 N 1 l 7 L 2 + m7 M 2 + n 7 N 2 l 7 L 3 + m7 M 3 + n 7 N 3
1
1
1
Now this expands as
.
(4.1.6)
(4.1.7)
(l6 m7 − m6 l7 )[L1 (M2 − M3 ) + L2 (M3 − M1 ) + L3 (M1 − M2 )]
+(m6 n7 − n6 m7 )[M1 (N2 − N3 ) + M2 (N3 − N1 ) + M3 (N1 − N2 )]
+(n6 l7 − l6 n7 )[N1 (L2 − L3 ) + N2 (L3 − L1 ) + N3 (L1 − L2 )].
We note that
L1 (M2 − M3 ) + L2 (M3 − M1 ) + L3 (M1 − M2 )
=(m2 n3 − n2 m3 )(M2 − M3 ) + (m3 n1 − n3 m1 )(M3 − M1 ) + (m1 n2 − n1 m2 )(M1 − M2 )
=n1 [m3 M3 − m3 M1 − m2 M1 + m2 M2 + m1 M1 − m1 M1 ] + n2 [−m3 M2 + m3 M3 + m1 M1
− m1 M2 + m2 M2 − m2 M2 ] + n3 [m2 M2 − m2 M3 − m1 M3 + m1 M1 + m3 M3 − m3 M3 ]
=n1 [det(M ) − (m1 + m2 + m3 )M1 ] + n2 [det(M ) − (m1 + m2 + m3 )M2 ]+
+ n3 [det(M ) − (m1 + m2 + m3 )M3 ]
=(n1 + n2 + n3 ) det(M ) − (m1 + m2 + m3 )(n1 M1 + n2 M2 + n3 M3 )
=(n1 + n2 + n3 ) det(M ).
By similar arguments we have that
M1 (N2 − N3 ) + M2 (N3 − N1 ) + M3 (N1 − N2 ) =(l1 + l2 + l3 ) det(M )
N1 (L2 − L3 ) + N2 (L3 − L1 ) + N3 (L1 − L2 ) =(m1 + m2 + m3 ) det(M ).
On inserting these, we see that (4.1.7) is det(M ) times the left-hand side of (4.1.5), and so by
(3.2.4) the two image lines in (4.1.6) are parallel.
For a further result, we suppose that l4 α + m4 β + n4 γ = 0 is the equation of a line; we suppose
that Z8 , Z9 are points on it and as the algebraic fractions in (4.1.1) are unaltered if we multiply
above and below by the same non-zero number, we may take l4 α + m4 β + n4 γ = δF (Z, Z8 , Z9 ).
Then
l4 = δF (Z1 , Z8 , Z9 ), m4 = δF (Z2 , Z8 , Z9 ), n4 = δF (Z3 , Z8 , Z9 ).
We also suppose that a line Z4 Z5 meets Z8 Z9 in the point Z6 and that (Z4 , Z5 , Z6 , Z7 ) is a
harmonic range. As points on the line Z4 Z5 have the form
Z=
µ
1
Z4 +
Z5 ,
1+µ
1+µ
we suppose for Z6 that µ = λ and then for Z7 we will have µ = −λ. Now
0 = δF (Z6 , Z8 , Z9 ) =
1
λ
δF (Z4 , Z8 , Z9 ) +
δF (Z5 , Z8 , Z9 ),
1+λ
1+λ
55
4.2. EQUIPOISED QUOTIENTS; CROSS-RATIO OF RANGES AND PENCILS
and so
λ=−
δF (Z4 , Z8 , Z9 )
.
δF (Z5 , Z8 , Z9 )
(4.1.8)
Under the transformation (4.1.1) we have that Z → W where
l1 α4 + m1 β4 + n1 γ4
,
l4 α4 + m4 β4 + n4 γ4
l1 α5 + m1 β5 + n1 γ5
α′5 =
,
l4 α5 + m4 β5 + n4 γ5
l1 α4 + m1 β4 + n1 γ4 + µ(l1 α5 + m1 β5 + n1 γ5 )
α′ =
.
l4 α4 + m4 β4 + n4 γ4 + µ(l4 α5 + m4 β5 + n4 γ5 )
α′4 =
From these we have that
α′ − α′4 =
(l1 α5 + m1 β5 + n1 γ5 )(l4 α4 + m4 β4 + n4 γ4 ) − (l1 α4 + m1 β4 + n1 γ4 )((l4 α5 + m4 β5 + n4 γ5 ))
,
(l4 α4 + m4 β4 + n4 γ4 )[l4 α4 + m4 β4 + n4 γ4 + µ(l4 α5 + m4 β5 + n4 γ5 )]
α′5 − α′4 =
µ
(l1 α5 + m1 β5 + n1 γ5 )(l4 α4 + m4 β4 + n4 γ4 ) − (l1 α4 + m1 β4 + n1 γ4 )(l4 α5 + m4 β5 + n4 γ5 )
,
(l4 α4 + m4 β4 + n4 γ4 )[l4 α4 + m4 β4 + n4 γ4 ]
and so α′ − α′4 = ν(α′5 − α′4 ) where
(l4 α5 + m4 β5 + n4 γ5 )µ
l4 α4 + m4 β4 + n4 γ4 + µ(l4 α5 + m4 β5 + n4 γ5 )
µδF (Z5 , Z8 , Z9 )
.
=
δF (Z4 , Z8 , Z9 ) + µδF (Z5 , Z8 , Z9 )
ν=
(4.1.9)
Similarly for the other coordinates.
Now for Z4 , µ = 0 and so for W4 , ν = 0; for Z5 , µ = ∞ and so for W5 , ν = 1. For Z7 we have
µ = −λ where λ is given in (4.1.8), and so for W7 we substitute this into (4.1.9) to get
ν=
δF (Z4 ,Z8 ,Z9 )
δF (Z5 ,Z8 ,Z9 ) δF (Z5 , Z8 , Z9 )
4 ,Z8 ,Z9 )
δF (Z4 , Z8 , Z9 ) + δδFF (Z
(Z5 ,Z8 ,Z9 ) δF (Z5 , Z8 , Z9 )
1
= .
2
It follows that W7 is the mid-point of W4 and W5 .
4.2
4.2.1
Equipoised quotients; cross-ratio of ranges and pencils
Equipoised quotients
If (α4 , β4 , γ4 ), (α5 , β5 , γ5 ), (α6 , β6 , γ6 ) are areal coordinates of points Z4 , Z5 , Z6 , respectively, then
under our general projective transformation their images W4 , W5 , W6 have areal coordinates which
56
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
satisfy
 
α′4
 β4′  = 
γ4′
 ′  
α5
 β5′  = 
γ5′
 ′  
α6
 β6′  = 
γ6′

and so
α′4

β4′
det
γ4′

α′5
β5′
γ5′
l1
l2
l3
m1
m2
m3
l1
l2
l3
m1
m2
m3
l1
l2
l3
m1
m2
m3

n1

n2  
n3

n1

n2  
n3

n1

n2  
n3
α4
l4 α4 +m4 β4 +n4 γ4
β4
l4 α4 +m4 β4 +n4 γ4
γ4
l4 α4 +m4 β4 +n4 γ4

α5
l4 α5 +m4 β5 +n4 γ5
β5
l4 α5 +m4 β5 +n4 γ5
γ5
l4 α5 +m4 β5 +n4 γ5

α6
l4 α6 +m4 β6 +n4 γ6
β6
l4 α6 +m4 β6 +n4 γ6
γ6
l4 α6 +m4 β6 +n4 γ6

,

,


,

α′6
1
β6′  =
•
(l
α
+
m
β
+
n
γ
)(l
α
+
m
4 4
4 4
4 4
4 5
4 β5 + n4 γ5 )(l4 α6 + m4 β6 + n4 γ6 )
γ6′




l 1 m1 n 1
α4 α5 α6
det  l2 m2 n2  det  β4 β5 β6  .
l 3 m3 n 3
γ4 γ5 γ6
Then by 3.1.2 we have
δF (W4 , W5 , W6 )
=
(4.2.1)
1
. det M.δF (Z4 , Z5 , Z6 ),
(l4 α4 + m4 β4 + n4 γ4 )(l4 α5 + m4 β5 + n4 γ5 )(l4 α6 + m4 β6 + n4 γ6 )
where M is the matrix of the transformation.
It follows from this that a quotient of a product of sensed areas of triangles over a product of
sensed areas of triangles, in which the set of vertices in the numerator is a permutation of the set
of vertices in the denominator, is a projective invariant. We refer to such an expression as an
equipoised quotient. This shows the significance of many of the expressions in Chapter 1, and
helps reveal the patterns.
In 1827 Möbius showed that
δF (A, B, C)δF (D, E, F )
(4.2.2)
δF (A, B, D)δF (C, E, F )
is a projective invariant, and this was noted too by Clifford in 1865. In 1833 Magnus dealt with
the less general
δF (A, C, D)δF (B, C, E)
.
δF (B, C, D)δF (A, C, E)
It also follows from (4.2.1) that the ratio of the sensed areas of any two triples of non-collinear
points is an affine invariant.
4.2.2
Values of permutations of Magnus quotient
An example of an equipoised quotient of the type mentioned in §4.2.1 as being a projective invariant
is the Magnus quotient
δF (Z, Z1 , Z3 )δF (Z, Z2 , Z4 )
.
(4.2.3)
δF (Z, Z1 , Z4 )δF (Z, Z2 , Z3 )
Suppose that the value of (4.2.3) is k and that we regard (4.2.3) as corresponding to the order
(Z1 , Z2 , Z3 , Z4 ) of the points.
4.2. EQUIPOISED QUOTIENTS; CROSS-RATIO OF RANGES AND PENCILS
57
First we note that we then have that
1
δF (Z, Z1 , Z4 )δF (Z, Z2 , Z3 )
= ,
δF (Z, Z1 , Z3 )δF (Z, Z2 , Z4 )
k
and this corresponds to the order (Z1 , Z2 , Z4 , Z3 ) of the points, so that the third and fourth have
been interchanged.
Secondly we note that we have that
δF (Z, Z1 , Z2 )δF (Z, Z3 , Z4 )
−1
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
δF (Z, Z1 , Z2 )δF (Z, Z3 , Z4 ) − δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
=
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
δF (Z2 , Z, Z1 )δF (Z4 , Z, Z3 ) − δF (Z4 , Z, Z1 )δF (Z2 , Z, Z3 )
=
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
δF [(Z2 , Z4 ), (Z, Z1 ), (Z, Z3 )]
=
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
δF [(Z, Z3 ), (Z2 , Z4 ), (Z, Z1 )]
=
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
δF (Z, Z2 , Z4 )δF (Z3 , Z, Z1 )
=
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
δF (Z, Z2 , Z4 )δF (Z, Z1 , Z3 )
=−
δF (Z, Z1 , Z4 )δF (Z, Z2 , Z3 )
= − k,
and so
δF (Z, Z1 , Z2 )δF (Z, Z3 , Z4 )
= 1 − k.
δF (Z, Z1 , Z4 )δF (Z, Z3 , Z2 )
This corresponds to the order (Z1 , Z3 , Z2 , Z4 ) of the points, so that the second and third in (4.2.3)
have been interchanged.
Finally we note that
δF (Z, Z2 , Z4 )δF (Z, Z1 , Z3 )
= k.
δF (Z, Z2 , Z3 )δF (Z, Z1 , Z4 )
This corresponds to the order (Z2 , Z1 , Z4 , Z3 ) of the points, so that the outer pairs in (4.2.3) have
both been interchanged.
If we combine the above permutations we obtain all 24 possible permutations of (Z1 , Z2 , Z3 , Z4 )
and the Magnus quotients have the possible values
k,
4.2.3
1
1
k−1
k
1
, 1 − k,
, 1− =
,
.
k
1−k
k
k
k−1
Cross-ratio of a range of collinear points
When Z1 , Z2 , Z3 , Z4 are collinear, by (1.1.12) the Magnus quotient (4.2.3) is equal to the quotient
of products of sensed or complex-valued distances,
Z1 Z3 Z2 Z4
.
Z1 Z4 Z2 Z3
This is called the cross ratio of the quadruple (Z1 , Z2 , Z3 , Z4 ), which we denote by cr(Z1 , Z2 , Z3 , Z4 ).
After a long build-up period this was first considered in its full signed form by Möbius in 1827. By
the foregoing it is a projective invariant.
58
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
If we take Z5 , Z6 as the basis for our line and r1 , r2 , r3 , r4 the corresponding parameters for
Z1 , Z2 , Z3 , Z4 , respectively, then
Z1 = (1 − r1 )Z5 + r1 Z6 , Z2 = (1 − r2 )Z5 + r2 Z6 , Z3 = (1 − r3 )Z5 + r3 Z6 , Z4 = (1 − r4 )Z5 + r4 Z6 ,
so that
Z3 − Z1 = (r1 − r3 )Z5 + (r3 − r1 )Z6 = (r3 − r1 )(Z6 − Z5 ),
and so similarly
Z3 − Z2 = (r3 − r2 )(Z6 − Z5 ),
and on combining these
Z3 − Z1 =
r3 − r1
(Z3 − Z2 ).
r3 − r2
Z4 − Z1 =
r4 − r1
(Z4 − Z2 ).
r4 − r2
Similarly
These give
Z1 Z3
r1 − r3 Z1 Z4
r1 − r4
=
=
,
.
r2 − r3 Z2 Z4
r2 − r4
Z2 Z3
From these we have
Z1 Z3 Z2 Z4
=
Z1 Z4 Z2 Z3
and these continue equal to
r1 −r3
r2 −r3
r1 −r4
r2 −r4
=
Z1 Z3
Z2 Z3
Z1 Z4
Z2 Z4
(r1 − r3 )(r2 − r4 )
.
(r1 − r4 )(r2 − r3 )
This enables us to calculate cross-ratios of ranges of points.
Z0
b
Z4
In Barry [2, 7.6.1] it was shown that {Z3 , Z4 } divide
{Z1 , Z2 } internally and externally in the same ratio
if and only if cr(Z1 , Z2 , Z3 , Z4 ) = −1. Such points
are said to form a harmonic range, and this is a
projective invariant.
b
Z3
Z2
b
Z1
b
b
b
W2
W4
Figure 3.1
b
4.2.4
b
b
W3
W1
Cross-ratio of a pencil of concurrent lines
Related to §4.2.3 consider a pencil of lines (Z0 Z1 , Z0 Z2 , Z0 Z3 , Z0 Z4 ) which we also denote by
Z0 (Z1 , Z2 , Z3 , Z4 ). For this consider
E=
δF (Z0 , Z1 , Z3 )δF (Z0 , Z2 , Z4 )
.
δF (Z0 , Z1 , Z4 )δF (Z0 , Z2 , Z3 )
Let an arbitrary line meet Z0 Z1 , Z0 Z2 , Z0 Z3 , Z0 Z4 in W1 , W2 , W3 , W4 , respectively. Then
Z1 =
ρ
1
σ
1
Z0 +
W1 , Z2 =
Z0 +
W2 ,
1+ρ
1+ρ
1+σ
1+σ
4.3. CONSTANT-SIZED ANGLE PROPERTY OF A CIRCLE
59
for some parameters, and so
δF (Z0 , Z1 , Z3 ) =
δF (Z0 , Z2 , Z3 ) =
ρ
ρ
δF (Z0 , W1 , Z3 ), δF (Z0 , Z1 , Z4 ) =
δF (Z0 , W1 , Z4 ),
1+ρ
1+ρ
σ
σ
δF (Z0 , W2 , Z3 ), δF (Z0 , Z2 , Z4 ) =
δF (Z0 , W2 , Z4 ).
1+σ
1+σ
Hence
E=
Now
W3 =
δF (Z0 , W1 , Z3 )δF (Z0 , W2 , Z4 )
.
δF (Z0 , W1 , Z4 )δF (Z0 , W2 , Z3 )
λ
1
µ
1
W1 +
W2 , W4 =
W1 +
W2 ,
1+λ
1+λ
1+µ
1+µ
for some parameters, so as Z0 , Z3 , W3 are collinear and so are Z0 , Z4 , W4 , we have
δF (Z0 , W1 , Z3 ) + λδF (Z0 , W2 , Z3 ) = 0, δF (Z0 , W1 , Z4 ) + µδF (Z0 , W2 , Z4 ) = 0.
Hence
E=
λδF (Z0 , W2 , Z3 )δF (Z0 , W2 , Z4 )
λ
= .
µδF (Z0 , W2 , Z4 )δF (Z0 , W2 , Z3 )
µ
But we recall that
cr(W1 , W2 , W3 , W4 ) =
and so
E=
λ
,
µ
δF (Z0 , Z1 , Z3 )δF (Z0 , Z2 , Z4 )
= cr(W1 , W2 , W3 , W4 ).
δF (Z0 , Z1 , Z4 )δF (Z0 , Z2 , Z3 )
Thus all the ranges of four points in which transversals cut the lines of the pencil have the same
cross-ratio. This common cross-ratio is defined to be the cross-ratio of the pencil, and denoted
by cr(Z0 Z1 , Z0 Z2 , Z0 Z3 , Z0 Z4 ) or by crZ0 (Z1 , Z2 , Z3 , Z4 ). As noted its value is E. This link
between the cross-ratio of a pencil and the cross-ration of a range of points cut by a transversal is
of fundamental importance.
If the cross-ratio of a concurrent pencil has the value −1, then it is called a harmonic pencil.
4.3
Constant-sized angle property of a circle
4.3.1
Given distinct points Z1 , Z2 , a point Z3 on the perpendicular bisector of [Z1 , Z2 ] will have coordinates of the form
x3 = 12 (x1 + x2 ) − s(y2 − y1 ), y3 = 12 (y1 + y2 ) + s(x2 − x1 ).
A variable point Z will lie on the circle C with centre Z0 and which passes through Z1 and Z2 if
|Z0 , Z|2 = |Z0 , Z1 |2 . If we insert this we find that
[x − 12 (x1 + x2 ) + s(y2 − y1 )]2 + [y − 12 (y1 + y2 ) − s(x2 − x1 )]2
=[x1 − 12 (x1 + x2 ) + s(y2 − y1 )]2 + [y1 − 21 (y1 + y2 ) − s(x2 − x1 )]2 .
If we solve this for s we find that
2s =
(x1 − x)(x2 − x) + (y1 − y)(y2 − y)
.
(x1 − x)(y2 − y) − (x2 − x)(y1 − y)
60
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
By Barry [2, 10.5.1(i)and(ii)] we have that
1
2 |Z , Z1 ||Z , Z2 | sin θ
1
2 |Z , Z1 ||Z , Z2 | cos θ
= 21 [(x1 − x)(y2 − y) − (x2 − x)(y1 − y)],
= 21 [(x1 − x)(x2 − x) + (y1 − y)(y2 − y)],
where θ = ∡F Z1 ZZ2 . It follows that
cos θ
.
sin θ
where θ = ∡F Z1 ZZ2 , and thus this expression is constant as Z varies on C.
We also note that
cos2 θ
1
=
1
+
= 1 + 4s2 ,
sin2 θ
sin2 θ
and so
1
sin θ = ± √
,
1 + 4s2
2s =
with the + or the − being taken on the different sides of the line Z1 Z2 .
By Barry [2, 10.5.2] we also note that
δF (Z1 ZZ2 ) = 12 [(x − x1 )(y2 − y1 ) − (x2 − x1 )(y − y1 )
4.3.2
Let Z1 , Z2 , Z3 , Z4 be distinct points on a circle C. From §4.3.1 we see that then for a point Z ∈ C
2δF (Z, Z1 , Z3 )
|Z, Z1 ||Z, Z3 |
2δF (Z, Z2 , Z4 )
|Z, Z2 ||Z, Z4 |
2δF (Z, Z1 , Z4 )
|Z, Z1 ||Z, Z4 |
2δF (Z, Z2 , Z3 )
|Z, Z2 ||Z, Z3 |
2δF (Z2 , Z1 , Z3 )
,
|Z2 , Z1 ||Z2 , Z3 |
2δF (Z1 , Z2 , Z4 )
=±
,
|Z1 , Z2 ||Z1 , Z4 |
2δF (Z2 , Z1 , Z4 )
=±
,
|Z2 , Z1 ||Z2 , Z4 |
2δF (Z1 , Z2 , Z3 )
=±
.
|Z1 , Z2 ||Z1 , Z3 |
=±
On dividing the product of the first two of these by the product of the third and fourth we find
that
|Z1 , Z3 ||Z2 , Z4 |
δF (Z1 , Z3 , Z)δF (Z2 , Z4 , Z)
=±
.
(4.3.1)
δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
|Z1 , Z4 ||Z2 , Z3 |
Now
δF (Z1 , Z3 , Z)δF (Z2 , Z4 , Z)
δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
(4.3.2)
has constant absolute value on C, and has constant value on some arcs. For the intersection of
the closed half-plane with edge Z1 Z3 which contains Z4 and the closed half-plane with edge Z1 Z4
which does not contain Z3 is an interior region. In it lies an arc of the circle with end-points Z1
and Z4 ; in it
δF (Z, Z1 , Z3 )
δF (Z, Z1 , Z4 )
has constant sign. Similarly the intersection of the closed half-plane with edge Z1 Z3 which does
not contain Z4 and the closed half-plane with edge Z1 Z4 which contains Z3 is the opposite interior
region. In it lies an arc of the circle with end-points Z1 and Z3 ; in it
δF (Z, Z1 , Z3 )
δF (Z, Z1 , Z4 )
61
4.4. PTOLEMY’S THEOREM
has constant sign. Then the arc of the circle which has end-points Z3 , Z4 and which contains Z1
lies in a duo-sector in which
δF (Z, Z1 , Z3 )
δF (Z, Z1 , Z3 )
has constant sign.
Thus two of the terms in (4.3.1) involve Z1 , and as Z passes through Z1 both of these change
sign so their quotient is of constant sign. A similar argument applies to Z2 , Z3 and Z4 , so (4.3.2)
has a constant sign and so a constant value for Z on the circle.
4.4
Ptolemy’s theorem
4.4.1
Z
b
Z2
b
Z1
b
Z2
b
Z3
b
Z3
b
Z1
b
b
b
Z4
Z4
Figure 3.2.
Ptolemy’s theorem (c.200 A.D.). Let Z1 , Z2 , Z3 , Z4 be distinct points on a circle C, such
that [Z1 , Z3 ] ∩ [Z2 , Z4 ] 6= ∅. Then
|Z1 , Z3 ||Z2 , Z4 | − |Z1 , Z4 ||Z2 , Z3 | = |Z1 , Z2 ||Z3 , Z4 |.
Proof.
For Z ∈ C
δF (Z1 , Z3 , Z)
δF (Z1 , Z4 , Z)
is of constant sign unless Z3 , Z4 are on opposite sides of Z1 Z and so unless Z1 , Z are on opposite
sides of Z3 Z4 . Thus it has constant sign on the arc with end-points Z3 , Z4 to which Z1 belongs.
We note that Z2 is on this arc. For a similar reason
δF (Z2 , Z4 , Z)
δF (Z2 , Z3 , Z)
has the same sign on this arc as its value at Z1 . On combining these we see that
δF (Z1 , Z3 , Z) δF (Z2 , Z4 , Z)
δF (Z1 , Z4 , Z) δF (Z2 , Z3 , Z)
has the same sign on this arc as
δF (Z1 , Z3 , Z2 ) δF (Z2 , Z4 , Z1 )
= 1,
δF (Z1 , Z4 , Z2 ) δF (Z2 , Z3 , Z1 )
62
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
and so is positive there. It follows by (4.3.1) that
|Z1 , Z3 ||Z2 , Z4 |
δF (Z1 , Z3 , Z)δF (Z2 , Z4 , Z)
=
.
δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
|Z1 , Z4 ||Z2 , Z3 |
On subtracting 1 from each side this gives
=
δF (Z1 , Z3 , Z)δF (Z2 , Z4 , Z) − δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
|Z1 , Z3 ||Z2 , Z4 | − |Z1 , Z4 ||Z2 , Z3 |
.
|Z1 , Z4 ||Z2 , Z3 |
But
δF (Z1 , Z3 , Z)δF (Z2 , Z4 , Z) − δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
=δF [(Z1 , Z2 ), (Z3 , Z), (Z4 , Z)] = δF (Z1 , Z2 , Z)δF (Z3 , Z4 , Z)
and by the same reasoning as above
|Z1 , Z2 ||Z3 , Z4 |
δF (Z1 , Z2 , Z)δF (Z3 , Z4 , Z)
=
.
δF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z)
|Z1 , Z4 ||Z2 , Z3 |
It follows that
|Z1 , Z3 ||Z2 , Z4 | − |Z1 , Z4 ||Z2 , Z3 | = |Z1 , Z2 ||Z3 , Z4 |.
This result is known as Ptolemy’s theorem.
4.5
4.5.1
Transformations in terms of equipoised quotients
Affine transformations
Let (Z1 , Z2 , Z3 ), (W1 , W2 , W3 ) be two triples of non- collinear points. We consider
δF (Z, Z2 , Z3 )
δF (W, W2 , W3 )
= k1
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
which maps lines parallel to Z2 Z3 to lines parallel to W2 W3 , and
δF (Z, Z3 , Z1 )
δF (W, W3 , W1 )
= k2
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
which maps lines parallel to Z3 Z1 to lines parallel to W3 W1 . Then we must further have
=
δF (W, W1 , W2 )
δF (Z, Z2 , Z3 )
δF (Z, Z3 , Z1 )
= 1 − k1
− k2
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z, Z3 , Z1 )
δF (Z, Z1 , Z2 )
δF (Z, Z2 , Z3 )
+ (1 − k2 )
+
.
(1 − k1 )
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 ) δF (Z1 , Z2 , Z3 )
Then lines parallel to Z1 Z2 map to lines parallel to W1 W2 if 1 − k1 = 1 − k2 = 0. Thus we have
the transformation
δF (W, W2 , W3 )
δF (Z, Z2 , Z3 )
=
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
δF (Z, Z3 , Z1 )
δF (W, W3 , W1 )
=
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
δF (Z, Z1 , Z2 )
δF (W, W1 , W2 )
=
,
δF (W1 , W2 , W3 )
δF (Z1 , Z2 , Z3 )
as an affine transformation under which Z1 , Z2 , Z3 map to W1 , W2 , W3 , respectively.
As an affine transformation preserves ratios of sensed areas, any such affine transformation
must have this form, and so it is uniquely determined.
63
4.6. PERSPECTIVITIES AND PROJECTIVITIES AS RESTRICTIONS
4.5.2
Projective transformations
Let (Z1 , Z2 , Z3 , Z4 ), (W1 , W2 , W3 , W4 ) be quadruples of points, each such that no triple of its points
is collinear. Consider
δF (W, W3 , W1 )
δF (Z, Z3 , Z1 )
= k1
,
δF (W, W2 , W3 )
δF (Z, Z2 , Z3 )
which maps lines through Z3 to lines through W3 , and
δF (Z, Z1 , Z2 )
δF (W, W1 , W2 )
= k2
,
δF (W, W2 , W3 )
δF (Z, Z2 , Z3 )
which maps lines through Z2 to lines through W2 . This defines a projective transformation, and
as for it
δF (W, W3 , W1 )
k1 δF (Z, Z3 , Z1 )
=
,
δF (W, W1 , W2 )
k2 δF (Z, Z1 , Z2 )
it also maps lines through Z1 to lines through W1 . Thus under it Z1 , Z2 , Z3 map to W1 , W2 , W3 ,
respectively.
We now wish to have Z4 map to W4 as well, and choose k1 , k2 for this purpose. This gives the
projective transformation specified by
δF (Z3 , Z, Z1 )δF (Z3 , Z4 , Z2 )
δF (W3 , W, W1 )δF (W3 , W4 , W2 )
=
,
δF (W3 , W, W2 )δF (W3 , W4 , W1 )
δF (Z3 , Z, Z2 )δF (Z3 , Z4 , Z1 )
δF (W2 , W1 , W )δF (W2 , W3 , W4 )
δF (Z2 , Z1 , Z)δF (Z2 , Z3 , Z4 )
=
.
δF (W2 , W1 , W4 )δF (W2 , W3 , W )
δF (Z2 , Z1 , Z4 )δF (Z2 , Z3 , Z)
It is also unique, as under any such projective transformation these equipoised quotients are
invariant.
4.6
Perspectivities and projectivities as restrictions
4.6.1
Perspectivities
Let Z1 Z2 , Z3 Z4 be lines, W1 a point on neither and take Z5 ∈ Z1 Z2 , Z6 ∈ Z3 Z4 so that W1 , Z5 , Z6
are collinear.
Z3
b
Z1
b
b
b
b
W1
b
Z6
Z5
Z5 Z2
b
Z6
b
b
b
b
W2 W1
Z4
b
b
b
b
b
b
b
b
Z1
Z2
b
b
b
W5
W6
b
Figure 3.3.
Suppose first that Z1 Z2 , Z3 Z4 intersect, at Z7 , say; take Z8 so that W1 , Z7 , Z8 are not collinear.
Consider
δF (Z, Z7 , W1 )
δF (W, Z7 , W1 )
= k1
.
δF (W, Z8 , Z7 )
δF (Z, Z8 , Z7 )
64
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
This maps lines through Z7 to lines through Z7 , and we wish Z1 Z2 to map to Z3 Z4 so we choose
k1 so that Z5 ∈ Z1 Z2 maps to Z6 ∈ Z3 Z4 . This gives
δF (W, Z7 , W1 ) δF (Z6 , Z8 , Z7 )
δF (Z, Z7 , W1 ) δF (Z5 , Z8 , Z7 )
=
.
δF (W, Z8 , Z7 ) δF (Z6 , Z7 , W1 )
δF (Z, Z8 , Z7 ) δF (Z5 , Z7 , W1 )
We also wish the lines through W1 to map to lines through W1 , so we take
δF (Z, Z7 , W1 )
δF (W, Z7 , W1 )
= k2
.
δF (W, W1 , Z8 )
δF (Z, W1 , Z8 )
We wish W1 Z5 to map to itself, so we choose k2 so that Z5 maps to Z6 . This gives
δF (Z, Z7 , W1 ) δF (Z5 , W1 , Z8 )
δF (W, Z7 , W1 ) δF (Z6 , W1 , Z8 )
=
.
δF (W, W1 , Z8 ) δF (Z6 , Z7 , W1 )
δF (Z, W1 , Z8 ) δF (Z5 , Z7 , W1 )
These relations combined define a projective transformation the restriction of which to Z1 Z2 is
the perspectivity from Z1 Z2 to Z3 Z4 with centre or vertex W1 .
Suppose next that Z1 Z2 is parallel to Z3 Z4 . Take Z7 so that W1 Z7 k Z1 Z2 and again take
Z8 6∈ W1 Z7 . We take
δF (W, Z7 , W1 ) = k1 δF (Z, Z7 , W1 ),
which maps lines parallel to W1 Z7 parallel to themselves. We wish Z1 Z2 to map to Z3 Z4 and for
this choose k1 so that Z5 maps to Z6 . This gives
k1 =
We also take
δF (Z6 , Z7 , W1 )
.
δF (Z5 , Z7 , W1 )
δF (W, W1 , Z8 )
δF (Z, W1 , Z8 )
= k2
,
δF (W, Z7 , W1 )
δF (Z, Z7 , W1 )
which maps lines through W1 to lines through W1 . We wish the line W1 Z5 to map to itself and so
choose k2 so that Z5 maps to Z6 . Then we have
δF (W, Z7 , W1 ) = k1 δF (Z, Z7 , W1 ),
δF (Z, W1 , Z8 )
δF (W, W1 , Z8 )
= k2
,
δF (W, Z7 , W1 )
δF (Z, Z7 , W1 )
on combining which we have
δF (W, Z7 , W1 ) = k1 δF (Z, Z7 , W1 ),
δF (W, W1 , Z8 ) = k1 k2 δF (Z, W1 , Z8 ).
Thus we have an affine transformation, the restriction of which to Z1 Z2 is the given parallel
perspectivity from Z1 Z2 to Z3 Z4 .
4.6.2
Projectivities
Given two lines, let them be Z5 Z6 and W5 W6 . Take distinct point Z1 , Z2 ∈ Z5 Z6 , distinct points
W1 , W2 ∈ W5 W6 , and Z3 6∈ Z5 Z6 , W3 6∈ W5 W6 . Then
δF (W, W1 , W2 )
δF (Z, Z1 , Z2 )
= k2
δF (W, W2 , W3 )
δF (Z, Z2 , Z3 )
maps Z1 Z2 to W1 W2 , and if we combine this with
δF (Z, Z3 , Z1 )
δF (W, W3 , W1 )
= k1
,
δF (W, W2 , W3 )
δF (Z, Z2 , Z3 )
(4.6.1)
4.6. PERSPECTIVITIES AND PROJECTIVITIES AS RESTRICTIONS
65
then we have a projective transformation.
Take a range of points on the line Z5 Z6 , of the form
Z=
where
so that
λ
1
Z5 +
Z6 ,
1+λ
1+λ
Z1 =
λ1
1
Z5 +
Z6 ,
1 + λ1
1 + λ1
Z2 =
1
λ2
Z5 +
Z6 ,
1 + λ2
1 + λ2
δF (Z, Z3 , Z1 )
λ − λ1
=
.
δF (Z, Z2 , Z3 )
λ2 − λ
Similarly let
W =
where
so that
µ
1
W5 +
W6 ,
1+µ
1+µ
W1 =
µ1
1
W5 +
W6 ,
1 + µ1
1 + µ1
W2 =
µ2
1
W5 +
Z6 ,
1 + µ2
1 + µ2
µ − µ1
δF (W, W3 , W1 )
=
.
δF (W, W2 , W3 )
µ2 − µ
Then from (4.6.1) we have
λ − λ1
µ − µ1
= k1
,
µ2 − µ
λ2 − λ
that is
(k1 − 1)λµ + (µ1 − k1 µ2 )λ + (λ2 − k1 λ1 )µ − µ1 λ2 + k1 λ1 µ2 = 0.
This has the form aλµ + bλ + cµ + d = 0.
In fact any non-degenerate relationship of this form can be found in this way. We seek
k1 − 1 = ka, µ1 − k1 µ2 = kb, λ2 − k1 λ1 = kc, k1 λ1 µ2 − λ2 µ1 = kd.
We take
k1 = 1 + ka, λ2 = kc + (1 + ka)λ1 , µ1 = kb + (1 + ka)µ2 ,
and take λ1 , µ2 to satisfy
a(1 + ka)λ1 µ2 + b(1 + ka)λ1 + c(1 + ka)µ2 + kbc + d = 0.
For this last to be non-degenerate we need
a(1 + ka)(kbc + d) − bc(1 + ka)2 6= 0,
and so
ad − bc 6= 0, 1 + ka 6= 0.
Thus any projectivity from Z5 Z6 to W5 W6 can be expressed as the restriction of a projective
transformation.
66
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
4.7
4.7.1
Quadrangle; diagonal triangle
Harmonic property of diagonal triangle
In §1.6.1 we seek to evaluate the Magnus quotient
δF (D1 , D2 , Z3 )δF (D1 , D3 , Z1 )
.
δF (D1 , D2 , Z1 )δF (D1 , D3 , Z3 )
(4.7.1)
We recall that
D1 = (1 − r)Z2 + rZ3 , D2 = (1 − s)Z3 + sZ1 , D3 = (1 − t)Z1 + tZ2 .
From this we have that
δF (D1 , D2 , Z3 ) = − (1 − r)sδF (Z1 , Z2 , Z3 ),
δF (D1 , D3 , Z1 ) = − rtδF (Z1 , Z2 , Z3 ),
δF (D1 , D2 , Z1 ) =(1 − r)(1 − s)δF (Z1 , Z2 , Z3 ),
δF (D1 , D3 , Z3 ) = − (1 − r)(1 − t)δF (Z1 , Z2 , Z3 ),
from which the Magnus quotient (4.7.1) is equal to
rst
(1 − r)(1 − s)(1 − t)
δF (Z1 , Z4 , Z2 ) δF (Z2 , Z4 , Z3 ) δF (Z3 , Z4 , Z1 )
=
δF (Z1 , Z4 , Z3 ) δF (Z2 , Z4 , Z1 ) δF (Z3 , Z4 , Z2 )
= − 1.
−
This shows that D1 (D2 , D3 , Z3 , Z1 ) is a harmonic pencil. Similar results hold for a pencil with
vertex D2 containing D3 , D1 and a pencil with vertex D2 containing D1 , D2 . This is known as the
harmonic property of a quadrangle and can be traced back to De La Hire in 1685.
4.7.2
The diagonal triple
The proofs in §1.6.1 and §4.7.1 are straightforward and easy but do not exactly reveal how the
diagonal points D1 , D2 , D3 are interlinked and why we should expect results involving them.
The point D1 on Z2 Z3 and Z1 Z4 has the form
D1 = (1 − r)Z2 + rZ3 = (1 − r′ )Z1 + r′ Z4 .
(4.7.2)
From this we have that
rZ3 + (r′ − 1)Z1 = (r − 1)Z2 + r′ Z4 ,
and so
r′ − 1
r−1
r′
r
Z
+
Z
=
Z
+
Z4 .
3
1
2
r + r′ − 1
r + r′ − 1
r + r′ − 1
r + r′ − 1
As the sum of the coefficients on each side here is 1 it represents in two ways the point of intersection
of Z3 Z1 and Z2 Z4 . This must be the point D2 and so we have that
D2 = (1 − s)Z3 + sZ1 = (1 − s′ )Z2 + s′ Z4 ,
where
r′ − 1
,
r + r′ − 1
Similarly from (4.7.2) we have that
s=
s′ =
r′
.
r + r′ − 1
(r′ − 1)Z1 + (1 − r)Z2 = −rZ3 + r′ Z4 .
67
4.8. SENSED-AREA AND SENSED-ANGLES
From this we get
1−r
r
r′
r′ − 1
Z
+
Z
=
−
Z
+
Z4 .
1
2
3
r′ − r
r′ − r
r′ − r
r′ − r
Then
D3 = (1 − t)Z1 + tZ2 = (1 − t′ )Z3 + t′ Z4 ,
where
t=
4.8
1−r
,
r′ − r
t′ =
r′
r′
.
−r
Sensed-area and sensed-angles
4.8.1
Suppose that Z1 Z2 , Z3 Z4 and Z5 Z6 are lines that all pass through a point W , so that
w = (1 − r)z1 + rz2 ,
Then
w = (1 − s)z3 + sz4 ,
z3 − w
s z4 − z3
,
=
z1 − w
r z2 − z1
and so
w = (1 − t)z5 + tz6 .
z4 − w
1 − s z4 − z3
,
=
z2 − w
1 − r z2 − z1
r 1 − s z2 − w
z1 − w
=
.
z3 − w
1 − r s z4 − w
By a similar argument we also have
z4 − w
z6 − w
z5 − w
z2 − w
=
=
(4.8.1)
1 − s t z3 − w
,
s 1 − t z5 − w
r z6 − w
t
.
1 − t 1 − r z1 − w
We recall from Barry [2, 10.4.1 and 10.5.1] that
z3 − z1
|Z1 , Z3 |
=
cis ∡F Z2 Z1 Z3 ,
z2 − z1
|Z1 , Z2 |
(4.8.2)
δF (Z1 , Z2 , Z3 ) = 21 |Z1 , Z2 ||Z1 , Z3 | sin ∡F Z2 Z1 Z3 ,
(4.8.3)
and
By (4.8.2) we have that
∡F Z 3 W Z 1
∡F Z 6 W Z 4
∡F Z 2 W Z 5
=
(
∡F Z 4 W Z 2
∡F Z4 W Z2 + 180F ′
if
if
r 1−s
1−r s
r 1−s
1−r s
> 0,
<0
=
(
∡F Z 5 W Z 3
∡F Z5 W Z3 + 180F ′
if
if
1−s t
s 1−t
1−s t
s 1−t
> 0,
<0
=
(
∡F Z 1 W Z 6
∡F Z1 W Z6 + 180F ′
if
if
r
t
1−t 1−r
t
r
1−t 1−r
> 0,
< 0,
where F ′ = tO,W (F ) is a translate of F . But the product of the three terms
r 1−s
,
1−r s
1−s t
,
s 1−t
r
t
1−t1−r
is positive, being equal to a square, so either all three are positive or one is positive (say the first)
and two are negative (then the second and third).
68
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
By (4.8.3) in the first case we have that
δF (W, Z3 , Z1 )
δF (W, Z4 , Z2 )
δF (W, Z6 , Z4 )
δF (W, Z5 , , Z3 )
δF (W, Z2 , Z5 )
δF (W, Z1 , Z6 )
=
=
=
|W , Z1 ||W , Z3 |
,
|W , Z2 ||W , Z4 |
|W , Z6 ||W , Z4 |
,
|W , Z5 ||W , Z3 |
|W , Z2 ||W , Z5 |
,
|W , Z1 ||W , Z6 |
and so by multiplication we have that
δF (W, Z3 , Z1 )δF (W, Z4 , Z2 )δF (W, Z6 , Z4 )
= 1.
δF (W, Z5 , Z3 )δF (W, Z2 , Z5 )δF (W, Z1 , Z6 )
(4.8.4)
In the second case we have that
δF (W, Z3 , Z1 )
δF (W, Z4 , Z2 )
δF (W, Z6 , Z4 )
δF (W, Z5 , , Z3 )
δF (W, Z2 , Z5 )
δF (W, Z1 , Z6 )
=
=
=
|W , Z1 ||W , Z3 |
,
|W , Z2 ||W , Z4 |
|W , Z6 ||W , Z4 |
−
,
|W , Z5 ||W , Z3 |
|W , Z2 ||W , Z5 |
−
,
|W , Z1 ||W , Z6 |
and so have (4.8.4) again. This improves on a result of Hoehn’s [14].
Exercises
3.1 Prove that each projective transformation has at least one fixed point.
3.2 Prove that for each projective transformation there is at least one line which is mapped to
itself.
3.3 Verify by computer algebra the identity
[δF (Z8 , Z9 , Z6 ) − δF (Z8 , Z9 , Z7 )]δF (Z, Z4 , Z5 )
=
3.4 Show that
+ [δF (Z4 , Z5 , Z8 ) − δF (Z4 , Z5 , Z9 )]δF (Z, Z6 , Z7 )
+ [δF (Z6 , Z7 , Z4 ) − δF (Z6 , Z7 , Z5 )]δF (Z, Z8 , Z9 )
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )].
δF (W, W3 , W1 )
= k1 δF (Z, Z3 , Z1 ),
δF (W, W2 , W3 )
δF (W, W1 , W2 )
= k2 δF (Z, Z1 , Z2 ),
δF (W, W2 , W3 )
yields a projective transformation.
3.5 Show that
δF (W, W3 , W1 )
k2
= k1 +
,
δF (W, W2 , W3 )
δF (Z, Z2 , Z3 )
δF (W, W2 , W3 ) = k3
yields a projective transformation.
δF (Z, Z2 , Z3 )
,
δF (Z, Z1 , Z2 )
69
4.8. SENSED-AREA AND SENSED-ANGLES
3.6 Prove the identity
δF (Z1 , Z2 , Z3 )δF [(Z4 , Z5 ), (Z6 , Z7 ), (U, V )] =
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z2 , Z3 )]δF (Z1 , U, V )
+ δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z3 , Z1 )]δF (Z2 , U, V )
+ δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z1 , Z2 )]δF (Z3 , U, V ).
3.7 (Pappus). For fixed non-collinear points Z1 , Z2 , Z3 , let
Z4 ∈ Z2 Z3 , Z5 ∈ Z3 Z1 , Z6 ∈ Z1 Z2
be such that
Z2 Z4
Z3 Z5
Z1 Z6
=
=
.
Z4 Z3
Z5 Z1
Z6 Z2
Show that then the centroid of [Z4 , Z5 , Z6 ] coincides with the centroid of [Z4 , Z5 , Z6 ].
Show too that the centroid of [Z1 , Z5 , Z6 ] lies on a fixed line.
3.8 (Bobilier) Let [Z1 , Z2 , Z3 , Z4 ] be a quadrangle and let Z5 ∈ Z1 Z4 , Z6 ∈ Z2 Z3 be such that
Z1 Z5
Z2 Z6
|Z1 , Z2 |
.
=
=
|Z
Z5 Z4
Z6 Z3
4 , Z3 |
Show that if Z1 Z2 and Z4 Z3 meet at a point Z0 , then Z5 Z6 is the mid-line of |Z1 Z0 Z4 .
3.9 On using the representations for W2 and W3 in 2.11.2, show that
Z3 W2 Z2 W1 W1 Z1 W2 Z4
= 1,
W2 Z1 W1 Z3 Z4 W1 Z2 W2
and find two similar identities.
3.10 Using the notation of 2.11 for a quadrangle and its diagonal triangle, show that
Z1
=
Z2
=
Z3
=
Z4
=
s1 + s2 − 1
W1 +
2(s1 − 1)
s1 + s2 − 1
W1 +
2s1
s1 + s2 − 1
W1 −
2(s2 − 1)
s1 + s2 − 1
W1 −
2s2
s1 − s2
1
W2 −
W3 ,
2(s1 − 1)
2(s1 − 1)
s1 − s2
1
W2 +
W3 ,
2s1
2s1
s1 − s2
1
W2 −
W3 ,
2(s2 − 1)
2(s2 − 1)
s1 − s2
1
W2 +
W3 .
2s2
2s2
Let Z2 Z3 meet W2 W3 at Z7 , Z2 Z4 meet W3 W1 at Z8 and Z3 Z4 meet W1 W2 at Z9 . By
expressing
λ
1
Z2 +
Z3 ,
1+λ
1+λ
in terms of W1 , W2 and W3 and choosing λ so that the term in W1 vanishes, show that for
Z7
s2 − 1
.
λ=−
s1
By a similar argument show that
Z8 =
1
µ
Z2 +
Z4 ,
1+µ
1+µ
70
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
Z9 =
1
ν
Z3 +
Z4 ,
1+ν
1+ν
where
µ=
s2
s1
, ν=
.
s2
s2 − 1
From Menelaus’ theorem, show that Z7 , Z8 , Z9 are collinear.
3.11 Suppose that W2 , W3 are distinct fixed points, and
Z1 =
λ
1
µ
1
W2 +
W3 , Z2 =
W2 +
W3 ,
1+λ
1+λ
1+µ
1+µ
variable points on the the line W2 W3 . For an arbitrary point U 6∈ W2 W3 and an arbitrary
line l through W2 , let U Z1 meet l at Z1′ , U W3 meet l at Z2′ , Z1′ W3 meet Z2′ Z2 at V , and U V
meet W2 W3 at the point Z3 where
Z3 =
ν
1
W2 +
W3 .
1+ν
1+ν
Prove that ν = λ + µ, so that the above construction of Z3 from Z1 and Z2 is independent
of the particular point U and line l used.
3.12 Suppose that W2 , W3 are distinct fixed points, W4 = 21 (W2 + W3 ), and
Z1 =
λ
1
µ
1
W2 +
W3 , Z2 =
W2 +
W3 ,
1+λ
1+λ
1+µ
1+µ
variable points on the the line W2 W3 . For an arbitrary point U 6∈ W2 W3 and an arbitrary
line l through W2 , let Z1 U meet l at Z1′ , and U W4 meet l at V . Suppose also that U W3 and
V Z2 meet at W , and W Z1′ meets W2 W3 at the point Z3 where
Z3 =
1
ν
W2 +
W3 .
1+ν
1+ν
Prove that ν = λµ, so that the above construction of Z3 from Z1 and Z2 is independent of
the particular point U and line l used.
3.13 (Pratt-Kasapi). Suppose that Z1 , Z2 , Z3 , Z4 , Z5 are the vertices of a pentagram and let
W1
=
W3
W5
=
=
Z1 Z3 ∩ Z2 Z5 , W2 = Z1 Z3 ∩ Z2 Z4 ,
Z2 Z4 ∩ Z3 Z5 , W4 = Z3 Z5 ∩ Z4 Z1 ,
Z4 Z1 ∩ Z5 Z2 .
Prove that
Z1 W1 Z2 W2 Z3 W3 Z4 W4 Z5 W5
= 1.
W2 Z3 W3 Z4 W4 Z5 W5 Z1 W1 Z2
[Suggestion. Let
W1
=
W2
=
W3
W4
=
=
W5
=
(1 − p)Z1 + pZ3 = (1 − y)Z5 + yZ2 ,
(1 − q)Z2 + qZ4 = (1 − u)Z1 + uZ3 ,
(1 − r)Z3 + rZ5 = (1 − v)Z2 + vZ4 ,
(1 − s)Z4 + sZ1 = (1 − w)Z3 + wZ5 ,
(1 − t)Z5 + tZ2 = (1 − x)Z4 + xZ2 .
4.8. SENSED-AREA AND SENSED-ANGLES
71
From
W1
W2
=
=
(1 − p)Z1 + pZ3 = (1 − y)Z5 + yZ2 ,
(1 − q)Z2 + qZ4 = (1 − u)Z1 + uZ3 ,
first eliminate Z1 and show that
p(1 − u) − u(1 − p)
y(1 − u) − (1 − q)(1 − p)
1−y
q
ρ
Z3 −
Z5 = ρ
Z2 −
Z4 ,
(1 − p)(1 − u)
1−p
(1 − p)(1 − u)
1−u
where ρ is chosen so that the sum of the coefficients on one side, and so on the other, is equal
to 1, and hence that
−(1 − y)(1 − u)
r
=
.
1−v
y(1 − u) − (1 − q)(1 − p)
Secondly from the equations for W1 and W2 eliminate Z2 and show that
uy − p(1 − q)
(1 − p)(1 − q) − y(1 − u)
1−y
q
σ
Z4 +
Z1 = σ
Z3 +
Z5 ,
1−q
y(1 − q
y(1 − q)
y
and hence that
(1 − p)(1 − q) − y(1 − u)
s
=
.
1−w
uy − p(1 − q)
Finally from the equations for W1 and W2 eliminate Z3 and show that
q
yu − p(1 − q)
(1 − p)u − (1 − u)p
1−y
Z5 +
Z2 = τ
Z4 +
Z1 ,
τ
p
pu
u
up
and hence that
t
yu − p(1 − q)
=
.
1−x
pq)
From these show that
pqrst
= 1.]
(1 − u)(1 − v)(1 − w)(1 − x)(1 − y)
3.14 As in §1.4.1, suppose that
Z=
λ
1
µ
1
Z1 +
Z2 , Z ′ =
Z3 +
Z4 .
1+λ
1+λ
1+µ
1+µ
Then by (1.4.1) Z ′ is the image of Z under central perspectivity in the point W1 if and only
if
aλµ + bλ + cµ + d = 0,
where
a = δF (W1 , Z2 , Z4 ), b = δF (W1 , Z2 , Z3 ), c = δF (W1 , Z1 , Z4 ), d = δF (W1 , Z1 , Z3 ).
Now we can express W1 = α′1 Z1 + β1′ Z2 + γ1′ Z3 where α′1 + β1′ + γ1′ = 1. From these show
that
a =δF (α′1 Z1 + β1′ Z2 + γ1′ Z3 , Z2 , Z4 ) = α′1 δF (Z1 , Z2 , Z4 ) + γ1′ δF (Z3 , Z2 , Z4 ),
b =δF (α′1 Z1 + β1′ Z2 + γ1′ Z3 , Z2 , Z3 ) = α′1 δF (Z1 , Z2 , Z3 ),
c =δF (α′1 Z1 + β1′ Z2 + γ1′ Z3 , Z1 , Z4 ) = β1′ δF (Z2 , Z1 , Z4 ) + γ1′ δF (Z3 , Z1 , Z4 ),
d =δF (α′1 Z1 + β1′ Z2 + γ1′ Z3 , Z1 , Z3 ) = β1′ δF (Z2 , Z1 , Z3 ).
72
CHAPTER 4. PROJECTIVE AND AFFINE TRANSFORMATIONS
Then
aδF (Z1 , Z2 , Z3 ) =bδF (Z1 , Z2 , Z4 ) + γ1′ δF (Z1 , Z2 , Z3 )δF (Z3 , Z2 , Z4 ),
cδF (Z1 , Z2 , Z3 ) = − dδF (Z2 , Z1 , Z4 ) + γ1′ δF (Z1 , Z2 , Z3 )δF (Z3 , Z1 , Z4 ).
From these
δF (Z1 , Z2 , Z3 )[aδF (Z3 , Z1 , Z4 ) − cδF (Z3 , Z2 , Z4 )]
=δF (Z1 , Z2 , Z4 )[bδF (Z3 , Z1 , Z4 ) − dδF (Z3 , Z2 , Z4 )].
This is a necessary condition for the coefficients in a central perspectivity. Show that it can
be rewritten as
δF (W1 , Z2 , Z4 )δF (Z1 , Z2 , Z3 )δF (Z3 , Z1 , Z4 ) δF (W1 , Z1 , Z4 )δF (Z1 , Z2 , Z3 )δF (Z3 , Z2 , Z4 )
−
δF (W1 , Z1 , Z3 )δF (Z1 , Z2 , Z4 )δF (Z3 , Z2 , Z4 ) δF (W1 , Z1 , Z3 )δF (Z1 , Z2 , Z4 )δF (Z3 , Z2 , Z4 )
δF (W1 , Z2 , Z3 )δF (Z1 , Z2 , Z4 )δF (Z3 , Z1 , Z4 )
= −1.
−
δF (W1 , Z1 , Z3 )δF (Z1 , Z2 , Z4 )δF (Z3 , Z2 , Z4 )
By (1.4.2) for a parallel perspectivity we have
a =δF (W1 , Z5 , Z2 ) − δF (W1 , Z5 , Z4 ), b = δF (W1 , Z5 , Z2 ) − δF (W1 , Z5 , Z3 ),
c = − δF (W1 , Z5 , Z4 ), d = −δF (W1 , Z5 , Z3 ).
By a similar method show that a − b + d − c = 0. Also verify this directly.
Chapter 5
Line equations of points; dual
concepts
5.1
Areal pair-coordinates; line equation of a point
COMMENT. The sequence of material in Chapter 2 is heading rapidly towards unmanageable
complication and ponderousness, so we now treat duality in connection with areal coordinates.
5.1.1
Line equation of a point
With a fixed point Z6 , on putting Z = Z6 the identity (3.2.1) becomes
δF (Z6 , U, V ) = δF (Z1 , U, V )α6 + δF (Z2 , U, V )β6 + δF (Z3 , U, V )γ6 .
(5.1.1)
On taking δF (Z6 , U, V ) = 0 we obtain as a condition that Z6 , U and V be collinear
α6 δF (Z1 , U, V ) + β6 δF (Z2 , U, V ) + γ6 δF (Z3 , U, V ) = 0.
(5.1.2)
We shall normally consider this when U 6= V when (5.1.2) is a condition that the line U V passes
through the fixed point Z6 .
Recalling the notation of (3.2.2)
l = δF (Z1 , U, V ), m = δF (Z2 , U, V ), n = δF (Z3 , U, V ),
as in §3.2 we call (l, m, n) areal pair-coordinates of a variable pair (U, V ), and the equation
(5.1.2) can be written as
α6 l + β6 m + γ6 n = 0,
(5.1.3)
We call this a line-equation of the point Z6 .
Note that in the line equation (5.1.3) the coefficients satisfy α6 + β6 + γ6 = 1. Consider as well
an equation
ul + vm + wn = 0,
(5.1.4)
where u + v + w 6= 0. We wish to show that there is a point Z6 such that (5.1.3) holds if and only
if (5.1.4) holds. In fact we show that there is a point Z6 and a number k 6= 0 such that
ul + vm + wn = k[α6 l + β6 m + γ6 n],
for all such triples (l, m, n). This requires that
uδF (Z1 , U, V ) + vδF (Z2 , U, V ) + wδF (Z3 , U, V )
=k[α6 δF (Z1 , U, V ) + β6 δF (Z2 , U, V ) + γ6 δF (Z3 , U, V )],
73
74
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
for all U and V . Putting U = Z2 , V = Z3 requires that u = kα6 , and by a similar argument we
need
v = kβ6 , w = kγ6 .
Then u + v + w = k(α6 + β6 + γ6 ) = k, and so
α6 =
v
w
u
, β6 =
, γ6 =
.
u+v+w
u+v+w
u+v+w
Conversely, given (5.1.4) we must have
ul + vm + wn = (u + v + w)
v
w
u
l+
m+
n ,
u+v+w
u+v+w
u+v+w
and since the sum of the coefficients in the square brackets is equal to 1 they must be the areal
coordinates of some point Z6 . With this we take k = u + v + w.
This correspondence between point and pair can clearly be exploited systematically to provide
dual results, and details of a slightly less explicit correspondence between point and line can be
found in a succession of textbooks on projective geometry from the mid-19th Century on. Line
coordinates which are based on trilinear coordinates have been in common use since the 19th
century, but I have not found any like ours which are based on areal coordinates.
5.1.1
Equation of a parallel pencil
We can re-write the fundamental identity (3.2.1) in the form
δF [(Z2 , Z3 ), (Z3 , Z1 ), (Z1 , Z2 )]δF (Z, U, V ) = δF [(Z3 , Z1 ), (Z1 , Z2 ), (U, V )]δF (Z, Z2 , Z3 )
+ δF [(Z1 , Z2 ), (Z2 , Z3 ), (U, V )]δF (Z, Z3 , Z1 ) + δF [(Z2 , Z3 ), (Z3 , Z1 ), (U, V )]δF (Z, Z1 , Z2 ),
as can be checked by expanding the dual sensed-areas. We re-write this as
δF (Z, U, V )
(5.1.5)
1
1
1
= δF [(Z3 , Z1 ), (Z1 , Z2 ), (U, V )]α + δF [(Z1 , Z2 ), (Z2 , Z3 ), (U, V )]β + δF [(Z2 , Z3 ), (Z3 , Z1 ), (U, V )]γ,
δ1
δ1
δ1
where δ1 = δF (Z1 , Z2 , Z3 ), so that we have the alternative expressions
1
δF [(Z3 , Z1 ), (Z1 , Z2 ), (U, V )],
δ1
1
m = δF [(Z1 , Z2 ), (Z2 , Z3 ), (U, V )],
δ1
1
n = δF [(Z2 , Z3 ), (Z3 , Z1 ), (U, V )].
δ1
l=
(5.1.6)
Suppose that we now start with Z4 Z5 k Z6 Z7 and take any (U, V ) such that Z4 Z5 k U V . Then
δF [(Z4 , Z5 ), (Z6 , Z7 ), (U, V )] = 0,
so that
δF (Z4 , Z6 , Z7 )δF (Z5 , U, V ) − δF (Z5 , Z6 , Z7 )δF (Z4 , U, V ) = 0.
On writing Z4 = α4 Z1 + β4 Z2 + γ4 Z3 , Z5 = α5 Z1 + β5 Z2 + γ5 Z3 , this expands to
δF (Z4 , Z6 , Z7 )(α5 l + β5 m + γ5 n) − δF (Z5 , Z6 , Z7 )(α4 l + β4 m + γ4 n) = 0.
(5.1.7)
On multiplying across by δF (Z1 , Z2 , Z2 ) in (5.1.7), we obtain an equation
ul + vm + wn = 0,
(5.1.8)
5.1. AREAL PAIR-COORDINATES; LINE EQUATION OF A POINT
75
with coefficient of l
u =δF (Z4 , Z6 , Z7 )δF (Z5 , Z2 , Z3 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z2 , Z3 )
=δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z2 , Z3 )],
and by a similar argument, coefficients of m and n, respectively,
v =δF (Z4 , Z6 , Z7 )δF (Z5 , Z3 , Z1 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z3 , Z1 )
=δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z3 , Z1 )],
and
w =δF (Z4 , Z6 , Z7 )δF (Z5 , Z1 , Z2 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z1 , Z2 )
=δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z1 , Z2 )].
Now the sum of these coefficients is
u + v + w =δF (Z4 , Z6 , Z7 )[δF (Z5 , Z2 , Z3 ) + δF (Z5 , Z3 , Z1 ) + δF (Z5 , Z1 , Z2 )]
−δF (Z5 , Z6 , Z7 )[δF (Z4 , Z2 , Z3 ) + δF (Z4 , Z3 , Z1 ) + δF (Z4 , Z1 , Z2 )]
=δF (Z1 , Z2 , Z2 )[δF (Z4 , Z6 , Z7 ) − δF (Z5 , Z6 , Z7 )]
=0,
as Z4 Z5 k Z6 Z7 .
Conversely, given u + v + w = 0, we ask whether any equation in the form (5.1.8) is the equation
of some parallel pencil. As
δF (Z4 , Z6 , Z7 ) = δF (Z5 , Z6 , Z7 ),
we need
δF (Z4 , Z6 , Z7 )[δF (Z5 , Z2 , Z3 ) − δF (Z4 , Z2 , Z3 )] =j1 u,
δF (Z4 , Z6 , Z7 )[δF (Z5 , Z3 , Z1 ) − δF (Z4 , Z3 , Z1 )] =j1 v,
δF (Z4 , Z6 , Z7 )[δF (Z5 , Z1 , Z2 ) − δF (Z4 , Z1 , Z2 )] =j1 w,
for some j1 6= 0. Thus we need
δF (Z5 , Z2 , Z3 ) − δF (Z4 , Z2 , Z3 )] =j2 u,
δF (Z5 , Z3 , Z1 ) − δF (Z4 , Z3 , Z1 )] =j2 v,
δF (Z5 , Z1 , Z2 ) − δF (Z4 , Z1 , Z2 )] =j2 w,
for some j2 6= 0, and so
α5 − α4 = j3 u,
β5 − β4 = j3 v,
γ5 − γ4 = j3 w,
for some j3 6= 0.
This requires that
Z5 − Z4 = j3 (uZ1 + vZ2 + wZ3 ),
(5.1.9)
which gives Z5 − Z4 to within a scalar multiple.
Conversely, supposed that (5.1.9) is satisfied, with j3 absorbed and taken to be 1. Then we
write
Z5 = Z4 − Z1 + (1 − v − w)Z1 + vZ2 + wZ3 ,
76
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
and have
δF (Z5 , Z2 , Z3 )
=δF Z4 − Z1 + (1 − v − w)Z1 + vZ2 + wZ3 , Z2 , Z3
=δF (1 − v − w)Z1 + vZ2 + wZ3 , Z2 − Z4 + Z1 , Z3 − Z4 + Z1
=(1 − v − w)δF Z1 , Z2 − Z4 + Z1 , Z3 − Z4 + Z1 + vδF Z2 , Z2 − Z4 + Z1 , Z3 − Z4 + Z1
+ wδF Z3 , Z2 − Z4 + Z1 , Z3 − Z4 + Z1
=(1 − v − w)δF Z4 , Z2 , Z3 + vδF Z2 + Z4 , Z2 + Z1 , Z3 + Z1 + wδF Z3 + Z4 , Z2 + Z1 , Z3 + Z1
=δF (Z4 , Z2 , Z3 ) + v[δF Z2 + Z4 , Z2 + Z1 , Z3 + Z1 − δF (Z4 , Z2 , Z3 )]
+ w[δF Z3 + Z4 , Z2 + Z1 , Z3 + Z1 − δF (Z4 , Z2 , Z3 )].
The coefficient of v here is equal to
δF (Z4 − Z1 , O, Z3 − Z2 ) − δF (Z4 , Z2 , Z3 ),
and that of w to
δF (Z4 − Z1 , Z2 − Z3 , O) − δF (Z4 , Z2 , Z3 ) = δF (Z4 − Z1 , O, Z3 − Z2 ) − δF (Z4 , Z2 , Z3 ),
and so we obtain that
δF (Z5 , Z2 , Z3 ) − δF (Z4 , Z2 , Z3 ) = −u[δF (Z4 − Z1 , O, Z3 − Z2 ) − δF (Z4 , Z2 , Z3 )].
(5.1.10)
By a similar argument, we write secondly Z5 = Z4 − Z2 + uZ1 + (1 − u − w)Z2 + wZ3 and obtain
that
δF (Z5 , Z3 , Z1 ) − δF (Z4 , Z3 , Z1 ) = −v[δF (Z4 − Z2 , O, Z1 − Z3 ) − δF (Z4 , Z3 , Z1 )],
(5.1.11)
and write thirdly Z5 = Z4 − Z3 + uZ1 + vZ2 + (1 − u − v)Z3 and obtain
δF (Z5 , Z1 , Z2 ) − δF (Z4 , Z1 , Z2 ) = −w[δF (Z4 − Z3 , O, Z2 − Z1 ) − δF (Z4 , Z1 , Z2 )].
(5.1.12)
Now the coefficients of −u, −v and −w in (5.1.10), (5.1.11) and (5.1.12) are equal. For the
coefficient of −v is equal to
δF (Z4 + Z3 − Z2 , Z3 , Z1 ) − δF (Z4 , Z3 , Z1 )
=δF (Z1 , Z4 + Z3 − Z2 , Z4 ) − δF (Z3 , Z4 + Z3 − Z2 , Z4 )
=δF (Z1 − Z4 , Z3 − Z2 , O) − δF (Z3 − Z4 , Z3 − Z2 , O)
=δF (Z4 − Z1 , O, Z3 − Z2 ) − δF (Z4 , Z2 , Z3 ),
and this is equal to the coefficient of −u.
By a similar argument, it can be shown that the coefficient of −w is equal to that of −v.
This gives the desired result, but we must exclude the case when u = v = w = 0 as we need to
have Z4 6= Z5 .
5.1.2
Linear expressions in areal pair-coordinates
We showed in §5.1.1, that given any u, v, w such that u + v + w 6= 0, then
uδF (Z1 , U, V ) + vδF (Z2 , U, V ) + wδF (Z3 , U, V )
=k[α6 δF (Z1 , U, V ) + β6 δF (Z2 , U, V ) + γ6 δF (Z3 , U, V )]
=kδF (Z6 , U, V ),
5.1. AREAL PAIR-COORDINATES; LINE EQUATION OF A POINT
77
where Z6 and k depend only on u, v and w. We now wish to consider this expression in the excluded
case when u + v + w = 0 but the trivial case u = v = w = 0 is excluded. Then we have
uδF (Z1 , U, V ) + vδF (Z2 , U, V ) + wδF (Z3 , U, V )
=uδF (Z1 , U, V ) + vδF (Z2 , U, V ) − (u + v)δF (Z3 , U, V )
=u[δF (Z1 , U, V ) − δF (Z3 , U, V )] + v[δF (Z2 , U, V ) − δF (Z3 , U, V )]
=uδF (Z1 − Z3 , U − V, O) + vδF (Z2 − Z3 , U − V, O)
=δF u(Z1 − Z3 ) + v(Z2 − Z3 ), U − V, O .
This takes all real-numbers as values unless u = v = 0 which would imply w = 0 and so reduce to
the trivial case.
These together show that there cannot be for l, m and n an identity of the form α + β + γ = 1.
We can easily show directly that there is no non-trivial identity of the form
dδF (Z1 , U, V ) + eδF (Z2 , U, V ) + f δF (Z3 , U, V ) = g.
For on putting U = Z2 , V = Z3 we would obtain
dδF (Z1 , Z2 , Z3 ) = g,
and by similar arguments
eδF (Z1 , Z2 , Z3 ) = g, f δF (Z1 , Z2 , Z3 ) = g.
We could have g = 0 but this entails only the trivial result that
0.δF (Z1 , U, V ) + 0.δF (Z2 , U, V ) + 0.δF (Z3 , U, V ) = 0.
If g 6= 0, we need
δF (Z1 , U, V ) + δF (Z2 , U, V )δF (Z3 , U, V ) = δF (Z1 , Z2 , Z3 ),
but putting U = V shows that this is false.
5.1.3
Determinant form of equation
Suppose that we are given pair-coordinates (l4 , m4 , n4 ), (l5 , m5 , n5 ) of two pairs which are both
collinear with a point Z6 . Then (α6 , β6 , γ6 ) is the unique solution of the simultaneous equations
l4 α + m4 β + n4 γ =0,
l5 α + m5 β + n5 γ =0,
α + β + γ =1.
It follows that

l4
det  l5
1
m4
m5
1

n4
n5  =
6 0.
1
(5.1.13)
If now (l, m, n) are coordinates of any line through Z6 , then we must have
lα6 + mβ6 + nγ6 =0,
l4 α6 + m4 β6 + n4 γ6 =0,
l5 α6 + m5 β6 + n5 γ6 =0,
and so

l
det  l4
l5
m
m4
m5

n
n4  = 0.
n5
(5.1.14)
78
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
This is a line equation of the point Z6 , as it has the correct form, and by (5.1.13) the sum of
the coefficients is non-zero. It generally will be only of the form (5.1.4) rather than (5.1.3), as we
see by §4.1.1. However let us take
(l4 , m4 , n4 ) = (l6,7 , m6,7 , n6,7 ),
(l5 , m5 , n5 ) = (l6,8 , m6,8 , n6,8 ),
for non-collinear points Z6 , Z7 , Z8 . Then
m6,7 n6,8 − n6,7 m6,8
=δF (Z2 , Z6 , Z7 )δF (Z3 , Z6 , Z8 ) − δF (Z3 , Z6 , Z7 )δF (Z2 , Z6 , Z8 )
=δF [(Z2 , Z3 ), (Z6 , Z7 ), (Z6 , Z8 )]
=δF [(Z6 , Z7 ), (Z6 , Z8 ), (Z2 , Z3 )]
=δF (Z6 , Z7 , Z8 )δF (Z6 , Z2 , Z3 )
=δ2 α6 ,
where δ2 = δF (Z6 , Z7 , Z8 ). By a similar argument
n6,7 l6,8 − l6,7 n6,8 = δ2 β6 , l6,7 m6,8 − m6,7 l6,8 = δ2 γ6 .
In this case, dividing the equation (5.1.14) by δ2 gives the form (5.1.3).
5.1.4
Line on two points
If
α4 l + β4 m + γ4 n = 0, α5 l + β5 m + γ5 n = 0,
are the line equations of two points Z4 , Z5 , then pair-coordinates of (Z4 , Z5 ) are found by solving
simultaneously
m
l
+ β4
n
n
m
l
α5 + β5
n
n
α4
=
−γ4 ,
=
−γ5 .
These equations have the solution
l = k(β4 γ5 − γ4 β5 ), m = k(γ4 α5 − α4 γ5 ), n = k(α4 β5 − β4 α5 ),
for some k.
From §3.1.2
δ1 (β4 γ5 − γ4 β5 ) = δF (Z1 , Z4 , Z5 ), δ1 (γ4 α5 − α4 γ5 ) = δF (Z2 , Z4 , Z5 ),
δ1 (α4 β5 − β4 α5 ) = δF (Z3 , Z4 , Z5 ),
and from the definition of areal coordinates, with U = Z4 , V = Z5 we have
l = δF (Z1 , Z4 , Z5 ), m = δF (Z2 , Z4 , Z5 ), n = δF (Z3 , Z4 , Z5 ).
Thus we must have k = δ1 .
In a number of applications, we shall need to use only some multiple of l, m, , n as in §5.1.8 to
come, which multiple being left undetermined.
79
5.1. AREAL PAIR-COORDINATES; LINE EQUATION OF A POINT
5.1.5
Parametric point equations
To deal with a pencil of pairs of points, we now take
(U, V ) = (1 − s)(U4 , V4 ) + s(U5 , V5 ),
s ∈ R.
By (2.1.2) and (2.1.2),
δF [(Z3 , Z1 ), (Z1 , Z2 ), (1 − s)(U4 , V4 ) + s(U5 , V5 )]
=(1 − s)δF [(Z3 , Z1 ), (Z1 , Z2 ), (U4 , V4 )] + sδF [(Z3 , Z1 ), (Z1 , Z2 ), (U5 , V5 )],
if and only if one of
U4 = U5 ,
V4 = V5 ,
U4 U5 k V4 V5 ,
(5.1.15)
holds. It follows by (5.1.6) that
l = (1 − s)l4 + sl5 ,
if and only if this condition (5.1.15) holds.
By a similar argument,
m = (1 − s)m4 + sm5 ,
n = (1 − s)n4 + sn5 ,
also hold when this condition obtains. Thus, under the condition (5.1.15), the pencil of pairs
collinear with the point of intersection of U4 V4 and U5 V5 when U4 V4 and U5 V5 are non-parallel,
and on lines parallel to U4 V4 and U5 V5 when the latter are parallel, have parametric equations
l = (1 − s)l4 + sl5 ,
5.1.6
m = (1 − s)m4 + sm5 ,
n = (1 − s)n4 + sn5 ,
s ∈ R.
(5.1.16)
Pair-coordinates under change of triple of reference
With the notation of §3.3, we have that
l′ =δF (Z4 , U, V )
=δF (α4 Z1 + β4 Z2 + γ4 Z3 , U, V )
=α4 δF (Z1 , U, V ) + β4 δF (Z2 , U, V ) + γ4 δF (Z3 , U, V )
=α4 l + β4 m + γ4 n,
and by a similar argument
m′ =α5 l + β5 m + γ5 n,
n′ =α6 l + β6 m + γ6 n.
The inverse transformation can be handled similarly.
5.1.7
Illustration using multiples of line-coordinates
In §5.1.4 we found a line equation of a point not precisely in the form (5.1.3) but only as some
non-zero multiple of it. It is often convenient to work with such multiples of line-equations.
As an illustration of how line coordinates of pairs of points and line equations of points can be
used to prove dual results by mimicing point coordinates and point equations of lines, we provide
a traditional proof of a case of the dual of Pappus’ theorem. Let g1 , g2 , g3 be lines through the
point Z2 and g4 , g5 , g6 lines through the distinct point Z3 . Let g2 meet g6 , g3 meet g5 , and h1 be
the line containing these points of intersection. Let g3 meet g4 , g1 meet g6 , and h2 be the line
containing these points of intersection. Let g1 meet g5 , g2 meet g4 , and h3 be the line containing
these points of intersection. We wish to show that h1 , h2 , h3 are concurrent or are all parallel.
80
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
As the line g2 with equation l2 α + m2 β + n2 γ = 0 passes through the point Z2 , we must have
m2 = 0 and so its equation is of the form
l2 α + n2 γ = 0.
Its line-coordinates are thus a multiple of (l2 , 0, n2 ). Similarly g6 has as line-coordinates some
multiple of (l6 , m6 , 0), and then the point of intersection of these has as areal coordinates some
multiple of (n2 m6 , −n2 l6 , −l2 m6 ). Similarly the point of intersection of intersection of g3 and
g5 has as areal coordinates some multiple of (n3 m5 , −n3 l5 , −l3 m5 ). From these the line h1 has
equation
(l3 l6 m5 n2 − l2 l5 m2 n3 )α + (l3 m5 m6 n2 − l2 m5 m6 n3 )β + (l6 m5 n2 n3 − l5 m6 n2 n3 )γ = 0.
On writing down similar equations for h2 and h3 , we find that h1 , h2 , h3 are concurrent or all
parallel if the following determinant of the multiples of the line-coordinates vanishes,
l 3 l 6 m5 n 2 − l 2 l 5 m2 n 3 l 3 m5 m6 n 2 − l 2 m5 m6 n 3 l 6 m5 n 2 n 3 − l 5 m6 n 2 n 3 l 1 l 4 m6 n 3 − l 3 l 6 m4 n 1 l 1 m6 m4 n 3 − l 3 m6 m4 n 1 l 4 m6 n 3 n 1 − l 6 m4 n 3 n 1 .
l 2 l 5 m4 n 1 − l 1 l 4 m5 n 2 l 2 m4 m5 n 1 − l 1 m4 m5 n 2 l 5 m4 n 1 n 2 − l 4 m5 n 1 n 2 In fact we can take g1 to be Z2 Z1 , so that l1 = 0, n1 = 1, and take g4 to be Z3 Z1 , so that
l4 = 0, m4 = 1. Then the above condition simplifies to
l 3 l 6 m5 n 2 − l 2 l 5 m2 n 3 l 3 m5 m6 n 2 − l 2 m5 m6 n 3 l 6 m5 n 2 n 3 − l 5 m6 n 2 n 3 = 0.
−l3 l6
−l3 m6
−l6 n3
l2 l5
l 2 m5
l 5 n2 −
This is clearly valid, as we see on adding m5 n2 times the second row and m6 n3 times the third
row to the first row.
5.2
5.2.1
Properties and identities for dual sensed-area
Equipoised quotients of dual sensed-areas
We now turn to the duals of some of our earlier material. The invariance of equipoised ratios,
noted in §4.2.1, is a very basic result. For a dual of it consider
=
=
δF [(Zi , Zj ), (Zk , Zl ), (Zm , Zn )]
δF (Zi , Zk , Zl )δF (Zj , Zm , Zn ) − δF (Zj , Zk , Zl )δF (Zi , Zm , Zn )
δF (Zi , Zk , Zl )δF (Zj , Zm , Zn )
−1 .
δF (Zj , Zk , Zl )δF (Zi , Zm , Zn )
δF (Zj , Zk , Zl )δF (Zi , Zm , Zn )
Now the term in square brackets here is a projective invariant. In a quotient of product of such total
terms, in which each point Zp appears exactly as often in the numerator as in the denominator, the
quotient of the product of such terms as those outside the square brackets will be an equipoised
quotient in the sense of §4.2.1, and so will be a projective invariant. Thus these dual equipoised
quotients are also projective invariants.
Material in Appendix C is included as preparation for §??; consideration of it could reasonably
deferred until then.
5.3
5.3.1
Parametric equations with pair-coordinates
Dual of parametric equations of a line
To pursue duality we need to identify the pair with pair-coordinates
(l, m, n) = (1 − s)(l6 , m6 , n6 ) + s(l9 , m9 , n9 ),
5.3. PARAMETRIC EQUATIONS WITH PAIR-COORDINATES
81
where (l6 , m6 , n6 ) and (l9 , m9 , n9 ) are pair coordinates of two given pairs of points. In §5.1.6 we
dealt with this under the condition (5.1.15) but now we wish to consider it more generally. We
recall the fundamental identity (3.2.1)
δF (Z1 , Z2 , Z3 )δF (Z, U, V ) = δF (Z1 , U, V )δF (Z, Z2 , Z3 )
+ δF (Z2 , U, V )δF (Z, Z3 , Z1 ) + δF (Z3 , U, V )δF (Z, Z1 , Z2 ),
(5.3.1)
which is the basis of our equations of lines in areal coordinates and also of our introduction of
pair-coordinates.
Let us take coordinates of a first pair (Z4 , Z5 ),
l6 = δF (Z1 , Z4 , Z5 ), m6 = δF (Z2 , Z4 , Z5 ), n6 = δF (Z3 , Z4 , Z5 ),
and of a second pair (Z7 , Z8 )
l9 = δF (Z1 , Z7 , Z8 ), m9 = δF (Z2 , Z7 , Z8 ), n9 = δF (Z3 , Z7 , Z8 ).
We now wish ((1 − s)l6 + sl9 , (1 − s)m6 + sm9 , (1 − s)n6 + sn9 ) to be pair-coordinates of some pair
(Us , Vs ) so that
δF (Z1 , Us , Vs ) =
(1 − s)δF (Z1 , Z4 , Z5 ) + sδF (Z1 , Z7 , Z8 ),
δF (Z3 , Us , Vs ) =
(1 − s)δF (Z3 , Z4 , Z5 ) + sδF (Z3 , Z7 , Z8 ).
δF (Z2 , Us , Vs ) =
(1 − s)δF (Z2 , Z4 , Z5 ) + sδF (Z2 , Z7 , Z8 )
If this is done then we will have, for all Z,
δF (Z1 , Z2 , Z3 )δF (Z, Us , Vs ) = δF (Z1 , Us , Vs )δF (Z, Z2 , Z3 )
+ δF (Z2 , Us , Vs )δF (Z, Z3 , Z1 ) + δF (Z3 , Us , Vs )δF (Z, Z1 , Z2 )
= [(1 − s)δF (Z1 , Z4 , Z5 ) + sδF (Z1 , Z7 , Z8 )]δF (Z, Z2 , Z3 )
+ [(1 − s)δF (Z2 , Z4 , Z5 ) + sδF (Z2 , Z7 , Z8 )]δF (Z, Z3 , Z1 )
+ [(1 − s)δF (Z3 , Z4 , Z5 ) + sδF (Z3 , Z7 , Z8 )]δF (Z, Z1 , Z2 )
= (1 − s)δF (Z1 , Z2 , Z3 )δF (Z, Z4 , Z5 ) + sδF (Z1 , Z2 , Z3 )δF (Z, Z7 , Z8 ).
Hence we need, for all Z,
δF (Z, Us , Vs ) = (1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 ).
(5.3.2)
This argument can be traced in reverse, so we will deal with our question in this latest form:
given points Z4 , Z5 , Z7 , Z8 and a real number s, can we find Us , Vs so that this identity holds.
5.3.2
The question rephrased
The question in this form leads first to the following. Given points W1 , W2 , W3 , W4 , under what
conditions is
δF (Z, W1 , W2 ) = δF (Z, W3 , W4 )
(5.3.3)
for all points Z in the plane? We considered this question already in §2.1.3, and found the answer
from (2.1.10) that
W4 − W3 = W2 − W1 and W3 ∈ W1 W2 .
As regards the question raised in §5.3.1, this shows that there never can be a unique solution
(Us , Vs ).
82
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
5.3.3
The question answered
Continuing with §5.3.1, given pairs (Z4 , Z5 ), (Z7 , Z8 ) we first suppose that Z4 Z5 and Z7 Z8 intersect
in a unique point; we denote this point of intersection by W1 . We take Z4 = (1 − s1 )Z5 + s1 W1 so
that
δF (Z, Z4 , Z5 ) = s1 δF (Z, W1 , Z5 ).
We also let Z7 = (1 − s2 )Z8 + s2 W1 so that
δF (Z, Z7 , Z8 ) = s2 δF (Z, W1 , Z8 ).
Then
=
=
=
(1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 )
(1 − s)s1 δF (Z, W1 , Z5 ) + ss2 δF (Z, W1 , Z8 )
ss2
(1 − s)s1
δF (Z, W1 , Z5 ) +
δF (Z, W1 , Z8 )
[(1 − s)s1 + ss2 ]
(1 − s)s1 + ss2
(1 − s)s1 + ss2
(1 − s)s1
ss2
[(1 − s)s1 + ss2 ]δF Z, W1 ,
Z5 +
Z8 .
(1 − s)s1 + ss2
(1 − s)s1 + ss2
We take Us = W1 and define Vs by
ss2
(1 − s)s1
Z5 +
Z8 . (5.3.4)
Vs = [1 − (1 − s)s1 − ss2 ]W1 + [(1 − s)s1 + ss2 ]
(1 − s)s1 + ss2
(1 − s)s1 + ss2
Then
δF (Z, W1 , Vs ) = [(1 − s)s1 + ss2 ]δF Z, W3 ,
(1 − s)s1
ss2
Z5 +
Z8
(1 − s)s1 + ss2
(1 − s)s1 + ss2
= (1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 ).
Now
Vs = [1 − (1 − s)s1 − ss2 ]W1 + (1 − s)s1 Z5 + ss2 Z8
so that
Vs − Us = Vs − W1
=
=
(1 − s)s1 (Z5 − W1 ) + ss2 (Z8 − W1 )
(1 − s)(Z5 − Z4 ) + s(Z8 − Z7 ).
We take W2 so that W2 − W1 = Z5 − Z4 so that W2 ∈ Z4 Z5 and tW1 ,Z4 (W2 ) = Z5 . We also
take W3 so that W3 − W1 = Z8 − Z7 so that W3 ∈ Z7 Z8 and tW1 ,Z7 (W3 ) = Z8 . Then we have that
Vs − Us =Vs − W1 = (1 − s)(W2 − W1 ) + s(W3 − W1 )
and so
Vs =(1 − s)W2 + sW3 .
This solves our problem in the case where Z4 Z5 , Z7 Z8 intersect.
(5.3.5)
83
5.3. PARAMETRIC EQUATIONS WITH PAIR-COORDINATES
b
Z8
b
Z7
W3
b
b
Vs
b
b
b
Us = W1
W2
Figure 5.1.
Z4
b
Z5
Turning now to the case of parallelism, with Z4 Z5 k Z7 Z8 we suppose that Z8 −Z7 = k1 (Z5 −Z4 )
where k1 6= 0. We first take Z4 Z5 6= Z7 Z8 . Then by (2.1.8)
δF (Z, Z7 , Z8 ) = k1 [δF (Z, Z4 , Z5 ) − δF (Z7 , Z4 , Z5 )].
(5.3.6)
Hence
(1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 ) = [1 − s + sk1 ]δF (Z, Z4 , Z5 ) − sk1 δF (Z7 , Z4 , Z5 ).
(5.3.7)
If s satisfies 1 − s + sk1 = 0 then our expression in (5.3.7) will be constant and we will be unable
to find such Us , Vs . Otherwise we suppose that Z5 − Z4 = k2 (Vs − Us ). Then
Vs − Us =
1
(Z5 − Z4 )
k2
(5.3.8)
so by (5.3.6)
δF (Z, Us , Vs ) =
1
[δF (Z, Z4 , Z5 ) − δF (Us , Z4 , Z5 )],
k2
and thus
δF (Z, Z4 , Z5 ) = k2 δF (Z, Us , Vs ) + δF (Us , Z4 , Z5 ).
On inserting this in (5.3.7) we have
=
(1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 )
k2 [1 − s + sk1 ]δF (Z, Us , Vs ) + [1 − s + sk1 ]δF (Us , Z4 , Z5 ) − sk1 δF (Z7 , Z4 , Z5 ).
We now choose Us so that
[1 − s + sk1 ]δF (Us , Z4 , Z5 ) − sk1 δF (Z7 , Z4 , Z5 ) = 0.
(5.3.9)
This will yield Us as any point on a specific line parallel to Z4 Z5 . Moreover we take k2 so that
k2 [1 − s + sk1 ] = 1.
With these choices we will have
δF (Z, Us , Vs ) = (1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 ).
(5.3.10)
84
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
Z7
Z8
b
Us
b
b
b
Vs
b
b
Z4
Z5
Figure 5.2.
To identify Us further in this case of parallelism, let
Us =
1
λ
Z4 +
Z7 .
1+λ
1+λ
Then we have that
[(1 − s + sk1 )
λ
− sk1 ]δF (Z7 , Z4 , Z5 ) = 0,
1+λ
from which we obtain that
λ
sk1
,
=
1+λ
1 − s + sk1
1
1−s
,
=
1+λ
1 − s + sk1
which yields
sk
1
1
Z4 + 1−ssk1 Z7 .
Us =
sk1
1 + 1−s
1 + 1−s
(5.3.11)
From (5.3.8) and (5.3.10) we have
Vs − Us = (1 − s + sk1 )(Z5 − Z4 ).
(5.3.12)
Note that by (5.3.12) we actually have that
Vs − Us = (1 − s)(Z5 − Z4 ) + s(Z8 − Z7 ).
If we change the notation to write
Z5 − Z4 = j1 (K2 − K1 ), Z8 − Z7 = j2 (K2 − K1 ),
then we will have k1 = j2 /j1 and so
Us
=
Vs − Us
=
(1 − s)j1
sj2
Z4 +
Z7 ,
(1 − s)j1 + sj2
(1 − s)j1 + sj2
[(1 − s)j1 + sj2 ](K2 − K1 ).
(5.3.13)
85
5.3. PARAMETRIC EQUATIONS WITH PAIR-COORDINATES
Z7
b
W2
Us
W1
W3
Z8
b
b
Z10
b
Z11
b
b
Z12
b
b
b
b
Vs
b
Z4
b
b
Z5
Z9
Figure 5.3.
A construction for locating Us and Vs in the case of parallelism is more complicated than in the
case of intersecting lines. We first take a point W1 ∈ Z4 Z7 so that W1 = (1 − s)Z4 + sZ7 , and then
W2 = sZ7 + (1 − s)Z4 , so that the mid-point of W1 and W2 is also the mid-point of Z7 and Z4 .
Note that
Z7 W2
s
.
=
1−s
Z7 W1
Let the line through W2 parallel to W2 Z8 meet the line Z7 Z8 at the point W3 . Then we have that
Z7 W3 =
s
s
k1 Z4 Z5 .
Z7 Z8 =
1−s
1−s
We take Z9 ∈ Z4 Z5 so that W3 Z9 k Z7 Z5 , and Z10 , Z11 ∈ Z7 Z8 so that Z5 Z10 k Z4 Z7 , Z9 Z11 k
Z4 Z7 . Then
s
k1 Z4 Z5 .
Z10 Z11 = Z5 Z9 =
1−s
We choose Us ∈ Z4 Z7 so that Z10 Us k Z11 Z4 . Then we have that
Z4 Us
Z11 Z10
Z9 Z5
s
=
=
=
k1 .
1−s
Us Z7
Z10 Z7
Z5 Z4
This thus yields an appropriate choice of Us .
To specify a corresponding Vs take the point Z12 = (1 − s)Z5 + sZ8 so that
Z12 − W1 = (1 − s)(Z5 − Z4 ) + s(Z8 − Z7 ) = (1 − s + k1 s)(Z5 − Z4 ) = Vs − Us ,
so that W1 Z12 k Z4 Z5 , and in fact [Us , W1 , Z12 , Vs ] is a parallelogram.
To find the locus of Vs as the real number s varies, without loss of generality we suppose that
Z4 Z5 ⊥ Z4 Z7 and then we introduce a frame of reference ([Z4 , Z5 , [Z4 , Z7 ) and take Cartesian
coordinates of points with respect to this. Then we have as coordinates (0, 0) for Z4 , (x5 , 0) for
Z5 for some x5 6= 0, (0, y7 ) for Z7 for some y7 6= 0, and (k1 x5 , y7 ) for Z8 . Then for Us and Vs we
have coordinates
k1 y7 s
k1 y7 s
, [1 + (k1 − 1)s]x5 ,
,
0,
1 + (k1 − 1)s
1 + (k1 − 1)s
If we write
x = [1 + (k1 − 1)s]x5 ,
y=
k1 y7 s
,
1 + (k1 − 1)s
(5.3.14)
for these coordinates of Vs , then as we are supposing initially that k1 6= 1, by eliminating s we find
that
k1 y7 x − x5
y=
.
(5.3.15)
k1 − 1 x
86
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
As we shall see in Chapter 7?, this locus of Vs is a hyperbola H1 , although we make no use of this
fact; it has as asymptotes the axis Z4 Z7 and the line with equation
y=
k1 y7
.
k1 − 1
It can be checked by substitution of the coordinates for Z5 and Z8 in Appendix D (?4.6.11) that
the hyperbola passes through these points.
Z8
Z7
b
b
b
b
Z4
Z5
Figure 5.4
When k1 = 1, by (5.3.11) we have that
Us = (1 − s)Z4 + sZ7 = U.
By (5.3.2) Vs − Us = Z5 − Z4 and so
Vs =(1 − s)Z4 + sZ7 + Z5 − Z4
=Z5 + s(Z7 − Z4 )
=Z5 + s(Z8 − Z5 )
=(1 − s)Z5 + sZ8
=V.
Thus in this case we have that Vs is on the line Z5 Z8 .
Z7
Z8
b
Us
b
b
b
Vs
b
Z4
b
Z5 Figure 5.5.
For our final case we take Z4 Z5 = Z7 Z8 and continue with Z8 − Z7 = k1 (Z5 − Z4 ), now with
5.3. PARAMETRIC EQUATIONS WITH PAIR-COORDINATES
k1 6= 1. Then
(1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 )
=(1 − s)δF (Z, Z4 , Z5 ) + sk1 δF (Z, Z7 , Z8 ),
=(1 − s + k1 s)δF (Z, Z4 , Z5 ),
by (2.1.7)????
and this is equal to δF (Z, Us , Vs ) if Us ∈ Z4 Z5 and
Vs − Us = (1 − s + k1 s)(Z5 − Z4 ) = (1 − s)(Z5 − Z4 ) + s(Z8 − Z7 ).
87
88
CHAPTER 5. LINE EQUATIONS OF POINTS; DUAL CONCEPTS
Chapter 6
Grassmann properties of rotors;
geometrical applications
6.1
6.1.1
Rotors
Definition of rotors
The material in §5.3.1, §5.3.2 and §5.3.3 gives us what we need to use in handling pair-coordinates,
but we can develop it to a much greater extent.
Our approach has given a natural introduction to pairs of points (Z4 , Z5 ) and (Z7 , Z8 ) such
that all four points are collinear. This corresponds to the informal concept of vectors which are
anchored to one line or sliding along one line, i.e. the two vectors have just the one line as a carrier
for both. This concept was used by Grassmann, and an informal term for such an anchored vector
is a rotor.
Given ordered pairs of points (Z4 , Z5 ) and (Z7 , Z8 ) we write
(Z4 , Z5 ) ≡ (Z7 , Z8 )
if δF (Z, Z4 , Z5 ) = δF (Z, Z7 , Z8 ) for all points Z ∈ Π. If Z5 = Z4 this is the case if and only
if Z8 = Z7 ; if Z5 6= Z4 , it is equivalent to having Z4 , Z5 , Z7 , Z8 all lie on some line and
Z5 − Z4 = Z8 − Z7 . We have earlier encountered this type of material in §2.1.3 and §3.2.3.
Then clearly ≡ is an equivalence relation, and we denote by hZ4 , Z5 i the unique equivalence
class which contains (Z4 , Z5 ) and call it a rotor.
When Z5 = Z4 we have δF (Z, Z7 , Z8 ) = δF (Z, Z4 , Z4 ) = 0 for all points Z if and only if
Z8 = Z7 . We call hZ4 , Z4 i the null rotor.
HISTORICAL NOTE.
For a treatment of rotors see Neville [21, pages 65–92]. Forder [9, pages 1–71] in 1941 and
Fearnley-Sander [7] in 1980 were intended as illuminating introductions to the concepts in Grassmann’s Calculus of Extension [13]. The fact that such introductions were judged appropriate after
a hundred years suggests that Grassmann’s material is difficult to penetrate, and in fact his book
is generally regarded as very difficult. The approach in this chapter may be a help in approaching
this material as it introduces a major concept from more conventional material.
6.1.2
Multiplication of a rotor by a number
We first consider multiplying a rotor by a scalar or number.
Given any rotor hZ4 , Z5 i and real number k we define the multiple khZ4 , Z5 i as follows. We
consider when we have
δF (Z, Z7 , Z8 ) = kδF (Z, Z4 , Z5 ),
89
90CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
for all points Z. When k = 0 or Z4 = Z5 we must have Z7 = Z8 . When k 6= 0 and Z4 6= Z5 , we
must have Z4 , Z5 , Z7 , Z8 collinear, and on taking
Z7 = (1 − s)Z4 + sZ5 , Z8 = (1 − t)Z4 + tZ5 ,
we find that
δF (Z, Z7 , Z8 ) = (t − s)δF (Z, Z4 , Z5 ),
so that we must take k = t − s and so, as well as having collinearity of the four points, we must
also have
Z8 − Z7 = k(Z5 − Z4 ).
In these cases we write
hZ7 , Z8 i = khZ4 , Z5 i.
It is straightforward to check that the outcome is unaltered if we replace (Z4 , Z5 ) by a pair
(Z10 , Z11 ) such that (Z4 , Z5 ) ≡ (Z10 , Z11 ), or if we replace (Z7 , Z8 ) by an equivalent pair.
We note that we can deal with the question of §5.3.1 when Z4 = Z5 , as we are asking if, given
s ∈ R, we can find Us and Vs such that
δF (Z, Us , Vs ) = (1 − s)δF (Z, Z4 , Z4 ) + sδF (Z, Z7 , Z8 ) = sδF (Z, Z7 , Z8 ),
for all Z. Similarly if Z7 = Z8 .
6.1.3
Associativity
It is straightforward to check in §6.1.2 that we then have
k2 [k1 hZ4 , Z5 i] = [k2 k1 ]hZ4 , Z5 i,
in all cases. This is an associative law.
6.1.4
Addition of rotors
We now turn to addition of rotors. We remark that in §5.3.3, given (Z4 , Z5 ) and (Z7 , Z8 ), we can
find U1/2 , V1/2 so that
1
2 δF (Z, Z4 , Z5 )
+ 12 δF (Z, Z7 , Z8 ) = δF (Z, U1/2 , V1/2 ),
for all Z, except when Z8 − Z7 = −(Z5 − Z4 ) and Z4 Z5 and Z7 Z8 are parallel to, and distinct
from, each other.
b
Z8
b
Z7
W3
b
b
W4
V1/2
b
b
U1/2 = W1
b
W2
Figure 4.6.
Z4
b
Z5
91
6.1. ROTORS
When Z4 Z5 and Z7 Z8 meet at a point W1 we take U1/2 = W1 and then V1/2 is the mid-point
of the points W2 and W3 in (4.3.5)???. We now take W4 so that V1/2 is the mid-point of U1/2
and W4 , making [W1 , W2 , W4 , W3 ] a parallelogram. We define the sum hZ4 , Z5 i + hZ7 , Z8 i to be
any rotor equal to hU1/2 , W4 i. This then is a rotor with carrier the line W1 V1/2 and with signed
magnitude 2[V1/2 − U1/2 ]. Note that
W4 − U1/2 = W2 − W1 + (W3 − W1 ) = Z5 − Z4 + (Z8 − Z7 ).
Then for all Z we have that
δF (Z, Z4 , Z5 ) + δF (Z, Z7 , Z8 ) = 2δF (Z, U1/2 , V1/2 ) = δF (Z, W1 , W4 ).
Z7
U1/2
Z10
Z8
b
b
b
b
b
b
b
b
W4
V1/2
b
b
Z11
Z4
b
b
Z5
Z9
Figure 4.7.
When Z4 Z5 and Z7 Z8 are parallel, but are not distinct lines with Z8 − Z7 = Z5 − Z4 , we let the
line through Z8 parallel to Z7 Z5 meet Z4 Z5 at Z9 , the line through Z5 parallel to Z4 Z6 meet Z7 Z8
at Z10 , the line through Z9 parallel to Z4 Z7 meet Z7 Z8 at Z11 , the line through Z10 parallel to
Z4 Z11 meet Z4 Z7 at U1/2 , and the line through U1/2 parallel to Z4 Z5 meet Z9 Z11 at W4 . Now
U1/2 divides (Z4 , Z7 ) in the ratio k1 : 1, as
U1/2 =
k1
1
Z4 +
Z7 .
1 + k1
1 + k1
Moreover [Z4 , Z9 , Z11 , Z7 ] is a parallelogram, and the line through the point of intersection of its
diagonals which is parallel to Z4 Z7 contains V1/2 , as
V1/2 − U1/2 = ( 12 +
k1
2 )(Z5
− Z4 ) = 12 [Z5 − Z4 + Z8 − Z7 ].
Here again W4 is taken so that V1/2 is the mid-point of U1/2 and W4 , and we define the sum
hZ4 , Z5 i + hZ7 , Z8 i to be any rotor equal to hU1/2 , W4 i. Again we have that for all Z,
δF (Z, Z4 , Z5 ) + δF (Z, Z7 , Z8 ) = 2δF (Z, U1/2 , V1/2 ) = δF (Z, U1/2 , W4 ).
Thus again
W4 − U1/2 = Z5 − Z4 + (Z8 − Z7 ).
In summary,
hZ4 , Z5 i + hZ7 , Z8 i = hZ9 , Z10 i
when
δF (Z, Z4 , Z5 ) + δF (Z, Z6 , Z7 ) = δF (Z, Z9 , Z10 ),
for all Z ∈ Π. We note that then the line Z9 Z10 has the equation
δF (Z, Z4 , Z5 ) + δF (Z, Z7 , Z8 ) = 0.
92CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
The above depends on having Z4 6= Z5 and Z7 6= Z8 . When Z4 = Z5 , we have that
δF (Z, Z4 , Z4 ) + δF (Z, Z7 , Z8 ) = δF (Z, Z7 , Z8 ),
for all Z and so we take
hZ4 , Z4 i + hZ7 , Z8 i = hZ7 , Z8 i.
Similarly if we have Z7 = Z8 .
It is straightforward to check that all this depends just on the equivalence classes and not on
the particular representatives of them chosen. Thus, except in one clearly specified case, we can
form the sum of two rotors.
In the one exceptional case when Z4 , Z5 , Z7 , Z8 are not collinear and Z8 − Z7 = −(Z5 − Z4 ),
with s = 12 in §5.3.3 we have that
1
2 δF (Z, Z4 , Z5 )
+ 21 δF (Z, Z7 , Z8 ) = 12 δF (Z7 , Z4 , Z5 ),
for all Z. The sum of these two rotors is not defined.
6.1.5
Commutativity of addition
If either of
hZ4 , Z5 i + hZ7 , Z8 i,
hZ7 , Z8 i + hZ4 , Z5 i
exists, then so does the other and their values are equal. For suppose that
hZ4 , Z5 i + hZ7 , Z8 i = hZ10 , Z11 i.
Then
δF (Z, Z4 , Z5 ) + δF (Z, Z7 , Z8 ) = δF (Z, Z10 , Z11 ),
for all Z ∈ Π and so
δF (Z, Z7 , Z8 ) + δF (Z, Z4 , Z5 ) = δF (Z, Z10 , Z11 ),
for all Z ∈ Π.
6.1.6
Additive identity
We note that the null-rotor is an additive identity, as
hZ4 , Z4 i + hZ7 , Z8 i = hZ7 , Z8 i,
holds in all cases. Combined with commutativity, this gives the result.
6.1.7
Addition associative
Given rotors hZ4 , Z5 i, hZ7 , Z8 i, hZ10 , Z11 i, if both of
[hZ4 , Z5 i + hZ7 , Z8 i] + hZ10 , Z11 i,
hZ4 , Z5 i + [hZ7 , Z8 i + hZ10 , Z11 i],
exist, then they are equal in value. To establish this, suppose that
hZ4 , Z5 i + hZ7 , Z8 i = hZ13 , Z14 i,
so that
δF (Z, Z4 , Z5 ) + δF (Z, Z7 , Z8 ) = δF (Z, Z13 , Z14 ),
for all Z. Suppose also that
hZ13 , Z14 i + hZ10 , Z11 i = hZ16 , Z17 i,
93
6.1. ROTORS
so that
δF (Z, Z13 , Z14 ) + δF (Z, Z10 , Z11 ) = δF (Z, Z16 , Z17 ),
and so
[δF (Z, Z4 , Z5 ) + δF (Z, Z7 , Z8 )] + δF (Z, Z10 , Z11 ) = δF (Z, Z16 , Z17 ),
for all Z.
On re-inserting the brackets, we also have that
δF (Z, Z4 , Z5 ) + [δF (Z, Z7 , Z8 ) + δF (Z, Z10 , Z11 )] = δF (Z, Z12 , Z13 ),
and so
[hZ4 , Z5 i + hZ7 , Z8 i] + hZ10 , Z11 i = hZ4 , Z5 i + [hZ7 , Z8 i + hZ10 , Z11 i],
when both of these exist. This is an associative law.
Z8
Z7
b
b
b
b
b
b
Z11
Z10
Z4
Z5
Figure 4.8.
Note that the existence of one of these triple sums does not imply the existence of the other. For
take distinct points Z4 , Z5 , points Z7 , Z8 not on Z4 Z5 but with Z8 − Z7 = −(Z5 − Z4 ), Z10 ∈ Z4 Z5
and Z11 − Z10 = Z4 − Z5 = Z8 − Z7 . Then
hZ4 , Z5 i + (hZ7 , Z8 i + hZ7 , Z10 i) = hZ4 , Z5 i + hZ7 , Z11 i
is defined but
hZ4 , Z5 i + hZ7 , Z8 i
is not.
6.1.8
Subtraction of rotors
On combining §6.1.2 and §6.1.4, we can define the difference of two rotors by
hZ7 , Z8 i − hZ4 , Z5 i = hZ7 , Z8 i + (−1)hZ4 , Z5 i,
except when Z4 , Z5 and Z7 , Z8 are on different parallel lines and Z5 − Z4 = Z8 − Z7 .
When Z4 Z5 and Z7 Z8 meet at a point W1 , using the notation of §6.1.4 we take W5 so that
W1 is the mid-point of W2 and W5 , we take W3 as before and W6 so that [W1 , W5 , W6 , W3 ] is a
parallelogram. Then
hZ7 , Z8 i − hZ4 , Z5 i = hW1 , W6 i.
Note that
W3 − W1 = Z8 − Z7 ,
W5 − W1 = Z4 − Z5 ,
W6 − W1 = W3 − W1 + (W5 − W1 ).
The carrier of this difference of rotors is the line W1 W6 and we note that in fact the line W2 W3 is
parallel to this.
94CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
b
Z8
b
Z7
W6
b
b
W3
b
b
b
b
b
W2
U1/2 = W1
W5
Z5
Z4
Figure 4.9.
When Z4 Z5 is parallel to Z7 Z8 we suppose that Z8 − Z7 = k3 (Z4 − Z5 ) so that k3 = −k1 .
We take W1 ∈ Z5 Z7 so that W1 = 21 Z5 + 12 Z7 and so W1 is the mid-point of Z5 and Z7 . Now
as 1 − s = s we will have W2 = W1 and W3 = Z8 . We take Z9 ∈ Z5 Z4 so that Z8 Z9 k Z7 Z4
and Z10 , Z11 ∈ Z7 Z8 so that Z4 Z10 k Z9 Z11 k Z5 Z7 . Then Z11 − Z10 = Z8 − Z7 . We choose
U1/2 ∈ Z5 Z7 so that Z10 U1/2 k Z11 Z5 . Moreover we take Z12 to be the mid-point of Z4 and
Z8 and choose V1/2 so that [U1/2 , W1 , Z12 , V1/2 ] is a parallelogram. Finally we take W4 so that
W4 − U1/2 = 2[V1/2 − U1/2 ].
b
Z7
Z11
b
Z8 = W3
b
b
Z10
Z12
b
b
b
W1 = W2
b
Z4
b
Z9
Z5
b
b
b
W4
V1/2
U1/2
Figure 4.10.
As the case of subtraction of rotors on parallel carriers is so important, we add further detail
on this case and provide another diagram. As
hZ7 , Z8 i − hZ4 , Z5 i = hZ7 , Z8 i + hZ5 , Z4 i,
we are reconsidering §5.3.3, now taking Z8 − Z7 = k3 (Z4 − Z5 ). With s =
that
1
k3
U1/2 =
Z5 +
Z7 ,
1 + k3
1 + k3
so will need
Z5 U1/2
= k3 .
U1/2 Z7
1
2
in (5.3.11) we have
Supposing that k3 6= 1, let Z5 Z7 and Z4 Z8 meet at a point W4 . Suppose that
Z7 = (1 − t1 )Z5 + t1 W4 ,
Z8 = (1 − t2 )Z4 + t2 W4 .
Then
Z8 − Z7 =(1 − t2 )Z4 − (1 − t1 )Z5 + (t2 − t1 )W4 ,
k3 (Z5 − Z4 ) =(1 − t2 )Z4 − (1 − t1 )Z5 + (t2 − t1 )W4 ,
(t2 − t1 )W4 =(k3 − 1 + t2 )Z4 + (1 − t1 − k3 )Z5 .
(6.1.1)
95
6.1. ROTORS
Now the sum of the coefficients on the right in the last line here is equal to t2 − t1 so this must be
equal to 0 as W4 6∈ Z4 Z5 . Then
Z7 = (1 − t1 )Z5 + t1 W4 ,
Z8 = (1 − t1 )Z4 + t1 W4 ,
so that Z8 − Z7 = (1 − t1 )(Z4 − Z5 ) and so 1 − t1 = k1 . Now Z7 − W4 = (1 − t1 )(Z5 − W4 ) so that
W4 Z7
= −k3 .
Z5 W4
Then
Z5 W4
1
=− .
k3
W4 Z7
We take W5 so that (Z5 , Z7 , W4 , W5 ) is a harmonic range, so that
Z5 W5
1
= ,
k
W5 Z6
3
and now if we take U1/2 so that Z5 U1/2 = U1/2 Z7 we also have that Z7 U1/2 = U1/2 Z5 and the
mid-point of W4 and W5 will also be the mid-point of Z5 and Z7 . We will now have (6.1.1) as
required.
By (5.3.13) we also have that
V1/2 − U1/2 = 12 [(Z4 − Z5 ) + (Z8 − Z7 )].
b
(6.1.2)
W5
b
Z8
Z7
b
b
W4
b
b
Z4
Figure 4.11.
Z5
b
b
V1/2
U1/2
When we write
we have in each case
and have
hZ7 , Z8 i − hZ4 , Z5 i = hZ10 , Z11 i,
W4 − U1/2 = Z8 − Z7 − (Z5 − Z4 ),
δF (Z, Z7 , Z8 ) − δF (Z, Z4 , Z5 ) = δF (Z, Z10 , Z11 ),
for all points Z.
In the exceptional case when Z4 , Z5 and Z7 , Z8 are on different parallel lines and Z5 − Z4 =
Z8 − Z7 , we have that
δF (Z, Z7 , Z8 ) − δF (Z, Z4 , Z5 ) = δF (Z5 , Z4 , Z7 ).
96CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
6.1.9
First distributive law
Consider now
k(hZ4 , Z5 i + hZ7 , Z8 i).
First suppose that k 6= 0, and let
hZ4 , Z5 i + hZ7 , Z8 i =
khZ4 , Z5 i = hZ13 , Z14 i,
hZ10 , Z11 i,
khZ7 , Z8 i = hZ16 , Z17 i.
Then it is not the case that Z4 Z5 and Z7 Z8 are parallel and distinct and Z8 − Z7 = −(Z5 − Z4 ).
It follows that it is not the case that (kZ4 )(kZ5 ) and (kZ7 )(kZ8 ) are parallel and distinct and
kZ8 − kZ7 = −(kZ5 − kZ4 ).
Hence
hZ13 , Z14 i + hZ16 , Z17 i = hZ19 , Z20 i
is defined. Then for all Z we have
δF (Z, Z19 , Z20 )
= δF (Z, Z13 , Z14 ) + δF (Z, Z16 , Z17 )
= kδF (Z, Z4 , Z5 ) + kδF (Z, Z7 , Z8 )
= kδF (Z, Z10 , Z11 ).
It follows that
k[hZ4 , Z5 i + hZ7 , Z8 i] =
=
khZ10 , Z11 i = hZ19 , Z20 i
hZ13 , Z14 i + hZ16 , Z17 i = khZ4 , Z5 i + khZ7 , Z8 i.
When k = 0 each side is the null-rotor. This is a distributive law.
6.1.10
Second distributive law
Consider now (k1 + k2 )hZ4 , Z5 i. For this let
k1 hZ4 , Z5 i
k1 hZ4 , Z5 i + k2 hZ4 , Z5 i
= hZ7 , Z8 i, k2 hZ4 , Z5 i = hZ10 , Z11 i,
= hZ7 , Z8 i + hZ10 , Z11 i = hZ13 , Z14 i.
Then for all Z,
δF (Z, Z13 , Z14 )
= δF (Z, Z7 , Z8 ) + δF (Z, Z10 , Z11 )
= k1 δF (Z, Z4 , Z5 ) + k2 δF (Z, Z4 , Z5 )
= [k1 + k2 ]δF (Z, Z4 , Z5 ),
and so
(k1 + k2 )hZ4 , Z5 i = k1 hZ4 , Z5 i + k2 hZ4 , Z5 i.
This is a second distributive law.
6.1.11
Moment of a force
If we have a force of the magnitude and sense Z2 Z3 acting along the line Z2 Z3 , then its moment
about the point Z1 is given by 2δF (Z1 , Z2 , Z3 ). Thus corresponding to our geometrical treatment
of rotors there is similar well-known material in mechanics.
All these concepts, introduced in a somewhat different way and with different notation, were
treated in Grassmann’s Calculus of Extension [13] and later presentations of the material in it.
6.2. DUAL SENSED-AREA AND ROTORS
6.1.12
97
Carrier of a rotor
In §5.3.3 we then have that
Vs − Us = (1 − s)(Z5 − Z4 ) + s(Z8 − Z7 ),
and that
(1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 ) = δF (Z, Us , Vs ),
for all Z ∈ Π. Thus we have that
(1 − s)hZ4 , Z5 i + shZ7 , Z8 i = hUs , Vs i.
We note too that the line Us Vs has the equation
(1 − s)δF (Z, Z4 , Z5 ) + sδF (Z, Z7 , Z8 ) = 0.
6.2
6.2.1
Dual sensed-area and rotors
Dual sensed-area and rotors
We now continue with some more specifically dual material. We first note that the expression for
dual sensed-area in (1.1.17) depends essentially on rotors. For suppose that the rotor hZ7 , Z8 i is
equal to the rotor hZ5 , Z6 i. Then we have that
δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )]
=δF (Z1 , Z3 , Z4 )δF (Z2 , Z7 , Z8 ) − δF (Z2 , Z3 , Z4 )δF (Z1 , Z7 , Z8 )
=δF (Z1 , Z3 , Z4 )δF (Z2 , Z5 , Z6 ) − δF (Z2 , Z3 , Z4 )δF (Z1 , Z5 , Z6 )
=δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )].
6.2.2
Areal pair-coordinates and rotors
We recall the definitions of areal pair-coordinates in (3.2.2):
l = δF (Z1 , U, V ), m = δF (Z2 , U, V ), n = δF (Z3 , U, V ).
By (3.2.3) we see that the pairs (U, V ) and (U ′ , V ′ ) have the same areal pair-coordinates if and
only if hU, V i = hU ′ , V ′ i. Thus areal pair-coordinates depend essentially on rotors.
6.2.3
Dual sensed-area and product of a rotor by a scalar
We next note that dual sensed-area in (1.1.17) is compatible with multiplication of a rotor by a
scalar. Suppose that hZ10 , Z11 i = khZ8 , Z9 i. Then
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z10 , Z11 )]
= δF (Z4 , Z6 , Z7 )δF (Z5 , Z10 , Z11 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z10 , Z11 )
= δF (Z4 , Z6 , Z7 )kδF (Z5 , Z8 , Z9 ) − δF (Z5 , Z6 , Z7 )kδF (Z4 , Z8 , Z9 )
= kδF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )].
6.2.4
Dual sensed-area and addition of rotors
We next note that dual sensed-area is compatible with addition of two rotors. Suppose that
hZ8 , Z9 i = hZ10 , Z11 i + hZ12 , Z13 i.
98CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
Then we have that
=
=
=
6.2.5
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )]
δF (Z4 , Z6 , Z7 )δF (Z5 , Z8 , Z9 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z8 , Z9 )
δF (Z4 , Z6 , Z7 )[δF (Z5 , Z10 , Z11 ) + δF (Z5 , Z12 , Z13 )]
− δF (Z5 , Z6 , Z7 )[δF (Z4 , Z10 , Z11 ) + δF (Z4 , Z12 , Z13 )]
δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z10 , Z11 )] + δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z12 , Z13 )].
A basic feature of dual sensed-area extended
We can now extend (2.1.1) as follows. Suppose that
k1 hZ5 , Z6 i + k2 hZ7 , Z8 i = hU, V i,
so that by §6.2.4 and §4.5.3,
k1 δF (Z, Z5 , Z6 ) + k2 δF (Z, Z7 , Z8 ) = δF (Z, U, V )
for all Z. Then, as in (1.1.7), we have
δF [(Z1 , Z2 ), (Z3 , Z4 ), (U, V )]
=δF (Z1 , Z3 , Z4 )δF (Z2 , U, V ) − δF (Z2 , Z3 , Z4 )δF (Z1 , U, V )
=δF (Z1 , Z3 , Z4 )[k1 δF (Z2 , Z5 , Z6 ) + k2 δF (Z2 , Z7 , Z8 )]
− δF (Z2 , Z3 , Z4 )[k1 δF (Z1 , Z5 , Z6 ) + k2 δF (Z1 , Z7 , Z8 )]
=k1 [δF (Z1 , Z3 , Z4 )δF (Z2 , Z5 , Z6 ) − δF (Z2 , Z3 , Z4 )δF (Z1 , Z5 , Z6 )]
+ k2 [δF (Z1 , Z3 , Z4 )δF (Z2 , Z7 , Z8 ) − δF (Z2 , Z3 , Z4 )δF (Z1 , Z7 , Z8 )]
=k1 δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] + k2 δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z7 , Z8 )].
We mainly apply this with k1 = 1 − s, k2 = s.
6.3
6.3.1
Pencils of rotors
Definition
In generalization of §2.1.2, given Z4 6= Z5 and Z7 6= Z8 we consider the rotors
hU, V i = (1 − r)hZ4 , Z5 i + rhZ7 , Z8 i,
for real numbers r. If hZ7 , Z8 i = khZ4 , Z5 i, then
hU, V i = [1 + (k − 1)r]hZ4 , Z5 i,
and this is a constant rotor if k = 1. Thus we assume that hZ7 , Z8 i 6= hZ4 , Z5 i; we then say that
these hU, V i form a pencil of rotors. Different values of the parameter give rise to different rotors
in the pencil.
By §5.3.3, when Z4 Z5 6k Z7 Z8 the line U V must pass through the unique point of intersection
of Z4 Z5 and Z7 Z8 , and when Z4 Z5 k Z7 Z8 the line U V is parallel to both of these lines.
6.3.2
Amenable pairs of pairs of points
We now consider two rotors hZ4 , Z5 i, hZ7 , Z8 i, where (Z4 , Z5 ) and (Z7 , Z8 ) are amenable pairs,
and show that if (U, V ) = (1 − s)(Z4 , Z5 ) + s(Z7 , Z8 ) as in §2.1.2, then we must have that
hU, V i = (1 − s)hZ4 , Z5 i + shZ7 , Z8 i.
99
6.3. PENCILS OF ROTORS
To see this, first suppose that in §5.3.3 Z4 Z5 and Z7 Z8 meet at a unique point W1 , and define
W2 , W3 , Us and Vs as there. In the case when Z7 = Z4 = W1 we have that U = Us , V = Vs , so the
conclusion is immediate. When Z7 = W1 but Z4 6= W1 , as Z4 Z7 k Z5 Z8 we must have Z8 ∈ Z4 Z5
which would make Z4 Z5 = Z7 Z8 and so is excluded. When either one or both of Z5 , Z8 is W1 the
case is not greatly different, so we may suppose that Z4 , Z5 , Z7 , Z8 are all different from W1 . As
Z4 Z7 k Z5 Z8 we have that Z8 − Z5 = k1 (Z7 − Z4 ) for some k1 6= 0; we cannot have k1 = 1 as that
would imply that Z8 − Z7 = Z5 − Z4 and so Z4 Z5 k Z7 Z8 .
Z5
b
b
Z4
V
b
b
U
b
b
Vs
W2
b
Z8
b
b
b
Z7
W3
W1 = Us
Figure 4.12
Now Z5 − Z4 = W2 − W1 , Z8 − Z7 = W3 − W1 and so
Vs − Us =(1 − s)(W2 − W1 ) + s(W3 − W1 )
=(1 − s)(Z5 − Z4 ) + s(Z8 − Z7 )
=[(1 − s)Z5 + sZ8 ] − [(1 − s)Z4 + sZ7 ]
=V − U.
Incidentally
W3 − W2 =(W3 − W1 ) − (W2 − W1 )
=(Z8 − Z7 ) − (Z5 − Z4 )
=Z8 − Z5 − (Z7 − Z4 )
=(k1 − 1)(Z7 − Z4 ),
and so W2 W3 k Z4 Z7 .
We have Z4 = (1 − t)W1 + tW2 , Z7 = (1 − t′ )W1 + t′ W3 , for some real numbers t and t′ . Then
Z7 − Z4 =(t − t′ )W1 + t′ W3 − tW2 ,
If t 6= t′ , we would have
1
(W3 − W2 ) =(t − t′ )W1 + t′ W3 − tW2 .
k1 − 1
W1 =
1
t − t′
t−
1
k1 − 1
W2 +
1
− t′ W3 ;
k1 − 1
as the sum of the coefficients here is equal to 1 we would have W1 ∈ W2 W3 which gives a contradiction. Hence we must have t = t′ . Then
U − W1 =(1 − s)(Z4 − W1 ) + s(Z7 − W1 )
=t[(1 − s)(W2 − W1 ) + s(W3 − W1 )]
=t(Vs − W1 ),
and so U ∈ W1 Vs = Us Vs . We conclude that
hU, V i = hUs , Vs i.
100CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
Z7
Z8
b
U = Us
b
b
b
V = Vs
b
b
Z4
Z5 Figure 4.13.
As a second major case we suppose that Z4 Z5 and Z7 Z8 are parallel and distinct. Then as in
§5.3.3 Z8 − Z7 = k1 (Z5 − Z4 ) for some k1 6= 0. We also have Z4 Z7 k Z5 Z8 so that [Z4 , Z5 , Z8 , Z7 ]
is a parallelogram and we must have k1 = 1. By (3.4.11???) with k1 = 1 we have that
Us = (1 − s)Z4 + sZ7 = U.
By (4.3.2???) Vs − Us = Z5 − Z4 and so
Vs =(1 − s)Z4 + sZ7 + Z5 − Z4
=Z5 + s(Z7 − Z4 )
=Z5 + s(Z8 − Z5 )
=(1 − s)Z5 + sZ8
=V.
Thus in this case we have that U = Us , V = Vs .
For our final case we suppose that Z4 Z5 = Z7 Z8 . For this take W1 6= W2 and
W3 =(1 − r3 )W1 + r3 W2 ,
Z7 =(1 − r7 )W1 + r7 W2 ,
Z4 = (1 − r4 )W1 + r4 W2 ,
Z8 = (1 − r8 )W1 + r8 W2 ,
Z5 = (1 − r5 )W1 + r5 W2 ,
where
r5 − r4 = 1,
r8 − r7 = r3 .
With U = (1 − s)Z4 + sZ7 , V = (1 − s)Z5 + sZ8 , we have that
V − U = (1 − s)(Z5 − Z4 ) + s(Z8 − Z7 ) = (1 − s + sr3 )(W2 − W1 ),
while with Us = W1 , Vs = (1 − s)W2 + sW3 , we have that
Vs − Us = (1 − s)(W2 − W1 ) + s(W3 − W1 ) = (1 − s + sr3 )(W2 − W1 ).
6.3.3
Rotors common to two pencils
For non-null rotors hZ4 , Z5 i, hZ7 , Z8 i, hZ10 , Z11 i, hZ13 , Z14 i, with hZ4 , Z5 i =
6 hZ7 , Z8 i and hZ10 , Z11 i 6=
hZ13 , Z14 i, let P1 be the rotor pencil on hZ4 , Z5 i and hZ7 , Z8 i, and P2 be the rotor pencil on
hZ10 , Z11 i and hZ13 , Z14 i. For any r ∈ R, suppose that
hUr , Vr i = (1 − r)hZ4 , Z5 i + rhZ7 , Z8 i,
(6.3.1)
so that hUr , Vr i is in P1 . Then
0 =δF [(Z10 , Z11 ), (Z13 , Z14 ), (Ur , Vr )]
=(1 − r)δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] + rδF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )]
=δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )]
+r δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] .
(6.3.2)
101
6.3. PENCILS OF ROTORS
From (6.3.2) we see that
Z10 Z11 , Z13 Z14 , Ur Vr
are concurrent or all parallel
(6.3.3)
if and only if
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )]
+r δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )]
=0.
(6.3.4)
Now we see that (6.3.4) has a unique solution in r if and only if
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] 6= 0,
(6.3.5)
(6.3.4) has no solution in r if and only if
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] = 0,
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] 6= 0,
(6.3.6)
and (6.3.4) has a solution for all r if and only if
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] = 0,
δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] = 0.
(6.3.7)
Now hUr , Vr i in (6.3.1) is in the pencil P1 and (6.3.3) is a necessary condition for this rotor also
to be in the second pencil P2 . We see that (6.3.6) is a sufficient condition for there to be no rotor
common to P1 and P2 .
There is detailed consideration of rotors common to two pencils in Appendix D.
6.3.4
Dual-parallelism of pencils
In seeking to choose a fruitful definition of when two pencils P1 and P2 of rotors are dual-parallel
we should like to have as close an analogue of parallelism of lines as we can. Now P1 and P2 having
no rotors in common does not give much control, as in Case E(i) of Appendix D, and the difference
between having none and a unique one in common can be very small, as in Case A(i) and Case
C(ii) there. Thus we need to take a different approach.
To help motivation we start by noting two properties. With the notation of §6.3.3, suppose
first that at least two rotors in the first pencil are also in the second pencil, that is that
hZ4 , Z5 i + r1 [hZ7 , Z8 i − hZ4 , Z5 i] =hZ10 , Z11 i + r1′ [hZ13 , Z14 i − hZ10 , Z11 i],
hZ4 , Z5 i + r2 [hZ7 , Z8 i − hZ4 , Z5 i] =hZ10 , Z11 i + r2′ [hZ13 , Z14 i − hZ10 , Z11 i],
for some real numbers r1 , r2 , r1′ and r2′ . Then by subtraction we have that
hZ7 , Z8 i − hZ4 , Z5 i = j[hZ13 , Z14 i − hZ10 , Z11 i],
where
j=
Then for any r ∈ R,
r2′ − r1′
.
r2 − r1
hZ4 , Z5 i + r[hZ7 , Z8 i − hZ4 , Z5 i]
=hZ10 , Z11 i + (r − r1 )[hZ7 , Z8 i − hZ4 , Z5 i] + r1′ [hZ13 , Z14 i − hZ10 , Z11 i]
=hZ10 , Z11 i + [j(r − r1 ) + r1′ ][hZ13 , Z14 i − hZ10 , Z11 i],
(6.3.8)
102CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
so that every rotor in the first pencil is also in the second pencil.
Suppose, secondly, that one rotor in the first pencil is also in the second and that (6.3.13) holds
for some j 6= 0. Taking it that
hZ4 , Z5 i + r1 [hZ7 , Z8 i − hZ4 , Z5 i] = hZ10 , Z11 i + r1′ [hZ13 , Z14 i − hZ10 , Z11 i],
for all r ∈ R we have that
hZ4 , Z5 i + r[hZ7 , Z8 i − hZ4 , Z5 i]
=hZ10 , Z11 i + [j(r − r1 ) + r1′ ][hZ13 , Z14 i − hZ10 , Z11 i],
and so again every rotor in the first pencil is also in the second pencil.
We now make the definition that the rotor-pencil P1 based on hZ4 , Z5 i and hZ7 , Z8 i, where
hZ4 , Z5 i 6= hZ7 , Z8 i, is dual-parallel to the rotor-pencil P2 based on hZ10 , Z11 i and hZ13 , Z14 i,
where hZ13 , Z14 i 6= hZ13 , Z14 i, if (4.6.13) holds for some real number j 6= 0. We write this as
P1 k P2 .
If
hUr , Vr i = (1 − r)hZ4 , Z5 i + rhZ7 , Z8 i, hUr′ ′ , Vr′′ i = (1 − r′ )hZ4 , Z5 i + r′ hZ7 , Z8 i,
then by (6.3.3) and (6.3.4) Z10 Z11 , Z13 Z14 , Ur Vr are concurrent or all parallel, and Z4 Z5 , Z7 Z8 , Ur′ ′ Vr′′
are concurrent or all parallel, if and only if
J1 = 0
and J2 = 0,
(6.3.9)
where
J1 =δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z7 , Z8 )] − δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )],
J2 =δF [(Z4 , Z5 ), (Z7 , Z8 )], (Z13 , Z14 )] − δF [(Z4 , Z5 ), (Z7 , Z8 )], (Z10 , Z11 )].
We wish to consider (6.3.14) in relation to (6.3.13).
In Case A or Case B in Appendix D, for the first pencil we take the point W5 so that
−hZ4 , Z5 i = hZ5 , Z4 i = hW1 , W5 i,
and then choose W6 so that
hW1 , W6 i = hW1 , W3 i + hW1 , W5 i = hZ7 , Z8 i − hZ4 , Z5 i.
Then we note that W6 is chosen so that [W1 , W2 , W3 , W6 ] is a parallelogram. Similarly for the
second pencil we take the point W11 so that
−hZ10 , Z11 i = hZ11 , Z10 i = hW7 , W11 i,
and then choose W12 so that
hW7 , W12 i = hW7 , W9 i + hW7 , W11 i = hZ13 , Z14 i − hZ10 , Z11 i.
Then we note that W12 is chosen so that [W7 , W8 , W9 , W12 ] is a parallelogram.
Then
hZ7 , Z8 i − hZ4 , Z5 i =hW1 , W6 i,
hZ13 , Z14 i − hZ10 , Z11 i =hW7 , W12 i.
If (6.3.14) holds, we must have that W7 ∈ W1 W6 and W1 ∈ W7 W12 . Thus W1 , W6 , W7 and W12
are collinear and so we have that
hW7 , W12 i = j[hW1 , W6 i],
so that condition (4.6.13) holds. The converse also holds, and so (6.3.13) and (6.3.14) are equivalent
in CASES A and B.
103
6.3. PENCILS OF ROTORS
b
b
Z8
Z7
b
b
W3
W6
b
Z5
b
b
Z4
b
W2
b
b
W1
b
W5
Z14
b
W12
Z13
W9
b
b
b
b
W7
W11
b
b
b
W8
Z10
Z11
b
Figure 4.31
b
Z14
b
Z8
b
Z13
b
Z7
W12
W9
W6
W3
b
b
W11
b
b
b
b
Z4
b
W2
W1 = W7
b
Z5
b
b
b
b
W8
Z10
Z11
W5
Figure 4.32
Next we suppose that the first pencil is as above but in the second Z10 Z11 and Z13 Z14 are
distinct parallel lines. Then taking again
hZ7 , Z8 i − hZ4 , Z5 i =hW1 , W6 i,
suppose that
hZ13 , Z14 i − hZ10 , Z11 i =hU ′ , W ′ i.
If (6.3.14) holds, we must have that U ′ W ′ k W1 W6 and W1 ∈ U ′ W ′ . Thus W1 , W6 , U ′ and
W are collinear and so we have that
′
hU ′ , W ′ i = j[hW1 , W6 i],
so that condition (6.3.13) holds. The converse also holds, and so (6.3.13) and (6.3.14) are equivalent
in CASE C.
104CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
b
b
W3
W6
b
Z5
b
Z4
b
W2
b
Z14
Z7
b
b
b
Z8
W1
b
W5
b
b
Z13
Z11
U′
b
W′
b
b
Figure 4.33
Z10
Finally suppose that Z4 Z5 and Z7 Z8 are distinct parallel lines, and Z10 Z11 and Z13 Z14 are
distinct parallel lines. Then suppose that
hZ7 , Z8 i − hZ4 , Z5 i =hU, W i,
suppose that
hZ13 , Z14 i − hZ10 , Z11 i =hU ′ , W ′ i.
If (6.3.14) holds, we only have that U W k U ′ W ′ , whereas for (6.3.13) we need U W = U ′ W ′ .
Thus (6.3.14) is weaker than (6.3.13) in this case, and that is why we take (6.3.13) instead of
(6.3.14) in our definition of dual-parallelism.
W
Z4
Z8
b
b
Z5
b
b
b
Z14
Z7
U
b
b
Z13
Z11
b
U′
b
W′
b
Figure 4.34.
6.3.5
Z10
Formula for a rotor common to two pencils of rotors
As in §2.1.3, if the two pencils of rotors in §6.3 are not dual-parallel, the rotor of the first pencil
which can also be in the second pencil is given by
hUr , Vr i = (1 − r)hZ4 , Z5 i + rhZ7 , Z8 i,
6.4. ROTORS AS DUALS OF POINTS
105
as in (6.3.1), where r is chosen to satisfy (6.3.4).
6.4
Rotors as duals of points
We continue our own development by regarding a pair (Z2 , Z3 ) of distinct points, a line Z2 Z3 and
a rotor hZ2 , Z3 i as all in some sense a dual of the concept of a point. We now concentrate on rotors
as duals of points. For simplicity, we shall concentrate on the cases where the rotors we operate
on have non-parallel carriers, in our diagrams and applications.
6.4.1
Sensed distance and distance from one rotor to another
From §6.1.8, when it exists the difference of two rotors
hZ6 , Z7 i − hZ4 , Z5 i
is a rotor of the form
hW1 , W1 + Z7 − Z6 − (Z5 − Z4 )i,
where W1 is any point on the carrier line of the rotor.
When this exists we define
hZ4 , Z5 i, hZ6 , Z7 i = hZ6 , Z7 i − hZ4 , Z5 i
as the sensed-difference of two rotors; we regard this as an analogue of our earlier concept of
complex-valued distance or sensed-distance, and we treat it as such.
When Z4 Z5 and Z6 Z7 meet at a point W1 , in the notation of §6.1.8 we note that this senseddistance is equal to W6 − W1 and also to W3 − W2 ; we shall generally use the latter of these for
convenience. We also note that W2 W3 k W1 W6 and so the line W2 W3 is parallel to the carrier line
of hZ6 , Z7 i − hZ4 , Z5 i.
In the general situation we would expect to have in particular
hZ4 , Z5 i, hZ6 , Z7 i + hZ6 , Z7 i, hZ8 , Z9 i = hZ4 , Z5 i, hZ8 , Z9 i.
This is indeed the case when both sides are defined. To prove this, suppose that
hZ6 , Z7 i − hZ4 , Z5 i =hZ10 , Z11 i,
hZ8 , Z9 i − hZ6 , Z7 i =hZ12 , Z13 i,
hZ8 , Z9 i − hZ4 , Z5 i =hZ14 , Z15 i.
Then for all Z ∈ Π,
δF (Z, Z6 , Z7 ) − δF (Z, Z4 , Z5 ) =δF (Z, Z10 , Z11 ),
δF (Z, Z8 , Z9 ) − δF (Z, Z6 , Z7 ) =δF (Z, Z12 , Z13 ),
δF (Z, Z8 , Z9 ) − δF (Z, Z4 , Z5 ) =δF (Z, Z14 , Z15 ).
Thus
δF (Z, Z8 , Z9 ) − δF (Z, Z4 , Z5 ) = δF (Z, Z10 , Z11 ) + δF (Z, Z12 , Z13 ),
and so
δF (Z, Z10 , Z11 ) + δF (Z, Z12 , Z13 ) = δF (Z, Z14 , Z15 ).
As
hZ10 , Z11 i + hZ12 , Z13 i = hZ16 , Z17 i,
is defined, we have that
δF (Z, Z10 , Z11 ) + δF (Z, Z12 , Z13 ) = δF (Z, Z16 , Z17 ).
106CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
Thus
δF (Z, Z14 , Z15 ) = δF (Z, Z16 , Z17 ),
and so
hZ10 , Z11 i + hZ12 , Z13 i = hZ14 , Z15 i.
When hZ6 , Z7 i − hZ4 , Z5 i exists, we also define the distance from the rotor hZ4 , Z5 i to the rotor
hZ6 , Z7 i to be
|hZ4 , Z5 i, hZ6 , Z7 i| =
p
[x7 − x6 − (x5 − x4 )]2 + [y7 − y6 − (y5 − y4 )]2 ,
which is non-negative and symmetrical. This distance is the magnitude of the sensed-distance
above. We note that it is equal to 0 if and only if hZ6 , Z7 i = hZ4 , Z5 i. For it is equal to 0 only
when Z7 − Z6 = Z5 − Z4 and since the sensed-difference is defined and Z4 Z5 k Z6 Z7 we cannot
have Z4 Z5 6= Z6 Z7 . It follows that Z4 , Z5 , Z6 , Z7 are collinear and so hZ6 , Z7 i = hZ4 , Z5 i.
When the lines Z4 Z5 and Z6 Z7 meet at a point W1 , then with the notation of §6.1.8 this
distance is equal to
|W1 , W6 | = |W2 , W3 |.
6.4.2
Division in a given ratio
Given a pair of rotors hZ4 , Z5 i, hZ6 , Z7 i, as in §5.3.3 we can consider any rotor equal to hUs , Vs i
as dividing the initial pair of rotors in the ratio of s to 1 − s. This is an analogue of the division
by the point W3 = (1 − s)W1 + sW2 of the pair of points (W1 , W2 ) in the ratio of s to 1 − s.
In particular we call hU1/2 , V1/2 i the mid-rotor of the pair of rotors hZ4 , Z5 i, hZ6 , Z7 i.
In having
hUs , Vs i = (1 − s)hZ4 , Z5 i + shZ6 , Z7 i,
we also have
hUs , Vs i − hZ4 , Z5 i = s(hZ6 , Z7 i − hZ4 , Z5 i).
Thus the sensed-distance from hZ4 , Z5 i to hUs , Vs i is s times the sensed-distance from hZ4 , Z5 i to
hZ6 , Z7 i, when these exist.
6.4.3
Dual of line, half-line and segment
Given distinct rotors hZ4 , Z5 i and hZ6 , Z7 i, we consider the set of rotors
(1 − s)hZ4 , Z5 i + shZ6 , Z7 i,
(s ∈ R),
as a dual of line. By §5.3.3, when Z4 Z5 , Z6 Z7 intersect at a point W1 all the carriers pass through
W1 ; when Z4 Z5 k Z6 Z7 all the carriers are parallel to these lines.
We consider this set of rotors for s ≥ 0 as a dual of half-line, and this set of rotors for 0 ≤ s ≤ 1
as a dual of segment. It is complicated to draw even partial diagrams to illustrate these concepts.
6.4.4
Sensed-angle in dual situation
The concept of a dual sensed-angle between two co-initial duals of half-lines, seems difficult to
visualize but we can at least extract some information on angles from this situation as follows.
107
6.4. ROTORS AS DUALS OF POINTS
b
Z9
W15
b
b
Z8
W14
Z7
b
b
b
θ
Z6
b
b
W13
W12
b
W9
b
W6
b
b
b
b
W3
b
b
W5
W11
b
b
W1
Z4
Z5
W7
Figure 4.35.
Suppose that the lines Z4 Z5 and Z6 Z7 meet at a point W1 . We use the notation W2 , W3 , W4 , W5 , W6
of §6.3.4, and then have
hZ6 , Z7 i − hZ4 , Z5 i = hW1 , W6 i.
Suppose also that Z4 Z5 and Z8 Z9 meet at a point W7 ; then we define W8 so that W8 − W7 =
Z5 − Z4 , W9 so that W9 − W7 = Z9 − Z8 , W11 so that W11 − W7 = Z4 − Z5 and W12 so that
[W7 , W11 , W12 , W9 ] is a parallelogram. We then have
hZ8 , Z9 i − hZ4 , Z5 i = hW7 , W12 i.
Suppose further that the carrier lines W1 W6 and W7 W12 meet at a point W13 , and define
W14 , W15 so that
W14 − W13 = W6 − W1 , W15 − W13 = W12 − W7 .
We denote the sensed-angle ∡F W14 W13 W15 by θ. Then on using complex-coordinates we have
successively
w6 − w1 = z7 − z6 − (z5 − z4 ), w12 − w7 = z9 − z8 − (z5 − z4 ),
and
w14 − w13 = w6 − w1 ,
Hence
w15 − w13 = w12 − w7 .
w15 − w13
|W13 , W15 |
z9 − z8 − (z5 − z4 )
=
cis θ =
.
w14 − w13
|W13 , W14 |
z7 − z6 − (z5 − z4 )
(6.4.1)
We shall refer to θ as the intrinsic angle associated with the triple (hZ6 , Z7 i, hZ4 , Z5 i, hZ8 , Z9 i).
The intrinsic angle’s being 90F is equivalent to having
[x7 − x6 − (x5 − x4 )][x9 − x8 − (x5 − x4 )] + [y7 − y6 − (y5 − y4 )][y9 − y8 − (y5 − y4 )] = 0.
This corresponds to having W1 W6 perpendicular to W7 W12 , or equivalently W2 W3 parallel to
W8 W9 . ?????????????
108CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
On utilizing complex coordinates, this condition can be written as
z9 − z8 − (z5 − z4 ) = ri[z7 − z6 − (z5 − z4 )],
for some non-zero real number r.
b
Z7
b
W14
b
θ
b
b
Z6
W15
W13
b
b
W11
b
W12
Z11
b
Z10
b
W9
b
W7
b
Z8
b
Z9
W6
W5
b
b
b
b
W3
Z4
Z5
b
b
W1
Figure 4.36.
More generally consider hZ6 , Z7 i − hZ4 , Z5 i and hZ10 , Z11 i − hZ8 , Z9 i. Suppose that the lines
Z4 Z5 and Z6 Z7 meet at a point W1 . We use the notation W2 , W3 , W4 , W5 , W6 of §6.3.4, and then
have
hZ6 , Z7 i − hZ4 , Z5 i = hW1 , W6 i.
Suppose also that Z8 Z9 and Z10 Z11 meet at a point W7 ; then we define W9 so that W9 −
W7 = Z11 − Z10 , W11 so that W11 − W7 = −(Z9 − Z8 ) and W12 so that [W7 , W11 , W12 , W9 ] is a
parallelogram. We then have
hZ10 , Z11 i − hZ8 , Z9 i = hW7 , W12 i.
Suppose further that the carrier lines W1 W6 and W7 W12 meet at a point W13 , and define
W14 , W15 so that
W14 − W13 = W6 − W1 , W15 − W13 = W12 − W7 .
We denote the sensed-angle ∡F W14 W13 W15 by θ. Then on using complex-coordinates we have
successively
w6 − w1 = z7 − z6 − (z5 − z4 ),
w12 − w7 = z11 − z10 − (z9 − z8 ),
and
Hence
w14 − w13 = w6 − w1 ,
w15 − w13 = w12 − w7 .
|W13 , W15 |
z11 − z10 − (z9 − z8 )
w15 − w13
=
cis θ =
.
w14 − w13
|W13 , W14 |
z7 − z6 − (z5 − z4 )
(6.4.2)
We shall refer to θ as the intrinsic angle from the rotor hZ6 , Z7 i − hZ4 , Z5 i to the rotor
hZ10 , Z11 i − hZ8 , Z9 i.
109
6.5. GEOMETRICAL APPLICATIONS OF ROTORS
6.5
Geometrical applications of rotors
Neville [21, page 66] wrote
The study of vectors localised in lines is incomparably more important than the study of vectors
associated with points; mathematics as a whole has suffered much from the appropriation of this
subject by particular branches of applied mathematics when it should be regarded from the beginning
as a fundamental part of pure mathematics, available for investigations that have no concern with
statics or dynamics just as Cartesian coordinates may be used in work that has little relation to
analytical geometry.
I now give a sample of results which can be viewed as being in accordance with Neville’s
programme.
6.5.1
Analogue of isosceles
triangle
′
W2
b
b
V4′
W3
Z1
b
b
W5
b
b
W6
b
W2
b
b
W12
b
b
Z2
W1
b
Z3
W9
b
b
b
b
W11
W1′
b
W1′′
W7
b
W3′
Figure 4.37.
We start with a skeleton figure as follows. On the sidelines of a triangle [Z1 , Z2 , Z3 ] we take
points W1 , W1′ ∈ Z2 Z3 , points W2 , W2′ ∈ Z3 Z1 and points W3 , W3′ ∈ Z1 Z2 so that
W1′ − Z3 = W1 − Z2 ,
W2′ − Z1 = W2 − Z3 ,
W3′ − Z2 = W3 − Z1 .
Suppose that V4′ is the mid-point of W2′ and W3 . We assume that Z1 V4′ meets Z2 Z3 at a point W7 ,
and choose W1′′ , W11 ∈ Z2 Z3 so that
W1′′ − W7 = W1 − Z2 , W7 − W11 = W1 − Z2 .
We also choose W9 ∈ Z1 W7 so that W9 −W7 = V4′ −Z1 , and choose W12 so that [W7 , W9 , W12 , W11 ]
is a parallelogram.
We now note that
′
1
2 (W2
− Z1 ) + 21 (W3 − Z1 ) − (W1 − Z2 ) =V4′ − Z1 − (W1′′ − W7 )
=W9 − W7 + (W11 − W7 )
=W12 − W7 .
110CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
It follows that W7 W12 is the carrier of
1
2 hZ3 , W2 i
+ 21 hZ1 , W3 i − hZ2 , W1 i,
and this is parallel to W1′′ W9 .
We further choose W5 ∈ Z1 Z3 so that W2′ − Z1 = Z1 − W5 , and then choose W6 so that
[Z1 , W3 , W6 , W5 ] is a parallelogram. Then
W6 − Z1 = W3 − Z1 − (W2′ − Z1 ) = W3 − Z1 − (W2 − Z3 ).
It follows that Z1 W6 is the carrier of hZ1 , W3 i − hZ3 , W2 i and is parallel to W2′ W3 .
Now, using scalar or dot products of vectors, we have that
[ 12 (W2′ − Z1 + W3 − Z1 ) − (W1′′ − W7 )] • [W3 − Z1 − (W2′ − Z1 )]
= 12 [W2′ − Z1 − (W1′′ − W7 )] + 21 [W3 − Z1 − (W1′′ − W7 )] •
W3 − Z1 − (W1′′ − W7 ) − [W2′ − Z1 − (W1′′ − W7 )]
= 21 [W3 − Z1 − (W1′′ − W7 )] • [W3 − Z1 − (W1′′ − W7 )]
− 21 [W2′ − Z1 − (W1′′ − W7 )] • [W2′ − Z1 − (W1′′ − W7 )].
This is equal to 0, and so the lines W7 W12 and Z1 W6 are perpendicular, if and only if the distances
from hZ1 , W3 i and hZ3 , W2 i, respectively, to hZ2 , W1 i are equal, that is the distances |W2 , W1′ |
and |W3′ , W1 | are equal.
We thus have proved the following result:- Suppose that the rotors hZ4 , Z5 i, hZ6 , Z7 i and
hZ8 , Z9 i have as carriers the sidelines Z2 Z3 , Z3 Z1 , Z1 Z2 of a triangle, respectively, and that the
carrier of hZ8 , Z9 i − hZ6 , Z7 i intersects the line Z2 Z3 . Then the carrier of
1
2 hZ6 , Z7 i
+ 21 hZ8 , Z9 i − hZ4 , Z5 i
is perpendicular to the carrier of hZ8 , Z9 i − hZ6 , Z7 i if and only if hZ4 , Z5 i is equidistant from
hZ6 , Z7 i and hZ8 , Z9 i.
This follows from the foregoing on selecting W1 so that W1 − Z2 = Z5 − Z4 , W2 so that
W2 − Z3 = Z7 − Z6 and W3 so that W3 − Z1 = Z9 − Z8 .
This generalizes material on isosceles triangles.
111
6.5. GEOMETRICAL APPLICATIONS OF ROTORS
6.5.2
Dual of Pythagoras’ theorem′
W2
b
b
W3
Z1
b
W5
b
b
W6
W18
W11
b
W2
b
W12
b
b
b
b
b
Z2
b
b
W1
W17
Z3
W1′
b
W3′
Figure 4.38.
We start with a skeleton figure as follows. On the sidelines of a triangle [Z1 , Z2 , Z3 ] we take
points
W1 , W1′ , W17 ∈ Z2 Z3 , W2 , W2′ , W5 ∈ Z3 Z1 , W3 , W3′ , W11 ∈ Z1 Z2 ,
so that
W1′ − Z3 =W1 − Z2 ,
W2′
W3′
− Z1 =W2 − Z3 ,
− Z2 =W3 − Z1 ,
Z3 − W17 = W1 − Z2 ,
Z1 − W5 = W2 − Z3 ,
Z2 − W11 = W3 − Z1 .
Choose W6 so that [Z1 , W5 , W6 , W3 ] is a parallelogram. Then
W3 − Z1 − (W2′ − Z1 ) = W3 − Z1 + (W5 − Z1 ) = W6 − Z1 ,
so that Z1 W6 is the carrier of hZ1 , W3 i − hZ1 , W2′ i and |hZ1 , W2′ i, hZ1 , W3 i| = |W2′ , W3 |.
Choose W12 so that [Z2 , W1 , W12 , W11 ] is a parallelogram. Then
W1 − Z2 − (W3′ − Z2 ) = W1 − Z2 + W11 − Z2 = W12 − Z2 ,
so that Z2 W12 is the carrier of hZ2 , W1 i − hZ1 , W3 i and |hZ2 , W1 i, hZ1 , W3 i| = |W3′ , W1 |.
Choose W18 so that [Z3 , W2 , W18 , W17 ] is a parallelogram. Then
W2 − Z3 − (W1′ − Z3 ) = W2 − Z3 + W17 − Z3 = W18 − Z3 ,
so that Z3 W18 is the carrier of hZ3 , W2 i − hZ2 , W1 i and |hZ2 , W1 i, hZ3 , W2 i| = |W1′ , W2 |.
Now
[W3 − Z1 − (W2 − Z3 )] • [W3 − Z1 − (W2 − Z3 ] − [W2 − Z3 − (W1 − Z2 )] • [W2 − Z3 − (W1 − Z2 )]
− [W1 − Z2 − (W3 − Z1 )] • [W1 − Z2 − (W3 − Z1 )]
= − 2(W3 − Z1 ) • (W2 − Z3 ) + 2(W2 − Z3 ) • (W1 − Z2 )
− 2(W1 − Z2 ) • (W1 − Z2 ) + 2(W1 − Z2 ) • (W3 − Z1 )
= − 2[W3 − Z1 − (W1 − Z2 )] • [W2 − Z3 − (W1 − Z2 )],
112CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
so that
|Z1 , W6 |2 = |Z2 , W12 |2 + |Z3 , W18 |2 ,
or equivalently
2
2
2
|W2′ , W3 | = |W3′ , W1 | + |W1′ , W2 | ,
if and only if
[W3′ − Z2 − (W1 − Z2 )] • [W2 − Z3 − (W1′ − Z3 )] = (W12 − Z2 ) • (W18 − Z3 ) = 0,
that is the lines Z2 W12 and Z3 W18 are perpendicular, or equivalently W1′ W2 and W1 W3′ are perpendicular.
Thus we have proved the following:- Suppose that we have three rotors
hZ4 , Z5 i, hZ6 , Z7 i, hZ8 , Z9 i,
on the sidelines Z2 Z3 , Z3 Z1 , Z1 Z2 , respectively, of a triangle [Z1 , Z2 , Z3 ]. Then
|hZ6 , Z7 i, hZ8 , Z9 i|2 = |hZ4 , Z5 i, hZ6 , Z7 i|2 + |hZ8 , Z9 i, hZ4 , Z5 i|2
if and only if the carriers of hZ6 , Z7 i − hZ4 , Z5 i and hZ8 , Z9 i − hZ4 , Z5 i are perpendicular to each
other.
6.5.3
Dual parallelograms
In the case of an ordinary parallelogram [Z1 , Z2 , Z3 , Z4 ] we start with the definition that
Z3 Z4 k Z1 Z2 ,
Z2 Z3 k Z1 Z4 ,
Z4 6∈ Z1 Z4 ,
and then for a development by means of vectors note that this yields
Z3 − Z4 = k1 (Z2 − Z1 ),
Z3 − Z2 = k2 (Z4 − Z1 ),
for some real numbers k1 and k2 . By subtraction it follows that
Z4 − Z2 = k2 (Z4 − Z1 ) − k1 (Z2 − Z1 ),
and hence
Z4 − Z1 − (Z2 − Z1 ) = k2 (Z4 − Z1 ) − k1 (Z2 − Z1 ).
From this it follows that k2 = k1 = 1 unless we have Z4 − Z1 = s(Z2 − Z1 ), for some real number
s, and this is ruled out as it would imply that Z4 ∈ Z1 Z2 . It thus follows that we have
Z3 − Z4 = Z2 − Z1 ,
(6.5.1)
Z3 − Z2 = Z4 − Z1 .
(6.5.2)
from which it follows that
To obtain a dual of this material, we start with rotors
hZ4 , Z5 i, hZ7 , Z8 i, hZ10 , Z11 i, hZ13 , Z14 i
and suppose that
hhZ10 , Z11 i, hZ13 , Z14 ii khhZ4 , Z5 i, hZ7 , Z8 ii
hhZ7 , Z8 i, hZ10 , Z11 ii khhZ4 , Z5 i, hZ13 , Z14 ii,
and that hZ13 , Z14 i is not in the rotor pencil based on hZ4 , Z5 i and hZ7 , Z8 i. We wish to have
hZ10 , Z11 i − hZ13 , Z14 i = hZ7 , Z8 i − hZ4 , Z5 i,
(6.5.3)
6.5. GEOMETRICAL APPLICATIONS OF ROTORS
113
but need to add conditions to ensure this.
In the first place we are taking it that
hZ10 , Z11 i − hZ13 , Z14 i =k1 [hZ7 , Z8 i − hZ4 , Z5 i],
hZ10 , Z11 i − hZ7 , Z8 i =k2 [hZ13 , Z14 i − hZ4 , Z5 i],
(6.5.4)
assuming that these four differences exist. From this we wish to deduce that
hZ10 , Z11 i−hZ7 , Z8 i−[hZ10 , Z11 i−hZ13 , Z14 i] = hZ13 , Z14 i−hZ4 , Z5 i]−[hZ7 , Z8 i−hZ4 , Z5 i]. (6.5.5)
. For this we need both sides to exist, as if
hZ10 , Z11 i − hZ7 , Z8 i − [hZ10 , Z11 i − hZ13 , Z14 i] = hZ16 , Z17 i,
and
hZ13 , Z14 i − hZ4 , Z5 i] − [hZ7 , Z8 i − hZ4 , Z5 i] = hZ19 , Z20 i,
then by §6.1.8 we have that, for all Z,
δF (Z, Z16 , Z17 )
=δF (Z, Z10 , Z11 ) − δF (Z, Z7 , Z8 ) − [δF (Z, Z10 , Z11 ) − δF (Z, Z13 , Z14 )]
=δF (Z, Z13 , Z14 ) − δF (Z, Z7 , Z8 ),
and
δF (Z, Z19 , Z20 )
=δF (Z, Z13 , Z14 ) − δF (Z, Z4 , Z5 ) − [δF (Z, Z10 , Z11 ) − δF (Z, Z7 , Z8 )]
=δF (Z, Z13 , Z14 ) − δF (Z, Z7 , Z8 ).
It follows that δF (Z, Z16 , Z17 ) = δF (Z, Z19 , Z20 ) for all Z and so hZ16 , Z17 i = hZ19 , Z20 i.
From (6.5.3) it follows that
hZ10 , Z11 i − hZ7 , Z8 i = hZ13 , Z14 i − hZ4 , Z5 i,
(6.5.6)
when the differences on the two sides exist.
In generalization of (6.5.1) and (6.5.2), we say that the rotors
hZ4 , Z5 i, hZ7 , Z8 i, hZ10 , Z11 i, hZ13 , Z14 i
(6.5.7)
form a dual parallelogram when (6.5.3) and (6.5.6) are satisfied and that hZ13 , Z14 i is not in
the rotor pencil based on hZ4 , Z5 i and hZ7 , Z8 i. We call the rotors in (6.5.7) the vertex rotors,
saying that hZ4 , Z5 i and hZ10 , Z11 i are opposite each other, as are hZ7 , Z8 i and hZ13 , Z14 i.
We can construct a skeleton diagram for a dual parallelogram as follows. Suppose that W1 , W6 , W7
and W12 are collinear points and that W12 − W7 = W6 − W1 . Choose W2 and W3 so that
[W1 , W2 , W3 , W6 ] is a parallelogram, from which it follows that hW1 , W3 i − hW1 , W2 i = hW1 , W6 i.
Similarly choose W9 and W8 so that [W7 , W9 , W8 , W12 ] is a parallelogram, from which it follows
that hW7 , W8 i − hW7 , W9 i = hW7 , W12 i. Suppose that W1 W2 and W7 W9 meet at W13 and choose
W14 ∈ W1 W2 and W15 ∈ W7 W9 so that W14 − W13 = W2 − W1 and W15 − W13 = W9 − W7 ; then
we have that
hW1 , W2 i = hW13 , W14 i, hW7 , W9 i = hW13 , W15 i.
We further choose W18 so that [W13 , W14 , W15 , W18 ] is a parallelogram, so that
hW7 , W9 i − hW1 , W2 i = hW13 , W15 i − hW13 , W15 i = hW13 , W18 i.
Finally we suppose thatW1 W3 and W7 W8 meet at W19 and choose W20 ∈ W1 W3 and W21 ∈ W7 W8
so that W20 − W19 = W3 − W1 and W21 − W19 = W8 − W7 ; then we have that
hW1 , W3 i = hW19 , W20 i,
hW7 , W8 i = hW19 , W21 i.
114CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
We further choose W24 so that [W19 , W20 , W21 , W24 ] is a parallelogram, so that
hW7 , W8 i − hW1 , W3 i = hW19 , W21 i − hW19 , W20 i = hW19 , W24 i.
From our earlier analysis it must happen that W19 and W24 must lie on the line W13 W18 and that
W24 − W19 = W18 − W13 .
Then we have a dual parallelogram, with vertex rotors
hW1 , W2 i, hW1 , W3 i, hW7 , W8 i, hW7 , W9 i.
b
W21
b
W20
W24
W6
W3
b
b
b
W19
b
b
W1
b
b
W2
W13
b
W15
b
W14
W8
b
W12
b
W18
b
b
b
W7
W9
Figure 4.39.
6.5.4
Mid-rotors of opposite pairs of vertex rotors
By (6.5.3) we have that
1
2 [hZ4 , Z5 i
+ hZ10 , Z11 i] = 21 [hZ7 , Z8 i + hZ13 , Z14 i],
provided sides exist. This shows that the mid-rotors of opposite pairs of vertex rotors in a dual
parallelogram coincide when they exist.
For the skeleton diagram in §6.5.3 we have that
1
2 [hW1 , W2 i
+ hW7 , W8 i] = 21 [hW1 , W3 i + hW7 , W9 i].
For this, suppose that W1 W2 and W7 W8 meet at W25 and take W26 ∈ W1 W2 and W27 ∈ W7 W8
so that
W26 − W25 = W2 − W1 , W27 − W25 = W8 − W7 ,
and choose V28 to be the mid-point of W26 and W27 . Then
1
2 [hW1 , W2 i
+ hW7 , W8 i] = hW25 , V28 i.
Suppose also that W1 W3 and W7 W9 meet at W31 and take W32 ∈ W1 W3 and W33 ∈ W7 W9 so
that
W32 − W31 = W3 − W1 , W33 − W31 = W9 − W7 ,
115
6.5. GEOMETRICAL APPLICATIONS OF ROTORS
and choose V34 to be the mid-point of W32 and W33 . Then
1
2 [hW1 , W3 i
+ hW7 , W9 i] = hW31 , V34 i.
From the general argument, W25 , V28 , W31 and V28 must be collinear and V28 −W25 = V34 −W31 .
b
W21
b
W20
W24
W6
W3
b
b
b
W19
b
b
W27
V28
W26
b
W1
b
b
b
b
W18
W2 = W25
b
W12
b
b
b
W15
b
W13
W8
b
b
W7
W32
b
W9
Figure 4.40.
b
V34
b
W33
b
W31
6.5.5
Menelaus’ theorem for rotors and pencils
Suppose that we are given three rotors hZ4 , Z5 i, hZ7 , Z8 i, hZ10 , Z11 i, with the carriers not concurrent or all parallel, so that δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] 6= 0. We take rotors as follows in
the three pencils based on these two-by-two,
1
λ
hZ7 , Z8 i +
hZ10 , Z11 i,
1+λ
1+λ
1
µ
hU2 , V2 i =
hZ10 , Z11 i +
hZ4 , Z5 i,
1+µ
1+µ
1
ν
hU3 , V3 i =
hZ4 , Z5 i +
hZ7 , Z8 i.
1+ν
1+ν
hU1 , V1 i =
We then ask when U1 V1 , U2 V2 , U3 V3 are concurrent or all parallel, that is
δF [(U1 , V1 ), (U2 , V2 ), (U3 , V3 )] = 0.
Now
(1 + λ)(1 + µ)(1 + ν)δF [(U1 , V1 ), (U2 , V2 ), (U3 , V3 )] = (λµν + 1)δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )]
and so our condition is λµν = −1.
W14
116CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
Wν
b
Z17
b
Zλ
b
b
b
Z1
Z19
b
Wµ
b
Z16
b
b
Z2
Z20
b
b
b
b
b
Z13
Z3
Zµ
Zν
b
Z14
b
Wλ
Figure 4.41.
To interpret this further we suppose that Z4 , Z5 are on the side-line Z2 Z3 , Z7 , Z8 are on the
side-line Z3 Z1 and Z10 , Z11 are on the side-line Z1 Z2 , respectively, of a triangle. We form a skeleton
diagram as follows. Take
Z13 , Z14 ∈ Z2 Z3 , Z16 , Z17 ∈ Z3 Z1 , Z19 , Z20 ∈ Z1 Z2 ,
so that
Z5 − Z4 =Z13 − Z2 = Z14 − Z3 ,
Z8 − Z7 =Z16 − Z3 = Z17 − Z1 ,
Z11 − Z10 =Z19 − Z1 = Z20 − Z2 .
Then
λ
1
hZ1 , Z17 i +
hZ1 , Z19 i,
1+λ
1+λ
1
λ
=hZ1 ,
Z17 +
Z19 i,
1+λ
1+λ
1
µ
hU2 , V2 i =
hZ2 , Z20 i +
hZ2 , Z13 i,
1+µ
1+µ
µ
1
Z20 +
Z13 i,
=hZ2 ,
1+µ
1+µ
1
ν
hU3 , V3 i =
hZ3 , Z14 i +
hZ3 , Z16 i
1+ν
1+ν
1
ν
=hZ3 ,
Z14 +
Z16 .
1+ν
1+ν
hU1 , V1 i =
We write
Zλ =
1
λ
1
µ
1
ν
Z17 +
Z19 , Zµ =
Z20 +
Z13 , Zν =
Z14 +
Z16 i.
1+λ
1+λ
1+µ
1+µ
1+ν
1+ν
Let us take
Z13 = (1 − r13 )Z2 + r13 Z3 ,
so that then
Z14 = −r13 Z2 + (1 + r13 )Z3 ,
117
6.6. DUAL OF ANGLE PROPERTY OF CIRCLES
and similarly
Z16 =(1 − s16 )Z3 + s16 Z1 , Z17 = −s16 Z3 + (1 + s16 )Z1 ,
Z19 =(1 − t19 )Z1 + t19 Z2 , Z20 = −t19 Z1 + (1 + t19 )Z2 .
Now take Z = (1 − r)Z1 + rZλ and choose r so that Z lies on Z2 Z3 ; in this way we find that
the point Wλ in which Z1 Zλ meets Z2 Z3 is given by
Wλ =
t19 λ
s16
Z2 −
Z3 .
t19 λ − s16
t19 λ − s16
It follows that
Z2 Wλ
s16
=−
.
t19 λ
Wλ Z3
By a similar argument we have that if Z2 Zµ meets Z3 Z1 at Wµ , and Z3 Zν meets Z1 Z2 at Wν ,
then
t19
Z3 Wµ
=−
,
r13 µ
Wµ Z1
Z1 Wν
r13
=−
.
s16 ν
Wν Z2
The product of these sensed-ratios is equal to
−
1
,
λµν
and so the lines Z1 Wλ , Z2 Wµ , Z3 Wν are concurrent or all parallel if and only if this product of
sensed-ratios is equal to 1. Thus Ceva’s theorem for points and lines is a consequence of Menelaus’
theorem for rotors and pencils.
6.6
6.6.1
Dual of angle property of circles
Original property of circle
We recall a very well-known property of circles. Suppose that C = C(O; k) is the circle with centre
the origin and length of radius k. Let Z4 and Z6 be fixed points on the circle and Z a variable
point on it. Supposing that Z4 , Z6 , Z have complex coordinates z4 , z6 , z, respectively, we note that
2
2
k
k
z6 − z z¯6 − z̄
= k2
z4 − z z¯ − kz̄2
4
z¯4 z̄ − z¯6
=
z¯6 z̄ − z¯4
z¯4 z¯6 − z̄
,
=
z¯6 z¯4 − z̄
and hence
z6 − z
z4 − z
2
=
z¯4
z¯6
But if θ is the sensed-angle ∡Z4 ZZ6 we have that
z6 − z 2
z4 − z .
z6 − z z6 − z
,
= cis θ z4 − z
z4 − z 118CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
and so
cis 2θ =
z¯4
= cis ı(θ6 − θ4 ),
z¯6
where θ4 , θ6 are the polar angles of Z4 and Z6 , respectively. It follows that
θ=
(
1
2 (θ6
1
2 (θ6
− θ4 )
− θ4 ) + 180F
if δF (Z4 , Z6 , Z) > 0,
if δF (Z4 , Z6 , Z) < 0.
Thus the variable sensed-angle has constant magnitude on each of two arcs of the circle.
6.6.2
Point of contact of tangents to a circle
For convenience we now take the unit circle with centre the origin and radius length 1, and a point
Z4 exterior to it. We wish to find a formula for the point of contact W4 of a tangent from Z4 to
2
2
the circle. For this we seek w4 ∈ C so that |w4 | = 1 and |z4 − w4 | + 1 = |z4 | . Thus we need
(z4 − w4 )(z¯4 − w¯4 ) + 1 =z4 z¯4 ,
−z4 w¯4 − w4 z¯4 + w4 w¯4 + 1 =0,
−z4 w¯4 − w4 z¯4 + 2 =0,
1
z¯4 w4 + z4
− 2 =0,
w4
z¯4 w42 − 2w4 + z4 =0
and so
√
1 ± i z4 z¯4 − 1
w4 =
.
z¯4
As we shall apply this when |z4 | = k, we note that this becomes
√
1 ± i k2 − 1
z4
w4 =
k 2√
√
x4 − y4 k 2 − 1
x4 k 2 − 1 + y4
=
±i
.
k2
k2
When we take the +i in this , so that
√
1 + ı k2 − 1
z4 ,
w4 =
k2
it will be found that
δF (O, Z4 , W4 ) =
√
k2 − 1
> 0,
2k 2
so that the triple (O, Z4 , W4 ) is anticlockwise in orientation. When we take −i instead, we obtain
the second point of contact, with orientation clockwise this time.
119
6.6. DUAL OF ANGLE PROPERTY OF CIRCLES
6.6.3
Dual of angle property
b
O
W
φ
b
W6
b
b
b
W4
b
Z
θ
b
Z6
Z4
Figure 4.42.
Suppose that we have two concentric circles, which without loss of generality we may take
to be the unit circle C1 and the circle C2 = C(O; k) where k > 1. We take pairs of points
(Z4 , Z5 ), (Z6 , Z7 ), (Z, Z ′ ) all on the outer circle C2 so that the lines Z4 Z5 , Z6 Z7 and ZZ ′ are all
tangent to the inner circle C1 . We also suppose that the triples (O, Z4 , Z5 ), (O, Z6 , Z7 ), (O, Z, Z ′ )
are similarly oriented, all anti-clockwise or all anticlockwise; we suppose that the orientation is
anti-clockwise, as the other case can be treated similarly. If W4 is the point of contact of Z4 Z5
then W4 is the mid-point of Z4 and Z5 and so (O, Z4 , W4 ) is anticlockwise. It follows that
√
1 + i k2 − 1
z4 .
(6.6.1)
w4 =
k2
By a similar argument we have that
√
1 + i k2 − 1
z6 ,
w6 =
k2
√
1 + i k2 − 1
w=
z,
k2
and so
◦
◦
w6 − w
z6 − z
=
.
z4 − z
w4 − w
It follows that |θ| = |φ| , where θ is the sensed-angle θ = ∡Z4 ZZ6 and φ is the sensed-angle
φ = ∡W4 W W6 .
But by §6.6.1 θ has constant magnitude when (Z4 , Z6 , Z) is anti-clockwise and so φ has constant
magnitude in this case. Similarly θ has constant magnitude when (Z4 , Z6 , Z) is clockwise and so
φ has constant magnitude in this case.
By (6.6.1)
p
z = (1 − i k 2 − 1)w,
120CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
and so
w−z =i
6.7
6.7.1
p
k 2 − 1w.
(6.6.2)
Double rotors
Equality of two couples of rotors
In generalization of §5.3.2, given pairs of points (Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 ), (Z13 , Z14 ) we ask
when we have
δF [(U, V ), (Z4 , Z5 ), (Z7 , Z8 )] = δF [(U, V ), (Z10 , Z11 ), (Z13 , Z14 )]
for all (U, V ). We use the notation of §6.3.3 and S 6.3.4 and suppose that Case A or Case B holds
for both pairs of pairs. Then
δF [(U, V ), (Z4 , Z5 ), (Z7 , Z8 )] = 0,
for all (U, V ) such that U V passes through the point of intersection W1 of Z4 Z5 and Z7 Z8 . Hence
δF [(U, V ), (Z10 , Z11 ), (Z13 , Z14 )] = 0,
for all such U and V , so that all such lines U V also pass through the point of intersection W7 of
Z10 Z11 and Z13 Z14 . It follows that we must have W1 = W7 .
Now
δF [(U, V ), (Z4 , Z5 ), (Z7 , Z8 )]
=δF [(U, V ), (W1 , W2 ), (W1 , W3 )]
=δF [(W1 , W2 ), (W1 , W3 ), (U, V )]
= − δF (W2 , W1 , W3 )δF (W1 , U, V ),
and by a similar argument
δF [(U, V ), (Z10 , Z11 ), (Z13 , Z14 )] = −δF (W8 , W1 , W9 )δF (W1 , U, V ).
Thus we must have
W1 = W7 ,
δF (W1 , W2 , W3 ) = δF (W1 , W8 , W9 ).
(6.7.1)
In this case we say that the pairs of rotors (hZ4 , Z5 i, hZ7 , Z8 i) and (hZ10 , Z11 i, hZ13 , Z14 i) are
equivalent. We call the set of pairs of rotors thus equivalent to a pair of rotors (hZ4 , Z5 i, hZ7 , Z8 i)
a double rotor and denote it by hhZ4 , Z5 i, hZ7 , Z8 ii.
6.7.2
Scalar multiplication of a double rotor
With the notation of §6.7.1, we ask when we have
δF [(U, V ), (Z10 , Z11 ), (Z13 , Z14 )] = kδF [(U, V ), (Z4 , Z5 ), (Z7 , Z8 )],
for all (U, V ), for some non-zero real number k. The analysis closely follows that in §6.7.1 and we
find that W1 = W7 and
δF (W1 , W8 , W9 ) = kδF (W1 , W2 , W3 ).
We then write
hhZ10 , Z11 i, hZ13 , Z14 ii = khhZ4 , Z5 i, hZ7 , Z8 ii.
121
6.7. DOUBLE ROTORS
6.7.3
Affine line property for double rotors
With the notation of §6.7.1, we ask if given s ∈ R we have
(1 − s)δF [(U, V ), (Z4 , Z5 ), (Z7 , Z8 )] + sδF [(U, V ), (Z10 , Z11 ), (Z13 , Z14 )]
=δF [(U, V ), (Zs , Us ), (Vs , Ws )],
for some points Zs , Us , Vs , Ws to be found. Because of §6.2.1 this is equivalent to having
δF [(U, V ), (Zs , Us ), (Vs , Ws )]
(6.7.2)
=(1 − s)δF [(U, V ), (W1 , W2 ), (W1 , W3 )] + sδF [(U, V ), (W7 , W8 ), (W7 , W9 )]
=(1 − s)δF [(W1 , W2 ), (W1 , W3 ), (U, V )] + sδF [(W7 , W8 ), (W7 , W9 ), (U, V )]
=(1 − s)δF (W1 , W2 , W3 )δF (W1 , U, V ) + sδF (W7 , W8 , W9 )δF (W7 , U, V ).
(6.7.3)
We now divide into two cases, first of all taking W7 = W1 . Then we need
[(1 − s)δF (W1 , W2 , W3 ) + sδF (W1 , W8 , W9 )]δF (W1 , U, V )
=δF [(U, V ), (Zs , Us ), (Vs , Ws )]
and as for all points U, V collinear with W1 the left-hand side must be equal to 0, on the righthand side the point of intersection of Zs Us and Vs Ws must be W1 . Accordingly we may take
Zs = Vs = W1 and so as
δF [(W1 , Us ), (W1 , Ws ), (U, V )] = δF (W1 , Us , Ws )δF (W1 , U, V ),
we need
(1 − s)δF (W1 , W2 , W3 ) + sδF (W1 , W8 , W9 ) = δF (W1 , Us , Ws ).
W3
b
l
W1
b
b
Figure 4.43
W4
b
W2
b
W5
Now taking any convenient line l, we suppose that the line through W3 which is parallel to
W1 W2 meets l at W4 , so that δF (W1 , W2 , W3 ) = δF (W1 , W2 , W4 ). We next suppose that the line
through W2 which is parallel to W1 W4 meets l at W5 , so that δF (W4 , W1 , W2 ) = δF (W4 , W1 , W5 ).
It follows that δF (W1 , W2 , W3 ) = δF (W1 , W5 , W4 ). We similarly take W10 , W11 ∈ l so that
δF (W1 , W8 , W9 ) = δF (W1 , W11 , W10 ). Then we need
(1 − s)δF (W1 , W5 , W4 ) + sδF (W1 , W11 , W10 ) = δF (W1 , Us , Ws ),
and so need
(1 − s)hW5 , W4 i + shW11 , W10 i = hUs , Vs i.
(6.7.4)
For our second case we suppose that W1 6= W7 . By (6.7.3) when U, V are points on the line
W1 W7 we must have
δF [(U, V ), (Zs , Us ), (Vs , Ws )] = 0.
122CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
Thus U V, Zs Us , Vs Ws must be concurrent or all parallel, so in the case of concurrency by §6.2.1
we may take Zs = Vs ∈ W1 W7 , and have
δF [(Zs , Us ), (Zs , Ws ), (U, V )] = δF (Zs , Us , Ws )δF (Zs , U, V ).
If we take Zs = (1 − t)W1 + tW7 , in (4.10.3) and (4.10.4) we need for all U and V
δF (Zs , Us , Ws )[(1 − t)δF (W1 , U, V ) + tδF (W7 , U, V )]
=(1 − s)δF (W1 , W2 , W3 )δF (W1 , U, V ) + sδF (W7 , W8 , W9 )δF (W7 , U, V ),
and so
(1 − t)δF (Zs , Us , Ws ) =(1 − s)δF (W1 , W2 , W3 ),
tδF (Zs , Us , Ws ) =sδF (W7 , W8 , W9 ).
(6.7.5)
By division we need
s δF (W7 , W8 , W9 )
t
=
.
1−t
1 − s δF (W1 , W2 , W3 )
This determines Zs uniquely on W1 W7 .
W3
b
l
W1
b
b
W4
b
Zs
b
W2
W5
b
W9
b
W7
b
W10
b
W11
b
Figure 4.44
b
W8
By addition in (6.7.5) we also need
δF (Zs , Us , Ws ) = (1 − s)δF (W1 , W2 , W3 ) + sδF (W7 , W8 , W9 ).
We now take our line l parallel to W1 W7 and on it take W4 , W5 , W10 , W11 as above so that
δF (W1 , W2 , W3 ) = δF (W1 , W5 , W4 ),
δF (W1 , W8 , W9 ) = δF (W1 , W11 , W10 ).
We now need
δF (Zs , Us , Ws ) =(1 − s)δF (W1 , W5 , W4 ) + sδF (W7 , W11 , W10 )
=(1 − s)δF (Zs , W5 , W4 ) + sδF (Zs , W11 , W10 ),
123
6.7. DOUBLE ROTORS
and this is ensured if we take
hUs , Ws i = (1 − s)hW5 , W4 i + shW11 , W10 i.
We denote the double rotor hhZs , Us i, hVs , Ws ii in (6.7.2) by
(1 − s)hhZ4 , Z5 i, hZ7 , Z8 ii + shhZ10 , Z11 )i, hZ13 , Z14 ii.
By a combination of §6.7.2 and this section, we can define the sum of two double rotors as
hhZ4 , Z5 i, hZ7 , Z8 ii + hhZ10 , Z11 i, hZ13 , Z14 ii
=2 12 hhZ4 , Z5 i, hZ7 , Z8 ii + 21 hhZ10 , Z11 i, hZ13 , Z14 ii ,
and the difference of double rotors by
hhZ10 , Z11 i, hZ13 , Z14 ii − hhZ4 , Z5 i, hZ7 , Z8 ii
=hhZ10 , Z11 i, hZ13 , Z14 ii + hhZ7 , Z8 i, hZ4 , Z5 ii.
6.7.4
Affine line property for doubly dual sensed-area
By §1.2.3 we can deduce a property for doubly dual sensed area similar to that in §6.7.3, that with
the same notation
δF (Z16 , Z17 ), (Z19 , Z20 ) , (Z22 , Z23 ), (Z25 , Z26 ) , (Zs , Us ), (Vs , Ws )
=δF [(Z16 , Z17 ), (Z22 , Z23 ), (Z25 , Z26 )]δF [(Z19 , Z20 ), (Zs , Us ), (Vs , Ws )]
6.7.5
−δF [(Z19 , Z20 ), (Z22 , Z23 ), (Z25 , Z26 )]δF [(Z16 , Z17 ), (Zs , Us ), (Vs , Ws )]
=δF [(Z16 , Z17 ), (Z22 , Z23 ), (Z25 , Z26 )] (1 − s)δF [(Z19 , Z20 ), (Z4 , Z5 ), (Z7 , Z8 )]
+sδF [(Z19 , Z20 ), (Z10 , Z11 ), (Z13 , Z14 )]
−δF [(Z19 , Z20 ), (Z22 , Z23 ), (Z25 , Z26 )] (1 − s)δF [(Z16 , Z17 ), (Z4 , Z5 ), (Z7 , Z8 )]
+sδF [(Z16 , Z17 ), (Z10 , Z11 ), (Z13 , Z14 )]
=(1 − s) δF [(Z16 , Z17 ), (Z22 , Z23 ), (Z25 , Z26 )]δF [(Z19 , Z20 ), (Z4 , Z5 ), (Z7 , Z8 )]
−δF [(Z19 , Z20 ), (Z22 , Z23 ), (Z25 , Z26 )]δF [(Z16 , Z17 ), (Z4 , Z5 ), (Z7 , Z8 )]
+s δF [(Z16 , Z17 ), (Z22 , Z23 ), (Z25 , Z26 )]δF [(Z19 , Z20 ), (Z10 , Z11 ), (Z13 , Z14 )]
−δF [(Z19 , Z20 ), (Z22 , Z23 ), (Z25 , Z26 )]δF [(Z16 , Z17 ), (Z10 , Z11 ), (Z13 , Z14 )]
=(1 − s)δF (Z16 , Z17 ), (Z19 , Z20 ) , (Z22 , Z23 ), (Z25 , Z26 ) , (Z4 , Z5 ), (Z7 , Z8 )
+sδF (Z16 , Z17 ), (Z19 , Z20 ) , (Z22 , Z23 ), (Z25 , Z26 ) , (Z10 , Z11 ), (Z13 , Z14 ) .
Menelaus’ theorem for double rotors and ranges
Given double rotors
hhZ4 , Z5 i, hZ7 , Z8 ii, hhZ10 , Z11 i, hZ13 , Z14 ii, hhZ16 , Z17 i, hZ19 , Z20 ii,
such that
δF
let us take
(Z4 , Z5 ), (Z7 , Z8 ) , (Z10 , Z11 ), (Z13 , Z14 ) , (Z16 , Z17 ), (Z19 , Z20 ) 6= 0,
λ
1
hhZ10 , Z11 i, hZ13 , Z14 ii +
hhZ16 , Z17 i, hZ19 , Z20 ii,
1+λ
1+λ
1
µ
hhH2 , U2 i, hV2 , W2 ii =
hhZ16 , Z17 i, hZ19 , Z20 ii +
hhZ4 , Z5 i, hZ7 , Z8 ii,
1+µ
1+µ
1
ν
hhH3 , U3 i, hV3 , W3 ii =
hhZ4 , Z5 i, hZ7 , Z8 ii +
hhZ10 , Z11 i, hZ13 , Z14 ii,
1+ν
1+ν
hhH1 , U1 i, hV1 , W1 ii =
124CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
Then by §6.7.4
δF
(H1 , U1 ), (V1 , W1 ) , (H2 , U2 ), (V2 , W2 ) , (H3 , U3 ), (V3 , W3 ) = 0,
if and only if λµν = −1. When H1 U1 , V1 W1 meet in a unique point K1 , H2 U2 , V2 W2 meet in a
unique point K2 , and H3 U3 , V3 W3 meet in a unique point K3 , this is the condition that K1 , K2 , K3
be collinear.
6.7.6
Ceva’s theorem for rotors and pencils
As we saw in §6.5.5 that Ceva’s theorem for points and lines is a consequence of Menelaus’ theorem
for rotors and pencils, then there should be a Ceva’s theorem for rotors and pencils, and it should
be a consequence of Menelaus’ theorem for double-rotors and ranges. The latter is a complicated
route to it however and we give a more direct derivation as follows.
Suppose that hZ4 , Z5 i, hZ7 , Z8 i, hZ10 , Z11 i are rotors such that
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] 6= 0.
Suppose that
λ
1
hZ7 , Z8 i +
hZ10 , Z11 i,
1+λ
1+λ
1
µ
hZ16 , Z17 i =
hZ10 , Z11 i +
hZ4 , Z5 i,
1+µ
1+µ
1
ν
hZ19 , Z20 i =
hZ4 , Z5 i +
hZ7 , Z8 i.
1+ν
1+ν
hZ13 , Z14 i =
Then we have that
δF (Z4 , Z5 ), (Z13 , Z14 ) , (Z7 , Z8 ), (Z16 , Z17 ) , (Z10 , Z11 ), (Z19 , Z20 )
=δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z16 , Z17 )]δF [(Z13 , Z14 ), (Z10 , Z11 ), (Z19 , Z20 )]
−δF [(Z13 , Z14 ), (Z7 , Z8 ), (Z16 , Z17 )]δF [(Z4 , Z5 ), (Z10 , Z11 ), (Z19 , Z20 )]
1
1
=
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )]
δF [(Z7 , Z8 ), (Z10 , Z11 ), (Z19 , Z20 )]
1+µ
1+λ
λ
ν
−
δF [(Z10 , Z11 ), (Z7 , Z8 ), (Z16 , Z17 )]
δF [(Z4 , Z5 ), Z10 Z11 , (Z7 , Z8 )]
1+λ
1+ν
1
1
1
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )]
δF [(Z7 , Z8 ), (Z10 , Z11 ), (Z4 , Z5 )]
=
1+µ
1+λ1+ν
µ
λ
ν
−
δF [(Z10 , Z11 ), (Z7 , Z8 ), (Z4 , Z5 )]
δF [(Z4 , Z5 ), Z10 Z11 , (Z7 , Z8 )]
1+λ1+µ
1+ν
1 − λµν
=
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )]2 .
(1 + λ)(1 + µ)(1 + ν)
This then is equal to 0 if and only if λµν = 1.
We enclose a skeleton diagram covering the case of a dual centroid, when λµν = 1, having
1
λ
W2′ +
W3 i,
1+λ
1+λ
1
µ
hZ16 , Z17 i =hZ2 ,
W3′ +
W1 i,
1+µ
1+µ
1
ν
hZ19 , Z20 i =hZ3 ,
W′ +
W2 i.
1+ν 1 1+ν
hZ13 , Z14 i =hZ1 ,
125
6.8. MISCELLANEOUS TOPICS
b
b
W2′
b
b
b
Z1
b
W3
b
W2
Z2
W3′
b
b
b
b
b
b
Z3
b
b
W1′
W1
Figure 4.45.
6.8
Miscellaneous topics
6.8.1
Dual of transversals of a pencil
As in §4.2.4, consider
E=
δF [(U, V ), (Z4 , Z5 ), (Z8 , Z9 )]δF [(U, V ), (Z6 , Z7 ), (Z10 , Z11 )]
,
δF [(U, V ), (Z4 , Z5 ), (Z10 , Z11 )]δF [(U, V ), (Z6 , Z7 ), (Z8 , Z9 )]
when the lines Z4 Z5 , Z6 Z7 , Z8 Z9 , Z10 Z11 are concurrent in some point W0 . We take points
Z12 , Z13 , Z14 , Z15 so that
Z5 − Z4 = Z12 − W0 , Z7 − Z6 = Z13 − W0 , Z9 − Z8 = Z14 − W0 , Z11 − Z10 = Z15 − W0 ,
and let the line Z12 Z13 meet W0 Z14 at Z16 , and meet W0 Z15 at Z17 , so that
Z14 − W0 = k1 (Z16 − W0 ), Z15 − W0 = k1 (Z17 − W0 ),
for some non-zero numbers k1 and k2 . Suppose that
Z16 =
1
λ
1
µ
Z12 +
Z13 , Z17 =
Z12 +
Z13 .
1+λ
1+λ
1+µ
1+µ
Then
δF [(U, V ), (Z4 , Z5 ), (Z8 , Z9 )] = δF [(U, V ), (W0 , Z12 ), (W0 , Z14 )]
= δF (U, W0 , Z12 )δF (V, W0 , Z14 ) − δF (V, W0 , Z12 )δF (U, W0 , Z14 )
= δF (U, W0 , Z12 )k1 δF (V, W0 , Z16 ) − δF (V, W0 , Z12 )k1 δF (U, W0 , Z16 )
k1
=
δF (U, W0 , Z12 )[δF (V, W0 , Z12 ) + λδF (V, W0 , Z13 )
1+λ
− δF (V, W0 , Z12 )[δF (U, W0 , Z12 ) + λδF (U, W0 , Z13 )
λ
= k1
[δF (U, W0 , Z12 )δF (V, W0 , Z13 ) − δF (V, W0 , Z12 )δF (U, W0 , Z13 )]
1+λ
λ
δF [(U, V ), (W0 , Z12 ), (W0 , Z13 )].
= k1
1+λ
126CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
By a similar argument, we have that
δF [(U, V ), (Z6 , Z7 ), (Z10 , Z11 )] =
δF [(U, V ), (Z4 , Z5 ), (Z10 , Z11 )] =
δF [(U, V ), (Z6 , Z7 ), (Z8 , Z9 )] =
1
δF [(U, V ), (W0 , Z13 ), (W0 , Z12 )],
1+µ
µ
δF [(U, V ), (W0 , Z12 ), (W0 , Z13 )],
k2
1+µ
1
k1
δF [(U, V ), (W0 , Z13 ), (W0 , Z12 )].
1+λ
k2
On inserting these, we find that
E=
6.8.2
λ
= crW0 (Z12 , Z13 , Z14 , Z15 ).
µ
Quadrilaterals
As a development of §2.2.1 we consider again a quadruple
(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 ), (Z7 , Z8 ),
of distinct pairs of distinct points. With these, we suppose first that in the rotor pencil on
(hZ3 , Z4 i, hZ5 , Z6 i) there is a rotor
hD1 , D2 i = (1 − r)hZ3 , Z4 i + rhZ5 , Z6 i,
such that D1 D2 , Z1 Z2 , Z7 Z7 are concurrent or all parallel. Then we have that
r=
1
1
δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z3 , Z4 )], 1 − r = − δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z5 , Z6 )],
J1
J1
where
J1 = δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z9 , Z10 )],
and
hZ9 , Z10 i = hZ3 , Z4 i − hZ5 , Z6 i.
We suppose secondly that there is a rotor
hD3 , D4 i = (1 − s)hZ5 , Z6 i + shZ1 , Z2 i,
in the rotor pencil on (hZ5 , Z6 i, hZ1 , Z2 i) such that the lines D3 , D4 , Z3 Z4 , Z7 , Z8 are concurrent
or all parallel. Then we have that
s=
1
1
δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z5 , Z6 )], 1 − s = − δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z1 , Z2 )],
J2
J2
where
J2 = δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z11 , Z12 )]
and
hZ11 , Z12 i = hZ5 , Z6 i − hZ1 , Z2 i.
We suppose thirdly that there is a rotor
hD5 , D6 i = (1 − t)hZ1 , Z2 i + thZ3 , Z4 i,
in the rotor pencil on (hZ1 , Z2 i, hZ3 , Z4 i) so that the lines D5 D6 , Z5 Z6 , Z7 Z8 are concurrent or all
parallel. Then
t=
1
1
δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z1 , Z2 )], 1 − t = − δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z3 , Z4 )],
J3
J3
127
6.9. IDENTITIES
where
J3 = δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z13 , Z14 )]
and
hZ13 , Z14 i = hZ1 , Z2 i − hZ3 , Z4 i.
b
Z8
D4
b
b
Z7
b
Z6
b
Z5
D3
b
b
b
b
b
b
b
b
b
Z1
Z2
D5
b
b
D2
D1
b
b
b
b
Z3
Z4
Figure 4.46.
We can then have a result and diagram as in §2.2.1, without necessarily having the restrictions incurred by stipulating amenable pairs of points. We retain the terminology that the lines
Z1 Z2 , Z3 Z4 , Z5 Z6 , Z7 Z8 form a quadrilateral and we refer to the lines D1 D2 , D3 D4 , D5 D6 as
its diagonal lines.
The result of §2.2.2 also generalizes.
6.8.3
Sensed area associated with a quadruple of points
We recall from §1.5 the concept of sensed area δF (Z1 , Z2 , Z3 , Z4 ) associated with an ordered quadruple (Z1 , Z2 , Z3 , Z4 ) of points. We note that
δF (Z1 , Z2 , Z3 , Z) =δF (Z2 , Z3 , Z, Z1 )
=δF (Z2 , Z3 , Z) + δF (Z2 , Z, Z1 )
=δF (Z2 , Z3 , Z) − δF (Z2 , Z1 , Z)
=δF (Z2 , Z5 , Z),
for all Z, where hZ2 , Z5 i = hZ2 , Z3 i − hZ2 , Z1 i.
Note that this notation can be used in (1.1.8) and so for the denominator in (1.1.15) and so
subsequently.
6.9
Identities
6.9.1
The condition in Menelaus’ theorem can be written as
δF (Z7 , Z2 , Z4 ) δF (Z8 , Z3 , Z5 ) δF (Z9 , Z1 , Z6 )
= −1,
δF (Z7 , Z4 , Z3 ) δF (Z8 , Z5 , Z1 ) δF (Z9 , Z6 , Z2 )
b
b
D6
128CHAPTER 6. GRASSMANN PROPERTIES OF ROTORS; GEOMETRICAL APPLICATIONS
where Z7 is any point not on Z2 Z3 , Z8 6∈ Z3 Z1 and Z9 6∈ Z1 Z2 . We just made the selection
Z7 = Z1 , Z8 = Z2 , Z9 = Z3 to keep down the complication. Another selection that we could make
is to take Z7 = Z8 = Z9 = Z and then we have the conclusion that
δF (Z, Z2 , Z4 ) δF (Z, Z3 , Z5 ) δF (Z, Z1 , Z6 )
= −1,
δF (Z, Z4 , Z3 ) δF (Z, Z5 , Z1 ) δF (Z, Z6 , Z2 )
(6.9.1)
for all points Z, and we could make a similar identity out of any of the results on sensed ratios in
the exercises in Chapter 3. We had a similar identity in (5.3.2) with
δF (Z, Z8 , Z9 ) = δF (Z, Z10 , Z11 )
(6.9.2)
for all Z when hZ8 , Z9 i and hZ10 , Z11 i are equal rotors.
These are different types, as in §(6.8.1)?? we have an equipoised quotient while in §(6.8.2)?? we
do not. In the first type we can work backwards to a result on sensed ratios if we have information
that certain triples of points are collinear, but in the second type we can do substantially more
and in view of the progress made with rotors it would seem worth exploring.
For example suppose that we take collinear points Z1 , Z2 , Z3 , Z4 and collinear points Z5 , Z6 , Z7 , Z8
such that
Z1 Z2
Z1 56
=k
.
Z3 Z4
Z7 Z8
Then we have for all Z
δF (Z, Z1 , Z2 )δF (Z, Z7 , Z8 )
= k.
δF (Z, Z3 , Z4 )δF (Z, Z5 , Z6 )
If we now start with this latter, on first putting Z = Z1 we deduce that Z1 ∈ Z3 Z4 or Z1 ∈ Z5 Z6 ;
similarly for Z2 , Z7 and Z8 . On putting Z = Z3 we have that Z3 ∈ Z1 Z2 or Z3 ∈ Z7 Z8 ; similarly
for Z4 , Z5 and Z6 . From these we can work out the various possible configurations.
As another example of the latter type, we note that we have
δF (Z, Z3 , Z4 ) = k1 [δF (Z, Z1 , Z2 ) + k2 ] ,
for all Z if and only if Z1 Z2 k Z3 Z4 and Z4 − Z3 = k1 (Z2 − Z1 ).
Exercises
4.1 In 13.1(a)??? we introduced areal coordinates, essentially from the identity
δF (Z1 , Z2 , Z3 )δF (Z, Z4 , Z5 ) = δF (Z1 , Z4 , Z5 )δF (Z, Z2 , Z3 )
+ δF (Z2 , Z4 , Z5 )δF (Z, Z3 , Z1 ) + δF (Z3 , Z4 , Z5 )δF (Z, Z1 , Z2 ).
A dual of this would be the identity
δF [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )]δF [(U, V ), (Z7 , Z8 ), (Z9 , Z10 )]
= δF [(Z1 , Z2 ), (Z7 , Z8 ), (Z9 , Z10 )]δF [(U, V ), (Z3 , Z4 ), (Z5 , Z6 )]
+ δF [(Z3 , Z4 ), (Z7 , Z8 ), (Z9 , Z10 )]δF [(U, V ), (Z5 , Z6 ), (Z1 , Z2 )]
+ δF [(Z5 , Z6 ), (Z7 , Z8 ), (Z9 , Z10 )]δF [(U, V ), (Z1 , Z2 ), (Z3 , Z4 )].
This is indeed correct and can be verified by computer software.
Chapter 7
Conic sections
In this chapter we introduce conic sections. We start with them in considerable generality and
gradually specialize to more specific and differentiated properties.
7.1
Definition of a conic section
7.1.1
Preparatory examples
Example 1
With a change of notation from §2.1.2 we consider two concurrent pencils of lines parametrized
by λ and µ, respectively, of the form
1
λ
1
µ
(Z1 , Z2 ) +
(Z1 , Z3 ),
(Z4 , Z5 ) +
(Z4 , Z6 ),
1+λ
1+λ
1+µ
1+µ
so that each is of the first type in Fig.
Z7 Z
2.1, and 1also a line
8 . Then by §1.1.3, a point Z on
µ
λ
1
Z2 + 1+λ
Z3 and Z4 1+µ
Z5 + 1+µ
Z6 is such that Z1 Z, Z4 Z, Z7 Z8 are
both of the lines Z1 1+λ
concurrent or Z1 Z = Z4 Z, only if
1
λ
1
µ
δF (Z1 ,
Z2 +
Z3 ), (Z4 ,
Z5 +
Z6 ), (Z7 , Z8 ) = 0.
1+λ
1+λ
1+µ
1+µ
This expands to
λµδF [(Z1 , Z3 ), (Z4 , Z6 ), (Z7 , Z8 )] + λδF [(Z1 , Z3 ), (Z4 , Z5 ), (Z7 , Z8 )]
+µδF [(Z1 , Z2 ), (Z4 , Z6 ), (Z7 , Z8 )] + δF [(Z1 , Z2 ), (Z4 , Z5 ), (Z7 , Z8 )] = 0,
which is a relationship of the form
a1 λµ + b1 λ + c1 µ + d1 = 0.
Now
δF Z 1 ,
and on expanding this we obtain
λ
1
Z2 +
Z3 , Z = 0,
1+λ
1+λ
λ=
Similarly
δF Z 4 ,
δF (Z, Z1 , Z2 )
.
δF (Z, Z3 , Z1 )
µ
1
Z5 +
Z6 , Z = 0,
1+µ
1+µ
129
130
CHAPTER 7. CONIC SECTIONS
and on expanding this we obtain
µ=
δF (Z, Z4 , Z5 )
.
δF (Z, Z6 , Z4 )
On inserting these and clearing of fractions we obtain the equation for Z
δF (Z, Z1 , Z2 )δF (Z, Z4 , Z5 )δF [(Z1 , Z3 ), (Z4 , Z6 ), (Z7 , Z8 )]
+δF (Z, Z1 , Z2 )δF (Z, Z6 , Z4 )δF [(Z1 , Z3 ), (Z4 , Z5 ), (Z7 , Z8 )]
+δF (Z, Z3 , Z1 )δF (Z, Z4 , Z5 )δF [(Z1 , Z2 ), (Z4 , Z6 ), (Z7 , Z8 )]
+δF (Z, Z3 , Z1 )δF (Z, Z6 , Z4 )δF [(Z1 , Z2 ), (Z4 , Z5 ), (Z7 , Z8 )] = 0.
The first two lines in the equation for Z can be manipulated as follows
Z(∼,1,2) Z(∼,4,5) Z[(1,3),(4,6),(7,8)] + Z(∼,1,2) Z(∼,6,4) Z[(1,3),(4,5),(7,8)]
=Z(∼,1,2) Z(∼,4,5) [Z(1,4,6) Z(3,7,8) − Z(3,4,6) Z(1,7,8) ]
+ Z(∼,1,2) Z(∼,6,4) [Z(1,4,5) Z(3,7,8) − Z(3,4,5) Z(1,7,8) ]
=Z(∼,1,2) Z(3,7,8) [Z(∼,4,5) Z(1,4,6) + Z(∼,6,4) Z(1,4,5) ]
− Z(∼,1,2) Z(1,7,8) [Z(∼,4,5) Z(3,4,6) + Z(∼,6,4) Z(3,4,5) ]
=Z(∼,1,2) Z(3,7,8) Z[(1,∼),(4,5),(6,4)] − Z(∼,1,2) Z(1,7,8) Z[(∼,3),(6,4),(4,5)]
=Z(∼,1,2) Z(3,7,8) Z[(6,4),(1,∼),(4,5)] − Z(∼,1,2) Z(1,7,8) Z[(4,5),(∼,3),(6,4)]
=Z(∼,1,2) Z(3,7,8) [−Z(4,1,∼) Z(6,4,5) ] − Z(∼,1,2) Z(1,7,8) [Z(4,∼,3) Z(5,6,4) ]
=Z(∼,1,2) Z(4,5,6) [Z(3,7,8) Z(∼,1,4) − Z(1,7,8) Z(∼,3,4) ].
Similarly the third and fourth lines in the equation for Z can be manipulated as
Z(∼,3,1) Z(∼,4,5) Z[(1,2),(4,6),(7,8)] + Z(∼,3,1) Z(∼,6,4) Z[(1,2),(4,5),(7,8)]
=Z(∼,3,1) Z(∼,4,5) [Z(1,4,6) Z(2,7,8) − Z(2,4,6) Z(1,7,8) ]
+ Z(∼,3,1) Z(∼,6,4) [Z(1,4,5) Z(2,7,8) − Z(2,4,5) Z(1,7,8) ]
=Z(∼,3,1) Z(2,7,8) [Z(∼,4,5) Z(1,4,6) + Z(∼,6,4) Z(1,4,5) ]
− Z(∼,3,1) Z(1,7,8) [Z(∼,4,5) Z(2,4,6) + Z(∼,6,4) Z(2,4,5) ]
=Z(∼,3,1) Z(2,7,8) Z[(∼,1),(4,5),(4,6)] + Z(∼,3,1) Z(1,7,8) Z[(∼,2),(4,5),(6,4)]
=Z(∼,3,1) Z(2,7,8) Z[(4,5),(∼,1),(6,4)] + Z(∼,3,1) Z(1,7,8) Z[(6,4),(∼,2),(4,5)]
=Z(∼,3,1) Z(2,7,8) [Z(4,∼),1 Z(5,6,4) ] + Z(∼,3,1) Z(1,7,8) [−Z(6,4,5) Z(4,∼,2) ]
=Z(∼,3,1) Z(4,5,6) [Z(2,7,8) Z(4,∼,1) − Z(1,7,8) Z(4,∼,2) ].
On adding together the two expressions thus obtained we get
Z(4,5,6) {Z(∼,1,4) [Z(∼,1,2) Z(3,7,8) + Z(∼,3,1) Z(2,7,8) ] + Z(1,7,8) [−Z(∼,1,2) Z(∼,3,4) − Z(∼,3,1) Z(∼,2,4) ]}
=Z(4,5,6) {Z(∼,1,4) Z[(3,2),(7,8),(∼,1)] + Z(1,7,8) [−Z(2,∼,1) Z(3,4,∼) + Z(3,∼,1) Z(2,4,∼) ]}
=Z(4,5,6) {Z(∼,1,4) Z[(3,2),(7,8),(∼,1)] + Z(1,7,8) Z[(3,2),(∼,1),(4,∼)]}
=Z(4,5,6) {Z(∼,1,4) Z[(∼,1),(3,2),(7,8)] + Z(1,7,8) Z[(4,∼),(3,2),(∼,1)]}
=Z(4,5,6) {Z(∼,1,4) Z[(∼,1),(3,2),(7,8)] − Z(1,7,8) Z(∼,3,2) Z(4,∼,1) }
=Z(4,5,6) Z(∼,1,4) Z(1,2,3) Z(∼,7,8)
=δF (Z1 , Z2 , Z3 )δF (Z4 , Z5 , Z6 )δF (Z, Z1 , Z4 )δF (Z, Z7 , Z8 ).
It follows that the locus of Z is the pair of lines Z1 Z4 and Z7 Z8 .
Example 2
131
7.1. DEFINITION OF A CONIC SECTION
b
Z2
Z
b
Wλ
Wµ
Z3
b
b
Z1
b
Z4
Figure 7.1.
As a second, more important example, if Z1 , Z2 , Z3 , Z4 are points of a circle, as Z varies on
1
λ
this circle let Z1 Z meet Z2 Z3 at the point Wλ = 1+λ
Z2 + 1+λ
Z3 , and let Z4 Z meet Z2 Z3 at the
µ
1
point Wµ = 1+µ Z2 + 1+µ Z3 . Then we have a pencil of lines Z1 Wλ through Z1 and a pencil of lines
Z4 Wµ through Z4 , and the locus of the point of intersection Z of Z1 Wλ and Z4 Wµ is the circle.
Using complex coordinates, we have
wλ =
1
λ
z2 +
z3 = z1 + r(z − z1 ),
1+λ
1+λ
for some real number r, so that
r(z − z1 ) =
λ
1
(z2 − z1 ) +
(z3 − z1 ).
1+λ
1+λ
On taking complex conjugates we also have
r(z̄ − z̄1 ) =
1
λ
(z̄2 − z̄1 ) +
(z̄3 − z̄1 ).
1+λ
1+λ
By division we can eliminate r and obtain
z − z1
z2 − z1 + λ(z3 − z1 )
=
.
z̄ − z̄1
z̄2 − z̄1 + λ(z̄3 − z̄1 )
From this we have that
λ=−
(z − z1 )(z̄2 − z̄1 ) − (z̄ − z̄1 )(z2 − z1 )
.
(z − z1 )(z̄3 − z̄1 ) − (z̄ − z̄1 )(z3 − z1 )
By a similar argument we obtain
µ=−
(z − z4 )(z̄2 − z̄4 ) − (z̄ − z̄4 )(z2 − z4 )
.
(z − z4 )(z̄3 − z̄4 ) − (z̄ − z̄4 )(z3 − z4 )
We now suppose that the circle has centre Z0 and radius length k. Then (z1 − z0 )(z̄1 − z̄0 ) = k 2
so that
k2
,
z̄1 − z̄0 =
z1 − z0
132
CHAPTER 7. CONIC SECTIONS
and similarly for the other points on the circle. On using these we obtain
(z − z1 )[z̄2 − z̄0 − (z̄1 − z̄0 )] − [z̄ − z̄0 − (z̄1 − z̄0 )](z2 − z1 )
(z − z1 )[z̄3 − z̄0 − [z̄1 − z̄0 )] − [z̄ − z̄0 − (z̄1 − z̄0 ](z3 − z1 )
2
2
2
k
k2
(z2 − z1 )
−
(z − z1 ) z2k−z0 − z1k−z0 − z−z
z
−z
0
1
0
=−
2
2
2
k2
− z1k−z0 (z3 − z1 )
(z − z1 ) z3k−z0 − z1k−z0 − z−z
0
λ=−
=−
1
(z1 −z0 )(z2 −z0 ) (z
1
(z3 −z0 )(z1 −z0 ) (z
=− =−
1
z2 −z0
−
1
z−z0
1
z3 −z0
−
1
z−z0
− z1 )(z1 − z2 ) −
− z1 )(z1 − z3 ) −
1
(z−z0 )(z1 −z0 ) (z1
1
(z−z0 )(z1 −z0 ) (z1
− z)(z2 − z1 )
− z)(z3 − z1 )
(z1 − z2 )
(z1 − z3 )
(z − z2 )(z3 − z0 )(z1 − z2 )
.
(z − z3 )(z2 − z0 )(z1 − z3 )
By a similar argument
µ=−
It follows that
(z − z2 )(z3 − z0 )(z4 − z2 )
.
(z − z3 )(z2 − z0 )(z4 − z3 )
(z1 − z2 )(z4 − z3 )
λ
=
= cr(z1 , z4 , z2 , z3 ).
µ
(z1 − z3 )(z4 − z2 )
Thus Z1 Wλ and Z4 Wµ are related lines in this pencil projectivity. This proves incidentally that
the value of this cross-ratio is real, which was also shown in Barry [2][p. 154].
7.1.2
Steiner’s definition of a conic
The person credited with first studying conics is Menaechmus, c.350 B.C., a pupil of Plato and
Eudoxus. Later relevant Greek authors were Aristaeus, Euclid, Archimedes and pre-eminently
Appolonius whose Conics rivalled Euclid’s The Elements. There have been many characterizations
of conics and we start with a modern definition on account of its flexibility and generality.
Following Steiner in c.1832, we generalize from the material in §7.1.1 and given any two pencils
in a projectivity we call the locus of the point of intersection Z of corresponding lines in the two
λ
1
Z2 + 1+λ
Z3 and as λ
pencils a conic or a conic section. We consider the line joining Z1 to 1+λ
µ
1
varies, this gives a pencil of lines through Z1 ; similarly the line joining Z4 to 1+µ Z5 + 1+µ
Z6 and
as µ varies, this gives a pencil of lines through Z4 . For the point of intersection Z of these two
lines we have
λ
1
Z2 +
Z3 = 0,
δF Z, Z1 ,
1+λ
1+λ
and so
λ=
δF (Z, Z1 , Z2 )
γ
= ,
δF (Z, Z3 , Z1 )
β
δF Z, Z4 ,
1
µ
Z5 +
Z6
1+µ
1+µ
µ=
γ′
δF (Z, Z4 , Z5 )
= ′,
δF (Z, Z6 , Z4 )
β
while similarly from
we have
= 0,
where the plain and primed areal coordinates refer to (Z1 , Z2 , Z3 ), (Z4 , Z5 , Z6 ), respectively. If
these are related by
a1 λµ + b1 λ + c1 µ + d1 = 0,
133
7.1. DEFINITION OF A CONIC SECTION
where the case a1 = b1 = c1 = d1 = 0 is excluded, we then have
a1
γ γ′
γ
γ′
+ b1 + c1 ′ + d1 = 0.
′
β β
β
β
Z
b
Z6
b
b
Z4
b
b
Z3
Z5
b
Z1
b
Z2
Figure 7.2.
b
In this we shall always multiply across by ββ ′ and take the locus as the set of Z satisfying the
equation
a1 γγ ′ + b1 γβ ′ + c1 γ ′ β + d1 ββ ′ = 0,
(7.1.1)
that is
a1 δF (Z, Z1 , Z2 )δF (Z, Z4 , Z5 ) + b1 δF (Z, Z1 , Z2 )δF (Z, Z6 , Z4 )
+ c1 δF (Z, Z4 , Z5 )δF (Z, Z3 , Z1 ) + d1 δF (Z, Z3 , Z1 )δF (Z, Z6 , Z4 ) = 0,
as satisfied by the point Z.
We note that this locus always contains the points Z1 and Z4 . It may consist of two lines,
either intersecting or parallel, or one line taken twice; in that case we shall call it an improper
conic. We dealt with that case in Example 1 in §7.1.1. When the locus does not consist of lines,
we shall refer to it as a proper conic. We dealt with the case of a circle in Example 2 in §7.1.1.
7.1.3
Ambiguous cases
In might seem as if in 7.1.2, by multiplying across by ββ ′ , and so by adding in the points for which
β = 0 and those for which β ′ = 0, we are inserting a lot of extra points. We look at these cases in
detail.
λ
1
Z2 + 1+λ
Z3 (λ ∈ R) of the line
The point Z3 is not included in the representation Z = 1+λ
′
Z2 Z3 , and we note that if we take λ = 1/λ then we have
Z=
1
λ′
Z
+
Z2 ,
3
1 + λ′
1 + λ′
which gives the points of the line Z2 Z3 based now on (Z3 , Z2 ) instead of (Z2 , Z3 ). This time Z2
is not included but we get Z3 by taking λ′ = 0. To avoid the awkwardness of having to do this
134
CHAPTER 7. CONIC SECTIONS
repeatedly, we will make the convention that the first representation includes Z3 on taking 1/λ = 0.
In the case of each line, we shall deal in this manner with the parametric representation of points
on it.
When a1 d1 − b1 c1 6= 0,
a1 λµ + b1 λ + c1 µ + d1 = 0
gives a 1:1 correspondence as then
µ=−
b1 λ + d1
.
a1 λ + c1
When a1 6= 0, no number µ corresponds to λ = −c1 /a1 and we make a convention that 1/µ = 0
corresponds to this. From
c1 µ + d1
,
λ=−
a1 µ + b 1
we see also that no number λ gives rise to µ = −b1 /a1 and we make the convention that 1/λ = 0
does.
Still continuing with a1 d1 − b1 c1 6= 0, when a1 = 0 we must have b1 6= 0, c1 6= 0. Now
λ=−
d1
c1
µ− ,
b1
b1
and this gives a bijection from R to itself. In this case we make the convention that 1/λ = 0
corresponds to 1/µ = 0.
Now turning to the case when a1 d1 − b1 c1 = 0, we note that when a1 6= 0 we can factorize the
expression in the relation as
(a1 λ + c1 )(a1 µ + b1 ) + a1 d1 − b1 c1 = (a1 λ + c1 )(a1 µ + b1 ) = 0,
and we take from this the equation
(a1 γ1 + c1 β)(a1 γ ′ + b1 β ′ ) = 0,
which represents a pair of straight lines. Similarly when d1 6= 0, we can factorize the expression in
the relation as
(a1 d1 − b1 c1 )λµ + (b1 λ + d1 )(c1 µ + d1 ) = (b1 λ + d1 )(c1 µ + d1 ) = 0.
and we take from this the equation
(b1 γ1 + d1 β)(c1 γ ′ + d1 β ′ ) = 0,
which again represents a pair of straight lines.
When a1 = c1 = 0 so that we have
b1 λ + d1 = 0,
we do not really have a relation between λ and µ, but in this case the equation we take is
(b1 γ + d1 β)β ′ = 0,
which represents a pair of straight lines.
When a1 = b1 = 0 so that we have
c1 µ + d1 = 0,
we do not really have a relation between λ and µ, but in this case the equation we take is
(c1 γ ′ + d1 β ′ )β = 0,
135
7.1. DEFINITION OF A CONIC SECTION
which represents a pair of straight lines.
Finally, when a1 = d1 = 0 so that our relation is
b1 λ + c1 µ = 0,
we can hardly speak of factorizing the expression involved, but again the equation we take is
b1 γβ ′ + c1 γ ′ β = 0,
and for a1 d1 − b1 c1 = 0 we must have b1 = 0 or c1 = 0, giving, respectively,
c1 γ ′ β = 0,
and
b1 γβ ′ = 0,
both representing pairs of lines.
With these conventions, the loci in the somewhat ambiguous cases all correspond to pairs of
lines.
7.1.4
Conic through the points of the triple of reference
If all the points of a conic lie on one line it must be improper, so we suppose that it contains three
non-collinear points Z1 , Z2 , Z3 .
Z
b
Z3
b
b
Z2
b
Z1
Figure 7.3.
We simplify the equation of the conic by considering the point of intersection Z of the line
µ
1
λ
1
joining Z1 to 1+λ
Z2 + 1+λ
Z3 and the line joining Z2 to 1+µ
Z3 + 1+µ
Z1 , where λ and µ have a
bilinear relation
2gλµ + cλ + 2hµ + 2f = 0.
We obtain the equation
cγ 2 + 2f βγ + 2gγα + 2hαβ = 0,
and since Z3 must satisfy this equation we must have c = 0. Thus we obtain
f βγ + gγα + hαβ = 0,
(7.1.2)
that is
=
f δF (Z, Z3 , Z1 )δF (Z, Z1 , Z2 ) + gδF (Z, Z1 , Z2 )δF (Z, Z2 , Z3 ) + hδF (Z, Z2 , Z3 )δF (Z, Z3 , Z1 )
0,
as the general form of the equation of a conic passing through Z1 , Z2 and Z3 . As it is to be proper,
each of f, g, h must be non- zero.
136
7.1.5
CHAPTER 7. CONIC SECTIONS
Two-term form of equation
Thus we have easily reduced the number of coefficients from four to three, and have the advantage
of now having a symmetrical relationship between Z1 , Z2 and Z3 . This is a big advance but it
still leads to complicated long drawn-out calculations, and so we take steps to reduce the number
of coefficients further.
From (7.1.1) we have that
a1 γγ ′ + b1 γβ ′ + c1 γ ′ β + d1 ββ ′ =0,
(a1 γ + c1 β)γ ′ + (b1 γ + d1 β)β ′ =0.
We already know that this passes through Z1 . Next we get it to pass through Z2 as well; this
requires
c1 γ2′ + d1 β2′ = 0,
so we may take c1 = β2′ , d1 = −γ2′ as we may re-scale the four coefficients. On insertion for c1 and
d1 this gives
(a1 γ + β2′ β)γ ′ + (b1 γ − γ2′ β)β ′ =0,
(a1 γ + β2′ β)(γ1′ α + γ2′ β + γ3′ γ) + (b1 γ − γ2′ β)(β1′ α + β2′ β + β3′ γ) =0.
(7.1.3)
In the expansion of (7.1.3) we first wish the coefficient of α2 to vanish; this is already achieved.
Secondly we wish the coefficient of β 2 to vanish; this coefficient is β2′ γ2′ − γ2′ β2′ so it is already
achieved. Thirdly we wish the coefficient of γα to vanish; this is a1 γ1′ + b1 β1′ = 0. Fourthly, we
wish the coefficient of βγ to vanish; that is a1 γ2′ + β2′ γ3′ + b1 β2′ − γ2′ β3′ = 0. Thus we need
γ1′ a1 + β1′ b1 =0,
γ2′ a1 + β2′ b1 =γ2′ β3′ − β2′ γ3′ .
(7.1.4)
Now
1
1
[Z(2,4,5) Z(3,6,4) − Z(3,4,5) Z(2,6,4) ] = 2
Z[(2,3),(4,5),(6,4)]
2
Z(4,5,6)
Z(4,5,6)
=
1
Z[(4,5),(6,4),(2,3)]
2
Z(4,5,6)
=−
Z(5,6,4) Z(4,2,3)
= −α4 ,
2
Z(4,5,6)
and the determinant of the array of coefficients on the left-hand side of (7.1.4) has the value
γ1′ β2′ − β1′ γ2′ = −γ4 . Then by Cramer’s rule, (7.1.4) has the solutions
a1 = −
α4 ′
β ,
γ4 1
b1 =
α4 ′
γ .
γ4 1
137
7.1. DEFINITION OF A CONIC SECTION
The coefficient of γ 2 is then
a1 γ3′ + b1 β3′
α4
= (γ1′ β3′ − γ3′ β1′ )
γ4
1 α4
= 2
[Z(1,4,5) Z(3,6,4) − Z(3,4,5) Z(1,6,4) ]
Z(4,5,6) γ4
=
1 α4
Z[(1,3),(4,5),(6,4)]
2
Z(4,5,6)
γ4
=
1 α4
Z[(4,5),(6,4),(1,3)]
2
Z(4,5,6)
γ4
=−
=
1
2
Z(4,5,6)
α4
Z(5,6,4) Z(4,1,3)]
γ4
α4
β4 .
γ4
Moreover the coefficient of αβ is
β2′ γ1′ − γ2′ β1′ = −γ4 .
Thus finally we obtain the equation
α4
β4 γ 2 − γ4 αβ = 0,
γ4
which we write as
γ2 −
γ42
αβ = 0.
α4 β4
(7.1.5)
It might be thought that Z4 might play a vital role in this equation because of the part it played
in its derivation, but if Z0 is any point on the proper conic which does not lie on any of the side
γ2
γ2
lines of the triple of reference (Z1 , Z2 , Z3 ), then α00β0 = α44β4 and so we also have the equation
γ2 −
γ02
αβ = 0.
α0 β0
(7.1.6)
This has the form
γ 2 − 4kαβ = 0,
(7.1.7)
where k 6= 0 is given by γ02 = 4kα0 β0 .
It can be checked that when (7.1.5) is rewritten as
γ 2 α4 β4
= 1,
γ42 αβ
the expression on the left is an equipoised quotient.
7.1.6
Parametric equations for a conic
For any proper conic take its equation in the form (7.1.6). For any point Z not on the line Z2 Z3 ,
we define t by
γ
γ0
t,
=
α
α0
so that
α=
α0
γ,
γ0 t
138
CHAPTER 7. CONIC SECTIONS
and then Z lies on the conic if and only if
γ2 =
and so β =
β0
γ0 tγ.
α0
γ0
γ02
β
γ=
βγ,
α0 β0 γ0 t
β0 t
Thus we have that
β=
From this we see that
β0 γ0
β0 2
t tα =
t α.
γ0 α0
α0
β0 2 γ0
t +
t ,
1=α+β+γ =α 1+
α0
α0
and so
α=
α0
α0 /β0
= 2 γ0
α0 + γ0 t + β0 t2
t + β0 t +
α0
β0
β=
β 0 t2
t2
=
γ
α0 + γ0 t + β0 t2
t2 + β00 t +
α0
β0
γ=
tγ0 /β0
γ0 t
= 2 γ0
α0 + γ0 t + β0 t2
t + β0 t +
α0
β0
,
,
.
(7.1.8)
Then as t varies through R we obtain all the points of the conic except on the line Z3 Z2 ; the sole
point Z2 of the conic on the latter is obtained by making t → ∞, or alternatively replacing t by
1/t′ , so as to get
α0 t′2
,
α0 t′2 + γ0 t′ + β0
β0
β=
,
′2
α0 t + γ0 t′ + β0
γ 0 t′
γ=
,
α0 t′2 + γ0 t′ + β0
α=
and now putting t′ = 0.
From (3.1.2) we can deduce the Cartesian form of these as
x=
x1 α0 + x3 γ0 t + x2 β0 t2
y1 α0 + y3 γ0 t + y2 β0 t2
,
y
=
,
α0 + γ0 t + β0 t2
α0 + γ0 t + β0 t2
(7.1.9)
from which graphs of individuals among these loci can be drawn by computer.
If we complete the square in the denominators in (7.1.8) we obtain
2
γ0
α0
4α0 β0 − γ02
1 γ0
t + t+
= t+ 2
+
β0
β0
β0
4β02
2
1 2
γ − γ2
γ0
= t + 21
+ k 0 2 0
β0
4β0
2
γ0
k(1 − k)γ02
= t + 21
.
+
β0
4k 2 β02
2
(7.1.10)
We obtain different patterns of graphs according as the denominator in (7.1.8) is the sum of two
squares, is itself a square, or is the difference of two squares, according as 0 < k < 1, k = 1 and
either k > 1 or k < 0. We call the conic an ellipse (when it is not a circle) when 0 < k < 1, a
parabola when k = 1 and a hyperbola when k(1 − k) < 0 so that either k > 1 or k < 0.
139
7.1. DEFINITION OF A CONIC SECTION
In choosing Z0 we need
γ02 = 4kβ0 (1 − β0 − γ0 ),
1 2
γ = 0,
β02 + (γ0 − 1)β0 +
4k 0
and so
β0 =
1
2
"
1 − γ0 ±
r
(γ0 −
1)2
#
1 2
− γ0 .
k
Thus we should choose (γ0 − 1)2 − k1 γ02 > 0.
As a convenient choice for examples, when k 6= 1 we take (γ0 − 1)2 − k1 γ02 = 1 and from this
γ0 =
k
1
−2k
, β0 =
, α0 =
.
1−k
1−k
1−k
When k = 1 we take 1 − 2γ0 = 9 and from this
γ0 = −4, β0 = 1, α0 = 4.
We also take Z1 ≡ (1, 0), Z2 ≡ (0, 1), Z3 ≡ (0, 0).
For an ellipse we make the further choice of k = 14 and then obtain the Cartesian parametric
equations
4
t2
x=
, y=
.
2
4 − 2t + t
4 − 2t + t2
By computer graphics we obtain the graph shown.
For a hyperbola we make the further choice of k = 4 and then obtain the Cartesian parametric
equations
1
t2
1
4t2
4
=
=
x=
, y=
.
3
2
2
2
1 − 8t + 4t
1 − 8t + 4t
(t − 1) − 4
(t − 1)2 − 34
These are inconvenient for computer graphics, so we take instead
√
1
3
= ,
t−1−
2
s
and then obtain the equations
h
√ i2
3
3
1
+
1
+
s
2
s
√ , y=
√
x=
.
4 + 4 3s
1 + 3s
With this we obtain the partial graph following.
140
CHAPTER 7. CONIC SECTIONS
For the parabola chosen we have the parametric equations
x=
t2
4
,
y
=
,
(t − 2)2
(t − 2)2
but as these are inconvenient for computer graphics, put
t−2=
1
,
s
and obtain
x = 4s2 , y = (1 + 2s)2 .
With these we obtain the partial graph shown.
Considering now a general ellipse, so that we have 0 < k < 1, in (7.1.10) we have for the
denominators in (7.1.8)
2
k(1 − k)γ02
1 γ0
t+ 2
+
,
β0
4k 2 β02
and as this is the sum of a non-negative term and a positive term it is always non-zero, positive in
fact. It follows that the ellipse is a continuous curve, except perhaps at Z2 which corresponds to
t → ∞; however as above we can cope with the curve near Z2 by writing t = 1/t′ and considering
values of t′ near 0. Thus an ellipse is a continuous closed curve. Moreover since for any j > 0
0<
s
1
1
1
1
≤ 2, − ≤ 2
≤ ,
s2 + j 2
j
2j
s + j2
2j
7.2. INTERSECTION OF A LINE AND A CONIC
141
for all real numbers s, |α| and |β| are bounded for points Z on the ellipse, and from this and
α + β + γ = 1 it follows that |γ| is bounded also. It follows that each ellipse is a bounded curve.
Turning next to a general parabola so that k = 1, in (7.1.10) we have for the denominators in
(7.1.8)
2
1 γ0
t+ 2
,
β0
and this is positive except for
t = − 21
γ0
.
β0
Thus a parabola is a continuous curve, as we can deal with Z2 (which corresponds to t → ∞) as
above. However a parabola is unbounded as |α|, |β| and |γ| all tend to ∞ as
t → − 12
γ0
.
β0
Considering now a general hyperbola, so that we have k(k − 1) > 0, in (7.1.10) we have for the
denominators in (7.1.8)
2
γ0
k(k − 1)γ02
t + 21
,
−
β0
4k 2 β02
which is equal to 0 when
t = − 21
We divide t into two ranges, first
− 21
equivalently
p
γ0 γ0
.
± 12 k(k − 1) β0
kβ0 p
p
γ0 γ0
< t < − 1 γ0 + 1 k(k − 1) γ0 ,
− 12 k(k − 1) 2
2
β0
kβ0
β0
kβ0 t+
1
2
γ0
β0
2
<
k(k − 1)γ02
,
4k 2 β02
and then
p
p
γ0 γ0 1 γ0
1
1
1 γ0
,
∪ [t :> − 2
− 2 k(k − 1) + 2 k(k − 1) t : t < −2
β0
kβ0 β0
kβ0 t+
1
2
γ0
β0
2
>
(7.1.11)
(7.1.12)
k(k − 1)γ02
.
4k 2 β02
In (7.1.11) Z traces out a continuous path, which is unbounded as t approaches either end of the
interval. In (7.1.12) Z traces out a continuous path as we can deal with Z2 , which corresponds
to t → ∞, as above; this path is unbounded as t approaches either end of its range. Thus each
hyperbola is the union of two continuous parts, both of which are unbounded. We call each of the
two continuous parts a branch of the hyperbola.
7.2
Intersection of a line and a conic
We now begin a study of the relationship of lines and conics. Although detailed and complicated,
the results are the basis on which the structures of conics are revealed. In this chapter we shall concentrate on establishing the most basic properties of proper conics and on discriminating between
the different types.
142
7.2.1
CHAPTER 7. CONIC SECTIONS
Intersection of a line and a conic; interior and exterior regions
Given a proper conic C with equation (7.1.7) γ 2 − 4kαβ = 0, we seek its intersection with a line
which passes through an arbitrary point Z4 with normalized areal coordinates (α4 , β4 , γ4 ), so that
the line has an equation of the form lα + mβ + nγ = 0, with the case l = m = n excluded, and
lα4 + mβ4 + nγ4 = 0.
We suppose first that γ4 6= 0. Then the line has equation γ4 (lα + mβ) = (lα4 + mβ4 )γ. We
assume as well that lα4 + mβ4 6= 0. For this line to meet the conic we need
(lα4 + mβ4 )2 γ 2 − 4k(lα4 + mβ4 )2 αβ =0,
so that
γ42 (lα + mβ)2 − 4k(lα4 + mβ4 )2 αβ =0,
and so
γ42 l2 α2 + 2[γ42 lm − 2k(lα4 + mβ4 )2 ]αβ + γ42 m2 β 2 = 0.
(7.2.1)
The discriminant of this quadratic equation (7.2.1) is
4[γ42 lm − 2k(lα4 + mβ4 )2 ]2 − 4γ44 l2 m2 = 16(lα4 + mβ4 )2 [−kγ42 lm + k 2 (lα4 + mβ4 )2 ],
(7.2.2)
which is a positive multiple of
k 2 α24 l2 + (2k 2 α4 β4 − kγ42 )lm + k 2 β42 m2 .
(7.2.3)
The discriminant of (7.2.3) is in turn
(2k 2 α4 β4 − kγ42 )2 − 4k 4 α24 β42 = k 2 γ42 (γ42 − 4kα4 β42 ).
(7.2.4)
− 4kα4 β42 > 0.
< 0 or
−
For points Z4 not on the conic C we must have either
2
2
When γ4 − 4kα4 β4 < 0, (7.2.3) will be positive for all (l, m) other than (0, 0), and so (7.2.1) will
have a real solution for all such l and m, and so every line through the point Z4 will meet the
conic. We call the set of such points Z4 the interior region of the conic C. On the other hand if
γ42 − 4kα4 β42 > 0, (7.2.3) will be positive for some (l, m) other than (0, 0) and negative for others,
and so (7.2.1) will have a real solution for some but not all such l and m, and so some but not
all lines through the point Z4 will meet the conic. We call the set of such points Z4 the exterior
region of the conic C. It is hoped that this terminology will not lead to confusion with the concepts
in Barry [2, §2.3].
Still retaining γ4 6= 0, we now suppose that lα4 + mβ4 = 0. Then the equation of the line is
lα + mβ = 0, and as γ4 n = 0 we have n = 0. As Z3 has α = β = 0, we note that it lies on this line.
If Z4 6= Z3 , then at least one of α4 , β4 is non-zero; suppose that β4 6= 0. Then m = − αβ44 l and the
line has equation l(β4 α − α4 β) = 0. We cannot have l = 0 as then we would have l = m = n = 0,
and so the line has equation β4 α − α4 β = 0. Thus we have just one line Z4 Z3 , and not a pencil
of lines with vertex Z4 . The remaining case here is when Z4 = Z3 and the line has equation
lα + mβ = 0. We cannot have l = m = 0, so suppose that m 6= 0. Then on the line we have
β = − ml α, and so for a point of intersection we have
γ42
γ 2 = −4k
4kα4 β42
γ42
l 2
α .
m
q
If k > 0 then we have solutions γ = ±2 −k ml α for all l/m < 0 and no solution for any l/m > 0.
Conversely if k < 0, we have these solutions for all l/m > 0 and no solution for any l/m < 0. Thus
Z3 is an exterior point and we note that for it we have γ 2 − 4kαβ = 1 > 0.
143
7.2. INTERSECTION OF A LINE AND A CONIC
The above analysis was based on the assumption that γ4 6= 0. To cover the remaining cases we
suppose that γ4 = 0; then we cannot have α4 = β4 = 0 as well, so without loss of generality we
may suppose that β4 6= 0. As our line passes through Z4 , now l and m must satisfy lα4 + mβ4 = 0
and our equation of the line becomes β4 nγ = −β4 lα + α4 lβ. For this to meet the conic we consider
β42 n2 γ 2 − 4kβ42 n2 αβ = 0, so that we need (−β4 lα + α4 lβ)2 − 4kβ42 n2 αβ = 0, which leads to
β42 l2 α2 − 2(α4 β4 l2 + 2kβ42 n2 )αβ + α24 l2 β 2 = 0.
(7.2.5)
The discriminant in (7.2.5) is
4(α4 β4 l2 + 2kβ42 n2 )2 − 4α24 β42 l4
=16β42 n2 (kα4 β4 l2 + k 2 β42 n2 ),
which is a positive multiple of
kα4 β4 l2 + k 2 β42 n2 .
(7.2.6)
The discriminant of (7.2.6) is
(−4kα4 β4 )k 2 β42 .
(7.2.7)
Now (7.2.7) is negative if and only if −4kα4 β4 < 0, which is our earlier condition, now with γ4 = 0.
The earlier argument still applies, now with (7.2.5), (7.2.6) and (7.2.7) instead of (7.2.2), (7.2.3)
and (7.2.4).
7.2.2
Parametric equation for the line; equation of incidence
We consider again the proper conic with equation (7.1.7) and now seek its intersection with the line
Z4 Z5 , for distinct points Z4 , Z5 which have normalised areal coordinates (α4 , β4 , γ4 ), (α5 , β5 , γ5 ),
respectively, taking parametric equations for this line.
b
b
b
Z7
Z5
Z6
Figure 7.7.
b
Z4
The points on Z4 Z5 can be expressed in either of the forms
Z = (1 − t)Z4 + tZ5
or
Z=
1
λ
Z4 +
Z5 .
1+λ
1+λ
144
CHAPTER 7. CONIC SECTIONS
We start with the first of these and so Z has areal coordinates (α, β, γ) of the form
α = α4 + t(α5 − α4 ), β = β4 + t(β5 − β4 ), γ = γ4 + t(γ5 − γ4 ).
On inserting this into the equation of the conic to find points of intersection, we obtain the quadratic
equation
[γ4 + t(γ5 − γ4 )]2 − 4k[α4 + t(α5 − α4 )][β4 + t(β5 − β4 )] = 0,
that is
[(γ5 − γ4 )2 − 4k(α5 − α4 )(β5 − β4 )]t2 + γ42 − 4kα4 β4
+ 2{γ4 (γ5 − γ4 ) − 2k[α4 (β5 − β4 ) + β4 (α5 − α4 )]}t = 0.
(7.2.8)
We shall refer to this quadratic equation as an equation of incidence for a conic and a line.
It is a quadratic equation in t and so generally has two roots, real and distinct or coincident, or
non-real.
The quadratic equation of incidence (7.2.8) reduces to one of lower order when the coefficient
of t2 vanishes, that is
(γ5 − γ4 )2 − 4k(α5 − α4 )(β5 − β4 ) = 0.
(7.2.9)
In this case we say that Z4 Z5 is a line of (singly) deficient incidence with the conic. This
condition is
[δF (Z5 , Z1 , Z2 ) − δF (Z4 , Z1 , Z2 )]2 − 4k[δF (Z5 , Z2 , Z3 ) − δF (Z4 , Z2 , Z3 )][δF (Z5 , Z3 , Z1 ) − δF (Z4 , Z3 , Z1 )]
= 0.
By (1.1.8)?? this is equivalent to
[δF (Z1 , Z4 , Z5 ) − δF (Z2 , Z4 , Z5 )]2 − 4k[δF (Z2 , Z4 , Z5 ) − δF (Z3 , Z4 , Z5 )][δF (Z3 , Z4 , Z5 ) − δF (Z1 , Z4 , Z5 )]
= 0,
and so in the notation of §3.2.2
[l4,5 − n4,5 − (m4,5 − n4,5 )]2 + 4k(m4,5 − n4,5 )(l4,5 − n4,5 ) =0,
(l4,5 − n4,5 )2 + 2(2k − 1)(l4,5 − n4,5 )(m4,5 − n4,5 ) + (m4,5 − n4,5 )2 =0.
For this to have one or two real solutions we need k(1 − k) ≤ 0, so that with k 6= 0, either k = 1
which corresponds to a parabola, or k(1 − k) < 0 which corresponds to a hyperbola.
We can also insert −(β5 − β4 ) = α5 − α4 + γ5 − γ4 in condition (7.2.9) and so obtain
4k(α5 − α4 )2 + 4k(α5 − α4 )(γ5 − γ4 ) + (γ5 − γ4 )2 = 0.
(7.2.10)
In the case of deficient incidence, if the coefficient of t in (7.2.8) is non-zero then Z4 Z5 meets the
conic in a unique point given by
t=−
γ42 − 4kα4 β4
1
.
2 γ4 (γ5 − γ4 ) − 2k[α4 (β5 − β4 ) + β4 (α5 − α4 )]
(7.2.11)
If the coefficients of t2 and t both vanish in the incidence equation (7.2.8), we say that Z4 Z5 is
a line of doubly deficient incidence. In the condition
γ4 (γ5 − γ4 ) − 2k[α4 (β5 − β4 ) + β4 (α5 − α4 )] = 0,
(7.2.12)
for the coefficient of t in (7.2.8) to vanish we insert −(γ5 − γ4 ) = α5 − α4 + β5 − β4 , and obtain
(γ4 + 2kβ4 )(α5 − α4 ) + (γ4 + 2kα4 )(β5 − β4 ) = 0.
(7.2.13)
7.2. INTERSECTION OF A LINE AND A CONIC
145
For doubly deficient incidence, this condition (7.2.13) is to apply in addition to the condition
(7.2.10) for deficiency.
We cannot have all three of the coefficients in the equation of incidence (5.2.8) vanish in the case
of a proper conic, as otherwise the equation of incidence would be satisfied by every real number
t and so every point of the line Z4 Z5 would lie on the conic, which would make it improper.
It might seem as if the foregoing definition of deficient incidence depends not on the line Z4 Z5
but on the particular pair of points Z4 , Z5 used to obtain the equation of incidence
pt2 + qt + r = 0,
(7.2.14)
as a shortened form of (7.2.8). The equation of incidence based on any other points of Z4 Z5 would
involve a change of parameter t = gs + h, where g 6= 0, and from (7.2.14) we would obtain
p(gs + h)2 + q(gs + h) + r = 0,
so that the second equation of incidence would have the form
pg 2 s2 + (2pgh + qg)s + ph2 + qh + r = 0.
(7.2.15)
Clearly pg 2 = 0 in (7.2.15) if and only if p = 0 in (7.2.14); when p = 0, clearly 2pgh + qg = 0 in
(7.2.15) if and only if q = 0 in (7.2.14).
7.2.3
Tangent to a conic
If Z4 is a point on a proper conic C, we say that the line Z4 Z5 is a tangent to C at Z4 if Z4 Z5
has only one point in common with C and is not a line of deficient incidence. With the notation of
§7.2.2 the equation of incidence (7.2.8) is
(γ5 −γ4 )2 −4k(α5 −α4 )(β5 −β4 ) t2 +2 γ4 (γ5 −γ4 )−2k α4 (β5 −β4 )+β4 (α5 −α4 ) t = 0. (7.2.16)
As this has the root t = 0, to avoid having another root the coefficient of t must vanish and the
condition for this is
γ4 (γ5 − γ4 ) − 2k[α4 (β5 − β4 ) + β4 (α5 − α4 )] = 0,
which reduces to
−2kβ4 α5 − 2kα4 β5 + γ4 γ5 = 0.
(7.2.17)
With this condition, (7.2.16) becomes
(γ5 − γ4 )2 − 4k(α5 − α4 )(β5 − β4 ) t2 = 0,
and the coefficient of t2 in this cannot be equal to 0, because if it were, all three coefficients in the
equation of incidence (7.2.8) would be equal to 0 and this cannot occur for a proper conic. Thus
the equation of incidence must reduce to the equation t2 = 0.
On replacing the fixed point Z5 by the variable point Z, the equation (7.2.17) becomes
−2kβ4 α − 2kα4 β + γ4 γ = 0.
(7.2.18)
Except in theoretically possible cases which we shall immediately consider, this is the equation of
a line, so that there is a unique tangent line to the proper conic at the point Z4 .
This equation (7.2.18) would fail to be the equation of a line if we had
−2kβ4 = −2kα4 = γ4 .
We have k 6= 0 as (7.1.7) is the equation of a proper conic, and thus (7.2.19) yields that
1
γ4 .
α4 + β4 + γ4 = 1 −
k
(7.2.19)
146
CHAPTER 7. CONIC SECTIONS
As α4 + β4 + γ4 = 1 6= 0, this would yield a contradiction if γ4 = 0. When γ4 6= 0, it gives an
immediate contradiction when k = 1, and when k 6= 1,
γ42 = 4k 2 α4 β4 ,
γ42 = 4kα4 β4 ,
give a contradiction.
If we had taken Z4 Z5 to be a line of deficient incidence, by (7.2.9) we would have taken
[α5 − α4 + (1 − 2k)(β5 − β4 )]2 − 4k(k − 1)(β5 − β4 )2 = 0,
instead of (7.2.17), and by (7.2.11) Z4 would be a unique point of intersection. On replacing Z5
by Z, this equation
2
α − α4 + (1 − 2k)(β − β4 ) − 4k(k − 1)(β − β4 )2 = 0,
(7.2.20)
would factorise to be the equation of a pair of lines (perhaps one line taken twice) if we had
k(k − 1) ≥ 0. We must guard against confusing these lines with the tangent in these cases.
Our definition of tangent is equivalent to stating that Z4 Z is a line which has exactly one point,
i.e. Z4 , in common with the conic but which has not got a single deficiency. Alternatively our
equation of incidence must be genuinely quadratic and have a double root. In fact we cannot have
a line of double deficiency which contains a point Z4 of a proper conic, as then this line would
have triple deficiency which we ruled out. Thus a tangent can never coincide with one of the lines
of singly deficient incidence mentioned in (7.2.20).
If Z6 is on the conic (7.1.7), in the equation (7.2.18) of the tangent at Z4 we may replace k by
γ62 /4α6 β6 and thus obtain for any point Z on the tangent
γ 2 α4 β
γ62 β4 α
+ 6
= 2,
γ4 γα6 β6
γ4 γα6 β6
which is condensed notation for
δF (Z6 , Z1 , Z2 )2 δF (Z4 , Z3 , Z1 )δF (Z, Z2 , Z3 )
δF (Z4 , Z1 , Z2 )δF (Z, Z1 , Z2 )δF (Z6 , Z2 , Z3 )δF (Z6 , Z3 , Z1 )
δF (Z6 , Z1 , Z2 )2 δF (Z4 , Z2 , Z3 )δF (Z, Z3 , Z1 )
= 2.
+
δF (Z4 , Z1 , Z2 )δF (Z, Z1 , Z2 )δF (Z6 , Z2 , Z3 )δF (Z6 , Z3 , Z1 )
(7.2.21)
This is a sum of two equipoised quotients.
b
Z2
b
Z3
b
Z1
Figure 7.8.
If we take Z4 = Z1 ≡ (1, 0, 0), we see that the tangent at Z1 is the line with equation β = 0, that
is the line Z1 Z3 . Similarly if we take Z4 = Z2 ≡ (0, 1, 0), we find that the tangent at Z2 is the line
with equation α = 0, that is the line Z2 Z3 . Thus we see that our selection in §7.1.5 involves the
points Z1 and Z2 being on the conic, and the side-lines Z1 Z3 , Z2 Z3 of the triple of reference being
tangents at Z1 and Z2 .
7.2. INTERSECTION OF A LINE AND A CONIC
7.2.4
147
Some aspects of tangents
If Z is a point on the tangent at Z4 to the conic (7.1.7), then by (7.2.18) when γ4 6= 0
4k 2
(β4 α + α4 β)2 − 4kαβ
γ42
4k 2
= 2 (β4 α − α4 β)2 ,
γ4
γ 2 − 4kαβ =
and this is positive unless δF (Z3 , Z4 , Z) = 0. This will happen only if Z4 is either Z1 or Z2 and so
γ4 = 0.
Turning next to the case when γ4 = 0, by (7.1.18) the tangent has equation β4 α + α4 β = 0.
Then one but not both of α4 and β4 equal to 0, so suppose that β4 6= 0. Then Z4 = Z2 and the
tangent there is α = 0. Then on the tangent we have γ 2 − 4kαβ = γ 2 and this is positive unless
γ = 0, i.e. at Z2 .
Thus we verify what was implied in §7.2.1, that every point of a tangent other than the point
of contact lies in the exterior region.
As a further detail, suppose that a proper conic is parametrised as in §7.1.6, that we have
α = α(t), β = β(t), γ = γ(t),
for differentiable functions α(t), β(t), γ(t), as in (5.1.5), with
γ(t)2 − 4kα(t)β(t) = 0.
On differentiating this, we have identically
γ(t)γ ′ (t) − 2k[α(t)β ′ (t) + α′ (t)β(t)] = 0.
From α(t) + β(t) + γ(t) = 1 we also have that
α′ (t) + β ′ (t) + γ ′ (t) = 0.
On solving between these we obtain
2kα(t) − γ(t) ′
γ (t)),
2k(α(t) − β(t))
2kβ(t) + γ(t) ′
β ′ (t) =
γ (t),
2k(α(t) − β(t))
α′ (t) = −
on putting t = t4 in which yields
2kα4 − γ4 ′
γ (t4 ),
2k(α4 − β4 )
2kβ4 + γ4 ′
β ′ (t4 ) =
γ (t4 ).
2k(α4 − β4 )
α′ (t4 ) = −
If we now consider the line with parametric equations
α =α4 + 2k(α4 − β4 )α′ (t4 )s = α4 − (2kα4 + γ4 )γ ′ (t4 )s,
β =β4 + 2k(α4 − β4 )β ′ (t4 )s = β4 + (2kβ4 + γ4 )γ ′ (t4 )s,
γ =γ4 + 2k(α4 − β4 )γ ′ (t4 )s,
on eliminating γ ′ (t4 )s from these we find that we have the line with equation
γ4 γ − 2k[β4 α + α4 β] = 0,
and this is the tangent to the conic at Z4 .
Thus our definition of a tangent to a proper conic agrees with the more general definition in
calculus.
148
7.2.5
CHAPTER 7. CONIC SECTIONS
Second parametric form for an incident line; polar of a point
We now return to the situation in §7.2.2 and consider the intersection of Z4 Z5 with C again, but
now take points Z on the line Z4 Z5 to have the form
α=
1
λ
1
λ
1
λ
α4 +
α5 , β =
β4 +
β5 , γ =
γ4 +
γ5 .
1+λ
1+λ
1+λ
1+λ
1+λ
1+λ
Then for a point of intersection we need λ to satisfy
(λγ5 + γ4 )2 − 4k(λα5 + α4 )(λβ5 + β4 ) = 0,
which expands to
(γ52 − 4kα5 β5 )λ2 + 2[γ4 γ5 − 2k(α4 β5 + β4 α5 )]λ + γ42 − 4kα4 β4 = 0.
(7.2.22)
This type of quadratic equation of incidence needs more careful handling than the earlier one,
as λ = −1 does not correspond to a point on the line Z4 Z5 and we need to take λ = ∞ or 1/λ = 0
to include Z5 .
If Z4 Z5 meets C in points Z6 and Z7 , we consider the situation where (Z4 , Z5 , Z6 , Z7 ) is a
harmonic range. For this we need the parameters of Z6 and Z7 to be of the form λ, −λ, and so
the sum of the roots of the quadratic equation (7.2.22) must be equal to 0. The condition for this
is
γ4 γ5 − 2k(α4 β5 + β4 α5 ) = 0,
On replacing the fixed point Z5 by a variable point Z, this condition is
−2kβ4 α − 2kα4 β + γ4 γ = 0.
(7.2.23)
This is generally the equation of a line, and when it is the line is called the polar of the point Z4
with respect to the conic C. This was originally considered by Desargues in 1639.
b
b
Z4
b
b
Z6
b
Z7
Z
b
Figure 7.9.
b
This is not to claim that for every point Z on this line, Z4 Z meets the conic in two points Z6
and Z7 ; Z4 Z might meet C in no point or just one point. But when it does, (Z4 , Z, Z6 , Z7 ) must
be a harmonic range. If there is a line of deficient incidence in this case, the sole point Z6 of Z4 Z
on the conic will have parameter λ = 1 (as λ = −1 does not correspond to a point), and so Z6 will
be the mid-point of Z4 and Z.
When Z4 is on C we see that the line with equation (7.2.23) is also the line with equation
(7.2.18) and so is the tangent to C at Z4 . To fit in with this, we shall interpret (Z4 , Z, Z4 , Z4 ) as a
149
7.2. INTERSECTION OF A LINE AND A CONIC
harmonic range in an extended sense, for all points Z. Thus for a point Z4 on the conic, a tangent
at Z4 is the polar with respect to Z4 .
For a second use of this type of parametrization, take points Z4 and Z5 on the conic with
equation (7.1.7) and we use the parametric representation (7.1.8) for the conic. Then for points
Z(λ) with areal coordinates as above, we have that
2λ
[γ4 γ5 − 2k(α4 β5 + β4 α5 )],
(1 + λ)2
γ02 t4 t5 − 2kα0 β0 (t24 + t25 )
2λ
=
(1 + λ)2 (α0 + γ0 t4 + β0 t24 )(α0 + γ0 t5 + β0 t25 )
(t5 − t4 )2
2λ
.
= − 12 γ02
(1 + λ)2 β02 ( αβ00 + βγ00 t4 + t24 )( αβ00 + βγ00 t5 + t25 )
γ(λ)2 − 4kα(λ)β(λ) =
Now, as in §7.1.6, when the conic is an ellipse or parabola, the denominator in this last expression
is always positive and so the whole expression is negative for λ > 0 that is for the points of the
segment [Z4 , Z5 ], other than the end-points Z4 and Z5 . Thus for a conic which is either an ellipse
or a parabola, if Z4 , Z5 are any distinct points on the conic, then every point of [Z4 , Z5 ] \ {Z4 , Z5 }
is in the interior region, and all points of Z4 Z5 \ [Z4 , Z5 ] are in the exterior region.
On the other hand, if Z4 , Z5 are distinct points on a hyperbola, then αβ00 + βγ00 t4 + t24 and
γ0
α0
2
β0 + β0 t5 + t5 have the same sign if Z4 and Z5 are on the one branch, and so the whole expression
is negative for λ > 0 that is for the points of the segment [Z4 , Z5 ], other than the end-points Z4
and Z5 . Moreover αβ00 + βγ00 t4 + t24 and αβ00 + βγ00 t5 + t25 have opposite signs if Z4 and Z5 are on different
branches, and so the whole expression is negative for λ < 0 that is for the points of the line Z4 Z5
not on the segment [Z4 , Z5 ].
Thus if Z4 , Z5 are any distinct points on a hyperbola, when they are on the same branch, every
point of [Z4 , Z5 ] \ {Z4 , Z5 } is in the interior region, and all points of Z4 Z5 \ [Z4 , Z5 ] are in the
exterior region, while when Z4 , Z5 are on different branches, every point of [Z4 , Z5 ] \ {Z4 , Z5 } is
in the exterior region, and all points of Z4 Z5 \ [Z4 , Z5 ] are in the interior region.
7.2.6
Centre of a conic
The equation (5.2.19)
−2kβ4 α − 2kα4 β + γ4 γ = 0,
will fail to represent a line, for a given point Z4 and a given proper conic C, if
α4 = −
1
γ4 ,
2k
β4 = −
1
γ4 ,
2k
with α4 + β4 + γ4 = 1. This would imply that
α4 + β4 + γ4 =
1
1−
k
γ4 .
Now γ4 6= 0 so that we have this failure when k 6= 1 but not when k = 1 and so our conic is a
parabola.
Thus when k 6= 1 and so our conic is either an ellipse/circle or a hyperbola,
1
1
k
Z4 ≡
,
(7.2.24)
,
,−
2(1 − k) 2(1 − k) 1 − k
is the only point for which there is not a polar line with respect to C. We call this unique point
the centre of C, and refer to each ellipse/circle or hyperbola as a central conic. Note that this
fits in with terminology for a circle, as with Z4 its centre, each line through Z4 meets the circle in
points Z6 and Z7 and Z4 is the mid-point of Z6 and Z7 ; thus there is no point Z ∈ Π such that
(Z4 , Z, Z6 , Z7 ) is a harmonic range. Thus its centre has no polar with respect to a circle.
150
CHAPTER 7. CONIC SECTIONS
Every point Z4 will have a polar line with respect to a parabola C so a parabola has no centre.
We shall refer to a parabola as a non-central conic.
When Z4 is the centre of an ellipse or hyperbola, we have that
γ42 − 4kα4 β4 =
k(k − 1)
,
(k − 1)2
and this is negative or positive according as k(k − 1) is. Thus the centre of an ellipse is in its
interior region, and the centre of a hyperbola is in its exterior region.
For an ellipse or hyperbola, in §7.2.1 take Z4 to be the centre (7.2.24). Then for any Z5 6= Z4 ,
in the incidence equation (7.2.8) the coefficient of 2t is equal to
γ4 γ5 − 2k(β4 α5 + α4 β5 ) − γ42 + 4kα4 β4
4k 2
4k
1
2kγ5 − 2k(α5 + β5 ) −
+
=
2(1 − k)
2(1 − k) 2(1 − k)
=0.
Thus if t6 , t7 are the roots of the equation of incidence we must have t6 + t7 = 0, and so if Z6 is
any point on the conic so must be Z7 where the centre Z4 is the mid-point of Z6 and Z7 . It follows
that each ellipse and hyperbola is symmetrical about its centre.
b
b
Z4
b
Figure 7.10. Centre of an ellipse.
Since the centre of a hyperbola is an exterior point, it follows from §7.2.5 that for these Z5 and
Z6 , one of them must be on each branch. Thus under central symmetry in the centre of a hyperbola
each branch is mapped onto the other. We can verify this last result directly as follows. Suppose
that points Z6 , Z7 on a hyperbola with parametric equations (7.1.8) have values of the parameter
t6 and t7 respectively. Then the line Z6 Z7 will pass through the centre of the hyperbola if and
only if
α0 β0 t26 γ0 t6 0 = α0 β0 t27 γ0 t7 1
1
−2k 2
α0
β 0 t6
γ 0 t6
= 2
2
0 β0 (t7 − t6 ) γ0 (t7 − t6 ) 1
1
−2k
2
α0
β 0 t6
γ0 t6 =(t7 − t6 ) .
0 β0 (t7 + t6 ) γ0 1
1
−2k 151
7.2. INTERSECTION OF A LINE AND A CONIC
The coefficient of −(t7 − t6 ) here is
β0 γ0 t6 t7 + 2kα0 β0 (t6 + t7 ) + γ0 α0 ,
in which we can replace 2kα0 β0 by 21 γ0 and divide across by γ0 to obtain
β0 t6 t7 + 12 γ0 (t6 + t7 ) + α0 = 0,
and so
t7 = −
γ0 t6 + 2α0
.
2β0 t6 + γ0
From this we find that
α0 + γ0 t7 + β0 t27 = (α0 + γ0 t6 + β0 t26 )
so that
α0
γ0
γ02
α0
γ0
+ t7 + t27 = −k(k − 1) 2
(
+ t6 + t26 ).
2
β0
β0
k (2β0 t6 + γ0 ) β0
β0
As k(k − 1) > 0, we see that αβ00 + βγ00 t6 + t26 and αβ00 +
and Z7 are on different branches of the hyperbola.
7.2.7
4α0 β0 − γ02
(2β0 t6 + γ0 )2
γ0
β 0 t7
+ t27 have opposite signs, and thus Z6
A symmetrical relation; conjugate points
If Z4 and Z8 both have polars with respect to C, then Z8 will lie on the polar with respect to the
conic of Z4 if
−2kβ4 α8 − 2kα4 β8 + γ4 γ8 = 0,
(7.2.25)
and this is precisely the condition that Z4 lie on the polar of Z8 . Thus Z8 lies on the polar of Z4
if and only if Z4 is on the polar of Z8 . Such Z4 and Z8 are called conjugate points with respect
to this conic.
7.2.8
Centre not on a polar
It is related to §7.2.7 that the centre of a central conic cannot lie on any polar, as we verify by
noting that
−2kβ4
1
k
k
1
− 2kα4
−
γ4 = −
(β4 + α4 + γ4 ) 6= 0,
2(1 − k)
2(1 − k) 1 − k
1−k
in connection with (7.2.24) and (7.2.25).
7.2.9
What lines are polars? Poles
Given a line with equation l4 α + m4 β + n4 γ = 0, we ask when this line is the polar of some point
Z4 with respect to the conic (7.1.7). By (7.2.25), for this we need
−2kβ4 = jl4 , −2kα4 = jm4 , γ4 = jn4 ,
for some j 6= 0. This would require
l4
m4
−
+ n4 ,
α4 + β4 + γ4 = j −
2k
2k
so that we need
−
l4
m4
−
+ n4 6= 0.
2k
2k
(7.2.26)
152
CHAPTER 7. CONIC SECTIONS
In the case of a central conic, by §7.2.8 the line must not pass through the centre, and when that
is the case, by (7.2.24) the condition (7.2.26) is satisfied, and we obtain as a unique solution
α4 = −
l4 /2k
n4
m4 /2k
, β4 = − m 4
, γ4 = m 4
.
l4
l4
l4
4
−
+
n
−
+
n
−
−m
−
−
4
4
2k
2k
2k
2k
2k
2k + n4
(7.2.27)
Thus in the case of a central conic, every line which does not pass through the centre is a polar.
When Z4 is not the centre Z0 of the conic, we have that
j=
=
4
−m
2k
1
−2k
=
l4
l4 + m4 − 2kn4
− 2k + n4
1
2(1−k) l4
+
2k
− 2(1−k)
1
2(1−k) m4
−
2k
2(1−k) n4
,
Then (7.2.27) becomes
jl4
=
β4 = −
2k
α4 = −
1
2(1−k) l4
jm4
=
2k
γ4 =jn4 =
+
1
2(1−k) l4
1
2(1−k) l4
+
1
2(1−k) l4
1
2(1−k) m4
+
−
2k
2(1−k) n4
,
1
2(1−k) m4
,
1
2k
2(1−k) m4 − 2(1−k) n4
−2k
2(1−k) n4
1
2(1−k) m4
−
2k
2(1−k) n4
.
If we take our initial line to be Z5 Z6 and as in §3.2.2 take
l4 = δF (Z1 , Z5 , Z6 ), m4 = δF (Z2 , Z5 , Z6 ), n4 = δF (Z3 , Z5 , Z6 ),
by (3.2.1) these formulae can be written as
β4 =
β0 δF (Z2 , Z5 , Z6 )
γ0 δF (Z3 , Z5 , Z6 )
α0 δF (Z1 , Z5 , Z6 )
, α4 =
, γ4 =
.
δF (Z0 , Z5 , Z6 )
δF (Z0 , Z5 , Z6 )
δF (Z0 , Z5 , Z6 )
(7.2.28)
This deals with central conics.
Turning now to parabolas, when k = 1, for the given line to be the polar of some point Z4 , by
(7.2.26) we need
1
1
− l4 − m4 + n4 6= 0.
(7.2.29)
2
2
By (7.2.9), a line Z4 Z5 of deficient incidence when k = 1 is one for which
(γ5 − γ4 )2 + 4(α5 − α4 )(α5 − α4 + γ5 − γ4 ) =0,
2
2(α5 − α4 ) + γ5 − γ4 =0,
so that we have γ5 − γ4 = −2(α5 − α4 ), and hence β5 − β4 = α5 − α4 . Now the line Z4 Z5 in general
has the equation
α
β
γ
= 0,
α
β
γ
4
4
4
α5 − α4 β5 − β4 γ5 − γ4 so for Z4 Z5 to be a line of deficient incidence this will be of the form
α β
γ α4 β4 γ4 = 0,
1
1 −2 153
7.3. LINES OF DEFICIENT INCIDENCE; ASYMPTOTES OF HYPERBOLA
and so we may take Z4 Z5 to have an equation of the form l4 α + m4 β + n4 γ = 0 where
l4 = −2β4 − γ4 , m4 = γ4 + 2α4 , n4 = α4 − β4 .
(7.2.30)
From this we note that
1
1
− l4 − m4 + n4 = 0,
2
2
which is inconsistent with (7.2.28) and so must be ruled out for a polar. Thus for a parabola, a
line of deficient incidence cannot be a polar.
Conversely, if we have a line Z4 Z5 which for a parabola is not of deficient incidence, it will have
an equation l4 α + m4 β + n4 γ = 0 where
l4 = β4 (γ5 − γ4 ) − γ4 (β5 − β4 ), m4 = γ4 (α5 − α4 ) − α4 (γ5 − γ4 ), n4 = α4 (β5 − β4 ) − β4 (α5 − α4 ),
and β5 − β4 6= α5 − α4 . Then
− 21 l4 − 12 m4 + n4
=(− 12 β4 + 12 α4 )[−(β5 − β4 ) − (α5 − α4 )] + [− 21 γ4 − β4 ](α5 − α4 ) + [ 21 γ4 + α4 ](β5 − β4 )
1
=(α5 − α4 )[ 12 β4 − 21 α4 − 21 γ4 − β4 ] + (β5 − β4 )[ 12 β4 − 21 α4 + γ4 + α4 ]
2
= 21 [(β5 − β4 ) − (α5 − α4 )] 6= 0.
Then this line will be the polar of the point Z4 with areal coordinates α4 , β4 , γ4 as given in (7.2.27)
but now with k = 1. Thus for a parabola, a line is a polar if and only if it is not of deficient
incidence.
It has been shown incidentally in this section that when a line is a polar with respect to a proper
conic, it is the polar of a unique point Z4 . We call this Z4 the pole of that line.
7.3
Lines of deficient incidence; asymptotes of hyperbola
We recall from §7.2.2 the conditions for a line of singly and doubly deficient incidence, the equation
(7.2.9)
4k(α5 − α4 )2 + 4k(α5 − α4 )(γ5 − γ4 ) + (γ5 − γ4 )2 = 0,
for deficient incidence, and in addition the equation (7.2.12)
(γ4 + 2kβ4 )(α5 − α4 ) + (γ4 + 2kα4 )(β5 − β4 ) = 0.
for double deficiency.
We now go on to consider these in turn, in the case of an ellipse/circle, a parabola and a
hyperbola.
7.3.1
No lines of deficient incidence for an ellipse/circle
When k 6= 1 in the condition (7.2.10) for Z4 Z5 to be a line of deficient incidence, there will be real
solutions in (α5 − α4 )/(γ5 − γ4 ) if and only if k(k − 1) > 0. This will not be so in the case of an
ellipse or circle. Thus there are no lines of deficient incidence in the case of an ellipse or circle.
7.3.2
Lines of deficient incidence for a parabola
We saw in (7.2.29) that a line Z4 Z5 of deficient incidence when k = 1 is one which has an equation
−(γ4 + 2β4 )α + (γ4 + 2α4 )β + (α4 − β4 )γ = 0.
(7.3.1)
This shows that for a parabola, through each point Z4 of the plane a unique line of deficient
incidence passes.
154
CHAPTER 7. CONIC SECTIONS
Moreover if Z8 Z9 is any other such line of deficient incidence, it will have an equation
−(γ8 + 2β8 )α + (γ8 + 2α8 )β + (α8 − β8 )γ = 0,
and by §3.2.6 these two lines are parallel to each other. For
−(γ4 + 2β4 ) γ4 + 2α4 α4 − β4 −(γ8 + 2β8 ) γ8 + 2α8 α8 − β8 1
1
1
γ8 − γ4 + 2(β8 − β4 ) −(γ8 − γ4 ) − 2(α8 − α4 )
−(γ8 + 2β8 )
γ8 + 2α8
= 1
1
−(α8 − α4 ) + β8 − β4
α8 − β8
1
on subtracting the second row from the first, and as
,
−(γ8 − γ4 ) = α8 − α4 + β8 − β4 ,
this is equal to
−(α8 − α4 ) + (β8 − β4 ) −(α8 − α4 ) + (β8 − β4 ) −(α8 − α4 ) + β8 − β4
−(γ8 + 2β8 )
γ8 + 2α8
α8 − β8
1
1
1
=0,
as the first row is a multiple of the third. This identifies the lines of single deficient incidence for
a parabola.
Figure 7.13. A line of deficient incidence for a parabola.
The conditions for Z4 Z5 to be a line of doubly deficient incidence are
β5 − β4 = α5 − α4 ,
and
(α4 + β4 + γ4 )(α5 − α4 ) = 0.
As Z4 6= Z5 this cannot happen, so there is no line of doubly deficient incidence in the case of a
parabola.
7.3. LINES OF DEFICIENT INCIDENCE; ASYMPTOTES OF HYPERBOLA
7.3.3
155
Lines of deficient incidence for a hyperbola; asymptotes
In the case of a hyperbola, when k(k −1) > 0, the condition (7.2.9) for Z4 Z5 to be a line of deficient
incidence has solutions
p
−k ± k(k − 1)
(γ5 − γ4 ),
α5 − α4 =
2k
and hence, on using β5 − β4 = −(α5 − α4 ) − (γ5 − γ4 ),
p
−k ∓ k(k − 1)
β5 − β4 =
(γ5 − γ4 ).
2k
On taking the + in α5 − α4 , the line Z4 Z5 has equation
α
β
γ
α
β
γ
4
4
4
p
p
−k + k(k − 1) −k − k(k − 1) 2k
= 0,
(7.3.2)
and as Z4 varies all these lines are parallel to each other, as the analogous equation for Z8 instead
of Z4 has equation
α
β
γ α8
γ8 = 0,
p
pβ8
−k + k(k − 1) −k − k(k − 1) 2k and by §3.2.6 these are parallel.
On taking the − in α5 − α4 we have instead
α
β
α
β
4
p
p4
−k − k(k − 1) −k + k(k − 1)

α
β
γ
α
β
γ
det 
8
8
8
p
p
−k − k(k − 1) −k + 2 k(k − 1) 2k
γ
γ4
2k

=0,
 = 0,
(7.3.3)
and these lines are parallel to each other, for the same reason.
Thus there are two sets of lines of deficient incidence for each hyperbola, each set being a family
of parallel lines. Through every point Z4 of the plane there are two lines of deficient incidence, one
in each set.
Figure 7.14. Two lines of deficient incidence for a hyperbola.
156
CHAPTER 7. CONIC SECTIONS
For Z4 Z5 to be a line of doubly deficient incidence for this hyperbola, we have the earlier
condition for deficiency now combined with the condition
(γ4 + 2kβ4 )(α5 − α4 ) + (γ4 + 2kα4 )(β5 − β4 ) = 0.
For double deficiency, we thus obtain on dividing across by γ5 − γ4 ,
p
p
−k ± k(k − 1)
−k ∓ k(k − 1)
− 2kβ4
= 0,
γ4 − 2kα4
2k
2k
and so
γ4 + k(α4 + β4 ) ±
Then we have
p
k(k − 1)(α4 − β4 ) = 0.
p
k(k − 1)(α4 − β4 ) =0,
p
γ4 + k(α4 + β4 ) − k(k − 1)(α4 − β4 ) =0.
γ4 + k(α4 + β4 ) +
(7.3.4)
As Z4 varies, these are the equations of two lines. Thus there are two lines of doubly deficient
incidence for a hyperbola and we call these its asymptotes.
As simultaneous equations, the equations (7.3.4) of the asymptotes have the same solution as
p
k(k − 1)(α4 − β4 ) = 0,
γ4 + k(α4 + β4 ) = 0,
and so they have a point of intersection given by
β4 = α4 , γ4 = −2kα4 .
Thus the coordinates (α4 , β4 , γ4 ) of the point of intersection are proportional to (1, 1, −2k) and
so it must be the centre of the hyperbola. Hence, the asymptotes pass through the centre of the
hyperbola, and so each asymptote is symmetrical about the centre.
As we saw that there is no line of triply deficient incidence for a proper conic, the asymptotes
have no point in common with the hyperbola.
Figure 7.15. The asymptotes of a hyperbola.
The asymptotes are among the lines of deficient incidence that we noted above for the hyperbola,
one in each set. The asymptotes have equations
p
γ + k(α + β) + k(k − 1)(α − β) =0,
p
γ + k(α + β) − k(k − 1)(α − β) =0.
(7.3.5)
7.3. LINES OF DEFICIENT INCIDENCE; ASYMPTOTES OF HYPERBOLA
157
For the incidence equation for the line Z4 Z5 and a hyperbola with equation (7.1.7), where Z4
is the centre of the hyperbola, and noted in §7.2.6 that
γ4 (γ5 − γ4 ) − 2k[α4 (β5 − β4 ) + β4 (α5 − α4 ] = 0,
which is (7.2.12), and γ42 − 4kα4 β4 = k(k − 1)/(k − 1)2 , so that the incidence equation (7.2.8)
becomes
k(k − 1)
.
(7.3.6)
[(γ5 − γ4 )2 − 4k(α5 − α4 )(β5 − β4 )]t2 = −
(k − 1)2
We now note that the coefficient of t2 equals
2
k
1
1
γ5 +
β5 −
− 4k α5 −
1−k
2(1 − k)
2(1 − k)
2
k −k
2k
(γ5 + α5 + β5 ) +
=γ52 − 4kα5 β5 +
1−k
(1 − k)2
k
=γ52 − 4kα5 β5 +
,
1−k
and so we have real roots of (7.3.6) if and only if
γ52 − 4kα5 β5 +
k
< 0.
1−k
Now we see that
[γ5 + k(α5 + β5 )]2 − k(k − 1)(β5 − α5 )2
=γ52 + 2kγ5 (α5 + β5 ) + [k 2 − k(k − 1)](α5 + β5 )2 + 4k(k − 1)α5 β5
=γ52 + 2kγ5 (α5 + β5 ) + k(α5 + β5 )2 + 4k(k − 1)α5 β5
=γ52 − 4kα5 β5 + k[2γ5 (α5 + β5 ) + (α5 + β5 )2 + 4kα5 β5 ]
=γ52 − 4kα5 β5 + k[(α5 + β5 + γ5 )2 − γ52 + 4kα5 β5 ]
k
2
=(1 − k) γ5 − 4kα5 β5 +
1−k
When k > 1 we have that 1 − k < 0 and so for (7.3.6) to have real roots we need
γ5 + k(α5 + β5 )]2 − k(k − 1)(β5 − α5 )2 > 0.
(7.3.7)
On the other hand when k < 0 we have that 1 − k > 0 and so for (7.3.6) to have real roots we need
γ5 + k(α5 + β5 )]2 − k(k − 1)(β5 − α5 )2 < 0.
(7.3.8)
Now the expression in (7.3.7) can be factorized as the product of two linear expressions which
when equated to 0, give the equations for Z5 of the two asymptotes respectively, and so (7.3.7) is a
condition that Z5 lies in a duo-sector with edges the asymptotes, in the sense of Barry [2, 10.10.2].
A similar argument applies for (7.3.8).
As Z5 can be taken to be any point of the hyperbola, we deduce that the hyperbola lies in one
of the two duo-sectors with edges the asymptotes.
7.3.4
Intersection of a line and a proper conic
In a somewhat cautionary subsection we again consider the intersection of the conic in (7.1.7) with
a line, but now with the line having equation lα + mβ + nγ = 0. We first suppose that n 6= 0 and
consider the simultaneous equations
n2 γ 2 − 4kn2 αβ = 0,
−nγ = lα + nβ.
158
CHAPTER 7. CONIC SECTIONS
These lead to
(lα + mβ)2 − 4kn2 αβ = 0,
and so to
l2 α2 + 2(lm − 2kn2 )αβ + m2 β 2 = 0.
For this to have a repeated root we need to have (lm − 2kn2 )2 − l2 m2 = 0 which reduces to
4kn2 (kn2 − lm) = 0, and so
kn2 − lm = 0.
(7.3.9)
This requires that l 6= 0 and m 6= 0. In this case there is a unique point of intersection of the conic
and line, given by
α4
2kn2 − lm
m
=
= ,
β4
l2
l
so that
α4 = jm, β4 = jl,
for some j 6= 0. We still need
0 = α4 l + β4 m + γ4 n = 2jlm + γ4 n,
so that γ4 n = −2jlm = −2jkn2 and so γ4 = −2jkn. Then
1 = α4 + β4 + γ4 = j(l + m − 2kn),
and so
j=
1
,
l + m − 2kn
α4 =
m
l
−2kn
, β4 =
, γ4 =
.
l + m − 2kn
l + m − 2kn
l + m − 2kn
Now the equation of the tangent to the conic at Z4
2kβ4 α + 2kα4 β − γ4 γ = 0,
becomes
2k
(lα + mβ + nγ) = 0.
l + m − 2kn
Thus the intersecting line is the tangent at Z4 in this case.
To complete this argument we need to consider the case when n = 0. Then we have
lα + mβ = 0,
γ 2 − 4kαβ = 0,
and as α and β are on the same footing so are l and m. Without loss of generality we may then
suppose that m 6= 0. Then we consider
mγ 2 − 4kαmβ = 0,
−mβ = lα,
from which 4klα2 + mγ 2 = 0. This has a repeated root if and only if lm = 0 which is the
condition (7.3.9), and so l = 0. The equation then becomes mγ 2 = 0 and so for the unique point
of intersection we have γ4 = 0. We also have mβ4 = lα4 + mβ4 = 0, and so β4 = 0. Then Z4 is the
point Z1 ≡ (1, 0, 0). The tangent there is β = 0 which is the equation of our original line. Thus
the intersecting line is the tangent at Z4 again.
It should be noted that this procedure produces tangents and does not detect lines of deficient
incidence.
We have standardized our equation for proper conics in the form γ 2 −4kαβ = 0 with k 6= 0. This
is convenient for establishing what is enjoyed in common by different types of conics but it makes
difficulties when trying to deal with lines of deficient incidence from the equation lα+mβ +nγ = 0.
For, when n 6= 0, for points of incidence we obtain the equation
l2 α2 + 2(lm − 2kn2 )αβ + m2 β 2 = 0.
(7.3.10)
7.3. LINES OF DEFICIENT INCIDENCE; ASYMPTOTES OF HYPERBOLA
159
For this to reduce to a linear equation in α we need l = 0; alternatively for it to reduce to a linear
equation in β we need m = 0, both of which are very restrictive.
To deal with this difficulty we note that (7.3.10) yields just the point Z3 or a pair of lines
passing through Z3 . For lα + mβ + nγ = 0 to be a line of deficient incidence, meeting the conic in
a single point W1 , we ask when (7.3.10) contains the line Z3 W1 and the line through Z3 which is
parallel to lα + mβ + nγ = 0. This parallel line would have an equation of the form
lα + mβ + nγ = j(α + β + γ),
and as it passes through Z3 ≡ (0, 0, 1) we would have n = j and so (l − n)α + (m − n)β = 0. If we
solve this for β and insert in (7.3.10) we obtain
l2 (m − n)2 − 2(lm − 2kn2 )(l − n)(m − n) + m2 (l − n)2 = 0.
Now
[l(m − n) − m(l − n)]2 = [n(m − l)]2 ,
and so the equation becomes in turn
n2 (m − l)2 + 4kn2 (l − n)(m − n) =0,
[m − n − (l − n)]2 + 4k(l − n)(m − n) =0,
(m − n)2 + 2(2k − 1)(m − n)(l − n) + (l − n)2 =0.
For this to have real roots we need k(1 − k) ≤ 0 so that we have either a parabola with k = 1 or a
hyperbola. We then obtain the solutions
p
m − n = [1 − 2k ± 2 k(k − 1)](l − n).
The parallel line through the origin thus has the equation
p
α + [1 − 2k ± 2 k(k − 1)]β = 0,
so the original line through Z4 and parallel to this has equation
n
o
o
n
p
p
α + [1 − 2k ± 2 k(k − 1)]β (α4 + β4 + γ4 ) = α4 + [1 − 2k ± 2 k(k − 1)]β4 (α + β + γ),
p
(α + β)(α4 + β4 + γ4 ) − (α4 + β4 )(α + β + γ) + 2[−k ± k(k − 1)][(α4 + β4 + γ4 )β − β4 (α + β + γ)] = 0,
p
γ4 (α + β) − γ(α4 + β4 ) + 2[−k ± k(k − 1)][β(α4 + γ4 ) − β4 (α + γ) = 0.
Now
p
−k ∓ k(k − 1)
1
p
,
=
k
−k ± k(k − 1)
so
2k[β(α4 + γ4 ) − β4 (α + γ)] + [−k ∓
This simplifies to
p
k(k − 1)][γ4 (α + β) − γ(α4 + β4 )] = 0.
k[(2β4 + γ4 )α − (2α4 + γ4 )β + (β4 − α4 )γ] ±
p
k(k − 1)[γ4 (α + β) − γ(α4 + β4 )] = 0.
This coincides with the equations (7.3.2) and (7.3.3) for lines of deficient incidence of a hyperbola,
and when k = 1 with (7.3.1) for a parabola.
In considering when a line lα + mβ + nγ = 0 is a tangent at some point to our conic, when
n 6= 0 we obtained the necessary condition kn2 − lm = 0 in (7.3.9). The further analysis there
showed that we must avoid the case when l + m − 2kn = 0. In this exceptional case we have
1
(l + m) = n,
2k
160
CHAPTER 7. CONIC SECTIONS
so on inserting this into (7.3.9) we must have
l2 + 2(1 − 2k)lm + m2 = 0.
For this to have a real root we need k(k − 1) ≥ 0. When k = 1, for a parabola, we obtain the
solution l = m = n and this is ruled out for a line. When k 6= 1, for a hyperbola, we have the
solutions
h
i
p
l = −1 + 2k ± 2 k(k − 1) m,
from which
n=
Hence
i
p
1h
1
k ± k(k − 1) m.
(l + m) =
2k
k
k
p
n
k ± k(k − 1)
p
=[k ∓ k(k − 1)]n,
m=
The equations for asymptotes (7.3.4)
l =2kn − m
h
i
p
= k ± k(k − 1) n.
γ + k(α + β) ±
can be rewritten as
which has the form
where
h
k∓
p
k(k − 1)(β − α) = 0
i
h
i
p
p
k(k − 1) α + k ± k(k − 1 β + γ = 0,
l
m
α + β + γ = 0,
n
n
i
h
p
l = k ∓ k(k − 1) n,
h
i
p
m = k ± k(k − 1) n,
and these are the equations just obtained.
1
1
−2k
We recall that a hyperbola has centre with coordinates 2(1−k)
, 2(1−k)
, 2(1−k)
. Note that the
line lα + mβ + nγ = 0 passes through the centre if and only if l + m − 2kn = 0, which gives us
another characterization of the exceptional case in the last paragraph.
Thus the tangential equation (7.3.9) includes all the tangents and, in the case of a hyperbola,
the asymptotes as well.
7.3.5
More on tangents
We saw in §7.2.4 that every point of a tangent, other than the point of contact, lies in the exterior
region. We now show that the converse of this is true also, yielding a characterization of tangents
like the classical Greek one.
To prove this consider lines of single deficient incidence. For a parabola, by what is between
(7.2.28) and (7.2.29) that for it we must have
γ5 − γ4 = −2(α5 − α4 ), β5 − β4 = α5 − α4 ,
so that we have
α = α4 + t(α5 − α4 ) = α4 + s, β = β4 + t(α5 − α4 ) = β4 + s, γ = γ4 − 2t(α5 − α4 ) = γ4 − 2s,
161
7.3. LINES OF DEFICIENT INCIDENCE; ASYMPTOTES OF HYPERBOLA
and then
γ 2 − 4αβ
=(γ4 − 2s)2 − 4(α4 + s)(β4 + s)
=γ42 − 4α4 β4 − 4(α4 + β4 + γ4 )s
=γ42 − 4α4 β4 − 4s,
and this is negative for some values of s, so that some points of the line lie in the interior region.
For a hyperbola, in §7.3.3 for a line of singly deficient incidence we have
α = α4 + [−k ±
Then
γ 2 − 4kαβ
p
p
k(k − 1)]s, β = β4 + [−k ∓ k(k − 1)]s, γ = γ4 + 2ks.
=(γ4 + 2ks)2 − 4k[α4 − ks ±
p
p
k(k − 1)][β4 − ks ∓ k(k − 1)]
=γ42 + 4kγ4 s + 4k 2 [(α4 − ks)(β4 − ks) − k(k − 1)s2 ± (β4 − ks − α4 + ks)s
p
=γ42 − 4kα4 β4 = 4k[γ4 + k(α4 + β4 ) ± (α4 − β4 ) k(k − 1)]s.
p
k(k − 1)]
As Z4 Z5 is not an asymptote, by (7.3.4) the coefficient of s is non-zero. Thus this expression is
negative for some values of s and so some of its points lie in the interior region.
7.3.6
Intersection of a polar with a conic
Continuing with the proper conic with equation (7.1.7), we ask for what points Z4 does the polar
of Z4 meet the conic in two points. By (7.2.33) the conic and polar have equations, respectively,
γ 2 − 4kαβ = 0, −2kβ4 α − 2kα4 β + γ4 γ = 0,
so for a point of intersection we need to solve
4kαβγ42 = γ42 γ 2 = (2kβ4 α + 2kα4 β)2 ,
and so
kβ42 α2 + (2kα4 β4 − γ42 )αβ + kα24 β 2 = 0.
(7.3.11)
The equation (7.2.28) will have two real roots for α as a multiple of β if the discriminant is positive,
which reduces to
γ42 − 4kα4 β4 > 0.
By §7.2.1 the set of such points Z4 is the exterior region of the conic. Thus a necessary condition
for the polar of a point Z4 with respect to a conic to meet that conic in two points is that Z4 be in
the exterior region of the conic.
If the polar of Z4 meets the conic in two points Z8 , Z9 , say, then by §7.2.7, Z4 is on the polar
of Z8 , that is on the tangent at Z8 . Similarly Z4 is on the tangent at Z9 , so Z4 Z8 , Z4 Z9 are a
pair of tangents from Z4 to the conic.
162
CHAPTER 7. CONIC SECTIONS
b
Z9
b
Z4
b
Z8
Figure 7.12.
b
Continuing with the equation (7.2.28), we obtain
α
−(2kα4 β4 − γ42 ) ± γ4
=
β
2kβ42
p
γ42 − 4kα4 β4
.
We are supposing that Z4 is an exterior point so that γ42 − 4kα4 β4 > 0. Then we have that
q
α = 2 −2kα4 β4 + γ42 ± γ4 γ42 − 4kα4 β4 j, β = 4kβ42 j,
for some j 6= 0. Now we note that we have
γ4 γ =2k(β4 α + α4 β)
q
=4kjβ4 γ42 γ4 ± γ42 − 4kα4 β4 ,
and so
Moreover
q
γ = 4kjβ4 γ4 ± γ42 − 4kα4 β4 .
q
α =j −4kα4 β4 + γ42 ± 2γ4 γ42 − 4kα4 β4 + γ42
2
q
2
=j γ4 ± γ4 − 4kα4 β4 .
By addition and using the fact that α + β + γ = 1 we deduce that
2
q
q
1
= γ4 ± γ42 − 4kα4 β4 + 4kβ42 + 4kβ4 γ4 ± γ42 − 4kα4 β4
j
2
q
2
= γ4 ± γ4 − 4kα4 β4 + 2kβ4 + 4k(1 − k)β42 .
This gives the coordinates of the points of intersection when j 6= 0.
7.3. LINES OF DEFICIENT INCIDENCE; ASYMPTOTES OF HYPERBOLA
163
A tangent to the conic with equation γ 2 − 4kαβ = 0 will pass through a given point Z0 if
γ4 γ0 −2k(β4 α0 +α4 β0 ) = 0 for some point Z4 with coordinates (α4 , β4 , γ4 ) such that γ42 −4kα4 β4 =
0. This requires
γ42 γ02 =4kα4 β4 γ02 ,
4k 2 (β4 α0 + α4 β0 )2 =4kα4 β4 γ02 ,
kβ02 α24 + (2kα0 β0 − γ02 )α4 β4 + kα20 β42 =0.
For this to have a real solution in α4 , β4 it is necessary and sufficient that
(γ02 − 2kα0 β0 )2 ≥ 4k 2 α20 β02 ,
and supposing that γ0 6= 0 and that Z0 is not on the conic, the condition becomes γ02 − 4kα0 β0 > 0
so that Z0 is an exterior point. We then have the solutions
p
γ02 − 2kα0 β0 ± γ02 − 4kα0 β0
β4 ,
α4 =
2kβ02
so that two tangents pass through Z0 .
But there is a case when these algebraic solutions do not produce a geometrical solution, that
is when α4 + β4 + γ4 = 0. On combining this with γ42 = 4kα4 β4 we obtain
(α4 + β4 )2 =4kα4 β4 ,
α24 + 2(1 − 2k)α4 β4 + β42 =0.
As this is to have a real solution we need k(k − 1) ≥ 0. A first possibility is that k = 1 so that
we have a parabola. Then the equation becomes α24 − 2α4 β4 + β42 = 0, so that α4 = β4 and then
γ4 = −α4 − β4 so that α4 = β4 = − 21 γ4 . On inserting this the defective tangent equation becomes
γ0 − 2(− 12 α0 − 21 β0 ) = γ0 + α0 + β0 = 0, which cannot arise for a point Z0 . Thus there is no
exception in the case of a parabola.
In the case of a hyperbola, we have the solutions
i
h
p
α4 = −1 + 2k ± 2 k(k − 1) β4 ,
and the defective tangent equation γ4 γ0 − 2k(β4 α0 + α4 β0 ) = 0 with γ4 = −α4 − β4 becomes
(β4 + α4 )γ0 + 2k(β4 α0 + α4 β0 ) =0,
h
i
n
o
p
p
γ0 2k ± 2 k(k − 1) β4 + 2k α0 + β0 [−1 + 2k ± 2 k(k − 1)] β4 =0,
i
i
h
h
p
p
γ0 k ± k(k − 1) + k α0 − β0 + 2(k ± k(k − 1))β0 =0,
k
p
(α0 − β0 ) =0,
k ± k(k − 1)
p
γ0 + 2kβ0 + (k ∓ k(k − 1))(α0 − β0 ) =0,
p
γ0 + k(α0 + β0 ) ∓ k(k − 1)(α0 − β0 ) =0.
γ0 + 2kβ0 +
But these are the equations of the asymptotes. Thus from an exterior point Z0 of a hyperbola
there fails to be a pair of tangents to the hyperbola when Z0 is the centre, in which case there are
no tangents passing through it, or when Z0 is a point on an asymptote but not the centre, in which
case only one tangent passes through it.
164
7.3.7
CHAPTER 7. CONIC SECTIONS
Parallel polars
We now study when the polars with respect to the proper conic with equation γ 2 − 4kαβ = 0 with
respect to two points Z4 , Z8 are parallel to each other. By (7.2.23) these polars have equations
−2kβ4 α − 2kα4 β + γ4 γ
−2kβ8 α − 2kα8 β + γ8 γ
=
0,
=
0,
and so, by §3.2.6, are parallel if and only if

−2kβ4
det  −2kβ8
1
−2kα4
−2kα8
1

γ4
γ8  = 0.
1
This expands to
(β4 γ8 − γ4 β8 ) + (γ4 α8 − α4 γ8 ) − 2k(α4 β8 − β4 α8 ) = 0.
(7.3.12)
In the case of a central conic, by (7.2.24) the equation (7.3.9) involves the multiple (1, 1, −2k) of
the coordinates of the centre, and so we find that the line Z4 Z8 passes through the centre. Thus in
the case of a proper central conic, the polars of two points are parallel when the line joining these
points passes through the centre. It follows that the tangents to a proper central conic at points on
a line which does not pass through the centre must intersect.
Going on to the case of a parabola, as k = 1 the equation (7.3.9) becomes
(β4 γ8 − γ4 β8 ) + (γ4 α8 − α4 γ8 ) − 2(α4 β8 − β4 α8 ) = 0,
whereas (7.3.1) can be re-arranged as
β4 γ − γ4 β + γ4 α − α4 γ − 2(α4 β − αβ4 ) = 0.
Thus (7.3.9) is the equation of a line Z4 Z5 of defective incidence. This shows that for a parabola,
the polars of two points are parallel when the line joining these points is of deficient incidence. As
a deduction from this, consider distinct points Z4 and Z8 on the parabola; the tangents at these
points are the polars of these points and if they were parallel to each other, the line Z4 Z8 would
be of deficient incidence. This cannot be the case as Z4 Z8 meets the parabola in two points. Thus
there cannot be parallel distinct tangents to a parabola.
7.3.8
Self-conjugate triples or triangles
Given a proper conic C we can construct what is called a self-conjugate triple or triangle
(Z5 , Z6 , Z7 ) with respect to it as follows.
Suppose first that C is a parabola. For any point Z5 consider a line Z5 Z4 which is of deficient
incidence. By §7.2.9 the polar of Z5 is not parallel to Z4 Z5 and so meets it in a unique point. We
let Z6 be any point on the polar other than this unique point of intersection. Now Z5 Z6 is not
parallel to a line of deficient incidence and so is the polar of a unique point Z7 . Then Z5 Z6 is the
polar of Z7 and by §7.2.7 Z6 Z7 is the polar of Z5 and Z7 Z5 is the polar of Z6 .
Suppose secondly that C is a central conic and denote by Z4 the centre. Now let Z5 be any
point other than Z4 such that Z5 Z4 is not a line of deficient incidence. Then we note that the
polar of Z5 meets Z4 Z5 at a unique point. For the polar of Z5 and the line Z4 Z5 have, respectively
as equations
−2kβ5 α − 2kα5 β + γ5 γ = 0, −(2kβ5 + γ5 )α + (2kα5 + γ5 )β + (α5 − β5 )γ = 0.
165
7.4. CONJUGATE DIAMETRAL LINES
For a point of intersection we need to solve these simultaneously with α + β + γ = 1. The the
determinant of the matrix of coefficients on the left-hand side is equal to
−2kβ5
−2kα5
γ5
−2kβ5 − γ5 2kα5 + γ5 α5 − β5 1
1
1
= − 2γ52 − 2k[α25 + β52 − 2α5 β5 + 2β5 γ5 + 2γ5 α5 ] − 8k 2 α5 β5
= − 2γ52 − 2k[(α5 + β5 + γ5 )2 − γ52 − 4α5 β5 ] − 8k 2 α5 β5
k
.
=2(k − 1) γ52 − 4kα5 β5 −
k−1
Now by §7.3.3
k
,
k−1
and this is non-zero as Z4 Z5 is not of deficient incidence. The value of the determinant is thus not
equal to 0.
We take as Z6 any point on the polar of Z5 other than this unique point of intersection. Now
Z5 Z6 is not a diametral line and so is the polar of a unique point Z7 . Then Z5 Z6 is the polar of
Z7 , and by §7.2.7, Z6 Z7 is the polar of Z5 and Z7 Z5 is the polar of Z6 ..
In each case, if Z7 ∈ Z5 Z6 then Z5 Z6 would be a tangent, Z7 the point of contact and so Z7
on the conic. For similar reasons we have that Z5 = Z6 = Z7 . Thus in addition to our initial
specification we take it that none of Z5 , Z6 , Z7 is on the conic.
7.4
7.4.1
(γ5 − γ4 )2 − 4k(α5 − α4 )(β5 − β4 ) = γ52 − 4kα5 β5 −
Conjugate diametral lines
Locus of mid-points of chords on parallel lines
Now given the equation of a fixed line
l4 α + m4 β + n4 γ = 0,
(7.4.1)
we consider the mid-points of the points of intersection, with the proper conic with equation (7.1.7),
of lines which are parallel to this fixed one. In view of (7.3.1), (7.3.2) and (7.3.3), we take this
initial fixed line (7.4.1) as not being parallel to the lines of deficient incidence in the case of a
parabola or hyperbola.
By §3.2.6 a line through the point Z ′ with normalised areal coordinates (α′ , β ′ , γ ′ ), parallel to
this line, has parametric equations
α = α′ + (m4 − n4 )t, β = β ′ + (n4 − l4 )t, γ = γ ′ + (l4 − m4 )t,
t ∈ R.
On inserting these in the equation (7.1.7) of the conic, the condition that Z ′ be the mid-point of
the points of intersection is that the coefficient of t vanish, that is
2k(n4 − l4 )α′ + 2k(m4 − n4 )β ′ − (l4 − m4 )γ ′ = 0.
(7.4.2)
As Z ′ varies, this is a linear equation for its locus. This locus is generally a line, but would fail to
be so if we had
2k(n4 − l4 ) = d, 2k(m4 − n4 ) = d, −(l4 − m4 ) = d,
for some d. This would imply
l4 − m4 = −d, m4 − n4 =
so that by addition d( k1 − 1) = 0.
d
d
, n4 − l 4 =
,
2k
2k
166
CHAPTER 7. CONIC SECTIONS
As k 6= 1 for a central conic this would imply that d = 0 so that l4 = m4 = n4 , which gives a
contradiction. Thus (7.4.2) is always the equation of some line for a central conic. On substituting
the coordinates of the centre, we get
2k(n4 − l4 )
=
1
k
1
+ 2k(m4 − n4 )
+ (l4 − m4 )
2(1 − k)
2(1 − k)
1−k
k
(n4 − l4 + m4 − n4 + l4 − m4 ) = 0.
1−k
Thus this line always contains the centre, in the case of an ellipse/circle or hyperbola.
We note that (7.4.2) has the form of an equation of a polar with respect to C of a point with
coordinates proportional to (m4 − n4 , n4 − l4 , l4 − m4 ) but there is no such point in Π as the sum
of the coordinates is equal to 0.
Going on to a parabola, with k = 1, the equation for the locus degenerates if
d
d
, n4 − l 4 = .
2
2
Now d may or may not be 0. The case d = 0 can be ruled out as for a central conic, as we would
have l4 = m4 = n4 and would not have a proper equation of the initial line (7.4.1). For d 6= 0, our
initial line would have had to have the equation
l4 − m4 = −d, m4 − n4 =
d
d
(n + )α + (n − )β + nγ = 0.
2
2
But this would make the initial line (7.4.1) parallel to the lines of deficient incidence, which we
ruled out.
We note too that each line of deficient incidence for a parabola, and so with equation (7.3.1),
has an equation of the form (7.4.2) and so is a locus of mid-points of chords on parallel lines. For
this we take as the equation
β4 α + α4 β − 21 γ4 = 0,
where (α4 , β4 , γ4 ) are areal coordinates of some point Z4 . This is the equation of some line as we
cannot have
β4 = α4 = − 12 γ4
as this would imply α4 + β4 + γ4 = 0. On now putting
l4 = β4 , m4 = α4 , n4 = − 21 γ4 ,
the equation (7.4.2) becomes (7.3.1).
b
b
b
Figure 7.16.
167
7.4. CONJUGATE DIAMETRAL LINES
Any such line which contains the mid-points of chords of a conic which are on parallel lines, is
called a diametral line of the conic. For a central conic all the diametral lines pass through the
centre. For a parabola all the diametral lines are parallel to each other, being in fact the lines of
deficient incidence.
7.4.2
Incidence of diametral line with central conic
Continuing with diametral lines in the case of an ellipse or hyperbola, we now start by supposing
that the initial line with equation l4 α + m4 β + n4 γ = 0 passes through the centre. Then the
equation is satisfied by coordinates proportional to (1, 1, −2k) so that we have
l4 + m4 − 2kn4 = 0,
(7.4.3)
and so the equation of the line can be written as
2k(l4 α + m4 β) + (l4 + m4 )γ = 0.
(7.4.4)
This line has parametric equations
1
1
+ (2km4 − l4 − m4 )t, β =
+ (l4 + m4 − 2kl4 )t,
2(1 − k)
2(1 − k)
k
γ=−
+ 2k(l4 − m4 )t, t ∈ R,
1−k
α=
(7.4.5)
and on inserting these in the equation γ 2 − 4kαβ = 0, we find that for points of intersection with
the conic we need the incidence equation
[k(l4 − m4 )2 + (l4 + m4 − 2kl4 )(l4 + m4 − 2km4 )]t2 +
1
= 0,
4(k − 1)
which can be rewritten as
[l42 + 2(1 − 2k)l4 m4 + m24 ]t2 =
1
.
4(k − 1)2
(7.4.6)
In the equation (7.4.4) of our initial line, we are excluding the degenerate case when 2kl4 =
2km4 = l4 + m4 and in particular m4 = 0, and we note that the equation
l42 + 2(1 − 2k)l4 m4 + m24 = 0
has no real solutions in l0 /m0 when (1 − 2k)2 − 1 < 0 i.e. k(k − 1) < 0, that is in the case of an
ellipse. The coefficient of t2 in (7.4.6) is then always positive and so this equation has two real
roots. This re-establishes the result that each line through the centre of an ellipse meets the curve
in two distinct points.
In the case of a hyperbola, the coefficient of t2 in (7.4.6) is
[l4 + (1 − 2k)m4 ]2 − 4k(k − 1)m24
and as k(k − 1) > 0 this is clearly sometimes positive, when the line through the centre will meet
the hyperbola in two distinct points, and sometimes negative when the line will not meet the
hyperbola at all.
7.4.3
Conjugate diametral lines
To continue the work of §7.4.1 in the case of an ellipse or hyperbola, we start by supposing as in
§7.4.2 that the initial line with equation l4 α + m4 β + n4 γ = 0 passes through the centre, so that
168
CHAPTER 7. CONIC SECTIONS
(7.4.3) is satisfied and the line has equation (7.4.4) and parametric equations (7.4.5). On applying
(7.4.3) to the equation (7.4.2), we find that the equation for the locus of Z ′ now takes the form
[(1 − 2k)l4 + m4 ]α′ + [−l4 + (2k − 1)m4 ]β ′ − (l4 − m4 )γ ′ = 0.
(7.4.7)
Then, on excluding the case of starting with an asymptote in the case of a hyperbola, this is the
equation of a diametral line. We again reach the conclusion of §5.4.1, but now the two equations
are related to the centre.
We now wish to apply this process twice, and take this locus of mid-points as a new starting
line. For this we write the equation (7.4.4) of the original line in the form
2k
m4
l4
α + 2k
β + γ = 0,
l 4 + m4
l 4 + m4
(7.4.8)
and the equation (7.4.7) of the diametral line corresponding to this as
l4 − (2k − 1)m4
(2k − 1)l4 − m4
α+
β + γ = 0.
l 4 − m4
l 4 − m4
(7.4.9)
We write the second of these, (7.4.9), in the form of the first of them, (7.4.8), by taking it to be
2k
l5
m5
α + 2k
β + γ = 0,
l 5 + m5
l 5 + m5
(7.4.10)
with
l5
(2k − 1)l4 − m4
m5
l4 − (2k − 1)m4
=
, 2k
=
.
l 5 + m5
l 4 − m4
l 5 + m5
l 4 − m4
By division in (7.4.11), this identification involves having
m5
m4
m5 m4
− (2k − 1)[
+
] + 1 = 0.
l5 l4
l5
l4
2k
(7.4.11)
(7.4.12)
This relationship (7.4.12) is symmetrical in m5 /l5 and m4 /l4 , so when it is applied twice we get
back to the initial diametral line.
b
b
b
Figure 7.17. Conjugate diametral lines for an ellipse.
169
7.4. CONJUGATE DIAMETRAL LINES
Thus for an ellipse/circle or hyperbola, diametral lines occur in pairs, each containing midpoints of chords of the conic on lines parallel to the other. We call two thus related diametral lines
conjugate. This concept can be traced back to Appolonius.
The above analysis does not apply to the asymptotes of a hyperbola, but it can be checked that
each of them is self-corresponding under the correspondence of lines through the centre given by
the formula in (7.4.12). For we can take the equation of an asymptote in (7.3.5) in the form
p
p
(k + k(k − 1α + (k − k(k − 1β + γ = 0,
in the form (7.4.8) by taking
p
l4
=k + k(k − 1),
l 4 + m4
p
p
(k − k(k − 1))l4 =(k + k(k − 1)),
p
k − k(k − 1)
m4
p
.
=
l4
k + k(k − 1)
2k
p
4
= k − k(k − 1) leads to the same conclusion. We note that
Taking 2k l4 m
+m
4
p
2 k(k − 1). Now in (7.4.12) take ml55 = ml44 and we have
m4
l4
2
m4
l4
= 2k − 1 −
m4
+1
l4
p
p
=(2k − 1)2 − 4(2k − 1) k(k − 1) + 4k(k − 1) − 2(2k − 1)(2k − 1 − 2 k(k − 1)) + 1
− 2(2k − 1)
= 0.
Moreover the coefficient of t2 in (7.4.6) is
l42 + 2(1 − 2k)l4 m4 + m24 ,
and on inserting from (7.4.11) we find that
l52 + 2(1 − 2k)l5 m5 + m25 =
l52
4k(1 − k)
[l2 + 2(1 − 2k)l4 m4 + m24 ].
l42 (2k − 1 − l4 /m4 )2 4
As these coefficients of t2 in the two equations of incidence have different signs in the case of a
hyperbola, we conclude that just one of any pair of conjugate diametral lines of a hyperbola meets
the hyperbola, whereas every diametral line of an ellipse/circle meets the locus in two points.
It follows from the above that every line through the centre is a diametral line in the case of an
ellipse, and every line through the centre with the exception of the asymptotes, is a diametral line
in the case of a hyperbola.
On putting together the results of §7.3 and §7.4, we have shown that for each parabola, every
line in the plane is either a polar or a diametral line, for each ellipse every line in the plane is
either a polar or a diametral line, and for each hyperbola each line in the plane with the exception
of the asymptotes, is either a polar or a diametral line.
We saw in §7.4.1 that for lines parallel to a fixed line with equation l4 α + m4 β + n4 γ = 0 and
which meet the conic in pairs of points, the locus of the mid-points Z ′ of these pairs is a diameter
with equation (7.4.2). Now the polar of Z ′ has equation −2kβ ′ α − 2kα′ β + γ ′ γ = 0. Now by
(3.2.4) this polar is parallel to the fixed line l4 α + m4 β + n4 γ = 0 as the condition amounts to
(7.4.2). When the point Z ′ on the diametral line is an exterior point for the conic, there will be
two tangents to the conic which pass through it and the polar is the line through the points of
contact of these tangents. It follows that the polar is parallel to the conjugate diametral line and
that the mid-point of the points of contact lie on the original diameter. Thus if Z ′ is any point
exterior to a central conic (except one on an asymptote in the case of a hyperbola), there are two
tangents from Z ′ to the conic and the mid-point of the points of contact of the tangents lies on the
line joining Z ′ to the centre.
170
7.5
7.5.1
CHAPTER 7. CONIC SECTIONS
Newton’s property of conics
Newton’s property
We now start to study other properties of conics, with a view to further discrimination between
them.
Z7
b
W4
Z9
b
W3
W6
Z8
Z6
b
W5
b
W2
b
b
W1
Figure 7.18.
W
We first take W3 , W4 to be fixed points with |W3 , W4 | = 1, with areal coordinates (α′3 , β3′ , γ3′ )
and (α′4 , β4′ , γ4′ ), respectively. For a variable point W with areal coordinates (α′ , β ′ , γ ′ ), we take a
point W1 so that W1 − W = W4 − W3 . Then points on the line W W1 , which is parallel to W3 W4 ,
have the form
W + t(W1 − W ) = W + t(W4 − W3 ), t ∈ R,
and for this point to be on the conic with equation γ 2 − 4kαβ = 0, we have the incidence equation
[γ ′ + t(γ4′ − γ3′ )]2 − 4k[α′ + t(α′4 − α′3 )][β ′ + t(β4′ − β3′ )] = 0,
which expands to a modification of (7.2.8). If W W1 meets the conic in points Z6 , Z7 which have
parameters t6 , t7 , respectively, then we have that
t6 t7 =
γ ′2 − 4kα′ β ′
.
(γ4′ − γ3′ )2 − 4k(α′4 − α′3 )(β4′ − β3′ )
To use this we recall that
2
W Z6 W Z7 = t6 t7 W W1 = t6 t7 |W , W1 |2 = t6 t7 |W3 , W4 |2 = t6 t7 .
Hence
W Z6 W Z7 =
(γ4′
−
γ3′ )2
γ ′2 − 4kα′ β ′
.
− 4k(α′4 − α′3 )(β4′ − β3′ )
(7.5.1)
7.6. CLASSICAL GREEK PROPERTY OF CONICS
171
Similarly we take W5 , W6 to be fixed points with |W5 , W6 | = 1, with areal coordinates (α′5 , β5′ , γ5′ )
and (α′6 , β6′ , γ6′ ), respectively. With W as before, we take W2 so that W2 − W = W6 − W5 . Then
points on the line W W2 have the form
W + s(W2 − W ) = W + s(W6 − W5 ),
and if W W2 meets the conic in the points Z8 , Z9 , respectively, with values s8 , s9 of the parameter,
then by the above we also have that
W Z 8 W Z 9 = s8 s9 =
γ ′2 − 4kα′ β ′
.
(γ6′ − γ5′ )2 − 4k(α′6 − α′5 )(β6′ − β5′ )
(7.5.2)
On combining these we have that
W Z8 W Z9 = k1 W Z6 W Z7 ,
(7.5.3)
where
k1 =
(γ4′ − γ3′ )2 − 4k(α′4 − α′3 )(β4′ − β3′ )
.
(γ6′ − γ5′ )2 − 4k(α′6 − α′5 )(β6′ − β5′ )
(7.5.4)
In this, W W1 is a variable line, always parallel to the fixed line W3 W4 and cutting the conic
in the variable points Z6 , Z7 , while W W2 is a variable line, always parallel to the fixed line W5 W6
and cutting the conic in the variable points Z8 , Z9 . In (7.5.3) the product of the second pair of
sensed distances is a constant k1 times the product of the first pair of sensed distances, as k1 in
(7.5.4) depends only on the fixed points W3 , W4 , W5 and W6 .
In the case of a hyperbola and a parabola, neither W3 W4 nor W5 W6 is to be parallel to a line
of deficient incidence.
This is a celebrated result on conics due to Isaac Newton.
7.6
7.6.1
Classical Greek property of conics
Classical Greek property for an ellipse or hyperbola
We now apply the result of §7.5.1 with the variability of W, W1 and W2 reduced. We take Z6 , Z7
to be fixed points of a central conic such that Z6 Z7 passes through the centre and so is a diametral
line. We take fixed points W3 , W4 ∈ Z6 Z7 so that |W3 , W4 | = 1, and then take W, W1 on the line
−−−→
Z6 Z7 with W1 − W = W4 − W3 . Thus W W1 is of magnitude 1 and slides along the fixed line Z6 Z7 .
We take W5 , W6 on the conjugate diametral line for Z6 Z7 , with |W5 , W6 | = 1. With W a
variable point on the line Z6 Z7 , let the line through the variable point W on Z6 Z7 and which is
parallel to the line W5 W6 meet the conic at the points Z8 , Z9 . Then by Newton’s result,
W Z8 W Z9 = k1 W Z6 W Z7 ,
for some constant k1 . But the line Z8 Z9 is parallel to the conjugate diametral line for Z6 Z7 and
so the mid-point of Z8 and Z9 is on Z6 Z7 . It follows that W is the mid-point of Z8 and Z9 and
hence, as W Z9 = −W Z8 we have that
2
|W , Z8 | = k1 Z6 W W Z7 .
(7.6.1)
172
CHAPTER 7. CONIC SECTIONS
Z8
b
b
b
Z6
W6
W
b
W5
b
b
b
b
Z7
b
Z9
b
Figure 7.19.
This is an ancient Greek characterization of an ellipse and a hyperbola.
7.6.2
Classical Greek property for a parabola
The result of §7.6.1 does not apply directly to a parabola as we are lacking a line playing the
precise role of Z6 Z7 .
Now, corresponding to §7.5.1, we take W3 , W4 to be fixed points with |W3 , W4 | = 1 and the
line W3 W4 to be a diametral line of the parabola with equation γ 2 − 4αβ = 0, which we recall
from §7.4.1 is also a line of deficient incidence. For a variable point W ∈ W3 W4 we take a point
−−−→
W1 such that W1 − W = W4 − W3 , and so too W1 ∈ W3 W4 . Thus W W1 is of magnitude 1 and
slides along the line W3 W4 . Then the line W W1 is also a line of deficient incidence and so meets
the parabola in a unique point, which we denote by Z6 . Points on the line W W1 have the form
W + t(W1 − W ) = W + t(W4 − W3 ),
t ∈ R,
and we suppose that Z4 has parameter t6 .
The line W W1 has parametric equations
α = α′ + t(α′4 − α′3 ), β = β ′ + (β4′ − β3′ ), γ = γ ′ + t(γ4′ − γ3′ ,
t ∈ R,
so for the parameter t of a point on the parabola we have the incidence equation
[γ ′ + t(γ4′ − γ3′ )]2 − 4[α′ + t(α′4 − α′3 )][β ′ + t(β4′ − β3′ )] = 0.
As this is of deficient incidence we must have
(γ4′ − γ3′ ) = 4(α′4 − α′3 )(β4′ − β3′ ),
(7.6.2)
and corresponding to (7.2.11) have
t6 = −
γ ′ (γ4′
−
γ3′ )
γ ′2 − 4α′ β ′
.
− 2[α′ (β4′ − β3′ ) + β ′ (α′4 − α′3 )]
(7.6.3)
We wish to express the denominator in (7.6.3) in a form which does not involve the areal coordinates
of W . For this we note that as
−(γ4′ − γ3′ ) = α′4 − α′3 + (β4′ − β3 ),
173
7.6. CLASSICAL GREEK PROPERTY OF CONICS
by (7.6.2) we have β4′ − β3′ = α4 − α′3 , from which it follows that γ4′ − γ3′ = −2(α′4 − α′3 ). Hence
the denominator is equal to
− 2(α′4 − α′3 )γ ′ − 2(α′4 − α′3 )α′ − 2(α′4 − α′3 )β ′
= − 2(α′4 − α′3 )(α′ + β ′ + γ ′ )
= − 2(α′4 − α′3 ).
It follows that
W Z 6 = t6 = −
1 γ ′2 − 4α′ β ′
.
2 −2(α′4 − α′3 )
We take W5 W6 to be a line which has W3 W4 as its corresponding diametral line, and with
|W5 , W6 | = 1. Then the line through W which is parallel to W5 W6 will meet the parabola in
distinct points Z8 and Z9 , and by (7.5.2) we have
W Z8 W Z9 =
γ ′2 − 4α′ β ′
.
(γ6′ − γ5′ )2 − 4(α′6 − α′5 )(β6′ − β5′ )
By division we then have
W Z8 W Z9 = k1 W Z6 ,
where
k1 =
(γ6′
−
γ5′ )2
−2(α′4 − α′3 )
.
− 4k(α′6 − α′5 )(β6′ − β5′ )
In this, W, W1 are variable points on the fixed line W3 W4 which is parallel to the diametral lines
of the parabola, and cutting the parabola in the point Z6 , while W W2 is a variable line, always
parallel to the fixed line W5 W6 , the diametral line of W5 W6 being W3 W4 , and cutting the parabola
in the variable points Z8 , Z9 , and the product of the pair of sensed distances is a constant k1 times
the first sensed distance.
W4
b
b
W3
b
b
W4
Z8
Z9
b
Z6
b
b
W3
W6
b
b
Z8
b
W6
b
W5
b
W5 W1
b
b
b
Z6
W
b
b
Z9
W
b
Figure 7.20.
But the mid-point of Z8 and Z9 is on W3 W4 and so W is the mid-point of Z8 and Z9 . Hence
we have that
|W, Z8 |2 = k1 Z6 W .
This is the ancient Greek characterization of a parabola.
174
CHAPTER 7. CONIC SECTIONS
7.7
An affine characterisation of conics
7.7.1
A re-working of the Greek properties
We now ask whether it is possible in §7.6.1 to express k1 Z6 W W Z7 , which is to be non-negative,
in the form
k1 Z6 W W Z7 = −k2 |W, F1 |2 + k3 |W, G1 |2 ,
for some positive constants k2 , k3 and some fixed points F1 , G1 ∈ W3 W4 and, separately, in §7.6.2
to express
k1 Z6 W = −k2 |W, F1 |2 + k3 |W, G1 |2 .
In the first of these, appropriate for a central conic, we take the centre W0 as the origin
of a coordinate system on Z6 Z7 , measuring sensed distance towards Z7 . Then we take Z6 ≡
(−d, 0), Z7 ≡ (d, 0) for some d > 0. We also take F1 ≡ (−p, 0), G1 ≡ (−q, 0) and will choose p, q to
be both positive if possible.
Then with W ≡ (u, 0), we need
−k2 (u + p)2 + k3 (u + q)2 = k1 (d2 − u2 ).
On equating coefficients of like powers of u, we need
−k2 + k3 = −k1 , −pk2 + qk3 = 0, −k2 p2 + k3 q 2 = k1 d.
On solving the first two of these equations we obtain
k2 =
k1 p
k1 q
, k3 =
,
q−p
q−p
and on inserting these into the third equation we have pq = d2 .
We are able to choose p and q to be both positive. When k1 > 0 we need q > p and use the
above formulae; when k1 < 0 we need p > q and use instead
k2 =
−k1 q
−k1 p
, k3 =
.
p−q
p−q
As W0 is the mid-point of Z6 and Z7 , F1 and G1 are on the one side of W0 and |W0 , F1 ||W0 , G1 | =
|W0 , Z6 |2 , we know that (Z6 , Z7 , F1 , G1 ) must be a harmonic range.
H1
b
b
b
b
G1
Z6
b
F1
b
b
b
b
W
b
b
Figure 7.21.
Z7
Z
7.7. AN AFFINE CHARACTERISATION OF CONICS
175
In the case of a parabola, we take Z6 ≡ (0, 0), F1 ≡ (−p, 0), G1 ≡ (−q, 0), and measure sensed
distance from Z6 towards F1 . With W ≡ (u, 0), we need
−k2 (u + p)2 + k3 (u + q)2 = k1 u,
and so on equating coefficients of like powers of u require that
−k2 + k3 = 0, −2pk2 + 2qk3 = k1 , −p2 k2 + q 2 k3 = 0.
From the first two of these we get
k2 =
k1
,
2q − 2p
k3 =
k1
.
2q − 2p
On inserting these into the third equation, we obtain
1
2 k1 (p
+ q) = 0.
Then on taking q > 0, we have p = −q < 0 and finally
k2 = k3 =
7.7.2
k1
.
4q
An affine characterization of conics
In §5.7.1 we have found another way of specifying conics, but we modify it slightly so as to get an
affine invariant form. This yields the following. Let F1 , G1 , H1 be non-collinear points and j1 , j2
positive real numbers. For a variable point Z, let W be the image of Z under projection onto
F1 G1 , parallel to G1 H1 , so that W ∈ F1 G1 with W = Z when Z ∈ F1 G1 and ZW k G1 H1 when
Z 6∈ F1 G1 . Then the locus of Z as it varies so that, when non-empty,
δF (G1 , F1 , Z)2 + j1 δF (F1 , W, H1 )2 = j2 δF (G1 , W, H1 )2 ,
is a conic.
To see this, we take G1 = Z1 , F1 = Z2 , H1 = Z3 as the vertices of our triangle of reference.
Then if Z has areal coordinates (α, β, γ), as Z3 Z1 has equation β ′ = 0, by §3.2.6 the line through
Z which is parallel to Z3 Z1 has equation
α′ + β ′ + γ ′
β′
=
.
β
α+β+γ
As the line Z1 Z2 has equation γ ′ = 0 we find that W has areal coordinates (γ + α, β, 0). Then
from §3.1.2 we obtain the equation
j1 α2 − j2 β 2 + (1 + j1 )γ 2 + 2j1 γα = 0,
for the conic.
As in §3.4.1 we find the discriminant of this to be
j1
0
j1
0
∆ = 0 −j2
j1
0
1 + j1
equal to
= −j1 j2 6= 0,
so that the conic is proper. The polar of Z4 has equation
j1 α4 α − j2 β4 β + (1 + j1 )γ4 )γ + j1 (γ4 α + α4 γ = 0,
which can be tidied up as
j1 (γ4 + α4 )α − j2 β4 β + [γ4 + j1 (γ4 + α4 )]γ = 0.
176
CHAPTER 7. CONIC SECTIONS
This fails to be an equation of a line when the coefficients are all equal, for which we need γ4 = 0
and hence
α4 + β4 = 1, (j1 − j2 )α4 = −j2 .
When j1 − j2 = 0 this implies that j2 = 0 which gives a contradiction. Thus in this case there is
no failure, every point has a polar, there is no centre and we are dealing with a parabola.
On the other hand when j1 − j2 6= 0 we have that
α4 =
j2
,
j2 − j1
β4 = −
j1
,
j2 − j1
and so there is a centre with coordinates
j1
j2
,−
,0 .
j2 − j1
j2 − j1
Going back we write our original equation as
j1 (γ + α)2 − j2 β 2 + γ 2 =0,
j1 (1 − β)2 − j2 β 2 + γ 2 =0,
(j1 − j2 )β 2 − 2j1 β + γ 2 + j1 =0,
2
j1 j2
j1 − j2 2
j1
γ =
.
+
β
2
j1 − j2
(j1 − j2 )
(j1 − j2 )2
When j1 > j2 we have a bounded locus and so an ellipse or a circle. When j1 < j2 we have an
unbounded locus and so a hyperbola.
Thus we find that it is an ellipse/circle, a parabola or hyperbola according as j1 − j2 is positive,
zero or negative.
This gives an affine invariant characterization of conics.
7.8
7.8.1
Axes; standard forms of equation
Perpendicular conjugate diametral lines
Areal coordinates in isolation do not suit purely metric concepts such as distance and perpendicularity, and we now revert to Cartesian coordinates to complete our introduction to conics. We
recall that in §7.7.2 we took non-collinear points F1 , G1 , H1 and positive real numbers j1 , j2 . For
a variable point Z, W was the image of Z under projection onto F1 G1 , parallel to G1 H1 , so that
W ∈ F1 G1 with W = Z when Z ∈ F1 G1 and ZW k G1 H1 when Z 6∈ F1 G1 . Then the locus of Z
as it varies so that
δF (G1 , F1 , Z)2 + j1 δF (F1 , W, H1 )2 = j2 δF (G1 , W, H1 )2 ,
is a conic. If we take
G1 ≡ (0, 0), F1 ≡ (a1 , 0), H1 ≡ (c1 , b1 ),
we obtain as equation for the locus
(j1 − j2 )b21 x2 + [a21 + (j1 − j2 )c21 ]y 2 − 2(j1 − j2 )b1 c1 xy − 2j1 a1 b21 x
+ 2j1 a1 b1 c1 y + j1 a21 b21 = 0.
For a point Z ′ ≡ (x′ , y ′ ), we take a line through Z ′ with parametric equations
x′′ = x′ + t(x1 − x0 ), y ′′ = y ′ + t(y1 − y0 ),
(7.8.1)
177
7.8. AXES; STANDARD FORMS OF EQUATION
so that as Z ′ varies we are considering parallel lines. This line meets the conic above in points for
which the parameters are given by the incidence equation
(j1 − j2 )b21 [x′ + t(x1 − x0 )]2 + [a21 + (j1 − j2 )c21 ][y ′ + t(y1 − y0 )]2
− 2(j1 − j2 )b1 c1 [x′ + t(x1 − x0 )][y ′ + t(y1 − y0 )] − 2j1 a1 b21 [x′ + t(x1 − x0 )]
+ 2j1 a1 b1 c1 [y ′ + t(y1 − y0 )] + j1 a21 b21
= 0.
We wish Z ′ to be the mid-point of these points of intersection, so its coordinates must satisfy the
equation
(j1 − j2 )b1 [b1 (x1 − x0 ) − c1 (y1 − y0 )]x′′
+ {−(j1 − j2 )b1 c1 (x1 − x0 ) + [a21 + (j1 − j2 )c21 ](y1 − y0 )}y ′′
− j1 a1 b21 (x1 − x0 ) + j1 a1 b1 c1 (y1 − y0 ) = 0,
which is the equation of a line.
The original line through Z ′ had equation
(y1 − y0 )(x′′ − x′ ) − (x1 − x0 )(y ′′ − y ′ ) = 0,
and so for the original line and its conjugate diametral line to be perpendicular it is necessary and
sufficient that
−(j1 −j2 )b1 c1 (y1 −y0 )2 +[(j1 −j2 )(b21 −c21 )−a21 ](y1 −y0 )(x1 −x0 )+(j1 −j2 )b1 c1 (x1 −x0 )2 = 0. (7.8.2)
If we divide across in this by (x1 − x0 )2 we have a quadratic polynomial in (y1 − y0 )/(x1 − x0 ),
and the discriminant is
[(j1 − j2 )(b21 − c21 ) − a21 ]2 + 4(j1 − j2 )2 b21 c21 .
As we are taking a1 > 0, b1 > 0, this discriminant can vanish only in the case when c1 =
0, (j1 − j2 )b21 = a21 . Otherwise, there are two real roots to the equation (7.8.2), in effect giving
two ratios (y1 − y0 )/(x1 − x0 ); as the product of the roots for these is −1, there is just one pair
of perpendicular lines involved. Thus, apart from the degenerate case, there is a unique pair of
diametral lines which are perpendicular to each other.
7.8.2
Standard forms of equation
To utilize the material in §7.8.1 in the simplest context, we take G1 F1 and G1 H1 parallel to the
particular pair of perpendicular lines identified, so that we have F1 G1 ⊥ G1 H1 . Then c1 = 0 in
§7.8.1 and we have a rectangular coordinate system. The case excluded there now corresponds to
(j1 − j2 )b21 6= a21 .
Continuing now with this specialized case, the equation (7.8.1) takes the form
1
a1 y 2 = 2b21 j1 (x − a1 ),
2
when j1 = j2 , and
(x −
j1 a1 2
j1 −j2 )
j1 j2 a21 /(j1
− j2
)2
+
y2
j1 j2 b21 /(j1
− j2 )
= 1.
otherwise.
As in §7.7.2 this is an ellipse/circle, a parabola or a hyperbola according as j1 − j2 is positive,
zero or negative. The excluded case is when (j1 − j2 )b21 = a21 and this corresponds to a circle.
178
CHAPTER 7. CONIC SECTIONS
Z
b
b
H1
b
b
b
b
b
b
G1 Z6 F1
b
b
W
Z7
b
Figure 7.22.
If we translate the origin of our frame of reference so as to replace (x − 12 a1 , y) by (x, y) in the
case of the parabola, we see that its equation has the form
y 2 = 4ax,
where a > 0.
a1
, y) by (x, y) in
If we translate the origin of our frame of reference so as to replace (x − jj11−j
2
the case of an ellipse or hyperbola, we see that an ellipse has an equation of the form
y2
x2
+ 2 = 1,
2
a
b
where a > b > 0, and that a hyperbola has an equation of the form
x2
y2
− 2 = 1,
2
a
b
where a > 0, b > 0.
These are the standard equations of an ellipse, hyperbola and parabola as used in a Euclidean
context. In particular they enable us to sketch the shapes of these conics by the ordinary methods
of calculus, as well as using the parametric equations in §7.1.6.
179
7.8. AXES; STANDARD FORMS OF EQUATION
7.8.3
Equation of a circle
Z2
b
b
Z3
Z
Z0
b
b
Figure 7.23
b
Z1
We should now like to make a direct connection between equations of conics in standard Cartesian form as in §7.8.2 and our two-term form in (7.1.7) for normalized areal coordinates. We start
with the case of a circle C with equation x2 + y 2 = a2 , where a > 0. We take a fixed number k1
with 0 < k1 < 1 and Z3 ≡ (− ka1 , 0). Then Z3 has as polar the line with equation − ka1 x + 0.y = a2 ,
p
2
whichis x = −ak1 . This meets the circle when x = −ak
1 , y = ±a 1 − k1 , and from this we take
p
p
Z1 ≡ −ak1 , −a 1 − k12 , Z2 ≡ −ak1 , a 1 − k12 . Then
−ak1
2δF (Z1 , Z2 , Z3 ) = −ak1
−a
k1
p
−ap 1 − k12
a 1 − k12
0
and for Z ≡ (x, y)
so that
2δF (Z, Z2 , Z3 ) = 2δF (Z, Z3 , Z1 ) = 2δF (Z, Z1 , Z2 ) = x
−ak1
− ka1
x
− ka1
−ak1
x
−ak1
−ak1
py
a 1 − k12
0
y
0
p
−a 1 − k12
py
−ap 1 − k12
a 1 − k12
1
1
1
1
1
1
2a2
=
(1 − k12 )3/2 ,
k1
q
q
= a 1 − k 2 [k1 x − 1 − k 2 y + a],
1
1
k1
1 q
q
a
1 =
1 − k12 [k1 x + 1 − k12 y + a],
k1
1 1 q
1 = −2a 1 − k12 [x + ak1 ],
1 p
p
k1 x + 1 − k12 y + a
2k1 x + 2ak12
k1 x − 1 − k12 y + a
,
β
=
,
γ
=
−
.
α=
2a(1 − k12 )
2a(1 − k12 )
2a(1 − k12 )
Note from this that
x=
1
1
[a(1 − k12 )(α + β) − a], y = p
a(1 − k12 )(β − α).
k1
1 − k12
180
CHAPTER 7. CONIC SECTIONS
Hence
x2 + y 2 − a2
1 = 2 [a(1 − k12 )(α + β) − a]2 + a2 k12 (1 − k12 )(β − α)2 − a2 k12
k1
a2 (1 − k12 ) =
(1 − k12 )(α + β)2 + k12 (β − α)2 − 2(α + β) + 1
2
k1
2
a (1 − k12 ) =
(α + β − 1)2 − 4k12 αβ
2
k1
2
a (1 − k12 ) 2
=
γ − 4k12 αβ ,
2
k1
and √
so Z is in C if and only if γ 2 − 4k12 αβ = 0. We obtain the exact form (7.1.7) if we now take
k = k1 and have 0 < k < 1.
7.8.4
Equations of ellipse and hyperbola
We start with the equation
(1 − e2 )x2 + y 2 − a2 (1 − e2 ) = 0,
for a conic C where either 0 < e < 1 so that we have an ellipse, or e > 1 so that we have a hyperbola.
We take Z3 ≡ (− ae , 0), the polar of which has equation (1 − e2 )(− ae )x + 0.y − a2 (1 − e2 ) = 0, that
is x = −ae. This line meets the conic when y = ±a(1 − e2 ), and so we take Z1 ≡ (−ae, −a(1 −
e2 )), Z2 ≡ (−ae, a(1 − e2 )). Then
and for Z ≡ (x, y)
−ae −a(1 − e2 ) 1 2a2
(1 − e2 )2 ,
2δF (Z1 , Z2 , Z3 ) = −ae a(1 − e2 ) 1 =
e
a
−
0
1 e
2δF (Z, Z2 , Z3 ) = 2δF (Z, Z3 , Z1 ) = 2δF (Z, Z1 , Z2 ) = From these it follows that
α=
x
y
1 a(1 − e2 )
−ae a(1 − e2 ) 1 =
(ex − y + a),
e
− ae
0
1 x
y
1 a(1 − e2 )
− ae
0
1 =
(ex + y + a),
e
−ae −a(1 − e2 ) 1 x
y
1 −ae −a(1 − e2 ) 1 = −2a(1 − e2 )(x + ae).
−ae a(1 − e2 ) 1 ex − y + a
ex + y + a
e(x + ae)
, β=
, γ=−
.
2
2
2a(1 − e )
2a(1 − e )
a(1 − e2 )
Hence
y = a(1 − e2 )(β − α),
x = −ae −
a(1 − e2 )
γ.
e
7.8. AXES; STANDARD FORMS OF EQUATION
181
It follows that
(1 − e2 )x2 + y 2 − a2 (1 − e2 )
2
1 − e2
=(1 − e2 )a2 e +
γ + a2 (1 − e2 )2 (β − α)2 − a2 (1 − e2 )
e
(1 − e2 )2 2
2
2
=a2 (1 − e2 ) e2 + 2(1 − e2 )γ +
γ
+
(1
−
e
)[(α
+
β)
−
4αβ]
−
1
e2
(1 − e2 )2 2
2
2
2
γ
+
(1
−
e
)(1
−
γ)
−
4(1
−
e
)αβ
=a2 (1 − e2 ) e2 − 1 + 2(1 − e2 )γ +
e2
(1 − e2 )2
2 2
2
=a2 (1 − e2 ) [
+
1
−
e
]γ
−
4(1
−
e
)αβ
e2
1 − e2
=a2 (1 − e2 )2 ( 2 + 1)γ 2 − 4αβ
e
2
2 2
a (1 − e ) 2
[γ − 4e2 αβ],
=
e2
and this is equal to 0 if and only if γ 2 − 4e2 αβ = 0. This has the form (7.1.7) with k = e2 so that
0 < k < 1 for an ellipse and k > 1 for a hyperbola, as expected.
7.8.5
Equation of parabola
We now start with the equation y 2 − 4ax = 0 for a parabola and take Z3 ≡ (−a, 0). The polar of
Z3 has equation y.0 − 2a(x + x3 ) = 0 and so x = −x3 = a. This line meets the parabola when
y = ±2a. Accordingly we take Z1 ≡ (a, −2a), Z2 ≡ (a, 2a). Then
a −2a 1 2a 1 = 8a2 ,
2δF (Z1 , Z2 , Z3 ) = a
−a
0
1 and for Z ≡ (x, y)
Then
2δF (Z, Z2 , Z3 ) = 2δF (Z, Z3 , Z1 ) = 2δF (Z, Z1 , Z2 ) = α=
y 1 2a 1 = 2a(x − y + a),
0 1 x
y
1 −a
0
1 = 2a(x + y + a),
a −2a 1 x
y
1 a −2a 1 = −4a(x − a).
a 2a 1 x
a
−a
x−y+a
x+y+a
a−x
, β=
, γ=
.
4a
4a
2a
From these we have that
x = a(1 − 2γ),
On making use of these we obtain
y = 2a(β − α).
y 2 − 4ax =4a2 (β − α)2 − (1 − 2γ)
=4a2 (α + β)2 − 4αβ − 1 + 2γ
=4a2 (1 − γ)2 − 4αβ − 1 + 2γ
=4a2 (γ 2 − 4αβ),
and so as expected the parabola has equation γ 2 − 4αβ = 0.
182
7.9
7.9.1
CHAPTER 7. CONIC SECTIONS
Images of conics under projective and affine transformations
Image of a conic under projective and affine transformations
We recall from §7.1.5, §7.2.1 and §7.2.3, that the points Z1 and Z2 lie on the conic with equation
(7.1.7), and the tangents to the conic at Z1 and Z2 meet at Z3 . Then Z1 , Z2 , Z3 are not collinear
as the tangent Z1 Z3 meets the conic only at Z1 . Now let Z4 be any point on the conic other than
Z1 and Z2 . Then Z1 , Z2 , Z4 are not collinear, as a line cannot meet the proper conic in more
than two points. For the same reason neither Z1 , Z3 , Z4 nor Z2 , Z3 , Z4 are collinear. Thus no
three of Z1 , Z2 , Z3 , Z4 are collinear. As Z4 is on the conic we must have
δF (Z4 , Z1 , Z2 )2
= 4k,
δF (Z4 , Z2 , Z3 )δF (Z4 , Z3 , Z1 )
and hence
δF (Z, Z1 , Z2 )2 δF (Z4 , Z2 , Z3 )δF (Z4 , Z3 , Z1 )
= 1.
δF (Z, Z2 , Z3 )δF (Z, Z3 , Z1 )δF (Z4 , Z1 , Z2 )2
As by §4.2.1 the left-hand side here is an equipoised quotient, this is in projectively invariant form.
Thus if W1 , W2 , W3 , W4 , W are the images of Z1 , Z2 , Z3 , Z4 , Z under a projective transformation,
we have that
δF (W, W1 , W2 )2 δF (W4 , W2 , W3 )δF (W4 , W3 , W1 )
= 1.
δF (W, W2 , W3 )δF (W, W3 , W1 )δF (W4 , W1 , W2 )2
Hence the image of a proper conic under a projective transformation is a proper conic, except that
as in §4.1.4 if the transformation is not affine, some line of the domain will have no image. Then
if this line has no point in common with the original conic, the image of the conic will be a conic;
if this line has one point in common with the original conic, then this point will have no image; if
this line has two points in common with the original conic, then these two will have no images.
Conversely, if W1 , W2 , W3 , W4 are points no three of which are collinear, then by §4.5.2 there
is a projective transformation which maps Z1 , Z2 , Z3 , Z4 to these, respectively. Then the proper
conic based on Z1 , Z2 , Z3 , Z4 as above is mapped onto the conic based similarly on W1 , W2 , W3 , W4 .
Hence, apart from points as noted in §4.1.4, any proper conic can be mapped projectively to any
other.
By a similar argument, if W1 , W2 , W3 are the images of Z1 , Z2 , Z3 under an affine transformation, then by §4.2.1 the equation
δF (Z, Z1 , Z2 )2
= 4k,
δF (Z, Z2 , Z3 )δF (Z, Z3 , Z1 )
converts to
δF (W, W1 , W2 )2
= 4k,
δF (W, W2 , W3 )δF (W, W3 , W1 )
and so to a proper conic based on W1 , W2 , W3 . As the type of this depends just on which of the
conditions for the values k above that it satisfies, the image conic must be of the same type as the
original.
Given any non-collinear points W1 , W2 , W3 , by §4.5.1 there is an affine transformation which
maps Z1 , Z2 , Z3 , respectively, to these. Hence, given any two proper conics of the same type, there
is an affine transformation which maps the first to the second.
7.9.2
Conics through four or five points
Suppose Z1 , Z2 , Z3 , Z4 , Z5 are five points, no three of which are collinear. We wish to examine if
a proper conic C passes through them. We take (Z1 , Z2 , Z3 ) as the triple of reference, so that by
7.9. IMAGES OF CONICS UNDER PROJECTIVE AND AFFINE TRANSFORMATIONS 183
(7.1.2) we are looking for f, g, h, none equal to 0, such that
f β4 γ4 + gγ4 α4 + hα4 β4
=
0,
f β5 γ5 + gγ5 α5 + hα5 β5
=
0.
We rewrite these equations as
f
β4 γ4 +
h
f
β5 γ5 +
h
g
γ4 α4 = − α4 β4 ,
h
g
γ5 α5 = − α5 β5 .
h
By Cramer’s rule we obtain the unique solutions
−α4 β4 γ4 α4 f −α5 β5 γ5 α5 = ,
h
β4 γ4 γ4 α4 β5 γ5 γ5 α5 β4 γ4 −α4 β4 g β5 γ5 −α5 β5 .
= h
β4 γ4 γ4 α4 β5 γ5 γ5 α5 We can write the common denominator in this as
γ4 γ5 [β4 α5 − α4 β5 ]
=γ4 γ5 [δF (Z4 , Z3 , Z1 )δF (Z5 , Z2 , Z3 ) − δF (Z5 , Z3 , Z1 )δF (Z4 , Z2 , Z3 )]
=γ4 γ5 δF [(Z4 , Z5 ), (Z3 , Z1 ), (Z2 , Z3 )]
=γ4 γ5 δF [(Z2 , Z3 ), (Z4 , Z5 ), (Z3 , Z1 )]
=γ4 γ5 δF (Z3 , Z5 , Z4 )δF (Z2 , Z3 , Z1 ).
By a similar calculation, the numerators in f /h and g/h can be shown to be, respectively,
α4 α5 δF (Z1 , Z5 , Z4 )δF (Z2 , Z3 , Z1 ),
β4 β5 δF (Z2 , Z5 , Z4 )δF (Z2 , Z3 , Z1 ).
From this we obtain the equation
α4 α5 δF (Z1 , Z5 , Z4 )βγ + β4 β5 δF (Z2 , Z5 , Z4 )γα + γ4 γ5 δF (Z3 , Z5 , Z4 )αβ = 0.
(7.9.1)
From this we obtain the equation
δF (Z4 , Z2 , Z3 )δF (Z5 , Z2 , Z3 )δF (Z1 , Z4 , Z5 )δF (Z, Z1 , Z2 )
δF (Z4 , Z1 , Z2 )δF (Z5 , Z1 , Z2 )δF (Z3 , Z4 , Z5 )δF (Z, Z2 , Z3 )
δF (Z4 , Z3 , Z1 )δF (Z5 , Z3 , Z1 )δF (Z2 , Z4 , Z5 )δF (Z, Z1 , Z2 )
+ 1 = 0.
+
δF (Z4 , Z1 , Z2 )δF (Z5 , Z1 , Z2 )δF (Z3 , Z4 , Z5 )δF (Z, Z3 , Z1 )
This proves that a unique proper conic passes through these points. Thus given any five points,
no three of which are collinear, a unique proper conic passes through them all.
Next we consider initially just four points Z1 , Z2 , Z3 , Z4 , no three of which are collinear. For
any constant k consider the locus with equation
δF (Z, Z1 , Z3 )δF (Z, Z2 , Z4 ) − kδF (Z, Z1 , Z4 )δF (Z, Z2 , Z3 ) = 0.
(7.9.2)
Clearly this is the equation of a conic which passes through all of these four points. The value of
k will determine whether such a conic is proper or not.
184
CHAPTER 7. CONIC SECTIONS
We wish to establish also a converse of this, namely that every proper conic through all of these
points has this form for some value of k. For this suppose that C is a proper conic which passes
through these four points. Let Z5 be any point on C other than these four points. Then no three
of Z1 , Z2 , Z3 , Z4 , Z5 are collinear and we can choose k so that (7.9.2) is satisfied by Z5 also, by
taking
δF (Z5 , Z1 , Z3 )δF (Z5 , Z2 , Z4 ) − kδF (Z5 , Z1 , Z4 )δF (Z5 , Z2 , Z3 ) = 0.
Now (7.9.2) is the equation of a proper conic which passes through these five points, and since
there is unique such proper conic, this must be the equation of C. Thus C has an equation of the
form (7.9.2).
We can also write (7.9.2) in the form
δF (Z, Z1 , Z3 )δF (Z, Z2 , Z4 )
= k,
δF (Z, Z1 , Z4 )δF (Z, Z2 , Z3 )
(7.9.3)
in the general situation, with the choice
k=
δF (Z5 , Z1 , Z3 )δF (Z5 , Z2 , Z4 )
,
δF (Z5 , Z1 , Z4 )δF (Z5 , Z2 , Z3 )
when including Z5 as well.
By §7.1.1, Example 2 in the case of a circle we have
k=±
|Z1 , Z3 ||Z2 , Z4 |
.
|Z1 , Z4 ||Z2 , Z3 |
The definition of a conic in §7.1.2 seems to give a special relationship to it to the two points
Z1 and Z4 on it. However (7.9.2) shows that, as regards this definition, a conic has the same
relationship to any pair of points on it.
Suppose now that C1 is a proper conic and f a projective transformation. Let C2 be the conic
obtained by completing f (C1 ). Take points Z1 and Z2 on C1 such that the tangents at these points
are not parallel to each other, and let Z3 be their point of intersection. On taking (Z1 , Z2 , Z3 ) as
our triple of reference, C1 will have an equation of the form γ 2 − 4kαβ = 0 as in (7.1.7) and its
exterior points will be those for which γ 2 − 4kαβ > 0 by §7.2.1. Take Z4 to be any other point on
γ2
C1 so that γ42 − 4kα4 β4 = 0 and then a general point Z is an exterior point if γ 2 > α44β4 αβ and
hence when γ 6= 0,
γ42 αβ
α4 β4 γ 2
< 1. This expands as
δF (Z4 , Z1 , Z2 )2 δF (Z, Z2 , Z3 )δF (Z, Z3 , Z1 )
< 1.
δF (Z4 , Z2 , Z3 )δF (Z4 , Z3 , Z1 )δF (Z, Z1 , Z2 )2
On the left-hand side of this is an equipoised quotient which is thus projectively invariant. We
deduce that the exterior region for C1 is mapped into the exterior region for C2 , with the possible
exception of those exterior points on Z1 Z2 . We can take similar points Z1′ , Z2′ on C1 with Z1 Z2 k
Z1′ Z2′ and apply this with the possible exception of exterior points on Z1′ Z2′ . Thus there are no
exceptional points.
It follows that if a tangent to C1 has an image under f then the line which is the completion of
that image is a tangent to C2 .
7.9.3
The Chasles-Steiner theorem
By §4.2.4 we can express the equation (7.9.3) in the form
crZ(Z1 , Z2 , Z3 , Z4 ) = k.
7.9. IMAGES OF CONICS UNDER PROJECTIVE AND AFFINE TRANSFORMATIONS 185
Z4
Z
b
Z1
b
b
b
Z2
b
Z3
Fig. 7.24
We now note that the result of (7.9.3) can be worded as follows. If Z is a variable point on a
proper conic which passes through the points Z1 , Z2 , Z3 , Z4 , no three of which are collinear, then
the pencil cross-ratio crZ(Z1 , Z2 , Z3 , Z4 ) is constant.
This is known as the Chasles-Steiner theorem, dating from c. 1830.
7.9.4
Image of a circle under some projective transformations
Consider the unit circle C1 with Cartesian equation x2 + y 2 − 1 = 0 and for fixed positive real
numbers p and q consider the transformation
x
,
x−p
x′ =
with inverse transformation
x=
px′
,
−1
x′
qy
,
x−p
(7.9.4)
p y′
.
q x′ − 1
(7.9.5)
y′ =
y=
Then the image of the unit circle has equation
(p2 − 1)x′2 + 2x′ − 1 +
p2 ′2
y = 0.
q2
(7.9.6)
We first suppose that p = 1 so that x − p = 0 is a tangent to the unit circle. Now the image
has equation
y ′2 = −2q 2 (x′ − 21 ).
As q varies we obtain parabolas of all magnitudes in this way.
When p 6= 1 (7.9.6) can be written as
x′ +
1
p2 −1
p2
(p2 −1)2
2
+
y ′2
q2
p2 −1
= 1.
(7.9.7)
Secondly we suppose that p > 1 so that the line x − p = 0 does not intersect the unit circle.
Then the image has equation
2
x′ + p21−1
y ′2
+
= 1,
(7.9.8)
a2
b2
186
CHAPTER 7. CONIC SECTIONS
where
a2 =
(p2
Then
p2
,
− 1)2
b2 =
q2
.
−1
p2
q 2 (p2 − 1)
b2
=
.
2
a
p2
We write
p
,
q = q1 p
2
p −1
where 0 < q1 < 1 so that we have 0 < b < a and
1−
b2
= 1 − q12 ,
a2
so that as p and q1 vary we obtain ellipses of all sizes and relevant eccentricities.
Thirdly we suppose that 0 < p < 1 so that the line x − p = 0 intersects the unit circle in two
points. Then the image has equation
2
1
x′ − 1−p
2
y ′2
−
= 1,
(7.9.9)
a2
b2
where
a2 =
Then
p2
,
(1 − p2 )2
1+
and if we take
b2 =
q2
.
1 − p2
q 2 (1 − p2 )
b2
=1+
,
2
a
p2
p
1 − p2
,
q1 =
p
q
as p and q1 vary we obtain hyperbolas of all sizes and relevant eccentricities.
Continuing with the hyperbola with equation (7.9.9), with fixed m′ and variable c′ we consider
the parallel pencil of lines with equations
y ′ = m′ x′ + c′ .
(7.9.10)
The diametral line in this pencil is the one with equation
1
y ′ = m′ x′ −
,
1 − p2
(7.9.11)
and as the conjugate with respect to the hyperbola
(y − y0 )2
(x − x0 )2
−
=1
a2
b2
2
of the diametral line y − y0 = m(x − x0 ) is the diametral line y − y0 = ab2 m (x − x0 ), the conjugate
diametral line with respect to our image hyperbola of (7.9.9) is the line with equation
1
q 2 /(1 − p2 )
′
x
−
,
y′ = 2 ′
p m /(1 − p2 )2
1 − p2
which simplifies to
−(1 − p2 )x′ +
p2 ′ ′
m y + 1 = 0.
q2
(7.9.12)
7.9. IMAGES OF CONICS UNDER PROJECTIVE AND AFFINE TRANSFORMATIONS 187
Now by (7.9.4) the lines (7.9.10) have as pre-images
the lines with equation qy = m′ x + c′ (x − p)
p ′
all of which pass through the point Z4 ≡ p, q m . We note that Z4 is on the line with equation
x = p, which has no image. Note too that Z4 has as polar with respect to the unit circle the line
with equation
p
px + m′ y = 1,
q
and by (7.9.5) the image of this polar is the line (7.9.12). Thus Z4 is a point on the line with no
image, and the pencil of lines concurrent at Z4 is mapped onto a pencil of parallel lines; furthermore,
the polar of Z4 with respect to the circle is mapped onto the conjugate diametral line for the hyperbola
which is associated with the parallel pencil.
In the opposite direction consider a parallel pencil of lines with equations
y = mx + c,
(7.9.13)
where m is fixed and c varies; the member of this pencil which passes through the centre of the
unit circle has equation y = mx, and the line through the centre which is perpendicular to this
1
line has equation y = − m
x. By (7.9.5) the image of this perpendicular diametral line is
my ′ = −qx′ .
(7.9.14)
By (7.9.5) the image of (7.9.13) has equation
p ′
y = mpx′ + c(x′ − 1),
q
all of which pass through the point Z5′ ≡ (1, mq). Since the polar of Z6 ≡ (x6 , y6 ) with respect to
the hyperbola
(x − x0 )2
y2
− 2 =1
2
a
b
has equation
y6
x2
x0 x6
x6 − x0
x − 2 y + 20 − 2 − 1 = 0,
2
a
b
a
a
the polar of Z5′ with respect to our hyperbola has equation
1
1
1
1 − 1−p
2
mq
(1−p2 )2 − 1−p2
′
′
x
−
− 1 = 0,
y
+
p2 /(1 − p2 )2
q 2 /(1 − p2 )
p2 /(1 − p2 )2
and this simplifies to (7.9.14). Thus Z5′ is a point on the line which has no pre-image and the
pencil of lines through Z5′ is the image of a parallel pencil of lines; furthermore the polar of Z5′ with
respect to this hyperbola is the image of the line which passes through the centre of the unit circle
and is perpendicular to the lines of the parallel pencil.
Exercises
p
5.1 Suppose that 0 < b < a and e = 1 − (b/a)2 . Let C1 ≡ (−ae, 0) and l be the line with equation x = −a/e. For any point Z on the circle C(O; k), let W be the foot of the perpendicular
from Z to l. Prove that
|Z, C1 |2
a2 − b 2
,
=
|Z, W |2
a2 cos2 θ + b2 sin2 θ
where
θ = ∡F OC1 Z.
188
CHAPTER 7. CONIC SECTIONS
5.2 Given fixed distinct points Z2 and Z3 , and a fixed line l with equation ax + by + c = 0, what
is the locus of points Z such that the mid-line of |Z2 ZZ3 is parallel to l?
[SUGGESTION. Let W ≡ (x + b, y − a) so that ZW k l and
Z2 z2 , Z3 z3 , Z z, W z + b − ai.
Then
z3 − z
(z + b − ai) − z
(z + b − ai) − z
z2 − z
=
=
|Z, Z3 | i∡F W ZZ3
,
e
|Z, W |
|Z, W | i∡F Z2 ZW
e
,
|Z, Z2 |
and ZW is the mid-line of |Z2 ZZ3 if and only if ∡F W ZZ3 and ∡F Z2 ZW are equal in
magnitude. Show that this is the case only if
(z3 − z)(z2 − z)
= j,
(b − ai)2
for some j ∈ R. By eliminating j, show that this is equivalent to
2ab[(x3 − x)(x2 − x) − (y3 − y)(y2 − y)] + (b2 − a2 )[(x3 − x)(y2 − y) + (y3 − y)(x2 − x)] = 0.
Show that this is a hyperbola, and that its asymptotes are parallel to the lines with equations
ax + by = 0, ay − bx = 0.]
As a further exercise, find for what Z the mid-line of |Z2 ZZ3 is l.
5.3 Let Z1 , Z2 , Z3 , Z4 , Z5 be five points, no three of which are collinear. Prove that a unique
proper conic passes through these points.
[SUGGESTION. Take (Z1 , Z2 , Z3 ) as the triple of reference, so that we are looking for f, g, h,
none equal to 0, such that
f β4 γ4 + gγ4 α4 + hα4 β4
f β5 γ5 + gγ5 α5 + hα5 β5
= 0,
= 0.
From this obtain the equation
δF (Z4 , Z2 , Z3 )δF (Z5 , Z2 , Z3 )δF (Z1 , Z4 , Z5 )δF (Z, Z1 , Z2 )
δF (Z4 , Z1 , Z2 )δF (Z5 , Z1 , Z2 )δF (Z3 , Z4 , Z5 )δF (Z, Z2 , Z3 )
δF (Z4 , Z3 , Z1 )δF (Z5 , Z3 , Z1 )δF (Z2 , Z4 , Z5 )δF (Z, Z1 , Z2 )
+ 1 = 0.
+
δF (Z4 , Z1 , Z2 )δF (Z5 , Z1 , Z2 )δF (Z3 , Z4 , Z5 )δF (Z, Z3 , Z1 )
Chapter 8
Dual conics
8.1
8.1.1
Point-pair and rotor conics
Dual of definition of a conic
Dually to §7.1.2, we parametrise the points Pλ on a line Z1 Z2 by taking Pλ to be the point of
1
λ
intersection of the line Z1 Z2 and the line joining Z3 to 1+λ
Z4 + 1+λ
Z5 . Similarly we parametrise
the points Qµ on a line Z6 Z7 by taking Qµ to be the point of intersection of the line Z6 Z7 and the
µ
1
Z9 + 1+µ
Z10 . We suppose that λ and µ correspond in a projectivity
line joining Z8 to 1+µ
a1 λµ + b1 λ + c1 µ + d1 = 0.
Z4
b
Z1
b
b
Z5
Z9
b
b
Z7
b
b
b
Z3
b
Z2
b
b
b
Z10
U
b
Z6
b
Z8
b
b
V
Figure 8.1.
1
If U and V lie on the line of the corresponding points Pλ and Qµ , then since Z1 Z2 , Z3 ( 1+λ
Z4 +
λ
1+λ Z5 ) and U V are concurrent, we have that
1
λ
= 0.
δF (U, V ), (Z1 , Z2 ), Z3 ,
Z4 +
Z5
1+λ
1+λ
µ
1
Similarly, since Z6 Z7 , Z8 ( 1+µ
Z9 + 1+µ
Z10 ) and U V are concurrent, we have that
1
µ
= 0.
δF (U, V ), (Z6 , Z7 ), Z8 ,
Z9 +
Z10
1+µ
1+µ
By §1.1.3, from the first of these we have
1
1
λ
λ
δF (U, Z1 , Z2 )δF V, Z3 ,
Z4 +
Z5 − δF (V, Z1 , Z2 )δF U, Z3 ,
Z4 +
Z5 = 0,
1+λ
1+λ
1+λ
1+λ
189
190
CHAPTER 8. DUAL CONICS
from which we obtain that
λ=
δF [(U, V ), (Z1 , Z2 ), (Z3 , Z4 )]
.
δF [(U, V ), (Z3 , Z5 ), (Z1 , Z2 )]
Similarly, from the second of these we have
µ=
δF [(U, V ), (Z6 , Z7 ), (Z8 , Z9 )]
.
δF [(U, V ), (Z8 , Z10 ), (Z6 , Z7 )]
On inserting these values, we obtain
a1 δF [(U, V ), (Z1 , Z2 ), (Z3 , Z4 )]δF [(U, V ), (Z6 , Z7 ), (Z8 , Z9 )]
+ b1 δF [(U, V ), (Z1 , Z2 ), (Z3 , Z4 )]δF [(U, V ), (Z8 , Z10 ), (Z6 , Z7 )]
+ c1 δ[(U, V ), (Z3 , Z5 ), (Z1 , Z2 )]δF [(U, V ), (Z6 , Z7 ), (Z8 , Z9 )]
+ d1 δF [(U, V ), (Z3 , Z5 ), (Z1 , Z2 )]δF [(U, V ), (Z8 , Z10 ), (Z6 , Z7 )]
=0.
(8.1.1)
For pairs (U, V ) of distinct points, this then is the equation of a dual of a conic; we refer to it as
a point-pair conic. It contains the pairs (Z1 , Z2 ) and (Z6 , Z7 ).
Suppose however that hU ′ , V ′ i = hU, V i where the pair (U, V ) satisfies (8.1.1). Then δF (Z, U ′ , V ′ ) =
δF (Z, U, V ) for all points Z, and so the pair (U ′ , V ′ ) also satisfies (8.1.1).
More generally, suppose that hU ′ , V ′ i = jhU, V i for some j 6= 0 in R. Then δF (Z, U ′ , V ′ ) =
jδF (Z, U, V ) for all points Z, and so the pair (U ′ , V ′ ) also satisfies (8.1.1).
Thus our dual of a conic with equation (8.1.1) relates to rotors as much as to pairs of points.
8.1.2
Point-pair conic containing pairs of vertices of triple of reference
λ
1
Z2 + 1+λ
Z3
As in §7.1.4, we obtain a simpler version of the material in §8.1.1 if we take Pλ = 1+λ
µ
1
and require that U V, Z2 Z3 , Z1 Pλ be concurrent, and also take Qµ = 1+µ Z3 + 1+µ Z1 and require
that U V, Z3 Z1 , Z2 Qµ be concurrent.
Now our equation for the point-pair conic turns out to be
a1 δF [(U, V ), (Z2 , Z3 ), (Z1 , Z2 )]δF [(U, V ), (Z3 , Z1 ), (Z2 , Z3 ), ]
+ b1 δF [(U, V ), (Z2 , Z3 ), (Z1 , Z2 )]δF [(U, V ), (Z3 , Z1 ), (Z1 , Z2 )]
+ c1 δF [(U, V ), (Z3 , Z1 ), (Z2 , Z3 )]δF [(U, V ), (Z2 , Z3 ), (Z3 , Z1 )]
=
+ d1 δF [(U, V ), (Z2 , Z3 ), (Z3 , Z1 )]δF [(U, V ), (Z3 , Z1 ), (Z1 , Z2 )]
0.
This set of pairs contains (Z2 , Z3 ) and (Z3 , Z1 ); if we also make it contain (Z1 , Z2 ) we will have
c1 = 0, and so our equation will have the form
f δF [(U, V ), (Z2 , Z3 ), (Z1 , Z2 )]δF [(U, V ), (Z3 , Z1 ), (Z2 , Z3 ), ]
+ gδF [(U, V ), (Z2 , Z3 ), (Z1 , Z2 )]δF [(U, V ), (Z3 , Z1 ), (Z1 , Z2 )]
=
8.1.3
+ hδF [(U, V ), (Z2 , Z3 ), (Z3 , Z1 )]δF [(U, V ), (Z3 , Z1 ), (Z1 , Z2 )]
0.
Two-term form of equation
We modify the material in §8.1.1 somewhat by taking
Pλ =
λ
1
Z2 +
Z3 ,
1+λ
1+λ
Qµ =
1
µ
Z5 +
Z6 ,
1+µ
1+µ
191
8.1. POINT-PAIR AND ROTOR CONICS
and supposing that U V, Z2 Z3 , Z1 Pλ are concurrent, and that U V, Z5 Z6 , Z4 Qµ are concurrent. Then
we have that
λ=
δF [(U, V ), (Z2 , Z3 ), (Z1 , Z2 )]
δF [(Z2 , Z3 ), (Z1 , Z2 ), (U, V )]
δF (Z2 , U, V )
=
=−
,
δF [(U, V ), (Z2 , Z3 ), (Z3 , Z1 )]
δF [(Z2 , Z3 ), (Z3 , Z1 ), (U, V )]
δF (Z3 , U, V )
and similarly
µ=
δF [(Z5 , Z6 ), (Z4 , Z5 ), (U, V )]
δF (Z5 , U, V )
δF [(U, V ), (Z5 , Z6 ), (Z4 , Z5 )]
=
=−
.
δF [(U, V ), (Z5 , Z6 ), (Z6 , Z4 )]
δF [(Z5 , Z6 ), (Z6 , Z4 ), (U, V )]
δF (Z6 , U, V )
For a projectivity a1 λµ + b1 λ + c1 µ + d1 = 0 this yields the equation
a1 δF (Z2 , U, V )δF (Z5 , U, V ) − b1 δF (Z2 , U, V )δF (Z6 , U, V )
−c1 δF (Z3 , U, V )δF (Z5 , U, V ) + d1 δF (Z3 , U, V )δF (Z6 , U, V ) = 0.
(8.1.2)
Now (U, V ) = (Z2 , Z3 ) clearly satisfies this. We also wish to have (U, V ) = (Z3 , Z1 ) satisfy it;
this requires that
a1 δF (Z5 , Z3 , Z1 ) − b1 δF (Z6 , Z3 , Z1 ) = 0,
and we may take
a1 = δF (Z6 , Z3 , Z1 ),
b1 = δF (Z5 , Z3 , Z1 ),
as the original coefficients may be re-normalized. Then (8.1.2) becomes
δF (Z6 , Z3 , Z1 )δF (Z2 , U, V ) − c1 δF (Z3 , U, V ) δF (Z5 , U, V )
+ −δF (Z5 , Z3 , Z1 )δF (Z2 , U, V ) + d1 δF (Z3 , U, V ) δF (Z6 , U, V )
=0.
On multiplying across by δF (Z1 , Z2 , Z3 ) and inserting expansions of δF (Z5 , U, V ) and δF (Z6 , U, V ),
we obtain from this that
δF (Z6 , Z3 , Z1 )δF (Z2 , U, V ) − c1 δF (Z3 , U, V ) •
δF (Z5 , Z2 , Z3 )δF (Z1 , U, V ) + δF (Z5 , Z3 , Z1 )δF (Z2 , U, V ) + δF (Z5 , Z1 , Z2 )δF (Z3 , U, V )
+ −δF (Z5 , Z3 , Z1 )δF (Z2 , U, V ) + d1 δF (Z3 , U, V ) •
δF (Z6 , Z2 , Z3 )δF (Z1 , U, V ) + δF (Z6 , Z3 , Z1 )δF (Z2 , U, V ) + δF (Z6 , Z1 , Z2 )δF (Z3 , U, V )
=0.
(8.1.3)
In (8.1.3) we wish to retain just the terms in δF (Z3 , U, V )2 and δF (Z1 , U, V )δF (Z2 , U, V ), so as
to have
−c1 δF (Z5 , Z1 , Z2 ) + d1 δF (Z6 , Z1 , Z2 ) δF (Z3 , U, V )2
+ δF (Z5 , Z2 , Z3 )δF (Z6 , Z3 , Z1 ) − δF (Z6 , Z2 , Z3 )δF (Z5 , Z3 , Z1 ) δF (Z1 , U, V )δF (Z2 , U, V )
=0.
(8.1.4)
There is no term in δF (Z1 , U, V )2 ; to make the term in δF (Z2 , U, V )2 vanish we need
δF (Z5 , Z3 , Z1 )δF (Z6 , Z3 , Z1 ) − δF (Z6 , Z3 , Z1 )δF (Z5 , Z3 , Z1 ) = 0,
which is automatically satisfied; to make the term in δF (Z2 , U, V )δF (Z3 , U, V ) vanish we need
δF (Z6 , Z3 , Z1 )δF (Z5 , Z1 , Z2 ) − c1 δF (Z5 , Z3 , Z1 )
−δF (Z5 , Z3 , Z1 )δF (Z6 , Z1 , Z2 ) + d1 δF (Z6 , Z3 , Z1 ) = 0;
to make the term in δF (Z3 , U, V )δF (Z1 , U, V ) vanish we need
−c1 δF (Z5 , Z2 , Z3 ) + d1 δF (Z6 , Z2 , Z3 ) = 0.
(8.1.5)
192
CHAPTER 8. DUAL CONICS
In (8.1.5) we take to the right-hand side the terms independent of c1 and d1 and obtain there
δF [(Z5 , Z6 ), (Z3 , Z1 ), (Z1 , Z2 )] = δF [(Z3 , Z1 ), (Z1 , Z2 ), (Z5 , Z6 )] = δF (Z3 , Z1 , Z2 )δF (Z1 , Z5 , Z6 ),
and the matrix of coefficients of c1 and d1 has determinant equal to
δF [(Z5 , Z6 ), (Z2 , Z3 ), (Z3 , Z1 )] = δF [(Z2 , Z3 ), (Z3 , Z1 ), (Z5 , Z6 )] = δF (Z2 , Z3 , Z1 )δF (Z3 , Z5 , Z6 ).
We then obtain the solutions
c1 =
δF (Z1 , Z5 , Z6 )
δF (Z1 , Z5 , Z6 )
δF (Z6 , Z2 , Z3 ), d1 =
δF (Z5 , Z2 , Z3 ).
δF (Z3 , Z5 , Z6 )
δF (Z3 , Z5 , Z6 )
On inserting these, (8.1.4) becomes successively
δF (Z1 , Z5 , Z6 ) δF (Z5 , Z1 , Z2 )δF (Z6 , Z2 , Z3 ) − δF (Z6 , Z1 , Z2 )δF (Z5 , Z2 , Z3 ) δF (Z3 , U, V )2
δF (Z3 , Z5 , Z6 )
+δF [(Z5 , Z6 ), (Z2 , Z3 ), (Z3 , Z1 )]δF (Z1 , U, V )δF (Z2 , U, V ) = 0,
−
δF (Z1 , Z5 , Z6 )
δF [(Z5 , Z6 ), (Z1 , Z2 ), (Z2 , Z3 )]δF (Z3 , U, V )2
δF (Z3 , Z5 , Z6 )
+δF [(Z5 , Z6 ), (Z2 , Z3 ), (Z3 , Z1 )]δF (Z1 , U, V )δF (Z2 , U, V ) = 0,
−
δF (Z1 , Z5 , Z6 )
δF [(Z1 , Z2 ), (Z2 , Z3 ), (Z5 , Z6 )]δF (Z3 , U, V )2
δF (Z3 , Z5 , Z6 )
+δF [(Z2 , Z3 ), (Z3 , Z1 ), (Z5 , Z6 )]δF (Z1 , U, V )δF (Z2 , U, V ) = 0,
−
and this simplifies to
δF (Z3 , U, V )2 −
δF (Z3 , Z5 , Z6 )2
δF (Z1 , U, V )δF (Z2 , U, V ) = 0.
δF (Z1 , Z5 , Z6 )δF (Z2 , Z5 , Z6 )
(8.1.6)
This has the form
n2 − 4k1 lm = 0,
(8.1.7)
in terms of pair-coordinates.
8.1.4
Equation of point-pairs in a tangent line to a conic
Our initial attention was focused on conics, and more recently it was on point-pair conics. Now
we note an intimate connection between the two. For if we write
4k1 =
1
,
k
we see that the equation kn2 − lm = 0 in (7.3.9) for point-pairs in the tangents to the conic with
equation γ 2 − 4kαβ = 0, and in either asymptote in the case of a hyperbola, is the equation (8.1.7)
just derived. This is a very important feature, an aspect of duality and we consider alternative
derivations of it.
Generalizing from §6.6.3, we recall from (7.2.18) that the tangent to the conic C1 with equation
γ ′2 − 4kα′ β ′ = 0 at the point Z ′ with normalized areal coordinates (α′ , β ′ , γ ′ ) has equation γ ′ γ =
2k(α′ β + β ′ α) so that
γ ′ α − α′ γ = γ ′ α −
Similarly
β′γ − γ ′β =
2kα′ ′
2kα′ ′
′
(α
β
+
β
α)
=
−
(α β − β ′ α).
γ′
γ′
2kβ ′ ′
2kβ ′ ′
(α β + β ′ α) − γ ′ β =
(α β − β ′ α).
′
γ
γ′
193
8.1. POINT-PAIR AND ROTOR CONICS
Moreover
γ ′2 γ 2 = 4k 2 (α′ β + β ′ α)2 , 4kα′ β ′ γ 2 = 4k 2 (α′ β + β ′ α)2 ,
so that
k(α′ β + β ′ α)2 =α′ β ′ γ 2 ,
k(α′ β − β ′ α)2 =α′ β ′ γ 2 − 4kα′ β ′ αβ = α′ β ′ (γ 2 − 4kαβ),
(α′ β − β ′ α)2 =
γ ′2
α′ β ′ 2
(γ − 4kαβ) = 2 (γ 2 − 4kαβ).
k
4k
We introduce a parameter s by taking γ 2 − 4kαβ = 4k 2 s2 , and then α′ β − β ′ α = ±γ ′ s. Without
loss of generality we take the + sign so that we have
α′ β − β ′ α = γ ′ s, β ′ γ − γ ′ β = −2kβ ′ s, γ ′ α − α′ γ = −2kα′ s.
(8.1.8)
From these we have that
l = δF (Z1 , Z ′ , Z) = −kβ ′ s, m = δF (Z2 , Z ′ , Z) = −kα′ s, n = δF (Z3 , Z ′ , Z) = 12 γ ′ s.
(8.1.9)
It follows from these that point-pairs in the tangents to the conic C1 satisfy equation (7.3.9).
For a less geometrical and more algebraic approach with a more general formula, we note that
the equation for a conic obtained in 5.1.2 is not in the most general form as Z1 and Z4 must be
on the locus. If we change the triangle of reference it will be found that the most general form is
aα2 + bβ 2 + cγ 2 + 2f βγ + 2gγα + 2hαβ = 0.
This has as tangent at the point W0 ≡ (α0 , β0 , γ0 ) on it, the line with equation
(aα0 + hβ0 + gγ0 )α + (hα0 + bβ0 + f γ0 )β + (gα0 + f β0 + cγ0 )γ = 0.
The line with equation
lα + mβ + nγ = 0
is a tangent if it has this form for some W0 . For this we need
aα0 + hβ0 + gγ0 − jl
hα0 + bβ0 + f γ0 − jm
= 0,
= 0,
lα0 + mβ0 + nγ0
= 0,
gα0 + f β0 + cγ0 − jn
= 0,
for some j 6= 0. The necessary condition for this is that
a h g l h b f m g f c n = 0.
l m n 0 This expands in the form
Al2 + Bm2 + Cn2 + 2F mn + 2Gnl + 2Hlm = 0,
where the capital letters denote the co-factors, that is appropriately signed minors, of the elements
in the determinant
a h g h b f ,
g f c 194
CHAPTER 8. DUAL CONICS
with the corresponding lower case letters, so that
h b h f b f 2
= f h − bg,
= f g − ch, G = = bc − f , H = − A =
g f g c f c a g a h a h 2
= ab − h2 .
B =
= ac − g , F = − = gh − af, C = g c g f h b Figure 4.19.
We refer to the point-pair conic with equation (7.3.9) as the dual of the point conic with
equation (7.1.7).
8.1.5
Sketching point-pair conics
By the general result in §8.1.4, the point conic in §7.7.2 with equation
j1 α2 − j2 β 2 + (1 + j1 )γ 2 + 2j1 γα = 0
has as dual the point-pair with equation
−j2 (1 + j1 )l2 + j1 m2 − j1 j2 n2 − 2j1 j2 nl = 0.
(8.1.10)
We recall that in §7.8.1??? we had
Z1 ≡ (0, 0),
Z2 ≡ (a1 , 0),
Z3 ≡ (0, b1 ).
We consider the pair (Z4 , Z5 ) where Z4 and Z5 have areal coordinates (1 − β4 , β4 , 0) and
(1 − γ5 , 0, γ5 ). Then for these we have that
l4,5 = δ1 β4 γ5 ,
m4,5 = −δ1 (1 − β4 )γ5 ,
n4,5 = −δ1 β4 (1 − γ5 ).
Then for (Z4 , Z5 ) to be in the dual conic we need
−j2 β42 γ52 + j1 (1 − β4 )2 γ52 − j1 j2 β42 (1 − γ5 )2 + 2j1 j2 β42 (1 − γ5 )γ5 = 0,
which can be re-written as
[−4j1 j2 β42 + (j1 − j2 )β42 − 2j1 β4 + j1 ]γ52 + 4j1 j2 β42 γ5 − j1 j2 β42 = 0.
This connects the areal coordinates of the point Z4 on Z1 Z2 and the point Z5 on Z1 Z3 where Z1 Z2
are in the conic envelope. Moreover if Z4 and Z5 have Cartesian coordinates (x4 , 0) and (0, y5 ),
respectively, we find that
x4
y5
β4 =
, γ5 = .
a1
b1
8.2. INCIDENCE OF A PENCIL AND A POINT-PAIR CONIC
195
On inserting these in the last equation we obtain
−4j1 j2
y2
x4
x2 y5
x2
x2
x24
+ (j1 − j2 ) 24 − 2j1
+ j1 25 + 4j1 j2 24
− j1 j2 24 = 0.
2
a1
a1
a1
b1
a1 b 1
a1
(8.1.11)
Depending on the positivity of the discriminant of this, for appropriate values of x4 we have
two corresponding values of y5 and so have two lines Z4 Z5′ , Z4 Z5′′ in the conic envelope. With
the selections of j1 , j2 , a1 , b1 in §7.8.2 we have the point conics in standard position (apart from a
translation along the x-axis for an ellipse or hyperbola), and (8.1.11) enables the tangents in their
duals to be identified from their intercepts on the two axes. In this way we encounter all the types
of proper point-pair conics.
8.2
Incidence of a pencil and a point-pair conic
8.2.1
Intersection of a pencil and a point-pair conic; interior and exterior
regions
Generalizing from §7.2.1, consider the intersection of kn2 − lm = 0 with lα + mβ + nγ = 0 where
l6 α + m6 β + n6 γ = 0 and the case α + β + γ = 0 is excluded. Suppose that n6 6= 0. Then
n6 (lα + mβ) = −n6 nγ = n(l6 α + m6 β),
and on substituting this into the equation of the point-pair conic, we obtain
kn26 (lα + mβ)2 = (l6 α + m6 β)2 kn2 =(l6 α + m6 β)2 lm,
kn26 α2 l2 + [2kn26 αβ − (l6 α + m6 β)2 ]lm + kn26 β 2 m2 =0.
(8.2.1)
The discriminant of this quadratic equation (8.2.1) is
[2kn26 αβ − (l6 α + m6 β)2 ]2 − 4k 2 n46 α2 β 2 = (l6 α + m6 β)2 [(l6 α + m6 β)2 − 4kn26 αβ]
(8.2.2)
which is a positive multiple of
l62 α2 + 2(l6 m6 − 2kn26 )αβ + m26 β 2 .
(8.2.3)
The discriminant of (8.2.3) is in turn
1
(l6 m6 − 2kn26 )2 − l62 m26 = 4k 2 n26 n26 − l6 m6 .
k
(8.2.4)
For pairs (Z4 , Z5 ), with dual coordinates (l6 , m6 , n6 ), which are not in the dual point-pair conic
we must have either n26 − k1 l6 m6 > 0 or n26 − k1 l6 m6 < 0. When n26 − k1 l6 m6 < 0, (8.2.2) will be
positive for all (α, β) other than (0, 0) and so (8.2.1) will have a real solution in l and m for all such
α and β. Thus every point Z on the line Z4 Z5 will be the vertex of pair in the point-pair conic
and so there will be a pair of tangents/asymptote from Z to the point conic. Every such point Z
will have to be an exterior point for the point conic C; thus the line Z4 Z5 must have no points in
common with the point conic C. We call any such pair (Z4 , Z5 ) interior for the point-pair conic.
When n26 − k1 l6 m6 > 0, (8.2.2) will be positive for some (α, β) other than (0, 0) and negative for
others. Then (8.2.1) will have a real solution in l and m for some such α and β and not for others.
Thus some points Z on the line Z4 Z5 will be the vertex of pair in the point-pair conic and so there
will be a pair of tangents/asymptote from Z to the point conic, while other points Z on Z4 Z5 will
not have this property. Some such points Z will have to be exterior points for the point conic C
while others will be interior points. Thus the line Z4 Z5 must cut the point conic C and contain
interior and exterior points for it. We call any such pair (Z4 , Z5 ) exterior for the point-pair conic.
196
CHAPTER 8. DUAL CONICS
8.2.2
Equation of incidence
We recall the equation (8.1.7) for a point-pair conic,
n2 − 4k1 lm = 0,
and, as in §5.1.6, consider a pencil of point-pairs
(U, V ) = (1 − s)(Z4 , Z5 ) + s(Z7 , Z8 ),
(s ∈ R),
satisfying, as in (5.1.15), one of the conditions
Z4 = Z7 , Z5 = Z8 , Z4 Z7 k Z5 Z8 ,
so that (Z4 , Z5 ), (Z7 , Z8 ) is an amenable pair of pairs of points. Then as in §5.1.6, with
l6 =δF (Z1 , Z4 , Z5 ), m6 = δF (Z2 , Z4 , Z5 ), n6 = δF (Z3 , Z4 , Z5 ),
l9 =δF (Z1 , Z7 , Z8 ), m9 = δF (Z2 , Z7 , Z8 ), n9 = δF (Z3 , Z7 , Z8 ),
l =δF (Z1 , U, V ), m = δF (Z2 , U, V ), n = δF (Z3 , U, V ),
by (5.1.16) we have parametric equations
l = (1 − s)l6 + sl9 , m = (1 − s)m6 + sm9 , n = (1 − s)n6 + sn9 .
On inserting this into the equation (8.1.7), so as to find what pairs of the pencil are in the point-pair
conic, we obtain the equation of incidence
2
(1 − s)n6 + sn9 − 4k1 (1 − s)l6 + sl9 (1 − s)m6 + sm9 = 0,
which we re-write as
(n9 − n6 )2 − 4k1 (l9 − l6 )(m9 − m6 ) s2
+2 n6 (n9 − n6 ) − 2k1 l6 (m9 − m6 ) + m6 (l9 − l6 ) s
+n26 − 4k1 l6 m6 = 0.
(8.2.5)
If hZ7 , Z8 i − hZ4 , Z5 i = hZ10 , Z11 i, then
δF (Z, Z7 , Z8 ) − δF (Z, Z4 , Z5 ) = δF (Z, Z10 , Z11 ),
for all Z, and so
l9 − l6 = δF (Z1 , Z10 , Z11 ), m9 − m6 = δF (Z2 , Z10 , Z11 ), n9 − n6 = δF (Z3 , Z10 , Z11 ).
(8.2.6)
If we denote these by l12 , m12 , n12 , respectively, then the coefficient of s2 in (8.2.5) is equal to
n212 − 4k1 l12 m12 . We say that we have deficient incidence if this is equal to 0, and by (8.1.8)
we see that this happens if and only if the pair (Z10 , Z11 ) is in the point-pair conic. This is our
condition for deficient incidence.
Suppose now that the pair (Z4 , Z5 ) is in the point-pair conic, so that the constant term in
(8.2.5) is equal to 0. Then we say that a point Z13 is a contact if it is the vertex of a concurrent
pencil containing (Z4 , Z5 ) and for this pencil the equation of incidence is equivalent to s2 = 0, so
that we have not got deficient incidence and the coefficient of s is equal to 0, that is
n6 (n9 − n6 ) − 2k1 l6 (m9 − m6 ) + m6 (l9 − l6 ) = 0.
If we use the fact that n26 − 4k1 l6 m6 = 0 and replace the fixed (Z7 , Z8 ) by the variable (U, V ) this
becomes
n6 n − 2k1 (m6 l + l6 m) = 0.
(8.2.7)
8.2. INCIDENCE OF A PENCIL AND A POINT-PAIR CONIC
197
A contact can arise as follows. As (Z4 , Z5 ) is in the point-pair conic, the line Z4 Z5 is tangent
to the point conic at some point, Z13 say. Let Z7 Z8 pass through Z13 but not coincide with Z4 Z5 .
Then for all s ∈ R, (1 − s)hZ4 , Z5 i + shZ7 , Z8 i passes through Z13 and the only tangent among
them, Z4 Z5 , corresponding to s = 0. Now let hZ7 , Z8 i − hZ4 , Z5 i = hZ10 , Z11 i where without loss
of generality we may take Z10 = Z13 . Then Z13 Z11 clearly does not lie along Z13 Z5 and so is not
a tangent. Thus there is not deficiency for this pencil and Z13 is a contact.
Conversely, when n6 − 2k1 (l6 + m6 ) 6= 0, (8.2.7) will be the equation of a point Z13 , the
normalized areal coordinates of which will satisfy
α13 = −2k2 k1 m6 , β13 = −2k2 k1 l6 , γ13 = k2 n6 ,
where k2 is chosen so that the sum of these is equal to 1. As
α13 l6 + β13 m6 + γ13 n6 = k2 (n26 − 4k1 l6 m6 ) = 0,
Z13 is on the line Z4 Z5 as expected. We also note that
2
γ13
−
1
1
α13 β13 = k22 n26 − 4k12 l6 m6 = 0.
k1
k1
With 1/k1 = 4k this shows that Z13 is on the point conic with equation γ 2 − 4kαβ = 0, and in
fact the line Z4 Z5 is the tangent to this conic at the point Z13 . The remainder must then turn out
as in the last paragraph.
8.2.3
Poles
Suppose now that for our point conic C with equation γ 2 −4kαβ = 0, the line Z4 Z5 is not diametral.
Then by §7.3.4 and §7.4.1 the polars of Z4 and Z5 with respect to C are not parallel and so intersect
at a unique point Z13 . By §7.2.7 Z4 Z5 is the polar of Z13 . If now Z10 is an exterior point of the conic
which lies on Z4 Z5 , the polar of Z10 passes through Z13 , cuts the conic at points Z14 and Z15 , the
tangents at which pass through Z10 , and meets Z4 Z5 in a point Z16 such that (Z13 , Z16 , Z14 , Z15 )
is a harmonic range. It follows that Z10 (Z13 , Z16 , Z14 , Z15 ) is a harmonic pencil.
Suppose that we now take our pencil in §8.2.2 in the form
(U, V ) =
1
λ
(Z4 , Z5 ) +
(Z7 , Z8 ),
1+λ
1+λ
where hZ7 , Z8 i − hZ4 , Z5 i = hZ10 , Z13 i. Our parametric equations now are
l=
λ
1
λ
1
λ
1
l6 +
l9 , m =
m6 +
m9 , n =
n6 +
n9 .
1+λ
1+λ
1+λ
1+λ
1+λ
1+λ
On substituting this into the equation (8.1.7) we obtain the incidence equation
(n6 + λn9 )2 − 4k1 (l6 + λl9 )(m6 + λm9 ) = 0.
(8.2.8)
We wish to have a harmonic pencil and so the sum of the roots of this equation in λ to be equal
to 0, that is the coefficient of λ to vanish. This gives
n6 n9 − 2k1 (m6 l9 + l6 m9 ) = 0.
On replacing the fixed (Z7 , Z8 ) by the variable (U, V ) this becomes −2k1 m6 l − 2k1 l6 m + n6 n = 0.
This must be a line equation of the point Z13 , so that we have a line pencil concurrent at the point
Z13 which is the pole of Z4 Z5 . We call this concurrent pencil dual polar of the line Z4 Z5 with
respect to this point-pair conic; for it we must have
−2k1 m6 − 2k1 l6 + n6 6= 0.
198
CHAPTER 8. DUAL CONICS
On taking 4k1 = 1/k, we have that
m6 l + l6 m − 2kn6 n = 0,
(8.2.9)
with
l6 + m6 − 2kn6 6= 0,
which verifies that Z4 Z5 does not pass through the point with normalized areal coordinates
1
1
−2k
,
,
,
2(1 − k) 2(1 − k) 2(1 − k)
when k 6= 1, i.e. Z4 Z5 does not pass through the centre in the case of a central conic.
We note that (8.2.9) is the same equation as (8.2.7), and so a contact is the pole of Z4 Z5 when
(Z4 , Z5 ) is in the point-pair conic.
In §7.2.9 we asked when a line Z4 Z5 with equation l6 α + m6 β + n6 γ = 0 is the polar of some
point Z ′ with respect to γ 2 − 4kαβ = 0. We found it to be
α′ =
l6
2(1−k)
m6 /2(1 − k)
, β′ =
m6
2kn6
+ 2(1−k)
− 2(1−k)
l6
2(1−k)
l6 /2(1 − k)
m6
+ 2(1−k)
−
2kn6
2(1−k)
, γ′ =
−2kn6 /2(1 − k)
m6
2kn6
+ 2(1−k)
− 2(1−k)
l6
2(1−k)
by (7.2.27) having by (7.2.26)
l6 + m6 − 2kn6 6= 0,
(8.2.10)
so that when the conic is central the line must not pass through the centre. Now in §8.2.2 we
have seen that the pole of the line with equation l6 α + m6 β + n6 γ = 0 is given by (8.2.9) and on
replacing 4k1 by 1/k this becomes
−
1
1
m6 l −
l6 m + n6 n = 0,
2k
2k
1
1
with − 2k
m6 − 2k
l6 m + n6 6= 0. Now (8.2.10) is the equation of the point Z ′ above. Thus a point
and its polar in §7.2.5 correspond to a pole and its originating line in §8.2.2.
We saw in (7.2.28) that the coordinates for Z ′ can be given as
β′ =
βc δF (Z1 , Z4 , Z5 ) ′
γc δF (Z1 , Z4 , Z5 )
αc δF (Z1 , Z4 , Z5 ) ′
, α =
, γ =
,
δF (Zc , Z4 , Z5 )
δF (Zc , Z4 , Z5 )
δF (Zc , Z4 , Z5 )
where Zc is the centre of the conic. To interpret this more fully we continue as follows. We note
first of all that the line through the point Z3 which passes through the mid-point of Z1 and Z2 ,
that is the median line, through the vertex Z3 , of the triangle [Z1 , Z2 , Z3 ] passes through the centre
Zc as
0 0
1 1 1
0 = 0.
2 2
1 1 −2k Any point Z ′ in the plane can be expressed in the form
s
Z ′ = (1 − s)Z3 + (Z1 + Z2 ) − t(Z2 − Z1 ) = ( 12 s + t)Z1 + ( 12 s − t)Z2 + (1 − s)Z3 ,
2
and then the image of Z ′ under oblique symmetry in the median line Z3 Zc , parallel to the line
Z1 Z2 is given by
s
Z ′∗ = (1 − s)Z3 + (Z1 + Z2 ) + t(Z2 − Z1 ).
2
We show that
α′ = β ′∗ , β ′ = α′∗ .
8.2. INCIDENCE OF A PENCIL AND A POINT-PAIR CONIC
199
For
δF (Z ′ , Z2 , Z3 )
s
=δF ((1 − s)Z3 + (Z1 + Z2 ) − t(Z2 − Z1 ), Z2 , Z3 )
2
1
=δF (s
(Z1 + Z2 ) − Z3 − t(Z2 − Z1 ), Z2 − Z3 , O)
2
1
=sδF ( (Z1 + Z2 ) − Z3 , Z2 − Z3 , O) − tδF (Z2 − Z1 , Z2 − Z3 , O)
2
1
= sδF (Z1 + Z2 − 2Z3 , Z2 − Z3 , O) − tδF (Z1 − Z2 , Z3 − Z2 , O)
2
1
= s[δF (Z1 − Z3 , Z2 − Z3 , O) + δF (Z2 − Z3 , Z2 − Z3 , O)] − tδF (Z1 , Z3 , Z2 )
2
1
= sδF (Z1 , Z2 , Z3 ) + tδF (Z1 , Z2 , Z3 )
2
1
=[ s + t]δF (Z1 , Z2 , Z3 ).
2
Similarly
δF (Z ′∗ , Z3 , Z1 )
s
=δF ((1 − s)Z3 + (Z1 + Z2 ) + t(Z2 − Z1 ), Z3 , Z1 )
2
1
=δF (s
(Z1 + Z2 ) − Z3 + t(Z2 − Z1 ), O, Z1 − Z3 )
2
1
=sδF ( (Z1 + Z2 ) − Z3 , O, Z1 − Z3 ) + tδF (Z2 − Z1 , O, Z1 − Z3 )
2
1
= sδF (Z1 + Z2 − 2Z3 , O, Z1 − Z3 ) − tδF (Z2 − Z1 , O, Z3 − Z1 )
2
1
= s[δF (Z1 − Z3 , O, Z1 − Z3 ) + δF (Z2 − Z3 , O, Z1 − Z3 )] − tδF (Z2 , Z1 , Z3 )
2
1
= sδF (Z2 , Z3 , Z1 ) + tδF (Z1 , Z2 , Z3 )
2
1
=[ s + t]δF (Z1 , Z2 , Z3 ).
2
Thus α′ = β ′∗ . On replacing t by −t we deduce that α′∗ = β ′ .
We now have the formula
α′∗ =
8.2.4
αc δF (Z1 , Z4 , Z5 ) ′∗
βc δF (Z1 , Z4 , Z5 ) ′∗ γc δF (Z1 , Z4 , Z5 )
, β =
, γ =
.
δF (Zc , Z4 , Z5 )
δF (Zc , Z4 , Z5 )
δF (Zc , Z4 , Z5 )
(8.2.11)
Deficient incidence in dual situation
Suppose that Z4 is an exterior point for the point conic C. Then by (5.2.23) and §7.3.5, through
Z4 pass either two tangents, or one tangent and one asymptote, or two asymptotes to C, it being
a hyperbola in the second and third cases. We suppose that Z4 Z5 is one tangent/asymptote and
that Z5 Z8 is parallel to the other tangent/asymptote but distinct from it. Then with Z7 = Z4 ,
hZ7 , Z8 i − hZ4 , Z5 i = hZ10 , Z11 i where Z10 = Z4 and Z4 Z11 is tangent/asymptote to the conic.
Now the pencil
(1 − s)hZ4 , Z5 i + shZ7 , Z8 i = (1 − s)(Z4 , Z5 ) + s(Z4 , Z8 ) = (Z4 , (1 − s)Z5 + sZ8 )
has deficient incidence as it contains only one point-pair since
hZ7 , Z8 i − hZ4 , Z5 i = hZ4 , Z8 i − hZ4 , Z5 i = hZ4 , Z11 i = hZ10 , Z11 i
200
CHAPTER 8. DUAL CONICS
and Z10 Z11 is a tangent/asymptote. Thus there are just two tangent/asymptotes through Z4 , one
Z4 Z5 is in the pencil, the other Z4 Z11 is not in the pencil, as we noted in §2.1.2, so there is only
one in the pencil.
We can construct an example along these lines as follows. Take Z4 = Z3 ≡ (0, 0, 1), Z5 = Z2 ≡
(0, 1, 0), Z7 = Z3 , Z8 = Z2 + Z1 − Z3 . Then in the pencil
(1−s)(Z4 , Z5 )+s(Z7 , Z8 ) = (1−s)(Z4 , Z2 )+s(Z4 , Z8 ) = (Z4 , (1−s)Z2 +sZ8 ) = (Z4 , Z2 +s(Z1 −Z3 )),
the point Z2 + s(Z1 − Z3 ) has coordinates (s, 1, −s) so the line joining this to Z4 has equation
α − sβ = 0 so that for it l = 1, m = −s, ; n = 0. When we insert this in kn2 − lm = 0 we obtain
the equation of incidence k.02 − 1(−s) = 0, which simplifies to s = 0 and so has deficient incidence.
To deal with double deficiency, we take Z4 to be a point on C, Z5 to be a point not on the
tangent at Z4 , Z7 = Z4 and Z8 6= Z5 to be a point on the line through Z5 which is parallel to the
tangent. Then the pencil
(1 − s)(Z4 , Z5 ) + s(Z7 , Z8 ) = (1 − s)(Z4 , Z5 ) + s(Z4 , Z8 ) = (Z4 , (1 − s)Z5 + sZ8 ),
which has vertex Z4 does not contain the tangent at Z4 . As a specific example we take Z4 = Z7 = 1,
so that Z3 is on the tangent at Z4 , and Z5 = Z2 , Z8 = Z2 + Z3 − Z1 . Then Z5 + s(Z8 − Z5 ) =
Z2 + s(Z3 − Z1 ) has coordinates (−s, 1, s), and the line joining it to Z1 has equation −sβ + γ = 0,
so that for it l = 0, m = −s, γ = 1 and on insertion into kn2 − lm = 0 we obtain the equation
of incidence k.12 − 0.(−s) = 0 which simplifies to 0.s2 + 0.s + k = 0, yielding doubly deficient
incidence.
Deficient incidence can occur in this way for the dual of any type of point conic, ellipse/circle,
parabola, hyperbola, and so cannot contribute to distinguishing between the types of duals. We
can distinguish between them intrinsically as follows. We saw in (7.2.23) and §7.3.4 that equation
(7.3.9),kn2 −lm = 0, is a necessary condition for lα+mβ +nγ = 0 to be a tangent to γ 2 −4kαβ = 0,
but when l + m − 2kn = 0 there is not such a tangent. For a parabola we have k = 1 and this
condition is the equation of a parallel pencil. When k 6= 1 we have a central conic and this condition
is the equation of a concurrent pencil. For an ellipse/circle we can check that this condition is
incompatible with kn2 − lm = 0 and there is no exception, whereas for a hyperbola there is an
exception.
8.3
Duals of diametral lines
8.3.1
Analogue of diametral lines
Our derivation of (8.2.9) as an equation of the pole of Z4 Z5 was based on an initial assumption
that Z4 Z5 is not a diametral line of our point conic C.
We take a fixed hZ10 , Z11 i such that Z10 Z11 is a diametral line of C; then we have that m12 +
l12 − 2kn12 = 0. We take hU, V i such that U V k Z10 Z11 but U V 6= Z10 Z11 and consider the pencil
of rotors
hU ′ , V ′ i = (1 − s)hZ10 , Z11 i + s(hZ10 , Z11 i + hU, V i) = hZ10 , Z11 i + shU, V i.
We have a parallel pencil for each hU, V i . By §8.1.2 and §8.1.4 we have that
δF (Z, U ′ , V ′ ) = δF (Z, Z10 , Z11 ) + sδF (Z, U, V ),
for all Z. In particular this yields the parametric equations
l′ = l12 + sl, m′ = m12 + sm, n′ = n12 + sn,
where l12 = δF (Z1 , Z10 , Z11 ) and so on. When we insert this into the equation
kn′2 − l′ m′ = 0,
201
8.3. DUALS OF DIAMETRAL LINES
we obtain the incidence equation for s
k(n12 + sn)2 − (sl12 + sl)(m12 + sm) =0,
(kn2 − lm)s2 + (2kn12 n − m12 l − l12 m)s + kn212 − l12 m12 =0.
We are supposing that hU, V i is not in the set of tangents to C so that kn2 − lm 6= 0. As
U V k Z10 Z11 , by §5.1.2 we have
m12 l + l12 m − 2kn12 n = 0,
(8.3.1)
as the sum of the coefficients of l, m and n is equal to 0. Thus the coefficient of s is equal
to 0, and if there are real roots s1 , s2 we have s1 + s2 = 0. Thus the mid-rotor of the rotors
hU ′ , V ′ is1 , hU ′ , V ′ is2 is hZ10 , Z11 i.
8.3.2
Dual of Newton property
Z17
b
b
Z16
b
Z15
b
Z14
b
P2
V13
V12
W7
W6
b
b
b
b
b
b
b
b
W3
b
W2
b
b
b
b
V11
b
P1
b
b
b
b
V10
b
b
W0
[U10 = U11 = P1 , U12 = U13 = P2 ]
b
W1
Figure 8.3. Dual of the Newton property.
We now consider an analogue of the Newton property of §7.5.1 for the point-pair conic with
equation
kn2 − lm = 0.
We first take rotors of the form
hU, V i = (1 − s)hW0 , W1 i + shW2 , W3 i,
and write hW2 , W3 i − hW0 , W1 i = hW4 , W5 i. Similarly we take rotors of the form
hU, V i = (1 − s)hW0 , W1 i + shW6 , W7 i,
and write hW6 , W7 i − hW0 , W1 i = hW8 , W9 i.
202
CHAPTER 8. DUAL CONICS
Then by (8.2.5) and (8.2.6) we obtain in the point-pair conic two rotors of the form
hU10 , V10 i =hW0 , W1 i + s10 hW4 , W5 i,
hU11 , V11 i =hW0 , W1 i + s11 hW4 , W5 i,
where
s10 s11 =
kδF (Z3 , W0 , W1 )2 − δF (Z1 , W0 , W1 )δF (Z2 , W0 , W1 )
.
kδF (Z3 , W4 , W5 )2 − δF (Z1 , W4 , W5 )δF (Z2 , W4 , W5 )
Similarly we obtain in the point-pair conic two rotors of the form
hU12 , V12 i =hW0 , W1 i + s12 hW8 , W9 i,
hU13 , V13 i =hW0 , W1 i + s13 hW8 , W9 i,
where
s12 s13 =
Hence
kδF (Z3 , W0 , W1 )2 − δF (Z1 , W0 , W1 )δF (Z2 , W0 , W1 )
.
kδF (Z3 , W8 , W9 )2 − δF (Z1 , W8 , W9 )δF (Z2 , W8 , W9 )
s12 s13
kδF (Z3 , W4 , W5 )2 − δF (Z1 , W4 , W5 )δF (Z2 , W4 , W5 )
=
.
s10 s11
kδF (Z3 , W8 , W9 )2 − δF (Z1 , W8 , W9 )δF (Z2 , W8 , W9 )
To apply this, we take fixed points Z14 , Z15 such that |Z14 , Z15 | = 1, and fixed points Z16 , Z17
such that |Z16 , Z17 | = 1. For a variable pair (W0 , W1 ), we choose W2 , W3 so that
hW2 , W3 i = hW0 , W1 i + hZ14 , Z15 i,
and thus
hW2 , W3 i − hW0 , W1 i = hZ14 , Z15 i.
We use this as the basis of the first pencil above and then s10 , s11 yield two tangents. We also
choose W6 , W7 so that
hW6 , W7 i = hW0 , W1 i + hZ16 , Z17 i,
and thus
hW6 , W7 i − hW0 , W1 i = hZ16 , Z17 i.
We use this as the basis of the second pencil above and then s12 , s13 yield two tangents.
We thus have
s12 s13
kδF (Z3 , Z14 , Z15 )2 − δF (Z1 , Z14 , Z15 )δF (Z2 , Z14 , Z15 )
=
,
s10 s11
kδF (Z3 , Z16 , Z17 )2 − δF (Z1 , Z16 , Z17 )δF (Z2 , Z16 , Z17 )
and this is equal to the quotient of products of sensed-distances
hW0 , W1 ihU12 , V12 i hW0 , W1 ihU13 , V13 i
hW0 , W1 ihU10 , V10 i hW0 , W1 ihU11 , V11 i
.
This last is thus constant as (W0 , W1 ) varies.
8.4
8.4.1
Normalised areal line coordinates
Definition of normalized point-pair coordinates; identities
To facilitate the establishment of dual results we should like manipulations involving areal point
coordinates to be convertible automatically to manipulations involving areal pair-coordinates. We
203
8.4. NORMALISED AREAL LINE COORDINATES
recall the fundamental identity (3.2.1) which we wrote in the forms
δF (Z, Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z, Z3 , Z1 )
δF (Z, Z1 , Z2 )
+δF (Z2 , U, V )
+ δF (Z3 , U, V )
,
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z, U, V ) =δF (Z1 , U, V )α + δF (Z2 , U, V )β + δF (Z3 , U, V )γ,
δF (Z, U, V ) =δF (Z1 , U, V )
and with the notation (3.2.2)
δF (Z, U, V ) = lα + mβ + nγ.
(8.4.1)
We note that (8.4.1) is perfectly symmetrical as between (α, β, γ) and (l, m, n), and so it seems
perfectly suited as a bridge between point and pair coordinates. But there is a significant difference
in that, as noted in §5.1.3, we do not have for the latter anything comparable to the relation
α + β + γ = 1. The precise location of the divergence is in the respective parametric equations, in
§3.2.8 and §5.1.6. In §3.2.8 we had
α = α4 + t(α5 − α4 ), β = β4 + t(β5 − β4 ), γ = γ4 + t(γ5 − γ4 ),
and the relation
(α5 − α4 ) + (β5 − β4 ) + (γ5 − γ4 ) = 0
means that in the three coefficients of t just two are independent. By contrast in §5.1.6 we had
l = l4 + t(l5 − l4 ),
m = m4 + t(m5 − m4 ),
n = n4 + t(n5 − n4 ),
and by §5.1.3 we have not got a similar relationship for these pair-coordinates. In order to have
such a relation for some pair-coordinates we now take a step which has far-reaching consequences.
DEFINITION For a a distinguished point Z0 in the plane Π and a pair (U, V ) of points in
Π \ {Z0 }, we say that the pair (U, V ) is Z0 -non-aligned with respect to Z0 if the points Z0 , U
and V are not collinear. We also call Z0 -non-incident a line U V which does not contain the
point Z0 .
To introduce coordinates, we take the point Z0 , with normalized areal coordinates (α0 , β0 , γ0 ),
not to be on any of the side-lines of the triple of reference. Then for a pair (U, V ) of Z0 -non-aligned
points, we take the fundamental identity in the form
δF (Z1 , U, V )
δF (Z2 , U, V )
δF (Z3 , U, V )
δF (Z, U, V )
=
α+
β+
γ.
δF (Z0 , U, V )
δ(Z0 , U, V )
δF (Z0 , U, V )
δF (Z0 , U, V )
(8.4.2)
From this, for a Z0 -non-aligned pair (U, V ) we introduce the notation
δF (Z1 , U, V )
δF (Z1 , U, V )
= α0
δF (Z0 , U, V )
α0 δF (Z1 , U, V ) + β0 δF (Z2 , U, V ) + γ0 δF (Z3 , U, V )
α0 l
,
=
α0 l + β0 m + γ0 n
δF (Z2 , U, V )
β0 δF (Z2 , U, V )
q =β0
=
,
δF (Z0 , U, V )
α0 δF (Z1 , U, V ) + β0 δF (Z2 , U, V ) + γ0 δF (Z3 , U, V )
β0 m
=
,
α0 l + β0 m + γ0 n
δF (Z3 , U, V )
γ0 δF (Z3 , U, V )
r =γ0
=
δF (Z0 , U, V )
α0 δF (Z1 , U, V ) + β0 δF (Z2 , U, V ) + γ0 δF (Z3 , U, V )
γ0 n
=
.
α0 l + β0 m + γ0 n
p =α0
(8.4.3)
204
CHAPTER 8. DUAL CONICS
We call (p, q, r) areal pair-coordinates, normalized with respect to the point Z0 , of the pair (U, V ).
Clearly p + q + r = 1. We now have the identity (8.4.2) in the form
δF (Z, U, V )
1
1
1
pα + qβ + rγ.
=
δF (Z0 , U, V )
α0
β0
γ0
(8.4.4)
We note that in (8.2.11) the normalized areal coordinates for Z ′∗ are the areal pair-coordinates,
normalized with respect to the centre Zc , of the pair (Z4 , Z5 ) on the polar of Z ′ .
8.4.2
Dependence of normalized pair-coordinates on lines
Suppose that for a Z0 -non-aligned pair (U, V ) we take distinct points U ′ , V ′ ∈ U V , both different
from Z0 . Then for some real numbers s1 , s2 we have that
U ′ = (1 − s1 )U + s1 V, V ′ = (1 − s2 )U + s2 V.
Now for all Z we have that
δF (Z, U ′ , V ′ ) = (s2 − s1 )δF (Z, U, V )
and so
δF (Z1 , U, V )
δF (Z1 , U ′ , V ′ )
=α0
,
δF (Z0 , U ′ , V ′ )
δF (Z0 , U, V )
δF (Z2 , U ′ , V ′ )
δF (Z2 , U, V )
β0
=β0
,
δF (Z0 , U ′ , V ′ )
δF (Z0 , U, V )
δF (Z3 , U, V )
δF (Z3 , U ′ , V ′ )
=γ0
.
γ0
δF (Z0 , U ′ , V ′ )
δF (Z0 , U, V )
α0
Thus the normalized point-pair coordinates of (U ′ , V ′ ) are the same as those of (U, V ).
Conversely suppose that (8.4.5) holds. Then immediately we have that
δF (Z1 , U, V )
δF (Z0 , U, V )
=
,
δF (Z1 , U ′ , V ′ ) δF (Z0 , U ′ , V ′ )
δF (Z0 , U, V )
δF (Z2 , U, V )
=
,
δF (Z2 , U ′ , V ′ ) δF (Z0 , U ′ , V ′ )
δF (Z0 , U, V )
δF (Z3 , U, V )
=
.
δF (Z3 , U ′ , V ′ ) δF (Z0 , U ′ , V ′ )
Thus
δF (Z3 , U, V )
δF (Z2 , U, V )
=
,
′
′
δF (Z2 , U , V )
δF (Z3 , U ′ , V ′ )
and so δF [(Z2 , Z3 ), (U, V ), (U ′ , V ′ )] = 0; similarly
δF [(Z3 , Z1 ), (U, V ), (U ′ , V ′ )] = 0, δF [(Z1 , Z2 ), (U, V ), (U ′ , V ′ )] = 0.
Now by (c.0.5) in Appendix C, we have
δF [(Z2 , Z3 ), (Z3 , Z1 ), (Z1 , Z2 )]δF [(W1 , W2 ), (U, V ), (U ′ , V ′ )]
=δF [(Z2 , Z3 ), (U, V ), (U ′ , V ′ )]δF [(W1 , W2 ), (Z3 , Z1 ), (Z1 , Z2 )]
+δF [(Z3 , Z1 ), (U, V ), (U ′ , V ′ )]δF [(W1 , W2 ), (Z1 , Z2 ), (Z2 , Z3 )]
+δF [(Z1 , Z2 ), (U, V ), (U ′ , V ′ )]δF [(W1 , W2 ), (Z2 , Z3 ), (Z3 , Z1 )]
=0,
(8.4.5)
8.4. NORMALISED AREAL LINE COORDINATES
205
for all (W1 , W2 ). We show that it follows from this that U V = U ′ V ′ . We have that
δF (W1 , U, V )δF (W2 , U ′ , V ′ ) − δF (W2 , U, V )δF (W1 , U ′ , V ′ ) = 0.
We first take W1 on U V but W2 not on U V . Then δF (W1 , U, V ) = 0 and δF (W2 , U, V ) 6= 0, so
that W1 ∈ U ′ V ′ . The same holds for a second point W1 on U V , and so U V = U ′ V ′ .
This shows that normalized point-pair coordinates are determined by any pair of points on a
line and so relate essentially to a line.
8.4.3
Equation of a line; coefficient triples
Given a Z0 -non-aligned pair (Z4 , Z5 ) of points, we suppose that (p6 , q6 , r6 ) are the normalized
areal pair-coordinates in §8.4.1 for (U, V ) = (Z4 , Z5 ). Then by (8.4.4) the variable point Z with
normalized areal point-coordinates (α, β, γ) will lie on the Z0 -non-incident line Z4 Z5 if and only if
q6
r6
p6
α + β + γ = 0.
(8.4.6)
α0
β0
γ0
We call this a point equation of the line Z4 Z5 . Note that since it involves normalized paircoordinates this process does not give us an equation of any line which passes through Z0 . The
equation (8.4.6) can be deduced from the equation (3.2.3) by the correspondence in the definitions
(8.4.3), as p, q and r have a common denominator. We shall prefer to work with equations of lines
in the form (3.2.3) whenever we can, rather than use the form (8.4.6), but in §§8.4.4 to 8.4.6 we
provide some detail on the latter so that it will be seen what pattern to expect.
As in (3.2.3) we had to exclude the case when l6 = m6 = n6 , so in (8.4.6) we must exclude the
case when
p6
q6
r6
=
= .
(8.4.7)
α0
β0
γ0
8.4.4
Equation of a line through two points
Given normalized areal coordinates (α4 , β4 , γ4 ), (α5 , β5 , γ5 ) for points in a Z0 -non-aligned pair
(Z4 , Z5 ), we wish to find directly an equation of the Z0 -non-incident line Z4 Z5 . For this we need
to solve for p6 , q6 , r6 the system
p6 + q6 + r6 =1,
β4
γ4
α4
p6 + q6 + r6 =0,
α0
β0
γ0
β5
γ5
α5
p6 + q6 + r6 =0.
α0
β0
γ0
These have the unique solutions
α0 (β4 γ5 − β5 γ4 )
,
α0 (β4 γ5 − β5 γ4 ) + β0 (γ4 α5 − γ5 α4 ) + γ0 (α4 β5 − α5 β4 )
β0 (γ4 α5 − γ5 α4 )
,
q6 =
α0 (β4 γ5 − β5 γ4 ) + β0 (γ4 α5 − γ5 α4 ) + γ0 (α4 β5 − α5 β4 )
γ0 (α4 β5 − α5 β4 )
.
r6 =
α0 (β4 γ5 − β5 γ4 ) + β0 (γ4 α5 − γ5 α4 ) + γ0 (α4 β5 − α5 β4 )
p6 =
8.4.5
Point of intersection of two lines
Suppose that
p6
α+
α0
p9
α+
α0
q6
β+
β0
q9
β+
β0
r6
γ =0,
γ0
r9
γ =0,
γ0
206
CHAPTER 8. DUAL CONICS
are equations of two Z0 -non-incident lines Z4 Z5 and Z7 Z8 . Then the coordinates of a point of
intersection is given by solving simultaneously these equations and the third equation
α + β + γ = 1.
These have the unique solution
α0 (q6 r9 − r6 q9 )
,
α0 (q6 r9 − r6 q9 ) + β0 (r6 p9 − p6 r9 ) + γ0 (p6 r9 − r6 p9 )
β0 (r6 p9 − p6 r9 )
β=
,
α0 (q6 r9 − r6 q9 ) + β0 (r6 p9 − p6 r9 ) + γ0 (p6 q9 − q6 p9 )
γ0 (p6 q9 − q6 p9 )
,
γ=
α0 (q6 r9 − r6 q9 ) + β0 (r6 p9 − p6 r9 ) + γ0 (p6 r9 − r6 p9 )
α=
provided that the common denominator here is non-zero.
8.4.6
Parallel lines
Suppose that
p6
α+
α0
p9
α+
α0
q6
β+
β0
q9
β+
β0
r6
γ =0,
γ0
r9
γ =0,
γ0
are equations of Z0 -non-incident lines. By §8.4.5 these are parallel if and only if
α0 (q6 r9 − r6 q9 ) + β0 (r6 p9 − p6 r9 ) + γ0 (p6 q9 − q6 p9 ) = 0.
We can write this condition as

and so in turn as
det 

p6
α0
p9
α0
q6
β0
q9
β0
r6
γ0
r9
γ0
1
1
1
p6
det  p9
α0
q6
q9
β0

 = 0,

r6
r9  = 0.
γ0
(8.4.8)
We remark that this is also the condition that the equations
p6 α + q6 β + r6 γ =0,
p9 α + q9 β + r9 γ =0,
α0 α + β0 β + γ0 γ =0,
in arbitrary triples of numbers (i.e. not necessarily with sum equal to 1) have a solution other than
(α, β, γ) = (0, 0, 0).
8.4.7
Equation of point-pairs in a concurrent or parallel pencil
We recall the equation
ul + vm + wn = 0,
in (5.1.4). By §5.1.1, when u + v + w 6= 0, this is an equation satisfied by pairs (U, V ) of points
collinear with some point Z6 , using non-normalized dual areal coordinates. On inserting normalized
dual areal coordinates from (8.4.3), we derive another equation for Z6 ,
v
w
u
p + q + r = 0,
α0
β0
γ0
207
8.4. NORMALISED AREAL LINE COORDINATES
and in turn
u′ p + v ′ q + w′ r = 0,
(8.4.9)
where
α0 u′ + β0 v ′ + γ0 w′
(8.4.10)
u
v
w
is not equal to 0. By §5.1.1 in all this u+v+w
must not be coordinates of Z0 .
, u+v+w
, u+v+w
Similarly by §5.1.2, when u + v + w = 0 but the case u = v = w = 0 is excluded, (8.4.10) is
an equation satisfied by pairs (U, V ) of points on lines in some parallel pencil, and now (8.4.9) is
equal to 0. Now the case u′ = v ′ = w′ = j 6= 0 must be excluded, for it would imply that
v
w
u
=
=
= j,
α0
β0
γ0
so that u = jα0 , v = jβ0 , w = jγ0 and then the original equation would be j(α0 l+β0 m+γ0 n) = 0,
which is precluded for Z0 -nonaligned pairs.
8.4.8
Pair of points common to two pencils
Suppose that we take two equations in §8.4.7,
u′1 p + v1′ q + w1′ r =0,
u′2 p + v2′ q + w2′ r =0,
of pencils. Then for normalized dual areal coordinates (p, q, r) of any pair of points in both pencils
we solve these equations simultaneously along with the third equation p + q + r = 1.
In particular, suppose that we take equations in §8.4.7 of pencils concurrent at the points Z6
and Z9 ,
α6
p+
α0
α9
p+
α0
β6
q+
β0
β9
q+
β0
γ6
r =0,
γ0
γ9
r =0.
γ0
Then for (p, q, r) to satisfy these equations simultaneously with p + q + r = 1, we must have
1 β6 γ9 − γ6 β9
,
β0 γ0
D
1 γ6 α9 − α6 γ9
q=
,
γ0 α0
D
1 α6 β9 − β6 α9
r=
,
α0 β0
D
p=
where
8.4.9

α6
1
det  α9
D=
α0 β0 γ0
α0
β6
β9
β0

γ6
γ9  .
γ0
Dual-parallelism of concurrent pencils
From §8.4.8 we see that pencils concurrent at
(U, V ) in common if and only if

α6
det  α9
α0
the points Z6 and Z9 have no Z0 -non-aligned pair
β6
β9
β0

γ6
γ9  = 0.
γ0
208
CHAPTER 8. DUAL CONICS
This is equivalent to δF (Z6 , Z9 , Z0 ) = 0. Thus we can introduce the concept of pencils concurrent
at the points Z6 and Z9 being dual-parallel if Z6 , Z9 and Z0 are collinear. This is consistent with
the fact that when we utilize normalized pair-coordinates we are excluding lines through Z0 from
our consideration.
8.4.10
Determinant form of equation of pencil
Suppose that as in §8.4.7 we have an equation of Z0 -non-aligned pairs (U, V ) of points in a pencil,
u′ p + v ′ q + w′ r = 0. Suppose also that (p6 , q6 , r6 ) and (p9 , q9 , r9 ) are normalized dual areal
coordinates of two pairs, (Z4 , Z5 ) and (Z7 , Z8 ) respectively, which are in this pencil but not all
collinear. Then
u′ p6 + v ′ q6 + w′ r6 = 0,
u′ p9 + v ′ q9 + w′ r9 = 0,
so that

p
det  p6
p9
and hence
q
q6
q9

r
r6  = 0,
r9
(q6 r9 − r6 q9 )p + (r6 p9 − p6 r9 )q + (p6 q9 − q6 p9 )r = 0.
(8.4.11)
This has the form (8.4.9), and (8.4.10) is equal to

α0
det  p6
p9
β0
q6
q9

γ0
r6  .
r9
In the case of a concurrent pencil, Z4 Z5 and Z7 Z8 are not concurrent at Z0 and so (8.4.10) is
non-zero as expected.
In the case of a parallel pencil, by (8.4.8) we have that (8.4.10) is equal to 0. We must still
check that the coefficients are not all equal to 0. For
r4 p5 − p4 r5 − (p4 q5 − q4 p5 ) = p5 (r4 + q4 ) − p4 (r5 + q5 )
=p5 (1 − p4 ) − p4 (1 − p5 ) = p5 − p4 ,
and similarly
p4 q5 − q4 p5 − (q4 r5 − r4 q5 ) =q5 − q4 ,
q4 r5 − r4 q5 − (r4 p5 − p4 r5 ) =r5 − r4 .
If the coefficients were equal to 0, this would make
p5 − p4 = 0,
q5 − q4 = 0,
r5 − r4 = 0.
By §8.4.2 the lines Z4 Z5 and Z7 Z8 would then coincide with each other, and this gives a contradiction.
8.4.11
Parametric equations of point-pairs in a concurrent pencil
We wish to establish parametric equations
p = p4 + t(p5 − p4 ), q = q4 + t(q5 − q4 ), r = r4 + t(r5 − r4 ),
(t ∈ R),
209
8.4. NORMALISED AREAL LINE COORDINATES
for Z0 -non-aligned pairs (U, V ) collinear with the point Z6 in §8.4.10. We first note that if (p, q, r)
have this form for any real number t, then


p q r
det  p4 q4 r4 
p5 q5 r5


p4 + t(p5 − p4 ) q4 + t(q5 − q4 ) r4 + t(r5 − r4 )

p4
q4
r4
= det 
p5
q5
r5




p4 q4 r4
p5 − p4 q5 − q4 r5 − r4

p4
q4
r4
= det  p4 q4 r4  + t det 
p5 q5 r5
p5
q5
r5
=0,
so that (p, q, r) satisfy (8.4.11).
Conversely suppose that (p, q, r) satisfy (8.4.11). To start we note that we cannot have two of
p5 − p4 = 0,
q5 − q4 = 0,
r5 − r4 = 0,
holding as then all three would hold and our original Z0 -non-aligned pairs of points collinear with
Z6 would coincide. Without loss of generality we may then suppose that p5 − p4 6= 0. We choose
t to satisfy
p = p4 + t(p5 − p4 ),
that is we take
t=
p − p4
.
p5 − p4
On inserting this into (6.4.11) we find that
(q4 r5 − r4 q5 )(p5 − p4 )t + (r4 p5 − p4 r5 )(q − q4 ) + (p4 q5 − q4 p5 )(r − r4 ) = 0.
(8.4.12)
Now (p4 , q4 , r4 ) and (p5 , q5 , r5 ) satisfy (8.4.11), so that
p4 (q4 r5 − r4 q5 ) + q4 (r4 p5 − p4 r5 ) + r4 (p4 q5 − q4 p5 ) =0,
p5 (q4 r5 − r4 q5 ) + q5 (r4 p5 − p4 r5 ) + r5 (p4 q5 − q4 p5 ) =0,
(8.4.13)
and so by subtraction
(p5 − p4 )(q4 r5 − r4 q5 ) + (q5 − q4 )(r4 p5 − p4 r5 ) + (r5 − r4 )(p4 q5 − q4 p5 ) = 0.
(8.4.14)
By (6.4.14)
(p5 − p4 )(q4 r5 − r4 q5 ) = −(q5 − q4 )(r4 p5 − p4 r5 ) − (r5 − r4 )(p4 q5 − q4 p5 )
and on inserting this into (8.4.12) we get that
[−(q5 − q4 )(r4 p5 − p4 r5 ) − (r5 − r4 )(p4 q5 − q4 p5 )]t + (r4 p5 − p4 r5 )(q − q4 ) + (p4 q5 − q4 p5 )(r − r4 ) = 0,
that is
(r4 p5 − p4 r5 )[q − q4 − (q5 − q4 )t] + (p4 q5 − q4 p5 )[r − r4 − (r5 − r4 )t] = 0.
Then we have that
q − q4 − (q5 − q4 )t = u(p4 q5 − q4 p5 ), r − r4 − (r5 − r4 )t = −u(r4 p5 − p4 r5 ),
210
CHAPTER 8. DUAL CONICS
for some u ∈ R. Now
1 =p + q + r
=p4 + q4 + r4 + t(p5 − p4 + q5 − q4 + r5 − r4 ) + u[p4 q5 − q4 p5 − (r4 p5 − p4 r5 )]
so that
u[p4 q5 − q4 p5 − (r4 p5 − p4 r5 )] = 0.
(8.4.15)
If we had
p4 q5 − q4 p5 = r4 p5 − p4 r5
then by §8.4.10 we would have p5 − p4 = 0, which was ruled out by our initial choice. It follows
from (8.4.15) that u = 0 and so the desired form is established.
Note that we have not now got a restriction like (5.1.15).
8.4.12
Correspondence of our two parametrisations
Suppose that we also consider parametric equations for Z0 -non-aligned pairs (U, V ) of points
collinear with a point Z4 , in terms of un-normalized line coordinates
l = l4 + s(l5 − l4 ), m = m4 + s(m5 − m4 ), n = n4 + s(n5 − n4 ),
where (l4 , m4 , n4 ) correspond to (Z4 , Z5 ) and (l5 , m5 , n5 ) correspond to (Z4 , Z6 ). Then as in §5.1.6,
s will be the ordinary parameter we have been using from the start. For corresponding normalized
pair-coordinates we have that
p=
=
α0 l
α0 l + β0 m + γ0 n
α0 l4 + sα0 (l5 − l4 )
α0 l4 + β0 m4 + γ0 n4 + s[α0 (l5 − l4 ) + β0 (m5 − m4 ) + γ0 (n5 − n4 )]
and
p4 =
α0 l4
.
α0 l4 + β0 m4 + γ0 n4
By subtraction
p − p4 =
sα0
β0 (l5 m4 − m5 l4 ) + γ0 (n4 l5 − l4 n5 )
(α0 l4 + β0 m4 + γ0 n4 ) {α0 l4 + β0 m4 + γ0 n4 + s[α0 (l5 − l4 ) + β0 (m5 − m4 ) + γ0 (n5 − n4 )]}
while
p5 − p4 = α0
β0 (l5 m4 − m5 l4 ) + γ0 (n4 l5 − l4 n5 )
,
(α0 l4 + β0 m4 + γ0 n4 )(α0 l5 + β0 m5 + γ0 n5 )
and so
p − p4
= t,
p5 − p4
where
t=s
α0 l5 + β0 m5 + γ0 n5
.
α0 l4 + β0 m4 + γ0 n4 + s[α0 (l5 − l4 ) + β0 (m5 − m4 ) + γ0 (n5 − n4 )]
By a similar argument we also obtain
q − q4
= t,
q5 − q4
r − r4
= t.
r5 − r4
(8.4.16)
211
8.4. NORMALISED AREAL LINE COORDINATES
Thus we have that
p = p4 + t(p5 − p4 ),
q = q4 + t(q5 − q4 ),
r = r4 + t(r5 − r4 ).
(8.4.17)
Now
α0 l4 + β0 m4 + γ0 n4 = α0 δF (Z1 , Z4 , Z5 ) + β0 δF (Z2 , Z4 , Z5 ) + γ0 δF (Z3 , Z4 , Z5 )
=δF (Z0 , Z4 , Z5 )
and similarly
α0 l5 + β0 m5 + γ0 n5 = δF (Z0 , Z4 , Z6 ),
so that (8.4.16) can be rewritten as
t=s
δF (Z0 , Z4 , Z6 )
.
δF (Z0 , Z4 , Z5 ) + s[δF (Z0 , Z4 , Z6 ) − δF (Z0 , Z4 , Z5 )]
From this
1 − t = (1 − s)
and so we obtain
δF (Z0 , Z4 , Z5 )
,
δF (Z0 , Z4 , Z5 ) + s[δF (Z0 , Z4 , Z6 ) − δF (Z0 , Z4 , Z5 )]
s δF (Z0 , Z4 , Z6 )
t
=
.
1−t
1 − s δF (Z0 , Z4 , Z5 )
This is a key result.
8.4.13
(8.4.18)
Parametric equations of pairs in a parallel pencil
Given a line Z4 Z5 with equation l6 α + m6 β + n6 γ = 0, any parallel line will have an equation of
the form
l6 α + m6 β + n6 γ = s = s(α + β + γ)
and so
(l6 − s)α + (m6 − s)β + (n6 − s)γ = 0.
Thus the line with equation lα + mβ + nγ = 0 is parallel to Z4 Z5 if and only if
l = l6 − s,
m = m6 − s,
n = n6 − s,
for some s ∈ R. This gives parametric equations for point-pairs in a parallel pencil.
With normalized dual areal coordinates, the first of these becomes
α0 (l6 − s)
α0 (l6 − s) + β0 (m6 − s) + γ0 (n6 − s)
α0 l6 − α0 s
.
=
α0 l6 + β0 m6 + γ0 n6 − s
p=
Now s = 0 gives
p6 =
α0 l6
,
α0 l6 + β0 m6 + γ0 n6
and so
α0 l6 − α0 s
α0 l6
−
α0 l6 + β0 m6 + γ0 n6 − s α0 l6 + β0 m6 + γ0 n6
α0 l6 + β0 m6 + γ0 n6 − l6
= − sα0
(α0 l6 + β0 m6 + γ0 n6 )(α0 l6 + β0 m6 + γ0 n6 − s)
α0 − p6
.
=−s
α0 l6 + β0 m6 + γ0 n6 − s
p − p6 =
(8.4.19)
212
CHAPTER 8. DUAL CONICS
By a similar argument
β0 − q6
,
α0 l6 + β0 m6 + γ0 n6 − s
α0 − p6
r − r6 = − s
.
α0 l6 + β0 m6 + γ0 n6 − s
q − q6 = − s
(8.4.20)
We can make the change of variable
t=−
s
,
α0 l6 + β0 m6 + γ0 n6 − s
(8.4.21)
q = q6 + t(β0 − q6 ),
(8.4.22)
to obtain other parametric equations
p = p6 + t(α0 − p6 ),
r = r6 + t(γ0 − r6 ).
Note that
1−t=1+
and so
By (8.4.22) we have that
α0 l6 + β0 m6 + γ0 n6
s
=
,
α0 l6 + β0 m6 + γ0 n6 − s
α0 l6 + β0 m6 + γ0 n6 − s
t
s
.
=−
1−t
α0 l6 + β0 m6 + γ0 n6
(8.4.23)
(β0 − q6 )p − (α0 − p6 )q =p6 (β0 − q6 ) − (α0 − p6 )q6
=[p6 (β0 − q6 ) − (α0 − p6 )q6 ](p + q + r),
so that
[(1 − p6)(β0 − q6 ) + q6 (α0 − p6 )]p − [(1 − q6 )(α0 − p6 ) + p6 (β0 − q6 )]q − [p6 (β0 − q6 ) − (α0 − p6 )q6 ]r = 0.
With this as the equation (8.4.9), we have that (8.4.10) is equal to
α0 [(1 − p6 )(β0 − q6 ) + q6 (α0 − p6 )] − β0 [(1 − q6 )(α0 − p6 ) + p6 (β0 − q6 )] − γ0 [p6 (β0 − q6 ) − (α0 − p6 )q6 ],
and it is easily checked that this is equal to 0.
If in (8.4.21) we put t = 1 we find that there is no corresponding value of s as a number;
symbolically we use s = ∞ in this case.
8.5
8.5.1
Further dual concepts
Dual harmonic pencil and dual mid-pair
With the notation of §8.4.12, pairs with parameters s = s1 , s = s2 will form a harmonic pencil
with Z4 Z5 and Z4 Z6 when
s2
s1
=−
,
1 − s2
1 − s1
and if these have corresponding parameters of the other type, t = t1 and t = t2 , the pencil will be
dual harmonic when
t2
t1
=−
.
(8.5.1)
1 − t2
1 − t1
But by (8.4.18) these conditions are equivalent and so for pairs in a concurrent pencil, the concept
of dual-harmonic coincides with the concept of harmonic.
We recall that the condition (8.5.1) can be expressed in terms of cross-ratios as cr(0, 1, t1 , t2 ) =
−1. Now (8.4.23) is bilinear in s and t and so cross-ratios are invariant under it. Moreover s = 0
213
8.5. FURTHER DUAL CONCEPTS
corresponds to t = 0, and s = ∞ corresponds to t = 1. Then in a parallel pencil as in §8.4.13, for
a dual-harmonic range we need
t1
−1 = cr(0, ∞, t1 , t2 ) = .
t2
Turning next to an analogue of mid-points, to obtain a dual mid-pair of the Z0 -non-aligned pairs
(Z4 , Z5 ) and (Z4 , Z6 ) in a concurrent pencil we need to put t = 12 in (8.4.18). The corresponding
value of s is given by
δF (Z0 , Z4 , Z5 )
s
=
.
(8.5.2)
1−s
δF (Z0 , Z4 , Z6 )
It follows that the dual mid-pair of (Z4 , Z5 ) and (Z4 , Z6 ) in a concurrent pencil is (Z4 , Z7 ) where
Z7 ∈ Z5 Z6 and
Z5 Z7
δF (Z0 , Z4 , Z5 )
.
=
δF (Z0 , Z4 , Z6 )
Z7 Z6
It follows from this that if Z7′ is the point where Z0 Z4 meets Z5 Z6 , then (Z5 , Z6 , Z7′ , Z7 ) is a
harmonic range. For if Z7′ = (1 − s′ )Z5 + s′ Z6 , then
0 = δF (Z0 , Z4 , (1 − s′ )Z5 + s′ Z6 ) = (1 − s′ )δF (Z0 , Z4 , Z5 ) + s′ δF (Z0 , Z4 , Z6 ).
Thus
δF (Z0 , Z4 , Z5 )
Z5 Z7′
s′
=
−
=−
.
δF (Z0 , Z4 , Z6 )
1 − s′
Z7′ Z6
A second case occurs when Z0 ∈ Z5 Z6 but Z4 6∈ Z5 Z6 . Taking Z0 = (1 − s′ )Z5 + s′ Z6 we have
that
δF (Z0 , Z4 , Z5 ) δF ((1 − s′ )Z5 + s′ Z6 , Z4 , Z5 )
=
δF (Z0 , Z4 , Z6 ) δF ((1 − s′ )Z5 + s′ Z6 , Z4 , Z6 )
=−
s′
Z5 Z0
Z5 Z7
=−
=
.
′
1−s
Z0 Z6
Z7 Z6
Thus (Z5 , Z6 , Z7 , Z0 ) is a harmonic range.
For an analogue of mid-point in a parallel pencil, we put t =
s = −(α0 l6 + β0 m6 + γ0 n6 ).
8.5.2
1
2
in (8.4.23) and this yields
Dual mid-line
We found in §8.5.1 that the dual mid-pair of the Z0 -non-aligned pairs (Z4 , Z5 ) and (Z4 , Z6 ) in a
concurrent pencil is (Z4 , Z7 ) where Z7 = (1 − s)Z5 + sZ6 and s satisfies (8.5.2). Suppose that we
now take another point Z8 in Z4 Z5 , say with Z8 = (1 − u)Z4 + uZ5 . Then the dual mid-pair of
(Z4 , Z8 ) and (Z4 , Z6 ) will be a pair (Z4 , Z9 ) where Z9 = (1 − v)Z8 + vZ6 , and
v
δF (Z0 , Z4 , Z8 )
δF (Z0 , Z4 , (1 − u)Z4 + uZ5 )
=
=
1 − v δF (Z0 , Z4 , Z6 )
δF (Z0 , Z4 , Z6 )
s
uδF (Z0 , Z4 , Z5 )
=u
.
=
δF (Z0 , Z4 , Z6 )
1−s
We now wish to calculate
δF (Z4 , Z7 , Z9 ) =δF Z4 , (1 − s)Z5 + sZ6 , (1 − v)Z8 + vZ6
=(1 − s)vδF (Z4 , Z5 , Z6 ) + s(1 − v)δF (Z4 , Z6 , Z8 )
=(1 − s)vδF (Z4 , Z5 , Z6 ) + s(1 − v)δF Z4 , Z6 , (1 − u)Z4 + uZ5
=[(1 − s)v − s(1 − v)u]δF (Z4 , Z5 , Z6 )
=0.
This shows that the points Z4 , Z7 and Z9 are collinear and enables us to define the dual
mid-line of the lines Z4 Z5 and Z4 Z6 in a concurrent pencil as the line Z4 Z7 above.
214
8.5.3
CHAPTER 8. DUAL CONICS
Sensed-distance between point-pairs in a concurrent pencil
As in §8.4.10 consider fixed pairs (Z6 , Z7 ), (Z6 , Z8 ) and a variable pair (Z6 , Z), with normalized
pair-coordinates
(p4 , q4 , r4 ), (p5 , q5 , r5 ), (p, q, r),
and recall that in §8.4.11 we had the parameter t satisfy
t=
q − q4
r − r4
p − p4
=
=
.
p5 − p4
q5 − q4
r5 − r4
We use the first of these to define a dual sensed-ratio
(Z6 , Z7 ) (Z6 , Z)
(Z6 , Z7 ) (Z6 , Z8 )
p − p4
=
p5 − p4
=
(Z1 ,Z6 ,Z)
1 ,Z6 ,Z7 )
α0 δδFF (Z
− α0 δδFF (Z
(Z0 ,Z6 ,Z7 )
0 ,Z6 ,Z)
δF (Z1 ,Z6 ,Z7 )
1 ,Z6 ,Z8 )
α0 δδFF (Z
(Z0 ,Z6 ,Z8 ) − α0 δF (Z0 ,Z6 ,Z7 )
δF (Z1 , Z6 , Z)δF (Z0 , Z6 , Z7 ) − δF (Z0 , Z6 , Z)δF (Z1 , Z6 , Z7 ) δF (Z0 , Z6 , Z8 )
δF (Z1 , Z6 , Z8 )δF (Z0 , Z6 , Z7 ) − δF (Z0 , Z6 , Z8 )δF (Z1 , Z6 , Z7 ) δF (Z0 , Z6 , Z)
δF [(Z0 , Z1 ), (Z6 , Z7 ), (Z6 , Z)] δF (Z0 , Z6 , Z8 )
.
=
δF [(Z0 , Z1 ), (Z6 , Z7 ), (Z6 , Z8 )] δF (Z0 , Z6 , Z)
=
This should remain unchanged if we replace Z1 by Z2 or Z3 . Taking Z2 , consider the difference
δF (Z2 , Z6 , Z)δF (Z0 , Z6 , Z7 ) − δF (Z0 , Z6 , Z)δF (Z2 , Z6 , Z7 )
δF (Z2 , Z6 , Z8 )δF (Z0 , Z6 , Z7 ) − δF (Z0 , Z6 , Z8 )δF (Z2 , Z6 , Z7 )
δF (Z1 , Z6 , Z)δF (Z0 , Z6 , Z7 ) − δF (Z0 , Z6 , Z)δF (Z1 , Z6 , Z7 )
.
−
δF (Z1 , Z6 , Z8 )δF (Z0 , Z6 , Z7 ) − δF (Z0 , Z6 , Z8 )δF (Z1 , Z6 , Z7 )
If we express this as one algebraic fraction the numerator will be equal to
δF (Z0 , Z6 , Z7 )δF (Z2 , Z6 , Z) − δF (Z2 , Z6 , Z7 )δF (Z0 , Z6 , Z) •
δF (Z0 , Z6 , Z7 )δF (Z1 , Z6 , Z8 ) − δF (Z1 , Z6 , Z7 )δF (Z0 , Z6 , Z8 )
− δF (Z0 , Z6 , Z7 )δF (Z1 , Z6 , Z) − δF (Z1 , Z6 , Z7 )δF (Z0 , Z6 , Z) •
δF (Z0 , Z6 , Z7 )δF (Z2 , Z6 , Z8 ) − δF (Z2 , Z6 , Z7 )δF (Z0 , Z6 , Z8 ) .
The coefficient of δF (Z0 , Z6 , Z7 )2 in this big expression is
δF (Z2 , Z6 , Z)δF (Z1 , Z6 , Z8 ) − δF (Z1 , Z6 , Z)δF (Z2 , Z6 , Z8 )
=δF [(Z1 , Z2 ), (Z6 , Z8 ), (Z6 , Z)]
=δF [(Z6 , Z8 ), (Z6 , Z), (Z1 , Z2 )]
= − δF (Z8 , Z6 , Z)δF (Z6 , Z1 , Z2 ).
The coefficient of δF (Z1 , Z6 , Z7 )δF (Z2 , Z6 , Z7 ) in the big expression is equal to 0.
The coefficient of δF (Z0 , Z6 , Z7 )δF (Z1 , Z6 , Z7 ) in the big expression is equal to
− δF (Z2 , Z6 , Z)δF (Z0 , Z6 , Z8 ) + δF (Z0 , Z6 , Z)δF (Z2 , Z6 , Z8 )
= − δF [(Z0 , Z2 ), (Z6 , Z8 ), (Z6 , Z)]
= − δF [(Z6 , Z8 ), (Z6 , Z), (Z0 , Z2 )]
= − δF (Z8 , Z6 , Z)δF (Z6 , Z2 , Z0 ).
215
8.5. FURTHER DUAL CONCEPTS
Finally the coefficient of δF (Z0 , Z6 , Z7 )δF (Z2 , Z6 , Z7 ) in the big expression is equal to
− δF (Z0 , Z6 , Z)δF (Z1 , Z6 , Z8 ) + δF (Z1 , Z6 , Z)δF (Z0 , Z6 , Z8 )
=δF [(Z0 , Z1 ), (Z6 , Z8 ), (Z6 , Z)]
=δF [(Z6 , Z8 ), (Z6 , Z), (Z0 , Z1 )]
= − δF (Z8 , Z6 , Z)δF (Z6 , Z0 , Z1 ).
The sum of these terms is equal to
− δF (Z0 , Z6 , Z7 )δF (Z8 , Z6 , Z) δF (Z0 , Z6 , Z7 )δF (Z6 , Z1 , Z2 )
+δF (Z1 , Z6 , Z7 )δF (Z6 , Z2 , Z0 ) + δF (Z2 , Z6 , Z7 )δF (Z6 , Z0 , Z1 )
= − δF (Z0 , Z6 , Z7 )δF (Z8 , Z6 , Z)δF (Z0 , Z1 , Z2 )δF (Z6 , Z6 , Z7 )
=0,
on using §3.2.1. This shows that our sensed ratios are well-defined.
If we choose Z6 , Z7 , Z8 so that
δF [(Z0 , Z1 ), (Z6 , Z7 ), (Z6 , Z8 )]
= 1,
δF (Z0 , Z6 , Z8 )
we can define the sensed-distance
(Z6 , Z7 ) (Z6 , Z) =
8.5.4
δF [(Z0 , Z1 ), (Z6 , Z7 ), (Z6 , Z)]
.
δF (Z0 , Z6 , Z)
Parametric equations of dual-parallel concurrent pencils
Suppose that (p4 , q4 , r4 ) and (p5 , q5 , r5 ) are normalized coordinates of two Z0 -non-aligned pairs
(Z6 , Z7 ), (Z6 , Z8 ), respectively, where Z6 , Z7 and Z8 are non- collinear. Then by §8.4.10 we have
that
α6
p4 +
α0
α6
p5 +
α0
β6
q4 +
β0
β6
q5 +
β0
γ6
r4 =0,
γ0
γ6
r5 =0.
γ0
From these we have that
and so by subtraction
From this we have that
α6 β6
−
q4 +
β0
α0
α6 β6
−
q5 +
β0
α0
α6 γ6
α6
−
,
r4 = −
γ0
α0
α0
α6 γ6
α6
−
,
r5 = −
γ0
α0
α0
β6
α6 α6 γ6
−
−
(q5 − q4 ) +
(r5 − r4 ) = 0.
β0
α0
γ0
α0
γ6 α6
−
,
α0
γ0
α6 β6
−
r5 − r4 =u
,
β0
α0
q5 − q4 =u
216
CHAPTER 8. DUAL CONICS
and so
p5 − p4 =u
β6 γ6
−
,
γ0
β0
(8.5.3)
for some u ∈ R.
Suppose that similarly (p8 , q8 , r8 ) and (p9 , q9 , r9 ) here and following are normalized coordinates
of two pairs (Z10 , Z11 ), (Z10 , Z12 ), respectively, collinear with the point Z10 . Then by (8.5.3) we
also have that
p9 − p8 = v
γ10
β10 −
,
γ0
β0
q9 − q8 = v
γ10 α10
−
,
α0
γ0
r9 − r8 = v
α10 β10
−
,
β0
α0
for some v ∈ R.
We now suppose that the points Z6 and Z10 are dual-parallel, so that by §8.4.9, Z0 , Z6 and
Z10 are collinear. Then successively
α6 β6 γ6 α10 β10 γ10 =0,
α0 β0 γ0 β6
γ6 α6
α0
β0
γ0 α10 β10 γ10 =0,
α0
β0
γ0 1
1
1 β6
γ6
α6
α6
α6 α0
β0 − α0
γ0 − α0 α10 β10
γ10
α10
α10 =0,
α0
β0 − α0
γ0 − α0 1
0
0
and so
β6
α6 γ10
α10 α6 β10
α10 γ6
−
−
−
−
=
.
β0
α0 γ0
α0
γ0
α0 β0
α0
It follows from (8.5.3) and its analogue for Z10 that
(q5 − q4 )(r9 − r8 ) = (r5 − r4 )(q9 − q8 ),
so that
q9 − q8 =w(q5 − q4 ),
r9 − r8 =w(r5 − r4 ),
and so
p9 − p8 =w(p5 − p4 ),
for some w ∈ R.
We now introduce un-normalized pair-coordinates
l4 = δF (Z1 , Z6 , Z7 ), m4 = δF (Z2 , Z6 , Z7 ), n4 =δF (Z3 , Z6 , Z7 ),
l5 = δF (Z1 , Z6 , Z8 ), m5 = δF (Z2 , Z6 , Z8 ), n5 =δF (Z3 , Z6 , Z8 ),
so that
p4 =
α0 δF (Z1 , Z6 , Z8 )
α0 δF (Z1 , Z6 , Z7 )
, p5 =
,
δF (Z0 , Z6 , Z7 )
δF (Z0 , Z6 , Z8 )
(8.5.4)
217
8.5. FURTHER DUAL CONCEPTS
and so
δF (Z1 , Z6 , Z8 ) δF (Z1 , Z6 , Z7 ) −
δF (Z0 , Z6 , Z8 ) δF (Z0 , Z6 , Z7 )
α0
=
δF [(Z1 , Z0 ), (Z6 , Z8 ), (Z6 , Z7 )]
δF (Z0 , Z6 , Z7 )δF (Z0 , Z6 , Z8 )
α0
δF [(Z6 , Z8 ), (Z6 , Z7 ), (Z1 , Z0 )]
=
δF (Z0 , Z6 , Z7 )δF (Z0 , Z6 , Z8 )
δF (Z6 , Z7 , Z8 )δF (Z1 , Z0 , Z6 )
.
= − α0
δF (Z0 , Z6 , Z7 )δF (Z0 , Z6 , Z8 )
p5 − p4 =α0
By a similar argument
p9 − p8 = −α0
δF (Z10 , Z11 , Z12 )δF (Z1 , Z0 , Z10 )
.
δF (Z0 , Z10 , Z11 )δF (Z0 , Z10 , Z12 )
By division it follows that
p9 − p8 =kT6 ,T10 (p5 − p4 ),
and so by (8.5.4)
q9 − q8 =kT6 ,T10 (q5 − q4 ),
r9 − r8 =kT6 ,T10 (r5 − r4 ),
(8.5.5)
where
kT6 ,T10 =
δF (Z10 , Z11 , Z12 )δF (Z1 , Z0 , Z10 ) δF (Z0 , Z6 , Z7 )δF (Z0 , Z6 , Z8 )
,
δF (Z0 , Z10 , Z11 )δF (Z0 , Z10 , Z12 ) δF (Z6 , Z7 , Z8 )δF (Z1 , Z0 , Z6 )
(8.5.6)
is a non-zero number depending on Z6 , Z7 , Z8 , Z10 , Z11 and Z12 as well as Z0 and we use the
notation
T6 = (Z6 , Z7 , Z8 ), T10 = (Z10 , Z11 , Z12 ).
Although it ostensibly depends on Z1 as well, we can replace Z1 by any other point. In fact we
need to check that
δF (Z2 , Z0 , Z10 )
δF (Z3 , Z0 , Z10 )
δF (Z1 , Z0 , Z10 )
=
=
.
δF (Z1 , Z0 , Z6 )
δF (Z2 , Z0 , Z6 )
δF (Z3 , Z0 , Z6 )
Now the equality of the first two here corresponds to having
δF (Z1 , Z0 , Z10 )δF (Z2 , Z0 , Z6 ) − δF (Z2 , Z0 , Z10 )δF (Z1 , Z0 , Z6 )
=δF [(Z1 , Z2 ), (Z0 , Z10 ), (Z0 , Z6 )]
=δF [(Z0 , Z10 ), (Z0 , Z6 ), (Z1 , Z2 )]
=δF (Z0 , Z0 , Z6 )δF (Z10 , Z1 , Z2 ) − δF (Z10 , Z0 , Z6 )δF (Z0 , Z1 , Z2 )
=0,
as Z0 , Z6 , Z10 are collinear. The equality of the second and third follows similarly.
From (8.5.5) and (8.5.6), we have that corresponding to parametric equations
p = p4 + t(p5 − p4 ), q = q4 + t(q5 − q4 ), r = r4 + t(r5 − r4 ),
(t ∈ R),
for pairs collinear with Z6 , when Z10 is dual-parallel to Z6 we have the parametric equations
p = p8 + tkT6 ,T10 (p5 − p4 ), q = q8 + tkT6 ,T10 (q5 − q4 ), r = r8 + tkT6 ,T10 (r5 − r4 ),
for pairs collinear with Z10 .
(t ∈ R) (8.5.7)
218
8.5.5
CHAPTER 8. DUAL CONICS
Sensed-distances between point-pairs in a parallel pencil
In §8.4.13 for a pair (U, V ) in a parallel pencil based on (Z4 , Z5 ), we had the parametric equations
(8.4.22) for the normalized dual areal coordinates (p, q, r) of (U, V ). From these
t=
q − q6
r − r6
p − p6
=
=
.
α0 − p6
β0 − q6
γ0 − r6
(8.5.8)
From the first of these we define a sensed-distance
p − p6
(Z4 , Z5 ) (U, V ) =
α0 − p6
=
α0 δF (Z1 ,U,V )
α0 δF (Z1 ,Z4 ,Z5 )
δF (Z0 ,U,V ) − δF (Z0 ,Z4 ,Z5 )
1 ,Z4 ,Z5 )
α0 − αδ0FδF(Z(Z0 ,Z
4 ,Z5 )
δF (Z1 , U, V )δF (Z0 , Z4 , Z5 ) − δF (Z0 , U, V )δF (Z1 , Z4 , Z5 )
δF (Z0 , U, V )[δF (Z0 , Z4 , Z5 ) − δF (Z1 , Z4 , Z5 )]
1
δF [(Z0 , Z1 ), (Z4 , Z5 ), (U, V )]
=
.
δF (Z0 , Z4 , Z5 ) − δF (Z1 , Z4 , Z5 )
δF (Z0 , U, V )
=
8.5.6
(8.5.9)
Change of triple of reference
With the notation of §3.3.1, we note that
δF (α4 Z1 + β4 Z2 + γ4 Z3 , U, V )
δF (Z4 , U, V )
= α′0
δF (Z0 , U, V )
δF (Z0 , U, V )
δ
(Z
,
U,
V
)
δ
(Z
δF (Z3 , U, V )
F
1
F
2 , U, V )
=α′0 α4
+ β4
+ γ4
δF (Z0 , U, V )
δF (Z0 , U, V )
δF (Z0 , U, V )
δ
(Z
,
U,
V
)
δ
(Z
,
U,
V
)
β
γ4 δF (Z3 , U, V )
α
F
1
F
2
4
4
′
α0
+ β0
+ γ0
=α0
α0 δF (Z0 , U, V ) β0 δF (Z0 , U, V ) γ0 δF (Z0 , U, V )
α4
β4
γ4
=α′0
p+ q+ r .
α0
β0
γ0
p′ =α′0
Similarly we can find that
α5
β5
γ5
p+ q+ r ,
α0
β0
γ0
β6
γ6
α6
p+ q+ r .
r′ =γ0′
α0
β0
γ0
q ′ =β0′
8.6
8.6.1
Z0 -deleted point-pair conics
Deleted point-pair conics
In (8.1.8) we noted that the point-pairs on tangents/asymptotes to the proper conic with equation
γ 2 − 4kαβ = 0, have equation
1
n2 − lm = 0.
k
This uses un-normalized pair-coordinates, and by §8.4.1 the corresponding equation for normalized
pair-coordinates is
1 γ02
pq = 0.
(8.6.1)
r2 −
k α0 β0
In using this equation we are confining ourselves to Z0 -non-aligned pairs of points on tangents/asymptotes,
and so we are excluding from consideration any tangents or asymptotes which pass through the
point Z0 . We call such a point-pair conic Z0 -deleted.
8.6. Z0 -DELETED POINT-PAIR CONICS
8.6.2
219
Incidence equation; type of Z0 -deleted point-pair conic
Consider the Z0 -deleted point-pair conic with equation (8.6.1) and a concurrent or parallel pencil,
so that by §8.4.11 and §8.4.13 the point-pairs in the pencil have parametric equations
p = p4 + t(p5 − p4 ), q = q4 + t(q5 − q4 ), r = r4 + t(r5 − r4 ),
(t ∈ R).
Then for this pair to be in the Z0 -deleted point-pair conic it is necessary and sufficient that t
satisfy the equation of incidence
1 γ02
(p5 − p4 )(q5 − q4 ) t2
k α0 β0
1 γ02 1 γ02
+ 2r4 (r5 − r4 ) −
p4 q4 = 0.
p4 (q5 − q4 ) + q4 (p5 − p4 ) t + r42 −
k α0 β0
k α0 β0
(r5 − r4 )2 −
(8.6.2)
We say that we have a pencil with deficient incidence if the coefficient of t2 in (8.6.2)
vanishes, that is if
1 γ02
(r5 − r4 )2 −
(p5 − p4 )(q5 − q4 ) = 0.
k α0 β0
On inserting −(r5 − r4 ) = p5 − p4 + q5 − q4 in this, it becomes
1 γ02 (p5 − p4 )(q5 − q4 ) + (q5 − q4 )2 = 0.
(p5 − p4 )2 + 2 1 −
2k α0 β0
(8.6.3)
The discriminant of the quadratic expression in (8.6.3) is
1−
1 γ02 2
1 γ2
− 1 = 2 2 0 2 γ02 − 4kα0 β0 .
2k α0 β0
4k α0 β0
With fixed (p4 , q4 , r4 ) we allow (p5 , q5 , r5 ) to vary. Then for a given pair (Z4 , Z5 ) with normalized
pair-coordinates (p4 , q4 , r4 ), as the pair (U, V ) with normalized pair-coordinates (p5 , q5 , r5 ) varies,
the pencil will or will not have deficient incidence with the dual-conic.
When γ02 − 4kα0 β0 < 0 the equation (8.6.3) will have no real roots and so for the pair with
coordinates (p4 , q4 , r4 ) there will be no pair (p5 , q5 , r5 ) for which we have deficient incidence. In
this case we say that the Z0 -deleted point-pair conic is of dual-elliptic type; note by §7.2.1 that
for this, Z0 is to be in the interior region of the point conic with equation γ 2 − 4kαβ = 0.
When γ02 − 4kα0 β0 = 0 the equation (8.6.3) will have one real root and so for the pair with
coordinates (p4 , q4 , r4 ) there will be essentially one pair (p5 , q5 , r5 ) for which we have deficient
incidence. In this case we say that the Z0 -deleted point-pair conic is of dual-parabolic type;
note by §7.2.1 that for this, Z0 is to be a point of the point conic with equation γ 2 − 4kαβ = 0.
When γ02 − 4kα0 β0 > 0 the equation (8.6.3) will have two real roots and so for the pair with
coordinates (p4 , q4 , r4 ) there will be essentially two pairs (p5 , q5 , r5 ) for which we have deficient
incidence. In this case we say that the Z0 -deleted point-pair conic is of dual-hyperbolic type;
note by §7.2.1 that for this, Z0 is to be in the exterior region of the point conic with equation
γ 2 − 4kαβ = 0.
8.6.3
Contact for a Z0 -deleted point-pair conic
Suppose now that the pair (Z4 , Z5 ) with normalized pair-coordinates (p4 , q4 , r4 ) is in the Z0 deleted point-pair conic with equation (8.6.1). We say that a point Z8 on the line Z4 Z5 is a
contact of the dual-conic if for any pair (Z6 , Z7 ), different from (Z4 , Z5 ), with normalized paircoordinates (p5 , q5 , r5 ) and such that Z6 Z7 meets Z4 Z5 at Z8 , the equation of incidence (8.6.2) for
the concurrent pencil with vertex at Z8 has only the root t = 0 and there is not deficient incidence,
so that it does not reduce from a quadratic to a linear equation. For this we have
r42 −
1 γ02
p4 r4 = 0,
k α0 β0
220
CHAPTER 8. DUAL CONICS
and need the coefficient of t in (8.6.2) to vanish,
2r4 (r5 − r4 ) −
which reduces to
−
1 γ02 p4 (q5 − q4 ) + q4 (p5 − p4 ) = 0,
k α0 β0
1 γ02
1 γ02
q4 p5 −
p4 q5 + 2r4 r5 = 0.
k α0 β0
k α0 β0
On replacing the fixed (Z6 , Z7 ) with normalized pair-coordinates (p5 , q5 , r5 ) with the variable (U, V )
with normalized pair-coordinates (p, q, r), this becomes
−
8.6.4
1 γ02
1 γ02
q4 p −
p4 q + 2r4 r = 0.
k α0 β0
k α0 β0
(8.6.4)
Pole of a line
In §8.6.2 let us now take a parametrization of the other form
p=
1
λ
1
λ
1
λ
p4 +
p5 , q =
q4 +
q5 , r =
r4 +
r5 .
1+λ
1+λ
1+λ
1+λ
1+λ
1+λ
On insertion of this into (8.6.1) we obtain the second type of incidence equation
(r4 + λr5 )2 −
1 γ02
(p4 + λp5 )(q4 + λq5 ) = 0.
k α0 β0
(8.6.5)
To have a dual harmonic pencil we need the coefficient of λ to vanish, that is
2r4 r5 −
1 γ02
(q4 p5 + p4 q5 ) = 0.
k α0 β0
On replacing the fixed (p5 , q5 , r5 ) by the variable (p, q, r), this becomes
−
1 γ02
(q4 p + p4 q) + r4 r = 0,
2k α0 β0
or
q4 p + p4 q − 2k
(8.6.6)
α0 β0
r4 r = 0.
γ02
Except possibly in degenerate cases, this is the pair-equation of pairs in a concurrent pencil
with vertex a point Z8 ; we call Z8 the pole of the line Z4 Z5 . Note that this coincides with the
equation (8.6.4) so that a pole coincides with a contact when (p4 , q4 , r4 ) correspond to a pair in
the Z0 -deleted point-pair conic.
8.6.5
Dual-centre of a point-pair conic
We now undertake an analysis of when (8.6.6) fails to be an equation of a concurrent or parallel
pencil. This requires that
1 γ02
1 γ02
q4 = −
p4 = 2r4 .
−
k α0 β0
k α0 β0
This in turn would require that
q4 = p4 ,
r4 = −
and so in turn
p4 + q4 + r4 = −
1 γ02
p4 ,
2k α0 β0
p4
γ02 − 4kα0 β0 .
2kα0 β0
8.6. Z0 -DELETED POINT-PAIR CONICS
221
In the case where the Z0 -deleted point-pair conic is of parabolic type this will yield the contradiction
that p4 + q4 + r4 = 0, so in this case each Z0 -non-aligned line Z4 Z5 will yield a pencil.
When the dual-conic is of elliptic or hyperbolic type, (8.6.6) will fail to be an equation of a
pencil in the case
−2kα0 β0
−2kα0 β0
γ02
(p4 , q4 , r4 ) =
.
(8.6.7)
,
,
γ02 − 4kα0 β0 γ02 − 4kα0 β0 γ02 − 4kα0 β0
For a Z0 -deleted point-pair conic of elliptic or hyperbolic dual type this identifies a unique line
Z4 Z5 which does not yield a pencil.
Now taking the second form of (8.6.6) as (8.4.9), for it (8.4.10) will be equal to
α0 β0
r4
γ0
β0 α0 δF (Z1 , Z6 , Z7 ) + α0 β0 δF (Z2 , Z6 , Z7 ) − 2kα0 β0 δF (Z3 , Z6 , Z7 )
=
δF (Z0 , Z6 , Z7 )
α0 β0
[l4 + m4 − 2kn4 ].
=
δF (Z0 , Z6 , Z7 )
α0 q4 + β0 p4 − 2k
By §7.4.1 and (7.4.3) we see that this is equal to 0 when Z4 Z5 is a diametral line for the point
conic; thus for (8.6.6) to be an equation of a concurrent pencil, we need Z4 Z5 to be non-diametral.
In fact this line Z4 Z5 is the polar of Z0 with respect to the conic-locus with equation γ 2 −4kαβ =
0, that is the line with equation
−2kβ0 α − 2kα0 β + γ0 γ = 0,
(8.6.8)
in (7.2.23). For normalized pair-coordinates we should write this in the form
1
1
1
(−2kα0 β0 )α + (−2kα0 β0 )β + γ02 γ = 0,
α0
β0
γ0
so that we need
p4 = −2kα0 β0 j,
q4 = −2kα0 β0 j, quadr4 = γ02 j,
where
and so
1 = p4 + q4 + r4 = (γ02 − 4kα0 β0 )j,
j=
1
,
γ02 − 4kα0 β0
giving (8.6.7) as required.
We call this line the dual-centre of the Z0 -deleted point-pair conic.
8.6.6
Locus of mid-pairs of pairs in a point-pair conic which are dualparallel
With the notation of §8.5.4, we ask what triples (p, q, r) in (8.5.7) are in the Z0 -deleted point-pair
conic with equation (8.6.1). For this we obtain the equation of incidence
[r8 + tkT6 ,T10 (r5 − r4 )]2 −
1 γ02
[p8 + tkT6 ,T10 (p5 − p4 )][q8 + tkT6 ,T10 (q5 − q4 )] = 0.
k α0 β0
We ask that in this the sum of the roots be equal to 0 so that they are additive inverses, and so
(p8 , q8 , r8 ) corresponds to dual mid-pair of the two pairs in the dual-conic. For this we need the
coefficient of t to vanish, that is
2(r5 − r4 )r8 −
1 γ02
[(q5 − q4 )p8 + (p5 − p4 )q8 ] = 0.
k α0 β0
222
CHAPTER 8. DUAL CONICS
If we replace the fixed point Z10 in this with the variable point Z, we obtain for the concurrent
pencil with vertex Z the equation
−
1 γ02
1 γ02
(q5 − q4 )p −
(p5 − p4 )q + 2(r5 − r4 )r = 0.
k α0 β0
k α0 β0
(8.6.9)
Except possibly in degenerate cases, this is an equation of a concurrent pencil and we call
it a dual diametral concurrent pencil. It is natural to ask if this lies on the dual-centre
defined in §8.6.5 and we now show that this is indeed so. The dual-centre has un-normalized paircoordinates proportional to (l, m, n) = (−2kβ0 , −2kα0 , γ0 ), and so has normalized pair-coordinates
proportional to
(p, q, r) = (α0 l, β0 m, γ0 n) = (−2kα0 β0 , −2kα0 β0 , γ02 ).
If we insert these on the left-hand side in (8.6.9) we obtain
1 γ02
1 γ02
(q5 − q4 )(−2kα0 β0 ) −
(p5 − p4 )(−2kα0 β0 ) + 2(r5 − r4 )γ02
k α0 β0
k α0 β0
=2γ02 [q5 − q4 + p5 − p4 + r5 − r4 ]
−
=0.
Thus the point Z10 is on the dual-centre.
By (8.5.3) we note that we can rewrite (8.6.9) as
−
8.6.7
γ6 β6 α6 1 γ02 α6
1 γ02 γ6
β6
−
−
−
p−
q+2
r = 0.
k α0 β0 α0
γ0
k α0 β0 γ0
β0
β0
α0
(8.6.10)
Conjugate dual-diametral concurrent pencils
Take a point Z6 on the dual-centre, that is on the polar of Z0 , and as in §7.3.8 take the point Z11
so that the triple (Z0 , Z6 , Z11 ) is self-conjugate with respect to the point-conic. Take on Z0 Z6 any
point W which is an exterior point for the point conic. Then the concurrent pencils with vertices
Z6 and W are dual-parallel.
8.6. Z0 -DELETED POINT-PAIR CONICS
b
223
Z11
b
b
Z19
b
b
Z0
Z22
b
b
Z6
W
b
Z18
b
Figure 8.?. Corresponding dual-diametral concurrent pencil.
The polar of W will pass through Z11 ; let it meet the point conic at Z18 and Z19 , which will
then be points of contact of tangents from W ; let it also meet Z0 Z6 at Z22 . Now Z22 is on the
polar of W and Z11 so the polar of Z22 is W Z11 . It follows that (Z22 , Z11 , Z18 , Z19 ) is a harmonic
range. Thus the dual mid-pair of (W, Z18 ) and (W, Z19 ) is (W, Z11 ). Thus as W varies on Z0 Z6 ,
when there are tangents W Z18 , W Z19 to the point conic, the mid-pair of (W, Z18 ) and (W, Z19 )
always passes through the fixed point Z11 on the dual-centre. We say that the concurrent pencil
with vertex Z11 corresponds to the concurrent pencil with vertex Z6 .
As now the initial point Z6 in §8.6.6 is on the dual-centre in (8.6.7), its coordinates satisfy
equation (8.6.8). Then we must have
−2kβ0 α6 − 2kα0 β6 + γ0 γ6 = 0,
so that
γ6 =
2k
(α6 β0 + β6 α0 ).
γ0
(8.6.11)
Then the concurrent pencil with vertex our diametral point Z6 has equation
β6
2k
α6
p + q + 2 (α6 β0 + β6 α0 )r = 0.
α0
β0
γ0
(8.6.12)
Reverting to the notation of §8.6.6, to get the corresponding diametral point Z10 we insert the
condition (8.6.11) in the equation (8.6.10). This gives
1 1
β6 β6 α6
α6
+ 2kα0 β0
− (2kα0 β0 − γ02 )
(2kα0 β0 − γ02 )
p + −2kα0 β0
q
k α0 β0
α0
β0
α0
β0
α6
β6
−
r = 0.
(8.6.13)
+ 2
β0
α0
224
CHAPTER 8. DUAL CONICS
Thus for the initial diametral point Z6 with equation (8.6.12) we have the associated diametral
point Z10 with equation (8.6.13). To find the associated diametral point of Z10 we rewrite (8.6.12)
as
1
1
γ02
α6
γ2
β6
p+ 0
q + r = 0.
(8.6.14)
2k β0 α6 + α0 β6 α0
2k β0 α6 + α0 β6 β0
Similarly we rewrite (8.6.13) as
γ02
1
1
α10
γ2
β10
p+ 0
q + r = 0,
2k β0 α10 + α0 β10 α0
2k β0 α10 + α0 β10 β0
(8.6.15)
with
γ02
β6
2 α6
α10
1 (2kα0 β0 − γ0 ) α0 + 2kα0 β0 β0
1
=
,
β6
α6
β0 α10 + α0 β10 α0
α0 β0
β − α
0
γ02
1
β10
1
=
β0 α10 + α0 β10 β0
α0 β0
6
−2kα0 β0 α
α0
0
− (2kα0 β0 − γ02 ) ββ06
β6
β0
−
α6
α0
.
(8.6.16)
By division in (8.6.16) we have that
α6
− (2kα0 β0 − γ02 ) ββ06
−2kα0 β0 α
β10 /β0
0
=
,
β6
6
α10 /α0
(2kα0 β0 − γ02 ) α
α0 + 2kα0 β0 β0
and so
2kα0 β0
β10 /β0
β6 /β0 β10 /β0 β6 /β0
− (γ02 − 2kα0 β0 )
+
+ 2kα0 β0 = 0.
α10 /α0 α6 /α0
α10 /α0
α6 /α0
β6 /β0
10 /β0
This is symmetrical in αβ10
/α0 and α6 /α0 , so the point associated with Z10 is the original point Z6 .
We call the concurrent pencils with vertices Z6 and Z11 conjugate dual-diametral pencils.
8.6.8
Second dual of Newton’s property
We now set out to construct a second dual of Newton’s property. We take a triple of non-collinear
points (Z6 , Z7 , Z8 ), such that (Z6 , Z7 ) and (Z6 , Z8 ) are Z0 -non-aligned pairs and we let (p4 , q4 , r4 )
and (p5 , q5 , r5 ), respectively, be normalized coordinates for these pairs. Similarly we taken a triple
of non-collinear points (Z11 , Z12 , Z13 ), such that (Z6 , Z11 ), (Z11 , Z12 ) and (Z11 , Z13 ) are Z0 -nonaligned pairs and we let (p9 , q9 , r9 ) and (p10 , q10 , r10 ), respectively, be normalized coordinates for
the latter two pairs. We suppose the foregoing to be all fixed.
We further take variable points W and W ′ which are dual-parallel to Z6 and Z11 , respectively,
so that (Z6 , W ) and (Z11 , W ′ ) are both Z0 -aligned, and let (p′ , q ′ , r′ ) be normalized coordinates
of (W, W ′ ). We further take points Z15 and Z17 so that (W, Z15 ) and (W ′ , Z17 ) are both Z0 -nonaligned and write
T6 = (Z6 , Z7 , Z8 ), T11 = (Z11 , Z12 , Z13 ), TW = (W, W ′ , Z15 ), TW ′ = (W ′ , W, Z17 ).
By (8.5.6), corresponding to parametric equations
p = p4 + t(p5 − p4 ), q = q4 + t(q5 − q4 ), r = r4 + t(r5 − r4 ),
for normalized coordinates of pairs through Z6 , we have parametric equations
p = p′ + tkT6 ,TW (p5 − p4 ), q = q ′ + tkT6 ,TW (q5 − q4 ), r = r′ + tkT6 ,TW (r5 − r4 ),
for normalized coordinates of pairs collinear with W . On inserting these in the equation
r2 −
1 γ02
pq = 0,
k α0 β0
8.6. Z0 -DELETED POINT-PAIR CONICS
225
of the Z0 -deleted point-pair conic, we obtain the equation of incidence
2 1 γ02 ′
p + tkT6 ,TW (p5 − p4 ) q ′ + tkT6 ,TW (q5 − q4 ) = 0.
r′ + tkT6 ,TW (r5 − r4 ) −
k α0 β0
(8.6.17)
The product of the roots in this equals
r′2 −
kT2 6 ,TW (r5 − r4 )2 −
2
1 γ0
′ ′
k α0 β0 p q
2
1 γ0
k α0 β0 (p5 −
Similarly, corresponding to parametric equations
.
p4 )(q5 − q4 )
(8.6.18)
p = p9 + t(p10 − p9 ), q = q9 + t(q10 − q9 ), r = r9 + t(r10 − r9 ),
for normalized coordinates of pairs through Z11 , we have parametric equations
p = p′ + tkT11 ,TW ′ (p10 − p9 ), q = q ′ + tkT11 ,TW ′ (q10 − q9 ), r = r′ + tkT11 ,TW ′ (r10 − r9 ),
for normalized coordinates of pairs collinear with W ′ . On inserting these in the equation of the
dual-conic, we obtain the equation of incidence
2 1 γ02 ′
p + tkT11 ,TW ′ (p10 − p9 ) q ′ + tkT11 ,TW ′ (q10 − q9 ) = 0. (8.6.19)
r′ + tkT11 ,TW ′ (r10 − r9 ) −
k α0 β0
The product of the roots in this equals
r′2 −
kT2 11 ,TW ′ (r10 − r9 )2 −
2
1 γ0
′ ′
k α0 β0 p q
2
1 γ0
k α0 β0 (p10
The ratio of (8.6.20) to (8.6.18) is
(r5 − r4 )2 −
(r10 − r9 )2 −
.
− p9 )(q10 − q9 )
2
1 γ0
k α0 β0 (p5 − p4 )(q5 − q4 )
2
1 γ0
k α0 β0 (p10 − p9 )(q10 − q9 )
kT2 6 ,TW
.
kT2 11 ,TW ′
(8.6.20)
(8.6.21)
The first quotient in (8.6.21) is constant; we must arrange that the second one is so also.
By (8.5.5), kT6 ,TW is a constant multiple of
δF (W, W ′ , Z15 )δF (Z1 , Z0 , W )
,
δF (Z0 , W, W ′ )δF (Z0 , W, Z15 )
(8.6.22)
and kT11 ,TW ′ is a constant multiple of
δF (W ′ , W, Z17 )δF (Z1 , Z0 , W ′ )
.
δF (Z0 , W ′ , W )δF (Z0 , W ′ , Z17 )
We now suppose that
Z15 − W = Z7 − Z6 , Z17 − W ′ = Z12 − Z11 ,
which makes Z15 and Z17 variable points. Then
δF (W, W ′ , Z15 ) =δF (O, W ′ − W, Z15 − W )
=δF (O, W ′ − W, Z7 − Z6 ),
and similarly
δF (W ′ , W, Z17 ) =δF (O, W − W ′ , Z17 − W ′ )
=δF (O, W − W ′ , Z7 − Z6 ),
(8.6.23)
226
CHAPTER 8. DUAL CONICS
and the quotient of these two is equal to −1.
Suppose that W − Z0 = u(Z6 − Z0 ), as W ∈ Z0 Z6 . Then
δF (Z1 , Z0 , W ) =δF (Z1 − Z0 , O, W − Z0 )
=δF (Z1 − Z0 , O, u(Z6 − Z0 ))
=uδF (Z1 − Z0 , O, Z6 − Z0 )
=uδF (Z1 , Z0 , Z6 )
while
δF (Z0 , W, Z15 ) =δF (Z0 − W, O, Z15 − W )
=δF (Z0 − W, O, Z7 − Z6 )
=δF (u(Z0 − Z6 ), O, Z7 − Z6 )
=uδF (Z0 − Z6 , O, Z7 − Z6 )
=uδF (Z0 , Z6 , Z7 )
and the quotient of these is constant.
Suppose similarly that W ′ − Z0 = v(Z11 − Z0 ), as W ′ ∈ Z0 Z11 . Then
δF (Z1 , Z0 , W ′ ) =δF (Z1 − Z0 , O, W ′ − Z0 )
=δF (Z1 − Z0 , O, v(Z11 − Z0 ))
=vδF (Z1 − Z0 , O, Z11 − Z0 )
=vδF (Z1 , Z0 , Z11 )
while
δF (Z0 , W ′ , Z17 ) =δF (Z0 − W ′ , O, Z17 − W ′ )
=δF (Z0 − W ′ , O, Z12 − Z11 )
=δF (v(Z0 − Z11 ), O, Z12 − Z11 )
=vδF (Z0 − Z11 , O, Z12 − Z11 )
=vδF (Z0 , Z11 , Z12 )
and the quotient of these is constant.
It now follows that the quotient of (8.6.22) over (8.6.23) is constant and so (8.6.21) is constant.
8.6. Z0 -DELETED POINT-PAIR CONICS
Z12
Z17
b
Z0
b
b
Z6
227
b
Z11
b
Z7
b
Z21
b
b
b
W′
Z19
b
Z20
b
Z18
b
W
b
Z15
Figure 8.??. Second dual of the Newton property.
Suppose now that Z18 , Z19 are the points of contact of the tangents from W to the conic and
that Z20 , Z21 are the points of contact of the tangents from W ′ to the conic. Then we have that
(W, W ′ ), (W, Z18 ) (W, W ′ ), (W, Z19 )
(W ′ , W ), (W ′ , Z20 ) (W ′ , W ), (W ′ , Z21 )
is constant.
228
CHAPTER 8. DUAL CONICS
8.6.9
Dual of classical Greek property
b
Z6
b
b
Z ′′
b
b
Z0
Z ′′′
b
b
Z11
Z13
Z14
W
b
Z′
b
Figure 8.?. Dual of classical Greek property.
We start by taking Z6 and Z11 to be the vertices of conjugate dual-diametral concurrent pencils,
with Z6 an exterior point for the point conic. We suppose that Z13 and Z14 are the points of contact
of the tangents from Z6 to the point conic, so that Z13 Z14 is the polar of Z6 and contains Z0 and
Z11 . We take the range of points W = (1 − t)Z13 + tZ14 on Z13 Z14 and then the pencil of pairs
(Z6 , W ); in particular this pencil contains the pairs (Z6 , Z13 ) and (Z6 , Z14 ). When W is exterior
to the point conic, let Z ′ and Z ′′ be the points of contact of the tangents from W to the conic.
Then Z ′ Z ′′ is the polar of W and as such passes through Z6 and meets Z0 W in a point Z ′′′ such
that (Z ′ , Z ′′ , Z6 , Z ′′′ ) is a harmonic range. By §8.5.1, (W, Z6 ) is the dual mid-pair of (W, Z ′ ) and
(W, Z ′′ ). Then as in §8.6.8 we have that, as W varies,
(W, Z6 ), (W, Z ′ )
2
(Z6 , W ), (Z6 , Z13 ) (Z6 , W ), (Z6 , Z14 )
is constant.
Chapter 9
Affine methods and results
In this chapter, we focus attention on deriving properties of ellipses from known properties of circles.
This gives us a method of discovering new results for ellipses, and gives a unifying viewpoint. We
can then check by other methods whether analogous results hold for hyperbolas and parabolas.
9.1
9.1.1
Affine transformations
Affine transformations recalled
An affine transformation, as treated in §4.1.2 and §4.1.3, can be expressed in terms of Cartesian
coordinates as follows. It is a function f : Π → Π : Z → Z ′ such that if Z and Z ′ have coordinates
(x, y), (x′ , y ′ ), respectively, then
x′ =ax + by + k1 ,
y ′ =cx + dy + k2 ,
(9.1.1)
for some constants a, b, c, d, k1 , k2 for which
ad − bc 6= 0.
(9.1.2)
Studies of affine transformations were first published by the Swiss mathematician Euler in 1748.
If a curve C2 is the image of a curve C1 under such a function f , Euler used the terminology that
C1 and C2 have an affinity, and our term is descended from this.
We can write this transformation in the matrix form
′ x
x
=A
+ K,
y′
y
where
A=
a
c
b
d
,K =
k1
k2
.
We recall that det A = ad − bc 6= 0.
We can solve the above equations uniquely for x and y in terms of x′ and y ′ . For
′ x
x
A
=
− K,
y
y′
′ x
x
− A−1 K.
= A−1
y′
y
We recall that for
A=
a
c
229
b
d
230
CHAPTER 9. AFFINE METHODS AND RESULTS
we have
A−1 =
1
det A
d −b
−c a
.
Clearly this inverse transformation is an affine transformation as it has the appropriate form (9.1.1)
and, since det A. det(A−1 ) = 1, we have det(A−1 ) 6= 0 so that (9.1.2) also holds. We denote this
inverse transformation by f −1 .
Thus each affine transformation has an inverse which is also an affine transformation.
9.2
Geometrical affine invariants
We consider the images of basic geometrical sets under an affine transformation For each affine
transformation f , we have all the following properties.
9.2.1
Lines, segments and half-lines
For all distinct points Z1 , Z2 ∈ Π, f (Z1 Z2 ) = f (Z1 )f (Z2 ), so that each line is mapped onto a line.
If Z3 6∈ Z1 Z2 then f (Z3 ) 6∈ f (Z1 )f (Z2 ).
For if Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ) are distinct points, the line l = Z1 Z2 has parametric equations
x = x1 + t(x2 − x1 ), y = y1 + t(y2 − y1 ) where t ∈ R.
(9.2.1)
Then we have
x′ =a[x1 + t(x2 − x1 )] + b[y1 + t(y2 − y1 )] + k1
=ax1 + by1 + k1 + t[a(x2 − x1 ) + b(y2 − y1 )]
=ax1 + by1 + k1 + t[ax2 + by2 + k1 − (ax1 + by1 + k1 )]
=x′1 + t(x′2 − x′1 ),
y ′ =c[x1 + t(x2 − x1 )] + d[y1 + t(y2 − y1 )] + k1
=cx1 + dy1 + k2 + t[c(x2 − x1 ) + d(y2 − y1 )]
=cx1 + dy1 + k2 + t[cx2 + dy2 + k2 − (cx1 + dy1 + k2 )]
=y1′ + t(y2′ − y1′ ).
(9.2.2)
(9.2.3)
As
x′2 − x′1 =a(x2 − x1 ) + b(y2 − y1 ),
y2′ − y1′ =c(x2 − x1 ) + d(y2 − y1 ),
(9.2.4)
the equations
x′2 − x′1 =0,
y2′ − y1′ =0,
are equivalent to
a(x2 − x1 ) + b(y2 − y1 ) =0,
c(x2 − x1 ) + d(y2 − y1 ) =0.
As det A 6= 0, the only solution of these equations is x2 − x1 = y2 − y1 = 0, and this cannot happen
as Z1 6= Z2 . Thus f (Z1 ) 6= f (Z2 ). As t runs through R in (9.2.1), the point Z runs through all
the points on the line l; it follows that, as t runs through R, the points Z ′ ≡ (x′ , y ′ ) all lie on the
line Z1′ Z2′ = f (Z1 )f (Z2 ) and in fact fill out the whole of that line. It follows that f (l) is the line
f (Z1 )f (Z2 ).
9.2. GEOMETRICAL AFFINE INVARIANTS
231
On using (9.2.4) we see that (9.2.1)-(9.2.3) can be written neatly as
f [Z1 + t(Z2 − Z1 )] = f (Z1 ) + t[f (Z2 ) − f (Z1 )].
(9.2.5)
If Z3 6∈ Z1 Z2 , we must have f (Z3 ) 6∈ f (Z1 )f (Z2 ). For otherwise f (Z3 ) ∈ f (Z1 )f (Z2 ), and since
f −1 is an affine transformation, by what has just been proved f −1 [f (Z3 )] ∈ f −1 [f [(Z1 )]f −1 [f (Z2 )]
so that Z3 ∈ Z1 Z2 , which gives a contradiction.
Next we note that for all Z1 , Z2 ∈ Π, f ([Z1 , Z2 ]) = [f (Z1 ), f (Z2 )], so that each segment is
mapped onto a segment. For if, in (9.2.1), we take 0 ≤ t ≤ 1, then we have the points of the
segment [f (Z1 ), f (Z2 )]; hence f ([Z1 , Z2 ]) = [f (Z1 , f (Z2 )].
Finally for all distinct points Z1 , Z2 ∈ Π, f ([Z1 , Z2 ) = [f (Z1 ), f (Z2 ), so that each half-line is
mapped onto a half-line. For if, in (9.2.1), we take t ≥ 0, then we have the points of the halfline [Z1 , Z2 and as their images the points of the half-line [f (Z1 ), f (Z2 ); hence f ([Z1 , Z2 ) =
[f (Z1 , f (Z2 ) .
9.2.2
Pairs of lines
If l and m are intersecting lines, then f (l) and f (m) are intersecting lines. If l and m are parallel
lines, then f (l) and f (m) are parallel lines.
For, let l and m be lines and suppose first that they meet at a point Z. Then Z ∈ l, Z ∈ m so
f (Z) ∈ f (l), f (Z) ∈ f (m). Thus the lines f (l), f (m) have the point f (Z) in common.
Suppose next that l k m. Then either l = m or l ∩ m = ∅. If l = m then f (l) = f (m) and so
f (l) k f (m). Alternatively if l ∩ m = ∅ then we must have f (l) ∩ f (m) = ∅. For, since f −1 is a
function, if W ∈ f (l)∩f (m) we would have f −1 (W ) ∈ f −1 (f (l)) = l and f −1 (W ) ∈ ∩f −1 (f (m)) =
m. Thus we would have f −1 (W ) in both l and m, which gives a contradiction.
9.2.3
Closed half-planes
If H1 , H2 are closed half-planes with common edge l, then f (H1 ), f (H2 ) are closed half planes with
common edge f (l).
For let Z1 be a point of H1 which is not on the edge l. Let H3 , H4 be the half-planes with
common edge f (l), H3 the one containing f (Z1 ). Let Z ∈ H1 , and we first wish to show that then
f (Z) ∈ H3 . Suppose that on the contrary f (Z) 6∈ H3 . Then f (Z) is in H4 but not on its edge,
so there is some point W ∈ f (l) ∩ [f (Z1 ), f (Z)]. But f −1 [f (l)] = l, f −1 [f (Z1 ), f (Z)] = [Z1 , Z] so
f −1 (W ) ∈ l ∩ [Z1 , Z]. As f −1 (f (Z)) 6∈ l this implies that Z 6∈ H1 and so gives a contradiction.
Thus f (H1 ) ⊂ H3 .
Moreover, if W ∈ H3 , by what has just been proved applied to f −1 , we have f −1 (W ) is
in the half-plane with edge l which contains f −1 [f (Z1 )] = Z1 , that is f −1 (W ) ∈ H1 . Then
f [f −1 (W )] = W , so W is a value of f at some point of H1 . Thus f (H1 ) = H3 .
9.2.4
Triangles
If [Z1 , Z2 , Z3 ] is a triangle, then f ([Z1 , Z2 , Z3 ]) is the triangle
[f (Z1 ), f (Z2 ), f (Z3) ].
We first prove a more general result that f (E1 ∩ E2 ) = f (E1 ) ∩ f (E2 ) for any sets E1 and
E2 . First suppose that Z ∈ E1 ∩ E2 . Then Z ∈ E1 so f (Z) ∈ f (E1 ). Similarly f (Z) ∈ f (E2 )
so f (Z) inf (E1 ) ∩ f (E2 ). Thus f (E1 ∩ E2 ) ⊂ f (E1 ) ∩ f (E2 ). On the other hand suppose that
W ∈ f (E1 ) ∩ f (E2 ). Then W ∈ f (E1 ) so that W = f (Z1 ) for some Z1 ∈ E1 . Similarly W = f (Z2 )
for some Z2 ∈ E2 . Then f (Z1 ) = f (Z2 ). Now since f −1 is a function, f is 1:1 on Π. It follows
that Z2 = Z1 and so Z1 ∈ E1 ∩ E2 . Thus W is the value of f at some point in E1 ∩ E2 . It follows
that f (E1 ∩ E2 ) = f (E1 ) ∩ f (E2 ).
To apply this, we let H1 be the half-plane with edge Z2 Z3 in which Z1 lies, let H3 be the
half-plane with edge Z3 Z1 in which Z2 lies and H5 be the half-plane with edge Z1 Z2 in which Z3
232
CHAPTER 9. AFFINE METHODS AND RESULTS
lies. Then the triangle [Z1 , Z2 , Z3 ] is the intersection H1 ∩ H3 ∩ H5 , and the image of this under
f is f (H1 ) ∩ f (H3 ) ∩ H5 ), so the image of [Z1 , Z2 , Z3 ] is [f (Z1 ), f (Z2 ), f (Z3 )].
9.2.5
Parallelograms
If [Z1 , Z2 , Z3 , Z4 ] is a parallelogram, then
f ([Z1 , Z2 , Z3 , Z4 ]) = [f (Z1 ), f (Z2 ), f (Z3 ), f (Z4 )]
and this is also a parallelogram.
To see this, suppose that Z1 Z2 k Z3 Z4 , Z1 Z4 k Z2 Z3 and Z3 6∈ Z1 Z2 . Let H1 , H3 , H5 , H7
be the half-planes with edges Z1 Z2 , Z1 Z4 , Z2 Z3 , Z3 Z4 , respectively and containing the points
Z4 , Z3 , Z1 , Z2 , respectively. Then the parallelogram [Z1 , Z2 , Z3 , Z4 ] is the intersection H1 ∩ H3 ∩
H5 ∩ H7 , and
f (H1 ∩ H3 ∩ H5 ∩ H7 ) = f (H1 ) ∩ f (H3 ) ∩ f (H5 ) ∩ f (H7 ).
Moreover f (Z1 )f (Z2 ) k f (Z3 )f (Z4 ), f (Z1 )f (Z4 ) k f (Z2 )f (Z3 ) and f (Z3 ) 6∈ f (Z1 )f (Z2 ).
9.3
Behaviour of distance and sensed-distance
We now consider behaviour of distance and sensed-distance under an affine transformation.
9.3.1
Magnification ratio
Let f as in (91̇.1) be an affine transformation. Consider distinct points Z1 ≡F (x1 , y1 ), Z2 ≡F
(x2 , y2 ), where F = ([O, I , [O, J ) is our frame of reference. Consider the translated frame
of reference F1 = ([Z1 , I1 , [Z1 , J1 ) where I1 = tO,Z1 (I), J1 = tO,Z1 (J). Let θ be the polar
angle with support |I1 Z1 Z2 in A(F1 ). On using the notation |Z1 , Z2 | = r and recalling that
Z2 ≡F1 (x2 − x1 , y2 − y1 ), we recall that
x2 − x1 = r cos θ, y2 − y1 = r sin θ.
It follows that
x′1 = ax1 + by1 + k1 , x′2 = ax2 + by2 + k1 ,
so
x′2 − x′1 = a(x2 − x1 ) + b(y2 − y1 ) = r(a cos θ + b sin θ).
Similarly
y1′ = cx1 + dy1 + k2 , y2′ = cx2 + dy2 + k2 ,
so
y2′ − y1′ = c(x2 − x1 ) + d(y2 − y1 ) = r(c cos θ + d sin θ).
Then
where
|f (Z1 ), f (Z2 )|
= k(f, θ),
|Z1 , Z2 |
k(f, θ)2 =(a cos θ + b sin θ)2 + (c cos θ + d sin θ)2
=(a2 + c2 ) cos2 θ + (b2 + d2 ) sin2 θ + 2(ab + cd) cos θ sin θ
1
1
= (a2 + c2 + b2 + d2 ) + (a2 + c2 − b2 − d2 ) cos 2θ + (ab + cd) sin 2θ.
2
2
We refer to k(f, θ) as the magnification ratio. We note that this depends only on f (because
of a, b, c, d) and the angle of inclination θ of the half-line [Z1 , Z2 or the angle of inclination φ of
9.3. BEHAVIOUR OF DISTANCE AND SENSED-DISTANCE
233
half-line [Z2 , Z1 . Our detail above was based on using the half-line [Z1 , Z2 . If we interchange Z1
and Z2 , we would be considering instead the translated frame of reference F2 = ([Z2 , I2 , [Z2 , J2 )
where I2 = tO,Z2 (I), J2 = tO,Z2 (J), and φ, the polar angle with support |I2 Z2 Z1 in A(F2 ). Then
we would have Z1 ≡F2 (x1 − x2 , y1 − y2 ), and that
x1 − x2 = r cos φ, y1 − y2 = r sin φ.
This would yield
cos φ = − cos θ, sin φ = − sin θ,
so that
cos2 φ = cos2 θ, sin2 φ = sin2 θ, cos φ sin φ = cos θ sin θ.
It follows that k(f, φ) = k(f, θ), and so the magnification ratio depends only on f and the line
Z1 Z2 .
9.3.2
Distances on parallel lines
Suppose that we have parallel lines Z1 Z2 k Z3 Z4 . Then
x4 − x3 = j(x2 − x1 ), y4 − y3 − j(y2 − y1 ),
for some non-zero j. It follows that |Z3 , Z4 | = |j||Z1 , Z2 |, and if ψ is the polar angle of [Z3 , Z4 in
an appropriate frame of reference, as well as
x2 − x1 = |Z1 , Z2 | cos θ, y2 − y1 = |Z1 , Z2 | sin θ,
we have that
x4 − x3 = |Z3 , Z4 | cos ψ, y4 − y3 = |Z3 , Z4 | sin ψ,
and so
j(x2 − x1 ) = |j||Z1 , Z2 | cos ψ, j(y2 − y1 ) = |j||Z1 , Z2 | sin ψ.
It follows that
cos θ =
|j|
|j|
cos ψ = ± cos ψ, sin θ =
sin ψ = ± sin ψ,
j
j
and thus k(f, ψ) = k(f, θ).
As
|f (Z3 ), f (Z4 )|
|f (Z1 ), f (Z2 )|
= k(f, θ),
= k(f, ψ),
|Z1 , Z2 |
|Z3 , Z4 |
we have
|f (Z3 ), f (Z4 )|
|f (Z1 ), f (Z2 )|
=
,
|Z3 , Z4 |
|Z1 , Z2 |
from which it follows that
|f (Z3 ), f (Z4 )|
|Z3 , Z4 |
=
.
|Z1 , Z2 |
|f (Z1 ), f (Z2 )|
It follows that if f is an affine transformation, then the ratios of distances on parallel lines
(including the case of a line being parallel to itself) are invariant under f .
234
9.3.3
CHAPTER 9. AFFINE METHODS AND RESULTS
Sensed distances
In fact we can strengthen the result in §9.3.2 to ratios of sensed distances on parallel lines, and we
continue to use the notation there. If we consider the point Z5 such that Z5 − Z3 = Z2 − Z1 , then
[Z1 , Z2 , Z5 , Z3 ] is a parallelogram and Z5 ∈ Z3 Z4 . We take the natural order on Z1 Z2 in which
Z1 precedes Z2 and the natural order on Z3 Z4 under which Z3 precedes Z5 in specifying sensed
distances on these lines. Then Z1 Z2 = |Z1 , Z2 |, Z3 Z5 = |Z1 , Z5 |, and as Z4 − Z3 = j(Z2 − Z1 ) we
have that Z3 Z4 = jZ1 Z2 . We will always try to make this type of arrangement for sensed distances
on parallel lines.
As Z4 = Z3 + j(Z2 − Z1 ), the argument in §9.2.1 can be modified to show that Z4′ = Z3′ +
j(Z2′ − Z1′ ) and so
Z3′ Z4′
Z3 Z4
=j=
.
′
′
Z1 Z2
Z1 Z2
Thus the ratios of sensed distances on parallel lines (including the case of a line being parallel to
itself) are affine invariant. In fact this can bw deduced from (1.1.7).
To deal with sensed distance rather than sensed distances, for points Z on the line Z1 Z2 we
take
x − x1 = t cos θ, y − y1 = t sin θ, (t ∈ R),
thus replacing the non-negative r in §9.3.1 by the number t of arbitrary sign. Then we have that
x′ , y ′ ) = (x′1 , y1′ ) + t(a cos θ + b sin θ, c cos θ + d sin θ).
Now
|(a cos θ + b sin θ, c cos θ + d sin θ)| = k(f, θ),
and so
Z ′ = Z1′ + tk(f, θ)
a cos θ + b sin θ c cos θ + d sin θ
,
k(f, θ
k(f, θ)
and so there is a unique angle ψ ∈ A(F1 ) so that
Z ′ − Z1′ = tk(f, θ)(cos ψ, sin ψ), Z1 Z = tk(f, θ)(cos ψ, sin ψ).
This shows the behaviour of sensed distance.
9.3.4
Mid-points, centroids and harmonic ranges
If Z1 6= Z2 , and Z0 = mp(Z1 , Z2 ), that is Z0 is the mid-point of Z1 and Z2 , then
Z1 Z0
1
= .
2
Z1 Z2
For any affine transformation f it follows that
f (Z1 )f (Z0 )
f (Z1 )f (Z2 )
=
1
.
2
Hence f (Z0 ) = mp(f (Z1 ), f (Z2 )). This is obvious when Z1 = Z2 , so we see that being a mid-point
is affine invariant. This could also be proved using ratios of distances and the results in §9.2.1.
From this we see that if Ce is the centroid of the triangle [Z1 , Z2 , Z3 ], that is the point of
concurrency of the lines joining the vertices to the mid-points of the opposite sides, then f (Ce ) is
the centroid of the triangle [f (Z1 ), f (Z2 ), f (Z3 )]. For if Z4 is the mid-point of Z2 and Z3 , then
Ce ∈ Z1 Z4 . It follows that f (Ce ) ∈ f (Z1 Z4 ) = f (Z1 )f (Z4 ) and f (Z4 ) is the mid-point of f (Z2 )
and f (Z3 ). Thus f (Ce ) is on the median of the image triangle which passes through the vertex
f (Z1 ). It follows that f (Ce ) also has this property with respect to the other two medians, so that
235
9.3. BEHAVIOUR OF DISTANCE AND SENSED-DISTANCE
it is on all three medians and thus is their point of concurrency. This shows that being the centroid
of a triangle is affine invariant.
We also note that if (A, B, C, D) is a harmonic range, so that these points are collinear and
C, D, in some order, divide [A, B] internally and externally in the same ratio, then
AC
AD
=−
.
CB
DB
It follows for any affine transformation f that f (A), f (B), f (C), f (D) are collinear and
f (A)f (C)
f (C)f (B)
=−
f (A)f (D)
f (D)f (B)
.
Thus f (C), f (D), in some order, divide [f (A), f (B)] internally and externally in the same ratio.
So internal and external division of a segment in equal ratios , or a harmonic range, is affine
invariant.
9.3.5
Ratio of sensed-areas, orientation of triples of non-collinear points
We recall that for Z1′ = f (Z1 ), Z2′ = f (Z2 ), Z3′ = f (Z3 ), where f is our affine transformation,

 ′


x1 x2 x3
x1 x′2 x′3
1
1
δF (Z1 , Z2 , Z3 ) = det  y1 y2 y3  , δF (Z1′ , Z2′ , Z3′ ) = det  y1′ y2′ y3′  .
2
2
1
1
1
1
1
1
We wish to derive from this that
δF (Z1′ , Z2′ , Z3′ ) = det(A) δF (Z1 , Z2 , Z3 ).
This can be proved using matrix methods
mation, we have
 ′  
x
 y′  = 
1
and so
x′1
 y1′
1

as matrix multiplication
 ′
x1
det  y1′
1
x′2
y2′
1
(9.3.1)
as follows. With x, y, x′ , y ′ as in our affine transfora b
c d
0 0
 
x′3
a
y3′  =  c
1
0


k1
x
k2   y 
1
1

b k1
x1
d k2   y1
0 1
1
is row by column. Then


x′2 x′3
a
y2′ y3′  = det  c
1
1
0
x2
y2
1


b k1
d k2  det 
0 1

x1 x2 x3
= det(A) det  y1 y2 y3
1
1
1

x3
y3 
1
x1 x2
y1 y2
1
1


x3
y3 
1
,
as stated.
From this, if (Z1 , Z2 , Z3 ), (Z4 , Z5 , Z6 ) are triples of non-collinear points, then
δF (Z1′ , Z2′ , Z3′ ) = det(A) δF (Z1 , Z2 , Z3 ), δF (Z4′ , Z5′ , Z6′ ) = det(A) δF (Z4 , Z5 , Z6 ),
so by division
δF (Z4 , Z5 , Z6 )
δF (Z4′ , Z5′ , Z6′ )
=
.
δF (Z1′ , Z2′ , Z3′ )
δF (Z1 , Z2 , Z3 )
236
CHAPTER 9. AFFINE METHODS AND RESULTS
Hence, the ratio of the sensed areas of two triples of non-collinear points is affine invariant.
From (9.3.1) we see that if det(A) > 0 then the ordered triples of points (Z1 , Z2 , Z3 ) and
(Z1′ , Z2′ , Z3′ ) are both anti-clockwise or both clockwise, so that f preserves orientation, while if
det(A) < 0, one of (Z1 , Z2 , Z3 ) and (Z1′ , Z2′ , Z3′ ) is anti-clockwise and the other is clockwise, so
that f reverses orientation. Thus similarity of orientation of two ordered triples of non-collinear
points (Z1 , Z2 , Z3 ) and (Z1′ , Z2′ , Z3′ ) is affine invariant, and so is dis-similarity of orientation.
9.4
Deduction of properties of ellipses
We now apply affine transformations to deduce properties of ellipses from known properties of
circles.
9.4.1
Ellipse as image of a circle
As seen in §7.8.2, by suitably choosing the axes of coordinates any ellipse E can be taken to have
the equation
x′2
y ′2
+
= 1,
(9.4.1)
a2
b2
where 0 < b < a. Consider the transformation f :x′ = x,
i.e.
x′
y′
=
1
0
As
det
y′ =
1
0
0
b
a
0
b
a
=
b
y,
a
(9.4.2)
x
y
b
6= 0,
a
this is an affine transformation and we note that its inverse is
a
f −1 : x = x′ , y = y ′ .
b
Let C be the circle with equation
.
x2 + y 2 = a2 ,
(9.4.3)
(9.4.4)
which is called the auxiliary circle of E. If Z ≡ (x, y) is a point of C, so that x, y satisfy the
equation (9.4.4) of the circle, then x′ , y ′ satisfy the equation (9.4.1) of the ellipse and so Z ′ ∈ E.
Thus f (C) ⊂ E. Conversely, if Z ′ ∈ E so that x′ , y ′ satisfy its equation (9.4.1), then x, y satisfy the
equation (9.4.4) of the circle and so Z ∈ C, f (Z) = Z ′ . Thus f (C) = E.
Incidentally this shows that being a circle is not affine invariant.
9.4.2
Chord, centre, diameter
We now consider the intersection of the ellipse E and any line l′ . By §9.4.1 E is the image of some
circle C under some affine transformation f . Then l = f −1 (l′ ) is a line and we recall that l meets
C in no point, precisely one point, or precisely two points. Correspondingly l′ meets E in no point,
in precisely one point, or in precisely two points.
Next consider any distinct points P ′ , Q′ which belong to the ellipse E; then the segment [P ′ , Q′ ]
is called a chord of the ellipse E; the line P ′ Q′ meets E only in the points P ′ , Q′ .
Let O be the centre of the circle C in the second last paragraph and f be as there. Suppose that
P ′ and Q′ are points on the ellipse and that O′ = f (O) lies on the line P ′ Q′ . Then O = f −1 (O)
lies on the line P Q of the circle C, where P = f −1 (P ′ ), Q = f −1 (Q′ ), and P and Q lie on the circle.
As is well known, then O is the mid-point of P and Q, and so O′ is the mid-point of P ′ and Q′ .
237
9.4. DEDUCTION OF PROPERTIES OF ELLIPSES
Thus the point O′ is the mid-point of every chord of the ellipse which lies on a line which contains
O′ . There cannot be another point with this property of O′ , for if O1′ 6= O′ , then the line O′ O1′
can meet E in only two points P ′ , Q′ , and O′ and O1′ cannot both be mid-points of P ′ and Q′ . For
this reason O′ is called the centre of E, and every chord of E containing O′ is called a diameter
of E.
9.4.3
Tangent
When in §9.4.2 l meets C in precisely one point P , then l is the tangent to C at P ; in that case l′
meets E in precisely one point P ′ , and then by §7.2.4 and §7.3.5, l′ is the tangent to E at P ′ .
Now let [P ′ , Q′ ] be any diameter of the ellipse E, and let l′ , m′ be the tangents to E at P ′ and
′
Q , respectively. Then l = f −1 (l′ ) and m = f −1 (m′ ) are tangents to the circle C at the points P
and Q. But [P, Q] is a diameter of C, so the tangents l, m, being both perpendicular to P Q, are
parallel to each other. Now f maps parallel lines to parallel lines so l′ k m′ . Thus tangents to an
ellipse at the end-points of a diameter are parallel to each other.
For an ellipse E as in §9.4.2, let P ′ be any point in the plane which does not lie on the ellipse.
Then P = f −1 (P ′ ) does not lie on the circle C, so that by §7.2.1 either P is an interior point for
C and so every line l through P meets C in two points, or else P is an exterior point for C and
so some lines l through P meet C in two points while others meet it in just one point or not at
all. It follows that either every line l′ through P ′ meets E in two points and then we recall that P ′
is an interior point for E, or else some lines through P ′ meet E in two points while others meet
it in just one point or not at all, and then we recall that P ′ is an exterior point for E. Note that
in particular the centre is an interior point for an ellipse. We recall that if P is an exterior point
for a circle, then there are exactly two tangents to the circle which pass through P . By applying
our affine transformation, we deduce that if P ′ is an exterior point for an ellipse, then there are
exactly two tangents to the ellipse which pass through P ′ .
By modifying the reasoning in §9.4.1 we can see that for the ellipse with equation (9.4.1), the
′2
′2
interior points P ′ ≡ (x′ , y ′ ) are those for which xa2 + yb2 < 1 and the exterior points are those for
′2
′2
which xa2 + yb2 > 1.
We derive an equation for a tangent to the ellipse with equation (9.4.1) as follows. If Z1 ≡
(x1 , y1 ) is a point on the circle C with equation (7.4.4), then the tangent l1 to C at Z1 has the
equation
x1
y − y1 = − (x − x1 ),
y1
and so
xx1 + yy1 = x21 + y12 = a2 .
Then by (9.4.2) and (9.4.3)
a
l1′ = f (l1 ) ≡ x′ x1 + y ′ y1 = a2 .
b
But as Z1′ ≡ (x′1 , y1′ ) = f (Z1 ), we have that x1 = x′1 , y1 =
equation
a a
x′ x′1 + y ′ . y1′ = a2 ,
b b
and so
y ′ y1′
x′ x′1
+
= 1.
a2
b2
This is the equation of the tangent to E at Z1′ ≡ (x′1 , y1′ ) ∈ E.
Thus if Z1′ ≡ (x′1 , y1′ ) is a point of the ellipse E, so that
x′2
y1′2
1
+
= 1,
a2
b2
a ′
b y1
and Z1′ ∈ E. Thus l1′ has the
238
CHAPTER 9. AFFINE METHODS AND RESULTS
then the tangent to E at Z1′ is the line with equation
x′ x′1
y′y′
+ 2 1 = 1.
2
a
b
2
Note that both l1 and l1′ meet the major axis at the point S1 ≡ ( xa1 , 0).
9.5
Results for ellipses, based mainly on distance and sensed
distance
With this preparation we can deal with conjugate diameters.
9.5.1
Locus of mid-points
Let [P ′ , Q′ ] be any diameter of the ellipse E and consider the chords [U ′ , V ′ ] where U ′ , V ′ are on
a line which is parallel to P ′ Q′ . Then with the notation of §9.4.2, [U, V ] = [f −1 (U ′ ), f −1 (V ′ )] are
chords of the circle C, all on lines parallel to P Q. Let [R, S] is the diameter of the circle C such
that P Q is perpendicular to RS. Then U V meets RS in a point W . Now RO ⊥ P Q, P Q k U V
so RO ⊥ U V , i.e., OW ⊥ U V . Then by Pythagoras’ theorem applied twice,
|O, V |2 = |O, W |2 + |W, V |2 ,
|O, U |2 = |O, W |2 + |W, U |2 ,
so as |O, U |2 = |O, V |2 we have |W, U |2 = |W, V |2 and so |W, U | = |W, V |. As W ∈ U V it follows
that W is the mid-point of U and V , is inside the circle and so is on [R, S]. Conversely suppose
that W is any point of [R, S] other that an end-point or the centre O of the circle. Then the line
though W which is parallel to P Q meets the circle in points U and V , and by the above W is the
mid-point of U and V . Thus W = mp(U, V ) ∈ [R, S], and in fact such mid-points fill out all of
[R, S]. Then f [mp(U, V )] = mp(U ′ , V ′ ) ∈ [R′ , S ′ ], and in fact all such mid-points W ′ fill out the
diameter [R′ , S ′ ] of the ellipse. Thus the locus of mid-points of chords of an ellipse E which are on
parallel lines, is a diameter (less its end-points) of E.
9.5.2
Conjugate diameters
In fact in §9.5.1, [P, Q] also bisects all chords of the circle which are on lines parallel to RS; then
[P ′ , Q′ ] bisects all chords of the ellipse E which are on lines parallel to R′ S ′ . Thus given any
diameter [P ′ , Q′ ] of an ellipse E, there is a second diameter [R′ , S ′ ], such that [P ′ , Q′ ] bisects all
chords of E on lines parallel to R′ S ′ and [R′ , S ′ ] bisects all chords of E on lines parallel to P ′ Q′ .
We recall that such diameters [P ′ , Q′ ], [R′ , S ′ ] are called conjugate diameters of E. Note that
a pair of conjugate diameters of an ellipse are images of diameters of a circle, the latter lying on
perpendicular lines.
9.5.3
Elliptic analogue of Grassmann supplement for a position vector
Given any point Z1 6= O, we recall from Barry [2, p. 191] the concept of the Grassmann supplement
−−→
−−→
−−→⊥
OZ2 of the position vector OZ1 , denoted by OZ1 and specified by
Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), x2 = −y1 , y2 = x1 .
Then |O, Z1 | = |O, Z2 | and Z2 is obtained from Z1 by anti-clockwise rotation about O through a
right-angle.
If Z1 is any point on the circle C with equation (9.4.4) then Z2 also lies on this circle. If we
−−→ −−→ −−−→
define the point W1 by OZ1 + OZ2 = OW1 , then [O, Z1 , W1 , Z2 ] is a parallelogram which, since
adjacent sides have equal lengths and lie on perpendicular adjacent side-lines, must be a square.
9.5. RESULTS FOR ELLIPSES, BASED MAINLY ON DISTANCE AND SENSED DISTANCE239
Thus Z1 W1 is the tangent to C at the point Z1 and is parallel to the diametral line OZ2 ; similarly
Z2 W1 is the tangent to C at the point Z2 and is parallel to the diametral line OZ1 .
−−→
−−→ −−→
−−−→
−−→⊥
Continuing from this, suppose that OZ2 = OZ3 and that OZ2 + OZ3 = OW2 . Then Z3 is
the point on the circle diametrically opposite to Z1 . As above, Z2 W2 is the tangent at Z2 and is
parallel to OZ3 = Z1 Z3 ; also Z3 W2 is the tangent at Z3 and is parallel to OZ2 and so to Z1 W1 .
−−→ −−→
−−→⊥ −−→
We continue this process twice more, first supposing that OZ3 = OZ4 and that OZ3 + OZ4 =
−−−→
OW3 . Then Z4 is the point on the circle diametrically opposite to Z2 , Z3 W2 = Z3 W3 is the tangent
at Z3 and is parallel to OZ4 = Z2 Z4 ; also Z4 W3 is the tangent at Z4 and is parallel to OZ3 and
so to Z1 Z3 .
−−−→
−−→ −−→
−−→
−−→⊥
Finally we suppose that OZ4 = OZ5 and that OZ4 + OZ5 = OW4 . Then we find that
Z5 = Z1 so that we are back where we started, and Z4 W4 , Z1 W4 are the tangents at Z4 and Z1 ,
respectively. It follows that [W1 , W2 , W3 , W4 ] is a square circumscribed to the circle; the points
of contact (Z1 , Z3 ) and (Z2 , Z4 ) are at the end-points of diameters on perpendicular lines. The
square is the union of four non-overlapping congruent squares.
b
Z
b
Z′
b
S′
b
b
Z̄
θ
O
b
R′
b
Q′
Figure 9.1
If Z ′ is a point of the ellipse E with equation (9.4.1), let Z̄ be the foot of the perpendicular
from Z ′ to the major axis, and let Z be the point on the auxiliary circle C with equation (9.4.4)
such that Z ′ ∈ [Z, Z̄]. Then if θ is the polar angle of the point Z and Z ≡ (x, y), Z ′ ≡ (x′ , y ′ ),
then
x = a cos θ, y = a sin θ,
and so
x′ = a cos θ, y ′ = b sin θ.
Then θ is called the eccentric angle of the point Z ′ on the ellipse.
Now starting with any point Z1′ ≡ (x′1 , y1′ ), we first take the corresponding point Z1 ≡ x′1 , ab y1′ ;
−−→
−−→⊥
thus Z1 = f −1 (Z1′ ). Next we take the point Z2 ≡ (x2 , y2 ) = (− ab y1′ , x′1 ) so that OZ1 = OZ2 .
Finally we take the point Z2′ ≡ (x′2 , y2′ ) so that
a
b
b
x′2 = x2 = − y1′ , y2′ = y2 = x′1 ;
b
a
a
−−→⊥E −−→′
= OZ2 , and regard
thus Z2′ = f (Z2 ). In summary Z2′ ≡ − ab y1′ , − ab y1′ . With this we write OZ1′
this as an analogue in relation to the ellipse of Grassmann’s supplement in relation to the circle.
x′2
y ′2
We note that if a12 + b12 = 1, then
2
y2′2
1 a ′ 2
1 b ′
x′2
2
+ 2 = 2 − y1 + 2
x
a2
b
a
b
b
a 1
y ′2
x′2
= 12 + 12 = 1,
b
a
240
CHAPTER 9. AFFINE METHODS AND RESULTS
so that if Z1′ ∈ E then Z2′ ∈ E, and also that
0
0
1
1
′
′
y1′
1
δ(O, Z1 , Z2 ) = x′1
2 a ′ b ′
− b y1 a x1 1
1 b ′2 a ′2
=
x + y1
2 a 1
b
′2
1
y1′2
x1
= ab
+ 2 .
2
a2
b
This is always positive, and is equal to 12 ab when Z1′ ∈ E.
In fact we can characterize this relationship by
x′2
y1′2
x′2
y2′2
1
2
+
=
1,
+
= 1, δ(O, Z1′ , Z2′ ) = 21 (x′1 y2′ − y1′ x′2 ) = 21 ab.
a2
b2
a2
b2
For then
y2′ =
ab + x′2 y1′
,
x′1
and so
(ab + x′2 y1′ )2
x′2
2
+
2
2
a
x′2
1b
′2
2 2 2
′ ′
′2 ′2
b2 x′2
1 x2 + a (a b + 2abx2 y1 + x2 y1 )
=
′2
a2 b2 x1
′2 2 ′2
2 ′2
x (b x1 + a y1 ) + a4 b2 + 2a3 bx′2 y1′
= 2
a2 b2 x′2
1
a ′ ′
2
′2 2 2
4 2
x2 a b + a b + 2a3 bx′2 y1′ x′2
2 + 2 b x2 y1 + a
=
a2 b2 x′2
x21
1
1=
(x′ + a y ′ )2 + a2 (1 −
= 2 b 1 2
x1
(x′2 + ab y1′ )2 + x′2
1
=
,
x′2
1
y1′2
b2 )
so that x′2 + ab y1′ = 0, yielding x′2 = − ab y1′ and then y2′ = ab x′1 .
−−→ −−→⊥E
To apply all this consider a semi-diameter [O, Z1′ ] of E and suppose that OZ2′ = OZ1′ . Then
[O, Z2′ ] is another semi-diameter of E and Z1′ = f (Z1 ), Z2′ = f (Z2 ) where [O, Z1 ], [O, Z2 ] are semidiameters of C which are on perpendicular lines. It follows that [O, Z1′ ] and [O, Z2′ ] are conjugate
semi-diameters of E. This is an important way of handling conjugate diameters.
If [O, Z1′ ] and [O, Z2′ ] are conjugate semi-diameters of E, then we note that
|O, Z1′ |2
+
|O, Z2′ |2
=
x′2
1
+
y1′2
+
a
b
y2′
2
+
b ′
x
a 1
2
2
2
= (a + b )
y1′2
x′2
1
+
a2
b2
= a2 + b 2 .
It follows that the sum of the squares of the lengths of conjugate semi-diameters of an ellipse is
constant. This result can be traced back to Appolonius (c. 200 B.C.).
Given a semi-diameter [O, Z1′ ] of the ellipse E, we consider as well its pre-image [O, Z1 ] which
is a semi-diameter of the circle C. We saw above how to obtain a square [Z1 , Z2 , Z3 , Z4 ] inscribed
in the circle, and a square [W1 , W2 , W3 , W4 ] the side lines of which are tangent at the points
Z1 , Z2 , Z3 , Z4 , respectively; the diagonal lines Z1 Z3 , Z2 Z4 are perpendicular diametral lines. On
applying the affine transformation f , it follows that [Z1′ , Z2′ , Z3′ , Z4′ ] is a parallelogram inscribed
9.5. RESULTS FOR ELLIPSES, BASED MAINLY ON DISTANCE AND SENSED DISTANCE241
in the ellipse, and [W1′ , W2′ , W3′ , W4′ ] is a parallelogram the side lines of which are tangent to the
ellipse at the points Z1′ , Z2′ , Z3′ , Z4′ , respectively; the diagonals [Z1′ , Z3′ ] and [Z1′ , Z3′ ] are conjugate
diameters. Then
δF (W1′ , W2′ , W3′ , W4′ ) = 4δF (W1′ , Z2′ , O, Z1′ ) = 8δF (O, Z1′ , Z2′ ) = 4ab.
It follows that the tangents at the end-points of a pair of conjugate diameters of an ellipse cut off
a parallelogram of constant area.
This result can be traced back to Appolonius.
W1′
b
Z1′
b
W4′
Z2′
b
b
b
b
b
Z4′
W2′
b
Figure 9.2
Z3′
b
9.5.4
W3′
Classical Greek property
Let [R, S], [U, V ] be fixed diameters, of a circle C with centre O, which are on perpendicular lines.
For a variable point P ∈ C, let Q be the foot of the perpendicular from P onto RS. Then
|∠QRP |◦ + |∠QP R|◦ = |∠QRP |◦ + |∠QP S|◦ = 90,
so that |∠QRP |◦ = |∠QP S|◦ . By a similar argument |∠QSP |◦ = |∠QP R|◦ . Then the triangles
[R, P, Q] and [P, S, Q] are similar, as their third angles are also equal in magnitude. Then
|P, Q|
|R, Q|
=
,
|Q, S|
|P, Q|
so that |P, Q|2 = |R, Q||Q, S|. As Q ∈ [R, S] we have |R, Q||Q, S| = RQ QS. and so
|O, V |2
|P, Q|2
=
,
|O, S|2
RQ QS
which is constant. We deduce from this the following result.
Let [R′ , S ′ ], [U ′ , V ′ ] be fixed conjugate diameters of an ellipse E with centre O. For a variable
point P ′ ∈ E, let Q′ be the point where the line through P ′ which is parallel to U ′ V ′ meets R′ S ′ .
Then Q′ ∈ [R′ , S ′ ] and
|P ′ , Q′ |2
|O′ , V ′ |2
,
=
|O′ , S ′ |2
R ′ Q′ Q′ S ′
which is constant. This is an ancient Greek basic property for an ellipse, in relation to a pair of
conjugate diameters, noted in §7.6.
9.5.5
Axes of ellipse
With the notation of §9.5.3, we note the scalar product of vectors
a −−→′ −−→′
b ′
a2 − b 2 ′ ′
OZ1 • OZ2 = x′1 − y1′ + y1′
x1 = −
x1 y1 .
b
a
ab
242
CHAPTER 9. AFFINE METHODS AND RESULTS
This is equal to 0, and so we have perpendicular conjugate diametral lines if and only if y1′ = 0 or
x′1 = 0. These give respectively Z1′ to have coordinates (a, 0) or (−a, 0), and Z2′ to have coordinates
(0, b) or (0, −b), respectively.
Hence in an ellipse there is only one pair of conjugate diameters on perpendicular lines, and
these are the major and minor axes.
9.5.6
Chord joining endpoints of conjugate diameters
U′
P′
b
b
b
b
b
b
b
R′
Figure 9.3.
V′
In a circle C, let P, R be the end-points of diameters on perpendicular lines and [U, V ] a diameter
√
such that U V k P R. Then [P, R] is the side of a square inscribed in a circle, so that |P, R| = a 2
and so
1
|P, R|
= √ .
|U, V |
2
From this we deduce the following.
Let P ′ , R′ be end-points of conjugate diameters of an ellipse E, and [U ′ , V ′ ] be a diameter such
that U ′ V ′ k P ′ R′ . Then
1
|P ′ , R′ |
= √ .
′
′
|U , V |
2
9.5. RESULTS FOR ELLIPSES, BASED MAINLY ON DISTANCE AND SENSED DISTANCE243
9.5.7
Inscribed triangle with centroid at centre
Q′
b
b
b
b
O
b
b
R′
P′
b
Figure 9.4.
Given any point P of a circle C with centre O and radius length a, take a point S ∈ [P, O such
that |P, S| = 23 |P, O| = 32 a. Then O ∈ [P, S] and |O, S| = |P, S| − |P, O| = 23 a − a = 21 a, so that S
is an interior point of the circle, and so the line through S which is perpendicular to P S cuts C in
unique points Q and R. By Pythagoras’ theorem
|Q, S|2 = |O, Q|2 − |O, S|2 = a2 − ( 12 a)2 = 34 a2 ,
√
so that |Q, S| = 23 a. Similarly |R, S| =
By Pythagoras’ theorem again,
√
3
2 a
and as S ∈ [Q, R], |Q, R| = |Q, S| + |S, R| =
|P, Q|2 = |P, S|2 + |S, Q|2 =
2
3
2a
+
√
3
2 a
2
√
3a.
= 3a2 ,
√
√
so that |P, Q| = 3a. Similarly |P, R| = 3a and so the triangle [P, Q, R] is equilateral. Its
centroid is the point O, which is the centre of the circle. In fact the converse of this last property
is also true, that if the circumcentre of a triangle coincides with its centroid then the triangle must
be equilateral. As P varies, this inscribed equilateral triangle has constant area.
From this we deduce the following by applying an appropriate affine transformation. Let E be a
fixed ellipse and O its centre. For any point P ′ ∈ E there is a unique triangle [P ′ , Q′ , R′ ] inscribed
in E which has O as its centroid. As P ′ varies on E the area of such triangles is constant.
244
9.5.8
CHAPTER 9. AFFINE METHODS AND RESULTS
An elliptic Pythagoras property
S′
P′
b
b
b
U′
b
V′
O
b
R′
b
b
Figure 9.5.
b
Q′
T′
We obtain an elliptic analogue of Pythagoras’ theorem as follows. Let C be a circle , [P, Q] a
diameter and R a point on the circle. Let [S, T ], [U, V ] be diameters such that RP k ST, RQ k U V .
Then by Pythagoras’ theorem
|P, R|2
|Q, R|2
+
= 1.
|S, T |2
|U, V |2
By applying a transformation to this we deduce the following
Let E be an ellipse, [P ′ , Q′ ] a diameter and R′ a point on E . Let [S ′ , T ′ ], [U ′ , V ′ ] be diameters
such that R′ P ′ k S ′ T ′ , R′ Q′ k U ′ V ′ . Then
|Q′ , R′ |2
|P ′ , R′ |2
+
= 1.
|S ′ , T ′ |2
|U ′ , V ′ |2
9.5.9
Tangents from an exterior point
It is shown in Barry [2, pp. 101-102] that if P is an exterior point for a circle C there are two
tangents P T1 and P T2 , with points of contact T1 and T2 . Furthermore, if l is any line which passes
through P and cuts C in points R and S it will cut [T1 , T2 ] in a point Q such that (P, Q, R, S) is
a harmonic range.
T1′
b
P′
b
b
b
R′
b
Q
′
T2′
b
b
Figure 9.6.
S′
9.5. RESULTS FOR ELLIPSES, BASED MAINLY ON DISTANCE AND SENSED DISTANCE245
By applying an appropriate affine transformation to this, we deduce that if P ′ is an exterior
point for an ellipse E there are two tangents P ′ T1′ and P ′ T2′ , with points of contact T1′ and T2′ .
Furthermore, if l′ is any line which passes through P ′ and cuts E in points R′ and S ′ it will cut
[T1′ , T2′ ] in a point Q′ such that (P ′ , Q′ , R′ , S ′ ) is a harmonic range. Of course the line T1′ T2′ is the
polar of P ′ with respect to E.
Q1
S1
b
b
b
R1
b
T1
b
b
O
P
b
b
T2
b
S2
R2
Fig. 9.7.
Q2
b
For a circle C suppose that P is an exterior point and there are two tangents P T1 and P T2 ,
with points of contact T1 and T2 . Let the diametral line which is parallel to T1 T2 meet P T1 at Q1
and P T2 at Q2 , respectively. Let R1 be the mid-point of P and Q1 , and R2 be the mid-point of
P and Q2 . Then R2 O k P T1 and we denote by S1 the point in which R2 O meets the circle and
which lies on the opposite side of the line Q1 Q2 from P . Similarly R1 O k P T2 and we denote by
S2 the point in which R1 O meets the circle and which lies on the opposite side of the line Q1 Q2
from P .
To justify this we suppose that C has equation (9.4.4) and we take T1 ≡ (x
0 , y0 ), T2 ≡ (x0 , −y0 )
2
where x0 > 0. Then the tangent at T1 meets the x-axis at P ≡ xa0 , 0 and the y-axis at
2
2
2
a
Q1 ≡ 0, ay0 . For the mid-point of P and Q1 we have R1 ≡ 2x
, a . Similarly the tangent at
0 2y0
2
T2 meets the x-axis at P and the y-axis at Q2 ≡ 0, − ay0 . For the mid-point of P and Q2 we
2
a
a2
have R2 ≡ 2x
,
−
2y0 . Then OR2 has slope −x0 /y0 which is the same as that of P T1 ; similarly
0
OR1 k P T2 . Now
P T1
P T2
=
.
OS1
OS2
246
CHAPTER 9. AFFINE METHODS AND RESULTS
b
Q′1
b
T1′ = R1′
b
S1′
b
P′
O
b
b
T2′ = R2′
b
S2′
b
Q′2
Figure 9.8.
Suppose that E is an ellipse with centre O. Let T1′ , T2′ be the points of contact of tangents to
E from an exterior point P ′ . Let the diametral line which is parallel to T1′ T2′ meet P ′ T1′ at Q′1 and
P ′ T2′ at Q′2 , respectively. Let R1′ be the mid-point of P ′ and Q′1 , and R2′ be the mid-point of P ′
and Q′2 . Then R2′ O k P ′ T1′ and we denote by S1′ the point in which R2′ O meets the ellipse and
which lies on the opposite side of the line Q′1 Q′2 from P ′ . Similarly R1′ O k P ′ T2′ and we denote by
S2′ the point in which R1′ O meets the ellipse and which lies on the opposite side of the line Q′1 Q′2
from P ′ . Then
P ′ T1′
OS1′
9.5.10
=
P ′ T2′
OS2′
.
Products of lengths of segments
Let E be an ellipse with centre O, P ′ 6∈ E and lines through P ′ meet E at Q′ and R′ , and S ′ and
T ′ , respectively. Let U ′ , V ′ ∈ E be such that OU ′ k P ′ Q′ R′ and OV ′ k P ′ S ′ T ′ . Then
|O, U ′ |2
P ′ Q′ P ′ R ′
=
.
|O, V ′ |2
P ′S ′ P ′T ′
This property was noted more generally in §7.5.1 and was originally proved by Isaac Newton.
247
9.6. RESULTS ON ELLIPSES, BASED MAINLY ON ANGLE-MEASURE
9.5.11
A tangent cutting a pair of parallel tangents
b
S
P
b
R
b
b
S′
P′
b
b
b
R′
b
T
O
b
b
b
U′
T
b
b
O
′
b
Q′
b
Q
Figure 9.9.
b
Let [P, Q] be a diameter of a circle C with centre O, so that the tangents l, m at P and Q are
parallel. Let a tangent at the point R ∈ C meet l at S and m at T . Then [O, S is the mid-line
of |P OR and [O, T is the midline of |QOR so ∠SOT is a right angle. Also R is the foot of the
perpendicular from O to the hypotenuse. It follows that
|S, P ||T, Q| = |S, R||R, T | = |O, R|2 .
From this we deduce the following. Let [P ′ , Q′ ] be a diameter of an ellipse E with centre O, so
that the tangents l′ , m′ at P ′ and Q′ are parallel. Let a tangent at the point R′ ∈ E meet l′ at S ′
and m′ at T ′ . Let U ′ ∈ E be such that OU ′ k l′ . Then
|S ′ , P ′ ||T ′ , Q′ | = |O, U ′ |2 .
This result can be traced back to Appolonius.
9.6
9.6.1
Results on ellipses, based mainly on angle-measure
Behaviour of magnitude of an angle
In §9.5 our results dealt mainly with the ratios of lengths of segments on parallel lines. We now
try to establish results involving the behaviour of angles under affine transformations.
Consider an affine transformation which maps the origin to itself,
x′
= ax + by,
′
= cx + dy.
y
Then we have
y′
c + dy/x
=
,
′
x
a + by/x
so that the line y = mx maps to the line y ′ = m′ x′ where
m′ =
c + dm
.
a + bm
Thus we have that the slopes of these lines satisfy
bmm′ − dm + am′ − c = 0,
248
CHAPTER 9. AFFINE METHODS AND RESULTS
so that these are corresponding elements in a projectivity from a pencil of lines through the origin
to itself.
If we take a projective transformation which maps the origin to itself,
x′
=
y′
=
then we obtain
a1 x + b 1 y
,
a3 x + b 3 y + c3
a2 x + b 2 y
,
a3 x + b 3 y + c3
y′
a2 + b2 y/x
=
,
′
x
a1 + b1 y/x
so that
m′ =
a2 + b 2 m
.
a1 + b 1 m
This has the same form as that for affine transformations. Thus slopes of lines, and so angles, do
not seem to have any extra properties under affine transformations as compared with projective
transformations.
What we can use though is that affine transformations do not disrupt the following pattern.
Z1
b
Z1
Z6
b
b
b
Z2
b
Z5
Z2
b
Z3
b
b
Z4
Z3
b
b
Z4
b
Z5
b
Figure 9.10
Z6
Suppose that (Z1 , Z2 , Z3 ) are non-collinear points, that so are (Z4 , Z5 , Z6 ), that Z1 Z2 k Z4 Z5 , Z1 Z3 k
Z4 Z6 , and that (Z1 , Z2 , Z3 ) and (Z4 , Z5 , Z6 ) are similarly oriented. Then
tZ1 ,Z4 ([Z1 , Z2 ) = [Z4 , tZ1 ,Z4 (Z2 ) , tZ1 ,Z4 ([Z1 , Z3 ) = [Z4 , tZ1 ,Z4 (Z3 ) ,
and the angle ∠tZ1 ,Z4 (Z2 )Z4 tZ1 ,Z4 (Z3 ) has magnitude either that of the angle ∠Z5 Z4 Z6 or the
angle opposite to it, according as
tZ1 ,Z4 (Z2 ) ∈ [Z4 , Z5
or
tZ1 ,Z4 (Z2 ) ∈ Z4 Z5 \ [Z4 , Z5 .
It follows that the angles ∠Z2 Z1 Z3 and ∠Z5 Z4 Z6 are equal in magnitude.
If now f is an affine transformation then f (Z1 ), . . . , f (Z6 ) will satisfy the same conditions and
therefore the angles ∠f (Z2 )f (Z1 )f (Z3 ) and ∠f (Z5 )f (Z4 )f (Z6 ) are also equal in magnitude.
249
9.6. RESULTS ON ELLIPSES, BASED MAINLY ON ANGLE-MEASURE
9.6.2
Mid-line of an angle-support
b
Z1′
Z1
b
O
O
b
b
b
Z3′
Z4′
b
b
Z5′
b
b
Z2
Z2′
Z4
b
Figure 9.11
b
Z3
b
Z5
As a lead to diverse results we consider the behaviour under an affine transformation of the
mid-line of an angle-support |Z2 Z1 Z3 . Now there seems no ostensible reason why this mid-line
should be intimately connected with circles, but we consider the circumcircle of [Z1 , Z2 , Z3 ]. Let
O be the circumcentre and let Z4 be the mid-point of Z2 and Z3 . The diametral line OZ4 is
perpendicular to Z2 Z3 and it meets the circle at points Z5 , Z6 ; we choose the notation so that
[Z1 , Z5 ⊂ IR(|Z2 Z1 Z3 ). Now ∠Z2 OZ5 , ∠Z3 OZ5 have equal measures and these are respectively
double the measures of ∠Z2 Z1 Z5 , ∠Z3 Z1 Z5 . The latter angles thus have equal measures and
accordingly Z1 Z5 is the mid-line of |Z2 Z1 Z3 . We emphasize that Z5 Z6 is the diametral line which
is perpendicular to the diametral line which is parallel to Z2 Z3 .
Now consider non-collinear points Z1′ , Z2′ , Z3′ which lie on an ellipse E. These are the images
under an affine transformation of points Z1 , Z2 , Z3 on a circle. The line through the centre of the
ellipse and the mid-point Z4′ of Z2′ , Z3′ will meet the ellipse in a point Z5′ which is in the interior
region of |Z2′ Z1′ Z3′ . Then the line Z1′ Z5′ is the image of the mid-line of |Z2 Z1 Z3 . We note that this
line Z4′ Z5′ is the diametral line conjugate to the diametral line parallel to Z2 Z3 , with respect to
this ellipse.
9.6.3
Analogue of orthocentre
Z1′
b
b
b
O
b
Z2′
Figure 9.12
Z3′
250
CHAPTER 9. AFFINE METHODS AND RESULTS
Given the lead in §9.6.2, we now consider again non-collinear points Z1′ , Z2′ , Z3′ which lie on an
ellipse E. These are the images under an affine transformation of points Z1 , Z2 , Z3 on a circle.
The line through Z1 perpendicular to Z2 Z3 , the line through Z2 perpendicular to Z3 Z1 , and the
line through Z3 perpendicular to Z1 Z2 are concurrent, the point of concurrency being known as
the orthocentre of the triangle [Z1 , Z2 , Z3 ]. But the line through Z1 perpendicular to Z2 Z3 maps
onto the line through Z1′ which is parallel to the diametral line which is conjugate to the diametral
line which is parallel to Z2′ Z3′ . The images of the altitudes of [Z1 , Z2 , Z3 ] through Z2 and Z3
can be characterized similarly. These image lines must be concurrent. This gives an analogue of
orthocentre for [Z1′ , Z2′ , Z3′ ] in the context of this ellipse and affine transformation.
9.6.4
Lines parallel to conjugate diametral lines
It is convenient here to derive a formula for lines parallel to conjugate diametral lines of a central
conic. We take the conic to have equation
f1 βγ + g1 γα + h1 αβ = 0,
and take a line Z4 Z5 which then has parametric equations
α = α4 + l1 s, β = β4 + m1 s, γ = γ4 + n1 s,
where
l1 = α5 − α4 , m1 = β5 − β4 , n1 = γ5 − γ4 .
The parallel line through the point Z ′ will then have parametric equations
α = α′ + l1 s, β = β ′ + m1 s, γ = γ ′ + n1 s.
On substituting this in the equation for the conic we get
f1 (β ′ + m1 s)(γ ′ + n1 s) + g1 (γ ′ + n1 s)(α′ + l1 s) + h1 (α′ + l1 s)(β ′ + m1 s).
The condition that Z ′ is always the mid-point of the points of intersection of this line with the
conic is then
(g1 n1 + h1 m1 )α′ + (h1 l1 + f1 n1 )β ′ + (f1 m1 + g1 l1 )γ ′ = 0.
This then is the equation of the diametral line conjugate to the diametral line parallel to the
original line.
Now on replacing (α′ , β ′ , γ ′ ) by (α, β, γ), we recall from §3.2.6 that the line through the point
Z6 which is parallel to this line has equation
α+β+γ
(g1 n1 + h1 m1 )α + (h1 l1 + f1 n1 )β + (f1 m1 + g1 l1 )γ
=
.
(g1 n1 + h1 m1 )α6 + (h1 l1 + f1 n1 )β6 + (f1 m1 + g1 l1 )γ6
α6 + β6 + γ6
9.6. RESULTS ON ELLIPSES, BASED MAINLY ON ANGLE-MEASURE
9.6.5
251
Analogue of Euler line
Z1′
b
b
b
O
Z3′
b
Z2′
Figure 9.13
To apply §9.6.4 to the situation in §9.6.3, we take again non-collinear points Z1′ , Z2′ , Z3′ which
lie on a central conic. Taking areal coordinates with respect to (Z1′ , Z2′ , Z3′ ), the conic will have an
equation
f1 βγ + g1 γα + h1 αβ = 0.
If now in §9.6.4 we take Z4 Z5 = Z2′ Z3′ we have
l1 = 0, m1 = −1, n1 = 1,
and we take with this Z6 = Z1′ = (1, 0, 0). Thus we get that the line through Z1′ in the direction
conjugate to the direction containing Z2′ Z3′ has equation
(f1 − g1 + h1 )β + (−f1 − g1 + h1 )γ = 0.
Similarly the line through Z2′ in the direction conjugate to the direction containing Z3′ Z1′ has
equation
(f1 − g1 − h1 )α + (f1 + g1 − h1 )γ = 0.
If we solve between these, we find that our analogue of orthocentre has areal coordinates proportional to
((f1 + g1 − h1 )(f1 − g1 + h1 ), (f1 + g1 − h1 )(−f1 + g1 + h1 ), (−f1 + g1 + h1 )(f1 − g1 + h1 )).
Now the analogue of the circumcentre is the centre of the conic, and the point W0 ≡ (α0 , β0 , γ0 )
has as polar with respect to this conic the line
(h1 β0 + g1 γ0 )α + (h1 α0 + f1 γ0 )β + (g1 α0 + f1 β0 )γ = 0.
Then W0 is the centre if
0.α0 + hβ0 + gγ0
hα0 + 0.β0 + f γ0
=
=
p,
p,
gα0 + f β0 + 0.γ0
=
p,
for some number p. On solving these equations, we find that the coordinates of the centre are
proportional to
(f1 (−f1 + g1 + h1 ), g1 (f1 − g1 + h1 ), h1 (f1 + g1 − h1 )).
(9.6.1)
252
CHAPTER 9. AFFINE METHODS AND RESULTS
The image of the centroid of the pre-triangle is the centroid of [Z1′ , Z2′ , Z3′ ] and so our analogue
of centroid is still centroid. This has areal coordinates proportional to (1, 1, 1).
If we now enter the coordinates of these three points in a determinant, we find that they are
collinear. Thus we have a generalisation of the classical property of an Euler line of any triangle
(essentially related to the circumscribed circle from our point of view) to a triangle inscribed in a
central conic.
9.6.6
Analogue of Wallace-Simson line
b
Z1′
b
Z′
b
b
Z3′
b
b
Z2′
Figure 9.14
We take again non-collinear points Z1′ , Z2′ , Z3′ which lie on a central conic. Taking areal coordinates with respect to (Z1′ , Z2′ , Z3′ ), the conic will have equation
f1 βγ + g1 γα + h1 αβ = 0.
If now in §9.6.4 we take Z4 Z5 = Z2′ Z3′ we have
l1 = 0, m1 = −1, n1 = 1,
and we take with this a general point Z6′ = (α6 , β6 , γ6 ). From §9.6.4 we get that the line through
Z6′ in a direction conjugate to the direction containing Z2′ Z3′ has equation
α+β+γ
(g1 − h1 )α + f1 β − f1 γ
=
.
(g1 − h1 )α6 + f1 β6 − f1 γ6
α6 + β6 + γ6
We note that this meets the line Z2′ Z3′ in the point with areal coordinates proportional to
(0, f1 (α6 + β6 + γ6 ) + (g1 − h1 )α6 + f1 (β6 − γ6 ), f1 (α6 + β6 + γ6 ) − (g1 − h1 )α6 − f1 (β6 − γ6 ).
By a similar argument the line through Z6′ in a direction conjugate to the direction containing
Z3′ Z1′ meets Z3′ Z1′ in a point with areal coordinates proportional to
(g1 (α6 + β6 + γ6 ) − (h1 − f1 )β6 − g1 (γ6 − α6 ), 0, g1 (α6 + β6 + γ6 ) + (h1 − f1 )β6 + g1 (γ6 − α6 )),
and the line through Z6′ in a direction conjugate to the direction containing Z1′ Z2′ meets Z1′ Z2′ in
a point with areal coordinates proportional to
(h1 (α6 + β6 + γ6 ) + (f1 − g1 )γ6 + h1 (α6 − β6 ), h1 (α6 + β6 + γ6 ) − (f1 − g1 )γ6 − h(α6 − β6 ), 0).
9.6. RESULTS ON ELLIPSES, BASED MAINLY ON ANGLE-MEASURE
253
If we form the determinant with these three sets of coordinates as entries we find that its value
is
2(−f12 − g12 − h21 + 2g1 h1 + 2h1 f1 + 2f1 g1 )(α6 + β6 + γ6 )(f1 β6 γ6 + g1 γ6 α6 + h1 α6 β6 ).
The term in f1 , g1 , h1 is non-zero as the conic is central and so the sum of the coefficients in (9.6.1)
is non-zero. We thus obtain that these points are collinear if and only if
f1 β6 γ6 + g1 γ6 α6 + h1 α6 β6 = 0,
that is if and only if Z6′ is on the conic.
This generalizes to central conics a result for circles proved by Wallace, but usually named after
Simson.
254
CHAPTER 9. AFFINE METHODS AND RESULTS
9.6.7
Analogue of constant-sized angle property of a circle
Turning to somewhat different material, we now obtain as follows for an ellipse an analogue of the
result for a circle that a fixed arc subtends at the circumference an angle of fixed magnitude. Let
C be a circle with centre O. Let [Z1 , Z2 ] be fixed points of C and Z a variable point of C. Let the
half-line [O, tZ,O (Z1 ) meet the circle at W1 , and [O, tZ,O (Z2 ) meet the circle at W2 . Then
∠W1 OW2 = ∠tZ,O (Z1 )tZ,O (Z)tZ,O (Z2 ),
and this has the constant magnitude of ∠Z1 ZZ2 .
Thus as Z varies on the circle, δF (O, W1 , W2 ) is constant in magnitude, changing sign as Z
crosses Z1 Z2 . This corresponds to sin θ being constant on an arc, where θ = ∡F Z1 ZZ2 .
Z′
b
b
O
b
Z1′
b
Z2′
b
W1′
b
W2′
tZ ′ ,O (Z1′ )
tZ ′ ,O (Z2′ )
Figure 9.15
On applying an affine transformation we deduce the following. Let E be an ellipse with centre
O. Let [Z1′ , Z2′ ] be fixed points of E and Z ′ a variable point of E. Let the half-line [O, tZ ′ ,O (Z1′ )
meet the ellipse at W1′ , and [O, tZ ′ ,O (Z2′ ) meet the ellipse at W2′ . Then as Z ′ varies on the ellipse,
δF (O, W1′ , W2′ ) is constant in magnitude, changing sign as Z ′ crosses Z1′ Z2′ .
9.6.8
Analogue of incentre
We now go on to develop §9.6.2 further. Given a conic with equation
f1 βγ + g1 γα + h1 αβ = 0,
we recall from (9.6.1) that the coordinates of the centre are proportional to
(f1 (−f1 + g1 + h1 ), g1 (f1 − g1 + h1 ), h1 (f1 + g1 − h1 )).
For any point W1 6= W0 the line W0 W1 has parametric equations
α = α0 + t(α1 − α0 ), β = β0 + t(β1 − β0 ), γ = γ0 + t(γ1 − γ0 ),
and the equation of incidence with the conic is
f1 [β0 +t(β1 −β0 )][γ0 +t(γ1 −γ0 )]+g1 [γ0 +t(γ1 −γ0 )][α0 +t(α1 −α0 )]+h1 [α0 +t(α1 −α0 )][β0 +t(β1 −β0 )] = 0.
9.6. RESULTS ON ELLIPSES, BASED MAINLY ON ANGLE-MEASURE
255
Since W0 is the centre, and thus the mid-point of the points of intersection, we must have the roots
satisfy t1 + t2 = 0, and so the equation assumes the form
[f1 (β1 −β0 )(γ1 −γ0 )+g1 (γ1 −γ0 )(α1 −α0 )+h1 (α1 −α0 )(β1 −β0 )]t2 +f1 β0 γ0 +g1 γ0 α0 +h1 α0 β0 = 0.
As γ1 − γ0 = −(α1 − α0 ) − (β1 − β0 ), the coefficient of t2 can be written as
−g1 (α1 − α0 )2 + (−f1 − g1 + h1 )(α1 − α0 )(β1 − β0 ) − f1 (β1 − β0 )2 .
For the conic to be an ellipse the discriminant of this must be negative so that the expression does
not change sign, that is
(f1 + g1 − h1 )2 − 4f1 g1 < 0.
This can be re-written in both of the other forms
(−f1 + g1 + h1 )2 − 4g1 h1 < 0, (f1 − g1 + h1 )2 − 4h1 f1 < 0.
Thus for an ellipse we must have
f1 g1 > 0, g1 h1 > 0, h1 f1 > 0.
Thus f1 , g1 , h1 must be of the one sign, so without loss of generality we may take them all to be
positive.
The mid-point Z4 of Z2 and Z3 has areal coordinates proportional to (0, 1, 1), and so the line
W0 Z4 has equation
α
β
γ
= 0,
0
1
1
f1 (−f1 + g1 + h1 ) g1 (f1 − g1 + h1 ) h1 (f1 + g1 − h1 ) and this simplifies to
(g1 − h1 )α + f1 (β − γ) = 0.
To find where this line W0 Z4 meets the ellipse we substitute
α=−
f1
(β − γ),
g 1 − h1
into the equation
f1 βγ + g1 γα + h1 αβ = 0.
This yields
h1 β 2 − g1 γ 2 = 0.
One solution is then a point Z5 for which we have the form
p
√
β = j g 1 , γ = j h1 ,
with then
f1 j
√ .
α = −√
g 1 + h1
A second solution is the point Z6 with coordinates of the form
p
√
β = −j g1 , γ = j h1 .
The line Z1 Z5 has equation
α
1
− √ f1√
g + h
1
1
β
0
√
g1
γ
0
√
h1
= 0,
256
CHAPTER 9. AFFINE METHODS AND RESULTS
which simplifies to
p
√
h1 β − g1 γ = 0.
√
√ √
This
√ meets the line Z2 Z3 at the point with coordinates proportional to (0, g1 , h1 ). As g1 and
h1 are both positive, this point is on the segment [Z2 , Z3 ], and so the half-line [Z1 , Z5 lies in the
interior region IR(|Z2 Z1 Z3 ).
The line Z1 Z5 , with equation
p
√
h1 β − g1 γ = 0,
is the elliptic analogue of the mid-line of the angle-support |Z2 Z1 Z3 . The corresponding lines
through Z2 and Z3 have the equations, respectively,
p
p
p
√
h1 α − f1 γ = 0, g1 α − f1 β = 0.
The three lines are concurrent, the coordinates of the point of concurrency being proportional to
p √ p
( f1 , g1 , h1 ).
This point is the elliptic analogue of the incentre of the triangle [Z1 , Z2 , Z3 ].
9.7
Results on ellipses, based mainly on sensed-areas
We wish to continue this process of finding affine results, and now concentrate on the ratio of two
sensed areas.
9.7.1
Elliptic sine and cosine
We recall the approach to trigonometry in Barry [2, Chapter 9] and take a frame of reference
F = ([O, Q , [O, R ) where as usual OQ ⊥ OR and now we take |O, Q| = |O, R| = k. Let H1
be the closed half-plane with edge OQ which contains R and introduce Cartesian coordinates in
the standard way. Then Q ≡F (k, 0), R ≡F (0, k). For any Z such that |Z| = k, let θ be
the angle in A∗ (F ) with support |QOZ. Then Z ≡F (k cos θ, k sin θ). We let Z ∗ be the point
such that |Z ∗ | = k, OZ ⊥ OZ ∗ and (O, Z, Z ∗ ) is positively oriented with respect to F . Then
Z ∗ ≡ (−k sin θ, k cos θ). We find that
δF (O, Q, Z)
δF (O, Q, Z ∗ )
δF (O, Z, R)
= sin θ,
=
= cos θ.
δF (O, Q, R)
δF (O, Q, R)
δF (O, Q, R)
These are the basis of our generalization of trigonometric functions.
Z
b
R
θ
b
Z∗
b
b
O
Figure 9.16
b
Q
257
9.7. RESULTS ON ELLIPSES, BASED MAINLY ON SENSED-AREAS
Let [O, Q], [O, R] be conjugate semi-diameters of an ellipse E and let H5 be the closed halfplane with edge OQ in which R lies. For any Z on the ellipse consider the line joining the centre
of the ellipse to the mid-point of Q and Z, and the half-line of this which is a subset of H5 ; we
take this half-line as a modified indicator of this angle. For this let θ be the angle with support
|QOZ the associated region of which (interior, exterior or half-plane) contains this half-line, so
that its modified indicator lies in H5 . Let [O, Z ∗ ] be the semi- diameter conjugate to [O, Z] with
(O, Q, R) and (O, Z, Z ∗ ) similarly oriented. Then we define
sinE θ =
δF (O, Q, Z ∗ )
δF (O, Q, Z)
, cosE θ =
.
δF (O, Q, R)
δF (O, Q, R)
Then as these can be obtained from standard cosine and sine by means of an affine transformation, they have similar properties.
9.7.2
Sum of squares
In particular in §9.7.1 we obtain that
cosE (θ)2 + sinE (θ)2 = 1
and so have
δF (O, Q, Z)2 + δF (O, Q, Z ∗ )2 = δF (O, Q, R)2 .
This is another affine analogue of Pythagoras’ theorem.
9.7.3
Elliptic addition of angles
R
b
Z1
b
b
Z2
b
b
b
Q
O
b
b
Z3
Figure 9.17
We can derive a modified addition of angles θ as in §9.7.1 so as to have the standard addition
formulae for the modified trigonometric functions. Let α, β be angles as specified with supports
|QOZ1 , |QOZ2 , respectively. We let Z4 be the mid-point of Z1 and Z2 , and let Z3 be the point
at which the line through Q and parallel to Z1 Z2 meets the ellipse again; then OZ1 and QZ3 are
in conjugate directions. We define the elliptic sum α +E β to be the angle with support |QOZ3
which has its modified indicator in H5 .
9.7.4
The duplication formula
The standard trigonometric identities, suitably interpreted, apply. For example the identity
sinE 2θ = 2 sinE θ cosE θ yields the following. In §9.7.3 let Z1 = Z2 = Z and α = β = θ. Then Z3
258
CHAPTER 9. AFFINE METHODS AND RESULTS
has elliptic polar angle 2θ and so we have
δF (O, Q, Z) δF (O, Q, Z ∗ )
δF (O, Q, Z3 )
=2
.
δF (O, Q, R)
δF (O, Q, R) δF (O, Q, R)
9.7.5
Analogue of areal form of Pythagoras’ theorem
b
G
b
b
b
B
b
A
b
H
F
b
E
C
b
K
b
L
Figure 9.18
Let [B, C] be a diameter of a circle with centre E. Let A be another point on the circle, and
EL ⊥ BC with L on the circle and on the opposite side of BC from A. Let F, G be the mid-points
of {A, C}, {A, B}, respectively. Now |∠LAC|◦ = 45, the circle on [A, C] as diameter meets EF at
a point H such that |∠HAC|◦ = 45, and the circle on [A, B] as diameter meets EG at a point K
such that |∠KAB|◦ = 45. It follows that A, H, K, L are collinear. From Pythagoras’ theorem we
have that
δF (A, H, F ) + δF (A, G, K) = δF (E, L, C).
An affine transformation would now yield an analogue of this for ellipses.
9.8
Some results of Appolonius’ recalled
In fact the treatment of conics by Appolonius was largely affine and he generally related properties
to any pair of conjugate diameters rather than to the axes. His Conics has many many results
on areas in affine form. In the 19th century, elementary treatments of conics tended to emphasise
metrical properties instead, while advanced treatments emphasised projective properties, and the
affine ones suffered neglect. We select a few results from Appolonius to give a flavour. He dealt
with them for ellipses and hyperbolas as well.
9.8. SOME RESULTS OF APPOLONIUS’ RECALLED
9.8.1
b
Q2
b
P1,2
259
b
Z2
b
b
b
Z1
O
Q1
Figure 9.19
Let C be a circle, with center O, and with Z1 , Z2 two points on it. We let the tangents at Z1
and Z2 meet each other at the point P1,2 , the tangent at Z2 meet OZ1 at Q1 , and the tangent at
Z1 meet OZ2 at Q2 . Then
δF (P1,2 , Q1 , Z1 ) = δF (P1,2 , Z2 , Q2 ).
To see this, we take coordinates so that
O ≡ (0, 0), Z1 ≡ (−k, 0),
and OZ2 to have slope m so that
Z2 ≡
k
km
−√
.
, −√
1 + m2
1 + m2
It follows by easy calculations that
!
√
p
k( 1 + m2 − 1)
, Q1 ≡ (−k 1 + m2 , 0), Q2 ≡ (−k, −mk).
P1,2 ≡ −k, −
m
The equality of the sensed areas follow from these. In fact, in the case of a circle these triangles
are congruent.
9.8.2
Q2
b
b
b
Z2
b
S2
S1
b
O
b
b
b
b
R1
b
Z1
Q1
Figure 9.20
Z
Next let Z ≡ (x, y) be an arbitrary point on the circle, and let R1 , S2 be the points where
the line though Z parallel to the tangent at Z1 meets OZ1 and OZ2 , respectively. Similarly, let
b
260
CHAPTER 9. AFFINE METHODS AND RESULTS
R2 , S1 be the points where the line though Z parallel to the tangent at Z2 meets OZ2 and OZ1 ,
respectively. Then
δF (Z, R1 , S1 ) + δF (O, R1 , S2 ) = δF (O, Z1 , Q2 ),
the latter being constant as Z varies. For, continuing the notation of (i), we have
R1 ≡ (x, 0), S2 ≡ (x, mx), S1 ≡ (x + my, 0).
The given relationship follows on inserting the coordinates and simplifying.
As S2 ∈ ZR1 we have
δF (Z, R1 , S1 ) = δF (Z, S2 , S1 )δF (S2 , R1 , S1 ),
and as R1 ∈ OS1 we have
δF (S2 , O, R1 ) + δF (S2 , R1 , S1 ) = δF (S2 , O, S1 ).
On using these we find that
δF (Z, R1 , S1 ) + δF (O, R1 , S2 ) =
=
δF (S1 , Z, S2 ) + δF (S2 , O, S1 )
δF (S1 , Z, S2 , O),
and this quadrangular sensed-area is constant as Z varies on the circle.
9.8.3
Q2
b
T2
b
b
Z
S2
b
Z2
b
b
b
O
R1
b
b
Z1
S1
b
Q1
Figure 9.21
Now, additionally, let T1 be the point where the line ZR2 = ZS1 meets the tangent at Z1 , and
let T2 be the point where the line ZR1 = ZS2 meets the tangent at Z2 . Then
δF (T2 , Z2 , S2 ) = δF (T2 , Q1 , R1 ) − δF (Z, S1 , R1 ).
For this we note that
T2 ≡
!
√
x + k 1 + m2
x, −
,
m
and now insertion of the coordinates proves the given relationship.
As Z ∈ R1 T2 , S1 ∈ R1 Q1 , by §1.7.2 we have that
δF (R1 , T2 , Q1 ) − δF (R1 , Z, S1 ) = δF (Z, T2 , Q1 , S1 ),
and so
δF (T2 , Z2 , S2 ) = δF (Z, T2 , Q1 , S1 ).
261
9.9. ANALOGOUS RESULTS FOR HYPERBOLAS
9.9
Analogous results for hyperbolas
The results for ellipses in §§9.4, 9.5 and 9.6 suggest that similar results may be valid for hyperbolas.
We tackle these by coordinate methods.
9.9.1
Triangle with centroid at centre of hyperbola
Z2
b
b
Z4
Z1
b
b
Z3
Figure 9.22.
In §9.5.7 we looked at a triangle inscribed in an ellipse with its centroid at the centre of the
ellipse. For a corresponding result for a hyperbola H, let us take the equation of H to be
x2
y2
− 2 − 1 = 0,
2
a
b
and let us call the hyperbola H∗ with equation
y2
x2
−
+ 1 = 0,
a2
b2
its conjugate hyperbola. We start with any point Z1 ∈ H so that
y2
x21
− 21 − 1 = 0.
2
a
b
Corresponding to it we take a point Z4 such that
Z1 O
= 2,
OZ4
so that x4 = − 21 x1 , y4 = − 12 y1 . A line with equation lx + my + n = 0, which passes through Z4 ,
has parametric equations
1
1
x = − x1 + mt, y = − y1 − lt,
2
2
(t ∈ R),
and this line meets the conjugate hyperbola when
2
5
l2 2
mx1
ly1
m
t
−
t + = 0.
−
+
2
2
2
2
a
b
a
b
4
We wish Z4 to be the mid-point of these points of intersection and for this take
mx1
ly1
+ 2 = 0.
2
a
b
262
CHAPTER 9. AFFINE METHODS AND RESULTS
We then have
y1
x1
, m = −j 2 ,
a2
b
for some j =
6 0. On inserting these, we find that the incidence equation becomes
l = −j
j 2 t2 =
5 2 2
a b ,
4
and the solutions give unique points Z2 , Z3 on the conjugate hyperbola, with
√
√
1
5a
5b
1
y1 , y2 = − y1 +
x1 ,
x2 = − x1 +
2
2 b
2
2 a
√
√
1
5a
5b
1
x3 = − x1 −
y1 , y3 = − y1 −
x1 .
2
2 b
2
2 a
We note that
x1 + x2 + x3 = 0, y1 + y2 + y3 = 0,
and so O is the centroid of [Z1 , Z2 , Z3 ]. Moreover it can be checked that
√
5
.
δF (Z1 , Z2 , Z3 ) =
4ab
9.9.2
Hyperbolic theorem of Pythagoras
Z10
b
b
Z12
Z4
b
b
Z5
Z11
b
Z8
b
Z9
b
Fig. 9.23.
In §9.5.8 we proved an elliptic analogue of Pythagoras’ theorem and we now consider a hyperbolic version. For a hyperbola H, let us take its equation to be
y2
x2
−
− 1 = 0,
a2
b2
and then the conjugate hyperbola H∗ has equation
y2
x2
− 2 + 1 = 0.
2
a
b
We start with fixed points Z4 , Z5 ∈ H so that Z4 Z5 is a diametral line and so x5 = −x4 , y5 = −y4 .
We first identify the interior region for H. The polar with respect to it of a point Z0 ≡ (x0 , y0 )
has equation
yy0
xx0
− 2 = 1.
2
a
b
263
9.9. ANALOGOUS RESULTS FOR HYPERBOLAS
When y0 6= 0, this meets H at points the x-coordinates of which satisfy
(a2 y02 − b2 x20 )x2 + 2a2 b2 x0 x − a4 (b2 + y02 ) = 0,
and this equation has distinct real roots when
x20
y2
− 20 − 1 > 0.
2
a
b
This is equivalent to
x20
2
y02
>a ,
<b
2
x20
−1 .
a2
(9.9.1)
This then is the condition that Z0 ≡ (x0 , y0 ) lies in the interior region.
We recall from the end of §7.2.5 that the points Z4 and Z8 of H are on the one branch if every
point of [Z4 , Z8 ], except the end points, is in the interior region. Then with
x = (1 − t)x4 + tx8 ,
y = (1 − t)y4 + ty8 ,
from the second part of (9.9.1) we need
[(1 − t)y4 + ty8 ]2
[(1 − t)x4 + tx8 ]2
<
− 1,
b2
a2
(9.9.2)
for all t such that 0 < t < 1. Now (9.9.2) can be re-written as
2
2
y42
y82
2 x8
2 x4
− 2 +t
− 2 − 1 > 0,
(1 − t)
a2
b
a2
b
and using the fact that Z4 and Z8 are on the hyperbola, we need
hx x
y4 y8 i
4 8
(1 − t)2 + t2 − 1 + 2t(1 − t)
> 0,
−
a2
b2
that is
2t(1 − t)
for 0 < t < 1. Thus
i
hx x
y4 y8
4 8
−
−
1
> 0,
a2
b2
x4 x8
y4 y8
− 2 − 1 > 0,
(9.9.3)
2
a
b
is a necessary condition for the points Z4 and Z8 on H are on the one branch of it, and we note
from it that then Z5 and Z8 are not on the one branch.
The diametral line parallel to Z4 Z8 has parametric equations
x = t(x8 − x4 ),
y = t(y8 − y4 ),
and this meets the conjugate hyperbola when
t2 =
1
2
x4 x8
a2
−
1
x4 x8
a2
−
y4 y8
b2
−1
,
and so by (9.9.3) at points
Z9 ≡
s
Z10 ≡ −
1
2
s
1
1
2
y4 y8
b2
−1
(x8 − x4 , y8 − y4 ),
1
x4 x8
a2
−
y4 y8
b2
−1
(x8 − x4 , y8 − y4 ).
264
CHAPTER 9. AFFINE METHODS AND RESULTS
The diametral line parallel to Z5 Z8 has parametric equations
x = t(x8 − x5 ),
y = t(y8 − y5 ),
and this meets the hyperbola when
1
2
t2 =
and this is
−x5 x8
a2
1
2
t2 =
1
x4 x8
a2
This gives the points
−
y4 y8
b2
+1
.
s
1
1
(x8 − x4 , y8 − y4 ),
x4 x8
2 a2 − y4b2y8 + 1
s
1
1
≡−
(x8 − x4 , y8 − y4 ).
2 xa4 x2 8 − y4b2y8 + 1
Z11 ≡
Z12
1
,
+ y5b2y8 + 1
From these we conclude that
|Z4 , Z8 |2
|Z5 , Z8 |2
−
= 1.
2
|Z11 , Z12 |
|Z9 , Z10 |2
This is our analogue for a hyperbola of Pythagoras’ theorem.
9.9.3
Analogue for hyperbola of constant-sized angle property of circle
W1
b
Z4
Z
b
b
b
b
b
W2
b
b
Z5
Figure 9.24.
We now seek an analogue of §9.6.7 for a hyperbola, and lay out our initial calculations for a
rectangular hyperbola H with equation
x2 − y 2 = a2 ,
with conjugate hyperbola H∗ with equation
x2 − y 2 = −a2 .
9.9. ANALOGOUS RESULTS FOR HYPERBOLAS
265
Take points Z4 , Z5 , Z on one branch of H. Let W1 , W2 be the points in which the half-lines
[O, tZ,O (Z4 ) , [O, tZ,O (Z5 ) meet the conjugate hyperbola H∗ . For W1 we look for a point with
coordinates
x′ = t(x4 − x), y ′ = t(y4 − y),
satisfying x′2 − y 2 = −a2 , and thus obtain
a
(x4 − x, y4 − y),
W1 ≡ p
−(x4 − x)2 + (y4 − y)2
and similarly
a
W2 ≡ p
(x5 − x, y5 − y).
−(x5 − x)2 + (y5 − y)2
From these
δF (O, W1 , W2 ) =
(x4 − x)(y5 − y) − (x5 − x)(y4 − y)
1 2
p
a p
.
2
−(x4 − x)2 + (y4 − y)2 −(x5 − x)2 + (y5 − y)2
We wish this to be a constant k and then would have
2k
(x4 − x)(y5 − y) − (x5 − x)(y4 − y)
p
p
= 2,
2
2
2
2
a
−(x4 − x) + (y4 − y) −(x5 − x) + (y5 − y)
and so
4k 2
[(x4 − x)(y5 − y) − (x5 − x)(y4 − y)]2
.
=
[−(x4 − x)2 + (y4 − y)2 ][−(x5 − x)2 + (y5 − y)2 ]
a4
But
[(x4 − x)(x5 − x) − (y4 − y)(y5 − y)]2 − [(x4 − x)(y5 − y) − (x5 − x)(y4 − y)]2
= 1,
[−(x4 − x)2 + (y4 − y)2 ][−(x5 − x)2 + (y5 − y)2 ]
so we need
and thus
4k 2
[(x4 − x)(x5 − x) − (y4 − y)(y5 − y)]2
,
=
1
+
[−(x4 − x)2 + (y4 − y)2 ][−(x5 − x)2 + (y5 − y)2 ]
a4
a4 + 4k 2
[(x4 − x)(x5 − x) − (y4 − y)(y5 − y)]2
=
.
[(x4 − x)(y5 − y) − (x5 − x)(y4 − y)]2
4k 2
This would give, with a possible variation in sign,
r
a4 + 4k 2
(x4 − x)(x5 − x) − (y4 − y)(y5 − y) =
[(x4 − x)(y5 − y) − (x5 − x)(y4 − y)],
4k 2
and so
x2 − [x4 + x5 + k1 (y4 − y5 )]x − y 2 + [y4 + y5 + k1 (x4 − x5 )]y + x4 x5 − y4 y5 − k1 (x4 y5 − x5 y4 ) = 0,
where
k1 =
r
a4 + 4k 2
.
4k 2
We wish this equation to be x2 − y 2 = a2 and so want
k1 = −
x4 + x5
y4 + y5
=−
.
y4 − y5
x4 − x5
These are compatible as
x24 − x25 = y42 − y52 ,
266
CHAPTER 9. AFFINE METHODS AND RESULTS
since x4 − y42 = x25 − y52 = a2 . We also want
k1 (x4 y5 − x5 y4 ) − x4 x5 + y4 y5 = a2 ,
and this follows from the above.
We thus need
a4 + 4k 2
=
4k 2
and so
x4 + x5
y4 − y5
2
a4 (y4 − y5 )2
.
+ x4 x5 + y4 y5 )
4k 2 =
2(a2
This gives the desired result. Other cases can be tackled similarly.
9.9.4
Hyperbolic sine and cosine
We now consider an analogue of the material in §9.7.1 for the hyperbola H with equation
x2
y2
−
= 1.
a2
b2
The conjugal diametral for the line with equation y = µx is the line with equation y = (b2 /a2 µ)x.
It is easy to verify that the right-hand branch has parametric equations
x = a cosh t, y = b sinh t,
(t ∈ R).
We take the fixed point Q ≡ (a cosh t0 , b sinh t0 ). The diametral line through Q has slope
µ0 =
b sinh t0
,
a cosh t0
and so the conjugate diametral line for this has equation
y=
b cosh t0
x.
a sinh t0
This does not meet the hyperbola but it does meet the conjugate hyperbola at the points with
coordinates (a sinh t0 , b cosh t0 ) and (−a sinh t0 , −b cosh t0 ). We take R ≡ (a sinh t0 , b cosh t0 ) as
then
1
δF (O, Q, R) = ab,
2
and so (O, Q, R) is positively oriented with respect to F . We now take the variable point Z ≡
(a cosh t, b sinh t) on the hyperbola and the concomitant point Z ∗ ≡ (a sinh t, b cosh t) which is
where the conjugate diametral line for OZ meets the conjugate hyperbola and has (O, Z, Z ∗ )
positively oriented. We then have
sinH ∠QOZ
=
cosH ∠QOZ
=
δF (Q, O, Z)
= sinh(t − t0 ),
δF (Q, O, R)
δF (Q, O, Z ∗ )
= cosh(t − t0 ).
δF (Q, O, R)
Thus the standard hyperbolic functions are the sine and cosine with respect to this hyperbola.
We note in particular that
(cosH ∠QOZ)2 − (sinH ∠QOZ)2 = 1,
which gives an analogue of Pythagoras’ theorem for a hyperbola.
267
9.9. ANALOGOUS RESULTS FOR HYPERBOLAS
9.9.5
Hyperbolic analogue of Grassmann’s supplement
We now turn to providing an analogue with respect to a hyperbola of the material in §7.5.3
on generalization to an ellipse of Grassmann’s supplement for a position vector. We consider a
hyperbola H with equation
y2
x2
−
= 1,
a2
b2
and its conjugate H∗ with equation
y2
x2
−
= −1.
a2
b2
For Z1 6= O, Z1 ≡ (x1 , y1 ) we define
−−→⊥1 −−→
OZ1 = OZ2 ,
Then
whereZ2 ≡ (x2 , y2 ) =
y22
a2 y12
b2 x21
x2
−
=
−
=−
a2
b2
b 2 a2
a2 b 2
a
b
y1 , x1 .
b
a
y2
x2
−
a2
b2
(9.9.4)
,
so that Z2 ∈ H∗ if and only if Z1 ∈ H.
Moreover
0
1
δF (O, Z1 , Z2 ) = x1
2 a
y1
b 2
x1
−
=ab
a2
0
1 x2 1 b
a x1 1
2
y1
.
b2
This is positive in the duo-sector with arms the asymptotes in which H lies. In particular when
Z1 ∈ H we have that δF (O, Z1 , Z2 ) = 21 ab. Note too that
|O, Z1 |2 − |O, Z2 |2 = (a2 − b2 )
x21
y2
− 21
2
a
b
,
and this is equal to a2 − b2 for Z1 ∈ H.
yy1
b2 x1
1
The tangent to H at Z1 has equation xx
a2 − b2 = 1 and so has slope a2 y1 , which is also
the slope of OZ2 . Thus this tangent is parallel to OZ2 . The tangent to H∗ at Z2 has equation
y1
yy2
xx2
b2 x2
a2 − b2 = −1 and so has slope a2 y2 = x1 , which is also the slope of OZ1 . Thus this tangent
is parallel to OZ1 . It follows that OZ1 and OZ2 are conjugate diametral lines for H. If we define
−−−→ −−→−−→
the point W1 by OW1 = OZ1 OZ2 , then [O, Z1 , W1 , Z2 ] is a parallelogram and so W1 is the point
of intersection of these tangents. Then
bx1 + ay1
b
a
(a, b),
W1 ≡ x1 + y1 , y1 + x1 =
b
a
ab
so W1 lies on the asymptote with equation y = ab x.
Similarly for Z7 6= O, Z7 ≡ (x7 , y7 ) we define
−−→⊥2 −−→
OZ7 = OZ8 ,
where
Z8 ≡ (x8 , y8 ) =
b
a
− y7 , − x7 .
b
a
(9.9.5)
Then Z8 ∈ H if and only if Z7 ∈ H∗ . Moreover δF (O, Z7 , Z8 ) is positive if Z7 is in the the
duo-sector with arms the asymptotes in which H∗ lies. In particular when Z7 ∈ H∗ we have that
δF (O, Z7 , Z8 ) = 12 ab. The lines OZ1 and OZ2 are conjugate diametral lines for H.
268
CHAPTER 9. AFFINE METHODS AND RESULTS
If now we start with a point Z1 ∈ H, take a point Z2 so that
−−→
=OZ2 ,
−−→
=OZ3 ,
−−→
=OZ4 ,
−−→
=OZ1 ,
−−→⊥1
OZ1
−−→⊥2
OZ2
−−→⊥1
OZ3
−−→⊥2
OZ4
−−→ −−→ −−−→
OZ1 + OZ2 = OW1 ,
−−→ −−→ −−−→
OZ2 + OZ3 = OW2 ,
−−→ −−→ −−−→
OZ3 + OZ4 = OW3 ,
−−→ −−→ −−−→
OZ4 + OZ1 = OW4 ,
then, as in S 9.5.3, [W1 , W2 , W3 , W4 ] is a parallelogram the sides of which are tangent to H or H∗
at Z1 , Z2 , Z3 , Z4 . Each vertex lies on one or other of the asymptotes. As Z1 varies the area of the
parallelogram is constant.
W1
b
Z2
b
W2
b
b
Z1
b
Z3
b
b
W4
b
Z4
b
W3
Figure 9.25.
9.10
Oblique analogues
9.10.1
Oblique axial symmetries
Consider the ellipse with equation
y2
x2
+ 2 = 1,
2
a
b
and the line l through the origin with slope µ. Then the conjugate diametral line m has slope
−b2 /µa2 . For any point Z ≡ (x, y), the line through Z parallel to m meets l in a point W , and
the point Z ′ such that W is the mid-point of Z and Z ′ has coordinates (x′ , y ′ ) which satisfy
x′
=
y′
=
b 2 − µ2 a 2
2µa2
x
+
y,
b 2 + µ2 a 2
b 2 + µ2 a 2
2µb2
b 2 − µ2 a 2
x
−
y.
b 2 + µ2 a 2
b 2 + µ2 a 2
This transformation has the nature of an oblique axial symmetry. We can express it in
matrix form as
!
′ 2µa2
b2 −µ2 a2
x
x
b2 +µ2 a2
b2 +µ2 a2
.
=
2
2 2
2µb2
y
y′
− b2 −µ2 a2
2
2 2
b +µ a
9.10.2
b +µ a
Oblique analogue of rotation
If we take the composition of two such oblique axial symmetries in §9.10.1, with µ equal to µ1
and µ2 in turn, we obtain an analogue of the notion of rotation about the origin. Its matrix is the
269
9.10. OBLIQUE ANALOGUES
product of the two matrices of the above type, and turns out to be
 2 2
2
2
2
2
2 2
2

This can be re-written as


where
b2 −µ23 a2
b2 +µ23 a2
2
3b
− b22µ
+µ23 a2
µ3 =
9.10.3
2a (µ1 −µ2 )(b +a µ1 µ2 )
(b2 +µ22 a2 )(b2 +µ21 a2 )
2
(b +a2 µ1 µ2 )2 −a2 b2 (µ1 −µ2 )2
(b2 +µ22 a2 )(b2 +µ21 a2 )
(b +a µ1 µ2 ) −a b (µ1 −µ2 )
(b2 +µ22 a2 )(b2 +µ21 a2 )
2b2 (µ1 −µ2 )(b2 +a2 µ1 µ2 )
− (b2 +µ2 a2 )(b2 +µ2 a2 )
2
1
2µ3 a2
b2 +µ23 a2
b2 −µ23 a2
b2 +µ23 a2

.

,
b2 (µ1 − µ2 )
.
b 2 + a 2 µ1 µ2
A three-axes result
If we post-multiply the matrix for an oblique axial symmetry with µ = µ4 by the matrix for an
oblique rotation in §9.10.2, we obtain the matrix of an oblique axial symmetry with
µ = µ5 =
b2 (µ3 + µ4 )
.
b 2 − a 2 µ3 µ4
Thus the composition of three oblique axial symmetries, in lines through the origin, with respect
to this ellipse, is also such an oblique axial symmetry. This yields a three-axes result.
9.10.4
Oblique frame of reference
We now introduce an oblique coordinate system. Take an ordered pair F1 = ([O, W1 , [O, W2 )
where W1 6= O, W2 6= O and O, W1 , W2 are non-collinear. We let H1′ , H2′ be the closed half-planes
with edge OW1 , with H1′ the one containing W2 , and H3′ , H4′ be the closed half-planes with edge
OW2 , with H3′ the one containing W1 .
Given any point Z ∈ Π, we seek to write
Z = x′ W1 + y ′ W2 .
Let Z ≡F (x, y), W1 ≡F (u1 , v1 ), W2 ≡F (u2 , v2 ). We then need
x
y
= u1 x′ + u2 y ′ ,
= v1 x′ + v2 y ′ .
As 2δF (O, W1 , W2 ) = u1 v2 − u2 v1 6= 0, we have the solutions
x′
=
y′
=
v2 x − u2 y
,
u1 v2 − u2 v1
−v1 x + u1 y
.
u1 v2 − u2 v1
If U ′ = x′ W1 then Z − U ′ = y ′ W2 , U ′ is on the line OW1 and the line ZU ′ is parallel to OW2 .
Thus U ′ is the point where the line through Z parallel to OW2 meets OW1 . Moreover
|O, U ′ |
= |x′ |.
|O, W1 |
A simple calculation shows that
x′ =
δF (O, Z, W2 )
δF (O, W1 , W2 )
270
CHAPTER 9. AFFINE METHODS AND RESULTS
and so x′ is positive when W ∈ H3′ and negative when W ∈ H4′ . Thus
x′ =
when W ∈ H3′ , and
|O, U ′ |
,
|O, W1 |
x′ = −
|O, U ′ |
,
|O, W1 |
when W ∈ H4′ .
By similar reasoning, if V ′ = y ′ W2 then Z − V ′ = x′ W1 , and V ′ is the point where the line
through Z parallel to OW1 meets OW2 . Moreover
y′ =
δF (O, W1 , Z)
δF (O, W1 , W2 )
and
y′ =
when W ∈ H1′ , and
|O, V ′ |
,
|O, W2 |
y′ = −
|O, V ′ |
,
|O, W2 |
when W ∈ H2′ .
We call (x′ , y ′ ) oblique coordinates of Z with respect to F1 .
It is easy to check that equations of the form l1 x′ + m1 y ′ + n1 = 0, where (l1 , m1 ) 6= (0, 0),
correspond to lines and that ratio results can be handled with these coordinates just as with rectangular coordinates. However these oblique coordinates do not suit distance or perpendicularity,
and we go on the consider appropriate analogues for these.
9.10.5
Conjugate directions
Let E be the ellipse with equation
y2
x2
+
= 1.
a2
b2
The diametral line with equation y = µx meets the ellipse at the point
!
ab
µab
W1 ≡F p
,
,p
b 2 + µ2 a 2
b 2 + µ2 a 2
2
b
and a diagonally opposite point. The conjugate diametral line has equation y = − µa
2 x, and this
meets the ellipse in the point
!
b2
µa2
,
,p
W2 ≡F − p
b 2 + µ2 a 2
b 2 + µ2 a 2
−−−→ −−−→⊥E
and a diagonally opposite point. It can be checked that OW2 = OW1 , as in §9.5.3.
We now take oblique coordinates (x′ , y ′ ) with respect to F1 = ([O, W1 , [O, W2 ). Then by
§9.10.4 we have
x =
y
=
abx′ − µa2 y ′
p
,
b 2 + µ2 a 2
µabx′ + b2 y ′
p
.
b 2 + µ2 a 2
9.11. GENERALISATION TO ELLIPSES NOT IN NORMAL POSITION
271
Now 2δF (O, W1 , W2 ) = ab, and so we have as well
x′
=
y′
=
From these we note that
b2 x + µa2 y
p
,
ab b2 + µ2 a2
−µabx + aby
p
.
ab b2 + µ2 a2
x2
y2
+ 2,
2
a
b
so it is this quadratic expression which is the analogue of square of distance. Thus loci
x′2 + y ′2 =
x′2 + y ′2 = k 2
are ellipses of the form
x2
y2
+ 2 2 = 1.
2
2
k a
k b
As an analogue let us consider lines with equations
y ′ = mx′ , y ′ = −
1 ′
x.
m
The first of these yields
m=
−µab + aby/x
,
b2 + µa2 y/x
and so corresponds to the line y = m1 x where
m=
−µab + abm1
.
b2 + µa2 m1
Similarly the second corresponds to the line y = m2 x where
−
1
−µab + abm2
.
= 2
m
b + µa2 m2
We multiply these to find the relationship of m1 and m2 . It reduces to
m1 m2 = −
b2
.
a2
These correspond to conjugate diametral lines for the ellipse, and so this is the analogue of perpendicularity.
9.11
Generalisation to ellipses not in normal position
9.11.1
Analogue of inner product and supplement
Consider an affine transformation given by (x′ , y ′ ) → (x, y),
x = ax′ + by ′ ,
y = cx′ + dy ′ ,
where ad − bc 6= 0. Then under it
x1 x2 + y1 y2 = (ax′1 + by1′ )(ax′2 + by2′ ) + (cx′1 + dy1′ )(cx′2 + dy2′ ),
and we use this last to define the analogue of an inner product,
(x′1 , y1′ ) ∗ (x′2 , y2′ ) = (ax′1 + by1′ )(ax′2 + by2′ ) + (cx′1 + dy1′ )(cx′2 + dy2′ ).
(9.11.1)
272
CHAPTER 9. AFFINE METHODS AND RESULTS
We take
(x′1 , y1′ ) ∗ (x′2 , y2′ ) = 0,
as an analogue of perpendicularity, and
(x′1 , y1′ ) ∗ (x′1 , y1′ ) = (ax′1 + by1′ )2 + (cx′1 + dy1′ )2
as an analogue of square of magnitude.
We also have the inverse transformation (x, y) → (x′ , y ′ ), given by
x′ =
1
(dx − by),
ad − bc
y′ =
1
(−cx + ay).
ad − bc
(9.11.2)
Now suppose that for the supplement (−y, x) of (x, y) we have (−y, x) → (x′′ , y ′′ ), so that
x′′
=
y ′′
=
1
(−bx − dy),
ad − bc
1
(ax + cy),
ad − bc
and thus, on using (9.11.1)
x′′
=
y ′′
=
−1
[(ab + cd)x′ + (b2 + d2 )y ′ ],
ad − bc
1
[(a2 + c2 )x′ + (ab + cd)y ′ ].
ad − bc
We regard this as a generalized supplement of (x′ , y ′ ) and it is straightforward to verify that
(x′′ , y ′′ ) ∗ (x′ , y ′ ) = 0,
so that we have a generalization of orthogonality, and that
(x′′ , y ′′ ) ∗ (x′′ , y ′′ ) = (ax′ + by ′ )2 + (cx′ + dy ′ )2 = (x′ , y ′ ) ∗ (x′ , y ′ ),
so that it has the same square of magnitude as the original.
9.11.2
Grassmann supplement as conjugate semi-diameter
With the transformation (9.11.1) we have
x2 + y 2 = (ax′ + by ′ )2 + (cx′ + dy ′ )2 ,
and so to the circle with equation
x2 + y 2 = k 2
we have corresponding the conic with equation
(a2 + c2 )x′2 + 2(ab + cd)x′ y ′ + (b2 + d2 )y ′2 − k 2 = 0.
As
(ab + cd)2 − (a2 + c2 )(b2 + d2 ) = −(ad − bc)2 < 0,
this is an ellipse.
Let us consider lines with parametric equations
x′ = x′′ + tx′1 ,
y ′ = y ′′ + ty1′ ,
(t ∈ R),
273
9.11. GENERALISATION TO ELLIPSES NOT IN NORMAL POSITION
which pass through the variable point Z ′′ ≡ (x′′ , y ′′ ) but are all parallel to the fixed line joining O
to Z1′ ≡ (x′1 , y1′ ). For points of intersection with the ellipse we have the equation of incidence
(a2 + c2 )[x′′ + tx′1 ]2 + 2(ab + cd)[x′′ + tx′1 ][y ′′ + ty1′ ] + (b2 + d2 )[y ′′ + ty1′ ]2 − k 2
=
0.
For Z ′′ to be the mid-point of the points of incidence we need the coefficient of t to vanish; this
gives
x′′ [(a2 + c2 )x′1 + (ab + cd)y1′ ] + y ′′ [(ab + cd)x′1 + (b2 + d2 )y1′ )] = 0.
This has parametric equations
x′′
y ′′
= −[(ab + cd)x′1 + (b2 + d2 )y1′ ]s,
= [(a2 + c2 )x′1 + (ab + cd)y1′ ]s, (s ∈ R).
Thus the generalized supplement of OZ1′ in §9.11.1 is along the conjugate diametral line with
respect to the ellipse.
9.11.3
Components along a vector and its generalized supplement
In Barry [2] we combined the components of a second vector along a first vector and the supplement
of the first vector with the use of complex numbers, by writing
z − z2 = (p + iq)(z3 − z2 ),
referring to (p, q) as mobile coordinates. On applying the affine transformation 7.11.1 to this
situation we have that
a(x′ −x′2 )+b(y ′ −y2′ )+ı[c(x′ −x′2 )+d(y ′ −y2′ )] = (p+ıq){a(x′3 −x′2 )+b(y3′ −y2′ )+ı[c(x′3 −x′2 )+d(y3′ −y2′ )]}.
On solving this for p and q, we obtain
p =
q
=
[a(x′ − x′2 ) + b(y ′ − y2′ )][a(x′3 − x′2 ) + b(y3′ − y2′ )] + [c(x′ − x′2 ) + d(y ′ − y2′ )][c(x′3 − x′2 ) + d(y3′ − y2′ )]
,
[a(x′3 − x′2 ) + b(y3′ − y2′ )]2 + [c(x′3 − x′2 ) + d(y3′ − y2′ )]2
[c(x′ − x′2 ) + d(y ′ − y2′ )][a(x′3 − x′2 ) + b(y3′ − y2′ )] − [a(x′ − x′2 ) + b(y ′ − y2′ )][c(x′3 − x′2 ) + d(y3′ − y2′ )]
.
[a(x′3 − x′2 ) + b(y3′ − y2′ )]2 + [c(x′3 − x′2 ) + d(y3′ − y2′ )]2
The numerator in p is a generalized inner product and the denominator is a generalized square of
distance. The numerator in q simplifies to
2(ad − bc)δF (Z2′ , Z3′ , Z ′ ).
9.11.4
Analogue of similar triangles
The methods of Barry [2] can be applied to these generalized vector products; in particular we
have the analogues of similar triangles under an affine transformation.
Exercises
9.1 Let C be the circle with centre O, with radius length a. Let Z1 , Z2 be points of the circle,
T1 the point where the tangent at Z1 meets OZ2 , and T2 the point where the tangent at Z2
meets OZ1 . Let W1 be the point on the circle such that OW1 ⊥ OZ1 and (O, Z1 , W1 ) is
positively oriented. Let W2 be similarly related to Z2 . Prove that
δF (O, Z1 , Z2 )
T 1 Z1
T 2 Z2
=−
,
=−
δF (O, Z1 , W2 )
OW1
OW2
and
T1 T2
a2
.
=
2δF (O, Z1 , W2 )
Z1 Z2
274
CHAPTER 9. AFFINE METHODS AND RESULTS
Chapter 10
Projective methods and results
In Chapter 4, we introduced affine and Euclidean concepts so as to be able to distinguish the types
of conics and to present classical results. We now revert to the state in Chapter 6 and deal with
projective concepts again; the results will generally apply to all proper conics.
10.1
Maclaurin property of conics
10.1.1
Let W1 , W2 , W3 be distinct fixed points and W4 W5 , W6 W7 distinct fixed lines. Consider a variable
triangle [X, Y, Z] such that the side-lines Y Z, ZX, XY pass through the fixed points W1 , W2 , W3 ,
respectively, and the vertices X, Y lie on the fixed lines W4 W5 , W6 W7 , respectively. Then
X=
λ
1
µ
1
W4 +
W5 , Y =
W6 +
W7 ,
1+λ
1+λ
1+µ
1+µ
for variable λ and µ. Now δF (Z, W2 , X) = 0, so
δF (Z, W2 , W4 ) + λδF (Z, W2 , W5 ) = 0.
Similarly δF (Z, W1 , Y ) = 0, so
δF (Z, W1 , W6 ) + µδF (Z, W1 , W7 ) = 0.
Moreover δF (W3 , X, Y ) = 0, so
λµδF (W3 , W5 , W7 ) + λδF (W3 , W5 , W6 ) + µδF (W3 , W4 , W7 ) + δF (W3 , W4 , W6 ) = 0.
On inserting for λ and µ we obtain
δF (Z, W2 , W4 )δF (Z, W1 , W6 )δF (W3 , W5 , W7 )
− δF (Z, W2 , W4 )δF (Z, W1 , W7 )δF (W3 , W5 , W6 )
− δF (Z, W1 , W6 )δF (Z, W2 , W5 )δF (W3 , W4 , W7 )
+ δF (Z, W2 , W5 )δF (Z, W1 , W7 )δF (W3 , W4 , W6 )
= 0.
Thus the locus of Z is a conic, and this characterization was due to Maclaurin (1721). In the
diagram we take W7 = W4 .
275
276
CHAPTER 10. PROJECTIVE METHODS AND RESULTS
Y
b
W4 = W7
b
W1
W6
b
b
b
b
b
Z
W2
X
b
W5
Figure 10.1
b
W3
10.2
Involution
10.2.1
Definition and some basic occurrences
A projectivity for a range of points as in (1.4.4) or for a pencil of pairs of points as in §2.3.1 is
called an involution if b = c, so that it has the form
aλµ + b(λ + µ) + d = 0,
(10.2.1)
and ad − b2 6= 0. If we take the correspondence λ → µ as defining a function, then the composition
of this function with itself will be the identity transformation, so that this function is its own
inverse.
There will be two self-corresponding elements with λ = µ if the equation
aλ2 + 2bλ + d = 0,
(10.2.2)
has two real roots, a necessary and sufficient condition for which is that b2 − ad > 0. If this is the
case and we denote by λ1 , λ2 the roots of (10.2.2), then when λ, µ satisfy (10.2.1) the cross-ratio
cr(λ, µ, λ1 , λ2 ) will be equal to −1, so that the elements corresponding to (λ, µ, λ1 , λ2 ) will form a
harmonic range or pencil. For
cr(λ, µ, λ1 , λ2 ) + 1
(λ − λ1 )(µ − λ2 )
+1
(λ − λ2 )(µ − λ1 )
λµ − λ1 µ − λ2 λ + λ1 λ2 + (λµ − λ2 µ − λ1 λ + λ1 λ2 )
=
λµ − λ2 µ − λ1 λ + λ1 λ2
2(λµ + λ1 λ2 ) − (λ1 + λ2 )(λ + µ)
=
.
λµ − λ2 µ − λ1 λ + λ1 λ2
=
277
10.2. INVOLUTION
On using (10.2.1) we see that the numerator in this is equal to
d
b
2 − (λ + µ) − + λ1 λ2 − (λ + µ)(λ1 + λ2 )
a
a
b
d
= − (λ + µ) 2 + λ1 + λ2 + 2 − + λ1 λ2 ,
a
a
and both terms here vanish because of (10.2.2).
As an illustration of the occurrence of an involution, on any line Z1 Z2 take the parametrisation
1
λ
Z = 1+λ
Z1 + 1+λ
Z2 . Then for points Z3 , Z4 , Z5 on this line,
Z3 Z4 =
λ4 − λ3
λ5 − λ3
(Z2 − Z1 ), Z3 Z5 =
(Z2 − Z1 ),
(1 + λ3 )(1 + λ4 )
(1 + λ3 )(1 + λ5 )
so that
Z3 Z4 Z3 Z4 =
(λ4 − λ3 )(λ5 − λ3 )
|Z1 , Z2 |2 .
(1 + λ3 )2 (1 + λ4 )(1 + λ5 )
Then for
Z3 Z4 Z3 Z4 = k 6= 0,
we have
(1 + λ3 )2
(λ4 − λ3 )(λ5 − λ3 )
= j 6= 0,
=k
(1 + λ4 )(1 + λ5 )
|Z1 , Z2 |2
so that
(10.2.3)
say,
(1 − j)λ4 λ5 − (λ3 + j)(λ4 + λ5 ) + λ23 − j = 0.
Here (1 − j)(λ23 − j) − (λ3 + j)2 = −j(1 + λ3 )2 6= 0. Thus for fixed Z3 , points Z4 and Z5 satisfying
(10.2.3) are mates in an involution.
10.2.2
Desargues’ involution theorem
For a result on involutions involving conics, we start with a conic C, with equation
δF (Z1 , Z3 , Z)δF (Z2 , Z4 , Z) − kδF (Z1 , Z4 , Z)δF (Z2 , Z3 , Z) = 0,
so that by (7.9.2) it is one of the pencil of conics through the points Z1 , Z2 , Z3 and Z4 , and proceed
as we did in §7.2.5. Consider the intersection with this conic of the line W0 W1 , the points of which
have the form
1
λ
Z=
W0 +
W1 .
1+λ
1+λ
Then as
δF (Z1 , Z3 ,
1
λ
1
λ
W0 +
W1 ) =
δF (Z1 , Z3 , W0 ) +
δF (Z1 , Z3 , W1 ),
1+λ
1+λ
1+λ
1+λ
and similarly for the other terms, for a point of intersection we have
[δF (Z1 , Z3 , W0 ) + λδF (Z1 , Z3 , W1 )][δF (Z2 , Z4 , W0 ) + λδF (Z2 , Z4 , W1 )]
− k[δF (Z1 , Z4 , W0 ) + λδF (Z1 , Z4 , W1 )][δF (Z2 , Z3 , W0 ) + λδF (Z2 , Z3 , W1 )] = 0,
and so obtain
λ2 {δF (Z1 , Z3 , W1 )δF (Z2 , Z4 , W1 ) − k[δF (Z1 , Z4 , W1 )δF (Z2 , Z3 , W1 )]}
+ λ{δF (Z1 , Z3 , W0 )δF (Z2 , Z4 , W1 )] + δF (Z1 , Z3 , W1 )δF (Z2 , Z4 , W0 )]
− k[δF (Z1 , Z4 , W0 )δF (Z2 , Z3 , W1 )] + δF (Z1 , Z4 , W1 )δF (Z2 , Z3 , W0 )]}
+ δF (Z1 , Z3 , W0 )δF (Z2 , Z4 , W0 ) − kδF (Z1 , Z4 , W0 )δF (Z2 , Z3 , W0 ) = 0.
278
CHAPTER 10. PROJECTIVE METHODS AND RESULTS
Figure 10.2.
The roots λ1 , λ2 of this quadratic equation satisfy
λ1 λ2 =
δF (Z1 , Z3 , W0 )δF (Z2 , Z4 , W0 ) − kδF (Z1 , Z4 , W0 )δF (Z2 , Z3 , W0 )
,
δF (Z1 , Z3 , W1 )δF (Z2 , Z4 , W1 ) − kδF (Z1 , Z4 , W1 )δF (Z2 , Z3 , W1 )
and −λ1 − λ2 equals the quotient of
δF (Z1 , Z3 , W0 )δF (Z2 , Z4 , W1 ) + δF (Z1 , Z3 , W1 )δF (Z2 , Z4 , W0 )
− k[δF (Z1 , Z4 , W0 )δF (Z2 , Z3 , W1 ) + δF (Z1 , Z4 , W1 )δF (Z2 , Z3 , W0 )]
by
δF (Z1 , Z3 , W1 )δF (Z2 , Z4 , W1 ) − kδF (Z1 , Z4 , W1 )δF (Z2 , Z3 , W1 ).
Thus we have a relationship
λ1 λ2 =
p + qk
r + sk
, λ1 + λ2 = −
.
t + uk
t + uk
On eliminating k between these, we obtain
(st − ur)λ1 λ2 + (qt − pu)(λ1 + λ2 ) + qr − ps = 0,
where the coefficients are independent of k.
This relationship of the points on W0 W1 to each other is an example of an involution and gives
us a fundamental result due to Desargues (1639), that the conics through four points Z1 , Z2 , Z3 , Z4
cut a fixed line in points which are pairs in an involution. This was a revolutionary concept which
he introduced for the study of conics.
279
10.2. INVOLUTION
10.2.3
The lines through a point, intersecting a conic
For a proper conic C we take its equation as in §7.1.6, and so have points Z on it have normalized
areal coordinates given as in (7.1.8),
α0
,
α0 + γ0 t + β0 t2
β 0 t2
β=
,
α0 + γ0 t + β0 t2
γ0 t
.
γ=
α0 + γ0 t + β0 t2
α=
For another point Z ′ on the conic we take these coordinates with the parameter t′ instead. We now
suppose that the line ZZ ′ always passes through a fixed point Z4 with coordinates (α4 , β4 , γ4 ).
Then we must always have δF (Z4 , Z, Z ′ ) = 0, so that
α4
β4
γ4 α0 β0 t2 γ0 t = 0,
α0 β0 t′2 γ0 t′ which simplifies to
(t − t′ )[α4 β0 γ0 tt′ − γ4 α0 β0 (t + t′ ) + β4 γ0 α0 ] = 0,
and so on dividing across by t − t′ , we obtain
α4 β0 γ0 tt′ − γ4 α0 β0 (t + t′ ) + β4 γ0 α0 = 0.
(10.2.4)
This has the form of an involution and for it
b2 − ad = γ42 α20 β02 − α4 β4 α0 β0 γ02 = α20 β02 (γ42 − 4kα4 β4 ),
since γ02 = 4kα0 β0 , and so we have an involution as long as Z4 does not lie on the conic. The
corresponding equation (10.2.2) is
α4 β0 γ0 t2 − 2γ4 α0 β0 t + β4 γ0 α0 = 0,
(10.2.5)
and the condition for this to have two self-corresponding pairs is α20 β02 (γ42 − 4kα4 β4 ) > 0, that is
for Z4 to be an exterior point.
The derivation of (10.2.5) becomes a bit hazy when t′ = t, but we can note explicitly that if Zt
is the point on the conic with coordinates (7.1.8), then the tangent at that point has equation
γ0 tγ − 2k(β0 t2 α + α0 β) = 0,
and this tangent passes through Z4 if and only if
2kα4 β0 t2 − γ4 γ0 t + 2kβ4 α0 = 0.
But
2kβ0 =
γ02
,
2α0
2kα0 =
γ02
,
2β0
so our equation becomes
γ0
[α4 β0 γ0 t2 − 2γ4 α0 β0 t + β4 γ0 α0 ] = 0,
2α0 β0
which is equivalent to (10.2.5). Thus the parameters for the points of contact of the tangents from
Z4 to the conic are the values of t which are self-corresponding. If we denote by t1 and t2 these
parameters for self-corresponding points, then by §10.2.1, cr(t, t′ , t1 , t2 ) = −1.
280
CHAPTER 10. PROJECTIVE METHODS AND RESULTS
10.2.4
Conjugate points lying on one line
Other illustrations of the occurrences of involutions are as follows. We recall from §7.2.7 that the
points W0 , W1 are conjugate with respect to the conic with equation γ 2 − 4kαβ = 0 if
γ1 γ0 − 2k(β1 α0 + α1 β0 ) = 0.
We now consider pairs of points on the line Z4 Z5 which are conjugates in this way. The points
Z=
λ
1
µ
1
Z4 +
Z5 , W =
Z4 +
Z5 ,
1+λ
1+λ
1+µ
1+µ
are conjugates when
1
λ
1
µ
1
λ
1
µ
γ4 +
γ5 )(
γ4 +
γ5 ) − 2k[(
α4 +
α5 )(
β4 +
β5 )
1+λ
1+λ
1+µ
1+µ
1+λ
1+λ
1+µ
1+µ
1
λ
1
µ
+(
β4 +
β5 )(
α4 +
α5 )] = 0.
1+λ
1+λ
1+µ
1+µ
(
This reduces to
λµ(γ52 − 4kα5 β5 ) + (λ + µ)[γ5 γ4 − 2k(α5 β4 + α4 β5 )] + γ42 − 4kα4 β4 = 0.
Thus this is an involution between points on a line.
10.2.5
Conjugate diametral lines
As a further example, we note that the condition in m1 /l1 and m0 /l0 in (7.4.12) for conjugate
diametral lines of a central conic gives an involution between lines through the centre.
10.3
Pascal’s and Carnot’s theorems
10.3.1
Pascal’s theorem
Pascal’s Theorem (1640). Let (Z1 , Z2 , Z3 , Z4 , Z5 , Z6 ) be distinct points of a conic C. Then :(i)if Z1 Z5 ∩ Z4 Z2 = {W ′ }, Z2 Z6 ∩ Z5 Z3 = {W ′′ }, Z3 Z4 ∩ Z6 Z1 = {W ′′′ }, the points
′
W , W ′′ , W ′′′ are collinear ;
(ii)if Z1 Z5 and Z4 Z2 are distinct and parallel and Z2 Z6 ∩Z5 Z3 = {W ′′ }, Z3 Z4 ∩Z6 Z1 = {W ′′′ },
then Z1 Z5 k W ′′ W ′′′ ;
(iii)if Z1 Z5 and Z4 Z2 are distinct and parallel, and so are Z2 Z6 and Z5 Z3 , then Z3 Z4 k Z6 Z1 .
Z2
b
Z3
b
b
Z1
b
b
b
Z6
b
b
Z5
b
Z4
Figure 10.3, Case (i).
281
10.3. PASCAL’S AND CARNOT’S THEOREMS
Proof:In (7.9.3) we start with the pencil conics through the points Z1 , Z2 , Z4 and Z5
δF (Z, Z1 , Z5 )δF (Z, Z4 , Z2 )
= k,
δF (Z, Z1 , Z4 )δF (Z, Z5 , Z2 )
and on making this pass through Z3 we have
δF (Z3 , Z1 , Z5 )δF (Z3 , Z4 , Z2 )
δF (Z, Z1 , Z5 )δF (Z, Z4 , Z2 )
=
,
δF (Z, Z1 , Z4 )δF (Z, Z5 , Z2 )
δF (Z3 , Z1 , Z4 )δF (Z3 , Z5 , Z2 )
and as this passes through Z6 we have
δF (Z6 , Z1 , Z5 )δF (Z6 , Z4 , Z2 ) δF (Z3 , Z1 , Z5 )δF (Z3 , Z4 , Z2 )
−
= 0.
δF (Z6 , Z1 , Z4 )δF (Z6 , Z5 , Z2 ) δF (Z3 , Z1 , Z4 )δF (Z3 , Z5 , Z2 )
(10.3.1)
Now
δF {[(Z1 , Z5 ), (Z4 , Z2 )], [(Z3 , Z4 ), (Z6 , Z1 )], [(Z2 , Z6 ), (Z5 , Z3 )]}
=δF [(Z1 , Z5 ), (Z3 , Z4 ), (Z6 , Z1 )]δF [(Z4 , Z2 ), (Z2 , Z6 ), (Z5 , Z3 )]
−δF [(Z4 , Z2 ), (Z3 , Z4 ), (Z6 , Z1 )]δF [(Z1 , Z5 ), (Z2 , Z6 ), (Z5 , Z3 )]
=δF (Z1 , Z3 , Z4 )δF (Z5 , Z6 , Z1 )δF (Z4 , Z2 , Z6 )δF (Z2 , Z5 , Z3 )
−δF (Z2 , Z3 , Z4 )δF (Z4 , Z6 , Z1 )δF (Z5 , Z2 , Z6 )δF (Z1 , Z5 , Z3 ).
By (11.3.1) this is equal to 0 and the remainder of the proof is as in that of Pappus’ theorem. In
Case (iii) we cannot have Z1 Z5 k Z2 Z6 as that would imply that Z4 Z2 k Z2 Z6 and so Z2 Z4 = Z2 Z6 .
Thus a straight line would meet a proper conic in more than two points.
In §11.7.1 we shall establish an extension of Pascal’s theorem involving two linked conics.
10.3.2
Carnot’s theorem (1803)
W3
Z6
b
b
Z5
b
Z7
b
b
Z4
b
b
Z8
b
Z9 W2
b
W1
Figure 10.4
We take the form (7.1.2) of equation for a conic. Suppose that for non-collinear points W1 , W2
and W3 , the lines W2 W3 , W3 W1 , W1 W2 cut the conic at the points Z4 , Z5 , Z6 , Z7 , Z8 , Z9 , respec-
282
CHAPTER 10. PROJECTIVE METHODS AND RESULTS
tively. Then we will have
λ1
1
W2 +
W3 ,
1 + λ1
1 + λ1
1
µ1
Z6 =
W3 +
W1 ,
1 + µ1
1 + µ1
1
ν1
Z8 =
W1 +
W2 ,
1 + ν1
1 + ν1
Z4 =
1
λ2
W2 +
W3 ,
1 + λ2
1 + λ2
1
µ2
Z7 =
W3 +
W1 ,
1 + µ2
1 + µ2
1
ν2
Z9 =
W1 +
W2 ,
1 + ν2
1 + ν2
Z5 =
for appropriate values of the parameters. If Z is a point on the line W2 W3 , with parameter λ, then
as in §7.2.5 Z will lie on the conic if
f δF (Z, Z3 , Z1 )δF (Z, Z1 , Z2 ) + gδF (Z, Z1 , Z2 )δF (Z, Z2 , Z3 ) + hδF (Z, Z2 , Z3 )δF (Z, Z1 , Z2 ) = 0,
and this expands to
f [δF (W2 , Z3 , Z1 ) + λδF (W3 , Z3 , Z1 )][δF (W2 , Z1 , Z2 ) + λδF (W3 , Z1 , Z2 )]
+g[δF (W2 , Z1 , Z2 ) + λδF (W3 , Z1 , Z2 )][δF (W2 , Z2 , Z3 ) + λδF (W3 , Z2 , Z3 )]
+h[δF (W2 , Z2 , Z3 ) + λδF (W3 , Z2 , Z3 )][δF (W2 , Z3 , Z1 ) + λδF (W3 , Z3 , Z1 )]
=0.
As λ1 and λ2 are the roots of this quadratic equation in λ, we have that
λ1 λ2 =
φ(W2 )
,
φ(W3 )
where
φ(W ) = f δF (W, Z3 , Z1 )δF (W, Z1 , Z2 )+gδF (W, Z1 , Z2 )δF (W, Z2 , Z3 )+hδF (W, Z2 , Z3 )δF (W, Z3 , Z1 ).
Similarly
µ1 µ2 =
φ(W3 )
,
φ(W1 )
ν1 ν2 =
φ(W1 )
,
φ(W2 )
and so
λ1 λ2 µ1 µ2 ν1 ν2 = 1.
However
λ1
δF (W1 , W2 , W3 ),
1 + λ1
1
δF (W1 , W2 , W3 ),
δF (W1 , Z4 , W3 ) =
1 + λ1
δF (W1 , W2 , Z4 ) =
and so
λ1 =
δF (W1 , W2 , Z4 )
W2 Z4
=
.
δF (W1 , Z4 , W3 )
Z4 W3
On using similar results for the other parameters, we obtain
δF (W1 , W2 , Z4 ) δF (W1 , W2 , Z5 ) δF (W2 , W3 , Z6 ) δF (W2 , W3 , Z7 ) δF (W3 , W1 , Z8 )
δF (W1 , Z4 , W3 ) δF (W1 , Z5 , W3 ) δF (W2 , Z6 , W1 ) δF (W2 , Z7 , W1 ) δF (W3 , Z8 , W2 )
δF (W3 , W1 , Z9 )
.
δF (W3 , Z9 , W2 )
W2 Z4 W2 Z5 W3 Z6 W1 Z7 W1 Z8 W1 Z9
Z4 W3 Z5 W3 Z6 W1 Z7 W2 Z8 W2 Z9 W2
=1.
=
10.4. IDENTIFICATION OF COEFFICIENTS IN EQUATIONS OF CONICS
283
A variant on this theorem is as follows. Suppose that for distinct points W2 and W3 we have
W2 W5 k W3 W4 and the lines W2 W3 , W3 W4 , W2 W5 cut the conic at the points Z4 , Z5 , Z6 , Z7 ,
Z8 , Z9 , respectively. Using primes for the coordinates of W -points, we re-use the material above
on the points Z4 and Z5 . Points Z on the line W3 W4 have the form
α = α′3 + s(α′4 − α′3 ), β = β3′ + s(β4′ − β3′ ), γ = γ3′ + s(γ4′ − γ3′ ),
and lie on the conic if s satisfies the equation of incidence
f [β3′ + s(β4′ − β3′ )][γ3′ + s(γ4′ − γ3′ )] + g[γ3′ + s(γ4′ − γ3′ )α′3 + s(α′4 − α′3 )]
+ h[α′3 + s(α′4 − α′3 )][β3′ + s(β4′ − β3′ )] = 0.
With s6 , s7 the values of the parameter, we then have that
f β3′ γ3′ + gγ3′ α′3 + hα′3 β3′
f (β4′ − β3′ )(γ4′ − γ3′ ) + g(γ4′ − γ3′ )(α′4 − α′3 ) + h(α′4 − α′3 )(β4′ − β3′ )
φ(W3 )
=
f (β4′ − β3′ )(γ4′ − γ3′ ) + g(γ4′ − γ3′ )(α′4 − α′3 ) + h(α′4 − α′3 )(β4′ − β3′ )
s6 s7 =
=
W3 Z6 W3 Z7
.
|W3 , W4 |2
As W2 W5 k W3 W4 we have that W5 − W2 = j(W4 − W3 ) for some j 6= 0, and with t8 , t9 being
parameters for Z8 and Z9 we have by a similar argument
t8 t9 =
j 2 [f (β4′
It follows that
10.4
−
β3′ )(γ4′
−
γ3′ )
+
g(γ4′
φ(W2 )
W2 Z8 W2 Z9
.
= 2
′
′
′
′
′
′
′
− γ3 )(α4 − α3 ) + h(α4 − α3 )(β4 − β3 )]
j |W3 , W4 |2
W2 Z4 W3 Z6 W3 Z7
= 1.
Z4 W3 W2 Z8 W2 Z9
Identification of coefficients in equations of conics
10.4.1
W2
b
Z1
Z3
b
b
b
W1
b
Z2
b
W3
Figure 10.5
284
CHAPTER 10. PROJECTIVE METHODS AND RESULTS
We take again the form (7.1.2)of the equation for a conic which passes through the points of
the triple of reference,
f βγ + gγα + hαβ = 0.
Now coefficients are not very geometrical objects so we seek to identify them in more geometrical
terms. Taking as usual (Z1 , Z2 , Z3 ) to be the triple of reference, we consider the poles W1 , W2 , W3
of Z2 Z3 , Z3 Z1 , Z1 Z2 , respectively.
As the polar of any point Z4 relative this conic has equation
f (β4 γ + γ4 β) + g(γ4 α + alpha4 γ) + h(α4 β + β4 α) = 0,
it is easy to find the equations of the tangents at Z2 and Z3 and that for their point of intersection
1
(−f, g, h). Similarly
we have W1 ≡ −f +g+h
W2 ≡
1
1
(f, −g, h), W3 ≡
(f, g, −h).
f −g+h
f +g−h
The lines Z1 W1 , Z2 W2 , Z3 W3 have equations, respectively,
0.α − hβ + gγ =0,
hα + 0.β − f γ =0,
−gα + f β + 0.γ =0,
and as the determinant of the array of coefficients has value 0, we have the well-known result that
these lines are concurrent.
Moreover
−f
δF (W1 , Z2 , Z3 )
=
,
δF (Z1 , Z2 , Z3 ) −f + g + h
g
δF (W1 , Z3 , Z1 )
=
,
δF (Z1 , Z2 , Z3 ) −f + g + h
δF (W1 , Z1 , Z2 )
h
=
.
δF (Z1 , Z2 , Z3 ) −f + g + h
Then the equation of the conic can be rewritten as
−δF (W1 , Z2 , Z3 )βγ + δF (W1 , Z3 , Z1 )γα + δF (W1 , Z1 , Z2 )αβ = 0.
This can be rewritten as
δF (W1 , Z3 , Z1 )δF (Z, Z2 , Z3 ) δF (W1 , Z1 , Z2 )δF (Z, Z2 , Z3 )
+
= 1.
δF (W1 , Z2 , Z3 )δF (Z, Z3 , Z1 ) δF (W1 , Z2 , Z3 )δF (Z, Z1 , Z2 )
This is projectively invariant.
Exercises
10.1 For non-collinear points Z1 , Z2 , Z3 , let a proper conic meet Z2 Z3 at Z4 and Z5 , meet Z3 Z1
at Z6 and Z7 , and meet Z1 Z2 at Z8 and Z9 . Prove that if Z1 Z4 , Z2 Z6 , Z3 Z8 are concurrent,
then so are Z1 Z5 , Z2 Z7 , Z3 Z9 .
Chapter 11
Similarity methods and results
11.1
Similarity transformations
11.1.1
Definition of similarity transformations
Those affine transformations f for which the magnification ratio k(f, θ) is constant, i.e. it is the
same on all lines in Π, are simpler than the general affine transformations, and are called similarity
transformations.
If f is a similarity transformation, with the constant magnification ratio k(f, θ) = j, and
[Z1 , Z2 , Z3 ] is any triangle, then
|f (Z2 ), f (Z3 )| = j|Z2 , Z3 |, |f (Z3 ), f (Z1 )| = j|Z3 , Z1 |, |f (Z1 ), f (Z2 )| = j|Z1 , Z2 |.
It follows by the cosine rule for a triangle that
|f (Z3 ), f (Z1 )|2 + |f (Z1 ), f (Z2 )|2 − |f (Z2 ), f (Z3 )|2
2|f (Z3 ), f (Z1 )||f (Z1 ), f (Z2 )|
2
|Z3 , Z1 | + |Z1 , Z2 |2 − |Z2 , Z3 |2
=
2|Z3 , Z1 ||Z1 , Z2 |
= cos ∠Z2 Z1 Z3 ,
cos ∠f (Z2 )f (Z1 )f (Z3 ) =
and so |∠f (Z2 )f (Z1 )f (Z3 )|◦ = |∠Z2 Z1 Z3 |◦ . Similarly the wedge-angles ∠f (Z3 )f (Z2 )f (Z1 ) and
∠Z3 Z2 Z1 have equal magnitudes, as have the wedge-angles ∠f (Z1 )f (Z3 )f (Z2 ) and ∠Z1 Z3 Z2 . It
follows that [Z1 , Z2 , Z3 ], [f (Z1 ), f (Z2 ), f (Z3 )] are similar triangles in the correspondence
(Z1 , Z2 , Z3 ) → (f (Z1 ), f (Z2 ), f (Z3 )).
This accounts for the name of similarity transformation.
11.1.2
Similarity invariants
In particular we have shown in §11.1 that each similarity transformation f maps each angle ψ
onto an angle of equal measure. As a special case of this, perpendicularity of two lines is such an
invariant.
As we know that the inverse of an affine transformation is an affine transformation, it follows
immediately from the definition in §11.1, that the inverse of each similarity transformation f is
also a similarity transformation.
If f is a similarity transformation, with its constant magnification ratio k(f, θ) = j, and Z1 6=
Z2 , Z3 6= Z4 , then
|Z1′ , Z2′ | = j|Z1 , Z2 |, |Z3′ , Z4′ | = j|Z3 , Z4 |,
285
286
CHAPTER 11. SIMILARITY METHODS AND RESULTS
and so
|Z1 , Z2 |
|Z1′ , Z2′ |
=
.
|Z3′ , Z4′ |
|Z3 , Z4 |
Thus the ratio of any two distances is an invariant under similarity transformations.
If Z1 , Z2 , Z3 are non-collinear points, let Ci be the circumcentre of [Z1 , Z2 , Z3 ]. If l is the
line which passes through the mid-point Z4 = mp(Z2 , Z3 ) and is perpendicular to Z2 Z3 , then
l′ = f (l) is a line, it contains Z4′ = mathrmmp(Z2′ , Z3′ ), and it is perpendicular to Z2′ Z3′ . Thus
the perpendicular bisector of one side of [Z1 , Z2 , Z3 ] maps into the perpendicular bisector of a
corresponding side of [Z1′ , Z2′ , Z3′ ]. Similarly for the perpendicular bisectors of the other two sides.
It follows that f (Ci ) is the circumcentre of [Z1′ , Z2′ , Z3′ ].
Next let H be the orthocentre of [Z1 , Z2 , Z3 ]. Now let l be the line through the vertex Z1
which is perpendicular to the opposite side-line Z2 Z3 . Then l′ = f (l) is a line, it contains Z1′
and it is perpendicular to Z2′ Z3′ . Thus the perpendicular from a vertex to the opposite side-line
of [Z1 , Z2 , Z3 ] maps into a perpendicular from a corresponding vertex to the opposite side-line of
Z1′ , Z2′ , Z3′ ]. Similarly for the other two such perpendiculars. It follows that f (H) is the orthocentre
of [Z1′ , Z2′ , Z3′ ].
Next let I be the incentre of [Z1 , Z2 , Z3 ]. If now l is the line Z1 I, then f (l) is the line Z1′ I ′ .
But the angles ∠IZ1 Z2 , ∠IZ1 Z3 are equal in size and so the angles ∠I ′ Z1′ Z2′ , ∠I ′ Z1′ Z3′ are equal
in size. It follows that l′ is the bisector of the wedge angle ∠Z2′ Z1′ Z3′ ] maps into the bisector of the
corresponding wedge angle of [Z1′ , Z2′ , Z3′ ]. Similarly for the two other angle-bisectors. It follows
that I ′ is the incentre of [Z1′ , Z2′ , Z3′ ].
Continuing with f a similarity transformation, with its constant magnification ratio k(f, θ) = j,
if |Z0 , Z| = k, then
|Z0′ , Z ′ | = j|Z0 , Z| = jk
and so the image under f of the circle C(Z0 ; k) is the circle C(Z0′ ; jk).
As every similarity transformation is an affine transformation, every affine invariant is also
an invariant under similarity transformations, as any property invariant under every affine transformation must also be invariant under a subset of these transformations. In addition to affine
invariants, the following are some of the invariants for similarity transformations:-
the ratio of any two distances,
the magnitude of an angle,
perpendicularity of two lines,
similarity of two triangles,
being an orthocentre of a triangle,
being a circumcentre of a triangle,
being an incentre of a triangle,
being a circle,
being a tangent to a circle,
being a circle circumscribed to a triangle,
being a circle inscribed in a triangle,
being a square,
being a rectangle,
being a right-angled triangle,
being an isosceles triangle,
being an equilateral triangle,
being the mid-line of an angle-support.
11.2. SPECIALISED SIMILARITY TRANSFORMATIONS
11.2
Specialised similarity transformations
11.2.1
Coefficients in a similarity transformation
287
As k(f, θ) is positive, it is constant when its square is positive. Now from (9.3.1)
k(f, θ)2 = 21 (a2 + c2 + b2 + d2 ) + 12 (a2 + c2 − b2 − d2 ) cos 2θ + (ab + cd) sin 2θ.
On putting θ equal to 0F and 45F we need a2 + c2 − b2 − d2 = 0 for k(f, θ)2 to be constant.
Similarly on putting θ equal to 45F and 135F we need ab + cd = 0 for k(f, θ)2 to be constant.
Thus for an affine transformation to be a similarity transformation we must have
a2 + c2 − b2 − d2 = 0, ab + cd = 0.
We note that we cannot have
b2 + d2 = 0,
as that would imply first of all b = 0, d = 0, and then it would imply
a2 + c2 = 0,
and so in turn a = c = 0. Then all entries in the matrix A would be equal to 0, which would make
det A = 0, which is ruled out for an affine transformation.
On multiplying through in the first equation by b2 we get
b2 a2 + b2 c2 = b2 (b2 + d2 ).
But by the second equation ab = −cd, so by squaring this
a2 b2 = c2 d2 .
On inserting this we get
c2 d2 + b2 c2 = b2 (b2 + d2 ),
and so
c2 (b2 + d2 ) = b2 (b2 + d2 ).
As b2 + d2 6= 0, this implies that
c2 = b 2 .
Thus
b = ±c.
Similarly on multiplying through in the first equation by d2 , we get
d = ±a.
Then we have one of the forms
b = −c,
b = c,
b = c,
b = −c,
d = a,
(11.2.1)
d = −a,
d = a,
(11.2.2)
(11.2.3)
d = −a.
(11.2.4)
But we can rule out (11.2.3) and (11.2.4) unless a = 0 or b = 0, in which cases (11.2.3) and (11.2.4)
can be put in the form (11.2.1) or (11.2.2). Thus the matrix A must have one or other of the forms
a −c
,
(11.2.5)
c a
a c
,
(11.2.6)
c −a
with a2 + c2 6= 0. The magnification ratio is then
1
k(f, θ) = j = (a2 + c2 + b2 + d2 ) = a2 + c2 .
2
288
11.2.2
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Isometries
We recall that transformations for which all distances are preserved are called isometries. By
the material in Barry [2, Chapters 8 and 10] these have the form of affine transformations, and so
are similarity transformations for which k(f, θ) = j = 1. Then the matrix A must have one of the
forms (11.2.5) or (11.2.6), now with the extra condition that a2 + c2 = 1, and so det A = ±1, The
point with coordinates (a, c) is on the unit circle, and so there is an angle α such that
a = cos α,
c = sin α.
When det A = 1, we then have the transformation
′ x
cos α − sin α
x
k1
=
.
.
+
y′
sin α cos α
y
k2
This preserves orientation, and the particular case when k1 = k2 = 0
′ x
cos α − sin α
x
=
.
,
y′
sin α cos α
y
corresponds to a rotation about the origin, anticlockwise through an angle α.
When det A = −1, we then have the transformation
′ x
cos α
sin α
x
k1
=
.
.
+
y′
sin α − cos α
k2
y
This reverses orientation, and the particular case when k1 = k2 = 0
′ x
cos α
sin α
x
=
.
,
y′
sin α − cos α
y
corresponds to axial symmetry in the line through the origin, with angle of inclination 12 α.
11.2.3
Similarity non-invariants which are metric invariants
As an isometry is an affine transformation with magnification-ratio equal to 1, it is a similarity
transformation. It follows that every invariant for similarity transformations is also an isometric
invariant. Isometric invariants which are NOT invariants for similarity transformations are
distance,
area,
congruence of triangles,
congruence of circles.
11.2.4
Dilatations
An affine transformation f is called a dilatation if for each line l, we have that f (l) k l. If l has
the equation l1 x + m1 y + n1 = 0, then by the formula for the inverse transformation we have
x
y
d
b
x′ −
y ′ + j1 ,
det(A)
det(A)
c
d
= −
x′ +
y ′ + j2 ,
det(A)
det(A)
=
for some numbers j1 and j2 . On inserting these into the equation of l, we get
l1 (dx′ − by ′ + j1 ) + m1 (−cx′ + dy ′ + j2 ) + n1 det(A) = 0,
289
11.2. SPECIALISED SIMILARITY TRANSFORMATIONS
and so
(l1 d − m1 c)x′ + (−l1 b + m1 a)y ′ + l1 j1 + m1 j2 + n1 det(A) = 0.
This is an equation for f (l), and so we will have f (l) k l if and only if
l1 (−l1 b + m1 a) − m1 (l1 d − m1 c) = 0,
that is
bl12 + (a − d)l1 m1 + cm21 = 0.
This is to hold for all lines l, and so for all values of l1 and m1 not simultaneously 0. It holds if
and only if
b = 0, a = d, c = 0,
and so
A=
a
0
0
a
=a
1
0
0
1
.
In this we must have a 6= 0 as det(A) 6= 0. Thus a dilation is a transformation of the form
x′
y′
=a
1
0
0
1
x
y
+
k1
k2
,
where a 6= 0. It preserves or reverses orientation according as a > 0 or a < 0.
11.2.5
Enlargements
A dilatation f for which there is a point Z1 such that f (Z1 ) = Z1 , is called an enlargement with
centre Z1 and scale factor a. For such a point Z1 we will have
k1
1 0
x1
x1
,
+
=a
k2
y1
0 1
y1
so by subtraction we have
′ 1
x
x1
=a
−
y1
y′
0
0
1
1 0
0 1
x
y
1
0
0
1
x − x1
y − y1
−a
x1
y1
,
and so
x′ − x1
y ′ − y1
=a
,
where a 6= 0. It preserves or reverses the orientation according as a > 0 or a < 0.
11.2.6
Translations
If a = 1, the dilatation becomes
x′
y′
=
=
1
0
x
y
0
1
+
and this is a translation. It preserves orientation.
x
y
k1
k2
+
,
k1
k2
,
290
CHAPTER 11. SIMILARITY METHODS AND RESULTS
11.3
Eccentricity, focus-directrix properties
11.3.1
Eccentricity, focus, directrix
We now return to the situation in §7.7.1 and ask that we take k2 = 1.
In the case of an ellipse, we then have 1 − k1 = p/q. As we want p/q > 0, this suggests that we
take 0 < k1 < 1. Now if [Z4 , Z5 ] is the diameter conjugate to [Z6 , Z7 ], then |W0 , Z4 |2 = k1 |W0 , Z6 |2
and so of a pair of conjugate diameters we are taking the longer as [Z6 , Z7 ]. Then we have the
unique solution
p
d
, k3 = 1 − k1 ,
p = d 1 − k1 , q = √
1 − k1
and
|Z, W |2 + |F1 , W |2 = (1 − k1 )|G1 , W |2 .
In the case of a hyperbola, we have
p
d
p = d 1 + k1 , q = √
, k3 = 1 + k1 ,
1 + k1
|Z, W |2 + |F1 , W |2 = (1 + k1 )|G1 , W |2 .
In the case of a parabola, we have
1
1
p = − k1 , q = k1 , k3 = 1,
4
4
and
|Z, W |2 + |F1 , W |2 = |G1 , W |2 .
H1
b
W6
b
Z
b
b
b
b
b
b
G1 Z6 F1
W0
b
b
W
Z7
b
W4
Figure 11.1.
Now in the case noted in §7.8.2, where the diameters are perpendicular to each other, by
Pythagoras’ theorem
|Z, W |2 + |F1 , W |2 = |F1 , Z|2 ,
and so in the three cases we have
|F1 , Z| = e|G1 , W |,
√
√
where we have e equal to 1 − k1 , 1 + k1 , 1, so that 0 < e < 1, e > 1, e = 1, respectively, in
the case of an ellipse, hyperbola or parabola.
The point F1 in this is called a focus and the number e is called the eccentricity of the conic.
The line G1 H1 perpendicular to G1 F1 is called the directrix corresponding to F1 ; the distance
|G1 , W | is equal to the perpendicular distance from Z to the directrix.
This is a similarity invariant characterization of conics, and can be traced back to the work of
Pappus c. 300 A.D. This gives a further interpretation of the notation used in §§7.8.4, 7.8.5.
291
11.3. ECCENTRICITY, FOCUS-DIRECTRIX PROPERTIES
11.3.2
Images of conics under similarity transformations
From the focus-directrix property, it is evident that the image of any proper conic under a similarity
transformation is a conic of the same eccentricity. Conversely if two proper conics have the same
eccentricity, one can be mapped onto the other by means of a similarity transformation. For we
can use a translation followed by a rotation to map the axes of the first to the axes of the second,
and then use an enlargement with centre the centre of a central conic or the vertex of a parabola
to map a focus and directrix of the first to a focus and directrix of the second.
11.3.3
Reference to other text-books
Very many detailed similarity invariant properties of conics have been derived from the focusdirectrix properties above and the standard equations in §7.8.2. We refer to traditional books on
coordinate geometry or geometrical conics for these. We have dealt rather with affine properties,
which are rather underdeveloped, and with the more general projective properties, and tried to
provide new results and approaches to discovering them.
T2
b
b
b
b
b
b
P1
b
D1 Z4 C1
R1
b
b
b
C2 Z5 D2
b
T1
Figure 11.2.
However we do prove one celebrated result in the case of an ellipse. Let E be the ellipse with
equation
x2
y2
+
= 1,
a2
b2
with foci the points C1 ≡ (−ae, 0), C2 ≡ (ae, 0), and D1 ≡ (− ae , 0), D2 ≡ ( ae , 0) the points where
the x-axis meets the directrices.
Let P1 ≡ (x1 , y1 ) be a point of the ellipse, R1 ≡ (x1 , 0) the foot of the perpendicular from P1 to
the major axis, and let the tangent l to E at P1 meet the x-axis at the point T1 , so that by §§7.4.3
2
T1 ≡ ( xa1 , 0).
Now when x1 > 0
|P1 , C2 |
|P1 , C1 |
=
=
e|R1 , D2 |
|a/e − x1 |
=
e|R1 , D1 |
|a/e + x1 |
|C2 , T1 |
|a2 /x1 − ae|
=
.
|a2 /x1 + ae|
|C1 , T1 |
From this we deduce that the tangent l makes equal angles with the focal lines, specifically
that |∠T1 P1 C2 |◦ = |∠T2 P1 C1 |◦ , where P1 is between T1 and T2 on line l. For [P1 , T1 must be the
external bisector of the angle ∠C1 P1 T1 as
|C2 , T1 |
|P1 , C2 |
=
.
|P1 , C1 |
|T1 , C1 |
292
CHAPTER 11. SIMILARITY METHODS AND RESULTS
We can also give a straight-forward proof of this using coordinate methods. The lines C1 P1 , C2 P1
have equations
(x1 + ae)y − y1 (x + ae) = 0, (x1 − ae)y − y1 (x − ae) = 0,
and so one of the mid-lines of this pair of lines has the equation
(x1 + ae)y − y1 (x + ae) (x1 − ae)y − y1 (x − ae)
p
p
−
= 0.
(x1 + ae)2 + y12
(x1 − ae)2 + y12
From the focus-directrix property this can be re-written as
(x1 + ae)y − y1 (x + ae) (x1 − ae)y − y1 (x − ae)
−
= 0.
e|x1 + a/e|
e|x1 − a/e|
But
x1 +
so this is
a
a
> 0, − x1 > 0,
e
e
(x1 + ae)y − y1 (x + ae) (x1 − ae)y − y1 (x − ae)
= 0.
−
ex1 + a
a − ex1
This simplifies to the equation of the tangent at P1 .
As a generalization of this, take points E1 ≡ (−c, 0), E2 ≡ (c, 0) on the major axis of this
ellipse. Then the harmonic conjugate of T1 with respect to (E1 , E2 ) is N1 ≡ (c2 x1 /a2 , 0). The
lines P1 T1 , P1 N1 have slopes
m1 = −
b2 x1
y1
, m2 =
,
a2 y 1
x1 − c2 x1 /a2
and so
m1 m2 = −
b2
.
a2 − c2
Thus, when 0 < c < a the lines P1 T1 , P1 N1 are in conjugate directions with respect to an ellipse
with the same axes as the given ellipse, and when c > a they are in conjugate directions with
respect to a hyperbola.
11.3.4
Sum or difference of focal distances
In the case of an ellipse or hyperbola in standard position, we had the equation
a
(x − ae)2 + y 2 = e2 (x + )2 ,
e
where the eccentricity e is less than 1 for an ellipse, and greater than 1 for a hyperbola. This
corresponds to a focus F1 ≡ (−ae, 0) and corresponding directrix with equation x = −a/e. Now
the above equation is easily seen to be equivalent to
a
(x + ae)2 + y 2 = e2 (x − )2 ,
e
and so there is a second focus F2 ≡ (ae, 0) and corresponding directrix the line with equation
x = a/e.
In the case of an ellipse we then have for any point Z on the curve,
a
a
|Z, F1 | + |Z, F2 | = e(x + ) + e( − x) = 2a,
e
e
and so the sum of the focal distances is constant.
11.3. ECCENTRICITY, FOCUS-DIRECTRIX PROPERTIES
293
Let us now consider a converse of this, that
p
p
(x − ae)2 + y 2 + (x + ae)2 + y 2 = 2a.
If we rewrite this as
p
p
(x − ae)2 + y 2 − a = a − (x + ae)2 + y 2 ,
and square each side, we obtain that
p
p
(x + ae)2 + y 2 − (x − ae)2 + y 2 = 2ex.
On combining this with the first line, we obtain that
p
(x − ae)2 + y 2 = a + ex,
which shows that the point is on the ellipse with focus F1 and directrix the line with equation
x = −a/e.
A similar result can be obtained in the case of a hyperbola, when e > 1, but now it is
|Z, F1 | − |Z, F2 | = 2a,
on one branch and
|Z, F2 | − |Z, F1 | = 2a,
on the other branch.
11.3.5
Another way of generating central conics
As the results of the last subsection are awkward to generalize directly, we add or subtract equal
collinear distances instead, as follows.
Z
b
b
b
b
b
F1
F2
Z2
W
b
b
W1
b
F1
b
b
2
Fig.W11.3
b
b
b
F2
Z1
294
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Let C1 be the fixed circle with centre the point F1 and radius length 2a, and let F2 be a second
fixed point, not on this circle. For a variable point W on this circle, let the perpendicular bisector
of the segment [F2 , W ] meet the line F1 W at the point Z. Then the point Z traces out a central
conic with foci F1 and F2 , an ellipse when F2 is inside the circle C1 , and a hyperbola when F2 is
outside this circle. In fact, the circle with centre Z and which passes through F1 , and the circle C1
will both pass through W and have the same tangent at that point.
11.3.6
An other characterization of a focus
In §7.7.2, with G1 = Z1 , F1 = Z2 , H1 = Z3 , we had the equation
j1 α2 − j2 β 2 + (1 + j1 )γ 2 + 2j1 γα = 0,
for the conic. From this the polar of the point W0 ≡ (α0 , β0 , γ0 ) has equation
j1 α0 α − j2 β0 β + (1 + j1 )γ0 γ + j1 (γ0 α + α0 γ) = 0.
In particular the polar of the focus F1 is the line H1 G1 , that is the corresponding directrix.
Now let C be a proper conic, W0 a point not on it and l the polar of W0 . Let W4 , W5 be points
on C, and let the tangents to C at Z4 and Z5 meet at Z6 . Let Z4 Z5 meet l at W1 and Z4 Z5 meet
W0 Z6 at W2 .
Now W1 is on the polar of Z6 , so by §5.2.7 Z6 is on the polar of W1 . Also W1 is on the polar l
of W0 , so W0 is on the polar of W1 . Thus the polar of W1 is W0 Z6 .
Then the line W1 W2 meets the conic in Z4 and Z5 , while W2 is on the polar of W1 . It follows
from the definition of polar, that then (W1 , W2 , Z4 , Z5 ) is a harmonic range.
We now specialize to the case when W0 is a focus of C; then as above, l is the corresponding
directrix. Let W4 , W5 be the feet of the perpendiculars from Z4 and Z5 , respectively, to l. Then
|Z5 , W0 |
|Z4 , W0 |
=e=
,
|Z4 , W4 |
|Z5 , W5 |
and so
|W0 , Z4 |
|Z4 , W4 |
=
.
|W0 , Z5 |
|Z5 , W5 |
As Z4 W4 k Z5 W5 , the latter is equal to
Hence
|W1 , Z4 |
.
|W1 , Z5 |
|Z4 , W1 |
|W0 , Z4 |
=
.
|W0 , Z5 |
|W1 , Z5 |
It follows that W0 W1 is the internal or external mid-line of |Z4 W0 Z5 . Since (Z4 , Z5 , W1 , W2 ) is a
harmonic range it follows that W0 W2 is the other mid-line of this angle support. It follows that
W0 W1 ⊥ W0 W2 .
This last is sometimes taken as a characterizing property of a focus.
11.4
Proto-foci
11.4.1
Proto-foci in the case of an ellipse
The pair of perpendicular conjugate diameters of a central conic noted and used in §7.8.1 are
called the axes of the conic. The standard equations for an ellipse or hyperbola in §7.8.2, and
the focus-directrix properties in §11.3, concentrate our attention very much on the axes, but the
ancient Greek treatment generally featured an analysis relative to any pair of conjugate diametral
lines. For this reason we call the point F1 in §7.7.1 a proto-focus and look at it in more detail in the
295
11.4. PROTO-FOCI
case of an ellipse. The line through G1 , parallel to Z6 Z7 , we call a corresponding proto-directrix ;
the distance to it from a variable point Z on the conic is measured is measured parallel to Z4 Z5 .
We now look at this in more detail in the case of an ellipse with equation
x2
y2
+
= 1,
a2
b2
where 0 < b < a. We start with the diametral line with equation y = mx; this meets the ellipse at
the points
ab
ab
,
, −m √
Z4 ≡ − √
b 2 + a 2 m2
b 2 + a 2 m2
ab
ab
Z5 ≡ √
.
, m√
b 2 + a 2 m2
b 2 + a 2 m2
Now the conjugate diametral line has equation
y=−
b2
x,
a2 m
and this meets the ellipse at the point
a2 m
b2
Z6 ≡ − √
.
,√
b 2 + a 2 m2
b 2 + a 2 m2
If through a point Z on the ellipse we draw a line parallel to the conjugate diametral line OZ6
it will meet the diameter [Z4 , Z5 ] at a point W , and from §7.6.1
|Z, W |2 = k1 Z4 W W Z5 .
We can find k1 by taking Z = Z6 , W = O so that
|O, Z6 |2 = k1 |O, Z4 |2 ,
and thus
k1 =
a 4 m2 + b 4
.
+ m2 )
a2 b2 (1
Let W ≡ (t, mt), F1 ≡ (−p, −mp), G1 ≡ (−q, −mq), so as to have
|Z, W |2 + |W, F1 |2 = k3 |W, G1 |2 ,
identically for some k3 , and thus
k1 Z4 W W Z5 + |W, F1 |2 = k3 |W, G1 |2 .
We then require
k1 (1 + m2 )
a2 b 2
− t2
2
b + a 2 m2
+ (1 + m2 )(t + p)2 = k3 (1 + m2 )(t + q)2 ,
as an identity and need for this
k3 − 1 = k1 , k3 − p = 0, k3 q 2 − p2 = k1
a2 b 2
.
b 2 + a 2 m2
For positive p and q we obtain the unique solutions
k3
=
q
=
(a2 − b2 )(b2 − a2 m2 )
, p=
a2 b2 (1 + m2 )
√
a 2 b 2 1 + m2
p
,
(a2 − b2 )(b4 − a4 m4 )
s
(a2 − b2 )(b2 − a2 m2 )
,
(1 + m2 )(b2 + a2 m2 )
296
CHAPTER 11. SIMILARITY METHODS AND RESULTS
and of course in this we need to keep a2 m2 < b2 .
Clearly
a2 b 2
,
b 2 + a 2 m2
pq =
so
G1 O F1 O = |O, Z4 |2 ,
and thus (Z4 , Z5 , F1 , G1 ) is a harmonic range.
b
G1
b
Z4
b
F1
b
b
O
b
Z5
Figure 11.4.
If we denote the coordinates by (x, y), then on eliminating m we find that the locus of F1 and
the corresponding proto-focus F2 ≡ (p, mp) has the equation
(x2 + y 2 )(b2 x2 + a2 y 2 ) − (a2 − b2 )(b2 x2 − a2 y 2 ) = 0.
This is a generalization of Bernoulli’s lemniscate. Similarly the locus of G1 and a corresponding
G2 ≡ (q, mq) has equation
(a2 − b2 )(b4 x4 − a4 y 4 ) − a4 b4 (x2 + y 2 ) = 0.
This seems to be a new quartic curve, quite like a hyperbola in shape.
11.5
Similar and similarly situated conics
If we deal with one conic at a time, we can take its equation in a specific simple form, as hitherto.
However if we wish to consider two conics simultaneously we must face more general equations.
One concept, involving two conics simultaneously, that we need is the following. Two conics
C1 and C2 are said to be similar and similarly situated (or homothetic from the Greek homo
thetikos = similarly placed ) if there is a dilatation f such that f (C1 ) = C2 .
11.5.1
Condition on coefficients
We start with a necessary and sufficient condition, involving the coefficients, for two conics to be
homothetic. We consider two proper conics, C1 with equation in Cartesian coordinates φ1 (x, y) = 0,
where
φ1 (x, y) = a1 x2 + 2h1 xy + b1 y 2 + 2g1 x + 2f1 y + c1 ,
and C2 with equation φ2 (x′ , y ′ ) = 0, where
φ2 (x′ , y ′ ) = a2 x′2 + 2h2 x′ y ′ + b2 y ′2 + 2g2 x′ + 2f2 y ′ + c2 ,
297
11.5. SIMILAR AND SIMILARLY SITUATED CONICS
and that g is a dilatation
x′
=
kx + k1 ,
y′
=
ky + k2 .
Then for Z ′ = g(Z),
φ2 (Z ′ ) = a2 (kx + k1 )2 + 2h2 (kx + k1 )(ky + k2 ) + b2 (ky + k2 )2 + 2g2 (kx + k1 ) + 2f2 (ky + k2 ) + c2
= k 2 [a2 x2 + 2h2 xy + b2 y 2 ] + 2k[(a2 k1 + h2 k2 + g2 )x + (h2 k1 + b2 k2 + f2 )y] + φ2 (k1 , k2 ).
Suppose now that g(C1 ) = C2 so that φ2 (g(Z)) = 0 for all Z such that φ1 (Z) = 0. Then for
each fixed y, the equations
a1 x2 + 2(h1 y + g1 )x + b1 y 2 + 2f1 y + c1 = 0,
k 2 a2 x2 + 2[k 2 h2 y + k(a2 k1 + h2 k2 + g2 )]x + k 2 b2 y 2 + 2k(h2 k1 + b2 k1 + f2 )y + φ2 (k1 , k2 ) = 0,
have the same roots when real. As C1 is a proper conic, there will be an interval of such y. For
these we have (from the product of the roots)
b1 y 2 + 2f1 y + c1
k 2 b2 y 2 + 2k(h2 k1 + b2 k1 + f2 )y + φ2 (k1 , k2 )
=
,
a1
k 2 a2
and (from the sum of roots)
k 2 h2 y + k(a2 k1 + h2 k2 + g2 )
h1 y + g 1
=
.
a1
k 2 a2
These are identities in y, so on equating coefficients of like powers, we must have
b2 f1
h2 k1 + b2 k2 + f2 c1
φ2 (k1 , k2 )
b1
= ,
=
,
=
,
a1
a2 a1
ka2
a1
k 2 a2
h1
h2 g 1
a2 k1 + h2 k2 + g2
=
,
=
.
a1
a2 a1
ka2
This implies that
h2
b2
a2 k1 + h2 k2 + g2
h2 k1 + b2 k2 + f2
φ2 (k1 , k2 )
a2
=
=
=
=
=
=j
a1
h1
b1
kg1
kf1
k 2 c1
say.
This gives, first,
a2 = ja1 , h2 = jh1 , b2 = jb1 ,
(11.5.1)
then
a1 k1 + h1 k2
=
h1 k1 + b1 k2
=
g2
,
j
f2
kf1 − ,
j
kg1 −
(11.5.2)
and finally
jk 2 c1 = φ2 (k1 , k2 ).
(11.5.3)
When the conic C1 is central, we have a1 b1 − h21 6= 0 and so, for any fixed j, can solve (11.5.2) for
k1 and k2 . Then we solve (11.5.3) for k. Thus the coefficients are not essentially restricted beyond
(11.5.1).
We note that
a2 (kx0 + k1 ) + h2 (ky0 + k2 ) + g2
=
jk(a1 x0 + h1 y0 + g1 ),
h2 (kx0 + k1 ) + b2 (ky0 + k2 ) + f2
=
jk(h1 x0 + b1 y0 + f1 ).
298
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Now Z0 will be a centre of C1 if the equations
a1 x0 + h1 y0 + g1
=
0,
h1 x0 + b1 y0 + f1
=
0,
are satisfied. In that case, f (Z0 ) is also the centre of C2 .
If we replace Cartesian coordinates (x, y) by homogeneous coordinates (ξ, η, ζ) the equations of
C1 and C2 become
a1 ξ 2 + 2h1 ξη + b1 η 2 + 2g1 ξζ + 2f1 ηζ + c1 ζ 2 = 0,
and
ja1 ξ 2 + 2jh1 ξη + jb1 η 2 + 2g2 ξζ + 2f2 ηζ + c2 ζ 2 = 0.
11.5.2
Images of two circles under one affine transformation
We next look at the special case of two ellipses being homothetic.
Let C1 and C2 be the circles with equations
(x′ − x′1 )2 + (y ′ − y1′ )2 = r12 , (x′ − x′2 )2 + (y ′ − y2′ )2 = r22 ,
respectively. Then under the inverse of the affine transformation
x′
= ax + by + k1 ,
′
= cx + dy + k2 ,
y
they map to the ellipses or circles E1 , E2 , respectively, with equations
(a2 + c2 )(x − x1 )2 + 2(ab + cd)(x − x1 )(y − y1 ) + (b2 + d2 )(y − y1 )2 = r12 ,
and
(a2 + c2 )(u − x2 )2 + 2(ab + cd)(u − x2 )(v − y2 ) + (b2 + d2 )(v − y2 )2 = r22 .
Now the transformation g : (x, y) → (u, v) for which
u − x2 =
r2
r2
(x − x1 ), v − y2 = (y − y1 ),
r1
r1
is a dilatation, and under it for a point Z ≡ (x, y) ∈ E1 we have
2
2
r1
r1
r1
r1
2
2
(u − x2 ) + 2(ab + cd) (u − x2 ) (v − y2 ) + (b + d )
(v − y2 ) = r12 ,
(a + c )
r2
r2
r2
r2
2
2
which gives
(a2 + c2 )(u − x2 )2 + 2(ab + cd)(u − x2 )(v − y2 ) + (b2 + d2 )(v − y2 )2 = r22 .
Thus (u, v) ∈ E2 and so g(E1 ) ⊂ E2 . Similarly g −1 (E2 ) ⊂ E1 and so g(E1 ) = E2
Thus the images of any two circles under any one affine transformation, when ellipses, are
similar and similarly situated.
11.6
Analogue of power property of a circle; radical axis
11.6.1
We now consider a proper conic C with equation in Cartesian coordinates φ(x, y) = 0, where
φ(x, y) = ax2 + 2hxy + by 2 + 2gx + 2f y + c,
11.6. ANALOGUE OF POWER PROPERTY OF A CIRCLE; RADICAL AXIS
299
and consider where the line Z0 Z1 meets the conic. The line has parametric equations
x = x0 + t(x1 − x0 ), y = y0 + t(y1 − y0 ),
and the parameters t of points of intersection are roots of the quadratic equation
a[x0 + t(x1 − x0 )]2 + 2h[x0 + t(x1 − x0 )][y0 + t(y1 − y0 )] + b[y0 + t(y1 − y0 )]2
+ 2g[x0 + t(x1 − x0 )] + 2f [y0 + t(y0 + t(y1 − y0 )] + c = 0.
On denoting by Z2 and Z3 these points of intersection, we note that their parameters t2 and t3
satisfy
ax20 + 2hx0 y0 + by02 + 2gx0 + 2f y0 + c
.
t2 t3 =
a(x1 − x0 )2 + 2h(x1 − x0 )(y1 − y0 ) + b(y1 − y0 )2
If we write
ψ(Z0 , Z) = a(x − x0 )2 + 2h(x − x0 )(y − y0 ) + b(y − y0 )2 ,
this can be written as
t2 t3 =
φ(Z0 )
.
ψ(Z0 , Z1 )
Now
x2 = x0 + t2 (x1 − x0 ), y2 = y0 + t2 (y1 − y0 ),
and so
x2 + ıy2 − (x0 + ıy0 ) = t2 [x1 + ıy1 − (x0 + ıy0 )], z2 − z0 = t2 (z1 − z0 ).
Thus the complex-valued distances satisfy
Z 0 Z 2 = t2 Z 0 Z 1 .
Similarly
Z 0 Z 3 = t3 Z 0 Z 1 ,
and so
2
Z 0 Z 2 Z 0 Z 3 = t2 t3 Z 0 Z 1 .
Z3
b
b
Z2
b
Z1
b
Z0
Figure 11.5.
We keep Z0 fixed but let Z1 vary so that ψ(Z0 , Z1 ) = k, for some k 6= 0. Then Z1 lies on the
conic Ck which has equation ψ(Z0 , Z) − k = 0. As φ(Z) and ψ(Z0 , Z) − k have the same quadratic
terms, C and Ck are similar and similarly situated. When C is central, Z0 is the centre of Ck .
Then with Z0 fixed, and Z1 varying on the conic Ck , if the variable line Z0 Z1 meets the conic
C in points Z2 and Z3 , then
Z0 Z2 Z0 Z3
Z0 Z1
2
300
CHAPTER 11. SIMILARITY METHODS AND RESULTS
is constant.
This is an affine invariant result which generalizes that for a circle. If C ′ is a conic similar and
similarly situated to C and these conics intersect in two points U2 , U3 , then for any fixed point Z0
on the line U2 U3 , the ratio
Z0 Z2 Z0 Z3
Z0 Z1
2
for intersections with Z0 Z1 , is the same for the two conics. This extends to similar and similarly
situated conics, the notion of the power of a point with respect to a circle, and the notion of
radical axis extends to the role of the line U2 U3 .
11.7
Miscellaneous results
11.8
Two-conic analogue of Pascal’s theorem
We take a line R1 R2 and points Z1 , Z2 , W1 and W2 not on it. For points Z and W we suppose
that the lines ZZ1 and W W1 meet on R1 R2 , and that the lines ZZ2 and W W2 also meet on R1 R2 .
Then we have that
δF [(R1 , R2 ), (Z1 , Z), (W1 , W )] = 0, δF [(R1 , R2 ), (Z2 , Z), (W2 , W )] = 0.
This expands as
δF (R1 , Z1 , Z)δF (R2 , W1 , W ) − δF (R2 , Z1 , Z)δF (R1 , W1 , W ) = 0,
δF (R1 , Z2 , Z)δF (R2 , W2 , W ) − δF (R2 , Z2 , Z)δF (R1 , W2 , W ) = 0.
It follows that
δF (R1 , Z1 , Z) δF (R1 , W1 , W )
=
,
δF (R2 , Z1 , Z) δF (R2 , W1 , W )
δF (R1 , Z2 , Z) δF (R1 , W2 , W )
=
.
δF (R2 , Z2 , Z) δF (R2 , W2 , W )
(11.8.1)
λ
1
R1 + 1+λ
R2
We can parametrise the pencil of lines through Z1 by taking the variable point 1+λ
on R1 R2 and taking the line through this point and Z1 . For Z to be on this line we need
λ
1
R1 +
R2 , Z1 , Z)
1+λ
1+λ
1
λ
=
δF (R1 , Z1 , Z) +
δF (R2 , Z1 , Z),
1+λ
1+λ
0 =δF (
so that
λ=−
δF (R1 , Z1 , Z)
.
δF (R2 , Z1 , Z)
1
R1 +
Similarly we can parametrise the pencil of lines through Z2 by taking the variable point 1+µ
µ
R
on
R
R
and
taking
the
line
through
this
point
and
Z
.
For
Z
to
be
on
this
line
we
need
1 2
2
1+µ 2
µ=−
δF (R1 , Z2 , Z)
.
δF (R2 , Z2 , Z)
If we now consider the locus of Z specified by
aλµ + bλ + cµ + d = 0,
301
11.8. TWO-CONIC ANALOGUE OF PASCAL’S THEOREM
where the coefficients satisfy ad − bc 6= 0, then by Steiner’s definition of conics, we have proper
conics C1 passing through Z1 and Z2 . By (11.8.1) there are corresponding conics C2 through W1
and W2 , related in this way.
For C1 , we have the equation
a
δF (R1 , Z1 , Z)
δF (R1 , Z2 , Z)
δF (R1 , Z1 , Z) δF (R1 , Z2 , Z)
−b
−c
+ d = 0.
δF (R2 , Z1 , Z) δF (R2 , Z2 , Z)
δF (R2 , Z1 , Z)
δF (R2 , Z2 , Z)
(11.8.2)
On clearing of fractions (11.8.2) becomes
aδF (R1 , Z1 , Z)δF (R1 , Z2 , Z) − bδF (R1 , Z1 , Z)δF (R2 , Z2 , Z)
−cδF (R1 , Z2 , Z)δF (R2 , Z1 , Z) + dδF (R2 , Z1 , Z)δF (R2 , Z2 , Z) =
0.
(11.8.3)
If we introduce Cartesian coordinates
Z ≡ (x, y), Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ), R1 ≡ (s1 , t1 ), R2 ≡ (s2 , t2 ),
the coefficients of x2 , y 2 , xy in (11.8.3), respectively, turn out to be
1
4 [a(t1
− y1 )(t1 − y2 ) − b(t1 − y1 )(t2 − y2 )
− c(t1 − y2 )(t2 − y1 ) + d(t2 − y1 )(t2 − y2 )],
1
4 [a(x1
− s1 )(x2 − s1 ) − b(x1 − s1 )(x1 − s2 )
− c(x2 − s1 )(x1 − s2 ) + d(x1 − s2 )(x2 − s2 )],
1
4 {a[(t1
− y1 )(x2 − s1 ) + (x1 − s1 )(t1 − y2 )]
− b[(t1 − y1 )(x2 − s2 ) + (x1 − s2 )(t2 − y2 )]
− c[(t1 − y2 )(x1 − s2 ) + (x2 − s1 )(t2 − y1 )]
+ d[(t2 − y1 )(x2 − s2 ) + (x1 − s2 )(t2 − y2 )]}.
We look to when C1 intersects the line R1 R2 . If Z = (1 − s)R1 + sR2 we have that
δF (R1 , Z1 , Z) =δF (R1 , Z1 , (1 − s)R1 + sR2 ) = sδF (R1 , Z1 , R2 ),
δF (R1 , Z2 , Z) =δF (R1 , Z2 , (1 − s)R1 + sR2 ) = sδF (R1 , Z2 , R2 ),
δF (R2 , Z1 , Z) =δF (R2 , Z1 , (1 − s)R1 + sR2 ) = (1 − s)δF (R2 , Z1 , R1 ),
δF (R2 , Z2 , Z) =δF (R2 , Z2 , (1 − s)R1 + sR2 ) = (1 − s)δF (R2 , Z2 , R1 ).
Now the left-hand side of (11.8.3) becomes
[as2 + (b + c)s(1 − s) + d(1 − s)2 ]δF (R2 , Z1 , R1 )δF (R2 , Z2 , R1 )
=[(a − b − c + d)s2 − (2d − b − c)s + d]δF (R2 , Z1 , R1 )δF (R2 , Z2 , R1 ).
As neither of Z1 and Z2 is on R1 R2 , points of C1 are on R1 R2 according as
(a − b − c + d)s2 − (2d − b − c)s + d = 0,
has real roots. This quadratic equation has discriminant (b + c)2 − 4ad. As the corresponding
condition for C2 to have a point on the line R1 R2 is obtained by replacing Z1 , Z2 above by W1 , W2 ,
the same quadratic equation in s will be involved. Thus any point common to R1 R2 and C1 will
also be common to R1 R2 and C2 .
In (11.8.1) we write temporarily
δF (R1 , W1 , W ) = k1 δF (R2 , W1 , W ), δF (R1 , W2 , W ) = k2 δF (R2 , W2 , W ),
where
k1 =
δF (R1 , Z1 , Z)
δF (R1 , Z2 , Z)
, k2 =
.
δF (R2 , Z1 , Z)
δF (R2 , Z2 , Z)
302
CHAPTER 11. SIMILARITY METHODS AND RESULTS
On using the Cartesian coordinates Z = (x, y), W = (u, v), R1 = (s1 , t1 ) etc., we have the
equations
[t1 − v1 − k1 (t2 − v1 )]u + [u1 − s1 − k1 (u1 − s2 )v =t1 u1 − s1 v1 + k1 (s2 v1 − t2 u1 ),
[t1 − v2 − k2 (t2 − v2 )]u + [u2 − s1 − k2 (u2 − s2 )v =t1 u2 − s1 v2 + k2 (s2 v2 − t2 u2 ).
By Cramer’s rule this will have solutions of the form u = fh , v = hg , where f, g, h are polynomials.
When we insert the values of k1 and k2 in these and simplify, we will obtain solutions of the form
u=
P (x, y)
,
R(x, y)
v=
Q(x, y)
,
R(x, y)
where each of P (x, y), Q(x, y), R(x, y) is a quadratic polynomial in x and y. This can be found
easily by using MAPLE but the print out of the formulae occupies several pages.
We recall that in Cartesian coordinates, a projective transformation is one of the form
u=
where
a1 x + b 1 y + c1
a2 x + b 2 y + c2
, v=
,
a3 x + b 3 y + c3
a3 x + b 3 y + c3

a1
det  a2
a3
b1
b2
b3

c1
c2  =
6 0.
c3
Thus the transformation (11.8.1) which maps C1 onto C2 is not a projective transformation.
The first identity in (11.8.1) can be rewritten in the form
δF (R1 , Z1 , Z)δF (R2 , W1 , W )
= 1,
δF (R2 , Z1 , Z)δF (R1 , W1 , W )
and as it is an equipoised quotient, it is projectively invariant. The same is true of the second
identity.
If we divide the second identity in (11.8.1) by the first, we obtain
δF (R1 , W2 , W )δF (R2 , W1 , W )
δF (R1 , Z2 , Z)δF (R2 , Z1 , Z)
=
.
δF (R1 , Z1 , Z)δF (R2 , Z2 , Z)
δF (R1 , W1 , W )δF (R2 , W2 , W )
On each side here we have an equipoised quotient. We conclude that
crZ(R1 , R2 , Z2 , Z1 ) = crW (R1 , R2 , W2 , W1 ),
(11.8.4)
which is a two-conic analogue of the Chasles-Steiner property of a conic.
By the Chasles-Steiner theorem, for the points Z ∈ C1 such that the lines ZR1 and ZR2 meet
C1 in second points which are distinct from Z1 and Z2 and from each other, the left-hand side
of (11.8.4) is constant, from which the right-hand side is also constant. In particular we need
the tangent to C1 at Z not to be parallel to R1 R2 . Otherwise, in the case of a parabola one of
ZR1 , ZR2 might not meet C1 again; in the case of a hyperbola one or both of ZR1 , ZR2 might
not meet C1 again.
As an example we take Z1 = (−5, 0), Z2 = (0, −4), Z3 = (5, 0), Z4 = (0, 4), Z5 = (−3, 16
5 , R1 =
),
W
=
(0,
6),
W
=
(6,
0).
Then
we
find
that
C
has
equation
(0, 0), R2 = (−5, 16
1
2
1
3
x2
y2
+
= 1,
25 16
and C2 has equation
−416u2 + 78uv − 405v 2 + 1056u + 990v + 8640 = 0.
303
11.8. TWO-CONIC ANALOGUE OF PASCAL’S THEOREM
8
y 4
0
-6
-4
-2
0
2
4
6
x
-4
-8
Figure 11.6.
In Barry [2, §10.11] we dealt with a case of Pascal’s theorem involving parallel lines and a circle,
and extended it to two linked circles. We have now considered this type of material more generally.
Instead of the awkward notation there we have used more straightforward notation now.
11.8.1
Locus with respect to four lines
Given an ordered pair of distinct points (Z1 , Z3 ), let l1 be a line which meets Z1 Z3 at a point W1
and take a point W2 ∈ l1 so that (Z1 , Z3 , W2 ) is positively oriented and so δF (Z1 , Z3 , W2 ) > 0.
Let α = ∡F Z1 W1 W2 . For a point Z 6∈ Z1 Z3 , let the line l5 through Z, parallel to l1 , meet Z1 Z3
at Z5 . Then tW1 ,Z5 (W2 ) ∈ l5 . We take the natural order ≤l5 for which Z5 ≤l5 tW1 ,Z5 (W2 ), and
base on this the sensed-distance Z5 Z.
Now consider 21 |Z1 , Z3 |Z5 Z, and note that |Z5 , Z|| sin α| is the perpendicular distance from Z
to Z1 Z3 . Thus δF (Z1 , Z3 , Z) and − 12 |Z1 , Z3 |Z5 Z sin α have the same magnitudes, and both change
sign as Z crosses Z1 Z3 . They are equal for Z = W2 and so throughout the plane. Hence
1
δF (Z1 , Z3 , Z) = − |Z1 , Z3 |Z5 Z sin α.
2
304
CHAPTER 11. SIMILARITY METHODS AND RESULTS
b
l2
Z6
b
b
b
b
b
Z1
b
b
Z4
Z
b
l3
Z7
b
b
b
l4
Z5
b
b
b
Z2
Z8
b
Z3
b
l1
W1
W2
Figure 11.7
b
α
Now let Z1 , Z2 , Z3 , Z4 be four points, no three of which are collinear. Then any proper conic
through these points has an equation
δF (Z, Z1 , Z3 )δF (Z, Z2 , Z4 )
= k.
δF (Z, Z1 , Z4 )δF (Z, Z2 , Z3 )
Let a line l1 meet Z1 Z3 at W1 and take W2 ∈ l1 so that (Z1 , Z3 , W2 ) is positively oriented. Let a
line l2 meet Z2 Z4 at W3 and take W4 ∈ l2 so that (Z2 , Z4 , W4 ) is positively oriented. Let a line l3
meet Z1 Z4 at W5 and take W6 ∈ l3 so that (Z1 , Z4 , W6 ) is positively oriented. Let a line l4 meet
Z2 Z3 at W7 and take W8 ∈ l4 so that (Z2 , Z3 , W8 ) is positively oriented. Let
α = ∡F Z1 W1 W2 , β = ∡F Z2 W3 W5 , γ = ∡F Z1 W5 W6 , δ = ∡F Z2 W7 W8 .
For a variable point Z, let the lines through Z, parallel to l1 , l2 , l3 , l4 , respectively, meet Z1 Z3 ,
Z2 Z4 , Z1 Z4 , Z2 Z3 at Z4 , Z5 , Z6 , Z7 , respectively. Then
|Z1 , Z3 ||Z2 , Z4 | sin α sin β Z5 Z Z6 Z
= k,
|Z1 , Z4 ||Z2 , Z3 | sin γ sin δ Z7 Z Z8 Z
and so
Z5 Z Z6 Z
|Z1 , Z4 ||Z2 , Z3 | sin γ sin δ
=k
.
|Z1 , Z3 ||Z2 , Z4 | sin α sin β
Z7 Z Z8 Z
The right-hand side here is constant as Z varies on the conic.
This is the noted locus with respect to four lines which so taxed the efforts of the ancient
Greeks, and which has been called Pappus’ theorem in some fairly modern books. It is the problem
on which Descartes was working when he was led to introduce coordinate geometry, his work being
published in 1637. The property is similarity invariant.
11.8. TWO-CONIC ANALOGUE OF PASCAL’S THEOREM
11.8.2
305
Point-pairs in lines tangent to a conic
In §7.8.4 starting with an equation
(1 − e2 )x2 + y 2 − a2 (1 − e2 ) = 0
for a central conic C, by taking normalized areal coordinates with respect to a suitably selected
triple of reference (Z1 , Z2 , Z3 ), we found that
(1 − e2 )x2 + y 2 − a2 (1 − e2 ) =
a2 (1 − e2 )2 2
[γ − 4e2 αβ].
e2
If we write e2 = k and turn to §8.1.4, we considered there a central conic C1 with equation γ ′2 −
4kα′ β ′ = 0 the tangent to which at the point Z ′ (α′ , β ′ , γ ′ ) had equation γ ′ γ ′′ = 2k(α′ β ′′ + β ′ α′′ ).
On introducing a parameter s such that γ ′′2 − 4kα′′ β ′′ = 4k 2 s2 , we obtained the parametric
equations (8.1.9)
α′ β ′′ − β ′ α′′ =γ ′ s,
β ′ γ ′′ − γ ′ β ′′ = − 2kβ ′ s,
γ ′ α′′ − α′ γ ′′ = − 2kα′ s,
which in (8.1.10) yielded parametric equations of point pairs (Z ′ , Z ′′ ) in the tangent to C1 . On
combining these we have that
a2 (1 − e2 )2 ′′2
[γ − 4kα′′ β ′′ ]
e2
a2 (1 − e2 )2 2 2
4k s ,
=
e2
(1 − e2 )x′′2 + y ′′2 − a2 (1 − e2 ) =
so that
x′′2
y ′′2
+
= 1.
a2 + 4k 2 a2 (1 − e2 )s2 /e2
a2 + 4k 2 a2 (1 − e2 )2 s2 /e2
Thus for each fixed s, the point Z ′′ lies on a conic C2 which is homothetic to C1 .
Now in seeking to generalize from §6.6.3 we take distinct points W4 , W5 , W6 , W7 , W ∈ C1 and
corresponding points Z4′′ , Z5′′ , Z6′′ , Z7′′ , Z ′′ ∈ C2 where the tangents to C1 at these points, respectively,
meet C2 . We wish to calculate first of all δF [(W4 , Z4′′ ), (W6 , Z6′′ ), (W, Z ′′ )]. For this note that with
δ(Z1 , Z2 , Z3 ) = δ1 , we have that
′
α4 β4′ γ4′ δ(W4 , W, Z ′′ ) =δ1 α′ β ′ γ ′ α′′ β ′′ γ ′′ =δ1 [α′4 (γ ′′ β ′ − β ′′ γ ′ ) + β4′ (α′′ γ ′ − γ ′′ α′ ) + γ4′ (β ′′ α′ − α′′ β ′ )]
=δ1 [α′4 (−2kβ ′ s) + β4′ (−2kα′ s) + γ4′ γ ′ s)]
=δ1 s[−2kα′4 β ′ − 2kβ4′ α′ + γ4′ γ ′ ],
and in particular
δ(W4 , W6 , Z6′′ ) = δ1 s[−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ ].
By a similar argument
δF (Z4′′ , W, Z ′′ ) = δ1 s[−2kα′′4 β ′ − 2kβ4′′ α′ + γ4′′ γ ′ ],
and in particular
δF (Z4′′ , W6 , Z6′′ ) = δ1 s[−2kα′′4 β6′ − 2kβ4′′ α′6 + γ4′′ γ6′ ].
306
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Then
1
[δF (W4 , W, Z ′′ )δF (Z4 , W6 , Z6′′ ) − δF (Z4′′ , W, Z ′′ )δF (W4 , W6 , Z6′′ )]
2
δ1 s2
=[−2kα′4 β ′ − 2kβ4′ α′ + γ4′ γ ′ ][−2kα′′4 β6′ − 2kβ4′′ α′6 + γ4′′ γ6′ ]
−[−2kα′′4 β ′ − 2kβ4′′ α′ + γ4′′ γ ′ ][−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ ]
=α′ {−2kβ4′ (−2kα′′4 β6′ − 2kβ4′′ α′6 + γ4′′ γ6′ ) + 2kβ4′′ (−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ }
+β ′ {−2kα′4 (−2kα′′4 β6′ − 2kβ4′′ α′6 + γ4′′ γ6′ ) + 2kα′′4 (−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ )}
+γ ′ {γ4′ (−2kα′′4 β6′ − 2kβ4′′ α′′6 + γ4′′ γ6′ ) − γ4′′ (−2kβ4′ α′6 + γ4′ γ6′ − 2kα′4 β6′ )} .
The coefficient of α′ here is equal to
2kα′6 [2kβ4′ β4′′ − 2kβ4′′ β4′ ] + 2kβ6′ [2kβ4′ α′′4 − 2kα′4 β4′′ ] + 2kγ6′ [−β4′ γ4′′ + β4′′ γ4′ ]
=4k 2 (−γ4′ s) + 2kγ6′ (2kβ4′ s)
=4k 2 s(β4′ γ6′ − γ4′ β6′ )).
By similar arguments the coefficients of β ′ and γ ′ are, respectively, 4k 2 s(γ4′ α′6 − α′4 γ6′ ) and
4k s(α′4 β6′ − β4′ α′6 ). Hence
2
δF [(W4 , Z4′′ ), (W6 , Z6′′ ), (W, Z ′′ )]
′
α β′ γ′ =δ12 s2 .4k 2 s α′4 β4′ γ4′ α′6 β6′ γ6′ =4k 2 s3 δ1 δF (W, W4 , W6 ).
From this we can write down that
δF [(W5 , Z5′′ ), (W7 , Z7′′ ), (W, Z ′′ )] =4k 2 s3 δ1 δF (W, W5 , W7 ),
δF [(W4 , Z4′′ ), (W7 , Z7′′ ), (W, Z ′′ )] =4k 2 s3 δ1 δF (W, W4 , W7 ),
δF [(W5 , Z5′′ ), (W6 , Z6′′ ), (W, Z ′′ )] =4k 2 s3 δ1 δF (W, W5 , W6 ),
On combining these we have that
δF [(W4 , Z4′′ ), (W6 , Z6′′ ), (W, Z ′′ )]δF [(W5 , Z5′′ ), (W7 , Z7′′ ), (W, Z ′′ )]
δF [(W4 , Z4′′ ), (W7 , Z7′′ ), (W, Z ′′ )]δF [(W5 , Z5′′ ), (W6 , Z6′′ ), (W, Z ′′ )]
δF (W, W4 , W6 )δF (W, W5 , W7 )
=
.
δF (W, W4 , W7 )δF (W, W5 , W6 )
This yields a dual of the Chasles-Steiner theorem.
11.8.3
Dual result for pairs on tangent lines to a conic again
In §6.6.3 we considered when a tangent at a point W to a circle C1 meets an outside concentric circle
C2 at a point Z, and established a result for a number of such pairs of points. We continued with
a generalization of this reasoning in §8.1.4, with W having normalized areal coordinates (α′ , β ′ , γ ′ )
and lying on a conic C1 with equation γ ′2 − 4kα′ β ′ = 0. With the point Z on the tangent at W
having coordinates (α, β, γ), we introduced a parameter s so that γ 2 − 4kαβ = 4k 2 s2 , and with
this had the relation (8.1.9)
α′ β − β ′ α = γ ′ s, β ′ γ − γ ′ β = −2kβ ′ s, γ ′ α − α′ β = −2kα′ s.
In §7.8.4, for a central conic with k = e2 where e > 0, and the transformation
α=
ex + y + a
e(x + ae
ex − y + a
, β=
, γ=−
,
2a(1 − e2 )
2a(1 − e2 )
a(1 − e2 )
11.8. TWO-CONIC ANALOGUE OF PASCAL’S THEOREM
we found that
(1 − e2 )x2 + y 2 − a2 (1 − e2 ) =
307
a2 (1 − e2 )2 2
[γ − 4e2 αβ],
e2
so that now we have that
(1 − e2 )x2 + y 2 − a2 (1 − e2 ) =
and so
a2 (1 − e2 )2 4 2
4e s ,
e2
y2
x2
+ 2
= 1 + 4e2 (1 − e2 )s2 .
2
a
a (1 − e2 )
Thus for each fixed s for which the right-hand side here is non-zero, the point Z lies on a conic Cs
which is homothetic to C1 .
We now consider five points W4 , W5 , W6 , W7 , W on C1 and the corresponding points Z4 , Z5 , Z6 , Z7 , Z
on Cs . We calculate initially the value of
δF [(W4 , Z4 ), (W6 , Z6 ), (W, Z)] = δF (W4 , W, Z)δF (Z4 , W6 , Z6 ) − δF (Z4 , W, Z)δF (W4 , W6 , Z6 ).
Now with δ1 = δF (Z1 , Z2 , Z3 ), by §3.1.2
′
α4 β4′
δF (W4 , W, Z) =δ1 α′ β ′
α β
γ4′
γ′
γ
=δ1 [α′4 (γβ ′ − βγ ′ ) + β4′ (αγ ′ − γα′ ) + γ4′ (βα′ − αβ ′ )]
=δ1 s[−2kα′4 β ′ − 2kβ4′ α′ + γ4′ γ ′ ],
and in particular
δF (W4 , W6 , Z6 ) = δ1 s[−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ ].
Similarly
α4
δF (Z4 , W, Z) =δ1 α′
α
β4
β′
β
γ4
γ′
γ
=δ1 [α4 (γβ ′ − βγ ′ ) + β4 (αγ ′ − γα′ ) + γ4 (βα′ − αβ ′ )]
=δ1 s[−2kα4 β ′ − 2kβ4 α′ + γ4 γ ′ ],
and in particular
δF (Z4 , W6 , Z6 ) = δ1 s[−2kα4 β6′ − 2kβ4 α′6 + γ4 γ6′ ].
Then
1
[δF (W4 , W, Z)δF (Z4 , W6 , Z6 ) − δF (Z4 , W, Z)δF (W4 , W6 , Z6 )]
δ12 s2
=[−2kα′4 β ′ − 2kβ4′ α′ + γ4′ γ ′ ][−2kα4 β6′ − 2kβ4 α′6 + γ4 γ6′ ]
−[−2kα4 β ′ − 2kβ4 α′ + γ4 γ ′ ][−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ ]
=α′ [−2kβ4′ (−2kα4 β6′ − 2kβ4 α′6 + γ4 γ6′ ) + 2kβ4 (−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ )]
+β ′ [−2kα′4 ((−2kα4 β6′ − 2kβ4 α′6 + γ4 γ6′ ) + 2kα4 (−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ )]
+γ ′ [γ4′ (−2kα4 β6′ − 2kβ4 α′6 + γ4 γ6′ ) − γ4 (−2kα′4 β6′ − 2kβ4′ α′6 + γ4′ γ6′ )] .
The coefficient of α′ is equal to
4k 2 α′6 [β4′ β4 − β4 β4′ ] + 4k 2 β6′ [β4′ α4 − α′4 β4 ] + 2kγ6′ [−β4′ γ4 + γ4′ β4 ]
=4k 2 β6′ (−γ4′ s) + 2kγ6′ (2kβ4′ s)
=4k 2 s(β4′ γ6′ − γ4′ β6′ ).
308
CHAPTER 11. SIMILARITY METHODS AND RESULTS
The coefficient of β ′ is equal to
4k 2 α′6 [α′4 β4 − α4 β4′ ] + 4k 2 β6′ [α′4 α4 − α4 α′4 ] + 2kγ6′ [−α′4 γ4 + γ4′ α4 ]
= − 4k 2 α′6 (−γ4′ s) − 2kγ6′ (2kα′4 s)
=4k 2 s(γ4′ α′6 − α′4 γ6′ ).
The coefficient of γ ′ is equal to
2kα′6 [−γ4′ β4 + γ4 β4′ ] + 2kβ6′ [−γ4′ α4 + α′4 γ4 ] + γ6′ [γ4′ γ4 − γ4′ γ4 ]
=2kα′6 (−2kβ4′ s) + 2kβ6′ (2kα′4 s)
=4k 2 s(α′4 β6′ − β4′ γ6′ ).
On combining these we have that
′
α
2 2 δF [(W4 , Z4 ), (W6 , Z6 ), (W, Z)] =4δ1 k s α′4
α′6
β′
β4′
β6′
γ′
γ4′
γ6′
=4k 2 s3 δ1 δF (W, W4 , W6 ).
On applying this results to three other pairs and combining the four results we find that
δF [(W4 , Z4 ), (W6 , Z6 ), (W, Z)]δF [(W5 , Z5 ), (W7 , Z7 ), (W, Z)]
δF [(W4 , Z4 ), (W7 , Z7 ), (W, Z)]δF [(W5 , Z5 ), (W6 , Z6 ), (W, Z)]
δF (W, W4 , W6 )δF (W, W5 , W7 )
.
=
δF (W, W4 , W7 )δF (W, W5 , W6 )
This can then be used in conjunction with the Chasles-Steiner theorem to derive a dual result
of the latter.
309
11.9. CARTESIAN AND PLÜCKER DUALITY
11.9
Cartesian and Plücker duality
11.9.1
Cartesian duality
As we wish to deal effectively with distances and angle-measures in this chapter, we give priority
to using (rectangular) Cartesian coordinates. With the identity (3.2.1), written now in the form
δF (Z, Z4 , Z5 ) = la α + ma β + na γ,
we have set up areal coordinates of a point Z and areal dual-coordinates of a pair (U, V ), and
based on the correspondence (α, β, γ) ↔ (la , ma , na ), we have developed what we shall refer to as
areal-coordinate duality.
Plücker c.1830 standardised the equations of a line which does not pass through the origin O
in the form ux + vy + 1 = 0. He called the pair (u, v), defined by (−1/u, 0), (0, −1/v) being
the coordinates of the points, respectively, in which the line cuts the x-axis and the y-axis, linecoordinates of this line. From an equation of a line in the form lx + my + n = 0 the triple
(l, m, n) are commonly called line coordinates; these are homogeneous in that for any k 6= 0 the
triple (kl, km, kn) represent the same line. When n 6= 0 this line has Plücker coordinates (u, v) =
(l/n, m/n). There are difficulties about coordinates of lines which pass through the origin or are
parallel to either the x-axis or the y-axis, and some authors refer to infinite coordinates. This
material is laid out in fair detail by Charlotte Angas Scott [?, pp. 15-43]. In an affine or Euclidean
context, taking homogeneous coordinates surrenders some information. Accordingly we take a
non-traditional approach.
Now we start our look at dual Cartesian coordinates corresponding to the dual areal ones above.
As in Barry [2, pp.74-75] we suppose that we have taken a frame of reference F = ([O, I , [O, J )
where OI ⊥ OJ and |O , I | = |O , J | = 1, so that I ≡F (1, 0), J ≡F (0, 1). We normalize the
equation of the line Z4 Z5 , based on (Z4 , Z5 ), by taking it to be
δF (Z, Z4 , Z5 ) = − 21 (y5 − y4 )x + 12 (x5 − x4 )y + 21 (x4 y5 − y4 x5 ) = 0,
on writing which as lC x + mC y + nC = 0 we note that we have the definition
lC = − 12 (y5 − y4 ), mC = 21 (x5 − x4 ), nC = 21 (x4 y5 − y4 x5 ).
(11.9.1)
We refer to (lC , mC , nC ) as dual Cartesian coordinates linked to the pair (Z4 , Z5 ) and the line
Z4 Z5 .
In the hope of throwing light on the relationship of Cartesian to areal coordinates, we note that
we can re-write the expression in §3.1.2 as
α4 β4 γ4 δF (Z4 , Z5 , Z6 ) =δF (Z1 , Z2 , Z3 ) α5 β5 γ5 α6 β6 γ6 α4 β4 α4 + β4 + γ4 (11.9.2)
=δF (Z1 , Z2 , Z3 ) α5 β5 α5 + β5 + γ5 .
α6 β6 α6 + β6 + γ6 Then we have from (11.8.2) for areal coordinates
α4
δF (Z4 , Z5 , Z6 ) = δF (Z1 , Z2 , Z3 ) α5
α6
β4
β5
β6
1
1
1
and if we identify a point Z by the pair (α, β) of its areal coordinates
α=
δF (Z, Z3 , Z1 )
δF (Z, Z2 , Z3 )
,β =
,
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
,
(11.9.3)
310
CHAPTER 11. SIMILARITY METHODS AND RESULTS
then as in §9.10.4 we have Cartesian coordinates x = α, y = β of Z with respect to Z3 = O, Z2 =
W2 , Z1 = W1 . We obtain rectangular Cartesian coordinates by selecting our frame of reference
with OI ⊥ OJ and |O, I| = |O, J| = 1; we can easily check that this process yields (11.8.1).
We now have our identity (3.2.1) in the form
δF (Z, Z4 , Z5 ) = lC x + mC y + nC ,
where lC , mC , nC are given by (11.8.1). We shall refer to the correspondence (x, y) ↔ (lC , mC , nC )
as Cartesian duality.
By expansion of the determinants involved it is easy to check that
lC =δF (I, Z4 , Z5 ) − δF (O, Z4 , Z5 ) = δF (O, I, Z4 ) − δF (O, I, Z5 ) = δF (O, I, Z4 − Z5 ),
mC =δF (J, Z4 , Z5 ) − δF (O, Z4 , Z5 ) = δF (O, J, Z4 ) − δF (O, J, Z5 )δF (O, J, Z4 − Z5 ),
nC =δF (O, Z4 , Z5 ).
(11.9.4)
In this we have made use of (1.1.4) and (1.1.9).
These evaluations can also be derived as
δF (Z, Z4 , Z5 ) = δF (xI + yJ, Z4 , Z5 )
=δF (x(I − Z4 ) + y(J − Z4 ) + (1 − x − y)(−Z4 , O, Z5 − Z4 )
=xδF (I − Z4 , O, Z5 − Z4 ) + yδF (J − Z4 , O, Z5 − Z4 ) + (1 − x − y)δF (−Z4 , O, Z5 − Z4 )
=xδF (I, Z4 , Z5 ) + yδF (J, Z4 , Z5 ) + (1 − x − y)δF (O, Z4 , Z5 ).
If we take a pair of distinct points Z7 and Z8 on Z4 Z5 , we have
Z7 = (1 − s)Z4 + sZ5 ,
Z8 = (1 − t)Z4 + tZ5 ,
for distinct numbers s and t, and as
δF (Z, Z7 , Z8 ) = (t − s)δF (Z, Z4 , Z5 ),
we see that the dual Cartesian coordinates of (Z7 , Z8 ) are obtained from those of (Z4 , Z5 ) on
multiplying by the non-zero number t − s. In fact the dual Cartesian coordinates lC , mC , nC
depend on a rotor hZ4 , Z5 i. For if hZ4 , Z5 i = hZ7 , Z8 i we have that
− 12 (y8 − y7 ) = − 12 (y5 − y4 ),
1
2 (x8
− x7 ) = 21 (x5 − x4 ),
and since δF (Z, Z4 , Z5 ) = δF (Z, Z7 , Z8 ) for all points Z, in particular
δF (O, Z7 , Z8 ) = δF (O, Z4 , Z5 ).
Conversely suppose that
− 12 (y8 − y7 ) = − 21 (y5 − y4 ),
1
2 (x8
− x7 ) = 21 (x5 − x4 ),
1
2 (x7 y8
− y7 x8 ) = 21 (x4 y5 − y4 x5 ).
It follows that x8 − x7 = x5 − x4 , y8 − y7 = y5 − y4 , so that Z8 − Z7 = Z5 − Z4 . Furthermore
x7 y8 − y7 x8 =x7 (y8 − y7 ) − y7 (x8 − x7 )
=x7 (y5 − y4 ) − y7 (x5 − x4 )
x7 y8 − y7 x8 =x4 y5 − y4 x5
=x4 (y5 − y4 ) − y4 (x5 − x4 ),
and so by subtraction
(x7 − x4 )(y5 − y4 ) − (y7 − y4 )(x5 − x4 ) = 0.
311
11.9. CARTESIAN AND PLÜCKER DUALITY
It follows that
x4
2δF (Z4 , Z5 , Z7 ) = x5
x7
y4
y5
y7
x4
= x5 − x4
x7 − x4
1
1
1
y4
y5 − y4
y7 − y4
1
0
0
= 0,
and so Z7 ∈ Z4 Z5 . We thus have the important result that our dual Cartesian coordinates (lC , mC , nC ) depend on rotors hZ4 , Z5 i.
In the foregoing we worked with δF (Z, Z4 , Z5 ) involving a fixed but arbitrary pair of distinct
point Z4 and Z5 , and a variable point Z. Then δF (Z, Z4 , Z5 ) = 0 is a point equation of the line
Z4 Z5 . Now dually we work with δF (Z6 , Z, Z ′ ) involving a variable pair of distinct points Z and
Z ′ , and an arbitrary fixed point Z6 . Then δF (Z6 , Z, Z ′ ) = 0 is a necessary and sufficient condition
that the line ZZ ′ passes through the point Z6 . This then is a line equation of the point Z6 . For
the latter, the variable dual Cartesian coordinates
lC = − 21 (y ′ − y), mC = 21 (x′ − x), nC = 21 (xy ′ − yx′ ),
satisfy x6 lC + y6 mC + nC = 0.
We turn now to a Cartesian equation for a line. Given an equation ax + by + c = 0, with
a = b = 0 excluded, we wish to express it in our standard format. For this we can use any pair of
distinct points Z4 and Z5 on the line. We subdivide into four cases as follows.
CASE 1. Suppose that a 6= 0, b 6= 0 and c 6= 0, so that the line is not parallel to either axis
and does not pass through the origin.
Then it meets the x-axis at a point Z4 ≡ − ac , 0 and meets
c
the y-axis at a point Z5 ≡ 0, − b . We then have
l=
1
2
c
, m=
b
1
2
c
, n=
a
1
2
c2
.
ab
We then rewrite the equation as
c
c
c2
x + 21 y + 21
= 0.
(11.9.5)
b
a
ab
CASE 2. Take a 6= 0, b = 0 so that our equation is ax + c = 0, which yields a line parallel
to the y-axis.
This meets the x-axis at the point Z4 ≡ − ac , 0 , and we take as well the point
Z5 ≡ − ac , 1 on it. Then we have that
1
2
c
l = − 21 , m = 0, n = − 12 ,
a
so we can rewrite the equation in standardized form as
− 21 x −
1
2
c
= 0.
a
CASE 3. Take a = 0, b 6= 0 so that our equation is by + c = 0, which yields a line parallel
c
to the x-axis.
This meets the y-axis at the point Z4 ≡ 0, − b , and we take as well the point
c
Z5 ≡ 1, − b on it. Then we have that
l = 0, m = 21 , n =
1
2
c
,
b
so we can rewrite the equation in standardized form as
1
2y
+
1
2
c
= 0.
b
CASE 4. Take a 6= 0, b 6= 0, c = 0, so that our equation is ax + by = 0, which yields a line
through the origin which is not either of the axes. On it we take the origin Z4 ≡ (0, 0) and the
point Z5 ≡ (b, −a). Then we have that
l = 12 a, m = 12 b, n = 0,
312
CHAPTER 11. SIMILARITY METHODS AND RESULTS
so we can rewrite the equation in standardized form as
1
2 ax
+ 21 by = 0.
To deal similarly with a line equation of a point, consider an equation al + bm + n = 0 in
homogeneous line coordinates. On multiplying across by a non-zero constant, this is equivalent to
alC + bmC + nC = 0.
′
Any line ZZ satisfying (11.8.6) for which mC =
to the y-axis. For it we must have
1
′
2 (x − x)
(11.9.6)
′
= 0 must have x = x and so be parallel
− 12 (y ′ − y)a + 21 (y ′ − y)x = 0,
and so as y ′ − y 6= 0 we must have x = a. Similarly for any line ZZ ′ satisfying (11.8.6) for which
lC = − 21 (y ′ − y) = 0 must have y ′ = y and so be parallel to the x-axis. For it we must have
′
1
2 (x
− x)b + 21 (x − x′ )y = 0,
and so as x′ − x 6= 0 we must have y = b. The lines with Cartesian equations x = a and y = b meet
at the point Z6 ≡ (a, b) and this must be the vertex of the pencil of lines represented by (11.8.6).
11.9.2
Relationship between Cartesian and areal dual coordinates
With the notation of §11.8.1, we note that
and similarly
x1
la = δF (Z1 , Z4 , Z5 ) = 12 x4
x5
ma = δF (Z2 , Z4 , Z5 ) = x2 lC + y2 mC + nC ,
y1
y4
y5
1
1
1
= x1 lC + y1 mC + nC ,
na = δF (Z3 , Z4 , Z5 ) = x3 lC + y3 mC + nC .
Conversely for the identity
δF (Z1 , Z2 , Z3 )δF (Z, Z4 , Z5 ) = δF (Z, Z2 , Z3 )la + δF (Z, Z3 , Z1 )ma + δF (Z, Z1 , Z2 )na ,
we note that
δF (Z, Z2 , Z3 ) = − 12 (y3 − y2 )x + 21 (x3 − x2 )y + 21 (x2 y3 − x3 y2 ),
δF (Z, Z3 , Z1 ) = − 12 (y1 − y3 )x + 21 (x1 − x3 )y + 21 (x3 y1 − x1 y3 ),
δF (Z, Z1 , Z2 ) = − 12 (y2 − y1 )x + 21 (x2 − x1 )y + 21 (x1 y2 − x2 y1 ).
Thus
la 1
[− (y3 − y2 )x + 21 (x3 − x2 )y + 21 (x2 y3 − x3 y2 )]
δ1 2
ma 1
+
[− 2 (y1 − y3 )x + 21 (x1 − x3 )y + 21 (x3 y1 − x1 y3 )]
δ1
na
+ [− 21 (y2 − y1 )x + 21 (x2 − x1 )y + 21 (x1 y2 − x2 y1 )]
δ1
=lC x + mC y + nC ,
δF (Z, Z4 , Z5 ) =
where
1 1
[− (y3 − y2 )la − 12 (y1 − y3 )ma − 21 (y2 − y1 )na ],
δ1 2
1
mC = [ 12 (x3 − x2 )la + 12 (x1 − x3 )ma + 21 (x2 − x1 )na ],
δ1
1
nC = [ 12 (x2 y3 − x3 y2 )la + 12 (x3 y1 − x1 y3 )ma + 21 (x1 y2 − x2 y1 )na ],
δ1
lC =
and δ1 = δF (Z1 , Z2 , Z3 ).
11.9. CARTESIAN AND PLÜCKER DUALITY
11.9.3
313
Derivation of formulae
With (l6 , m6 , n6 ) referring to hZ4 , Z5 i and (l9 , m9 , n9 ) referring to hZ7 , Z8 i, for many purposes we
need to evaluate l9 n6 − l6 n9 and m9 n6 − m6 n9 . For these we note that
l 9 n6 − l 6 n9
=[δF (I, Z7 , Z8 ) − δF (O, Z7 , Z8 )]δF (O, Z4 , Z5 ) − [δF (I, Z4 , Z5 ) − δF (O, Z4 , Z5 )]δF (O, Z7 , Z8 )
=δF (I, Z7 , Z8 )δF (O, Z4 , Z5 ) − δF (I, Z4 , Z5 )δF (O, Z7 , Z8 )]
=δF [(O, I), (Z4 , Z5 ), (Z7 , Z8 )],
and by a similar argument
m9 n6 − m6 n9 = δF [(O, J), (Z4 , Z5 ), (Z7 , Z8 )].
Thus we have
l9 n6 − l6 n9 = δF [(O, I), (Z4 , Z5 ), (Z7 , Z8 )],
m9 n6 − m6 n9 = δF [(O, J), (Z4 , Z5 ), (Z7 , Z8 )].
(11.9.7)
As an application, if we consider the simultaneous equations of lines
l 6 x + m6 y = − n 6 ,
l 9 x + m9 y = − n 9 ,
by Cramer’s rule we obtain the solution
l6 −n6 l9 −n9 δF [(O, I), (Z4 , Z5 ), (Z7 , Z8 )]
= 1
,
x=
l 6 m6 2 δF (O, Z4 − Z5 , Z7 − Z8 )
l 9 m9 −n6 m6 −n9 m9 δ [(O, J), (Z4 , Z5 ), (Z7 , Z8 )]
= F1
y = ,
l
m
6 6
2 δF (O, Z4 − Z5 , Z7 − Z8 )
l 9 m9 (11.9.8)
for the coordinates of the point of intersection.
11.9.4
Tangential equations for a circle and other conics
We consider the circle C(O; a) with equation x2 + y 2 = a2 , and a line with equation lx + my + n = 0
which is in standard form with respect to a pair of points Z4 and Z5 . When m 6= 0 we have
m2 x2 + m2 y 2 = m2 a2 , so on substituting from the equation of the line, we have for any points of
intersection
m2 x2 + (lx + n)2 =m2 a2 ,
(l2 + m2 )x2 + 2nlx + n2 − m2 a2 =0.
For this equation in x to have a repeated root, and so Z4 Z5 to be a tangent to the circle, it is
necessary and sufficient that n2 l2 = (l2 + m2 )n2 − m2 a2 , which simplifies to a2 (l2 + m2 ) = n2 .
This condition amounts to
a2 δF (Z4 , O, Z5 , I)2 + δF (Z4 , O, Z5 , J)2 = δF (O, Z4 , Z5 )2 .
An alternative method of deriving this is that the perpendicular distance from the origin to the
, and for a tangent to the circle this is equal to a, so
line with equation lx + my + n = 0 is √l2|n|
+m2
that
|n|
√
= a;
l 2 + m2
314
CHAPTER 11. SIMILARITY METHODS AND RESULTS
on squaring this yields
n2 = a2 (l2 + m2 ).
More generally, by the formulae of §11.8.2 any homogeneous quadratic equation in la , ma , na
transforms into a quadratic equation in lC , mC , nC , so we can derive from the results of Chapter
8 on dual conics analogous results involving Cartesian dual coordinates.
11.9.5
Plücker duality
For distinct points Z4 and Z5 such that the line Z4 Z5 does not pass through the origin O, we start
with
1
− 1 (y5 − y4 )
(x5 − x4 )
δF (Z, Z4 , Z5 )
x+ 1 2
y + 1,
(11.9.9)
= 1 2
δF (O, Z4 , Z5 )
2 (x4 y5 − x5 y4 )
2 (x4 y5 − x5 y4 )
which we write as
u6 x + v6 y + 1,
(11.9.10)
where
1
(−(y5 − y4 ), x5 − x4 )
x4 y5 − x5 y4
1
=
(Z5 − Z4 )⊥ .
2δF (O, Z4 , Z5 )
W6 ≡ (u6 , v6 ) =
Thus
u6 =
lC
,
nC
v6 =
(11.9.11)
mC
.
nC
We refer to (u6 , v6 ) as Plücker coordinates; note that u6 = v6 = 0 cannot occur. Then the line
Z4 Z5 has equation
u6 x + v6 y + 1 = 0.
(11.9.12)
We saw in §11.8.1 that if Z7 , Z8 are any distinct points on the line Z4 Z5 , with
Z7 = (1 − s)Z4 + sZ5 ,
Z8 = (1 − t)Z4 + tZ5 ,
for numbers s and t, then
− 21 (y8 −y7 ) = − 21 (t−s)(y5 −y4 ),
1
2 (x8 −x7 )
= 21 (t−s)(x5 −x4 ),
1
2 (x7 y8 −y7 x8 )
= 12 (t−s)(x4 y5 −y4 x5 ).
It follows that we must have
− 21 (y8 − y7 )
=
1
2 (x7 y8 − x8 y7 )
− 21 (y5 − y4 )
,
1
2 (x4 y5 − x5 y4 )
1
2 (x8
− x7 )
=
1
2 (x7 y8 − x8 y7 )
1
2 (x5
1
2 (x4 y5
− x4 )
.
− x5 y4 )
Conversely suppose that these last two equalities hold. Then
y8 − y7
x8 − x7
x7 y8 − x8 y7
=
=
.
y5 − y4
x5 − x4
x4 y5 − x5 y4
Then for some j 6= 0 we have
x8 − x7 = j(x5 − x4 ), y8 − y7 = j(y5 − y4 ), x7 y8 − x8 y7 = j(x4 y5 − x5 y4 ).
Then Z8 − Z7 = j(Z5 − Z4 ). Furthermore
x7 y8 − x8 y7 =x7 (y8 − y7 ) − y7 (x8 − x7 )
=j[x7 (y5 − y4 ) − y7 (x5 − x4 )].
315
11.9. CARTESIAN AND PLÜCKER DUALITY
But also
x7 y8 − x8 y7 = j(x4 y5 − x5 y4 ) = j[x4 (y5 − y4 ) − y4 (x5 − x4 )],
and so
(x7 − x4 )(y5 − y4 ) − (y7 − y4 )(x5 − x4 ).
Thus δF (Z4 , Z7 , Z5 ) = δF (O, Z7 − Z4 , Z5 − Z4 ) = 0 and so Z7 ∈ Z4 Z5 . Hence Z4 Z5 = Z7 Z8 . It
follows that our Plücker coordinates (u6 , v6 ) depend on lines Z4 Z5 .
As well as (11.8.9) for a fixed pair Z4 , Z5 of distinct points not collinear with O and a variable
point Z, we now consider analogously
1
− 1 (y ′ − y)
(x′ − x)
δF (Z6 , Z, Z ′ )
= x6 1 2 ′
+ y6 1 2 ′
+ 1,
′
′
′
δF (O, Z, Z )
2 (xy − x y)
2 (xy − x y)
(11.9.13)
involving a fixed point Z6 ≡ (x6 , y6 ) and distinct variable points Z and Z ′ not collinear with the
origin O. We write this as
x6 u + y6 v + 1,
(11.9.14)
where
1
(−(y ′ − y), x′ − x)
δF (O, Z, Z ′ )
1
= 12
(Z ′ − Z)⊥ ,
δF (O, Z, Z ′ )
W ≡ (u, v) = 12
(11.9.15)
are variable Plücker coordinates. The equation
x6 u + y6 v + 1 = 0,
(11.9.16)
now is satisfied by lines ZZ ′ which pass through the point Z6 but not through the origin O. Note
that Z6 = 0 cannot occur.
The line (11.8.2) is called the Plücker reciprocal of the point W6 in (11.8.11). As (11.8.12)
can be re-written as (−u6 )x + (−v6 )y = 1 this line is the polar of the point −W6 with respect to
the unit circle, as the latter has equation x2 + y 2 = 1. The line OW6 has equation v6 x − u6 y = 0
so thisis perpendicular to the polar, and the point of intersection of these two lines is the point
v6
1
∗
6
= − u2 +v
and
W6∗ ≡ − u2u+v
2 , − u2 +v 2
2 W6 . Then W6 ∈ [O, −W6
6
6
6
6
6
6
|O, W6∗ ||O, W6 | = |O, W6∗ ||O, −W6 | = 1.
The polar is then the line through the point W6∗ which is perpendicular to the line OW6 .
Given two lines Z4 Z5 and Z7 Z8 with equations u6 x + v6 y + 1 = 0 as in (11.8.12) and
u9 x + v9 y + 1 = 0,
(11.9.17)
the Plücker reciprocals of W6 ≡ (u6 , v6 ), W9 ≡ (u9 , v9 ), respectively, we note that they are parallel
if and only if u6 v9 − u9 v6 = 0, that is if δF (O, W6 , W9 ) = 0 so that O ∈ W6 W9 . When these lines
are not parallel their point of intersection is
1
u9 − u6
v9 − v6
=
,
(W9 − W6 )⊥ .
(11.9.18)
Z6,9 ≡ −
u6 v9 − u9 v6 u6 v9 − u9 v6
2δF (O, W6 , W9 )
By (11.8.11) and (11.8.18), the line W6 W9 is the Plücker reciprocal of the point Z6,9 . This
can also be seen by a more general argument as follows. By (11.8.12), u6 x12 + v6 y12 + 1 = 0 is a
condition that the point Z12 ≡ (x12 , y12 lies on the Plücker reciprocal of W6 and by the symmetry
of the expression is also a condition that the point W6 lies on the Plücker reciprocal of Z12 . As Z6,9
is on the the Plücker reciprocals of W6 and W9 , in turn W6 and W9 lie on the Plücker reciprocal
of Z6,9 which must thus be the line W6 W9 .
316
CHAPTER 11. SIMILARITY METHODS AND RESULTS
As Z6,9 ∈ Z4 Z5 and Z6,9 ∈ Z7 Z8 and the Plücker coordinates are independent of the particular
points used on a line,we have the following formulae which are often convenient to apply,
y5 − y6,9
x5 − x6,9
u6 = −
,
v6 =
,
x6,9 y5 − y6,9 x5
x6,9 y5 − y6,9 x5
x8 − x6,9
y8 − y6,9
,
v9 =
,
(11.9.19)
u9 = −
x6,9 y8 − y6,9 x8
x6,9 y8 − y6,9 x8
provided that neither Z5 nor Z8 is the point of intersection.
Given intersecting lines with equations (11.8.12) and (11.8.17), respectively, we note that for
any real number s the equations
(1 − s)[u6 x + v6 y + 1] + s[u9 x + v9 y + 1]
=[(1 − s)u6 + su9 ]x + [(1 − s)v6 + sv9 ]y + 1
=0
represent lines which pass through the point of intersection Z6,9 . We can write this as
(1 − s)
δF (Z, Z7 , Z8 )
δF (Z, Z4 , Z5 )
+s
= 0,
δF (O, Z4 , Z5 )
δF (O, Z7 , Z8 )
and, as in (11.8.19),
δF (Z, Z6,9 , Z5 )
δF (Z, Z6,9 , Z8 )
+s
= 0,
δF (O, Z6,9 , Z5 )
δF (O, Z6,9 , Z8 )
on the assumption that neither Z5 nor Z8 is the point of intersection. To handle this expression
we keep Z5 a fixed point of the first line but choose Z8 to be the point in which the line through
Z5 parallel to the line OZ6,9 meets the second line. Thus we have that Z5 Z8 k OZ6,9 and so
equivalently
δF (O, Z6,9 , Z5 ) = δF (O, Z6,9 , Z8 ).
(11.9.20)
(1 − s)
With this choice our equation can be written as
δF (Z, Z6,9 , (1 − s)Z5 + sZ8 )
= 0,
δF (O, Z6,9 , Z5 )
and this is equivalent to
δF (Z, Z6,9 , (1 − s)Z5 + sZ8 )
= 0.
δF (O, Z6,9 , (1 − s)Z5 + sZ8 )
This line has Plücker coordinates
(1 − s)x5 + sx8 − x6,9
(1 − s)y5 + sy8 − y6,9
, vs =
.
us = −
x6,9 [(1 − s)y5 + sy8 ] − y6,9 [(1 − s)x5 + sx8 ]
x6,9 [(1 − s)y5 + sy8 ] − y6,9 [(1 − s)x5 + sx8 ]
On putting s = 0 in this we see by (11.8.19) that us and vs become, respectively, u6 and v6 .
Similarly on putting s = 1 in this we see by (11.8.19) that us and vs become, respectively, u9 and
v9 .
Now if we have parametric equations of this sort
u=
a + bs
,
e + fs
v=
c + ds
,
e + fs
in which s = 0 yields the point W6 and s = 1 yields the point W9 , then
a
u6 = ,
e
(be − af )s
,
u − u6 =
e(e + f s)
be − af
u9 − u6 =
,
e(e + f )
s
u =u6 + (u9 − u6 )(e + f )
,
e + fs
317
11.9. CARTESIAN AND PLÜCKER DUALITY
s
and similarly v = v6 + (v9 − v6 )(e + f ) e+f
s . In our example we have
e = x6,9 y5 − y6,9 x5 =2δF (O, Z6,9 , Z5 ),
f = x6,9 (y8 − y5 ) − y6,9 (x8 − x5 ) =2δF (O, Z6,9 , Z8 − Z5 ) = 2[δF (O, Z6,9 , Z8 ) − δF (O, Z6,9 , Z5 )] = 0,
s
=s.
(e + f )
e + fs
Hence we have
us = u6 + (u9 − u6 )s,
vs = v6 + (v9 − v6 )s.
(11.9.21)
Thus Ws is a point on the line W6 W9 . This shows that the Plücker reciprocals of the points on
the line W6 W9 are the pencil of lines passing through the point Z6,9 , with the exception of the line
Z6,9 O. The union of these lines is (Π \ Z6,9 O) ∪ {Z6,9 }.
We can introduce the concept of a half-pencil [Z6,9 Z5 , Z6,9 Z8 by taking s ≥ 0 in the last
paragraph. This would yield as the Plücker reciprocal of the half-line [W6 , W9 the set of lines
through Z6,9 which intercept the half-line Z5 , Z8 ; their union would be a semi-open duo-sector
obtained by taking the duo-sector with arms Z6,9 Z5 and Z6,9 O, removing the arm Z6,9 O and then
restoring the vertex Z6,9 .
Continuing, suppose that as well as W6 and W9 as in the last paragraph, we consider a point W12
with Plücker reciprocal Z10 Z11 and that this third line meets the first one Z4 Z5 at Z6,12 . We shall
now identify the first line Z4 Z5 as Z6,12 Z5 and slide Z11 along the third line until Z5 Z11 k Z6,12 O.
We now take the half-pencil [Z6,12 Z5 , Z6,12 Z11 as the set of lines through Z6,12 which intersect the
half-line [Z5 , Z11 ; their union is a semi-open duo-sector with arms Z6,12 Z5 = Z4 Z5 and Z6,12 O.
On taking this half-pencil in conjunction with the half-pencil in the last paragraph, we have a pair
of co-initial half-pencils which are the Plücker reciprocal of the angle-support |W9 W6 W12 .
11.9.6
Features of Plücker reciprocation
We start by considering an item under the inverse transformation of Plücker reciprocation. Given
the lines (11.8.12) and (11.8.17), we note that these have slopes given by
µ6 = −
u6
,
v6
µ9 = −
u9
.
v9
Then the sensed duo-angle θd from (11.8.12) to (11.8.17) satisfies
µ9 − µ6
−u9 /v9 − (−u6 /v6 )
=
1 + µ6 µ9
1 + (−u6 /v6 )(−u9 /v9 )
u6 v9 − u9 v6
v9 /u9 − v6 /u6
=
=
= tan φd ,
u6 u9 + v6 v9
1 + (v6 /u6 )(v9 /u9 )
tan θd =
where φd is the sensed duo-angle from OW6 to OW9 . This induces for us an angular measure from
the point W6 to the point W9 . Hence θd and φd are equal in magnitude.
We get a more informative similar result on sensed angles as follows. If t 6= 0 the translation
u6
v6
1
1
′
,
y
=
y
+
1
−
,
x′ = x + 1 −
t u26 + v62
t u26 + v62
maps the line with equation (9.8.17) onto the line with equation
1
u6
v6
1
′
y
−
1
−
+
v
+ 1 =0,
u6 x′ − 1 −
6
t u26 + v62
t u26 + v62
1
u6 x′ + v6 y ′ + =0,
t
′
′
tu6 x + tv6 y + 1 =0,
318
CHAPTER 11. SIMILARITY METHODS AND RESULTS
and so for sensed angles with transversals, without loss of generalityp(11.8.12) may be replaced by
the line with equation tu6 x + tv6 y + 1 = 0. We do this with t = 1/ u26 + v62 , so that equivalently
we may retain (11.8.12) but now with the assumption that u26 + v62 = 1, so that W6 is on the unit
circle C(O; 1). Then −W6 is also on the unit circle C(O; 1), and its polar with respect to it is the
tangent at −W6 . Similarly we may assume that u29 + v92 = 1, with analogous consequences. Now
∗
Z6,9 is the point of intersection of the tangents at −W6 and −W9 , and if we take the point Z6,9
so
∗
that the mid-point of −W9 and it is Z6,9 , then Z6,9
is another point on the tangent at −W9 . Now
on using complex coordinates we have by (11.8.18) that
1
[−(v9 − v6 ) + i(u9 − u6 )]
u6 v9 − u9 v6
i
[u9 − u6 + i(v9 − v6 )]
=
u6 v9 − u9 v6
i
(w9 − w6 ).
=
u6 v9 − u9 v6
z6,9 =
(11.9.22)
Now
w9 u9 + iv9
(u9 + iv9 )(u6 − iv6 )
=
=
= u6 u9 + v6 v9 + i(u6 v9 − u9 v6 ),
w6 u6 + iv6
u26 + v62
w6 (u6 + iv6 )(u9 − iv9 )
=
= u6 u9 + v6 v9 − i(u6 v9 − u9 v6 ),
w9
u26 + v62
so that
2i(u6 v9 − u9 v6 ) =
Hence
z6,9 =
i
w92 −w62
2iw6 w9
w6
w2 − w62
w9
−
= 9
.
w6
w9
w6 w9
(w9 − w6 ) = −
2w6 w9
.
w6 + w9
∗
For the sensed angle θ = ∡(−W6 )Z6,9 Z6,9
, we have that
∗
z6,9
− z6,9
2z6,9 − (−w9 ) − z6,9
z6,9 + w9
=
=
−w6 − z6,9
−w6 − z6,9
−z6,9 − w6
=
=
6 w9
− w2w
+ w9
6 +w9
2w6 w9
w6 +w9
w9
.
w6
− w6
It follows that θ = φ where φ = ∡W6 OW9 .
As a preparation for several results, we note that (11.8.22) was derived before the restrictions
on W6 and W9 were utilised, and from it we have that
w9 − w6 = −iz6,9 (u6 v9 − u9 v6 ).
Now by (11.8.20),
−(y5 − y6,9 )(x8 − x6,9 ) + (x5 − x6,9 )(y8 − y6,9 )
(x6,9 y5 − y6,9 x5 )(x6,9 y8 − y6,9 x8 )
δF (O, Z5 − Z6,9 , Z8 − Z6,9 )
.
=
2δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Z8 )
u6 v9 − u9 v6 =
On combining these two formulae we obtain
w9 − w6 = −iz6,9
δF (Z6,9 , Z5 , Z8 )
.
2δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Z8 )
(11.9.23)
319
11.9. CARTESIAN AND PLÜCKER DUALITY
To turn now to Plücker reciprocal of complex-valued distance, we define (Z4 Z5 )(Z7 Z8 ) to be
W6 W9 = w9 − w6 and by (11.8.23) this is given by
(Z4 Z5 )(Z7 Z8 ) = W6 W9 = −iOZ6,9
δF (Z6,9 , Z5 , Z8 )
.
2δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Z8 )
(11.9.24)
We refer to this as the Plücker reciprocal complex-valued distance from the line Z4 Z5 to the
line Z7 Z8 .
On multiplying (11.8.24) by its complex conjugate, we define the reciprocal square of distance
between the lines Z4 Z5 and Z7 Z8 by
|Z4 Z5 , Z7 Z8 |2 =|W6 , W9 |2 = (u9 − u6 )2 + (v9 − v6 )2
δF (Z6,9 , Z5 , Z8 )2
1
.
= |O, Z6,9 |2
4
2δF (O, Z6,9 , Z5 )2 δF (O, Z6,9 , Z8 )2
(11.9.25)
By (11.8.7) we can also note the formula
|Z4 Z5 , Z7 Z8 |2 = (u9 − u6 )2 + (v9 − v6 )2
2 2
l9
m9
l6
m6
=
−
−
+
n9
n6
n9
n6
1
= 2 2 [(l9 n6 − l6 n9 )2 + (m9 n6 − m6 n9 )2 ]
n6 n9
1
δF [(O, I), (Z4 , Z5 ), (Z7 , Z8 )]2 + δF [(O, J), (Z4 , Z5 ), (Z7 , Z8 )]2 .
=
2
2
δF (O, Z4 , Z5 ) δF (O, Z7 , Z8 )
As well as the Plücker reciprocals Z4 Z5 and Z7 Z8 of W6 and W9 , respectively, as used in
(11.8.23), we consider the reciprocal Z10 Z11 of W12 , and suppose that Z10 Z11 meets Z7 Z8 at Z9,12 ,
and that Z10 Z11 meets Z4 Z5 at Z12,6 . Then as in (11.8.23)
w12 − w6 = −iz12,6
δF (Z12,6 , Z5 , Z11 )
.
2δF (O, Z12,6 , Z5 )δF (O, Z12,6 , Z11 )
To use these two formulae simultaneously, we note that in (11.8.23) we may replace Z5 by Z12,6
and Z8 by Z9,12 , giving
w9 − w6 = −iz6,9
δF (Z6,9 , Z12,6 , Z9,12 )
.
2δF (O, Z6,9 , Z12,6 )δF (O, Z6,9 , Z9,12 )
Similarly in the formula for w12 − w6 we may replace Z5 by Z6,9 and Z11 by Z9,12 , giving
w12 − w6 = −iz12,6
δF (Z12,6 , Z6,9 , Z9,12 )
.
2δF (O, Z12,6 , Z6,9 )δF (O, Z12,6 , Z9,12 )
By division, we obtain from these,
δF (O, Z6,9 , Z9,12 ) z12,6
w12 − w6
=
.
w9 − w6
δF (O, Z12,6 , Z9,12 ) z6,9
(11.9.26)
With the notation at the end of §11.8.5, we now define the Plücker reciprocal sensed angle
θ from the half-pencil [Z6,9 Z5 , Z6,9 Z8 to the half-pencil [Z6,12 Z5 , Z6,12 Z11 satisfying
w12 − w6
|W6 , W12 |
=
exp(iθ).
w9 − w6
|W6 , W9 |
It follows from (11.8.26) that eiθ = ±eiφ , where φ = ∡Z6,9 OZ12,6 , with the + or the − being
taken according as
δF (O, Z6,9 , Z9,12 )
δF (O, Z12,6 , Z9,12 )
320
CHAPTER 11. SIMILARITY METHODS AND RESULTS
is positive or negative. Then the magnitude of θ is equal to that of either φ, or φ plus a straight
angle, respectively.
Continuing with the reciprocal of measure of a sensed-angle, we note that
u12 − u6 + i(v12 − v6 )
u9 − u6 + i(v9 − v6 )
[u12 − u6 + i(v12 − v6 )][u9 − u6 − i(v9 − v6 )]
=
(u9 − u6 )2 + (v9 − v6 )2
(u9 − u6 )(u12 − u6 ) + (v9 − v6 )(v12 − v6 )
=
(u9 − u6 )2 + (v9 − v6 )2
(u9 − u6 )(v12 − v6 ) − (v9 − v6 )(u12 − u6 )
+i
(u9 − u6 )2 + (v9 − v6 )2
p
(u12 − u6 )2 + (v12 − v6 )2
= p
•
(u9 − u6 )2 + (v9 − v6 )2
"
(u9 − u6 )(u12 − u6 ) + (v9 − v6 )(v12 − v6 )
• p
[(u9 − u6 )2 + (v9 − v6 )2 ][(u12 − u6 )2 + (v12 − v6 )2 ]
(u9 − u6 )(v12 − v6 ) − (v9 − v6 )(u12 − u6 )
+i p
[(u9 − u6 )2 + (v9 − v6 )2 ][(u12 − u6 )2 + (v12 − v6 )2 ]
=r(cos φ + i sin φ),
#
(11.9.27)
where
r=
s
(u12 − u6 )2 + (v12 − v6 )2
.
(u9 − u6 )2 + (v9 − v6 )2
We are now in a position to develop a Plücker reciprocal of Pythagoras’ theorem. Continuing with the same notation, we consider
(u12 − u9 )2 + (v12 − v9 )2
=[(u12 − u6 ) − (u9 − u6 )]2 + [(v12 − v6 ) − (v9 − v6 )]2
and for this to be equal to
[(u12 − u6 )2 + (v12 − v6 )2 ] + [(u9 − u6 )2 + (v9 − v6 )2 ]
it is necessary and sufficient that
(u9 − u6 )(u12 − u6 ) + (v9 − v6 )(v12 − v6 ) = 0,
and by (11.8.27) this is equivalent to cos φ = 0 so that OZ6,9 ⊥ OZ6,12 .
To deal with Plücker sensed-ratios, we take Ws = (1−s)W6 +sW9 so that ws −w6 = s(w9 −w6 ).
Then, on the assumption (11.8.20), the Plücker reciprocal of Ws is the line Z6,9 Z ′ , where Z ′ =
(1 − s)Z5 + sZ8 . Now by (11.8.23),
i
δF (Z6,9 , Z5 , Zs′ )
ws − w6 = − z6,9
,
2
δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Zs′ )
δF (Z6,9 , Z5 , Z8 )
i
.
w9 − w6 = − z6,9
2
δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Z8 )
321
11.9. CARTESIAN AND PLÜCKER DUALITY
By division
δF (O, Z6,9 , Zs′ )δF (Z5 , Z6,9 , Z8 )
δF (O, Z6,9 , Z8 )δF (Z5 , Z6,9 , Zs′ )
′
δF (O, Z6,9 , Zs )δF (Z5 , Z6,9 , Z8 ) − δF (O, Z6,9 , Z8 )δF (Z5 , Z6,9 , Zs′ )
δF (O, Z6,9 , Z8 )δF (Z5 , Z6,9 , Zs′ )
δF [(O, Z5 ), (Z6,9 , Zs′ ), (Z6,9 , Z8 )]
δF (O, Z6,9 , Z8 )δF (Z5 , Z6,9 , Zs′ )
δF [(Z6,9 , Z8 ), (O, Z5 ), (Z6,9 , Zs′ )]
δF (O, Z6,9 , Z8 )δF (Z5 , Z6,9 , Zs′ )
δF (Z6,9 , O, Z5 )δF (Z8 , Z6,9 , Zs′ )
δF (Z5 , Z6,9 , Zs′ )δF (O, Z6,9 , Z8 )
δF (Z6,9 , Z5 , Zs′ )δF (Z6,9 , Z8 , O)
δF (Z6,9 , Z5 , O)δF (Z6,9 , Z8 , Zs′ )
and so
crZ6,9 (Z5 , Z8 , Zs′ , O) = −
1
= ,
s
1
= − 1,
s
1−s
=
,
s
1−s
=
,
s
1−s
=
,
s
s
=−
,
1−s
s
.
1−s
(11.9.28)
′
′
In particular, when s = 21 , Z6,9 (Z5 , Z8 , Z1/2
, O) is a harmonic pencil and this identifies Z6,9 Z1/2
as the Plücker reciprocal mid-line of the lines Z4 Z5 and Z7 Z8 .
Now above we have
i
δF (Z6,9 , Z5 , Zs′ )
ws − w6 = − z6,9
,
2
δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Zs′ )
δF (Z6,9 , Z5 , (1 − s)Z5 + sZ8 )
i
= − z6,9
2
δF (O, Z6,9 , Z5 )δF (O, Z6,9 , (1 − s)Z5 + sZ8 )
i
sδF (Z6,9 , Z5 , Z8 )
= − z6,9
.
2
δF (O, Z6,9 , Z5 )[(1 − s)δF (O, Z6,9 , Z5 ) + sδF (O, Z6,9 , Z8 )]
As cross-ratio is invariant under a bilinear transformation, we can conclude from this that for a
range of four points on a line W6 W9 and the pencil of four lines through the point Z6,9 which are
their Plücker reciprocals, the cross-ratio of the four lines in the pencil is equal to the cross-ratio of
the range of points.
To deal with Plücker reciprocation of parallelism, we take points W6 , W9 , W12 , W15 so that
W6 W9 k W12 W15 . We then have w12 − w15 = j(w9 − w6 ) for some non-zero real number j1 . We
suppose that the Plücker reciprocals of these points, are Z4 Z5 , Z7 Z8 , Z10 Z11 , Z13 Z14 , respectively.
Now by (11.8.19),
δ(Z6,9 , Z5 , Z8 )
i
,
w9 − w6 = − z6,9
2
δF (O, Z6,9 , Z5 )δF (O, Z6,9 , Z8 )
i
δ(Z12,15 , Z14 , Z11 )
w12 − w15 = − z12,15
.
2
δF (O, Z12,15 , Z14 )δF (O, Z12,15 , Z11 )
It follows from this that z12,15 = j2 z6,9 for some non-zero real number j2 , and so the points
O, Z6,9 , Z12,15 are collinear. We then say that the pencils on Z6,9 and Z12,15 are Plücker reciprocal
parallel.
To go on to the Plücker reciprocal of a parallelogram, we suppose in the last paragraph that
W6 W15 k W9 W12 . Then j1 = 1 in the foregoing and we also have w15 − w6 = w12 − w9 . It follows
that O, Z6,15 , Z9,12 are collinear, and we say that the pencils at the points Z6,9 , Z12,15 , Z6,15 , Z9,12
form a Plücker reciprocal parallelogram. As a feature of this we recall that the mid-point of W6
and W12 coincides with the mid-point of W9 and W15 . It follows that the mid-line of Z6,12 Z5 and
322
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Z6,12 Z11 coincides with the mid-line of Z9,15 Z8 and Z9,15 Z14 , so that each of these mid-lines is
Z6,12 Z9,15 . Hence each of
Z6,12 (Z5 , Z11 , Z9,15 , O),
Z9,15 (Z8 , Z14 , Z6,12 , O),
is a harmonic pencil. We note from §1.6.1 that (O, Z6,12 , Z9,15 ) is the diagonal triple of the
quadrangle (Z6,9 , Z6,15 , Z9,12 , Z12,15 ) and that this is the polar property of the diagonal triple
dealt with in §4.7.1.
To deal with the Plücker reciprocal of the perpendicular distance from a point to a line, we
start with a pencil, with line equation
x6 u + y6 v + 1 = 0,
(11.9.29)
and a line
Z10 Z11
with line-coordinates (u12 , v12 ).
(11.9.30)
Now a second pencil which is reciprocal perpendicular to (9.8.29) has equation
ty6 u − tx6 v + 1 =0,
1
y6 u − x6 v + =0,
t
for some t 6= 0, and one which contains (11.8.30) will satisfy
y6 u12 − x6 v12 +
1
= 0,
t
and so, on substituting for 1/t, be
y6 (u − u12 ) − x6 (v − v12 ) = 0.
Now (11.8.29) can be re-written as
x6 (u − u12 ) + y6 (v − v12 ) = −(x6 u12 + y6 v12 + 1).
Then for the unique solution in u and v for these equations we have
y62 (u − u6 )2 + x26 (v − v12 )2 − 2x6 y6 ((u − u12 )(v − v12 ) =0,
x26 (u − u6 )2 + y62 (v − v12 )2 + 2x6 y6 ((u − u12 )(v − v12 ) =(x6 u12 + y6 v12 + 1)2 .
By addition, we obtain from this that
p
|x6 u12 + y6 v12 + 1|
p
,
(u − u12 )2 + (v − v12 )2 =
x26 + y62
(11.9.31)
and this is the Plücker reciprocal perpendicular distance from the line (11.8.30) to the pencil
(11.8.21).
11.9.7
Plücker reciprocal of focus-directrix property of ellipse
As in the first argument in §11.8.4, we can find that the ellipse with equation
y2
x2
+ 2 = 1,
2
a
b
where 0 < b < a has tangential equation a2 u2 + b2 v 2 = 1 and so as b2 = a2 (1 − e2 ),
1
u2 + v 2 = e2 v 2 + 2 2 .
a e
(11.9.32)
323
11.9. CARTESIAN AND PLÜCKER DUALITY
As a reciprocal focus we take a line Z4 Z5 which is the reciprocal of the point W6 ≡ (u6 , v6 )
and so has equation (11.8.12), and as a reciprocal directrix we take the pencil through the point
Z9,12 ≡ (x9,12 , y9,12 ), so that it has line equation x9,12 u + y9,12 v + 1 = 0. Then by (11.8.31) a
variable line ZZ ′ with line-coordinates (u, v) satisfies the tangential equation if
(u − u6 )2 + (v − v6 )2 = e2
(x9,12 u + y9,12 v + 1)2
.
2
x29,12 + y9,12
(11.9.33)
We wish to choose our parameters so that this becomes the equation (11.8.32). We first take
x9,12 = u6 = 0 so that (11.8.33) reduces to
e2
e2
v + 2 − v62 .
u2 + v 2 = e2 v 2 + 2 v6 +
y9,12
y9,12
To reduce this further we take v6 = −e2 /y9,12 , giving
"
2
2
u +v =e
2
#
1 − e2
v + 2
.
y9,12
2
We need
1
1 − e2
= 2 2,
2
y9,12
a e
2
y9,12
=a2 e2 (1 − e2 ) = b2 e2 ,
and so obtain (11.8.32) when we make as our final choice y9,12 = be. Thus we have as our reciprocal
focus the fixed line with line coordinates
e
e2
e2
= 0, − ,
= 0, −
(u6 , v6 ) = 0,
y9,12
be
b
that is the line with equation 0.x − eb y + 1 = 0 which is y =
directrix, the pencil line equation
b
e.
Furthermore for the reciprocal
x9,12 u + y9,12 v + 1 = 0
becomes bev + 1 = 0, so the pencil has vertex Z9,12 ≡ (0, be).
We have the variable tangent with line coordinates (u, v) and need to identify the foot of the
reciprocal perpendicular from this to the pencil with vertex Z9,12 . Let Z15 be the point where the
tangent meets the major axis, i.e. the x-axis. Then the pencils on Z9,12 and Z15 are reciprocal
perpendicular as OZ9,12 ⊥ OZ15 . The variable tangent has point equation ux + vy + 1 = 0
1
and so the point in which it meets the x-axis is Z15
≡ − u , 0 . The line Z9,12 Z15 has equation
1
1
ux− be y + 1 = 0, and so has line-coordinates u, − be . It is the foot of the reciprocal perpendicular
from the variable line (u, v) to the reciprocal directrix. The reciprocal focus line with equation
− eb y + 1 = 0 has line coordinates (0, −e/b). Then the reciprocal focus-directrix property gives us
the equation
"
2 #
1
e 2
2
2
2
,
= e (u − u) + v +
(u − 0) + v +
b
be
which reduces to
1 − e2
1
2
2
2
u +v =e v + 2 2
=e v + 2 2 .
b e
a e
2
2
2
To reach the conclusion by using (11.8.31), we have the square of distance from the variable
tangent with line coordinates (u, v) to the reciprocal-focus, the line with coordinates 0, − eb , to
be (u − 0)2 + (v + e/b)2 . Similarly we have the square of the reciprocal perpendicular distance from
324
CHAPTER 11. SIMILARITY METHODS AND RESULTS
the variable tangent (u, v) to the reciprocal-directrix, the pencil with line equation bev + 1 = 0 to
be
(bev + 1)2
,
b2 e2
which yields the equation
e 2
(bev + 1)2
(u − 0)2 + v +
= e2
.
b
b2 e2
This is the same as before.
11.9.8
Plücker reciprocal of sensed area
Given a triple of lines (Z4 Z5 , Z7 Z8 , Z10 Z11 ) which are Plücker reciprocals of the points W6 ≡
(u6 , v6 ), W9 ≡ (u9 , v9 ), W12 ≡ (u12 , v12 ), respectively, we define a reciprocal sensed-area by
u
v6 1 1 6
δF (Z4 Z5 , Z7 Z8 , Z10 Z11 ) = u9 v9 1 .
4
u12 v12 1 We recall
y5 − y4
x5 − x4
, v6 =
,
x4 y5 − x5 y4
x4 y5 − x5 y4
y8 − y7
x8 − x7
u9 = −
, v9 =
,
x7 y8 − x8 y7
x7 y8 − x8 y7
y11 − y10
x11 − x10
u12 = −
, v12 =
.
x10 y11 − x11 y10
x10 y11 − x11 y10
u6 = −
From this we obtain
δF (Z4 Z5 , Z7 Z8 , Z10 Z11 ) =
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )]
1
,
2 δF (O, Z4 , Z5 )δF (O, Z7 , Z8 )δF (O, Z10 , Z11 )
(11.9.34)
by (1.1.20).
This is a variant of Clifford’s projector mentioned in §1.1.4.
11.9.9
Polar reciprocation
Plücker reciprocation is common, but common also is polar reciprocation respect to the unit
circle C1 with equation x2 + y 2 = 1 in which to any point W6 ≡ (u6 , v6 ), not the centre O, there is
made correspond the line which is its polar, that is the line with Cartesian equation
u6 x + v6 y − 1 = 0.
(11.9.35)
This is parallel to the Plücker reciprocal of W6 with equation (11.8.12), and we recall that early
in §11.8.6), what we noted for translations included the fact that, from the case when t = −1, the
translation
u6
v6
x′ = x + 2 2
, y′ = y + 2 2
,
2
u6 + v6
u6 + v62
maps the Plücker reciprocal (11.8.12) onto the polar reciprocal (11.8.35). In fact the polar reciprocal of W6 coincides with the Plücker reciprocal of −W6 , and formulae for polar reciprocation
similar to those for Plücker reciprocation in §§11.8.5 - 11.8.8 can readily be deduced from this fact.
To avoid confusion we adopt the following notation. We suppose that two points on the line
(11.8.35) are R4 ≡ (p4 , q4 ) and R5 ≡ (p5 , q5 ). Then on solving the simultaneous equations
u6 p4 + v6 q4 = 1,
u6 p5 + v6 q5 = 1,
325
11.9. CARTESIAN AND PLÜCKER DUALITY
we obtain
u6 =
q5 − q4
,
p4 q5 − q4 p5
v6 = −
p5 − p4
,
p4 q5 − q4 p5
(11.9.36)
as an analogue of (11.8.11). Similarly we take the line with equation
u9 x + v9 y = 1,
(11.9.37)
and suppose that the points R7 ≡ (p7 , q7 ) and R8 ≡ (p8 , q8 ) lie on it. Then by solving the
simultaneous equations (11.8.35) and (11.8.37), we find that when they are not parallel their point
of intersection is R6,9 ≡ (p6,9 , q6,9 ), where
p6,9 =
v9 − v6
u9 − u6
, q6,9 = −
.
u6 v9 − u9 v6
u6 v9 − u9 v6
(11.9.38)
This is an analogue of (11.8.18). Like (11.8.19) we also have
q5 − q6,9
,
p6,9 q5 − q6,9 p5
q8 − q6,9
u9 =
,
p6,9 q8 − q6,9 p8
u6 =
p5 − p6,9
,
p6,9 q5 − q6,9 p5
p8 − p6,9
v9 = −
.
p6,9 q8 − q6,9 p8
v6 = −
(11.9.39)
Suppose that ψd is the sensed duo-angle from (11.8.35) to (11.8.37). Then the initial argument
in §11.8.6, since it did not involve the constant terms in the equations (11.8.12) and (11.8.17),
also shows that ψd = φd . To deal with the more informative sensed-angles and using the notation
r6,9 = p6,9 + iq6,9 , we have the analogue of (11.8.23)
w9 − w6 = ir6,9
δF (R6,9 , R5 , R8 )
.
2δF (O, R6,9 , R5 )δF (O, R6,9 , R8 )
(11.9.40)
Corresponding to a third point W12 ≡ (u12 , v12 ) we introduce its polar with equation u12 x+ v12 y =
1, suppose that it contains the points R10 ≡ (p10 , q10 ) and R11 ≡ (p11 , q11 ). If we denote by
R6,12 ≡ (p6,12 , q6,12 ) the point of intersection of the polars of W6 and W12 , then from (11.8.40) we
have that
w12 − w6 = ir6,12
δF (R6,12 , R5 , R11 )
,
2δF (O, R6,12 , R5 )δF (O, R6,12 , R11 )
and by a similar argument we obtain as an analogue of (11.8.26) that
r6,12
δF (O, R12,6 , R9,12 ) w12 − w6
=
.
r6,9
δF (O, R6,9 , R9,12 ) w9 − w6
(11.9.41)
If ψ = ∡R6,9 OR12,6 and as before φ = ∡W9 W6 W12 , then we have that exp ψ = ± exp φ, with the
+ or the − according as the coefficient in (11.8.41) is positive or negative.
Thus the identity is proven.
326
11.9.10
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Conics and conic envelopes as images under polar reciprocation
b
b
C2
W
T
b
C1
C
O
Figure 11.8.
We now prove a result which has general application.
We let C1 be the circle with centre the origin O and radius-length 1, so that it has equation
x2 + y 2 = 1. Let C2 be the circle with centre Z1 ≡ (−b, 0) and radius-length c, where b > 0, so
that its equation is (x + b)2 + y 2 = c2 . The point W ≡ (u, v) has as polar p with respect to C1 the
line with equation
ux + vy = 1.
(11.9.42)
On substituting for y from this into the equation for C2 , we obtain the equation
(u2 + v 2 )x2 − 2(u − bv 2 )x + (b2 − c2 )v 2 + 1 = 0.
The condition that this quadratic equation in x will have a repeated root is
(u − bv 2 )2 = (u2 + v 2 )[(b2 − c2 )v 2 + 1].
This can be manipulated to the form
1
c2 (u2 + v 2 ) = (bu + 1)2 = b2 (u + )2 .
b
(11.9.43)
This is the equation of a conic C, with focus O, directrix the line with equation u = −1/b, and
eccentricity b/c. The eccentricity is less than, equal to or greater than 1 according as O is inside,
on, or outside C2 , and so the conic is an ellipse, parabola or hyperbola, respectively. Its equation
can also be expressed as
2
b
u − c2 −b
2
v2
+
= 1.
2
2
2
2
2
c /(c − b )
1/(c − b2 )
As there are no lines of deficient incidence for a circle, this shows that (11.8.42) is a tangent to C2 .
We can verify this as follows.
The repeated root of the quadratic equation gives as the unique point of intersection of the
polar p of W and the circle C2 the point T with coordinates (x, y) = (x′ , y ′ ) where
x′ =
v(1 + bu)
u − bv 2 ′
, y = 2
.
u2 + v 2
u + v2
327
11.9. CARTESIAN AND PLÜCKER DUALITY
Then the polar p (of the point W with respect to C1 ) must also be a tangent to C2 , with T the
point of contact. This can be verified as follows. C2 has equation x2 + y 2 + 2bx + b2 − c2 = 0, so a
point Z1′ ≡ (x′1 , y1′ ) has as polar the line with equation
x′1 x + y1′ y + b(x + x′1 ) + b2 − c2 = 0.
In particular for the point T ≡ (x′ , y ′ ) on C2 , the tangent at T has the equation
x′ x + y ′ y + b(x + x′ ) + b2 − c2 =0,
(1 + bu)v
u − bv 2
u − bv 2
+ b2 − c2 =0,
x
+
y
+
b
x
+
u2 + v 2
u2 + v 2
u2 + v 2
(u − bv 2 )x + (1 + bu)vy + b(u2 + v 2 )x + b(u − bv 2 ) + (b2 − c2 )(u2 + v 2 ) =0.
This simplifies to
ux + vy = − bu2 x − buvy − bu − b2 u2 + c2 (u2 + v 2 )
= − bu(ux + vy) − bu − b2 u2 + (bu + 1)2 ,
(1 + bu)(ux + vy) =1 + bu,
and so ux + vy = 1. Thus the images of the points of C under polar reciprocation in the unit circle
are the tangents to the circle C2 .
Now working in the opposite direction, we note from (11.8.43) that the conic C has equation
(c2 − b2 )x2 − 2bx + c2 y 2 − 1 = 0,
and so the tangent to C, at the point W on it, is the line with equation
(c2 − b2 )ux − b(x + u) + c2 vy − 1 = 0,
that is
[(c2 − b2 )u − b]x + c2 vy = 1 + bu.
Moreover, the polar with respect to the circle C1 of the point T ≡ (x′ , y ′ ) above, is the line with
equation
x′ x + y ′ y = 1,
that is
u − bv 2
v(1 + bu)
x+ 2
y = 1.
u2 + v 2
u + v2
As by (11.8.43)
u − bv 2
(c2 − b2 )u − b
−
= 0,
2
2
u +v
1 + bu
c2 v
v(1 + bu)
−
= 0,
2
2
u +v
1 + bu
this latter is the equation we have just noted for the tangent to C at W . Thus the polar of T with
respect to the circle C1 is also the tangent to the conic C at the point W . Hence the images of the
points of the circle C2 under polar reciprocation in the unit circle are the tangents to C.
328
11.9.11
CHAPTER 11. SIMILARITY METHODS AND RESULTS
A tangent to a conic meeting the directrix
R
b
b
C2
b
Z1
b
C1
W
T
b
O
C
Figure 11.9.
We now illustrate how reciprocation allows us to prove properties of conics. With the situation
as in §11.8.11, we note first that the polar of the centre Z1 of the circle C2 is the directrix of C.
Suppose that the tangent to the conic C at the point W meets the directrix at the point R. The
R is on the polar of Z1 and on the polar of T , so the polar of R is the line Z1 T . The polars of W
and R are perpendicular to each other so, by §11.8.5, OW and OR are perpendicular. Thus if the
tangent at W meets the directrix at R, the angle ∠W OR is a right-angle, O being the focus.
11.9.12
Analogue of constant-sized angle property of a circle
With the notation of §11.8.10, we take fixed points W9 and W12 , and a variable point W6 , all on a
circle C2 . Then, by [2, §10.9], the sensed angle ∡W9 W6 W12 is of constant magnitude as W6 varies
on C2 but does not cross the line W9 W12 . Suppose that the polars of W6 , W9 , W12 with respect to
the circle C1 are R4 R5 , R7 R8 , R10 R11 , respectively, and that R4 R5 and R7 R8 meet at R6,9 , R7 R8
and R10 R11 meet at R9,12 , and R10 R11 and R4 R5 meet at R12,6 . Then these polars of W6 , W9 , W12
are tangents to the conic C, with the variable one R4 R5 meeting the fixed ones R7 R8 and R10 R11
at the points R6,9 and R12,6 , respectively.
Now by (11.4.41)
δF (O, R12,6 , R9,12 ) w12 − w6
r12,6
=
,
r6,9
δF (O, R6,9 , R9,12 ) w9 − w6
so ∡R6,9 OR12,6 is constantly equal in measure to ∡W9 W6 W12 , or to ∡W9 W6 W12 plus a straightangle, provided that
δF (O, R12,6 , R9,12 )δF (O, R6,9 , R9,12 )
does not change sign.
Now as with Plücker reciprocation in §11.8.8,
δF (W6 , W9 , W12 ) =
1
δF [(R4 , R5 ), (R7 , R8 ), (R10 , R11 )]
,
2 δF (O, R4 , R5 )δF (O, R7 , R8 )δF (O, R10 , R11 )
11.10. NINE-POINT HOMOTHETIC CONIC
329
as we are essentially just multiplying each of the first two columns of the determinant by −1. We
may apply this with
(R4 , R5 ) = (R6,9 , R12,6 ), (R7 , R8 ) = (R9,12 , R6,9 ), (R10 , R11 ) = (R12,6 , R9,12 ),
and so obtain
1
δF [(R6,9 , R12,6 ), (R9,12 , R6,9 ), (R12,6 , R9,12 )]
2 δF (O, R6,9 , R12,6 )δF (O, R9,12 , R6,9 )δF (O, R12,6 , R9,12 )
δF (R6,9 , R12,6 , R9,12 )2
1
.
=−
2 δF (O, R6,9 , R12,6 )δF (O, R9,12 , R6,9 )δF (O, R12,6 , R9,12 )
δF (W6 , W9 , W12 ) =
Thus for W6 not to cross the line W9 W12 we need
δF (O, R6,9 , R12,6 )δF (O, R9,12 , R6,9 )δF (O, R12,6 , R9,12 )
not to change sign, and earlier we needed
δF (O, R12,6 , R9,12 )δF (O, R6,9 , R9,12 )
not to change sign. It follows that δF (O, R6,9 , R12,6 ) must not change sign. Then ∡R6,9 OR12,6 is
of constant measure.
11.9.13
Five tangents to a conic
Suppose that R4 R5 , R7 R8 , R10 R11 , R13 R14 are fixed tangents to a conic C and that RR′ is a
variable tangent to it. We suppose that we have the situation in §9.8.10 and that these have poles
W6 , W9 , W12 , W15 , W with respect to the circle C1 . Then these points all lie on the circle C2 and
so by (7.9.3) we have
δF (W, W6 , W12 )δF (W, W9 , W15 )
= k,
δF (W, W6 , W15 )δF (W, W9 , W12 )
for some constant k. Then by applying four times the analogue of §11.8.8 for polar reciprocation,
we can deduce that
δF [(R, R′ ), (R4 , R5 ), (R10 , R11 )]δF [(R, R′ ), (R7 , R8 ), (R13 , R14 )]
= k.
δF [(R, R′ ), (R4 , R5 ), (R13 , R14 )]δF [(R, R′ ), (R7 , R8 ), (R10 , R11 )]
11.10
Nine-point homothetic conic
11.10.1
Given a proper conic C0 , with Cartesian equation
a0 x2 + 2h0 xy + 2b0 y 2 + 2g0 x + 2f0 y + c0 = 0,
any conic with an equation of the form
a0 x2 + 2h0 xy + 2b0 y 2 + 2gx + 2f y + c = 0,
is homothetic to C0 .
Given any non-collinear points Z1 , Z2 , Z3 , it is clearly possible to choose g, f, c so that the
homothetic conic passes through these three points.
330
11.10.2
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Conjugate directions
Given the line l1 x′ + m1 y ′ = 0, the line through Z ≡ (x, y) and parallel to this has parametric
equations
x′ = x + m1 t, y ′ = y − l1 t, (t ∈ R).
This meets the conic C0 when
a0 (x + m1 t)2 + 2h0 (x + m1 t)(y − l1 t) + b0 (y − l1 t)2 + 2g0 (x + m1 t) + 2f0 (y − l1 t) + c0 = 0.
For Z to be the mid-point of the points of intersection we need the coefficient of t to vanish, that
is
(a0 m1 − h0 l1 )x + (h0 m1 − b0 l1 )y + g0 m1 − f0 l1 = 0.
This is a line as Z varies and in fact is a diameter. It is parallel to l2 x + m2 y = 0 when
a0 m1 m2 + b0 l1 l2 − h0 (l1 m2 + m1 l2 ) = 0.
This is the condition that the lines with equations
l1 x + m1 y = 0, l2 x + m2 y = 0,
are parallel to a pair of conjugate diametral lines of C0 . This condition depends only on a0 , b0 , h0
and so is the same for all conics homothetic to C0 .
11.10.3
Nine-point homothetic conics
To simplify the calculations, we now suppose the frame of reference chosen so that
Z1 ≡ (0, p), Z2 ≡ (q, 0), Z3 ≡ (r, 0).
The conic homothetic to C0 which passes through the mid- points of the sides of [Z1 , Z2 , Z3 ]
has Cartesian equation
1
1 a0 qr
1
2
2
a0 x + 2h0 xy + b0 y − (a0 q + a0 r + 2h0 p)x +
− 2h0 q − 2h0 r − b0 p y + h0 p(q + r) = 0.
2
2
p
2
λ
1
Z2 + 1+λ
Z3 so that the lines Z2 Z3 and Z1 Z4 are in conjugate directions
Now take Z4 = 1+λ
with respect to the conic. The condition for this is
a0 (q + rλ) + b0 .0 − h0 p(1 + λ) = 0,
so that
λ=
−a0 q + h0 p
.
a0 r − h0 p
x4 =
h0 p
, y4 = 0.
a0
Hence
µ
1
Also take Z5 = 1+µ
Z3 + 1+µ
Z1 so that the lines Z3 Z1 and Z2 Z5 are in conjugate directions
with respect to the conic. The condition for this is
a0 r(r − q − qµ) − b0 p2 µ − h0 [p(r − q − qµ) − rpµ = 0,
so that
µ=
Hence
x5 =
a0 rq + b0
p2
a0 r2 − a0 rq − h0 pr + h0 pq
.
a0 rq + b0 p2 − h0 pq − h0 pr
r(a0 rq + b0 p2 − h0 pq − h0 pr)
,
− h0 pq − h0 pr + a0 r2 − a0 rq − h0 pr + h0 pq
331
11.10. NINE-POINT HOMOTHETIC CONIC
y5 =
p(a0 r2 − a0 rq − h0 pr + h0 pq)
.
a0 rq + b0 p2 − h0 pq − h0 pr + a)r2 − a0 rq − h0 pr + h0 pq
ν
1
Z1 + 1+ν
Z2 so that the lines Z1 Z2 and Z3 Z6 are in conjugate
We could also take Z6 = 1+ν
directions with respect to the conic.
The points Z4 , Z5 , Z6 are the analogues of the feet of the perpendiculars from the vertices to
the opposite sides. It can be checked that all three lie on the homothetic conic above. It can be
checked that the lines Z1 Z4 , Z2 Z5 , Z3 Z6 are concurrent, yielding an analogue of the orthocentre
of the triangle. It can be checked that the point of intersection of Z1 Z4 and Z2 Z5 is the point Z7
with
x7 =
−a0 h0 rp + a20 rq + h20 p2 − a0 h0 pq)
h0 (a0 rq + b0 p2 − h0 pq − h0 pr)
,
y
=
−
.
7
p(−h20 + a0 b0 )
p(−h20 + a0 b0 )
It can now be checked that the mid-points of {Z1 , Z7 }, {Z2 , Z7 } and {Z3 , Z7 } are all on the homothetic conic.
Thus we have found nine points on this homothetic conic, in generalization of a classical result
on the nine-points circle.
11.10.4
Miquel’s theorem
We now generalize to homothetic conics Miquel’s theorem which was noted in Barry [2, Chapter
11]. Our method is one of brute force, using symbolic manipulation on a computer. As in the last
subsection, to simplify the calculations, we suppose the frame of reference chosen so that
Z1 ≡ (0, p), Z2 ≡ (q, 0), Z3 ≡ (r, 0),
and we take points
Z4 ≡ ((1 − u)q + ur, 0), Z5 ≡ ((1 − v)r, rp), Z6 ≡ (wq, (1 − w)p),
on the side-lines Z2 Z3 , Z3 Z1 , Z1 Z2 , respectively. The conic homothetic to C0 which passes through
the mid- points of the sides of [Z1 , Z2 , Z3 ] has Cartesian equation
1
1
a0 x + 2h0 xy + b0 y − (a0 q + a0 r + 2h0 p)x +
2
2
2
2
a0 qr
1
− 2h0 q − 2h0 r − b0 p y + h0 p(q + r) = 0.
p
2
The homothetic conic C1 passing through Z1 , Z5 , Z6 has coefficients
g1
f1
c1
1 −2h0 rvp + b0 vp2 + a0 r2 v − a0 r2 − b0 p2 + a0 wq 2 + 2h0 qp − 2h0 wqp + b0 p2 w
,
2
−r + q
1 −2h0 rvpq + b0 vp2 q + a0 r2 vq − a0 r2 q − 2b0 rp2 + a0 rq 2 + 2h0 rqp − 2h0 rwqp + b0 rp2 w + b0 p2 q
= −
,
2
p(−r + q)
−b0 rp2 − 2h0 rvpq + b0 vp2 q + a0 r2 vq − a0 r2 q + a0 rwq 2 + 2h0 rqp − 2h0 rwqp + b0 rp2 w
=
.
−r + q
= =−
Similarly the homothetic conic C2 passing through Z2 , Z4 , Z6 has coefficients
g2
=
f2
=
c2
=
1
1
= −a0 q + a0 qu − a0 ur,
2
2
1 a0 wq 2 + b0 p2 w − 2h0 wqp − b0 p2 − a0 q 2 + a0 q 2 u − a0 qur
,
2
p
a0 q 2 − a0 q 2 u + a0 qur.
332
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Similarly the homothetic conic C3 passing through Z3 , Z4 , Z5 has coefficients
g3
=
f3
=
c3
=
a0
(qu − ur − q − r),
2
1 a0 r2 − a0 r2 v − 2h0 rp + 2h0 rvp − b0 vp2 + a0 qur − a0 ur2 − a0 qr
,
2
p
a0 (−qur + ur2 + qr).
=
If we solve the equations of the common secant lines of C2 and C3 , and of C3 and C1 , we obtain
ponderous coordinates (x7 , y7 ) of their point of intersection Z7 . It can be verified that Z7 lies on
each of the homothetic conics.
11.10.5
First Lemoine circle
In the last quarter of the 19th century there was a flourishing of results concerning triangles and
circles. We continue our approach of showing that the natural generalization of those results is to
homothetic conics by generalizing the first Lemoine circle. For it we use a generalization of the
notion of isogonal conjugates as used in Barry [2, Chapter 11].
As before, to simplify the calculations we suppose the frame of reference chosen so that
Z1 ≡ (0, p), Z2 ≡ (q, 0), Z3 ≡ (r, 0).
Then the homothetic conic through the vertices of this triangle has Cartesian equation with coefficients
g1
=
f1
=
c1
=
a0
(q + r),
2
1 b0 p2 + a0 qr
,
−
2
p
a0 qr.
−
The line through Z1 and the mid-point W1 of Z2 and Z3 meets the conic again at the point
W2 ≡ (u2 , v2 ) and the line through W2 parallel to Z2 Z3 meets the conic again at the point
W3 ≡ (u3 , v3 ) where
u3
=
v3
=
b0 qp2 + b0 rp2 + a0 r2 q − 4h0 rpq + a0 rq 2
,
4b0 p2 − 4h0 pq − 4h0 pr + 2a0 qr + a0 q 2 + a0 r2
a0 p(r2 − 2qr + q 2 )
,
4b0 p2 − 4h0 pq − 4h0 pr + 2a0 qr + a0 q 2 + a0 r2
2
The line Z1 W3 is then the analogue of a symmedian through Z1 .
The line through Z2 and the mid-point W4 of Z3 and Z1 meets the conic again at the point
W5 ≡ (u5 , v5 ), and the line through W5 parallel to Z3 Z1 meets the conic again at the point
W6 ≡ (u6 , v6 ) where
u5
=
v5
=
2a0 rq 2 − 3a0 r2 q − b0 qp2 + a0 r3 + b0 rp2
,
4a0 q 2 − 4a0 qr + a0 r2 − 4h0 pq + 2h0 pr + b0 p2
p(2a0 q 2 − 2a0 qr − 2h0 pq + a0 r2 + b0 p2 )
,
4a0 q 2 − 4a0 qr + a0 r2 − 4h0 pq + 2h0 pr + b0 p2
and
u6 = u5 + rs2 , v6 = v5 − ps2 ,
where
s2 =
4a0
q2
a0 r2 − 2a0 qr + 2h0 pq − b0 p2
.
− 4a0 qr + a0 r2 − 4h0 pq + 2h0 pr + b0 p2
11.10. NINE-POINT HOMOTHETIC CONIC
333
Then Z2 W6 is an analogue of a symmedian through Z2 .
The line through Z3 and the mid-point W7 of Z1 and Z2 meets the conic again at the point
W8 ≡ (u8 , v8 ), and the line through W8 parallel to Z1 Z2 meets the conic again at the point
W9 ≡ (u9 , v9 ) where
u8
=
v8
=
2a0 r2 q − 3a0 rq 2 − b0 rp2 + a0 q 3 + b0 qp2
,
4a0 r2 − 4a0 qr + a0 q 2 − 4h0 pr + 2h0 pq + b0 p2
p(2a0 r2 − 2a0 qr − 2h0 pr + a0 q 2 + b0 p2 )
,
4a0 r2 − 4a0 qr + a0 q 2 − 4h0 pr + 2h0 pq + b0 p2
and
u9 = u8 + qs3 , v9 = v8 − ps3 ,
where
s3 =
4a0
r2
a0 q 2 − 2a0 qr + 2h0 pr − b0 p2
.
− 4a0 qr + a0 q 2 − 4h0 pr + 2h0 pq + b0 p2
Then Z3 W9 is an analogue of a symmedian through Z3 .
It can be checked that the lines Z1 W3 , Z2 W6 , Z3 W9 are concurrent at the point Z4 ≡ (x4 , y4 )
where
x4
=
y4
=
1 b0 qp2 + b0 rp2 + a0 r2 q − 4h0 rpq + a0 rq 2
,
2 b0 p2 − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
a0 p(r2 − 2qr + q 2 )
1
,
2 b0 p2 − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
which is then an analogue of the symmedian point.
The line through Z4 and parallel to Z2 Z3 meets Z3 Z1 in the point Z5 ≡ (x5 , y5 ) where
x5 = x4 + t1 , y5 = y4 ,
and
t1 =
1 −b0 qp2 + b0 rp2 − a0 r2 q + 2h0 rpq + a0 r3 − 2h0 pr2
.
2
b0 p2 − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
This same line meets the line Z1 Z3 in the point Z6 ≡ (x6 , y6 ) where
x6 = x4 + t2 , y6 = y4 ,
and
t2 = −
1 −b0 qp2 + b0 rp2 + a0 rq 2 − 2h0 rpq − a0 q 3 + 2h0 pq 2
.
2
b0 p2 − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
The line through Z4 and parallel to Z3 Z1 meets Z1 Z2 in the point Z7 ≡ (x7 , y7 ) where
x7 = x4 + rt3 , y7 = y4 − pt3 ,
and
t3 = −
a0 q 2 + b0 p2 − 2h0 pq
1
.
2
2 b0 p − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
This same line meets the line Z2 Z3 in the point Z8 ≡ (x8 , y8 ) where
x8 = x4 + rt4 , y8 = 0,
and
t4 =
a0 (r2 − 2qr + q 2 )
1
.
2
2 b0 p − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
334
CHAPTER 11. SIMILARITY METHODS AND RESULTS
The line through Z4 and parallel to Z1 Z2 meets Z2 Z3 in the point Z9 ≡ (x9 , y9 ) where
x9 = x4 + qt5 , y9 = 0,
and
t5 =
1
a0 (r2 − 2qr + q 2 )
.
2 b0 p2 − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
This same line meets the line Z3 Z1 in the point Z10 ≡ (x10 , y10 ) where
x10 = x4 + qt6 , y10 = y4 − pt6 ,
and
t6 = −
a0 r2 − 2h0 pr + p2
1
.
2
2 b0 p − h0 pr − h0 pq − a0 qr + a0 q 2 + a0 r2
By finding the equation of the homothetic conic which passes through Z5 , Z8 and Z9 , we can
verify that this conic also passes through Z6 , Z7 and Z10 . This homothetic conic is an analogue of
the first Lemoine circle.
11.10.6
Components along a vector and its supplement
Let us consider the situation in Barry [2, §11.6]
z − z2 = (p + qı)(z3 − z2 ),
where Z2 and Z3 are fixed, as a correspondence between points with coordinates (p, q) and points
with coordinates (x, y). Then
x − x2 + ı(y − y2 ) = (p + ıq)[x3 − x2 + ı(y3 − y2 )],
so that
x − x2
y − y2
=
=
(x3 − x2 )p − (y3 − y2 )q,
(y3 − y2 )p + (x3 − x2 )q.
Then by §9.2.1, the correspondence (p, q) → (x, y) has the format of a similarity transformation
which preserves orientation, from its form and as
x3 − x2 −(y3 − y2 ) y3 − y2 (x3 − x2 ) > 0.
We can thus treat (p, q) like Cartesian coordinates, and quote results involving similarity invariants for them.
Exercises
9.1 If a conic is inscribed in a triangle, the lines joining the vertices to the point of contact with
the opposite side-lines are concurrent.
9.2 The reciprocal of the ellipse
y2
x2
+
=1
a2
b2
with respect to the circle C(O; k) is the ellipse
a2 x2 + b2 y 2 = k 4 .
11.10. NINE-POINT HOMOTHETIC CONIC
335
9.3 If a variable tangent to a conic with focus F cut two fixed tangents at points P and Q, the
wedge-angle ∠P F Q has constant magnitude.
9.4 If from the focus O of a central conic, perpendiculars are drawn to two parallel tangents, the
product of their lengths is constant.
9.5 If from an interior point P of a circle perpendiculars are let fall on any two tangents, whose
chord of contact passes through P , the sum of the reciprocals of the perpendicular distances
is constant.
9.6 A proper conic which passes through the origin has the equation in Cartesian coordinates
ax′2 + 2hx′ y ′ + by ′2 + 2gx′ + 2f y ′ = 0.
If the line with equation y ′ = tx′ meets the conic again at the point with coordinates (x, y),
show that
g + ft
g + ft
, y = −2t
.
x = −2
2
a + 2ht + bt
a + 2ht + bt2
This yields parametric equations for the conic.
9.7 Show that any similarity transformation can be put in one of the forms
W − W1 = p(Z − Z1 ) + q(Z − Z1 )⊥ ,
or
W − W1 = p(Z − Z1 )⊢ + q(Z − Z1 )⊣ ,
on using the notation of exercise ????.
9.8 Suppose that Z2 and Z3 are fixed distinct points, and l is a fixed line with Cartesian equation
ax + by + c = 0. Find the locus of the point Z such that the mid-line of |Z2 ZZ3 is parallel
to l. Hence find for what Z this mid-line is l.
336
CHAPTER 11. SIMILARITY METHODS AND RESULTS
Chapter 12
The Complex Euclidean Plane
12.1
12.1.1
Complexification
For a point O ∈ RE2 = Π, let V(RE2 , O) be the set of all couples (O, Z) for Z ∈ RE2 . Then as
in Barry [2, pp. 174-178] the sum (O, Z1 ) + (O, Z2 ) = (O, Z3 ) and the product by a real number
t.(O, Z1 ) = (O, Z4 ) can be defined so that V(RE2 , O) is a vector space over the field R. Given a
framework F = ([O, I , [O, J ) for the plane RE2 , with Cartesian coordinates (x1 , y1 ), (x2 , y2 ) for
the points Z1 , Z2 , we have coordinates (x1 + x2 , y1 + y2 ), (tx1 , ty1 ) for Z3 and Z4 , respectively. We
−→
can use the notation OZ for (O, Z), calling this a position vector. Once O is understood we can
→
−
abbreviate this vector notation to Z , and it is also convenient to use the notation E for V(RE2 , O).
→ −
−
→
Now the Cartesian product E 2 = E × E is the set of ordered pairs ( Z , W ). To form what is called
the complexification E C of E, we first define vector addition on E 2 by
−
→ −→
→ −
→ −
−
→
→ −
−
→ −→
( Z , W ) + (Z ′ , W ′ ) = ( Z + Z ′ , W + W ′ ).
To start on the task of introducing complex number multiples of elements of E 2 , for each pair
→ −
−
→
(λ, µ) of real numbers and each ( Z , W ) ∈ E 2 , we define
→ −
−
→
→
−
−
→ −
→
→
−
(λ, µ)( Z , W ) = (λ Z − µW , λW + µ Z ),
which is an element of E 2 .
It is straightforward to establish two distributive properties, first
→ −
→ −
−
→ −→
(λ, µ)( Z + Z ′ , W + W ′ )
→
→
→ −
−
−
→ −→
−
→ −→
→ −
−
=(λ[ Z + Z ′ ] − µ[W + W ′ ], λ[W + W ′ ] + µ[ Z + Z ′ ])
−
→
−→ −→
−
→
→
−
−
→ −
→
→
−
=(λ Z − µW , λW + µ Z ) + (λZ ′ − µW ′ , λW ′ + µZ ′ )
−
→ −→
→ −
−
→
=(λ, µ)( Z , W ) + (λ, µ)(Z ′ , W ′ ),
so that
−
→ −→
−
→ −→
→ −
−
→
→ −
−
→
(λ, µ)[( Z , W ) + (Z ′ , W ′ )] = (λ, µ)( Z , W ) + (λ, µ)(Z ′ , W ′ ).
Secondly, with addition of pairs of real numbers defined by
(λ, µ) + (λ′ , µ′ ) = (λ + λ′ , µ + µ′ ),
337
(12.1.1)
338
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
we have that
→ −
−
→
[(λ, µ) + (λ′ , µ′ )]( Z , W )
→ −
−
→
=(λ + λ′ , µ + µ′ )( Z , W )
→
−
−
→
−
→
→
−
=([λ + λ′ ] Z − [µ + µ′ ]W , [λ + λ′ ]W + [µ + µ′ ] Z )
→
−
−
→
−
→
→
−
→
−
−
→ −
→
→
−
→W + µ′ Z )
=(λ Z − µW , λW + µ Z ) + (λ′ Z − µ′ W , λ′ −
→ −
−
→
→ −
−
→
=(λ, µ)( Z , W ) + (λ′ , µ′ )( Z , W ).
Moreover, from (12.1.1), for λ′ , µ′ ∈ R, we have that
→ −
−
→
(λ′ , µ′ )[(λ, µ)( Z , W )]
→
−
−
→ −
→
→
−
=(λ′ , µ′ )(λ Z − µW , λW + µ Z )
→
−
−
→
−
→
→
−
−
→
→
−
→
−
−
→
=(λ′ [λ Z − µW ] − µ′ [λW + µ Z ], λ′ [λW + µ Z ] + µ′ [λ Z − µW ])
→
−
−
→
→
−
−
→
=((λ′ λ − µ′ µ) Z − (λ′ µ + µ′ λ)W , (λ′ µ + µ′ λ) Z + (λ′ λ − µ′ µ)W )
→ −
−
→
=(λ′ λ − µ′ µ, λ′ µ + µ′ λ)( Z , W ).
If we make the definition
(λ′ , µ′ )(λ, µ) = (λ′ λ − µ′ µ, λ′ µ + µ′ λ),
(12.1.2)
then we have the associative law
→ −
−
→
→ −
−
→
(λ′ , µ′ )[(λ, µ)( Z , W )] = [(λ′ , µ′ )(λ, µ)]( Z , W ).
With the above addition of pairs of real numbers combined with (12.1.2) we have W. R. Hamilton’s construction of complex numbers. We recall that (λ, µ) = (λ, 0) + (0, µ) and by (12.1.1)
(0, µ) = (0, 1)(µ, 0), so that
(λ, µ) = (λ, 0) + (0, 1)(µ, 0),
while by (10.1.2) (0, 1)2 = (0, 1)(0, 1) = (−1, 0). Now couples of the form (λ, 0), where λ ∈ R, form
a field which is an isomorphic copy of R in the correspondence λ → (λ, 0). Then (λ, 0) + (0, 1)(µ, 0)
corresponds to λ + iµ where i2 = −1. Thus our pairs of real numbers (λ, µ), under the definition
of sums and products, constitute an isomorphic copy of the field C of complex numbers.
We can now conclude that E 2 with the defined vector addition and these multiples by λ + iµ is
→ −
−
→
a vector space over C. We denote it by E C . We shall henceforth use the notation (λ + iµ)( Z , W )
→ −
−
→
for (λ, µ)( Z , W ) in (12.1.1).
We can write the elements of E C as follows,
→ −
−
→
→ −
−
→
→ −
−
→
( Z , W ) =( Z , O ) + ( O , W )
→ −
−
→
−
→ −
→
=( Z , O ) + (0, 1)(W , O )
→ −
−
→
−
→ −
→
=( Z , O ) + i(W , O ).
→ →
−
−
→
−
This brings to the fore pairs of the form ( Z , O ) where Z ∈ E, the set of which we denote by F .
Now
→ −
−
→
→ −
−
→
(λ + iµ)( Z , O ) = (λ Z , µ Z ),
and these are not all in F , so that F is not a vector space over C. However
−
→ −
→ −
→ −
−
→
→
→ −
−
→
→ −
−
→
→ −
−
→
( Z , O ) + (Z ′ , O ) = ( Z + Z ′ , O ), and λ( Z , O ) = (λ Z , O ),
339
12.1.
→
−
→ −
for all Z , Z ′ ∈ E and all λ ∈ R, so that F is a vector space over R. Moreover the correspondence
−
→
→
−
Z → ( Z , O) is an isomorphism from E to F . In the light of this we are justified in identifying
→
−
→
−
( Z , O) with Z in this context. Thus we can take our elements of E C to be pairs
−
→
−
→
→
−
→
−
→
−
→
−
Z + iW =x I + y J + i(u I + v J )
→
−
→
−
=(x + iu) I + (y + iv) J .
If we take u = v = 0 we see that we obtain the objects of E so that E ⊂ E C . We then consider
adjoining objects to RE2 so that the new objects formally have coordinates (x + iu, y + iv) with
respect to F , where either u 6= 0 or v 6= 0 or both, and the whole ensemble of points and new objects
are to be manipulated as are the elements of E C . We call the ensuing set the complex Euclidean
plane and denote it by CE2 . We then have items which have complex numbers (x + iu, y + iv) as
coordinates with respect to F , all of which we think of as points in our new context. As RE2 ⊂ CE2
we call the points of RE2 real points and the new objects non-real points.
For complexification of a real vector space we refer to Halmos [pp. 150-151]. Frenkel [12] and
Berger [3] provide general comments on complexification and employ it in more general geometrical
contexts than ours.
With this preparation we thus have embarked on a major generalization. As the above extension
may seem complicated and un-necessary, we provide a re-capitulation as follows. Given a framework
F for the plane RE2 , we have had Cartesian coordinates (x, y) for all the points Z of RE2 . In this
x and y were real numbers, and we now consider pairs (x + iu, y + iv) of complex numbers and
regard each such pair as representing a point. The set of all such points we call the complex
Euclidean plane and we denote it by CE2 . New points, which are in CE2 but not in RE2 , we
shall call non-real points.
To fit in with common notation we now write, instead of the above, Z = (x, y) where x, y ∈ C
and Z ∈ E C . When we need to denote the real and imaginary parts of these complex coordinates
we write x = ℜx + iℑx. We proceed to introduce concepts which are analogous to those for RE2 .
In this we are guided by two principles. First when we put ℑx = ℑy = 0 so that we are reverting to
real points, our new concepts must be expected to reduce to familiar real concepts. Generally this
will be evident to the reader. However we must be prepared for some new unfamiliar features, and
attention will be drawn to these. Secondly, led by von Staudt, we must expect that our concepts
do not depend on a particular choice of a frame of reference F for the plane RE2 , and so we must
check that we are led to analogous concepts if we take any similarly-oriented frame of reference
F ′ for the plane RE2 , when we apply a displacement consisting of a rotation and a translation
which maps F onto F ′ . It is convenient not to perform this checking piecemeal as concepts are
introduced but to leave it to a retrospective checking in a late subsection in this chapter. For a
similar, somewhat terse, approach we refer to Darboux [6].
12.1.2
The trigonometric tangent function
To start with we need as a pre-requisite the treatment of the tangent function as commonly given
in courses in complex analysis. We recall that when tan z is considered there, z is a complex
number and not an angle, although there is a connection which traces back to radian measure of
real angles. We lay out as follows the detail which we need. Given an arbitrary complex number
u + iv, we wish to specify the complex number θ = φ + iψ from the equation
tan θ = u + iv,
(12.1.3)
where u v, φ and ψ are real numbers. Assuming an elementary knowledge of complex analysis on
the part of the reader, we recall that
tan θ =
sin θ
,
cos θ
sin θ =
1 iθ
e − e−iθ ,
2i
cos θ =
1 iθ
e + e−iθ ,
2
340
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
where ez = exp(z) is the complex exponential. Thus we need
eiθ − e−iθ
=i(u + iv),
eiθ + e−iθ
(eiθ )2 − 1
=i(u + iv),
(eiθ )2 + 1
1 + i(u + iv)
.
e2iθ =
1 − i(u + iv)
(12.1.4)
Now the final fraction here does not exist if its denominator is equal to 0, that is if u+iv = 1/i = −i.
Moreover we recall that the exponential function never takes the value 0 and so the numerator in
the final fraction cannot be equal to 0 either, that is we are not allowed to have u + iv = −1/i = i.
Thus neither of the numbers i and −i can occur as the value of the tangent function.
Continuing we have
1 − v + iu
.
(12.1.5)
e2i(φ+iψ) = e−2ψ e2iφ =
1 + v − iu
As a first application of (12.1.5) we equate the squares of moduli of the two sides to get
e−4ψ =
(1 − v)2 + u2
.
(1 + v)2 + u2
(12.1.6)
As a second application of (12.5) we note that
1 − v + iu 2ψ
(1 − v + iu)(1 + v + iu) 2ψ
e
e =
1 + v − iu
(1 + v)2 + u2
p
1 − v 2 − u2 + 2ui (1 + v)2 + u2
p
=
,
(1 + v)2 + u2
(1 − v)2 + u2
e2iφ =
from which we obtain
1 − v 2 − u2
p
,
cos 2φ = p
(1 + v)2 + u2 (1 − v)2 + u2
2u
p
sin 2φ = p
,
2
(1 + v) + u2 (1 − v)2 + u2
2u
tan 2φ =
.
1 − v 2 − u2
(12.1.7)
As tan(θ + π) = tan θ in all cases when it exists, for a given u + iv, if θ satisfies (12.1.3) then so
also does θ+nπ for every integer n. To obtain a unique θ from (12.1.3) it is necessary to restrict the
set of numbers that we allow for it. This can be done in many ways and our selection is based on
our aim of applying our material to the generalization of duo-angles from the real to the complex
case. One straightforward approach would be to take any φ1 ∈ R and then take θ = φ + iψ where
φ1 ≤ φ < φ1 + π, in effect restricting the domain of our tangent function in the complex plane to
a vertical strip of width π, closed on the left and open on the right. However for our geometry to
reduce to the real case we need a slightly more subtle approach as follows.
We recall the definition of lexicographic order ≺L on C whereby φ1 +iψ1 ≺L φ2 +iψ2 if either
φ1 < φ2 and ψ1 and ψ2 are unrestricted or φ1 = φ2 and ψ1 < ψ2 . We also write φ1 +iψ1 4L φ2 +iψ2
when either φ1 + iψ1 ≺L φ2 + iψ2 or φ1 + iψ1 = φ2 + iψ2 .
With this notation we define the set T of complex numbers φ + iψ as those which satisfy
0 4L φ + iψ ≺L π.
(12.1.8)
As tan θ is not defined when cos θ = 0 and the only such number in T is 12 π, we define
T ∗ = T \ { 21 π}. Then for θ = φ + iψ ∈ T ∗ we now study the values of tan θ from (12.1.7).
341
12.1.
We suppose first that u 6= 0 and so sin 2φ 6= 0. We deduce from (12.1.7) that
cos2 φ =1 + cos 2φ
1 − v 2 − u2
p
= 21 + 21 p
,
2
(1 + v) + u2 (1 − v)2 + u2
s
1 − v 2 − u2
p
.
cos φ = ± 12 + 21 p
(1 + v)2 + u2 (1 − v)2 + u2
Thus we see that we need to take

!
r

2 −u2

1
1
1−v

√
√
if 0 < φ < π2 ,

 arccos
2 + 2
(1+v)2 +u2 (1−v)2 +u2
!
φ=
r

2
2

1
1
1−v √
−u

if π2 < φ < π.

 arccos − 2 + 2 √
2
2
2
2
(1+v) +u
We combine this with
ψ = − 14 ln
(12.1.9)
(1−v) +u
(1 − v)2 + u2
,
(1 + v)2 + u2
(12.1.10)
from (12.1.6)
We next suppose that u = 0 and v 2 6= 1, and so sin 2φ = 0. For φ ∈ T we have that φ can be
one of 0, π2 , π. For φ = 0 we have
tan θ = tan iψ =
Thus v =
e2ψ −1
e2ψ +1
1 ei(iψ) − e−i(iψ)
e2ψ − 1
.
=
i
i ei(iψ) + e−i(iψ)
e2ψ + 1
and we note that as ψ increases on [0, ∞), v increases on [0, 1).
2ψ
−1
When φ = π we note that tan(π + iψ) = tan(iψ) = i ee2ψ +1
as with φ = 0. Again v =
as ψ increases on (−∞, 0), v increases on (−1, 0).
Finally when φ = π2 , we have that
tan θ =
e2ψ −1
e2ψ +1
and
sin( π2 + iψ)
e2ψ + 1
cos iψ
= i 2ψ
.
=
π
cos( 2 + iψ)
− sin iψ
e −1
2ψ
Thus v = ee2ψ +1
−1 and as ψ increases on (−∞, 0), v decreases on (−∞, −1). Similarly as ψ increases
on (0, ∞), v decreases on (1, ∞).
On combining these cases, we see that tan θ = u + iv is a bijection from T ∗ to C \ {−i, i}.
Some more detail is provided on the trigonometrical tangent function in Appendix E.
To begin to utilise the foregoing material we now define what is essentially addition mod(T ).
Given
θ1 = φ1 + iψ1 , θ2 = φ2 + iψ2 ,
which are both in T we have that 0 ≤ φ1 ≤ π, 0 ≤ φ2 ≤ π, and so 0 ≤ φ1 + φ2 ≤ 2π. Then
φ1 + φ2 = n1,2 π + r1,2 for some integer n1,2 which is 0, 1 or 2 and the remainder r1,2 is a real
number such that 0 ≤ r1,2 < π. In the following our first two cases are when r1,2 6= 0 and the
remaining four cases are when r1,2 = 0. Consider
θ1 + θ2 = r1,2 + i(ψ1 + ψ2 ),
θ1 + θ2 = π + r1,2 + i(ψ1 + ψ2 ),
θ1 + θ2 = 0 + i(ψ1 + ψ2 ),
θ1 + θ2 = π + i(ψ1 + ψ2 ),
θ1 + θ2 = π + i(ψ1 + ψ2 ),
θ1 + θ2 = 2π + i(ψ1 + ψ2 ),
where 0 < r1,2 < π,
(12.1.11a)
where 0 < r1,2 < π,
where ψ1 ≥ 0, ψ2 ≥ 0,
(12.1.11b)
(12.1.11c)
where ψ1 + ψ2 ≥ 0,
where ψ1 + ψ2 < 0,
(12.1.11d)
(12.1.11e)
where ψ1 < 0, ψ2 < 0,
(12.1.11f)
342
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Note that in (12.1.11c) we have φ1 = φ2 = 0, in (12.1.11d) and (12.1.11e) we have φ1 + φ2 = π,
and in (12.1.11f) we have φ1 = φ2 = π. We now define θ1 ⊕ θ2 as follows :θ1 ⊕ θ2 = r1,2 + i(ψ1 + ψ2 ) in (12.1.11a) so that θ1 ⊕ θ2 = θ1 + θ2 ,
θ1 ⊕ θ2 = r1,2 + i(ψ1 + ψ2 ) in (12.1.11b) so that θ1 ⊕ θ2 = θ1 + θ2 − π,
θ1 ⊕ θ2 = i(ψ1 + ψ2 ) in (12.1.11c) so that θ1 ⊕ θ2 = θ1 + θ2 ,
θ1 ⊕ θ2 = i(ψ1 + ψ2 ) in (12.1.11d) so that θ1 ⊕ θ2 = θ1 + θ2 − π,
θ1 ⊕ θ2 = π + i(ψ1 + ψ2 ) in (12.1.11e) so that θ1 ⊕ θ2 = θ1 + θ2 ,
θ1 ⊕ θ2 = π + i(ψ1 + ψ2 )
in (12.1.11f) so that θ1 ⊕ θ2 = θ1 + θ2 − π.
(12.1.12a)
(12.1.12b)
(12.1.12c)
(12.1.12d)
(12.1.12e)
(12.1.12f)
We note that in all cases θ1 ⊕ θ2 ∈ T and as the definition is symmetrical as to the roles of θ1
and θ2 , we have θ1 ⊕ θ2 = θ2 ⊕ θ1 so that the operation ⊕ is commutative on T . We note that in
(12.1.12a), (12.1.12c) and (12.1.12e), we have θ1 ⊕ θ2 = θ1 + θ2 , and we shall call these cases the
set A. Similarly in (12.1.12b), (12.1.12d) and (12.1.12f), we have θ1 ⊕ θ2 = θ1 + θ2 − π and we shall
call these cases the set B. We note that in each case θ1 ⊕ θ2 is either equal to θ1 + θ2 or differs
from it by π.
Given
θ1 = φ1 + iψ1 ,
θ2 = φ2 + iψ2 ,
θ3 = φ3 + iψ3 ,
all in T , then (θ1 ⊕ θ2 ) ⊕ θ3 lies in T and is either equal to (θ1 + θ2 ) + θ3 or differs from it by
either π or 2π. Similarly θ1 ⊕ (θ2 ⊕ θ3 ) lies in T and is either equal to θ1 + (θ2 + θ3 ) or differs from
it by either π or 2π. It follows that (θ1 ⊕ θ2 ) ⊕ θ3 and θ1 ⊕ (θ2 ⊕ θ3 ) are either equal or differ by
a non-zero integer multiple of π. But since they are both in T they cannot differ by a non-zero
multiple of π and so must be equal. Hence
(θ1 ⊕ θ2 ) ⊕ θ3 = θ1 ⊕ (θ2 ⊕ θ3 )
in all cases and so ⊕ is associative on T .
From the definition, clearly 0 is an identity for ⊕ on T . For θ1 ⊕ 0 = θ1 , by (12.1.12c) when
φ1 = 0, by (12.1.12a) when 0 < φ1 < π as then r1,2 = φ1 , and by (12.1.12e) when φ1 = π as then
ψ1 < 0.
For additive inverses we first note that if 0 < φ1 < π and we take θ3 = π − φ1 − iψ1 then θ3 ∈ T
and
θ1 + θ3 = (φ1 + iψ1 ) + (π − φ1 − iψ1 ) = π,
so that by (12.1.12d) we have θ1 ⊕ θ3 = 0.
Secondly consider when φ1 = 0 so that ψ1 ≥ 0. When ψ1 = 0 so that θ1 = 0 we take θ3 = 0
and then by (12.1.12c) we have θ1 ⊕ θ3 = 0. When ψ1 > 0 we take θ3 = π − iψ1 and then by
(12.1.12d) we have θ1 ⊕ θ3 = 0.
Finally when φ1 = π we have ψ1 < 0 and take θ3 = −iψ and apply (12.1.12d) again. Thus each
element in T has an additive inverse in T . To summarise the cases here we note the following
for 0 < φ1 < π
for
φ1 = 0, ψ1 = 0,
for φ1 = 0, ψ1 > 0,
for φ1 = π, ψ1 < 0,
take θ3 = π − φ1 − iψ1
take θ3 = 0
so that θ1 + θ3 = π,
so that θ1 + θ3 = 0,
take θ3 = π − iψ1 so that θ1 + θ3 = π,
take θ3 = −iψ1 so that θ1 + θ3 = π.
(12.1.13a)
(12.1.13b)
(12.1.13c)
(12.1.13d)
On combining the results above we can conclude that (T , ⊕) is a commutative group.
We now turn to the addition formulae for trigonometric functions. From Cases A and B in
343
12.1.
(12.1.12), for θ1 , θ2 ∈ T
when θ1 + θ2 ∈ T , we have θ1 ⊕ θ2 = θ1 + θ2
cos(θ1 ⊕ θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 ,
sin(θ1 ⊕ θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 ,
and
when θ1 + θ2 6∈ T , we have θ1 ⊕ θ2 = θ1 + θ2 − π
− cos(θ1 ⊕ θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 ,
(12.1.14a)
and
− sin(θ1 ⊕ θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 .
As
tan(θ1 ⊕ θ2 ) =
is equal to
(12.1.14b)
sin(θ1 ⊕ θ2 )
cos(θ1 ⊕ θ2 )
sin(θ1 + θ2 )
cos(θ1 + θ2 )
in (12.1.14a), and to
− sin(θ1 + θ2 )
− cos(θ1 + θ2 )
in (12.1.14b), we have that
tan θ1 + tan θ2
,
(12.1.15)
1 − tan θ1 tan θ2
in all cases for which tan θ1 tan θ2 6= 1. Thus the standard addition formula for the tangent function
is valid for our modified addition.
We denote by 2⊙θ1 the sum θ1 ⊕θ1 and look to finding duplication formulae. With θ1 = φ1 +iψ1
so that θ1 + θ1 = 2φ1 + i2ψ1 , we divide into the cases :tan(θ1 ⊕ θ2 ) =
0 < 2φ1 < π
π < 2φ1 < 2π
so that
0 < φ1 < 12 π,
(12.1.16a)
so that
1
2π
< φ1 < π,
(12.1.16b)
φ1 = 0, ψ1 ≥ 0,
(12.1.16c)
2φ1 = 0 so that
φ1 =
1
2 π,
1
2 π,
ψ1 ≥ 0,
2φ1 = π
so that
2φ1 = π
so that
φ1 =
2φ1 = 2π
so that
φ1 = π, ψ1 < 0.
ψ1 < 0,
(12.1.16d)
(12.1.16e)
(12.1.16f)
We then obtain the following
2 ⊙ θ1 = 2φ1 + i2ψ1 in (12.1.16a) so that 2 ⊙ θ1 = 2θ1 ,
2 ⊙ θ1 = 2φ1 − π + i2ψ1 in (12.1.16b) so that 2 ⊙ θ1 = 2θ1 − π,
(12.1.17a)
(12.1.17b)
2 ⊙ θ1 = π + i2ψ1 in (12.1.16e) so that 2 ⊙ θ1 = 2θ1 ,
2 ⊙ θ1 = π + i2ψ1 in (12.1.16f) so that 2 ⊙ θ1 = 2θ1 − π.
(12.1.17e)
(12.1.17f)
2 ⊙ θ1 = i2ψ1 in (12.1.16c) so that 2 ⊙ θ1 = 2θ1 ,
2 ⊙ θ1 = 2iψ1 in (12.1.16d) so that 2 ⊙ θ1 = 2θ1 − π,
(12.1.17c)
(12.1.17d)
Note that in (12.1.16a), (12.1.16c) and (12.1.16e) we have 0 4L θ1 ≺L 21 π and 2 ⊙ θ1 = 2θ1 , while
in (12.1.16b), (12.1.16d) and (12.1.16f) we have 12 π 4L θ1 ≺L π and 2 ⊙ θ1 = 2θ1 − π. Thus
when 0 4L θ1 ≺L 21 π,
when 0
when
when
when
4L θ1 ≺L 21 π,
1
2π
1
2π
1
2π
we have 2 ⊙ θ1 = 2θ1 ,
we have
cos(2 ⊙ θ1 ) = cos 2θ1 , sin(2 ⊙ θ1 ) = sin 2θ1 ,
(12.1.18b)
cos(2 ⊙ θ1 ) = cos(2θ1 − π) = − cos 2θ1 ,
(12.1.18d)
4L θ1 ≺L π,
we have 2 ⊙ θ1 = 2θ1 − π,
4L θ1 ≺L π,
we have
4L θ1 ≺L π,
we have
(12.1.18a)
sin(2 ⊙ θ1 ) = sin(2θ1 − π) = − sin 2θ1 .
(12.1.18c)
(12.1.18e)
344
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
We define θ2 ⊖ θ1 to be θ2 ⊕ θ3 where θ3 is the additive inverse of θ1 . In (12.1.13a), (12.1.13c)
and (12.1.13d), that is except for θ1 = 0, we have θ1 + θ3 = π and so
cos θ3 = cos(π − θ1 ) = − cos θ1 , sin θ3 = sin(π − θ1 ) = sin θ1 ,
(12.1.19)
and so tan θ3 = − tan θ1 . When θ1 = 0 we have θ3 = 0 and so cos θ3 = 1, sin θ3 = tan θ3 = 0; thus
only the second of the equalities in (12.1.19) holds in this case.
We note easily that θ2 ⊖ 0 = θ2 and in what follows we consider θ2 ⊖ θ1 for θ1 6= 0. By (12.1.13a)
for 0 < φ1 < π we took θ3 = π − φ1 − iψ1 so that
θ2 + θ3 = π + φ2 − φ1 + i(ψ2 − ψ1 ).
(12.1.20)
In (12.1.20) we have θ2 + θ3 6∈ T if either φ1 < φ2 or φ1 = φ2 , ψ1 ≤ ψ2 , when we have θ1 4L θ2 .
We have θ2 + θ3 ∈ T if either φ1 > φ2 or φ1 = φ2 , ψ2 < ψ1 , when we have θ1 ≻L θ2 . Thus
θ2 ⊕ θ3 = θ2 − θ1
if θ1 4L θ2 ,
θ2 ⊕ θ3 = θ2 − θ1 + π
if θ1 ≻L θ2 .
(12.1.21)
By (12.1.13c) for φ1 = 0, ψ1 > 0, we took θ3 = π − iψ1 so that
θ2 + θ3 = π + φ2 + i(ψ2 − ψ1 ).
(12.1.22)
In (12.1.22) we have θ2 + θ3 6∈ T if either φ2 > 0 or φ2 = 0, ψ2 ≥ ψ1 , when we have that θ1 4L θ2 .
We have θ2 + θ3 ∈ T if φ2 = 0, ψ1 > ψ2 , when we have that θ1 ≻L θ2 . Thus we have (12.1.21)
again.
By (12.1.13d) for φ1 = π, ψ1 < 0, we took θ3 = −iψ1 so that
θ2 + θ3 = φ2 + i(ψ2 − ψ1 ).
(12.1.23)
In (12.1.23) we have θ2 + θ3 6∈ T either if φ2 = π or φ2 = 0, ψ1 ≤ ψ2 , when we have that θ1 4L θ2 ,
or if φ2 = 0, ψ1 > ψ2 and this latter cannot happen as ψ2 ≥ 0, ψ1 < 0. In (12.1.23) we have
θ2 + θ3 ∈ T if either 0 < φ2 < π, or if φ2 = π, ψ1 > ψ2 , or if φ2 = 0, ψ1 ≤ ψ2 , when we have that
θ1 ≻L θ2 in each case. Thus we have (12.1.21) again. It follows that (12.1.21) holds in all cases
when θ1 , θ2 ∈ T and θ1 6= 0.
It follows that when θ1 , θ2 ∈ T and θ1 6= 0,
when θ1 4L θ2 , we have θ2 ⊖ θ1 = θ2 − θ1 and
cos(θ2 ⊖ θ1 ) = cos θ2 cos θ1 + sin θ2 sin θ1 ,
sin(θ2 ⊖ θ1 ) = sin θ2 cos θ1 − cos θ2 sin θ1 ,
when θ1 ≻L θ2 , we have θ2 ⊖ θ1 = θ2 + (π − θ1 ) and
− cos(θ2 ⊖ θ1 ) = − [cos θ2 cos(π − θ1 ) − sin θ2 sin(π − θ1 )]
= cos θ2 cos θ1 + sin θ2 sin θ1 ,
− sin(θ2 ⊖ θ1 ) = − [sin(θ2 − θ1 ) cos π + cos(θ2 − θ1 ) sin π]
= sin θ2 cos θ1 − cos θ2 sin θ1 .
(12.1.24a)
(12.1.24b)
(12.1.24c)
(12.1.24d)
Given θ1 , θ2 ∈ T with θ1 ≺L θ2 we define a lexicographic strip S[θ1 , θ2 ) to be
and S[θ2 , θ1 ) to be
{θ ∈ T : θ1 4L θ ≺L θ2 },
(12.1.25)
S[θ2 , π) ∪ S[0, θ1 ).
(12.1.26)
Given any θ2 ∈ T we seek θ1 ∈ T such that 2 ⊙ θ1 = θ2 . By (12.1.18a) when 0 4L θ1 ≺L 12 π we
have 2 ⊙ θ1 = 2θ1 so that we need 2θ1 = θ2 , that is θ1 = 12 θ2 . By (12.1.18b) when 21 π 4L θ1 ≺L π
we have 2 ⊙ θ1 = 2θ1 − π so that we need 2θ1 − π = θ2 , that is θ1 = 12 θ2 + 12 π.
12.2. INITIAL GEOMETRY OF THE COMPLEX EUCLIDEAN PLANE
12.2
Initial geometry of the complex Euclidean plane
12.2.1
Complex sensed-area and areal coordinates
345
A transformation of CE2 of the form Z → Z ′ where
x′ =x + k1 ,
y ′ =y + k2 ,
(12.2.1)
where k1 and k2 are any complex numbers, is called a complex translation of CE2 . Note that
for it we have
x =x′ − k1 ,
y =y ′ − k2 ,
so that each complex translation is a function which has an inverse which is also a complex
translation.
For points Z1 , Z2 , Z3 we define an analogue of real sensed-area by
x y1 1 1 1
δF (Z1 , Z2 , Z3 ) = x2 y2 1 .
2
x3 y3 1 An expansion of this into real and imaginary parts is straightforward.
The argument in Barry [2, §10.5.3] shows that
δF (Z1 , Z2 , (1 − s)Z3 + sZ4 ])
= (1 − s)δF (Z1 , Z2 , Z3 ) + sδF (Z1 , Z2 , Z4 ),
(12.2.2)
for all complex numbers s, and there is an analogous result for pZ3 + qZ4 + rZ5 for any complex
numbers p, q, r such that p + q + r = 1.
Sensed-area is invariant under each translation. For with the above notation,
′
x1 y1′ 1 1
δF (Z1′ , Z2′ , Z3′ ) = x′2 y2′ 1 2 ′
x3 y3′ 1 x + k1 y1 + k2 1 1 1
= x2 + k1 y2 + k2 1 2
x3 + k1 y3 + k2 1 x y1 + k2 1 1 k1 y1 + k2 1 1 1
= x2 y2 + k2 1 + k1 y2 + k2 1 2
2
x3 y3 + k2 1 k1 y3 + k2 1 x y1 1 1 x1 k2 1 1 1
= x2 y2 1 + x2 k2 1 2
2
x3 y3 1 x3 k2 1 =δF (Z1 , Z2 , Z3 ).
In this argument we have twice used the property of a determinant, that if one column is a constant
multiple of another, then the value of the determinant is equal to 0.
Given any points such that δF (Z1 , Z2 , Z3 ) 6= 0, any point Z of the complex plane can be
expressed in the form Z = pZ1 +qZ2 +rZ3 where p, q, r are complex numbers such that p+q+r = 1.
For, as in Barry [2, §11.4.5] we first seek q and r such that
q(x2 − x1 ) + r(x3 − x1 ) =x − x1 ,
q(y2 − y1 ) + r(y3 − y1 ) =y − y1 .
346
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
These equations have the solution
δF (Z, Z3 , Z1 )
,
δF (Z1 , Z2 , Z3 )
δF (Z, Z1 , Z2 )
,
r=
δF (Z1 , Z2 , Z3 )
q=
and we now take p = 1 − q − r so that
p=
δF (Z, Z2 , Z3 )
.
δF (Z1 , Z2 , Z3 )
We refer to (p, q, r) as the complex areal coordinates of the point Z with respect to the ordered
triple (Z1 , Z2 , Z3 ).
12.2.2
Complex lines
Given any distinct complex points Z2 and Z3 , the set of complex points Z satisfying
δF (Z, Z2 , Z3 ) = 0,
is called the complex line containing Z2 and Z3 . This is
x y 1 x2 y2 1 =0,
x3 y3 1 x − x2 y − y2 0 x2
y2
1 =0,
x3 − x2 y3 − y2 0 and so
−(y3 − y2 )(x − x2 ) + (x3 − x2 )(y − y2 ) = 0.
(12.2.3)
ax + by + c = 0,
(12.2.4)
This has the form
where a, b, c ∈ C and the case a = b = 0 is excluded. We call this a Cartesian equation of this
line.
Conversely any equation of the form (12.2.4), with the case a = b = 0 excluded, is a Cartesian
equation of some line. For take distinct points Z2 and Z3 the coordinates of which satisfy this
equation. We also denote the equation (12.2.3) in this case by a1 x + b1 y + c1 = 0. We first note
that both of these equations are satisfied by ((1 − s)x2 + sx3 , (1 − s)y2 + sy3 for all s ∈ C; on taking
a value of s other than 0 and 1 we have a third point which satisfies both equations. As
ax2 + by2 + c =0,
a1 x2 + b1 y2 + c1 =0,
ax3 + by3 + c = 0,
a1 x3 + b1 y3 + c1 = 0,
we have
a(x3 − x2 ) + b(y3 − y2 ) =0,
a1 (x3 − x2 ) + b1 (y3 − y2 ) =0.
As this pair of simultaneous equations has solutions other than (0, 0), we must have
a b a1 b1 = 0.
12.2. INITIAL GEOMETRY OF THE COMPLEX EUCLIDEAN PLANE
347
Thus the rows of this determinant must be linearly dependent, and as neither row consists of a
pair of zeros, we must have (a, b) = k(a1 , b1 ), for some non-zero complex number k. Then we have
a1 x2 + b1 y2 +
c
= 0,
k
a1 x2 + b1 y2 + c1 = 0,
and so c = kc1 as well. This proves the result.
Suppose that Z1 and Z2 are distinct points on a line with equation (12.2.4). Then we have that
ax1 + by1 + c =0,
ax2 + by2 + c =0,
so by addition,
a(x1 + x2 ) + b(y1 + y2 ) + 2c = 0,
and on division across by 2
a
y1 + y2
x1 + x2
+b
+ c = 0.
2
2
Thus the point Z3 such that
x3 = 12 (x1 + x2 ),
y3 = 12 (y1 + y2 ),
(12.2.5)
also lies on this line. We call this point Z3 the mid-point of the points Z1 and Z2 . We extend
this to the case of equal points by saying that the mid-point of Z and Z is Z.
12.2.3
Parallel lines
Consider two lines with equations of the form (12.2.4),
a1 x + b1 y + c1 =0,
(12.2.6)
a2 x + b2 y + c2 =0.
(12.2.7)
We say that these lines are parallel either if they have no point in common or else they coincide.
A necessary and sufficient condition for this is that
a1 b 1 a2 b2 = 0,
and thus
a1 b2 − a2 b1 = 0.
(12.2.8)
As the rows are thus linearly dependent but not consisting entirely of zeros, we have
a2 = ka1 ,
b2 = kb1 ,
(12.2.9)
for some non-zero complex number k. Thus in the case of parallelism the equation (12.2.7) can be
re-written as
c2
a1 x + b 1 y +
= 0.
(12.2.10)
k
If the lines are not parallel, then (12.2.8) is false and then lines have a unique point of intersection.
Given any line, with equation (12.2.6), and any point Z4 , there is a unique line parallel to the
given line and passing through the point. For any parallel line has the form (12.2.10) and to make
it pass through Z4 we just need to choose c2 so that
a1 x4 + b1 y4 +
x2
= 0.
k
Furthermore suppose that we have two lines both parallel to one line, (12.2.7) and
a3 x + b3 y + c3 = 0,
(12.2.11)
348
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
both parallel to (12.2.6). Then by (12.2.9) we have
a2 = ka1 , b2 = kb1 , a3 = k ′ a1 , b3 = k ′ b1 ,
for some non-zero numbers k and k ′ . Hence
a3 =
k′
a2 ,
k
b3 =
k′
b2 .
k
It follows that the lines with equations (12.2.7) and (12.2.11) are parallel to each other. Thus lines
which are parallel to the same line are parallel to each other.
The image of a complex line under a translation is a parallel line. For given a line with equation
(12.2.4), when we apply (12.2.1) we obtain
ax′ + by ′ − ak1 − bk2 + c = 0.
This is an equation of a line, which moreover is parallel to the original one.
Given any two parallel lines,
a1 x + b1 y + c1 =0,
a1 x + b1 y + c2 =0,
there is a translation which maps the first onto the second. For on applying (12.2.1) to the first of
these equations we obtain
a1 x′ + b1 y ′ + c1 − a1 k1 − b1 k2 = 0.
Thus we need to choose k1 and k2 so that
a1 k1 + b1 k2 = c1 − c2 .
(12.2.12)
As (a1 , b1 ) 6= (0, 0) this can be done in many ways.
Suppose now that distinct points Z2 and Z3 have images Z2′ and Z3′ , respectively, under the
translation (12.2.1). Then we have that
x′2 =x2 + k1 ,
y2′ = y2 + k2 ,
x′3 =x3 + k1 ,
y3′ = y3 + k2 .
It follows that
′
1
2 (x2
+ x′3 ) = 21 (x2 + x3 ) + k1 ,
1 ′
2 (y2
+ y3′ ) = 12 (y2 + y3 ) + k2 ,
so that the image of the mid-point is mid-point of the images. Hence mid-point is invariant under
translations. Continuing with this situation and notation, we have that
x′3 − x′2 = x3 − x2 , y3′ − y2′ = y3 − y2 .
Hence by (12.2.3) and (12.2.10) the lines Z2 Z3 and Z2′ Z3′ are parallel, which re-proves a result
above. But now we also have that
x′2 − x2 = x′3 − x3 , y2′ − y2 = y3′ − y3 .
It follows, again from (12.2.3) and (12.2.10), that the lines Z2 Z2′ and Z3 Z3′ are parallel. Thus
unless k1 and k2 are such that Z2′ , and so Z3′ , is on the line Z2 Z3 , the points Z2 , Z3 , Z3′ and Z2′
are the vertices of a parallelogram. We note moreover that
′
1
2 (x2 + x3 )
1 ′
2 (y2 + y3 )
= 12 (x2 + x′3 ),
= 12 (y2 + y3′ ).
Thus the mid-point of Z2′ and Z3 coincides with the mid-point of Z2 and Z3′ .
349
12.2. INITIAL GEOMETRY OF THE COMPLEX EUCLIDEAN PLANE
12.2.4
Perpendicular lines
Given any line with equation (12.2.6), any line with equation (12.2.7) is said to be perpendicular
to it if
a1 a2 + b1 b2 = 0.
(12.2.13)
It then follows that the first line is also perpendicular to the second. Now (12.2.13) can be expressed
as
b1 −a1 a2 b2 = 0,
so the rows of this determinant are linearly dependent; as neither row consists of two zeros we then
must have that
a2 = kb1 , b2 = −ka1 ,
(12.2.14)
for some non-zero complex number k, so the line with equation (12.2.7) is perpendicular to that
with equation (12.2.6) if and only if (12.2.7) can be re-written as
b 1 x − a1 y +
c2
k
= 0.
(12.2.15)
Given any line with equation (12.2.6) and any point Z4 , there is a unique line which passes
through this point and is perpendicular to the given line. For by (12.2.15) a line perpendicular to
(12.2.6) has the form b1 x − a1 y + c3 = 0, and this contains the point Z4 if and only if c3 is chosen
to b1 x4 − a1 y4 + c3 = 0.
Given any line with equation (12.2.6) and any line with equation (12.2.7) perpendicular to it,
any line with equation (12.2.11) is perpendicular to the first line if and only if it is parallel to the
second line. For by (12.2.14) we have
a2 = kb1 ,
b2 = −ka1 ,
for some non-zero complex number k. Then the third line is also perpendicular to the first if
a3 = k ′ b 1 ,
b3 = −k ′ a1 ,
and this is the case if and only if
a3 = k ′′ a2 ,
b3 = k ′′ b2 ,
for some non-zero number k ′′ = k ′ /k.
We ask whether a line can be perpendicular to itself. By (12.2.13) a line with equation (12.2.6)
is perpendicular to itself if and only if a21 + b21 = 0, or equivalently a21 = −b21 so that a1 = ±b1 i.
Now we must have b1 6= 0 and so (12.2.6) can be written in the form
y ± ix + c1 = 0.
Thus a line is perpendicular to itself if and only if it can be put in one of the forms
y − ix + c1 = 0,
y + ix + c1 = 0.
(12.2.16)
(12.2.17)
These lines were introduced in 1852 and 1853 by Laguerre, from factorising the expression for
square-of-distance
(x − x1 )2 + (y − y1 )2 = [y − y1 − i(x − x1 )][y − y1 + i(x − x1 )].
He returned to these lines in 1870, giving them the name isotropic, with those in (12.2.16) being
in the first system and with those in (12.2.17) in the second system. The name means of equal
turning and we shall refer later to his reason for this name.
350
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
We note that through any point Z4 of the plane, precisely one isotropic line in each system
passes. These are
y − y4 − i(x − x4 ) =0
y − y4 + i(x − x4 ) =0,
We note that all the lines in the first system are parallel to each other, and similarly all the lines
in the second system are parallel to each other.
We next ask whether two perpendicular lines must intersect. Taking (12.2.6) and a slightly
modified (12.2.15), we consider the simultaneous equations
a1 x + b 1 y = − c1 ,
b 1 x − a1 y = − c2 .
These have a unique solution if and only if a21 + b21 6= 0, that is when neither is an isotropic line.
Thus two perpendicular lines must intersect in a unique point when neither is an isotropic line.
When (12.2.6) is an isotropic line, a line perpendicular to it is parallel to it so we have no solution
unless we take the second line to coincide with the first.
Taking a line (12.2.6) and a point Z4 not on it, we ask when we can drop a perpendicular from
Z4 to this line. We take a line through this point and perpendicular to (12.2.6). This will meet
(12.2.6) in a unique point if and only if (12.2.6) is not isotropic. Thus we can drop a perpendicular
from a point to any non-isotropic line. The last pair of equations have the solution
x=−
−b1 c1 + a1 c2
a1 c1 + b 1 c2
, y=
,
2
2
a1 + b 1
a21 + b21
and for the line with modified equation (12.2.15) to pass through Z4 we need
−c2 = b1 x4 − a1 y4 ,
so that combining these gives the coordinates of the foot of the perpendicular from Z4 to the line
(12.2.6).
Any translation maps a pair of perpendicular lines onto a pair of perpendicular lines. For under
(12.2.1) the lines with equations
a1 x + b1 y + c1 = 0, −b1 x + a1 y + c2 = 0,
map to the ones with equations
a1 x′ + b1 y ′ + c1 − a1 k1 − b1 k2 = 0, −b1 x + a1 y) + c2 + b1 k1 − a1 k2 = 0.
It follows that under a translation an isotropic line of the first system maps to an isotropic line
of the first system, and similarly for those of the second system.
12.2.5
Complex axial symmetries
Consider a non-isotropic line with equation (12.2.4) ax + by + c = 0 and a point Z1 . Let W be the
foot of the perpendicular from this point to this line, and let Z2 be such that W is the mid-point of
Z1 and Z2 . Then we describe the correspondence Z1 → Z2 as complex axial symmetry of CE2
in the given line. By the calculations in Barry [2, p. 86, §6.6.1] applied with complex coordinates
we have that
2a
(ax1 + by1 + c),
a2 + b 2
2b
y2 =y1 − 2
(ax1 + by1 + c).
a + b2
x2 =x1 −
12.3. HESSE’S NORMAL FORM OF EQUATION FOR A LINE
351
We work mainly with the axis of symmetry passing through the origin, so that c = 0, as more
general axial symmetries are compositions of translations and these axial symmetries. Thus we
can take Z → Z ′ where
a2 − b 2
ab
x−2 2
y,
a2 + b 2
a + b2
a2 − b 2
ab
x+ 2
y.
y′ = − 2 2
2
a +b
a + b2
x′ = −
(12.2.18)
From the symmetry of the definition as between Z1 and Z2 , we know that an axial symmetry is
its own inverse. The determinant of the matrix of coefficients is equal to
−(a2 − b2 )2 − 4a2 b2
= −1.
(a2 + b2 )2
Take a line through the origin, say with equation
a1 x + b1 y = 0.
(12.2.19)
If we substitute in this from the inverse of (12.2.18) we have
2
ab
a − b2 ′
ab
a2 − b 2 ′
′
′
−2
+
b
= 0,
x
−
2
y
x
+
y
a1 − 2
1
a + b2
a2 + b 2
a2 + b 2
a2 + b 2
and so obtain
−[(a2 − b2 )a1 + 2abb1 ]x′ + [−2aba1 + (a2 − b2 )b1 ]y ′ = 0.
′
(12.2.20)
′
If we substitute in this for x and y from (12.2.18), we find that we recover the original equation
(12.2.19). It follows that the image of any line through the origin (12.2.19) under any axial
symmetry (12.2.18) is a line through the origin (12.2.20).
To cover the case where (12.2.19) is isotropic in the first system, we take b1 = 1, a1 = −i.
Then (12.2.20) becomes
(2abi + a2 − b2 )y ′ + i(a2 − b2 + 2abi)x′ =0,
(a + ib)2 (y ′ + ix′ ) =0,
and so, since a + ib 6= 0 as axial symmetry must be in a non-isotropic line,
y ′ + ix′ = 0.
Thus under an axial symmetry in a line which passes through the origin, the image of an isotropic
line of the first system which passes through the origin, is an isotropic line of the second system
which passes through the origin, and vice versa. Under the composition of any two axial symmetries
in lines through the origin an isotropic line maps onto itself and so is left invariant. These were
the types of aspects mentioned by Laguerre in 1870 when coining the name isotropic.
12.3
Hesse’s normal form of equation for a line
12.3.1
Difficulties due to Square Roots
Any non-zero complex number z has two square roots and as we vary z continuously it can be
difficult to distinguish one square root from the other. That is the main reason that we have not
introduced distance between two complex points so far, as the distance formula
p
(x2 − x1 )2 + (y2 − y1 )2
can have two values and so is ambiguous without further information.
352
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
A similar difficulty arises with axial symmetries. Given a line through the origin with equation
(12.2.19)
a1 x + b1 y = 0,
and a second line through the origin with equation
a2 x + b2 y = 0,
(12.3.1)
we ask if there is an axial symmetry (12.2.18) which maps the first of these onto the second. In
other words, can we put (12.3.1) into the form (12.2.20)?
To avoid difficult calculations we take (12.2.19) to represent the x-axis with equation y = 0, so
that we can take a1 = 0, b1 = 1. Now (12.2.20) becomes
−2
a2
a2 − b 2 ′
ab
x′ + 2
y = 0,
2
+b
a + b2
(12.3.2)
and we ask if we can find a and b so that
−2
a2
ab
= ja2 ,
+ b2
a2 − b 2
= jb2 ,
a2 + b 2
+ −2
2
for some j 6= 0. We then need
a2 − b 2
a2 + b 2
2
ab
a2 + b 2
and so
j2 =
Then in
= 1 = j 2 (a22 + b22 ),
1
.
a22 + b22
1
,
j = ±p 2
a2 + b22
we have an instance of the difficulty adverted to. Moreover on eliminating j we need
b 2 − a2
(b/a)2 − 1
b2
=
= ,
2ab
2b/a
a2
a quadratic equation with the solutions
b2
b
±
=
a
a2
s
b22
+ 1,
a22
which raises the difficulty again.
12.3.2
Hesse’s normal form of equation for a line
We deal with the serious problem discussed in §12.3.1 by managing to circumvent it as follows.
Consider now a point Z3 which is not on the y-axis, so that x3 6= 0, and is such that the line
OZ3 is non-isotropic. Then this line has equation
y3 x − x3 y = 0.
We can rewrite this in the form
y3
y
,
=
x
x3
(12.3.3)
12.4. INSERTION OF SUPPLEMENTARY MATERIAL FOR RE2
353
and note that as the line we are considering is non-isotropic the expression on the right-hand side
of (12.3.3) is not equal to either i or −i. We are then able to choose the unique number θ3 in T ∗
such that
sin θ3
y3
= tan θ3 =
.
(12.3.4)
x3
cos θ3
From this we adopt as a standard Cartesian equation for our non-isotropic line OZ3
x sin θ3 − y cos θ3 = 0,
(12.3.5)
which is commonly known as the Hesse normal form of the equation of a line through the origin.
We note that we can include the one excluded non-isotropic line, that is the y-axis, under the form
(12.3.5) by taking θ3 = 12 π.
12.4
Insertion of supplementary material for RE2
12.4.1
Use of polar coordinates in RE2
It is convenient to add a little detail on the polar form of parametric equations of a line to the
material in Barry [2, pp. 156-163]. For any line l through the origin O, we considered the duo-angle
α4d with side-lines OI and l such that the indicator m of α4d lies in the duo-sector D1 . We denoted
by DA(F ) the set of such duo-angles and said that they were in standard position with respect to
F . Then any point Z ≡F (x, y) lies on l if and only if
p
p
x = x2 + y 2 cos α4d , y = x2 + y 2 sin α4d ,
when y ≥ 0, and it follows from this that Z lies on l if and only if
p
p
x = − x2 + y 2 cos α4d , y = − x2 + y 2 sin α4d ,
when y ≤ 0. Thus we have a modification of the polar form, that Z ∈ l if and only if
x = t cos α4d , y = t sin α4d ,
for some t ∈ R.
As well as l we consider also the line l′ which makes a duo-angle α5d with OI and show that
the rotation
x′ = cos(α5d − α4d )x − sin(α5d − α4d )y,
y ′ = sin(α5d − α4d )x + cos(α5d − α4d )y,
maps l onto l′ . For points on l satisfy
x = t cos α4d , y = t sin α4d ,
(t ∈ R),
and from this
x′ =t[cos(α5d − α4d ) cos α4d − sin(α5d − α4d ) sin α4d ]
=t′ cos α5d ,
y ′ =t[sin(α5d − α4d ) cos α4d + cos(α5d − α4d ) sin α4d ]
=t′ sin α5d ,
where t′ = ±t, taking the + or the − according as |α5d − α4d |◦ + |α4d |◦ is less than or greater
than 180. See [2, §10.10.5]
354
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Now consider consider points W4 , W5 distinct from O with W4
(t5 cos α5d , t5 sin α5d ). Then
0
0
δF (O, W4 , W5 ) = 12 t4 cos α4d t4 sin α4d
t5 cos α5d t5 sin α5d
≡F (t4 cos α4d , t4 sin α4d ), W5 ≡F
1
1
1
,
= 12 t4 t5 [sin α5d cos α4d − cos α5d sin α4d ].
Now | − α4d |◦ = 180 − |α4d |◦ and
sin(α5d − α4d ) = sin[α5d + (−α4d )].
Also
sin α5d cos α4d − cos α5d sin α4d =
sin(α5d − α4d )
− sin(α5d − α4d )
if |α5d |◦ + 180 − |α4d |◦ < 180
.
if |α5d |◦ + 180 − |α4d |◦ > 180
It follows that
δF (O, W4 , W5 ) =
12.4.2
1
2 t4 t5
sin(α5d − α4d )
− sin(α5d − α4d )
if |α5d |◦ < |α4d |◦
.
if |α5d |◦ > |α4d |◦
Duo-sectors in RE2
In Barry [2, Chapter 10] we introduced the concept of duo-sector in the real Euclidean plane. We
first make an addition to the material there. Given a pair {l1,2 = W1 W2 , l3 = W1 W3 } of distinct
lines through the point W1 , we essentially characterised two duo-sectors with side-lines l1,2 and
l1,3 in RE2 as two sets both containing these sidelines and also containing the points W , in one
case such as satisfy
δF (W1 , W2 , W )
> 0,
(12.4.1)
δF (W1 , W, W3 )
which we denote by Dl∗1,2 ,l1,3 , and in the other case such as satisfy
δF (W1 , W2 , W )
< 0,
δF (W1 , W, W3 )
(12.4.2)
which we denote by Dl∗∗
, called supplementary to the other. As
1,2 ,l1,3
δF (W1 , W, W3 )
δF (W1 , W3 , W )
δF (W1 , W2 , W )
> 0,
> 0,
> 0,
δF (W1 , W, W3 )
δF (W1 , W2 , W )
δF (W1 , W, W2 )
are equivalent, these duo-sectors do not depend on the order in which we take the lines. We note
that
= {l1,2 , l1,3 },
Dl∗1,2 ,l1,3 ∩ Dl∗∗
1,2 ,l1,3
and that Dl∗1,2 ,l1,3 ∪ Dl∗∗
is the set of all lines passing through the point W1 .
1,2 ,l1,3
Now take
W2 ≡F (t2 cos θ2d , t2 sin θ2d ), W3 ≡F (t3 cos θ3d , t3 sin θ3d ), W ≡F (t cos θd , t sin θd ),
where t2 and t3 have the same sign as each other, t 6= 0 and |θ2d |r < |θ3d |r . Then by (12.3.15) we
have
t2 sin(θd − θ2d )
δF (O, W2 , W )
=
> 0,
(12.4.3)
δF (O, W, W3 )
t3 sin(θ3d − θd )
when |θ2d |r < |θd |r < |θ3d |r , and
t2 sin(θd − θ2d )
δF (O, W2 , W )
=−
< 0,
δF (O, W, W3 )
t3 sin(θ3d − θd )
when either |θd |r < |θ2d |r or |θ3d |r < |θd |r .
(12.4.4)
12.5. RETURN TO CE2
355
12.5
Return to CE2
12.5.1
Parametric equations of a line in CE2
Given any distinct points Z2 and Z3 , and any point Z1 not on the line Z2 Z3 , for any point Z we
have Z = pZ1 + qZ2 + rZ3 for some p, q, r ∈ C with p + q + r = 1. In particular we had
r=
δF (Z, Z1 , Z2 )
.
δF (Z1 , Z2 , Z3 )
Then Z is on the line Z2 Z3 if and only if p = 0, so that q = 1 − r and thus
Z = (1 − r)Z2 + rZ3 ,
(12.5.1)
for some r ∈ C. Conversely if Z has the form (12.5.1) for any complex number r,
δF (Z, Z2 , Z3 )
=(1 − r)δF (Z2 , Z2 , Z3 ) + rδF (Z3 , Z2 , Z3 )
=0,
and so the definition in §12.5.2 is satisfied. Thus in (12.5.1) we have parametric equations of
the line Z2 Z3 to accompany the Cartesian equation (12.2.3). There is a profound difference with
the real case however since, as is well known, the complex field cannot be made into an ordered
field. Thus there is difficulty defining half-lines and segments in our new environment. When we
take the case r = 21 in (12.5.1), the point Z obtained is the mid-point of Z2 and Z3 .
In (12.5.1) we have
δF (Z, Z1 , Z2 )
r=
,
(12.5.2)
δF (Z1 , Z2 , Z3 )
for any Z1 6∈ Z2 Z3 , and we shall provisionally refer to the expression on the right-hand side in this
as the parameter identifier. It follows that Z − Z2 = r(Z3 − Z2 ) so that for Z4 , Z5 ∈ Z2 Z3
Z4 − Z2 =r4 (Z3 − Z2 ), Z5 − Z2 = r5 (Z3 − Z2 ),
Z5 − Z4 =(r5 − r4 )(Z3 − Z2 ),
(x5 − x4 , y5 − y4 ) =(r5 − r4 )(x3 − x2 , y3 − y2 ),
x5 − x4 =(r5 − r4 )(x3 − x2 ), y5 − y4 = (r5 − r4 )(y3 − y2 ),
(x5 − x4 )2 + (y5 − y4 )2 =(r5 − r4 )2 [(x3 − x2 )2 + (y3 − y2 )2 ].
(12.5.3)
We refer to the ordered pair of points (Z2 , Z3 ) in (12.5.1) as the base of this representation.
Any pair of points on the line can be used in this way and we first note how to get from one base
to another. Suppose that Z4 , Z5 lie on this line, with
Z4 = (1 − r4 )Z2 + r4 Z3 ,
Z5 = (1 − r5 )Z2 + r5 Z3 .
Then on solving for Z2 and Z3 we obtain
r5
r4
Z4 −
Z5 ,
r5 − r4
r5 − r4
1 − r4
r5 − 1
Z4 +
Z5 .
Z3 =
r5 − r4
r5 − r4
Z2 =
Then for points on the line we have
(1 − r)r5 + r(r5 − 1)
−(1 − r)r4 + r(1 − r4 )
Z4 +
Z5 ,
r5 − r4
r5 − r4
r − r4
r5 − r
Z4 +
Z5
=
r5 − r4
r5 − r4
=(1 − r′ )Z4 + r′ Z5 , say.
Z=
(12.5.4)
356
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
We we wish to note that parametric equations of a line are invariant under translation. For on
applying (12.2.1) to (12.5.1) we have that if
x = (1 − r)x2 + rx3 ,
y = (1 − r)y2 + ry3 ,
by (12.2.1) we have
x′ − k1 = (1 − r)(x′2 − k1 ) + r(x′3 − k1 ),
y ′ − k2 = (1 − r)(y2′ − k2 ) + r(y3′ − k2 ),
and so
x′ = (1 − r)x′2 + rx′3 ,
y ′ = (1 − r)y2′ + ry3′ .
Thus parametric equations of collinear points are preserved under translations when we take the
base points for the second line to be the images of the base points in the first.
We recall the definition of a concurrent pencil of lines in §2.1.2. With an obvious modification
it extends immediately to the complex Euclidean plane. Given any non-collinear points Z4 , Z5 , Z8 ,
any line which passes through Z4 and is not parallel to Z5 Z8 will meet Z5 Z8 in a unique point
Zr = (1 − r)Z5 + rZ8 for some r ∈ C. Then all such lines are of the form Z4 Zr and are said to
form a concurrent pencil of lines with vertex Z4 . We showed in §4.2.3 that all ranges of four
points in which transversals cut the lines of a concurrent pencil have the same cross-ratio, and the
same holds true here with the same proof.
For Z4 , Z5 ∈ Z2 Z3 we have from (12.5.2) that
2(r5 − r4 )δF (Z1 , Z2 , Z3 ) =2δF (Z1 , Z2 , Z5 ) − 2δF (Z1 , Z2 , Z4 )
x1 y1 1 x1 y1 1 = x2 y2 1 − x2 y2 1 x5 y5 1 x4 y4 1 x1
y1
1 x2
y2
1 = x5 − x4 y5 − y4 0 x1
y1
1 = x2 − x1 y2 − y1 0 x5 − x4 y5 − y4 0 0
0
1 = x2 − x1 y2 − y1 1 x5 − x4 y5 − y4 1 =2δF (O, Z2 − Z1 , Z5 − Z4 )
and so
r5 − r4 =
2δF (O, Z2 − Z1 , Z5 − Z4 )
.
2δF (Z1 , Z2 , Z3 )
(12.5.5)
We apply this in a major way as follows. Let l be a non-isotropic line which passes through
through the origin. Then as in (12.3.4) there is a unique angle θ ∈ T ∗ such that the point U (θ)
with coordinates (cos θ, sin θ) lies on l. We also introduce the second point V (θ) with coordinates
specified as follows
V (θ) = (− sin θ, cos θ) if
V (θ) = (sin θ, − cos θ) if
0≤θ<
π
,
2
π
≤ θ < π.
2
(12.5.6a)
(12.5.6b)
Specialising we take Z2 = O ≡ (0, 0), Z3 = U (θ) as basis points for our parametric equations
12.5. RETURN TO CE2
357
for the line l and for Z4 , Z5 on l have
2δF (V (θ), O, Z5 − Z4 )
= 2δF (V (θ, O, Z5 − Z4 )
2δF (V (θ), O, U (θ)
−2δF (V (θ), O, Z5 − Z4 )
= −2δF (O, −V (θ), Z5 − Z4 )
r5 − r4 =
−2δF (V (θ), O, U (θ)
if
0≤θ<
π
,
2
(12.5.7a)
if
π
≤ θ < π.
2
(12.5.7b)
r5 − r4 =
The reason for our choice of U (θ) and V (θ) was to have the denominators in (12.5.7a) and (12.5.7b)
both to have the value 1.
We can define sensed-distance on this line by taking the sensed-distance from Z4 to Z5 , points
on the line with values of the parameter r4 , r5 , respectively, denoted by Z4 Z5 , to be r5 − r4 , that
is
Z4 Z5 = r5 − r4 .
In particular Z2 Z3 = 1.
These sensed distances, along any line through the origin, have the general properties
Z5 Z4 = −Z4 Z5 ,
Z4 Z5 + Z5 Z6 = Z4 Z6 .
(12.5.8)
We note that, with our choice, by (12.5.3) we have
2
Z4 Z5 = (r5 − r4 )2 = (x5 − x4 )2 + (y5 − y4 )2 ,
which we refer to as square-of-distance.
If we take points Z6 , Z7 , Z8 , Z9 on this line we can then deal with sensed-ratios
Z8 Z9
.
Z6 Z7
In the case where we take (Z6 , Z7 ) the basis for this line so that we have Z − Z6 = r(Z7 − Z6 ),
then it follows that
Z6 Z
r=
,
Z6 Z7
so that our parameter identifier is equal to the value of this sensed ratio.
Now, generalizing, take any non-isotropic line l in the plane and also any point Z2 ∈ l. We
note that there is a non-isotropic line m which passes through through the origin O and which is
parallel to l. Then there will be a point U (θ) lying on m for some θ ∈ T ∗ . We take the point
Z3 = Z2 + U (θ) lying on l and the point Z4 = Z2 + V (θ) not lying on l. With (Z2 , Z3 ) as our basis
for l we will have parametric equations
Z =Z2 + r(Z3 − Z2 ) = Z2 + rU (θ), r ∈ C,
x =x2 + r cos θ, y = y2 + r sin θ, r ∈ C,
(12.5.9)
for the line l.
For points Z5 , Z6 ∈ m we note that Z2 + Z5 , Z2 + Z6 ∈ m and by (12.5.5) we have
2r5 δF (Z4 , Z2 , Z3 ) =2r5 δF (Z2 + V (θ), Z2 , Z2 + U (θ)) = 2r5 δF (V (θ), O, U (θ))
=2δF (Z2 + V (θ), Z2 , Z2 + Z5 ) = 2δF (V (θ), O, Z5 ),
and so
2(r6 − r5 )δF (V (θ), O, U (θ)) =2δF (V (θ), O, Z6 ) − 2δF (V (θ), O, Z5 )
=2δF (V (θ), O, Z6 − Z5 ).
(12.5.10)
358
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Then from (12.5.7a) and (12.5.7b) we have
2δF (V (θ), O, Z6 − Z5
= 2δF (V (θ, O, Z6 − Z5 )
2δF (V (θd ), O, U (θ)
−2δF (V (θ), O, Z6 − Z5 )
= −2δF (O, −V (θ), Z6 − Z5 )
r6 − r5 =
−2δF (V (θ), O, U (θ)
r6 − r5 =
π
,
2
(12.5.11a)
π
≤ θ < π.
2
(12.5.11b)
if 0 ≤ θ <
if
We can define sensed distance on this line by taking the sensed distance from Z2 + Z5 to Z2 + Z6 ,
denoted by (Z2 + Z5 )(Z2 + Z6 ), to be r6 − r5 .
Thus we have specified complex sensed-distances along all non-isotropic lines in the complex
Euclidean plane. The properties listed in (12.5.8) extend to all such lines.
We note that we have established, for each non-isotopic line through the point Z2 , parametric
equations of the form
Z = Z2 + r(cos θ, sin θ), r ∈ C,
(12.5.12)
for some θ ∈ T ∗ . As Z2 (Z2 + U (θ)) = 1 we have Z2 Z = r and by (12.2.3) r2 = (x−x2 )2 +(y −y2)2 .
We refer to this as the polar form of parametric equations for l, and θ as the polar angle of l
with respect to F .
12.5.2
Use of polar parametric equations in CE2
Suppose now that OZ5 and OZ6 are both non-isotropic lines. By (12.5.12) we can take
x5 = r5 cos θ5 , y5 = r5 sin θ5 , x6 = r6 cos θ6 , y6 = r6 sin θ6 .
From this we note first by (12.1.24b) and (12.1.24d) that
δF (O, Z5 , Z6 )
= 12 (x5 y6 − x6 y5 )
= 21 r5 r6 (cos θ5 sin θ6 − cos θ6 sin θ5 )
= ± 12 OZ5 OZ6 sin(θ6 ⊖ θ5 ),
(12.5.13)
taking the + sign when θ5 4L θ6 , and taking the − sign when θ5 ≻L θ6 , and having in both
θ5 6= 0.
Similarly by (12.1.24a) and (12.1.24c)
x5 x6 + y5 y6
=r5 r6 (cos θ5 cos θ6 + sin θ5 sin θ6 )
= ± OZ5 OZ6 cos(θ6 ⊖ θ5 ),
(12.5.14)
taking the + sign when θ5 4L θ6 , and taking the − sign when θ5 ≻L θ6 , and having in both
θ5 6= 0.
We see from (12.5.14) that when the lines OZ5 and OZ6 are both non-isotropic, they are
perpendicular to each other when cos(θ6 ⊖ θ5 ) = 0. This happens iff θ6 ⊖ θ5 = 12 π and so if and
only if (θ6 ⊖ θ5 ) ⊕ θ5 = 21 π ⊕ θ5 , i.e. θ6 = 12 π ⊕ θ5 .
12.5.3
Duo-sectors in CE2
Now in the complex Euclidean plane CE2 , we consider a pair {l5 , l6 } of non-isotropic lines through
the origin, l5 = OZ5 , l6 = OZ6 , with equations of the form (12.5.2),
x sin θ5 − y cos θ5 =0,
x sin θ6 − y cos θ6 =0,
12.5. RETURN TO CE2
359
respectively, where θ5 4L θ6 . Then the set of lines with equation
x sin θ − y cos θ = 0,
(12.5.15)
where θ ∈ S[θ5 , θ6 or θ = θ6 as in (12.1.25) is called a duo-sector with edges l5 and l6 , which
we denote by Dl∗5 ,l6 , as is also the set of lines with equation (12.5.15) where θ ∈ S[θ6 , θ5 or θ = θ5
as in (12.1.26), called the supplementary duo-sector and denoted by Dl∗∗
. We obtain the same
5 ,l6
duo-sectors if we exchange the roles of l5 and l6 , as
= Dl∗5 ,l6 .
, Dl∗∗
Dl∗6 ,l5 = Dl∗∗
6 ,l5
5 ,l6
We note that
= {l5 , l6 },
Dl∗5 ,l6 ∩ Dl∗∗
5 ,l6
and that Dl∗5 ,l6 ∪ Dl∗∗
is the set of all non-isotropic lines passing through the origin O. We denote
5 ,l6
by DQ1 the duo-sector Dl∗5 ,l6 where OI = l5 , OJ = l6 , which we call the first duo-quadrant;
it consists of the non-isotropic lines (12.5.15) with 0 4L θ 4L 12 π. Its supplement we denote
by DQ2 and call the second duo-quadrant; it consists of the non-isotropic lines (12.5.15) with
1
2 π 4L θ ≺L π or θ = 0.
12.5.4
Duo-angles in CE2
If we compare our use of polar coordinates on the one hand in Barry [2, pp.156-163] and on the
other hand on §12.4.1 and (12.5.12)? here, it should be evident that the complex numbers θ which
we used for our trigonometry in §12.1.2 are analogous to complex radian measures of duo-angles
in CE2 . We need not fuss about formally and carefully defining duo-angles from the materials
in §12.5.3 but equivalently use the complex numbers in §12.1.2 instead as giving duo-angles in
standard position. For an ordered pair (θ1 , θ2 ) of such angles, we take θ2 ⊖ θ1 as a corresponding
sensed-angle.
More generally, given an ordered pair of intersecting lines (Z5 Z6 , Z5 Z7 ) we define the sensedangle ∢(Z5 Z6 , Z5 Z7 ) to be θ5,7 ⊖ θ5,6 , where θ5,6 and θ5,7 are the polar angles of these lines.
12.5.5
Complex Rotations
Let Z0 be an arbitrary point of CE2 and α an angle in T . Let Z 6= Z0 be a point with polar
coordinate Z ′ = Z0 + t(cos θ, sin θ), so that Z0 Z is a non-isotropic line. We define the point
Z ′ = Z0 + t(cos(θ ⊕ α), sin(θ ⊕ α)); we extend this definition to Z0′ = Z0 . Then the correspondence
Z → Z ′ is called complex rotation about the point Z0 through an angleα. The domain of
this function is the point Z itself and the set of all Z such that Z0 Z is not isotropic. Then we have
that
(x′ , y ′ ) − (x0 , y0 ) = t[±(cos θ cos α − sin θ sin α, sin θ cos α + cos θ sin α)],
where we take the + or − according as θ + α ∈ T or θ + α 6∈ T , respectively. From this, on first
taking the + we have
(x′ − x0 , y ′ − y0 ) =(cos α.t cos θ − sin α.t sin α, sin α.t cos θ + cos α.t sin θ)
=(cos α(x − x0 ) − sin α(y − y0 ), sin α(x − x0 ) + cos α(y − y0 )).
(12.5.16)
This is valid when 0 4L θ ≺L π − α.
Secondly on taking the − we have
(x′ − x0 , y ′ − y0 ) =(− cos α.t cos θ + sin α.t sin α, − sin α.t cos θ − cos α.t sin θ)
=(− cos α(x − x0 ) + sin α(y − y0 ), − sin α(x − x0 ) − cos α(y − y0 )).
(12.5.17)
This is valid when π − α 4L θ ≺L π. So for a fixed α ∈ T , we have a double-barrelled formula for
a rotation, with two regions for θ, each being a duo-sector less one of its sidelines..
360
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
We can use matrix notation to give tidier notation as follows. We have
cos α − sin α
x1 − x0
x2 − x0
,
=±
y1 − y0
sin α cos α
y2 − y0
where the + sign is taken when 0 4L θ ≺L π−α, and the minus sign is taken when π−α 4L θ ≺L π.
A rotation about a point Z0 maps any non-isotropic line through Z0 onto a non-isotropic line
through Z0 . For if l1 is a line through Z0 with polar angle θ1 , then the points Z on it have the form
Z = Z0 + t(cos θ1 , sin θ1 ) for all t ∈ C and then the images Z ′ = Z0 + t(cos(θ1 ⊕ α), sin(θ1 ⊕ α)),
for all t ∈ C, form a non-isotropic line through Z0 .
Moreover, if l1 and l2 are non-isotropic lines through Z0 , with equations
Z =Z0 + t(cos θ1 , sin θ1 )
Z =Z0 + t(cos θ2 , sin θ2 )
for all t ∈ C,
for all t ∈ C,
respectively, then there is a rotation about Z0 which maps l1 onto l2 . For with α = θ2 ⊖ θ1 we
have that α ⊕ θ1 = (θ2 ⊖ θ1 ) ⊕ θ1 = θ2 , by (12.1.13).
12.5.6
Dropping a perpendicular
Consider an arbitrary non-isotropic line l = Z0 (Z0 + Uθ ) and for an arbitrary point Z1 take the
line m = Z1 (Z1 + Vθ ) which passes through Z1 and is perpendicular to l. Then m will intersect l
in a unique point W of the form W = (1 − t)Z1 + t(Z1 + Vθ ) for some t ∈ C. To find this t we note
that we must have
0 =δF ((1 − t)Z1 + t(Z1 + Vθ ), Z0 , Z0 + Uθ )
=(1 − t)δF (Z1 , Z0 , Z0 + Uθ ) + tδF (Z1 + Vθ ), Z0 , Z0 + Uθ )
=(1 − t)δF (Z1 − Z0 , O, Uθ ) + tδF (Z1 − Z0 + Vθ ), O, Uθ )
=δF (Z1 − Z0 , O, Uθ ) + t[δF (Z1 − Z0 + Vθ ), O, Uθ ) − δF (Z1 − Z0 , O, Uθ )]
=δF (Z1 − Z0 , O, Uθ ) + tδF (Vθ , O, Uθ )
and so we have the solution
t=−
δF (Z1 − Z0 , O, Uθ )
.
δF (O, Uθ , Vθ )
(12.5.18)
. Now δF (O, Uθ , Vθ ) is equal to 12 or − 12 according as θ ≺L π2 or θ π2 . Then the complex
magnitude of the sensed distance Z1 W is ∓2δF (Z1 − Z0 , O, Uθ ) according as θ ≺L π2 or θ π2 .
12.5.7
Mid-lines
Given a pair of non-isotropic lines meeting at a point Z0 , l1 = Z0 (Z0 + Uθ1 ) and l2 = Z0 (Z0 + Uθ2 ),
we may assume that 0 F θ1 ≺F θ2 ≺F π. Then for a point Z the complex-magnitudes of the
sensed-distances ZW1 , ZW2 , where W1 , W2 are the feet of the perpendiculars from Z to l1 and l2 ,
respectively, are
∓2δF (Z − Z0 , O, Uθ1 ), ∓2δF (Z − Z0 , O, Uθ2 ),
and these are equal to each other or additive inverses of each other if
δF (Z − Z0 , O, Uθ2 ) − δF (Z − Z0 , O, Uθ1 ) = 0,
and
δF (Z − Z0 , O, Uθ2 ) + δF (Z − Z0 , O, Uθ1 ) = 0.
These are respectively,
δF (Z − Z0 , O, Uθ2 − Uθ1 ) = 0,
(12.5.19)
12.5. RETURN TO CE2
361
and
δF (Z − Z0 , O, Uθ2 + Uθ1 ) = 0,
(12.5.20)
respectively.
Now Uθ2 − Uθ1 = (cos θ2 − cos θ1 , sin θ2 − sin θ1 ), and Uθ2 − Uθ1 = (cos θ2 + cos θ1 , sin θ2 + sin θ1 ).
Then
cos θ2 − cos θ1 = cos[θ1 + (θ2 − θ1 )] − cos θ1
= cos θ1 cos(θ2 − θ1 ) − sin θ1 sin(θ2 − θ1 ) − cos θ1
=[cos θ1 cos(θ2 − θ1 ) − 1] − sin θ1 sin(θ2 − θ1 )
θ2 − θ1
θ2 − θ1
θ2 − θ1
] − 2 sin θ1 sin
cos
= cos θ1 [−2 sin2
2
2
2
θ2 − θ1
θ2 − θ1
θ2 − θ1
[cos θ1 sin
+ sin θ1 cos
]
= − 2 sin
2
2
2
θ2 − θ1
θ2 − θ1
[sin(
+ θ1 )]
= − 2 sin
2
2
θ2 − θ1
θ2 + θ1
= − 2 sin
sin
.
2
2
Similarly
sin θ2 − sin θ1 = sin[θ1 + (θ2 − θ1 )] − sin θ1
= sin θ1 cos(θ2 − θ1 ) + cos θ1 sin(θ2 − θ1 ) − sin θ1
= sin θ1 [cos(θ2 − θ1 ) − 1] + cos θ1 sin(θ2 − θ1 )
θ2 − θ1
θ2 − θ1
θ2 − θ1
= sin θ1 [−2 sin2
] + 2 cos θ1 sin
cos
2
2
2
θ2 − θ1
θ2 − θ1
θ2 − θ1
[− sin θ1 sin
+ cos θ1 cos
]
=2 sin
2
2
2
θ2 − θ1
θ2 − θ1
[cos(
+ θ1 )]
=2 sin
2
2
θ2 − θ1
θ2 + θ1
=2 sin
cos
.
2
2
Then equation (12.5.19) is equivalent to
(x − x0 ) sin
θ2 + θ1
θ2 + θ1
− (y − y0 ) cos
= 0,
2
2
which is the equation in Hesse form of a line through Z0 .
For a second analogous line we start with
cos θ2 + cos θ1 = cos[θ1 + (θ2 − θ1 )] + cos θ1
= cos θ1 cos(θ2 − θ1 ) − sin θ1 sin(θ2 − θ1 ) + cos θ1
= cos θ1 [cos(θ2 − θ1 ) + 1] − sin θ1 sin(θ2 − θ1 )
θ2 − θ1
= cos θ1 [2 cos2
] − sin θ1 sin θ2 − θ1 )
2
θ2 − θ1
θ2 − θ1
[cos θ1 cos
− sin θ1 sin(θ2 − θ1 )
=2 cos
2
2
θ2 − θ1
θ2 − θ1
[cos(
+ θ1 )]
=2 cos
2
2
θ2 + θ1
θ2 − θ1
cos
.
=2 cos
2
2
(12.5.21)
362
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
We continue with
sin theta2 + sin θ1 = sin[θ1 + (θ2 − θ1 )] + sin θ1
= sin θ1 cos(θ2 − θ1 ) + cos θ1 sin(θ2 − θ1 ) + sin θ1
= sin θ1 ([cos(θ2 − θ1 ) + 1] + cos θ1 sin(θ2 − θ1 )
θ2 − θ1
θ2 − θ1
θ2 − θ1
= sin θ1 [2 cos2
+ 2 cos θ1 sin
cos
2
2
2
θ2 − θ1
θ2 − θ1
θ2 − θ1
[sin θ1 cos
+ cos θ1 sin
]
=2 cos
2
2
2
θ2 + θ1
θ2 − θ1
sin
.
=2 cos
2
2
Then equation (12.5.20) is equivalent to
(x − x0 ) cos
θ2 + θ1
θ2 + θ1
+ (y − y0 ) sin
= 0,
2
2
and so in turn to
π
θ2 + θ1
π
θ2 + θ1
+ ) − (y − y0 ) cos(
+ ) = 0,
(12.5.22)
2
2
2
2
which is the equation in Hesse’s form of a second line through Z0 .
The lines with equations (12.5.21) and (12.5.22) are called the mid-lines of the original pair
l1 and l2 . They are perpendicular to each other.
(x − x0 ) sin(
12.5.8
Axial symmetries
Let l be the line Z0 (Z0 + Uθ ) and for a point Z1 let W be the foot of the perpendicular from Z1
onto l, and Z2 the image of Z1 under axial symmetry in l. Then Z2 = 2W − Z1 and from (12.5.18)
1 −Z0 ,O,Uθ )
. From this we have that
we recall that W = (1 − t)Z1 + t(Z1 + Vθ ) where t = − δFδ(Z
F (O,Uθ ,Vθ )
Z2 − Z0 = Z1 − Z0 − 2
δF (Z1 − Z0 , O, Uθ )
Vθ .
δF (O, Uθ , Vθ )
(12.5.23)
Now it is straightforward to check that 2δF (O, Uθ , Vθ ) = ±1 according as 0 L θ ≺L
π
2 L θ ≺L π. Then
π
π Z2 − Z0 =Z1 − Z0 − 2(±)[− sin θ(x1 − x0 ) + cos θ(y1 − y0 )] cos(θ ⊕ ), sin(θ ⊕ ) ,
2
2
From this we have that
π
π
x2 − x0 =[1 + 2(±) sin θ cos(θ ⊕ )](x1 − x0 ) − 2(±) cos θ cos(θ ⊕ )(y1 − y0 ),
2
2
π
π
y2 − y0 =2(±) sin θ sin(θ ⊕ )(x − x0 ) + [1 − 2(±) cos θ sin θ ⊕ )(y1 − y0 ).
2
2
.
When 0 L θ ≺L π2 this yields
π
π
x2 − x0 ={1 + 2 sin θ[cos(θ + )]}(x1 − x0 ) − 2 cos θ cos(θ + )(y1 − y0 ),
2
2
=(1 − 2 sin2 θ)(x1 − x0 ) + 2 cos θ sin θ(y1 − y0 )
= cos 2θ(x1 − x0 ) + sin 2θ(y1 − y0 )
= cos 2 ⊙ θ(x1 − x0 ) + sin 2 ⊙ θ(y1 − y0 ),
π
π
y2 − y0 =2 sin θ sin(θ + )(x − x0 ) + [1 − 2 cos θ sin(θ + )(y1 − y0 )
2
2
=2 sin θ cos θ(x1 − x0 ) + (1 − 2 cos2 θ)(y1 − y0 )
= sin 2θ(x1 − x0 ) − cos 2θ(y1 − y0 )
= sin 2 ⊙ θ(x1 − x0 ) − cos 2 ⊙ θ(y1 − y0 ).
π
2
or
12.5. RETURN TO CE2
When
π
2
363
L θ ≺L π we obtain
π
)](x1 − x0 ) + 2 cos θ cos(θ ⊕
2
π
=[1 − 2 sin θ cos(θ − )](x1 − x0 ) + 2 cos θ cos(θ −
2
=(1 − 2 sin2 θ)(x1 − x0 ) + 2 cos θ sin θ(y1 − y0 )
x2 − x0 =[1 − 2 sin θ cos(θ ⊕
π
)(y1 − y0 )
2
π
)(y1 − y0 ),
2
= cos 2θ(x1 − x0 ) + sin 2θ(y1 − y0 )
= − cos 2 ⊙ θ(x1 − x0 ) − sin 2 ⊙ θ(y1 − y0 ),
π
π
y2 − y0 = − 2 sin θ sin(θ − )(x1 − x0 ) + [1 + 2 cos θ sin(θ − )]
2
2
= − 2 sin θ(− cos θ)(x1 − x0 ) + (1 + 2 cos θ(− cos θ)(y1 − y0 )
= sin 2θ(x1 − x0 ) − cos 2θ(y1 − y0 )
= − sin 2 ⊙ θ(x1 − x0 ) + cos 2 ⊙ θ(y1 − y0 ).
We can put both of these results in one formula as
12.5.9
x2 − x0
y2 − y0
=±
cos 2 ⊙ θ
sin 2 ⊙ θ
sin 2 ⊙ θ
− cos 2 ⊙ θ
x1 − x0
y1 − y0
.
Composition of two axial symmetries
Apply a first axial symmetry
x2 − x0
y2 − y0
=±
cos 2 ⊙ θ1
sin 2 ⊙ θ1
sin 2 ⊙ θ1
− cos 2 ⊙ θ1
x1 − x0
y1 − y0
.
x2 − x0
y2 − y0
,
and follow this by applying a second axial symmetry
x3 − x0
y3 − y0
=±
cos 2 ⊙ θ2
sin 2 ⊙ θ2
sin 2 ⊙ θ2
− cos 2 ⊙ θ2
so that
x3 − x0
y3 − y0
cos 2 ⊙ θ2
=(±)(±)
sin 2 ⊙ θ2
sin 2 ⊙ θ2
− cos 2 ⊙ θ2
cos 2 ⊙ θ1
sin 2 ⊙ θ1
sin 2 ⊙ θ1
− cos 2 ⊙ θ1
x1 − x0
y1 − y0
The product matrix here is
cos 2 ⊙ θ2 cos 2 ⊙ θ1 + sin 2 ⊙ θ2 sin 2 ⊙ θ1
sin 2 ⊙ θ2 cos 2 ⊙ θ1 − sin 2 ⊙ θ2 cos 2 ⊙ θ1
cos 2 ⊙ θ2 sin 2 ⊙ θ1 − sin 2 ⊙ θ2 cos 2 ⊙ θ1
sin 2 ⊙ θ2 sin 2 ⊙ θ1 + cos 2 ⊙ θ2 cos 2 ⊙ θ1
,
so we end up with
x3 − x0
y3 − y0
= (±)(±)(±)
cos(2 ⊙ θ2 ⊖ cos(2 ⊙ θ1 )
sin(2 ⊙ θ2 ⊖ cos(2 ⊙ θ1 )
− sin(2 ⊙ θ2 ⊖ cos(2 ⊙ θ1 )
cos(2 ⊙ θ2 ⊖ cos(2 ⊙ θ1 )
x1 − x0
y1 − y0
,
where in one of the ± signs we have + or − according as θ1 is acute or right/obtuse, in a second of
the ± signs we have + or − according as θ2 is acute or right/obtuse, and the third of the ± signs
we have + or − according as 2 ⊙ θ1 is less than or greater than 2 ⊙ θ2 .
364
12.5.10
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Some material on triangles
Now consider a triangle with vertices at the points Z5 , Z6 and Z7 such that none of the sidelines
is isotropic. As the points O, Z6 − Z5 , Z6 , Z5 are the vertices of a parallelogram, we have that
Z5 Z6 = O(Z6 − Z5 ),
and similarly
Z6 Z7 = O(Z7 − Z6 ), Z7 Z5 = O(Z5 − Z7 ).
Suppose that θ5,6 , θ6,7 , θ7,5 are the polar angles of the lines Z5 Z6 , Z6 Z7 , Z7 Z5 , respectively, and
we use the further notation that
α =θ5,7 ⊖ θ5,6 = ∢(Z5 Z6 , Z5 Z7 ),
β =θ6,5 ⊖ θ6,7 = ∢(Z6 Z7 , Z6 Z5 ),
γ =θ7,6 ⊖ θ7,5 = ∢(Z7 Z5 , Z7 Z6 ),
for the sensed angles of the triangle. For precision we make the assumption that
θ7,5 ≺L θ6,7 ,
and θ6,7 ≺L θ5,6 .
(12.5.24)
It follows that θ7,5 ≺L θ5,6 .
Then by (12.5.13) we have
2δF (Z5 , Z6 , Z7 ) = − Z5 Z6 Z5 Z7 sin α,
2δF (Z6 , Z7 , Z5 ) =Z6 Z7 Z6 Z5 sin β,
2δF (Z7 , Z5 , Z6 ) =Z7 Z5 Z7 Z6 sin γ,
and as these three are equal we deduce the sine rule for a triangle
−
sin α
sin β
sin γ
=
=
.
Z6 Z7
Z7 Z5
Z5 Z6
(12.5.25)
Continuing with the above notation but now not making the assumption (12.5.24), by (12.5.14)
we have that
Z6 Z7
2
2
=(Z6 − Z5 )(Z7 − Z5 )
=[x7 − x5 − (x6 − x5 )]2 + [y7 − y5 − (y6 − y5 )]2
=[(x7 − x5 )2 + (y7 − y5 )2 ] + [(x6 − x5 )2 + (y6 − y5 )2 ]
− 2[(x7 − x5 )(x6 − x5 ) + (y7 − y5 )(y6 − y5 )]}
2
2
=Z5 Z6 + Z5 Z7 ∓ 2Z5 Z6 Z5 Z7 cos α,
(12.5.26)
where we take the − sign or the + sign according as θ5,6 ≺T θ7,5 or θ7,5 ≺T θ5,6 . This yields the
cosine rule for a triangle.
As a special case suppose that Z5 Z6 is perpendicular to Z5 Z7 so that |α|r = 12 π. Then cos α = 0,
and we deduce that
2
2
2
Z6 Z7 = Z5 Z6 + Z7 Z5 ,
which yields Pythagoras’ theorem.
Straightforward analogues of many results for the real Euclidean plane can be established for
the complex Euclidean plane. For instance by using areal coordinates as in Barry [2, p.179] we can
prove the theorems of Menelaus, Ceva, Desargues on perspective and Pappus, and deal with the
centroid of a triangle.
365
12.6. SQUARE OF DISTANCE, CIRCLES
We can also justify the use of mobile coordinates as in Barry [2, p.192] as follows. Given distinct
points Z6 and Z7 , for any Z we seek complex numbers p and q such that
(x − x6 , y − y6 )
=p(x7 − x6 , y7 − y6 ) + q(−y7 + y6 , x7 − x6 ).
(12.5.27)
Then we need
(x7 − x6 )p + (−y7 + y6 )q =x − x6 ,
(y7 − y6 )p + (x7 − x6 )q =y − y6 .
The determinant of the array of coefficients on the left-hand side here has value
2
(x7 − x6 )2 + (y7 − y6 )2 = Z6 Z7 ,
and this is non-zero for distinct Z6 and Z7 if and only if the line Z6 Z7 is non-isotropic. Under this
condition, by Cramer’s rule we obtain the unique solution
p=
(x − x6 )(x7 − x6 ) + (y − y6 )(y7 − y6 )
2
=±
Z6 , Z7
Z6 Z Z6 Z7 cos ∢(Z6 Z7 , Z6 Z)
Z6 Z7
2
,
Z6 Z
cos ∢(Z6 Z7 , Z6 Z),
Z6 Z7
(y − y6 )(x7 − x6 ) − (x − x6 )(y7 − y6 )
=±
q=
=∓
=∓
(12.5.28a)
2
Z6 Z7
Z6 Z Z6 Z7 sin ∢(Z6 Z7 , Z6 Z)
Z6 Z7
2
,
Z6 Z
sin ∢(Z6 Z7 , Z6 Z)
Z6 Z7
(12.5.28b)
where by (12.5.13) and (12.5.14) we take the + sign in (12.5.28a) and the − sign in (12.5.28b)
when θ6,7 ≺L θ6,− , θ6,7 and θ6,− being the polar angles of the lines O(Z7 − Z6 ) and O(Z − Z6 ),
respectively. These mobile coordinates can then be used to seek to prove analogues of the results
in Barry [2].
12.6
Square of distance, circles
12.6.1
Square of distance
We have defined square-of-distance between two complex points in a very straight-forward way,
by
|Z1 , Z2 |2 = (x2 − x1 )2 + (y2 − y1 )2 .
The square-of-distance from a point to itself is always equal to 0, but the converse is false as
the square-of-distance from Z1 to Z is equal to 0 if and only if
y − y1 = ±i(x − x1 ).
On taking the + and the special point K ≡ (1, i), this is equivalent to
δF (Z, Z1 + K, Z1 ) = 0,
so this is a line through Z1 . On taking the − and the special point L ≡ (1, −i), we obtain
δF (Z, Z1 + L, Z1 ) = 0,
so this is a line, also through Z1 . These lines are of course the isotropic ones through Z1 .
366
12.6.2
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Complex circle
We call a complex circle, with centre Z1 , a set of points Z such that
|Z1 , Z|2 = h2 ,
for some non-zero complex number h. Without loss of generality we always assume that then
h >L 0.
We ask if every line through the centre meets the circle at some point. The answer is no, as the
two isotropic lines through the centre do not; every point on both is at distance 0 from the centre.
However every other line Z1 Z2 through the centre meets each such circle in two points. For points
on the line Z1 Z2 have the form
Z = Z1 + r(Z2 − Z1 ),
r ∈ C,
so that
|Z1 , Z|2
if
r2 =
= r2 |Z1 , Z2 |2
= h2
h2
.
|Z1 , Z2 |2
12.7
Independence of particular frame of reference
12.7.1
Concepts based on sensed-area
As mentioned, the great German mathematician von Staudt pointed out that in using the real
Euclidean plane to construct the complex plane or the real projective plane by an approach based
on coordinates, it is essential that the process used should not depend on the particular frame of
reference taken. Given two frames of reference F = ([O, I , [O, J ), F ′ = ([O′ , I ′ , [O′ , J ′ ), where
OI ⊥ OJ, O′ I ′ ⊥ O′ J ′ , if a point Z has Cartesian coordinates (x, y), (x′ , y ′ ) with respect to these
frames, respectively, then the coordinates will be connected by a transformation of the form
x′ =ax + by + k1 ,
y ′ =cx + dy + k2 ,
where ad − bc = ±1. This will consist of the composition of a translation and either a rotation or
an axial symmetry. We wish to use the rotation and for this ad − bc = 1. It is convenient to use
matrix notation for this transformation,
 ′  


x
a b k1
x
 y ′  =  c d k2   y  .
1
0 0 1
1
We extend this to the complex plane, with
and consider the transformation
 ′  
x
a
 y′  =  c
1
0
the coordinates now all being complex numbers,


b k1
x
d k2   y  .
0 1
1
If we apply this in turn to the three points Z1 , Z2 , Z3 we obtain
 ′
 

x1 x′2 x′3
a b k1
x1 x2
 y1′ y2′ y3′  =  c d k2   y1 y2
1
1
1
0 0 1
1
1
(12.7.1)

x3
y3  .
1
367
12.7. INDEPENDENCE OF PARTICULAR FRAME OF REFERENCE
From this we deduce that
 ′
x1 x′2
det  y1′ y2′
1
1
and so


x′3
a
y3′  = det  c
1
0


b k1
x1
d k2  det  y1
0 1
1
x2
y2
1

x3
y3  ,
1
δF ′ (Z1 , Z2 , Z3 ) = (ad − bc)δF (Z1 , Z2 , Z3 ),
and in this ad − bc = 1. Thus sensed-area is invariant under the transformation (12.7.1).
From this, in §12.2.1 we see that complex areal coordinates of Z ′ with respect to (Z1′ , Z2′ , Z3′ )
are the same as the complex areal coordinates of Z with respect to (Z1 , Z2 , Z3 ). In §12.2.2 we see
that the complex line δF (Z, Z2 , Z3 ) = 0 maps onto the complex line δF ′ (Z ′ , Z2′ , Z3′ ) = 0. We also
see in §12.5.1?? that parametric equations of a line are invariant under (12.7.1) as in (12.5.2) we
have
δF (Z, Z2 , Z3 )
,
r=
δF (Z1 , Z2 , Z3 )
for any Z1 not on the line Z2 Z3 . In particular the case of r = 12 shows that the concept of mid-point
of two points is invariant. It also follows that complex sensed-distance is invariant.
As our transformation is 1:1 and onto, clearly the images of non-intersecting lines are nonintersecting and on putting this together with the case of a line being parallel to itself we see
that the images of two parallel lines are also parallel. We can also include this in the following
argument. By analogy with (11.9.1) and (11.9.4) we see that δF (Z, Z2 , Z3 ) = lx + my + n where
l = − 12 (y3 − y2 ) = δF (O, I, Z2 − Z3 ),
m = 12 (x3 − x2 ) = δF (O, J, Z2 − Z3 ),
n =δF (O, Z2 , Z3 ).
(12.7.2)
Then the lines Z2 Z3 and Z4 Z5 are perpendicular if and only if
δF (O, I, Z2 − Z3 )δF (O, I, Z4 − Z5 )
+ δF (O, J, Z2 − Z3 )δF (O, J, Z4 − Z5 ) = 0.
From this it follows that the image lines are also perpendicular to each other, as O′ I ′ ⊥ O′ J ′ and
|O′ , I ′ | = |O′ , J ′ | = 1.
We note moreover that
y3 − y2 ± i(x3 − x2 )
= − 2δF (O, I, Z2 − Z3 ) ± i2δF (O, J, Z2 − Z3 ),
so that the image of an isotropic line is an isotropic line of the same class. We also have that
(x3 − x2 )2 + (y3 − y2 )2
=4 δF (O, J, Z2 − Z3 )2 + δF (O, I, Z2 − Z3 )2 ,
so that square of distance is invariant.
To deal with Grassmann supplements, we note that by (12.7.2) we can write
(x3 − x2 , y3 − y2 ) and (−y3 + y2 , x3 − x2 )
as
(2δF (O, J, Z2 − Z3 ), −2δF (O, I, Z2 − Z3 )) and
(2δF (O, I, Z2 − Z3 ), 2δF (O, J, Z2 − Z3 )).
Thus Grassmann supplements are invariant.
(12.7.3)
368
12.7.2
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Duo-sectors
Suppose that for θ3 , θ4 , θ5 ∈ C we have θ3 4L θ5 4L θ4 . Then for all θ6 ∈ C, it follows that
θ3 + θ6 4 L θ5 + θ6 4 L θ4 + θ6 .
For with θ3 = φ3 + iψ3 , θ5 = φ5 + iψ5 , θ6 = φ6 + iψ6 , as θ3 4L θ5 we have either φ3 < φ5
or φ3 = φ5 , ψ3 ≤ ψ5 . In these cases we then have either φ3 + φ6 < φ5 + φ6 or φ3 + φ6 =
φ5 + φ6 , ψ3 + ψ6 ≤ ψ5 + ψ6 . Hence θ3 + θ6 4L θ5 + θ6 . On applying this to θ5 4L θ4 we also have
θ5 + θ6 4 L θ4 + θ6 .
Now suppose that θ3 , θ4 , θ5 , θ6 ∈ T , with θ3 ≺L θ4 , and to meet von Staudt’s point exactly
we suppose that θ6 = φ6 is a non-zero real number. We subdivide into three cases as follows.
Case A. Suppose that θ4 ≺L π − φ6 .
Case B. Suppose that π − φ6 4L θ3 .
Case C. Suppose that θ3 ≺L π − φ6 4L θ4 .
We work with Case A in detail and suppose first that θ3 4L θ5 4L θ4 . Then
θ3 + φ6 4L θ5 + φ6 4L θ4 + φ6 ≺L π,
so that
θ3 + φ6 ∈ T
θ5 + φ6 ∈ T
θ4 + φ6 ∈ T
and so θ3 ⊕ φ6 = θ3 + φ6 ,
and so θ5 ⊕ φ6 = θ5 + φ6 ,
and so θ4 ⊕ φ6 = θ4 + φ6 .
It follows that
θ3 ⊕ φ6 4L θ5 ⊕ φ6 4L θ4 ⊕ φ6 .
(12.7.4)
Continuing with Case A, we now wish to consider the two subcases
(a) θ4 4L θ5 ≺L π,
(b) 0 4L θ5 4L θ3 ,
and we further subdivide subcase (a) into
(a)(i) θ4 4L θ5 ≺L π − φ6 ,
(a)(ii) π − φ6 4L θ5 ≺L π.
In subcase (a)(i) we have
θ4 + φ6 4L θ5 + φ6 ≺L π,
so that in turn θ4 + φ6 , θ4 + φ6 ∈ T , θ4 ⊕ φ6 = θ4 + φ6 , θ5 ⊕ φ6 = θ5 + φ6 , and so
θ4 ⊕ φ6 4L θ5 ⊕ φ6 ≺L π.
(12.7.5)
In subcase (a)(ii) we have π 4L θ5 + φ6 so that θ5 + φ6 6∈ T . Then θ5 ⊕ φ6 = θ5 + φ6 − π and
0 4L θ5 ⊕ φ6 ≺L φ6 .
(12.7.6)
In subcase (b)
φ6 4L θ5 + φ6 4L θ3 + φ6 ≺L θ4 + φ6 ≺L π.
Then θ5 + φ6 ∈ T so that θ5 ⊕ φ6 = θ5 + φ6 and hence
φ6 4L θ5 ⊕ φ6 4L θ3 ⊕ φ6 .
(12.7.7)
12.7. INDEPENDENCE OF PARTICULAR FRAME OF REFERENCE
369
Thus in Case A we have either the conclusion (12.7.4) or the combined conclusions (12.7.5),
(12.7.6) and (12.7.7). In Cases B and C, by similar methods we obtain similar conclusions.
To apply these, starting with 0 4L θ3 ≺L θ4 , we consider the lines l3 = OZ3 , l4 = OZ4 , l5 =
OZ5 and assume that these have equations
x sin θ3 − y cos θ3 =0,
x sin θ4 − y cos θ4 =0,
x sin θ5 − y cos θ5 =0,
respectively. Then we know from §12.5.5 that under rotation about the origin through an angle
of magnitude θ6 , these lines have images l3′ , l4′ , l5′ with angles of inclination, in T , θ3 ⊕ θ6 , θ4 ⊕
θ6 , θ5 ⊕ θ6 , respectively.
In Case A, in first taking θ3 4L θ5 4L θ4 we are considering the duo-sector Dl∗3 ,l4 and see that
its lines map under the rotation into the duo-sector Dl∗′ ,l′ . In continuing with Case A, we showed
3 4
that the lines in the supplementary duo-sector Dl∗∗
map into the duo-sector Dl∗∗
As every
′ ,l′ .
3 ,l4
3 4
non-isotropic line through the origin is the image of some non-isotropic line under this rotation,
and Dl∗′ ,l′ ∩ Dl∗∗
′ ,l′ we can conclude that this mapping from duo-sector to duo-sector is onto.
3 4
3 4
We obtain similar conclusions in Cases B and C.
370
CHAPTER 12. THE COMPLEX EUCLIDEAN PLANE
Chapter 13
Extension of Euclidean to
projective planes
Motivation of the type that we are giving in Chapters 12, 13 and 14 is contained, on a smaller
scale, in Archbold [1, pages 1– 44], Modenov and Parkhomenko [20], Robinson [26, Chapter V],
Seidenberg [27, Chapter 1], and Walker [30, pages 13–66]
13.1
Homogeneous coordinates, the Real Projective Plane
13.1.1
Motivation
In the real Euclidean plane RE2 , in a number of instances against a background of generally
existing, a concept fails to exist in somewhat isolated circumstances. Thus distinct parallel lines
do not have a point of intersection, the mid-point of two points does not have a harmonic conjugate,
the bisector of a pair of intersecting lines does not have a harmonic conjugate, a parabola does not
have a centre, for a central conic the centre does not have a polar, for hyperbolas and parabolas
there are lines of deficient incidence.
On the other hand, the form of an equation of a diametral line of a conic was seen in §7.4.1 to
be the same as the form of the equation in §7.2.5 of the polar with respect to the conic of a point
Z with normalised areal coordinates (α, β, γ), but the former involves a triple (α, β, γ) for which
α + β + γ = 0, and so this is not a triple of areal coordinates for a point, for which we would have
to have α + β + γ = 1. This (and other material ?) raises the question of whether advantageous
use can be made, in our geometry, of triples (α, β, γ) for which α + β + γ = 0. The expression in
(11.8.2) would give us an opening to handle such triples, and we pursue such matters now.
In (3.1.2) we had
x = x1 α + x2 β + x3 γ, y = y1 α + y2 β + y3 γ,
(13.1.1)
where
δF (Z, Z3 , Z1 )
δF (Z, Z1 , Z2 )
δF (Z, Z2 , Z3 )
, β=
, γ=
,
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
δF (Z1 , Z2 , Z3 )
are normalised areal coordinates. This gives a relationship between areal and Cartesian coordinates.
We recall that for all points Z ∈ RE2 , α + β + γ = 1. We now note that if we write
α=
ξ = x1 α + x2 β + x3 γ,
η = y1 α + y2 β + y3 γ,
and
ξ
,
ζ
y=
η
,
ζ
x1 α + x2 β + x3 γ
,
α+β+γ
y=
y1 α + y2 β + y3 γ
,
α+β+γ
x=
then we have
x=
ζ = α + β + γ,
371
(13.1.2)
(13.1.3)
372
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
and when ζ 6= 0 or, equivalently, α + β + γ 6= 0, this yields the transformation (13.1.1) on RE2 .
If we extend to triples (ξ, η, ζ) with ζ = 0 or equivalently to triples (α, β, γ) with α + β + γ = 0,
(13.1.2) still gives a correspondence between them.
We remark that in RE2 the form (1 − s)Z1 + sZ2 , for parametric equations of a line, in the case
when 1 − s = −s does not generally give a point in RE2 as no s satisfies this. This case corresponds
to Z1 − Z2 or Z2 − Z1 , and the line O(Z2 − Z1 ) is parallel to Z1 Z2 . Thus Z2 − Z1 yields a point
of Z1 Z2 only when Z1 Z2 passes through the origin.
13.1.2
Use of Plücker’s homogeneous coordinates for projective points
We now take our biggest step, that of entering on plane projective geometry. Aspects of this topic
became familiar to the painters and architects of Western and Central Europe of Renaissance times,
through their consideration of perspective. The first to apply these concepts to a study of geometry
was Girard Desargues with a book published in 1639. He had one follower in Blaise Pascal but
then the subject languished until about 1790 when it was taken up by the students of Gaspard
Monge. The treatment of the subject was synthetic until 1831 when Julius Plücker introduced the
concept of homogeneous coordinates, the use of which made the subject much more approachable.
The initial idea is easy to explain. We are used to locating a point Z of the real Euclidean plane
by a pair of Cartesian coordinates (x, y) relative to some rectangular frame of reference F . We can
also locate Z in an unusual way by a triple (ξ, η, ζ) of real numbers such that ζ 6= 0, where these
are to be selected so that
η
ξ
x= , y= .
ζ
ζ
In this way Z would also be represented by a triple (ξ ′ , η ′ , ζ ′ ) with ζ ′ 6= 0 if and only if
ξ′
ξ η′
η
=
, ′ = .
′
ζ
ζ ζ
ζ
Now if we write ζ ′ = kζ we have k 6= 0 and
ξ′ =
ζ′
ζ′
ξ = kξ, η ′ = η = kη.
ζ
ζ
Thus the triple (ξ ′ , η ′ , ζ ′ ) represents the same point Z as the triple (ξ, η, ζ) if and only if
ξ ′ = kξ, η ′ = kη, ζ ′ = kζ,
for some k 6= 0. Because of this property these triples are called Plücker’s homogeneous
coordinates.
These triples are certainly more awkward to deal with than Cartesian coordinates but they allow
us to consider as well triples of the form (ξ, η, 0) which are not included above. We can regard this
as adding new points to the Euclidean plane and it opens up a whole field of exploration. To start
with, for consistency we shall regard (ξ, η, 0) and (ξ ′ , η ′ , 0) as representing the same new point if
and only if
ξ ′ = kξ, η ′ = kη,
for some k 6= 0. We do not however regard multiples of the triple (0, 0, 0) as representing a point
in this extended system.
To help in our development we introduce notation as follows. As well as writing RE2 for real
Euclidean two space, we use R2 for the set of all pairs of real numbers; the latter constitute of
course the rectangular Cartesian coordinates of points in RE2 , relative to some frame of reference.
Correspondingly we write RE3 for real Euclidean three space and R3 for the set of all triples of real
numbers; the latter constitute of course the rectangular Cartesian coordinates of points in RE3 ,
relative to some frame of reference.
13.1. HOMOGENEOUS COORDINATES, THE REAL PROJECTIVE PLANE
373
We recall that R2 is a vector space over R under the operations
(x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ),
k(x1 , y1 ) = (kx1 , ky1 ) for all k ∈ R,
and that also R3 is a vector space over R under the operations
(ξ1 , η1 , ζ1 ) + (ξ2 , η2 , ζ2 ) =(ξ1 + ξ2 , η1 + η2 , ζ1 + ζ2 ),
k(ξ1 , η1 , ζ1 ) =(kξ1 , kη1 , kζ1 ) for all k ∈ R.
(13.1.4)
We shall seek to make use as far as possible of the familiar vector properties in (13.1.4) subject to
the overriding stipulation that
for a given (ξ, η, ζ) 6= (0, 0, 0), the triples k(ξ, η, ζ) are to be interpreted
as representing one point for all non-zero k ∈ R.
(13.1.5)
For any triple (ξ1 , η1 , ζ1 ) 6= (0, 0, 0) in R3 we call the set of multiples k(ξ1 , η1 , ζ1 ) for k 6= 0 in R a
projective point or p-point, and denote it by [(ξ1 , η1 , ζ1 )]. We call (ξ1 , η1 , ζ1 ) a representative
of this p-point. The set of all such p-points we shall call the real projective plane and denote
it by RP2 .
Now if these representative triples are viewed as the Cartesian coordinates of points in RE3
relative to some frame of reference F , they correspond to points on a line OZ1 , where Z1 ≡F
(ξ1 , η1 , ζ1 ) is other than the origin O itself. We call such a set in RE3 an O-deleted line. Thus
each p-point in RP2 corresponds to an O-deleted line in RE3 . This gives us a model for p-points
in a familiar context.
To prepare to deal with pairs of p-points, we suppose first that representatives (ξ1 , η1 , ζ1 ) and
(ξ2 , η2 , ζ2 ) of p-points are linearly dependent in R3 , so that (ξ2 , η2 , ζ2 ) = j(ξ1 , η1 , ζ1 ), for some
j 6= 0. Then when we consider for non-zero p and q,
p(ξ1 , η1 , ζ1 ) + q(ξ2 , η2 , ζ2 ) = p(ξ1 , η1 , ζ1 ) + qj(ξ1 , η1 , ζ1 ) = (p + qj)(ξ1 , η1 , ζ1 ),
we see that we have multiples of some representative of just one p-point. Thus linearly dependent
representative triples refer to the same p-point and linear combinations of them just lead us to
deal with that one p-point. Because of this they are not of interest to us and we do not consider
them further.
Turning now to deal with pairs of p-points for which representatives (ξ1 , η1 , ζ1 ) and (ξ2 , η2 , ζ2 )
are linearly independent in R3 , we consider for real numbers p and q not both equal to 0,
p(ξ1 , η1 , ζ1 ) + q(ξ2 , η2 , ζ2 ) = (pξ1 + qξ2 , pη1 + qη2 , pζ1 + qζ2 ).
(13.1.6)
Now taking other representative triples for the p-points involved in (13.1.6), we consider
p(k1 ξ1 , k1 η1 , k1 ζ1 ) + q(k2 ξ2 , k2 η2 , k2 ζ2 ) = (pk1 ξ1 + qk2 ξ2 , pk1 η1 + qk2 η2 , pk1 ζ1 + qk2 ζ2 ), (13.1.7)
and look at the ratios
pξ1 + q kk12 ξ2
pk1 ξ1 + qk2 ξ2
=
,
pk1 ζ1 + qk2 ζ2
pζ1 + q kk12 ζ2
pη1 + q kk12 η2
pk1 η1 + qk2 η2
=
.
pk1 ζ1 + qk2 ζ2
pζ1 + q kk12 ζ2
(13.1.8)
We see that the ratios in (13.1.8) are constant, for constant ξ1 , η1 , ζ1 , ξ2 , η2 , ζ2 , p and q 6= 0, if
and only if the ratio k2 /k1 is constant. Thus the expressions in (13.1.7) do not represent p-points
directly. What can be done? The answer is that we regard (13.1.7) as producing a representative
triple of some p-point and now turn to seeing this in the simplest possible light.
Returning to (13.1.8), when p 6= 0 we write the ratios as
ξ1 +
pk1 ξ1 + qk2 ξ2
=
pk1 ζ1 + qk2 ζ2
ζ1 +
qk2
pk1 ξ2
,
qk2
pk1 ζ2
η1 +
pk1 η1 + qk2 η2
=
pk1 ζ1 + qk2 ζ2
ζ1 +
qk2
pk1 η2
,
qk2
pk1 ζ2
(13.1.9)
374
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
on keeping ξ1 , η1 , ζ1 , ξ2 , η2 , ζ2 fixed. Given any λ 6= 0 in R we can then obtain the representative
(ξ1 + λξ2 , η1 + λη2 , ζ1 + λζ2 ) from arbitrary non-zero k1 and k2 by suitably choosing p and q in
(13.1.9) and so are led to a p-point which we denote by
[(ξ1 + λξ2 , η1 + λη2 , ζ1 + λζ2 )].
(13.1.10)
Perhaps it should be pointed out that this selection depends not only on the p-points [(ξ1 , η1 , ζ1 )],
[(ξ2 , η2 , ζ2 )], but on the particular representatives (ξ1 , η1 , ζ1 ), (ξ2 , η2 , ζ2 ) that we use. In effect we
have performed a normalization. In our motivation, in (13.1.9) it was assumed that we did not
have ζ1 = ζ2 = 0, but for that case in (13.1.10) we still have
(pk1 ξ1 + qk2 ξ2 , pk1 η1 + qk2 η2 , 0) = pk1 (ξ1 + λξ2 , η1 + λη2 , 0),
where λ = qk2 /pk1 .
As well as this, for the case λ = 0 we take the original p-point
[(ξ1 , η1 , ζ1 )].
(13.1.11)
[(ξ2 , η2 , ζ2 )]
(13.1.12)
We can force the p-point
into the form (13.1.10) by taking
1
1
1
[(ξ1 + λξ2 , η1 + λη2 , ζ1 + λζ2 )] = [( ξ1 + ξ2 , η1 + η2 , ζ1 + ζ2 )],
λ
λ
λ
and replacing 1/λ by 0; we will say that this corresponds to taking λ = ∞.
13.1.3
Projective lines of p-points
We now call the set of p-points in (13.1.10) for λ ∈ R ∪ {∞} (so that it is extended to include
(13.1.11) and (13.1.12)) a projective line or p-line. Writing
ξ = ξ1 + λξ2 , η = η1 + λη2 , ζ = ζ1 + λζ2 ,
for λ ∈ R ∪ {∞},
we say that we have parametric equations for representative triples of p-points on a
p-line, which we abbreviate to parametric equations for a p-line.
To prepare for the next material, for two representative triples of p-points (ξ1 , η1 , ζ1 ) and
(ξ2 , η2 , ζ2 ) we define the determinants
ξ1 η1 ζ1 ξ1 η1 ζ1 = ξ1 η2 − ξ2 η1 .
= ζ1 ξ2 − ζ2 ξ1 , N1,2 = = η1 ζ2 − η2 ζ1 , M1,2 = L1,2 = ξ2 η2 ζ2 ξ2 η2 ζ2 (13.1.13)
We now prove the helpful result that
LEMMA. (ξ1 , η1 , ζ1 ) and (ξ2 , η2 , ζ2 ) are linearly dependent in R3 if and only each of the pairs
(η1 , ζ1 ) and (η2 , ζ2 ), (ζ1 , ξ1 ) and (ζ2 , ξ2 ), and (ξ1 , η1 ) and (ξ2 , η2 ) is linearly dependent in R2 , and
this holds if and only if all three of L1,2 , M1,2 , N1,2 have the value 0.
For this suppose first that (ξ1 , η1 , ζ1 ) and (ξ2 , η2 , ζ2 ) are linearly dependent in R3 . Then we
have that
k1 (ξ1 , η1 , ζ1 ) + k2 (ξ2 , η2 , ζ2 ) = (0, 0, 0),
for some (k1 , k2 ) 6= (0, 0). It immediately follows that
k1 (η1 , ζ1 ) + k2 (η2 , ζ2 ) = (0, 0),
so that these pairs are linearly dependent in R2 , and hence that L1,2 = 0. A similar argument
applies to the other two pairs.
Conversely suppose that L1,2 = M1,2 = N1,2 = 0. We subdivide into two cases.
13.1. HOMOGENEOUS COORDINATES, THE REAL PROJECTIVE PLANE
375
CASE 1. Suppose that at least one of (ξ1 , ξ2 ), (η1 , η2 ), (ζ1 , ζ2 ) is a pair of zeros. Taking first
that (ξ1 , ξ2 ) = (0, 0), we note that since L1,2 = 0 its rows are linearly dependent in R2 and so there
are numbers k1 and k2 , not both equal to 0 such that k1 (η1 , ζ1 ) + k2 (η2 , ζ2 ) = (0, 0). It follows
immediately that
k1 (0, η1 , ζ1 ) + k2 (0, η2 , ζ2 ) = (0, 0, 0),
and so our two triples are linearly dependent in R3 .
A similar argument applies to the other two cases.
CASE 2. Suppose that CASE 1 does not hold, i.e. that
(ξ1 , ξ2 ) 6= (0, 0), (η1 , η2 ) 6= (0, 0) and (ζ1 , ζ2 ) 6= (0, 0).
From M1,2 = N1,2 = 0 we have that
−ζ2 ξ1 + ζ1 ξ2 =0,
η2 ξ1 − η1 ξ2 =0.
The columns of L1,2 are linearly dependent since its value is equal to 0, and so from these equations
we have that either
or
−ζ2 ξ1 + ζ1 ξ2 =0,
−j1 (−ζ2 ξ1 + ζ1 ξ2 ) =0,
ζ1
η1
, for some j1 6= 0.
=j1
with
ζ2
η2
−j2 (η2 ξ1 − η1 ξ2 ) =0,
η2 ξ1 − η1 ξ2 =0,
η1
ζ1
, for some j2 6= 0.
=j2
with
η2
ζ2
(13.1.14a)
(13.1.14b)
(13.1.15a)
(13.1.15b)
In (13.1.14a) when ξ1 6= 0 we have ζ2 = ξξ21 ζ1 , and then on using (13.1.14b) twice we obtain
ξ2
ξ2
(ξ2 , η2 , ζ2 ) = ξ2 , η2 , ζ1 = ξ2 , j1 ζ2 , ζ1
ξ1
ξ1
ξ2
ξ2
ξ2
ξ2
= ξ2 , j1 ζ1 , ζ1 = (ξ1 , j1 ζ1 , ζ1 ) = (ξ1 , η1 , ζ1 ),
ξ1
ξ1
ξ1
ξ1
so that the two representative triples are linearly dependent.
When instead in (13.1.14a) we have ξ2 6= 0, we also have ζ1 = ξξ12 ζ2 , and then on using (13.1.14b)
twice we obtain
ξ1
ξ1
(ξ1 , η1 , ζ1 ) = ξ1 , η1 , ζ2 = ξ1 , j1 ζ1 , ζ2
ξ2
ξ2
ξ1
ξ1
ξ1
ξ1
= ξ1 , j1 ζ2 , ζ2 = (ξ2 , j1 ζ2 , ζ2 ) = (ξ2 , η2 , ζ2 ),
ξ2
ξ2
ξ2
ξ2
so that the two representative triples are linearly dependent.
A similar argument applies to (13.1.15a) and (13.1.15b). This establishes the Lemma.
We turn now to another form of equation for a p-line. If (ξ, η, ζ) is any representative of
(13.1.10) we have that for some k 6= 0 in R
ξ − kξ1 − kλξ2 =0,
η − kη1 − kλη2 =0,
ζ − kζ1 − kλζ2 =0.
376
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
As we have a non-trivial solution (1, −k, −kλ)

ξ
det  ξ1
ξ2
and so
for homogeneous equations here we must have

η ζ
(13.1.16)
η1 ζ1  = 0,
η2 ζ2
L1,2 ξ + M1,2 η + N1,2 ζ = 0.
(13.1.17)
This equation (13.1.17) is also satisfied for the case λ = ∞. Conversely suppose that (ξ, η, ζ) is a
representative triple which satisfies (13.1.17). Then the rows of the matrix in (13.1.16) are linearly
dependent in R3 ,
l(ξ, η, ζ) + m(ξ1 , η1 , ζ1 ) + n(ξ2 , η2 , ζ2 ) = (0, 0, 0),
for some (l, m, n) 6= (0, 0, 0). We cannot have l = 0 as that would imply that m(ξ1 , η1 , ζ1 ) +
n(ξ2 , η2 , ζ2 ) = (0, 0, 0), with (m, n) 6= (0, 0), and this is ruled out as these triples are linearly
independent in R3 . Thus l 6= 0 and so
(ξ, η, ζ) = −
m
n
(ξ1 , η1 , ζ1 ) − (ξ2 , η2 , ζ2 ).
l
l
Then −m/l and −n/l cannot both be equal to 0 as (ξ, η, ζ) = (0, 0, 0) is ruled out. Thus
(ξ, η, ζ) = p(ξ1 , η1 , ζ1 ) + q(ξ2 , η2 , ζ2 ),
for some p, q ∈ R, not both equal to 0.
The equation (13.1.17) has the form
aξ + bη + cζ = 0,
(13.1.18)
where the case a = b = c = 0 cannot occur, as we showed in the Lemma that L1,2 , M1,2 and N1,2
cannot all have the value 0. In fact every equation (13.1.18) with this restriction arises in this way.
For this, given (a, b, c) 6= (0, 0, 0), we need to find representative triples (ξ1 , η1 , ζ1 ), (ξ2 , η2 , ζ2 ),
which are linearly independent in R3 , such that
η1 ζ2 − ζ1 η2 =ka,
ζ1 ξ2 − ξ1 ζ2 =kb,
ξ1 η2 − η1 ξ2 =kc,
(13.1.19)
for some k 6= 0 in R. Suppose first that a 6= 0 and so η1 ζ2 − ζ1 η2 6= 0. Then from the second and
third equations in (13.1.19) we obtain that
ξ1 = −
bη1 + cζ1
bη2 + cζ2
, ξ2 = −
.
a
a
We can ensure that the triples are linearly independent in R3 by choosing ζ2 = ζ1 and η2 6= η1 .
This completes the first case, and the cases when b 6= 0 and c 6= 0 are dealt with similarly.
As an immediate application of this we see that the analogue in RE3 of a p-line is a plane which
passes through the origin O, less the point O itself. We shall call such a set a O-deleted plane.
We saw in (13.1.8) that we do not restrict obtaining (13.1.10) by choosing k1 and k2 arbitrarily
apart from being non-zero. We make use of this in (13.1.10) when ζ1 6= 0 and ζ2 6= 0. Then if for
distinct points Z1 ≡ (ξ1 , η1 ), Z2 ≡ (ξ2 , η2 ) in RE2 , we write
1 Z1
≡ (ξ1 , η1 , 1),
1 Z2
≡ (ξ2 , η2 , 1),
for the corresponding points on the plane with equation ζ = 1 in RE3 , then we are starting with
O-deleted lines from the lines O(1 Z1 ) and O(1 Z2 ) and in (13.1.10) obtain
1
λ
1
λ
[(ξ1 + λξ2 , η1 + λη2 , 1 + λ)] =
ξ1 +
ξ2 ,
η1 +
η2 , 1 .
(13.1.20)
1+λ
1+λ
1+λ
1+λ
13.1. HOMOGENEOUS COORDINATES, THE REAL PROJECTIVE PLANE
377
If we denote by 1 W the indicated representative in this, then we obtain in the plane with equation
ζ = 1 the ratio of sensed-distances
(1 Z1 )(1 W )
= λ,
(1 W )(1 Z2 )
and this is equal to
Z1 W
.
W Z2
This gives us an interpretation of the role of λ.
The p-points [(ξ, η, ζ)] with ζ 6= 0 are isomorphic in aggregate to RE2 and we refer to them
as ordinary p-points. The new p-points [(ξ, η, 0)] we shall refer to as ideal; as they are the ones
which satisfy ζ = 0 which is an equation of the form (13.1.18), they form a p-line which we refer
to as the ideal line. Older names for ideal p-points and the ideal line were points at infinity
and the line at infinity.
There are several reasons for excluding the case of the set of multiples of (0, 0, 0) being a p-point.
One very good one is that if it were allowed then the coordinates (ξ, η, ζ) = (0, 0, 0) would be valid
in (13.1.18) for all triples (a, b, c) 6= (0, 0, 0). Thus we would have a p-point which lies on every
p-line, so that all the p-lines would be concurrent. This would result in a geometrical pattern very
different from what we are generalizing.
Our task now is to generalize as far as we can our geometrical concepts from RE2 to RP2 in
the hope that there will be some advantages in working in the enlarged context. As always with
generalization, we must expect some properties to be restricted or not to survive. The first casualty
is that we do not have a realistic two-dimensional visualization or diagram for RP2 in the process.
See however Springer [28, pages 144–151].
13.1.4
Further properties of p-lines
Let us take two distinct p-points on the p-line with equation given by (13.1.16), with parameters
λ3 , λ4 ∈ R in (13.1.10). Then by (13.1.16) the p-line on these p-points has equation
ξ η ζ 0 = ξ3 η3 ζ3 ξ4 η4 ζ4 ξ
η
ζ
= ξ1 + λ3 ξ2 η1 + λ3 η2 ζ1 + λ3 η2 ξ1 + λ4 ξ2 η1 + λ4 η2 ζ1 + λ4 η2 ξ η ζ = ξ1 η1 ζ1 .
ξ1 η1 ζ1 This is the equation of the original p-line, and as this argument can be modified to include a
p-point with parameter λ = ∞, we can conclude that any p-line is determined uniquely by any two
of its points.
For any distinct points Z1 ≡ (x1 , y1 ), Z2 ≡ (x2 , y2 ) in RE2 , as in (13.1.20) we can take
representatives (x1 , y1 , 1), (x2 , y2 , 1) and then parametric equations
ξ = x1 + λx2 , η = y1 + λy2 , ζ = 1 + λ,
λ ∈ R ∪ {∞},
for the p-line containing them. On putting λ = −1 we see that there is a unique ideal p-point
on this p-line, with representative (x1 − x2 , y1 − y2 , 0). We call this p-line an extension of the
ordinary line Z1 Z2 and denote it by Z1 Z2e ; we also call Z1 Z2 the restriction of this p-line. We
denote by Z1 Z2∞ the unique ideal point on Z1 Z2e .
378
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
Let us now consider two ordinary lines Z1 Z2 and Z3 Z4 . Then the ideal points Z1 Z2∞ , Z3 Z4∞ on
their extensions Z1 Z2e and Z3 Z4e have representatives
(x1 − x2 , y1 − y2 , 0), (x3 − x4 , y3 − y4 , 0).
This is the same ideal point if and only if
x3 − x4 = k(x1 − x2 ), y3 − y4 = k(y1 − y2 ),
for some k 6= 0. But this happens if and only if Z1 Z2 k Z3 Z4 . Thus the ordinary lines Z1 Z2 and
Z3 Z4 are parallel if and only if their extensions Z1 Z2e and Z3 Z4e have an ideal point in common.
In a way, this is the most striking feature of the projective plane.
To deal with parametric equations for the ideal line let (ξ1 , η1 , 0), (ξ2 , η2 , 0) be representative
triples of distinct p-points Z1 and Z2 , so that ξ1 η2 − η1 ξ2 6= 0. Then for any (ξ, η, 0) 6= (0, 0, 0) as
representatives of a p-point Z we seek p, q ∈ R so that
pξ1 + qξ2 =ξ,
pη1 + qη2 =η.
These equations have the unique solutions
p=
−η1 ξ + ξ1 η
η2 ξ − ξ2 η
, q=
,
ξ1 η2 − η1 ξ2
ξ1 η2 − η1 ξ2
and this yields the representative (η2 ξ − ξ2 η)Z1 + (−η1 ξ + ξ1 η)Z2 for Z.
Consider now two (distinct) lines, with equations
a1 ξ + b1 η + c1 ζ = 0, a2 ξ + b2 η + c2 ζ = 0,
so that (a1 , b1 , c1 ) 6= (0, 0, 0), (a2 , b2 , c2 ) 6= (0, 0, 0) and these triples are not linearly dependent in
R3 . Then by the Lemma,
b 1 c2 − c1 b 2 , c1 a2 − a1 c2 , a1 b 2 − b 1 a2 ,
cannot all be equal to 0. Supposing that a1 b2 − b1 a2 6= 0 and considering
a1 ξ + b1 η = − c1 ζ,
a2 ξ + b2 η = − c2 ζ,
we obtain the solutions
b 1 c2 − c1 b 2
c1 a2 − a1 c2
ζ, η =
ζ,
a1 b 2 − b 1 a2
a1 b 2 − b 1 a2
and this yields a unique p-point as a solution. The pattern is similar in the other two cases. Thus
we can conclude that any two distinct p-lines meet in a unique p-point.
An other way of presenting some of the above is that given an equation aξ + bη + cζ = 0 of an
extended line, to find the ideal point on it we merely put ζ = 0 in it and then have aξ + bη = 0
from which we can read off the solution ξ = b, η = −a and so the ideal point [(b, −a, 0)].
Moreover if we have two distinct parallel lines, equations of their extensions can be put in the
form
aξ + bη + c1 ζ = 0, aξ + bη + c2 ζ = 0, (c1 6= c2 ).
ξ=
By subtraction we note that these hold simultaneously if and only (c1 − c2 )ζ = 0 and so ζ = 0, so
that they have an ideal point in common.
We can revisit parametric equations for the ideal line as follows. Given two triples (b1 , −a1 , 0), b2 , −a2 , 0)
which are linearly independent and so represent distinct ideal points, we seek to represent a general triple (b, −a, 0) with (a, b) 6= (0, 0) in the form (b, −a, 0) = λ(b1 , −a1 , 0) + µ(b2 , −a2 , 0). This
requires
b1 λ + b2 µ =b,
−a1 λ − a2 µ = − a,
13.1. HOMOGENEOUS COORDINATES, THE REAL PROJECTIVE PLANE
379
and has the unique solution
λ=
b 2 a − a2 b
,
b 2 a1 − a2 b 1
µ=
−b1 a + a1 b
.
b 2 a1 − a2 b 1
We can then pick the representative pair λ = b2 a − a2 b, µ = −b1 a + a1 b.
13.1.5
p-Pencils of p-lines
Given any p-point [(ξ3 , η3 , ζ3 )], the set of all p-lines passing through [(ξ3 , η3 , ζ3 )] is called a p-pencil
of p-lines with vertex [(ξ3 , η3 , ζ3 )].
Now take any p-points [(ξ1 , η, ζ1 )] and [(ξ2 , η2 , ζ2 )] so that these three p-points are not on
some p-line, i.e. they are not collinear. Then any p-line through the vertex will meet the p-line
[(ξ1 , η, ζ1 )][(ξ2 , η2 , ζ2 )] in a p-point [(ξ1 + λξ2 , η1 + λη2 , ζ1 + λζ2 ), for some s ∈ R ∪ {∞}. Then by
(13.1.16) this p-line will have the equation


ξ
η
ζ
 = 0,
det 
ξ3
η3
ζ3
ξ1 + λξ2 η1 + λη2 ζ1 + λζ2
so that
L3,1 ξ + M3,1 η + N3,1 ζ + λ(L3,2 ξ + M3,2 η + N3,2 ζ) = 0,
and so
(L3,1 + λL3,2 )ξ + (M3,1 + λM3,2 )η + (N3,1 + λN3,2 )ζ = 0.
(13.1.21)
Thus if the p-lines [(ξ3 , η3 , ζ3 )][(ξ1 , η, ζ1 )] and [(ξ3 , η3 , ζ3 )][(ξ1 , η, ζ1 )] have equations of the form
a1 ξ + b1 η + c1 ζ = 0,
a2 ξ + b2 η + c2 ζ = 0,
respectively, the p-line [(ξ3 , η3 , ζ3 )][(ξ1 + λξ2 , η1 + λη2 , ζ1 + λζ2 )] will have equation (13.1.18) with
a = a1 + λa2 , b = b1 + λb2 , c = c1 + λc2 .
13.1.6
Overview
We have introduced the equation (13.1.18) as having fixed coefficients (a, b, c) relating to a fixed
p-line, and the variable homogeneous coordinates (ξ, η, ζ) being for p-points which lie on this p-line.
But we can also reverse these roles and have ξ, η, ζ as fixed coefficients relating to a fixed p-point
and (a, b, c) as variable homogeneous coordinates for the p-lines which pass through this p-point.
This sets up a fundamental duality between p-points on a p-line and p-lines through a p-point.
Later on in this chapter we shall deal with this material when it is complex instead of real numbers
are what are involved in the homogeneous coordinates. This is the basis of the classical treatment
of plane projective geometry, an elegant, comprehensive and powerful topic. Accounts of this are
available in many books and we do not propose to follow exactly the course of this established
development. Instead, by means of our frame of reference, we shortly seek to see how far we can
identify which representative of a p-point we are using and extend our methods of Chapter 1. We
do this in two ways.
13.1.7
Cross-ratio of a range of points
We are not able to generalize the concept of distance to a situation which involves a single ideal
point, as
s
2 2
ξ2
η2
η1
ξ1
|Z1 , Z2 | =
−
−
+
ζ1
ζ2
ζ1
ζ2
380
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
and we may not put ζ1 = 0 in this as ζ1 occurs in a denominator. However, we can consider a ratio
of two distances which have a common ideal point. As square roots are awkward to handle, it is
more productive to deal with sensed-ratios or ratios of squares of distances. Dealing now with the
former of these, if in (13.1.10) we take the representatives (ξ1 , η1 , 1), (ξ2 , η2 , 1) of ordinary p-points
Z1 and Z2 , then as in (13.1.20) on taking our standard parametric equations for a p-line on these
p-points as
(ξ, η, ζ) = (ξ1 , η1 , 1) + λ(ξ2 , η2 , 1)
we noted the ratio of sensed distances
Z1 Z
.
(13.1.22)
ZZ2
Now in this parametrization we note that the values of λ corresponding to Z1 , Z2 and Z1 Z2∞ are
0, ∞ and −1 respectively, and so can evaluate the cross-ratio
λ=
cr(Z1 , Z2 , Z, Z1 Z2∞ ) = cr(0, ∞, λ, −1) = −λ.
(13.1.23)
Sensed ratios are not projectively invariant but cross-ratios are, so in this chapter we shall subsume
sensed ratios into our study of cross-ratios.
We note for further use that λ = 1 in (13.1.22) and (13.1.23) if and only if Z is the midpoint of Z1 and Z2 . From §13.1.3, for Z1 ≡ (ξ1 , η1 ), Z2 ≡ (ξ2 , η2 ), Z ≡ (ξ, η) we have 1 Z1 ≡
(ξ1 , η1 , 1), 1 Z2 ≡ (ξ2 , η2 , 1), 1 Z ≡ (ξ, η, 1) and when λ = 1 the line O(1 Z) in RE3 is the median
through O of the triangle with vertices O, 1 Z1 , 1 Z2 .
As in §13.1.4 we consider distinct representative triples
(ξ3 , η3 , ζ3 ) = (ξ1 , η1 , ζ1 ) + λ3 (ξ2 , η2 , ζ2 ),
(ξ4 , η4 , ζ4 ) = (ξ1 , η1 , ζ1 ) + λ4 (ξ2 , η2 , ζ2 ).
We have
(ξ4 , η4 , ζ4 ) − (ξ3 , η3 , ζ3 ) = (λ4 − λ3 )(ξ2 , η2 , ζ2 ),
and so
Thus
λ4 (ξ3 , η3 , ζ3 ) − λ3 (ξ4 , η4 , ζ4 ) = (λ4 − λ3 )(ξ1 , η1 , ζ1 ),
λ4 − λ
(ξ1 , η1 , ζ1 ) + λ(ξ2 , η2 , ζ2 ) =
λ4 − λ3
−λ3 + λ
(ξ4 , η4 , ζ4 ) .
(ξ3 , η3 , ζ3 ) +
λ4 − λ
[(ξ1 , η1 , ζ1 ) + λ(ξ2 , η2 , ζ2 )] =[(ξ3 , η3 , ζ3 ) + µ(ξ4 , η4 , ζ4 )],
−λ3 + λ
µ=
.
λ4 − λ
(13.1.24)
It follows readily that
(µ5 − µ7 )(µ6 − µ8 )
(λ5 − λ7 )(λ6 − λ8 )
=
.
(µ5 − µ8 )(µ6 − µ7 )
(λ5 − λ8 )(λ6 − λ7 )
It follows that this ratio depends only on the points and not the particular parametrization chosen
and it is taken to be the cross-ratio of a range of four collinear p-points.
In §4.2.3 we showed that all ranges of four points in which transversals cut the lines of a pencil
have the same cross-ratio. We now extend this to the ranges of p-points on transversals cut the
p-lines of a p-pencil. At this stage this involves extending the property to the ideal points involved.
Consider a pencil of lines (Z0 Z1 , Z0 Z2 , Z0 Z3 , Z0 Z4 ) where as Z1 , Z2 , Z3 , Z4 are collinear we have
Z3 = (1 − r3 )Z1 + r3 Z2 ,
Z4 = (1 − r4 )Z1 + r4 Z2 ,
r3 , r4 ∈ R.
Then
Z0 Z1∞ ∼ (x1 − x0 , y1 − y0 , 0), Z0 Z2∞ ∼ (x2 − x0 , y2 − y0 , 0),
Z0 Z3∞ ∼ ((1 − r3 )x1 + r3 x2 − x0 , (1 − r3 )y1 + r3 xy − y0 , 0),
Z0 Z4∞ ∼ ((1 − r4 )x1 + r4 x2 − x0 , (1 − r4 )y1 + r4 xy − y0 , 0).
13.1. HOMOGENEOUS COORDINATES, THE REAL PROJECTIVE PLANE
381
It follows from these that
Z0 Z3∞ = (1 − r3 )Z0 Z1∞ + r3 Z0 Z2∞ , Z0 Z4∞ = (1 − r4 )Z0 Z1∞ + r4 Z0 Z2∞ ,
and hence cr(Z1 , Z2 , Z3 , Z4 ) = cr(Z0 Z1∞ , Z0 Z2∞ , Z0 Z3∞ , Z0 Z4∞ ).
13.1.8
p-Segments
If in (13.1.20) we write
[(ξ1 , η1 , 1) + λ(ξ2 , η2 , 1)] = [(1 − t)(ξ1 , η1 , 1) + t(ξ2 , η2 , 1)],
t
we have 1−t
= λ. It is increasing or decreasing values of t that correspond to natural orders on
the line Z1 Z2 . Both t = ∞ and t = −∞ correspond to λ = −1 and so yield the ideal point Z1 Z2∞ .
From this we see that we do not have a natural order on the p-line, as we would have the p-points
in the order Z1 , Z2 , Z1 Z2∞ , Z1 , so that Z1 would both precede and succeed Z2 . All that can be
managed instead is a type of cyclic order on each p-line, which we do not detail. As natural orders
thus do not generalize to p-lines, we need to reconsider the concepts of segments and half-lines,
which were defined in terms of natural orders.
We recall from (13.4.1)? Barry [2] that the sensed-ratio
Z1 Z
ZZ2
has the value λ and that λ > 0 for Z in the segment [Z1 , Z2 ], while λ < 0 for Z otherwise on Z1 Z2 ,
but not Z1 or Z2 in either case. We now define two p-segments with end-points Z1 and Z2 ,in the
first instance Z1 and Z2 themselves and the points Z ∈ Z1 Z2e such that
cr(Z1 , Z2 , Z, Z1 Z2∞ ) < 0,
and in the second instance Z1 and Z2 themselves and the points Z ∈ Z1 Z2e such that
cr(Z1 , Z2 , Z, Z1 Z2∞ ) > 0.
Thus in the first of these we have Z1 and Z2 themselves and those Z on the line for which the
cross-ratio −λ is negative, and in the second instance those Z for which the cross-ratio is positive.
The ideal point Z1 Z2∞ lies in the second. The p-line is the union of these two p-segments, and their
intersection is the pair of points Z1 and Z2 . These two p-segments replace, in RP2 , the concept of
segment in RE2 .
Turning now to half-lines, for the opposite half-line to [Z1 , Z2 we have for Z 6= Z1 that
Z1 ∈ [Z, Z2 ] so that on interchanging Z and Z2 in (13.1.22) and (13.1.23), we have
0 < −cr(∞, λ, 0, −1) = −
λ+1
,
λ
which is equivalent to −1 < λ < 0. A little more complicatedly for Z, other than Z1 and Z2
in [Z1 , Z2 we can have that Z ∈ [Z1 , Z2 ], when λ > 0. Otherwise Z2 ∈ [Z1 , Z] so that on
interchanging Z and Z2 in (13.1.22) and (13.1.23) we have
0 < −cr(0, λ, ∞, −1) = −(λ + 1),
which is equivalent to λ < −1. Thus combined we have {λ : λ < −1} ∪ {λ : λ > 0}. We also
call p-segments first the set consisting of Z1 and Z1 Z2∞ and the p-points [(ξ, η, ζ)] for which
−1 < λ < 0, and secondly the set consisting of Z1 and Z1 Z2∞ and the p-points [(ξ, η, ζ)] for which
λ < −1 or λ > 0. The union of these p-segments is the p-line, and their intersection is the pair of
points Z1 and Z1 Z2∞ . Thus instead of opposite half-lines, we have a pair of p-segments with the
same end-points of the form Z1 and Z1 Z2∞ .
382
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
13.2
Selecting representatives of p-points
13.2.1
Use of Plücker coordinates; first method
For a p-point Z, with general representative (ξ, η, ζ), which is a Euclidean point with Cartesian
coordinates (x, y) relative to F , we select the representative (x, y, 1) as for it ζ 6= 0. When Z is
an ideal point its representatives are of the form (ξ, η, 0) with ξ = η = 0 excluded. As our main
η
ξ
, ξ+η
, 0) which has
case now we suppose that ξ + η 6= 0 and take the unique representative ( ξ+η
the form (1 − s, s, 0) where s ∈ R. For our subsidiary case here we take η = −ξ 6= 0 and select the
representative (1, −1, 0) for this remaining p-point.
With this convention we refer to such unique representatives as being relative to an extended
frame of reference on F . For three p-points Z1 , Z2 , Z3 with such unique representative triples
(ξ1 , η1 , ζ1 ), (ξ2 , η2 , ζ2 ),(ξ3 , η3 , ζ3 ), respectively, we define
ξ η1 ζ1 1 1
(13.2.1)
∂F (Z1 , Z2 , Z3 ) = ξ2 η2 ζ2 .
2
ξ3 η3 ζ3 as an analogue of sensed-area.
As (ξ, η, ζ) = ξ(1, 0, 0) + η(0, 1, 0) + ζ(0, 0, 1) we take the triple of reference
Z1 ∼ (1, 0, 0),
Z2 ∼ (0, 1, 0),
Z3 ∼ (0, 0, 1)
and form normalised areal coordinates from these. It is straightforward to check that for general
Z ∼ (ξ, η, ζ) we have that
α=
∂F (Z, Z2 , Z3 )
= ξ,
∂F (Z1 , Z2 , Z3
β=
∂F (Z, Z3 , Z1 )
= η,
∂F (Z1 , Z2 , Z3
γ=
∂F (Z, Z1 , Z2 )
= ζ.
∂F (Z1 , Z2 , Z3
Thus these normalized areal-coordinates reduce to Plücker coordinates.
Returning to the general situation we note that
2[∂F (Z, Z2 , Z3 ) + ∂F (Z, Z3 , Z1 ) + ∂F (Z, Z1 , Z2 )]
ξ η ζ ξ η ζ ξ η ζ = ξ2 η2 ζ2 − ξ1 η1 ζ1 + ξ1 η1 ζ1 ξ3 η3 ζ3 ξ3 η3 ζ3 ξ2 η2 ζ2 ξ
η
ζ
ξ
η
= ξ2 − ξ1 η2 − η1 ζ2 − ζ1 − ξ2 − ξ1 η2 − η1
ξ3
η3
ζ3
ξ1
η1
ξ
η
ζ
= ξ2 − ξ1 η2 − η1 ζ2 − ζ1 .
ξ3 − ξ1 η3 − η1 ζ3 − ζ1 ζ
ζ2 − ζ1
ζ1
When Z1 , Z2 , Z3 are Euclidean points we have ζ1 = ζ2 = ζ3 = 1 and this reduces to
ξ − ξ1 η2 − η1 ζ 2
ξ3 − ξ1 η3 − η1 which gives the expected values when ζ = 1 and when ζ = 0.
13.2.2
Extended dual sensed-area
If we take p-points Z6 ∼ (ξ6 , η6 , ζ6 ), Z7 ∼ (ξ7 , η7 , ζ7 ), we consider
(1 − s)Z6 + sZ7 ∼ ((1 − s)ξ6 + sξ7 , (1 − s)η6 + sη7 , (1 − s)ζ6 + sζ7 ),
(13.2.2)
383
13.2. SELECTING REPRESENTATIVES OF P-POINTS
for s ∈ R.
Case 1.
As a first case for the point (13.2.2) we suppose that it is a Euclidean point and so for it we
take the unique representative
(1 − s)ξ6 + sξ7 (1 − s)η6 + sη7
,
,1 .
(13.2.3)
(1 − s)ζ6 + sζ7 (1 − s)ζ6 + sζ7
Subcase 1,1.
As a first sub-case of this we suppose that ζ6 6= 0, ζ7 6= 0. Then from the representative (13.2.3)
we can write
(1 − s)ξ6 + sξ7
(1 − s)ζ6
ξ6
sζ7
ξ7
=
+
,
(1 − s)ζ6 + sζ7 (1 − s)ζ6 + sζ7 ζ6
(1 − s)ζ6 + sζ7 ζ7
(1 − s)ζ6
η6
sζ7
η7
(1 − s)η6 + sη7
=
+
.
(1 − s)ζ6 + sζ7 (1 − s)ζ6 + sζ7 ζ6
(1 − s)ζ6 + sζ7 ζ7
Thus we have
Z=
(1 − s)ζ6
(1 − s)ζ6 + sζ7
sζ7
ξ6 η6
ξ7 η7
, ,1 +
, ,1 .
ζ6 ζ6
(1 − s)ζ6 + sζ7 ζ7 ζ7
For this p-point to lie on a p-line Z4 Z5 we need
ξ4 η4 ζ4 sζ7
(1 − s)ζ6
ξ5 η5 ζ5 +
0=
(1 − s)ζ6 + sζ7 ξ6 η6 1 (1 − s)ζ6 + sζ7
ζ6
ζ6
Similarly for this p-point to lie on a p-line Z8 Z9 we need
ξ8 η8 ζ8 (1 − s)ζ6
sζ7
ξ9 η9 ζ9 +
0=
(1 − s)ζ6 + sζ7 ξ6 η6 1 (1 − s)ζ6 + sζ7
ζ6
ζ6
ξ4
ξ5
ξ7
η4
η5
ξ8
ξ9
ξ7
η8
η9
ζ7
ζ7
η7
ζ7
η7
ζ7
ζ4
ζ5
1
.
ζ8
ζ9
1
.
(13.2.4)
(13.2.5)
(13.2.6)
Then for the p-lines Z4 Z5 , Z6 Z7 , Z8 Z9 to be concurrent we need (13.2.5) and (13.2.6) to hold
simultaneously and so to have
∂F (Z6 , Z4 , Z5 )∂F (Z7 , Z8 , Z9 ) − ∂F (Z7 , Z4 , Z5 )∂F (Z6 , Z8 , Z9 ) = 0.
We write this as
∂F [(Z6 , Z7 ), (Z4 , Z5 ), (Z8 , Z9 )] = 0,
(13.2.7)
and have here an extension of dual sensed-area.
Subcase 1,2.
As a second subcase of this we take ζ6 6=, ζ7 = 0 and the by (13.2.3) we have
ξ6
s ξ7 η6
s ξ7
+
,
+
, 1, .
ζ6
1 − s ζ6 ζ6
1 − s ζ6
For this p-point to lie on a p-line Z4 Z5 we need
ξ4
η4
ξ5
η5
0 = ξ6 + s ξ7 η6 + s ξ7
ζ6
1−s ζ6
ζ6
1−s ζ6
ξ4 η4 ζ4 ξ4
= ξ5 η5 ζ5 + ξ5
ξ6 η6 1 s ξ7
ζ6
ζ6
1−s ζ6
ζ4
ζ5
1
η4
η5
s ξ7
1−s ζ6
ζ4
ζ5
0
.
(13.2.8)
384
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
Thus for Z ∈ Z4 Z5 we need
0 = ∂F (Z4 , Z5 , Z6 ) +
s 1
∂F (Z4 , Z5 , Z7 ).
1 − s ζ6
(13.2.9)
s 1
∂F (Z8 , Z9 , Z7 ).
1 − s ζ6
(13.2.10)
Similarly for Z ∈ Z8 Z9 we need
0 = ∂F (Z8 , Z9 , Z6 ) +
For Z4 Z5 , Z6 Z7 , Z8 Z9 to be concurrent we need (13.2.9) and (13.2.10) to hold simultaneously and
so (13.2.7) to hold again.
Subcase 1,3.
As a third subcase of this we take ζ6 = 0, ζ7 6= 0 and then by (13.2.3) we have
ξ7 1 − s η6
η7
1 − s ξ6
+ ,
+ ,1 .
(13.2.11)
Z=
s ζ7
ζ7
s ζ7
ζ7
For this p-point to lie on a p-line Z4 Z5 we need
ξ4
η4
ζ4
ξ5
η5
ζ5
0 = 1−s ξ6 + ξ7 1−s η6 + η7 1
s ζ7
ζ7
s ζ7
ζ7
ξ4
ξ4
η
ζ
4
4 ξ
η
ζ
=
5
5
5 + ξ5
ξ
η
1−s 6 1−s 6 0 ξ7
s ζ7
s ζ7
ζ7
and so we need
η4
η5
η7
ζ7
ζ4
ζ5
1
.
1−s
∂F (Z4 , Z5 , Z6 ) + ∂F (Z4 , Z5 , Z7 ).
s
Similarly for this p-point to lie on a p-line Z8 Z9 we need
0=
0=
1−s
∂F (Z8 , Z9 , Z6 ) + ∂F (Z4 , Z8 , Z9 ).
s
(13.2.12)
(13.2.13)
Then for the p-lines Z4 Z5 , Z6 Z7 , Z8 Z9 to be concurrent we need (13.2.12) and (13.2.13) to hold
simultaneously and so we obtain (13.2.7) again.
Case 2.
Now returning to (13.2.2) we suppose that we have an ideal point with (1 − s)ζ6 + sζ7 = 0, and
so have
Z = ((1 − s)ξ6 + sξ7 , (1 − s)η6 + sη7 , 0).
(13.2.14)
Subcase 2,1.
As a first subcase of this we suppose that (1 − s)(ξ6 + η6 ) + s(ξ7 + η7 ) 6= 0 and so have as unique
representative of this p-point we take
(1 − s)ξ6 + sξ7
(1 − s)η6 + sη7
Z=
,
,0 .
(13.2.15)
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 ) (1 − s)(ξ6 + η6 ) + s(ξ7 + η7 )
Subsubcase 2,1,1
As a first sub-sub-case of this we suppose that ξ6 + η6 6= 0, ξ7 + η7 6= 0. Then by (13.2.15) we
obtain
ξ6
η6
(1 − s)(ξ6 + η6 )
,
,0
Z=
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 ) ξ6 + η6 ξ6 + η6
s(ξ7 + η7 )
ξ7
η7
+
,
,0 .
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 ) ξ7 + η7 ξ7 + η7
385
13.2. SELECTING REPRESENTATIVES OF P-POINTS
Then for Z ∈ Z4 Z5 we need
s(ξ7 + η7 )
(1 − s)(ξ6 + η6 )
∂F (Z4 , Z5 , Z6 ) +
∂F (Z4 , Z5 , Z7 ).
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 )
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 )
(13.2.16)
Similarly Z ∈ Z8 Z9 we need
0=
s(ξ7 + η7 )
(1 − s)(ξ6 + η6 )
∂F (Z8 , Z9 , Z6 ) +
∂F (Z4 , Z8 , Z9 ).
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 )
(1 − s)(ξ6 + η6 ) + s(ξ7 + η7 )
(13.2.17)
On combining (13.2.16) and (13.2.17) , for Z4 Z5 , Z6 Z7 , Z8 Z9 to be concurrent we obtain (13.2.7)
again.
Subsubcase 2,1,2.
As a second sub-sub-case of this we suppose that ξ6 + η6 6= 0, ξ7 + η7 = 0. Then by (13.2.15)
we have
ξ7
η7
s
η6
s
ξ6
+
,
+
,0
Z=
ξ6 + η6
1 − s ξ6 + η6 ξ6 + η6
1 − s ξ6 + η6
ξ7
η6
s
η7
ξ6
,
,0 +
,
,0 .
=
ξ6 + η6 ξ6 + η6
1 − s ξ6 + η6 ξ6 + η6
0=
For Z ∈ Z4 Z5 we need
0 = ∂F (Z4 , Z5 , Z6 ) +
Similarly for Z ∈ Z8 Z9 we need
0 = ∂F (Z8 , Z9 , Z6 ) +
s
∂F (Z4 , Z5 , Z7 ).
1−s
(13.2.18)
s
∂F (Z8 , Z5 , Z9 ).
1−s
(13.2.19)
Then for Z4 Z5 , Z6 Z7 , Z8 Z9 to be concurrent we need (13.2.18 ) and (13.2.19 ) to hold simultaneously and so have (13.2.7) again.
Subsubcase 2,1,3.
As a third sub-sub-case of this we suppose that ξ6 + η6 = 0, ξ7 + η7 6= 0. Then by (13.2.15) we
must have
1 − s ξ6
ξ7
1 − s η6
η7
+
,
+
t, 0 .
(13.2.20)
s ξ7 + η7
ξ7 + η7
s ξ7 + η7
ξ7 + η7
This can be brought to the standard conclusion as was the last subsubcase.
Subcase 2,2.
As a second subcase here we suppose that (1 − s)(ξ6 + η6 ) + s(ξ7 + η7 ) = 0 and now have in
(13.2.2)
[(1 − s)ξ6 + sη6 ](1, −1, 0),
from which we have the unique representative (1, −1, 0). Thus in this case (13.2.2) represents a
p-point instead of a p-line and so we omit it in most of the analysis which follows.
13.2.3
A property of dual sensed-area
Here we extend the material §1.1.3. We form the product
4∂F (Z1 , Z3 , Z4 )∂F (Z2 , Z5 , Z6 )
ξ1 η1 ζ1 ξ2 η2 5 ζ2 = ξ3 η3 ζ3 ξ5 η5 ζ5 ξ4 η4 ζ4 ξ6 η6 ζ6 η3 ζ3 ζ3 ξ3 ξ
= ξ1 + η1 + ζ1 3
η4 ζ4
ζ4 ξ4
ξ4
η
η3 ξ2 5
η4
η6
ζ
ζ5 + η2 5
ζ6
ζ6
ξ
ξ5 + ζ2 5
ξ6
ξ6
η5 η6 386
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
On interchanging the subscripts 1 and 2 in this we deduce the following result.
4∂F (Z2 , Z3 , Z4 )∂F (Z1 , Z5 , Z6 )
η ζ ζ ξ ξ
= ξ2 3 3 + η2 3 3 + ζ2 3
η4 ζ4
ζ4 ξ4
ξ4
η
η3 ξ1 5
η4
η6
ξ3
ζ5 +
η
1
ζ6
ξ4
On subtracting the second of these from the first we obtain that
ξ5
η3 +
ζ
1
η4
ξ6
4∂F (Z1 , Z3 , Z4 )∂F (Z2 , Z5 , Z6 ) − 4∂F (Z2 , Z3 , Z4 )∂F (Z1 , Z5 , Z6 )
ζ ξ ξ η5 + η2 ζ1 ξ3 η3 ζ5 ξ5 =η1 ζ2 3 3 5
ζ4 ξ4
ξ6 η6
ξ4 η4
ζ6 ξ6 ζ ξ ξ η5 − η1 ζ2 ξ3 η3 ζ5 ξ5 −η2 ζ1 3 3 5
ζ4 ξ4
ξ6 η6
ξ4 η4
ζ6 ξ6 ξ η3 η5 ζ5 + ζ2 ξ1 η3 ζ3 ξ5 η5 +ζ1 ξ2 3
ξ4 η4
η6 ζ6
η4 ζ4
ξ6 η6 ξ η3 η5 ζ5 − ζ1 ξ2 η3 ζ3 ξ5 η5 −ζ2 ξ1 3
ξ4 η4
η6 ζ6
η4 ζ4
ξ6 η6 η ζ ζ ξ ζ ξ η ζ +ξ1 η2 3 3 5 5 + ξ2 η1 3 3 5 5 η4 ζ4
ζ6 ξ6
ζ4 ξ4
η6 ζ6
ζ3 ξ3 η5 ζ5 η3 ζ3 ζ5 ξ5 − ξ1 η2 −ξ2 η1 ζ4 ξ4 η6 ζ6 η4 ζ4 ζ6 ξ6 η1 ζ1 ζ1 ξ1 ξ1 η1 η2 ζ2 ζ2 ξ2 ξ2 η2 η3 ζ3 ζ3 ξ3 ξ3 η3 = η4 ζ4 ζ4 ξ4 ξ4 η4 .
η5 ζ5 ζ5 ξ5 ξ5 η5 η6 ζ6 ζ6 ξ6 ξ6 η6 η5 η6 (13.2.21)
From (13.2.21) we see that the value of ∂F [(Z1 , Z2 ), (Z3 , Z4 ), (Z5 , Z6 )] is unaltered if the ordered
couples of p-points are interchanged while preserving the cyclic order but the value is replaced by
its additive inverse if the cyclic order is altered.
13.2.4
Equation of a p-line in extended areal coordinates
As in §3.2.1 but now with p-points Z1 , Z2 , Z3 , Z, U, V all with unique representatives chosen as
in the first paragraph of §13.2.1, fixed non-collinear (Z1 , Z2 , Z3 ) and general Z, U, V , we prove an
analogue of a fundamental identity (3.2.1)
∂F (Z1 , Z2 , Z3 )∂F (Z, U, V )
=∂F (Z1 , U, V )∂F (Z, Z2 , Z3 ) + ∂F (Z2 , U, V )∂F (Z, Z3 , Z1 ) + ∂F (Z3 , U, V )∂F (Z, Z1 , Z2 ). (13.2.22)
For as in §3.2.1 we have
∂F (Z1 , Z2 , Z3 )∂F (Z, U, V ) − ∂F (Z1 , U, V )∂F (Z, Z2 , Z3 ) = ∂F [(Z1 , Z), (Z2 , Z3 ), (U, V )],
also
∂F (Z2 , U, V )∂F (Z, Z3 , Z1 ) − ∂F (Z3 , U, V )∂F (Z, Z1 , Z2 ) = ∂F [(Z2 , Z3 ), (U, V ), (Z1 , Z)],
and these are equal.
Use of the material in the last subsection and this should enable generalization of some of the
material in the early chapters.
387
13.2. SELECTING REPRESENTATIVES OF P-POINTS
13.2.5
Use of Plücker coordinates; second method
For a p-point Z, with general representative (ξ, η, ζ), which is a Euclidean point with Cartesian
coordinates (x, y) relative to F , as before we select the representative (x, y, 1) as for it ζ 6= 0. When
Z is an ideal point its representatives are of the form (ξ, η, 0) with ξ = η = 0 excluded. Then the
η
ξ
∗
has polar form
point Z ∗ ≡F ( ξ2 +η
2 , ξ 2 +η 2 ) lies on the unit circle and so the line OZ
x = t cos θd , y = t sin θd ,
t ∈ R.
We then select (cos θd , sin θd , 0) as our unique representative for the ideal point Z.
We consider Euclidean points
W2 ≡F (cos θ2d , sin θ2d ), W3 ≡F (cos θ3d , sin θ3d ), W ≡F (cos θd , sin θd ),
and seek numbers s and k such that
(1 − s) cos θ2d + s cos θ3d =k cos θd ,
(1 − s) sin θ2d + s sin θ3d =k sin θd ,
(13.2.23)
when |θ2d |r < |θd |r < |θ3d |r . By Cramer’s rule we have that
sin θ3d k cos θd − cos θ3d k sin θd
,
sin θ3d cos θ2d − cos θ3d sin θ2d
k sin θd cos θ2d − k cos θd sin θ2d
s=
,
sin θ3d cos θ2d − cos θ3d sin θ2d
1−s=
giving
sin θd cos θ2d − cos θd sin θ2d
s
.
=
1−s
sin θ3d cos θd − cos θ3d sin θd
(13.2.24)
Then s/(1 − s) has the values given in (12.4.3) and (12.4.4) when t2 = t3 = 1. Also we have
1 = (1 − s) + s =
k
[sin(θ3d − θd ) + sin(θd − θ2d )].
sin(θ3d − θ2d )
This identifies k.
Turning now to ideal points, on considering points
W2 ≡F (cos θ2d , sin θ2d , 0), W3 ≡F (cos θ3d , sin θ3d , 0), W ≡F (cos θd , sin θd , 0),
and seeking numbers s and k such that
(1 − s) cos θ2d + s cos θ3d =k cos θd ,
(1 − s) sin θ2d + s sin θ3d =k sin θd ,
(13.2.25)
when |θ2d |r < |θd |r < |θ3d |r , we obtain the same conclusions as for (13.2.23) since (1 − s)0 + s0 = 0
automatically.
13.2.6
p-Duo-sectors
Further concepts in RE2 depended on the notion of segment and must be considered as to whether
there is an extension of them. These include closed half-planes, interior and exterior regions.
We resume the notation of §12.4.2, starting with a pair {l1,2 = W1 W2 , l1,3 = W1 W3 } of distinct
lines through the point W1 . There we defined one duo-sector Dl∗1,2 ,l1,3 in RE2 as consisting of the
points of the arms l1,2 and l1,3 and of those points W for which by (12.4.1)
δF (W1 , W2 , W )
> 0,
δF (W1 , W, W2 )
388
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
and by (12.3.19), in the particular case when W1 = O, by |θ2d |r < |θd |r < |θ3d |r , and a second
duo-sector Dl∗∗
in RE2 as consisting of the points of the arms l1,2 and l1,3 and of those points
1,2 ,l1,3
W for which by (12.4.2)
δF (W1 , W2 , W )
< 0,
δF (W1 , W, W2 )
and by (12.4.4), in the particular case when W1 = O, by having either |θd |r < |θ2d |r or |θ3d |r < |θd |r .
In RP2 we extend each duo-sector in RE2 by adjoining to it, for each line W W1 in it, the ideal
point W W1inf on its extension. By §12.4.1 and (12.4.3) and (12.4.4) this corresponds to the polar
angle θd satisfying the same inequalities as hold in RE2 . We refer to these extended figures as
p-duo-sectors.
Recapping, to generalize the concept of duo-sector from the real Euclidean plane RE2 to the
extended real plane RP2 , we first proceed as follows. We take non-collinear Euclidean points
O, W2 , W3 with homogeneous representatives
(0, 0, 1), (t2 cos θ2d , t2 sin θ2d , 1), (t3 cos θ3d , t3 sin θ3d , 1),
where t2 and t3 have the same sign, and a point W not on either line OW2 , OW3 . When W is a
Euclidean point we take the representative triple (t cos θd , t sin θd , 1). Then we clearly have
δF (O, W2 , W )
∂F (O, W2 , W )
=
.
∂F (O, W, W3 )
δF (O, W, W3 )
(13.2.26)
When W is an ideal point we take the representative (cos θd , sin θd , 0) and assign W to the relevant extended duo-sector according to whether or not the first quotient in (13.2.26) is positive or
negative. It can be checked that this quotient is the same as those in (13.4.3) and (13.4.4) when
t2 = t3 = 1 and so has the same signs as there.
If we now specialize OW2∞ , OW3∞ , OW ∞ as for (13.2.23) with
OW2inf ≡F (cos θ2d , sin θ2d , 0), OW3inf ≡F (cos θ3d , sin θ3d , 0), OW inf ≡F (cos θd , sin θd , 0),
and seek to take parametric equations for the ideal line, the equations (13.2.23) suffice again as
s
still applies.
the third coordinates automatically fall into line. Thus the formula (13.2.24) for 1−s
We also consider a p-point W4 which does not lie on either of the side-p-lines. Then
∂F (W1 , W2 , W )∂F (W1 , W4 , W3 )
> 0.
∂F (W1 , W, W3 )∂F (W1 , W2 , W4 )
(13.2.27)
specifies the p-duo-sector with side-p-lines l1,2 and l1,3 and containing the p-point W4 . The supplementary p-duo-sector to this, with the same side-p-lines but not containing the p-point W4 has
instead that the inequality in (13.2.27)is reversed. The expression in (13.2.27) is an equipoised
quotient and so is a projective invariant.
We denote by DQ1 the p-duo-sector Dl∗2 ,l3 where OI = l2 , OJ = l3 , which we call the first
duo-quadrant; it consists of the p-lines through O in §13.2.5 with 0 ≤ |θ|r ≤ 12 π. Its supplement
we denote by DQ2 and call the second duo-quadrant; it consists of the p-lines through O in
§13.2.5 with 12 π ≤ |θ|r < π or |θ|r = 0.
13.2.7
Extended duo-angles
With any pair of extended lines W1 W2e and W1 W3e , we thus have two p-duo-sectors D∗ and D∗∗
with side-lines these extended lines, and we call each of the pairs
({W1 W2e , W1 W3e }, D∗ ), ({W1 W2e , W1 W3e }, D∗∗ ),
an extended duo-angle. Thus in RE2 we had duo-angles as optional extras to angles, but in
going to the extended plane we need to replace angles by extended duo-angles.
389
13.2. SELECTING REPRESENTATIVES OF P-POINTS
Given a pair of lines W1 W2e and W1 W3e , we take W4 6= W1 so that W1 ∈ [W2 , W4 ]. Then the
mid-lines of the angle-supports |W2 W1 W3 and |W4 W1 W3 , when extended, are called the p-midlines of the pair of p-lines {W1 W2e , W1 W3e }.
b
D∗∗
b
D∗
b
b
W3
b
b
W2
b
W1
W4
b
b
D∗
b
b
D∗∗
Figure 13.1.
b
For a pair of p-lines (l1 , l2 ) = (OW2e , OW3e ) with polar angles θ1 , θ2 , we can use (12.5.21) and
(12.5.22) to see that the ideal points on their mid-lines are (cos 12 (θ2 +θ1 ), sin 12 (θ2 +θ1 ), 0), (cos( 21 (θ2 +
θ1 ) + π2 ), sin( 21 (θ2 + θ1 ) + π2 ), 0). In order to distinguish between the two extended duo-angles with
the same side extended-lines, we note that, of the mid-lines of the pair of lines, one lies in each
of the extended duo-sectors involved and we use each mid-line as an indicator for the extended
duo-angle involving the extended duo-sector which contains that mid-line.
13.2.8
Extended orientation
In our generalization, we are faced with the question of whether we can extend the notion of
orientation. As we used the fact of δF (W, W1 , W2 )’s being positive or negative for positive or
negative orientation of the triple (W1 , W2 , W ) of non-collinear points in RE2 , we now use
∂F (W1 , W2 , W )
> 0,
∂F (W1 , W, W3 )
and
∂F (W1 , W2 , W )
< 0,
∂F (W1 , W, W3 )
as giving positive and negative, respectively, orientation of the ordered quadruple (W1 ; W2 , W3 ; W )
in the case of p-points W not on W1 W2 or W1 W3 . Thus for points W ∈ RE2 we have positive
orientation of (W1 ; W2 , W3 ; W ) if (W, W1 , W2 ) and (W, W3 , W1 ) are similarly oriented in RE2 , and
negative extended orientation if (W, W1 , W2 ) and (W, W3 , W1 ) are oppositely oriented in RE2 . It
can be checked that we have positive orientation in the case when W is the centroid of the triple
(W1 , W2 , W3 ).
We can extend the foregoing in a straightforward way to when W , or any of the others, is an
ideal point.
390
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
From (11.8.28) if
Ws =
1
σ
W6 +
W9 ,
1+σ
1+σ
then on the assumption (11.8.20), from the material preceding (11.8.28)
δF (Z6,9 , Z5 , O)
δF (Z6,9 , Z5 , Zs′ )
=
σ,
′
δF (Z6,9 , Zs , Z8 )
δF (Z6,9 , Z8 , O)
and so Plücker reciprocation maps the two extended segments with end-points W6 and W9 into
the extended duo-sectors with side extended-lines Z6,9 Z5e and Z6,9 Z8e . For direct correspondence
we need
δF (Z6,9 , Z5 , O)
< 0,
δF (Z6,9 , O, Z8 )
so that (Z6,9 ; Z5 , Z8 ; O) is negatively extended-oriented.
13.2.9
Extended duo-angles in standard position; addition
H4
D2
H3
H4
D1
J
b
b
b
b
O
I
D2
H1
H2
H3
D1
J
b
b
b
O
D1
b
I
D2
D1
H1
H2
D2
b
Figure 13.3.
b
By a slight modification of the material on duo-angles in Barry [2, 10.10.4] we introduce in what
follows the notion of extended duo-angles in standard position with respect to our frame of
reference.
We extend our frame of reference F by taking W1 = O ≡ (0, 0), W2 = I, W3 = J so that
W1 W2 ⊥ W1 W3 and we take a canonical pair of extended duo-sectors D1 and D2 , with (O; I, J; Z)
having positive extended orientation for Z in the interior of D1 . Then D1 is the union of the first
and third quadrants Q1 and Q3 as extended to include ideal points, while D2 is similarly related
to the second and fourth quadrants.
We now consider extended duo-angles in standard position, with side extended-lines W1 W2e =
OIe and W1 We = OWe , the indicators of which lie in D1 , so that for the points Z ≡ (x, y) of the
relevant midline acting as indicator, (O; I, J; Z) has positive extended orientation. This amounts
to
∂F (O, I, Z)
y
= > 0.
∂F (O, Z, J)
x
13.2. SELECTING REPRESENTATIVES OF P-POINTS
391
b
D2
b
W5
W6
D1
R
b
b
βd
J
b
b
b
S
b
αd
O I
b
W4
b
Q
b
b
b
D1
T
b
D2
b
Figure 13.4. Addition of extended duo-angles.
To deal with addition of extended duo-angles in standard position, let Q ≡ (k, 0), R ≡ (0, k)
for some k > 0, and αd have side extended-lines OQe and OW4e , and βd have side extended-lines
OQe and OW5e , where |O, W4 | = |O, W5 | = k and both have their indicators in D1 . We take it
that v4 > 0 and v5 > 0 unless W4e or W5e , respectively, is Q. Then the line through Q which is
parallel to W4 W5 will meet the circle C(O; k) in a second point, which we denote by W6 ≡ (u6 , v6 )
and find for (u6 , v6 ) that
u6 = k
(u5 − u4 )(v5 − v4 )
(v5 − v4 )2 − (u5 − u4 )2
, v6 = −2k
.
(u5 − u4 )2 + (v5 − v4 )2
(u5 − u4 )2 + (v5 − v4 )2
We define the sum
αd + βd = γd ,
where γd has side extended-lines OQe and OW6e and has its indicator in D1 . When W4 = W5 we
take QW6 as the line through Q which is parallel to the tangent to the circle at W4 .
13.2.10
Sine, cosine, tangent of extended duo-angles
We define trigonometric functions of the extended duo-angles in §13.2.9 as follows. First we
−−−−−−−−−→ −−−→
introduce the point r90F ;O (W4 ) ≡ (−v4 , u4 ), so that Or90F ;O (W4 ) = OW4 ⊥ . In the real Euclidean
plane we revise our definitions to
∂F (O, Q, W4 )
v4
= ,
∂F (O, Q, R)
k
∂F (O, W4 , R)
u4
∂F (O, Q, r90F ;O (W4 ))
=
=
.
cos αd =
δF (O, Q, R)
∂F (O, Q, R)
k
sin αd =
(13.2.28)
As in Barry [2, Chapter 8] these can be shown to depend only on αd .
For an ideal point OW4∞ ≡F ( √ ξ24 2 , √ η24 2 , 0) we have OQ∞ ≡F (1, 0, 0) and OR∞ ≡F
ξ4 +η4
ξ4 +η4
(0, 1, 0). It can be checked that
η4
ξ4
∂F (O, OW4∞ , OR∞ )
∂F (O, OQ∞ , OW4∞ )
p
=
=p 2
,
.
∞
∞
∞
∞
2
2
∂F (O, OQ , OR )
ξ4 + η4 ∂F (O, OQ , OR )
ξ4 + η42
(13.2.29)
392
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
Then the ratios for the sine and cosine of α for the ideal p-point in (13.2.29) is the same as that
for all the points on the arm in (13.2.28). Thus the sine and cosine of an extended duo-angle are
the same as those of the duo-angle.
Turning now to addition formulae, we recall that the sum γd = αd + βd has side-lines OQe and
OW6 , and when −(u5 − u4 )(v5 − v4 ) ≥ 0 we have
sin γd =
u6
v6
, cos γd =
.
k
k
It was shown in Barry [2, 10.10.5], that for the cases in (13.2.28),
sin(αd + βd ) = sin αd cos βd + cos αd sin βd , cos(αd + βd ) = cos αd cos βd − sin αd sin βd ,
while when −(u5 − u4 )(v5 − v4 ) < 0, we have
sin γd = −
v6
u6
, cos γd = − ,
k
k
and hence, as above,
− sin(αd + βd ) = sin αd cos βd + cos αd sin βd , − cos(αd + βd ) = cos αd cos βd − sin αd sin βd .
Thus the addition formulae for sine and cosine of extended duo-angles are more complicated than
those of angles. It follows that these also hold in the case (13.2.29).
Turning now to the tangent of an extended duo-angle, with O = W1 , Q = W2 , R = W3 , we
note that the ideal points on W1 W2e , W1 W3e , W1 W4e , have representative triples
u4
v4
(1, 0, 0), (0, 1, 0), ( p 2
,p 2
, 0),
ξ4 + η42
ξ4 + η42
respectively, and in (13.2.24) the ratio s/(1 − s) of W1 W4∞ equals
1 √ u4
2
2
ξ4 +η4 0 √ v4
ξ42 +η42 ∂ (W , W , W )
v
= 4 = F 1 2
.
√ u4
u
∂F (W1 , W, W3 )
4
ξ2 +η2 0 4
4
√ v4
ξ42 +η42 1 Accordingly we define the tangent of the extended duo-angle αd in standard position and with side
extended-lines W1 W2e = OQe and W1 W4e = OWe to be
tan αd =
∂F (W1 , W2 , W )
,
∂F (W1 , W, W3 )
with the above specific identification of W1 , W2 and W3 .
Then clearly
sin αd
,
tan αd =
cos αd
and as in Barry [2, 10.10.6] we have the addition formulae
tan(αd + βd ) =
tan αd + tan βd
,
1 − tan αd tan βd
provided that 0 does not occur in a denominator.
393
13.2. SELECTING REPRESENTATIVES OF P-POINTS
13.2.11
Sensed extended duo-angles
With extended duo-angles as in §13.2.9, now let ∼ W4 ≡ (−u4 , v4 ) and let ∼ αd = (OQe , O(∼
W4e )). Then, as above,
u4
v4
sin(∼ αd ) = , cos(∼ αd ) = − .
k
k
We define
βd − αd = βd + (∼ αd ),
and this is the extended duo-angle in standard position with side extended-lines OQe and OW7e ,
where W7 ≡ (u7 , v7 ) is the point where the line through Q and parallel to W5 (∼ W4 ) meets the
circle C(O; k) again. We call βd − αd the sensed extended duo-angle with side extended-lines
OW4e , OW5e and denote it by ∢F (OW4e , OW5e ).
We note that
sin(α̃d ) = sin αd , cos(α̃d ) = − cos αd .
From the formula for v6 in §13.2.9 we have
v7 = −2k
(u5 + u4 )(v5 − v4 )
.
(u5 + u4 )2 + (v5 − v4 )2
Then when −(u5 + u4 )(v5 − v4 ) ≥ 0 we have
sin(βd − αd )
cos(βd − αd )
= sin[βd + (∼ αd )]
= sin βd cos(∼ αd ) + cos βd cos(∼ αd )
= − sin βd cos αd + cos βd sin αd ,
= cos[βd + (∼ αd )]
= cos βd cos(∼ αd ) − sin βd sin(∼ αd )
= − cos βd cos αd − sin βd sin αd .
When −(u5 + u4 )(v5 − v4 ) < 0, we multiply both of the left-hand sides here by −1.
We also have
tan βd − tan αd
tan(βd − αd ) =
,
1 + tan αd tan βd
provided βd − αd is not equal to ∢F (OQe , ORe ).
13.2.12
Extended duo-angles with vertex on the ideal line
To specialize in studying tan θd , in §13.2.10 we can normalize by taking Q ≡F (1, 0), R ≡F (0, 1)
and W4 ≡F (k cos θ4d , k sin θ4d ). Then we have that
0
0
1 1
0
1 k
cos
θ
k
sin
θ
1
δF (O, Q, W4 )
4d
4d
= tan θ4d .
= δF (O, W4 , R)
0
0
1 k cos θ4d k sin θ4d 1 0
1
1 To generalize from this, for any Euclidean point W ≡F (u, v)
k cos θ4d k sin θ4d
1
0
u
v
∂F (OW4∞ , OQ∞ , W )
= ∞
∞
∂F (OW4 , W, OR )
k cos θ4d k sin θ4d
u
v
0
1
we note that
0 0 1 = tan θ4d ,
0 1 0 394
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
and we interpret the duo-angles with arms (OW4∞ )(OQ∞ ) and (OW4∞ )W , that is having arms the
ideal line and (OW4∞ )W , as having magnitude |θ4d |r . Note that this is constant as W varies and is
consistent with a family of parallel Euclidean lines making a constant duo-angle with a transversal.
Generalizing from this, let W1 6= O and take the representative triple for OW1∞ to be (cos θ1d , sin θ1d , 0).
Then
cos θ1d sin θ1d 0 v2
1 ∂F (OW1∞ , W2 , W3 ) = 12 u2
u3
v3
1 cos θ1d sin θ1d 0 u2
v2
1 = 12 u3 − u2 v3 − v2 0 cos θ1d
sin θ1d
0 u2
v2
1 = 12 t2,3 cos θ2,3d t2,3 sin θ2,3d 0 = − 21 t2,3 (sin θ2,3d cos θ1d − cos θ2,3d sin θ1d )
= ∓ 21 t2,3 sin(θ2,3d − θ1d ),
taking the − or the + according as |θ2,3d |r < |θ1d |r or |θ1d |r < |θ2,3d |r , respectively. Thus we have
that
θ2,3d − θ1d if |θ1d |r < |θ2,3d |r
∡F ((OW1∞ )W2 , (OW1∞ )W3 ) =
.
(13.2.30)
θ1d − θ2,3d if |θ2,3d |r < |θ1d |r
Taking a different situation now let O, W1 ,
triples for OW1∞ and OW2∞ , (cos θ1d , sin θ1d , 0)
Euclidean point W we have
∞
∞
1 ∂F (OW1 , OW2 , W ) = 2 W2 be distinct points and take as representative
and (cos θ2d , sin θ2d , 0), respectively. Then for any
cos θ1d
cos θ2d
u
sin θ1d
sin θ2d
v
0
0
1
= 12 (sin θ2d cos θ1d − sin θ1d cos θ2d )
=±
1
2
sin(θ2d − θ1d ),
taking the + or − according as |θ1d |r < |θ2d |r or |θ2d |r < |θ1d |r . Hence for all Euclidean points W
we have
θ2d − θ1d if |θ1d |r < |θ2d |r
∞
∞
∞
∡F ((OW1 )(OW2 ), (OW1 )W ) =
.
(13.2.31)
θ1d − θ2d if |θ2d |r < |θ1d |r
Thus we have an angle of constant magnitude as W varies. This is related to the fact that in RE2
opposite angles of a parallelogram are equal to one another in magnitude and also to the vertically
opposite angle for either of them.
13.3
Projective coordinates
13.3.1
Projective coordinates
Given a frame of reference F for RE2 , let (Z1 , Z2 , Z3 ) and Z4 be an ordered triple of non-collinear
points and a fourth point such that no three of the four points are collinear. With (Z1 , Z2 , Z3 )
as our triple of reference and Z4 as our unit point, we say that we have projective frame of
reference and we denote this by PF.
For a p-point Z with normalized areal coordinates
∂F (Z, Z2 , Z3 ) ∂F (Z, Z3 , Z1 ) ∂F (Z, Z1 , Z2 )
,
,
,
(α, β, γ) =
∂F (Z1 , Z2 , Z3 ) ∂F (Z1 , Z2 , Z3 ) ∂F (Z1 , Z2 , Z3 )
395
13.3. PROJECTIVE COORDINATES
we take
(θ, φ, ψ) =
α β γ
, ,
α4 β4 γ4
.
In terms of homogeneous coordinates this is linked to ( ψθ , ψφ , 1) which has equipoised quotients and
so is projectively invariant. From these we define
θ5 φ5 ψ5 ∂PF (Z5 , Z6 , Z7 ) = 12 θ6 φ6 ψ6 .
θ7 φ7 ψ7 We have α4 θ+β4 φ+γ4 ψ = α+β+γ = 1 in all cases here. The equation of a line lα+mβ+nγ = 0,
with l = m = n excluded, becomes
lα4 θ + mβ4 φ + nγ4 ψ = 0,
(13.3.1)
with l = m = n excluded.
We adjoin to RE2 new elements of the following form. We take all triples (θ, φ, ψ) of real
numbers such that
α4 θ + β4 φ + γ4 ψ = 0,
(13.3.2)
but with the triple (0, 0, 0) disallowed. Each allowed triple (θ, φ, ψ) and all multiples (kθ, kφ, kψ)
for k 6= 0 in R, represent one new element which we call an ideal point. As these homogeneous
coordinates satisfy an equation (13.3.2) which is of the form of the equation (13.3.1) but with
l = m = n 6= 0 now included, we say that the ideal points form a line which we call the ideal
line.
If we try to solve formally between the equations (13.3.1) and (13.3.2), as l = m = n = 0 is
excluded, suppose e.g. that n 6= 0. Then we have
m
l
α4 θ + β4 φ + γ4 ψ =0,
n
n
α4 θ + β4 φ + γ4 ψ =0,
so that
and thus we have a solution
m
l
− 1 α4 θ +
− 1 β4 φ = 0,
n
n
α4 θ =
m
− 1,
n
β4 = 1 −
l
,
n
provided we take
m
l
− .
n
n
Thus we obtain the unique ideal point with representative (m − n, n − l, l − m).
We adjoin this unique ideal point to the Euclidean line involved and say that we have a projective line, extended from the Euclidean line. We thus have these extended lines and the ideal
line.
When two distinct extended lines
γ4 ψ = −α4 θ − β4 ψ =
l1 α4 θ + m1 β4 φ + n1 γ4 ψ =0,
l2 α4 θ + m2 β4 φ + n2 γ4 ψ =0,
(13.3.3)
have an ideal point in common, these equations and (13.3.2) have solutions of the form (kθ, kφ, kψ)
where k ∈ R. Thus we must have
1
1
1 l1 m1 n1 = 0.
l 2 m2 n 2 396
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
It follows that the equations (13.3.3) and α4 θ + β4 φ + γ4 ψ = 1 have no solution. Thus these lines
have no point in common in RE2 and so are parallel. This means that distinct lines in RE2 are
parallel if and only if their extensions have an ideal point in common.
We shall not continue with this level of generality but introduce specialization so as to simplify
the calculations. First we take Z4 to be a unit point by choosing it so that α4 = β4 = γ1 giving
these have the common value 31 , and we have selected the centroid of [Z1 , Z2 , Z3 ]. Secondly we
take
Z1 ≡F (1, 0), Z2 ≡F (0, 1), Z3 ≡F (0, 0).
Then on taking coordinates (ξ, η, ζ) for Z we have
α = ξ, β = η, γ = −ξ − η + ζ,
so that α + β + γ = ζ. Then we have that
θ φ5
1 5
∂PF (Z5 , Z6 , Z7 ) = θ6 φ6
2
θ7 φ7
ξ5 η5
27 = ξ6 η6
2 ξ7 φ7
ξ5 η5
27 = ξ6 η6
2 ξ7 η7
ψ5
ψ6
ψ7
−ξ5 − η5 + ζ5
ξ6 − η6 + ζ6
−ξ7 η7 + ψ7
ζ5 ζ6 ,
ζ7 (13.3.4)
(13.3.5)
and each ζ equals 1 or 0 in value. This is formally the same as taking Plücker homogeneous
coordinates so we have linked up with those. We refer to this as having a specialized projective
frame of reference.
13.3.2
More detail in use of projective coordinates
Euclidean lines Z5 Z6 , where Z5 and Z6 have areal coordinates (α5 , β5 , γ5 ) and (α6 , β6 , γ6 ), respectively, have parametric equations
(α, β, γ) = (1 − s)(α5 , β5 , γ5 ) + s(α6 , β6 , γ6 ),
s ∈ R,
which yields in projective coordinates
(α4 θ, β4 φ, γ4 ψ) = (1 − s)(α4 θ5 , β4 φ5 , γ4 ψ5 ) + s(α4 θ6 , β4 φ6 , γ4 ψ6 ),
s ∈ R,
and so
(θ, φ, ψ) = (1 − s)(θ5 , φ5 , ψ5 ) + s(θ6 , φ6 , ψ6 ),
s ∈ R.
In §4.5.2 on taking two triples of reference (Z1 , Z2 , Z3 ) and (W1 , W2 , W3 ), we found that any
projective transformation can be expressed in the form
β
β′
= k1
′
α
α
γ′
γ
= k2 ,
′
α
α
397
13.3. PROJECTIVE COORDINATES
where k1 and k2 are non-zero constants. Then
β ′ + γ ′ k1 β + k2 γ
=
,
α′
α
k1 β + k2 γ
β′ + γ′
+1=
+ 1,
α′
α
1 α + k1 β + k2 γ
=
,
α′
α
α
α′ =
,
α + k1 β + k2 γ
k1 β
β′ =
,
α + k1 β + k2 γ
k2 γ
γ′ =
.
α + k1 β + k2 γ
Now we have
and so
β4′ φ′
β4 φ
= k1
,
′
′
α4 θ
α4 θ
φ′
φ
= j1 ′ ,
θ
θ
where
j1 =
1 α4 β4′
,
k1 β4 α′4
γ4 ψ
γ4′ ψ ′
= k2
,
′
′
α4 θ
α4 θ
ψ
ψ′
= j1 ′ ,
θ
θ
j2 =
(13.3.6)
1 γ4 α′4
.
k2 α4 γ4′
Arising from (13.3.1) we consider the linear equation
l 1 + m1
φ
ψ
+ n1 = 0.
θ
θ
On applying to it the transformation (13.3.6) we obtain the equation
l1 θ′ + j1 m1 φ′ + j2 n1 ψ ′ = 0
for the image line.
Similarly the conic with equation (10.5.7)
2
2
ψ
φψ
ψ
φ
φ
+ c1
+ 2f1 2 + 2g1 + 2h1 = 0,
a1 + b 1
θ
θ
θ
θ
θ
has as image under the transformation (13.3.6) the conic with equation
a1 θ′2 + b1 j12 φ′2 + c1 j22 ψ ′2 + 2f1 j1 j2 φ′ ψ ′ + 2g1 j2 ψ ′ θ′ + 2h1 j1 θ′ φ′ = 0.
13.3.3
Conics and the ideal line
We recall the equation γ 2 = 4kαβ for proper conics. The ideal line has equation α + β + γ = 0 so
to solve these simultaneously we need to solve
[α + (1 − 2k)β]2 = 4k(k − 1)β 2 .
From this we see that an ellipse does not meet the ideal line, a parabola (for which k = 1) meets
it in precisely one point (α, α, −2α). The polar of the point Z6 has equation
γγ6 = 2k(αβ6 + βα6 ),
398
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
and in the case of the parabola for the point Z5 the polar has equation
γ(−2α5 ) = 2(αα5 + βα5 ,
so that α + β + γ = 0 and so the polar of Z5 is the ideal line. Thus the ideal line is the tangent to
the parabola at the point Z5 on it.
Under the transformation (13.3.6) the conic with equation γ 2 = 4kαβ maps on to the conic
with equation
ψ 2 = 4k
α4 β4
θφ,
γ42
and the polar of the point Z6 maps on to the line with equation
ψψ6 =
13.4
2kα4 β4
(θφ6 + φθ6 ).
γ42
The theory independent of the frame of reference
Starting with any frame of reference F = ([O, I , [O, J ), for Z ≡F (x, y) we took x = ζξ , y = ηζ .
For a second frame of reference F ′ obtained from F by translation and rotation, let Z ≡F ′ (x′ , y ′ )
where
x′ = ax + by + k1 , y ′ = −bx + ay + k2 ,
and x′ =
ξ′
ζ′ ,
y′ =
η′
ζ′ .
with
a2 + b2 = 1,
Then we have that
ξ
ξ′
η
aξ + bη + k1 ζ
=a + b + k1 =
,
ζ′
ζ
ζ
ζ
η′
ξ
η
−bξ + aη + k2 ζ
= − b + a + k2 =
,
ζ′
ζ
ζ
ζ
which is achieved by the correspondence
ξ ′ =aξ + bη + k1 ζ,
η ′ = − bξ + aη + k2 ζ,
ζ ′ =ζ.
(13.4.1)
We start with a linear combination of triples (ξ, η, ζ) = p(ξ1 , η1 , ζ1 ) + q(ξ2 η2 , ζ2 ) so that ξ =
pξ1 + qξ2 , η = pη1 + qη2 , ζ = pζ1 + qζ2 . From this we have that
ξ ′ =a(pξ1 + qξ2 ) + b(pη1 + qη2 ) + k1 (pζ1 + qζ2 )
=p(aξ1 + bη1 + k1 ζ1 ) + q(−bξ1 + aη1 + k1 ζ1 )
=pξ1′ + qξ2′ ,
and similarly η ′ = pη1′ + qη2′ , ζ ′ = pζ1′ + qζ2′ .
For a p-point if ξ = ξ1 + λξ2 , η = η1 + λη2 , ζ = ζ1 + λζ2 , then it follows that ξ ′ = ξ1′ + λξ2′ , η ′ =
η1′ + λη2′ , ζ ′ = ζ1′ + λζ2′ .
13.4. THE THEORY INDEPENDENT OF THE FRAME OF REFERENCE
We further note that in all cases
′
ξ1 η1′ ζ1′ ′
ξ2 η2′ ζ2′ ′
ξ3 η3′ ζ3′ aξ1 + bη1 + k1 ζ1 −bξ1 + aη1 + k2 ζ1
= aξ2 + bη2 + k1 ζ2 −bξ2 + aη2 + k2 ζ2
aξ3 + bη3 + k1 ζ3 −bξ3 + aη3 + k2 ζ3
aξ1 + bη1 −bξ1 + aη1 ζ1 = aξ2 + bη2 −bξ2 + aη2 ζ2 aξ3 + bη3 −bξ3 + aη3 ζ3 ′
ξ1 η1′ ζ1′ =(a2 + b2 ) ξ2′ η2′ ζ2′ ξ3′ η3′ ζ3′ ′
ξ1 η1′ ζ1′ ′
= ξ2 η2′ ζ2′ .
ξ3′ η3′ ζ3′ ζ1
ζ2
ζ3
399
(13.4.2)
By (13.1.16) and (13.1.17) we see that our p-lines are independent of the frame of reference.
For the parametrization of p-lines, consider ξ = ξ1 + λξ2 , η = η1 + λη2 , ζ = ζ1 + λζ2 . Then from
above we have that ξ ′ = ξ1′ + λξ2′ , η ′ = η1′ + λη2′ , ζ ′ = ζ1′ + λζ2′ . Thus the value of the parameter is
preserved. It follows that in particular the cross-ratio of an ordered quadruple of collinear points
is independent of the particular frame of reference. Hence p-segments are independent of the
particular frame of reference.
From §13.2.1 it follows that normalized areal coordinates of a p-point depend only on the triple
of reference. From §13.2.4 p-duo-sectors are independent of the particular frame of reference, as
from §13.2.7 are extended duo-angles.
From §13.2.8 extended-orientation is similarly independent.
400
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
13.5
THE COMPLEX PROJECTIVE PLANE
13.5.1
Introduction
We now apply to the complex Euclidean plane CE2 the process which we applied from §13.1.2
onwards to the real Euclidean plane RE2 so as to then form the real projective plane RP2 . In a
similar way we form the complex projective plane CP2 . For this we need to replace the triples of
real numbers (ξ, η, ζ) 6= (0, 0, 0), which are representatives of homogeneous coordinates, by triples
of complex numbers. We use the same notation as before for these but refer to CE2 instead of RE2
for the material we work on.
Thus, then we started with a pair (x, y) of real numbers as Cartesian coordinates of any point
Z in the real Euclidean plane, with respect to some frame of reference. We then introduced triples
(ξ, η, ζ) of real numbers with ζ 6= 0 and
x=
ξ
,
ζ
y=
ξ
,
ζ
as a new type of coordinates for Z. This implies that (ξ ′ , η ′ , ζ ′ ) also represents this point Z if and
only if
ξ ′ = kξ, η ′ = kη, ζ ′ = kζ,
for some non-zero real number k. We can then also write the standard form of equation of a line
in RE2 ax + by + c = 0, (a, b) 6= (0, 0) in the form
aξ + bη + cζ = 0,
(a, b) 6= (0, 0).
This opened the way for us to adjoin ideal points with this new type coordinates of the form
(kξ, kη, 0), but not (0, 0, 0) itself, for all non-zero real numbers k with any given (ξ, η, 0) 6= (0, 0, 0),
and regard it as lying on an extension of the given line still with equation aξ +bη+cζ = 0, (a, b) 6=
(0, 0). As the ideal points, and only these, all have ζ = 0 and so fit into the equation aξ + bη + cζ =
0, (a, b, c) 6= (0, 0, 0) we say that they form an ideal line, and this led us to equation (13.1.18) as
the general form for extended lines in RP2 .
Now, generalizing, we start with a pair (x, y) of complex numbers as Cartesian coordinates of
any point Z in the complex Euclidean plane, with respect to some frame of reference. We now
introduce triples (ξ, η, ζ) of complex numbers with ζ 6= 0 and
x=
ξ
ξ
, quady = ,
ζ
ζ
as a new type of coordinates for Z. This implies that (ξ ′ , η ′ , ζ ′ ) also represents this point Z if and
only if
ξ ′ = kξ, η ′ = kη, ζ ′ = kζ,
for some non-zero complex number k. We can also write, from (12.2.4), the standard form of
equation of a line in CE2 ax + by + c = 0, (a, b) 6= (0, 0) in the form
aξ + bη + cζ = 0,
(a, b) 6= (0, 0).
This opens the way for us to adjoin ideal points with this new type coordinates of the form
(kξ, kη, 0), but not (0, 0, 0) itself, for all non-zero complex numbers k with any given (ξ, η, 0) 6=
(0, 0, 0), and regard it as lying on an extension of the given line, still with equation aξ + bη + cζ =
0, (a, b) 6= (0, 0). As the ideal points, and only these, all have ζ = 0 and so fit into the equation
aξ + bη + cζ = 0,
(a, b, c) 6= (0, 0, 0),
(13.5.1)
we say that they form an ideal line, and this leads us to equation (13.5.1) as the general form for
extended lines in CP2 .
We can re-interpret in this way the material of §13.1.2, §13.1.3 and §13.1.4. We note in particular
that an extended line will have an equation of the form aξ + bη + cζ = 0 and that this meets the
ideal line with equation ζ = 0 at the p-point with homogeneous coordinates (b, −a, 0).
401
13.5. THE COMPLEX PROJECTIVE PLANE
13.5.2
Development
We can now develop an interplay of the material following §12.2.1 with the material following §13.1
leading to complex analogues of the latter material.
From §13.1.4 we consider the intersection of two (distinct) p-lines with equations
a1 ξ + b1 η + c1 ζ = 0,
a2 ξ + b2 η + c2 ζ = 0,
(13.5.2)
where all the coordinates and all the coefficients are complex and the two triples of coordinates are
not linearly dependent in C3 . Then these have a unique p-point in common in the formula given
in §13.1.4.
Moreover if we have two distinct parallel lines, the corresponding p-lines have equations which
can be put in the form
aξ + bη + c1 ζ = 0,
aξ + bη + c2 ζ = 0,
(c1 6= c2 ).
By subtraction we note that these hold simultaneously if and only if (c1 − c2 )ζ = and so ζ = 0, so
that they have an ideal point in common.
We need to take up from §12.2.4 on the question of perpendicular lines. On replacing ordered
pairs by triples, we note that given p-lines l1 , l2 with equations
a1 ξ + b1 η + c1 ζ = 0,
a2 ξ + b2 η + c2 ζ = 0,
we say that these lines are perpendicular when
a1 a2 + b1 b2 = 0.
This is independent of the particular coefficients taken, as for
j1 a1 ξ + j1 b1 η + j1 c1 ζ = 0,
j2 a2 ξ + j2 b2 η + j2 c2 ζ =
0,
the condition is
j1 j2 (a1 a2 + b1 b2 ) = 0,
and this is equivalent.
Clearly, given any l1 there is a unique direction for extended lines perpendicular to it, given by
(a2 , b2 ) = j(−b1 , a1 ),
where j 6= 0.
Now take two isotropic p-lines. Suppose we denote the first by l1 and that it has equation
−iξ + η + c1 ζ = 0,
so that it meets the ideal line with equation ζ = 0 at the point with homogeneous coordinates
(1, i, 0). We denote this point by I.Suppose we denote the second by l1 and that it has equation
iξ + η + c1 ζ = 0,
so that it meets the ideal line with equation ζ = 0 at the point with homogeneous coordinates
(1, −i, 0). We denote this point by J . These two points I and J are called the ideal circular
points. We shall explain the reason for this name in the next chapter.
Referring to l1 above, for a perpendicular p-line l2 we have
(λ2 , µ2 ) = j(−1, −i) = −ji(−i, 1),
and so l2 is parallel to l1 . We reach a similar conclusion if we take (λ1 , µ1 ) = (i, 1). Thus an
isotropic p-line is perpendicular to the p-lines which are parallel to it.
402
13.5.3
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
Mid-lines
Now suppose that we have a non-isotropic line l1 with equation
λ1 x + µ1 y + ν1 = 0,
and a p-point Z0 not on it. Then there is a unique line through Z0 such that λ21 + µ21 6= 0, and
a point Z0 which is perpendicular to l1 , and it meets l1 at a unique point. Thus we can drop a
perpendicular from a point to a non-isotropic line.
Moreover, as in the real case, the square of perpendicular distance from Z0 to l1 is given by
[λ1 x0 + µ1 y0 + ν1 ]2
.
λ21 + µ21
From this, given two non-parallel and non-isotropic lines l1 , l2 with equations
λ1 x + µ1 y + ν1
λ2 x + µ2 y + ν2
=
=
0,
0,
the points Z, the perpendicular distances from which to l1 and l2 are equal, are those which satisfy
[λ2 x + µ2 y + ν2 ]2
[λ1 x + µ1 y + ν1 ]2
=
.
2
2
λ1 + µ1
λ22 + µ22
This is satisfied if and only if
λ1 (x + ıu) + µ1 (y + ıv) + ν1
λ2 (x + ıu) + µ2 (y + ıv) + ν2
p
p
+
= 0,
2
2
p.v. λ1 + µ1
p.v. λ22 + µ22
or
λ1 (x + ıu) + µ1 (y + ıv) + ν1
λ2 (x + ıu) + µ2 (y + ıv) + ν2
p
p
−
= 0,
2
2
p.v. λ1 + µ1
p.v. λ22 + µ22
where we are using an obvious notation for the principal value of a square root. These are the
equations of two lines. It can be checked that these are perpendicular and are non-isotropic.
We call these the mid-lines of the pair of lines l1 , l2 .
13.5.4
Further material on sensed duo-angles
when y ≤ 0. Thus we have a modification of the polar form, that Z ∈ l if and only if
x = t cos α4d , y = t sin α4d ,
for some t ∈ R.
In 1853 Laguerre considered a method of using cross ratios to handle duo-angles. Consider
cr(tan θd , 0, i, −i) = ν,
where i is the imaginary number such that i2 = −1 and ν = λ + iµ is a complex number. Then
we have that
1 + i tan θd
= ν,
1 − i tan θd
from which
1−ν
.
1+ν
When the expression on the right-hand side here is a real number, this will define a unique duoangle θd .
tan θd = i
403
13.5. THE COMPLEX PROJECTIVE PLANE
Now more generally consider duo-angles in standard position αd and βd . Then we have the
following
cr(tan αd , tan βd , i, −i)
(tan αd − i)(tan βd + i)
(tan αd + i)(tan βd − i)
tan αd tan βd + 1 + i(tan αd − tan βd )
=
tan αd tan βd + 1 − i(tan αd − tan βd )
=
=
tan αd −tan βd
1 + i 1+tan
αd tan βd
tan αd −tan βd
1 − i 1+tan
αd tan βd
1 + i tan(αd − βd )
1 − i tan(αd − βd )
1 + i tan θd
,
=
1 − i tan θd
=
where θd = αd − βd is a sensed duo-angle. For clarity we use the notation 2 ∗ θd = θd + θd in
connection with duo-angles, where addition is modulo a straight angle of course. Then we can
continue with
1 + i tan θd
1 − i tan θd
(1 + i tan θd )2
=
1 + tan2 θd
1 − tan2 θd + 2i tan θd
=
sec2 θd
sin2 θd
2 tan θd
2
= 1−
cos θd 1 + i
cos2 θd
1 − tan2 θd
=(cos2 θd − sin2 θd )(1 + i tan 2 ∗ θd )
cos2 θd − sin2 θd
(cos 2 ∗ θd + i sin 2 ∗ θd )
cos 2 ∗ θd
= ± (cos 2 ∗ θd + i sin 2 ∗ θd )
= ± (cos |2 ∗ θd |r + i sin |2 ∗ θd |r ),
=
with the + or the − according as |θd |r lies in (0, π/2) or (π/2, π).
When 0 < |θd |r < 12 π we have that |2 ∗ θd |r = 2|θd |r . Then
r
cr(tan θd , 0, i, −i) = ei2|θd | .
When 21 π < |θd |r < π we have that |2 ∗ θd |r = 2|θd |r − π. Then
cr(tan θd , 0, i, −i) = − ei(2|θd |
r
−π)
r
=ei2|θd | .
This is Laguerre’s formula, which can also be written in the form
|θd |r =
13.5.5
1
log cr(tan θd , 0, i, −i).
2i
Duo-angles
For convenience we take Z1 = O throughout this section; the more general situations can subsequently be handled by translation.
404
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
We let D1 be the closed duo-sector defined by
δF (O, I, Z + ıW )
>L 0,
δF (O, Z + ıW, J)
which works out at
y + ıv
>L 0.
x + ıu
If we write x + ıu and y + ıv in polar form as
p
p
x + ıu = x2 + u2 cis θd , y + ıv = y 2 + v 2 cis φd ,
then
p
y2 + v2
y + ıv
cis (φd − θd ).
= √
x + ıu
x2 + u2
Thus we need cos(φd − θd ) + ı sin(φd − θd ) to be lexicographically positive.
The equation of any line l through O, other than OJ, can be put in the form
y + ıv = (p + ıq)(x + ıu),
and we note that the condition that Z + ıW be in D1 is that p + ıq be lexicographically positive.
Correspondingly, we let a second closed duo-sector with the same side- lines D2 be the set of
points on the lines OI or OJ and the set of points Z + ıW such that
δF (O, I, Z + ıW )
<L 0,
δF (O, Z + ıW, J)
If the line l through O, other than OJ, with equation
y + ıv = (p + ıq)(x + ıu),
is non-isotropic, then the mid-lines of the pair OI and l are the lines
(a)
y + ıv =
(b)
p + ıq
p
(x + ıu),
1 + p.v. 1 + (p + ıq)2
p
1 + pv. 1 + (p + ıq)2
(x + ıu).
y + ıv = −
p + ıq
The slopes of these have the form
g + ık, −
1
−g + ık
,
= 2
g + ık
g + k2
and these have lexicographically different signs unless g = 0. We then ask when
p + ıq
p
1 ± p.v. 1 + (p + ıq)2
is purely real, equal to ık, say. This happens when
p + ıq =
Now
2k
ı.
+1
k2
2k
k2 + 1
405
13.5. THE COMPLEX PROJECTIVE PLANE
has absolute value at least 1, so the difficulty arises when p = 0 and |q| ≥ 1. We must exclude
the cases p + ıq = ±ı as these correspond to isotropic lines, and if we also exclude the cases when
p = 0, |q| > 1, then we will be able to discriminate between the two duo-sectors with arms OI and
l when the corresponding mid- line lies in D1 or D2 .
When l is the axis OJ, the corresponding mid-lines are the lines with equations
(a)
y + ıv = x + ıu,
(b)
y + ıv = −(x + ıu),
and it is (a) here that is in D1 .
However, identifying in practice a duo-sector by means of the midline lying in it leads to very
laborious calculations. Consequently we adopt the following alternative approach.
We let l4 be a line O(Z4 + ıW4 ), through the origin O but not being OJ. As we can also identify
it as O(−Z4 − ıW4 ), without loss of generality we may assume that
1
>L 0.
x4 + ıy4
We define a duo-region D4 with arms OI, l4 as follows:- if Z4 + ıW4 ∈ D1 , so that
δF (O, I, Z4 + ıW4 )
y4 + ıv4
>L 0,
=
δF (O, Z4 + ıW4 , J)
x4 + ıu4
let Z + ıW ∈ D4 if
δF (O, I, Z + ıW )
>L 0,
δF (O, Z + ıW, Z4 + ıW4 )
and if Z4 + ıW4 6∈ D1 , so that
y4 + ıv4
δF (O, I, Z4 + ıW4 )
<L 0,
=
δF (O, Z4 + ıW4 , J)
x4 + ıu4
let Z + ıW ∈ D4 if
δF (O, I, Z + ıW )
<L 0.
δF (O, Z + ıW, Z4 + ıW4 )
Note that we have J 6∈ D4 according as Z4 + ıW4 ∈ D1 , and conversely.
We add to the foregoing that when Z4 + ıW4 ∈ OJ we take D4 to be D1 itself.
Now we call any pair of lines {OI, l4 } through the origin, in association with the duo-sector D4 ,
a duo-angle in standardized position with respect to F , and denote it by (OI, l4 ). We emphasize
that l4 is not to be an isotropic line.
13.5.6
Tangent of a duo-angle
For a point Z4 + ıW4 6= O, we let αd be the duo-angle (OI, O(Z4 + ıW4 )). With
h2 = |O, Z4 + ıW4 |2 6= 0
we may assume that h is lexicographically positive. We let Q ≡ (h, 0), R ≡ (0, h) and as we just
need to identify the side-lines of a duo-angle, we may assume without loss of generality that the
real part of y4 + ıv4 is lexicographically positive.
When Z + ıW is not on OJ, so that x + ıu 6= 0, we define
tan αd =
y + ıv
.
x + ıu
406
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
This is well-defined, as the value of the ratio is unaltered if we replace (x + ıu, y + ıv) by (r(x +
ıu), r(y + ıv)) for any non-zero complex number r.
As in 8.1.11 we can define the sum of two such duo-angles so that addition formulae for the
tangent function holds. We take points Z4 + ıW4 ≡ (x4 + ıu4 , y4 + ıv4 ) and Z5 + ıW5 ≡ (x5 +
ıu5 , y5 + ıv5 ), neither on an isotropic line through the origin or on OJ, and so without loss of
generality may take them both to be at square-of-distance h2 from the origin. We suppose that
the real parts of h, y4 + ıv4 and y5 + ıv5 are lexicographically positive.
We let Q ≡ (h, 0) and then the line through Q, parallel to the line (Z4 + ıW4 )(Z5 + ıW5 ), meets
again in a point Z6 + ıW6 the circle with centre the origin and square-of-radius- length h2 . Then
as in 8.1.11 we have
x6 + ıu6
y6 + ıv6
[y5 + ıv5 − (y4 + ıv4 )]2 − [x5 + ıu5 − (x4 + ıu4 )]2
,
[x5 + ıu5 − (x4 + ıu4 )]2 + [y5 + ıv5 − (y4 + ıv4 )]2
[x5 + ıu5 − (x4 + ıu4 )][y5 + ıv5 − (y4 + ıv4 )]
.
= −2h
[x5 + ıu5 − (x4 + ıu4 )]2 + [y5 + ıv5 − (y4 + ıv4 )]2
= h
In these circumstances we define the sum of the duo-angles αd = (OI, O(Z4 + ıW4 )), βd =
(OI, O(Z5 + ıW5 )), in that order, to be the duo-angle γd = (OI, O(Z6 + ıW6 )).
When Z3 + ıW3 ≡ (0, h) and (Z4 + ıW4 )(Z5 + ıW5 ) is not parallel to (Z2 + ıW2 )(Z3 + ıW3 ),
Z6 + ıW6 will not be on OJ and so tan γd will be defined. Then, as in 8.1.11, we have the addition
formula
tan αd + tan βd
tan γd =
.
1 − tan αd tan βd
13.5.7
Sensed duo-angles
We also introduce the point Z4 + ıW4 ≡ (x4 + ıu4 , −y4 − ıv4 ) which is not on an isotropic line
through the origin as Z4 + ıW4 is not, and denote the duo-angle (OI, O(Z4 + ıW4 )) by −αd . Then
clearly
tan(−αd ) = − tan αd .
We define the difference βd − αd to be the sum βd + (−αd ). We call this the sensed duo-angle with
sides (O(Z4 + ıW4 ), O(Z5 + ıW5 )). When (Z4 + ıW4 )(Z5 + ıW5 ) is not parallel to (Z2 + ıW2 )(Z3 +
ıW3 ), then tan(βd − αd ) is defined and we then have the addition formula
tan(βd − αd ) =
tan βd − tan αd
.
1 + tan αd tan βd
If O(Z4 + ıW4 ) is perpendicular to O(Z5 + ıW5 ), then as in (ix) we can take
Z5 + ıW5 ≡ (x5 + ıu5 , y5 + ıv5 ) = (−(y4 + ıv4 ), x4 + ıv4 ).
Then the point Z7 + ıW7 for the duo-angle βd − αd ) has x7 + ıu7 = 0, y7 + ıv7 = h. Thus in this
case βd − αd is the duo-angle (OI, OJ) and so its tangent is not defined. We also note that in this
case tan αd tan βd = −1 so the right-hand side of the addition formula for tan(βd − αd ) also fails.
13.5.8
Complex-valued distance, square of distance
We note that
2
δF (Z + ıW, Z1 + ıW1 + K, Z1 + ıW1 ) =
ı
=
=
2
δF (Z + ıW − Z1 − ıW1 , K, O)
ı
x + ıu − (x1 + ıu1 ) + ı[y + ıv − (y1 + ıv1 )
h(cos αd + ı sin αd ),
where αd is the sensed duo-angle ((Z1 + ıW1 )(Z1 + ıW1 + I), (Z1 + ıW1 )(Z + ıW ). Thus this can
be interpreted as complex-valued distance (Z1 + ıW1 )(Z + ıW ).
407
13.6. EXTENSION OF RATIOS OF SENSED-AREAS
On taking a quotient of such terms, we have
δF (Z + ıW, Z1 + ıW1 + K, Z1 + ıW1 )
= cos 2αd + ı sin 2αd .
δF (Z + ıW, Z1 + ıW1 + L, Z1 + ıW1 )
We have similarly that
4δF (Z + ıW − (Z1 + ıW1 ), K, O)δF (Z + ıW − (Z1 + ıW1 ), L, O) = |Z1 + ıW1 , Z + ıW |2 .
Thus complex-valued distance, cis (αd ) and square-of-distance can be expressed in terms of
sensed area, and this is one of the features of Πc which guides us to generalization.
13.5.9
Another route to specialization
As indicated, very much of our geometry can be worked with in Πc , but with clear restrictions on
the generalization of some concepts. Such complex geometry is of interest for its own sake. But
we can also specialize from it in other ways. For instance we can represent points by coordinates
of the form (x, ıv), taking y = u = 0. Then we find that
x
1 1
δF (Z1 + ıW1 , Z2 + ıW2 , Z3 + ıW3 ) = ı x2
2
x3
v1
v2
v3
1
1
1
.
Now the square-of-distance from the origin to Z + ıW is equal to x2 − v 2 . On working with this we
find a geometry in which rectangular hyperbolas play the role of locus at unit square of distance
from the origin.
13.6
EXTENSION OF RATIOS OF SENSED-AREAS
13.6.1
Ratio of sensed-areas
As ratio of sensed areas is the simplest affine invariant, we look for more from the extension of it.
Consider the ratio of sensed areas
ξ4 η4 0 x5 y5 1 x6 y6 1 δF (OZ4∞ , Z5 , Z6 )
= δF (OZ4∞ , Z7 , Z8 )
ξ4 η4 0 x7 y7 1 x8 y8 1 x5 + ξ4 y5 + η4 1 x5
y5
1 x6
y6
1 =
x7 + ξ4 y7 η4 1 x7
y7
1 x8
y8
1 =
δF (Z5 + Z4+ , Z5 , Z6 )
,
δF (Z7 + Z4+ , Z7 , Z8 )
where we have introduced an auxiliary point Z4+ ≡ (ξ4 , η4 ).
408
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
13.6.2
Ratios of distances
In the extended complex plane Πce we can write
(Z4 + ıW4 )(Z5 + ıW5 )
(Z6 + ıW6 )(Z7 + ıW7
=
δF (Z4 + ıW4 , OL∞ , Z5 + ıW5 )
,
δF (Z6 + ıW6 , OL∞ , Z7 + ıW7 )
=
δF (Z4 + ıW4 , OL∞ , Z6 + ıW6 )
.
δF (Z4 + ıW4 , OL∞ , Z5 + ıW5 )
and in particular
(Z4 + ıW4 )(Z6 + ıW6 )
(Z4 + ıW4 )(Z5 + ıW5
Moreover, we also have
=
=
δF (Z4 + ıW4 , OL∞ , Z6 + ıW6 )δF (Z4 + ıW4 , OK ∞ , Z6 + ıW6 )
δF (Z4 + ıW4 , OL∞ , Z5 + ıW5 )δF (Z4 + ıW4 , OK ∞ , Z5 + ıW5 )
[x5 + ıu5 − x4 − ıu4 + ı(y5 + ıv5 − y4 − ıv4 )][x5 + ıu5 − x4 − ıu4 − ı(y5 + ıv5 − y4 − ıv4 )]
[x7 + ıu7 − x6 − ıu6 + ı(y7 + ıv7 − y6 − ıv6 )][x7 + ıu7 − x6 − ıu6 − ı(y7 + ıv7 − y6 − ıv6 )]
|Z4 + ıW4 , Z5 + ıW5 |2
,
|Z6 + ıW6 , Z7 + ıW7 |2
which is the ratio of square distances.
We shall use a special notation I for OL∞ and J for OK ∞ .
13.6.3
Complex extended conics
The definition of proper conics in 4.1.2 generalizes and it is desirable to go from that directly to
the characterization of conics in 4.7.2 in terms of sensed-area, that, given fixed non-collinear points
Z1 + iW1 , Z2 + iW2 , Z3 + iW3 , for a variable point Z + iW we let Z ′ + iW ′ be the projection of
Z + iW onto the line Z1 + iW1 Z2 + iW2 , parallel to the line Z2 + iW2 Z3 + iW3 . Then for fixed
non-zero complex numbers j1 , j2 , the locus of Z + iW such that
δ(Z2 +iW2 , Z1 +iW1 , Z+iW )2 +j1 δ(Z1 +iW1 , Z ′ +iW ′ , Z3 +iW3 )2 = j2 δ(Z2 +iW2 , Z ′ +iW ′ , Z3 +iW3 )2
is a complex conic.
Any point Z ∈ Z1 Z2 has the form Z = (1 − s)Z1 + sZ2 , and then for Z3 6∈ Z1 Z2
δF (Z1 , Z, Z3 ) = (1 − s)δF (Z1 , Z1 , Z3 ) + sδF (Z1 , Z2 , Z3 ) = sδF (Z1 , Z2 , Z3 ),
so the mid-point of Z1 and Z2 is the unique point in Z1 Z2 such that
δF (Z1 , Z, Z3 ) =
1
δF (Z1 , Z2 , Z3 ).
2
This definition generalizes to complex-valued coordinates, and so the concept of conjugate diametral lines does also.
Now under the algebraic correspondence by which a diametral line for a hyperbola maps to
its conjugate diametral line, each asymptote is self- corresponding. Similarly with complex affine
geometry, for any central conic there will be a pair of self-conjugate diametral lines. In particular,
in generalized perpendicularity we will have two self-perpendicular directions, called by Laguerre
isotropic. Thus the complex generalization resembles the hyperbolic case rather than the more
familiar elliptic/circular case.
409
13.6. EXTENSION OF RATIOS OF SENSED-AREAS
13.6.4
Generalization of distance
We now look generalization from the above concepts and relationships and start with the concept
of distance, first in Π. Given any a > 0 and b > 0, we can generalize complex-valued distance by
defining
x5 − x4
y5 − y4
Z4 Z5 E =
+i
.
a
b
This has the properties of complex-valued distance:
1. Z4 Z5 E = 0 if and only if Z4 = Z5 ;
2. Z4 Z5E = Z5 Z4E for all Z4 and Z5 ;
3. Z4 Z5 E + Z5 Z6 E = Z4 Z6 E for all Z4 , Z5 and Z6 .
We can also introduce a complex conjugate of Z4 Z5E , to wit
∗
Z4 Z5 E =
x5 − x4
y5 − y4
−i
.
a
b
From these we can define distance by
∗
|Z4 , Z5 |E
1
= [Z4 Z5 E Z4 Z5 E ] 2
r
y2
x2
+ 2.
=
2
a
b
This distance-function has properties similar to those of standard distance, and the locus of points
at unit E-distance from the origin is the ellipse with equation
x2
y2
+ 2 = 1.
2
a
b
Similarly, still with any a > 0 and b > 0, we can define
Z4 Z5H =
y5 − y4
x5 − x4
+
.
a
b
This has the properties of complex-valued distance:
1. Z4 Z5 H = 0 if Z4 = Z5 ;
2. Z4 Z5 H = Z5 Z4 H for all Z4 and Z5 ;
3. Z4 Z5H + Z5 Z6H = Z4 Z6H for all Z4 , Z5 and Z6 ,
but now (1) has a fuller form, that Z4 Z5 H = 0 if Z4 = Z5 or if Z4 Z5 is parallel to the line with
∗
equation y = −(b/a)x. We can also introduce a complex conjugate of Z4 Z5 H , to wit
∗
Z4 Z5 H =
x5 − x4
y5 − y4
−
.
a
b
From these we can define distance by
|Z4 , Z5 |H
∗
1
= [Z4 Z5H Z4 Z5 H ] 2
y2 1
x2
= [ 2 − 2 ]2 .
a
b
This distance-function has some properties similar to those of standard distance, and the locus of
points at unit H-distance from the origin is the hyperbola with equation
y2
x2
−
= 1.
a2
b2
410
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
However, if Z4 and Z5 are on the union of the asymptotes, this distance is equal to 0, and if Z5 is
in the region, bounded by the asymptotes, in which
y2
x2
−
< 0,
a2
b2
then the distance |O, Z5 |H is a purely imaginary number.
In a complex affine plane, any central conic has an equation which can be put in either of the
forms
z2
w2
+
= 1,
a2
b2
or
w2
z2
−
= 1,
a2
b2
but, whichever we take, the pattern of complex-valued distance and distance will be of the more
awkward hyperbolic type above.
13.6.5
Mapping an ellipse to a hyperbola
In complex affine geometry there are no bounded conics and so we lose the distinction between
ellipses and hyperbolas.
If we consider a transformation on the complex affine plane
z ′ = z, w′ = ıw,
where ı2 = −1, we note that it maps the locus with equation
w2
z2
+ 2 =1
2
a
b
onto the locus with equation
z ′2
w′2
− 2 = 1.
2
a
b
Moreover we note that
δF (Z + iW, Z4 + iW4 , Z5 + iW5 )
δF (Z ′ + iW ′ , Z4′ + iW4′ , Z5′ + iW5′ )
=
,
δF (Z ′ + iW ′ , Z6′ + iW6′ , Z7′ + iW7′ )
δF (Z + iW, Z6 + iW6 , Z7 + iW7 )
and so this ratio of sensed-areas is invariant under this transformation.
It follows that any property of an ellipse which is proved entirely by use of ratios of sensed-areas
is also valid for each hyperbola.
13.6.6
A result on sensed-areas
Now in Π let Z1 , Z2 , Z3 , Z4 be points on a circle, with centre O, such that Z1 Z2 k Z3 Z4 . Then the
line through O perpendicular to Z1 Z2 bisects [Z1 , Z2 ] and [Z3 , Z4 ], and if W1 , W2 are the feet of
the perpendiculars from Z1 , Z2 to Z3 Z4 , the triangles [Z1 , W1 , Z3 ] and [Z2 , W2 , Z4 ] are congruent.
In particular |Z1 , Z3 | = |Z2 , Z4 |. It follows that the triangles [O, Z1 , Z3 ], [O, Z2 , Z4 ] are congruent
and have equal sensed-area. Thus
δF (O, Z1 , Z3 )
= 1.
δF (O, Z4 , Z2 )
If we apply the affine transformation of 5.4.1, we deduce an analogue of this as follows. Now
let Z1′ , Z2′ , Z3′ , Z4′ be points on an ellipse, with centre O, such that Z1′ Z2′ k Z3′ Z4′ . Then
δF (O, Z1′ , Z3′ )
= 1.
δF (O, Z4′ , Z2′ )
If we apply now the special transformation in 8.5.5, we deduce that this result further holds for
a hyperbola.
411
13.7. EXTENDED REAL CONICS
Z4
b
b
Z2
b
Z3
Z1
b
b
b
Z2
Z4
b
b
b
Z0+
b
O
O
b
b
Z3
b
Z1
Figure 5.29
We seek a similar result for a parabola from 8.5.1. From this we find, that if Z1 , Z2 , Z3 , Z4
are variable points on the parabola with equation y 2 = 4ax and Z1 Z2 k Z3 Z4 , then for any point
Z0+ ≡ (x0 , 0) on the axis of the parabola we have
δF (Z1 + Z0+ , Z1 , Z3 )
= 1.
δF (Z4 + Z0+ , Z4 , Z2 )
13.7
EXTENDED REAL CONICS
13.7.1
Equations in areal coordinates
When working with areal coordinates we shall take triples (α, β, γ) 6= (0, 0, 0) such that α+β+γ = 0
as representing ideal points. Then α + β + γ = 0 is an equation of the ideal line.
Now look at the proper conic C in 4.2.1 with equation
γ 2 − 4kαβ = 0,
and extend this conic by adjoining the ideal points which satisfy this equation. In 4.2.3 and 4.2.4
we obtained the polar of the point W0 ≡ (α0 , β0 , γ0 ) as the line with equation
−2kβ0 α − 2kα0 β + γ0 γ = 0.
This failed to be a line when k 6= 1 if (α0 , β0 , γ0 ) are proportional to (1, 1, −2k). But in that case
the equation of the polar reduces to
−2kα − 2kβ − 2kγ = 0,
that is the equation of the ideal line. Then, by extension, we take the ideal line to be the polar of
the centre of an ellipse or hyperbola, and conversely the centre to be the pole of the ideal line.
13.7.2
Type of conic given by intersections with ideal line
The extended conic with equation γ 2 − 4kαβ = 0 meets the ideal line with equation α + β + γ = 0
when
(−α − β)2 − 4kαβ = 0,
412
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
and so when
α2 + 2(1 − 2k)αβ + β 2 = 0.
The discriminant of this is
4(1 − 2k)2 − 4 = 16k(k − 1).
Thus for a parabola, when k = 1, there is a unique point of intersection with coordinates
proportional to (1, 1, −2).
For a circle/ellipse, when k(k − 1) < 0, there are no real points of intersection.
For a hyperbola, when k(k − 1) > 0, we have
p
α = (2k − 1 ± 2 k(k − 1))β,
so there are two real points of intersection, with coordinates proportional to
p
p
p
p
(2k − 1 + 2 k(k − 1), 1, −2k − 2 k(k − 1)), (2k − 1 − 2 k(k − 1), 1, −2k + 2 k(k − 1)).
13.7.3
Centre of extended parabola
In the case of the parabola, the formal equation of the tangent at the unique point (with coordinates
proportional to (1, 1, −2)) of the parabola on the ideal line is
−2α − 2β − 2γ = 0,
and this is the equation of the ideal line. Thus the pole of the ideal line with respect to the parabola
is the ideal point with coordinates (1, 1, −2), and, by extension from 8.6.1, we call this point the
centre of the parabola.
13.7.4
Asymptotes
The formal tangents to the hyperbola, at its points of intersection with the ideal line noted in (ii),
are the lines with equations
p
p
−2kα − 2k(2k − 1 ± 2 k(k − 1))β + (−2k ∓ 2 k(k − 1))γ = 0.
By 4.4.3 we see that these are the equations of the asymptotes. Thus the asymptotes of a hyperbola
are the tangents at the points where the hyperbola meets the ideal line.
13.7.5
Diametral lines as polars
In 4.5.3 we found diametral lines to have equations of the form
2k(n0 − l0 )α + 2k(m0 − n0 )β − (l0 − m0 )γ = 0.
Now we note that this is the formal polar of the ideal point with coordinates (n0 − m0 , l0 − n0 , m0 −
l0 ). Thus diametral lines are polars of ideal points.
13.7.6
Identification of the foci
For the ellipse in standard position
x2
y2
+ 2 = 1,
2
a
b
where 0 < b < a, the focus C1 satisfies
x = −ae = −
p
a2 − b2 , y = 0.
413
13.7. EXTENDED REAL CONICS
This is the only real root of
(x +
p
a2 − b2 )2 + y2 = 0,
but if we introduce complex numbers this can be written as
p
y = ±ı(x + a2 − b2 ),
in homogeneous coordinates
η = ±ı(ξ + ζ
p
a2 − b2 ).
On putting ζ = 0, we get η = ±ıξ. Thus we have two non-real lines C1 I and C1 J , and the focus
is their point of intersection.
Can we find some other property to identify these lines? On substituting
p
y = ±ı(x + a2 − b2 ),
into the equation of the ellipse, we obtain
p
(x a2 − b2 + a2 )2 = 0.
This quadratic equation has a repeated root, so this line is a tangent to the ellipse.
Thus this focus can be identified as the
√ point of intersection of tangents to the ellipse from the
points I and J . The other focus C2 ≡ ( a2 − b2 , 0) can be found by the same process.
Now on re-writing the equation of the ellipse as
b2
y2
x2
+ 2 = 1,
2
2
− (b − a )
b
we obtain
b2 x2 + b2 y 2 + b2 (b2 − a2 ) = (b2 − a2 )y 2 + b4 ,
and, on completing the square in y on the left,
2
p
b2
a2 − b 2
√
y
+
a2 − b2 )2 = −
.
b2
ı a2 − b 2
√
Thus we have an imaginary focus at (0, −ı a2 − b2 ), with corresponding directrix
x2 + (y + ı
b2
.
y=− √
ı a2 − b 2
The eccentricity now is
√
a2 − b 2
ı
.
b
There is a second non-real focus symmetrically placed with respect to the x-axis.
All four foci of the ellipse turn up as the points of intersection of tangents from the points I
and J .
13.7.7
Plücker reciprocation, areal line coordinates
In 7.7.7 we considered reciprocation with respect to any conic. Now we note that in 8.2.1 in
associating with the point with Cartesian coordinates (l, m) the line with Plücker line coordinates
(l, m), i.e. the line with equation lx + my + 1 = 0, we are making the point correspond to its polar
with respect to the conic with equation in Cartesian coordinates
x2 + y 2 + 1 = 0.
414
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
This conic does not contain any points in Π but does in Πc .
We can deal similarly with our line coordinates as given in 3.1.1. We note that with respect to
the conic with equation
α2 + β 2 + γ 2 = 0,
the point with areal coordinates (α0 , β0 , γ0 ) has as polar the line with equation
α0 α + β0 β + γ0 γ = 0.
From this, the point with areal coordinates
δF (Z1 , Z2 , Z3 )δF (Z1 , U, V )
,
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
δF (Z1 , Z2 , Z3 )δF (Z2 , U, V )
,
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
δF (Z1 , Z2 , Z3 )δF (Z3 , U, V )
,
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
has as polar the line with equation
δF (Z1 , Z2 , Z3 )
[δF (Z1 , U, V )α + δF (Z1 , U, V )β + δF (Z3 , U, V )γ] = 0.
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
This identifies the expression in the identity in 3.??.
Thus for the line λα + µβ + νγ = 0, we now have the line coordinates
λ
=
µ
=
ν
=
δ1 δF (Z1 , U, V )
,
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
δ1 δF (Z2 , U, V )
,
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
δ1 δF (Z3 , U, V )
.
δF (Z1 , U, V ) + δF (Z2 , U, V ) + δF (Z3 , U, V )
In the other direction, take the equations of two lines
δF (Z1 , U1 , V1 )α + δF (Z2 , U1 , V1 )β + δF (Z3 , U1 , V1 )γ
δF (Z1 , U2 , V2 )α + δF (Z2 , U2 , V2 )β + δF (Z3 , U2 , V2 )γ
α+β+γ
= 0,
= 0,
= δ1 .
On solving these we obtain
α =
β
=
γ
=
δ1 δF [(Z2 , Z3 ), (U1 , V1 ), (U2 , V2 )]
,
δF [(Z2 , Z3 ), (U1 , V1 ), (U2 , V2 )] + δF [(Z3 , Z1 ), (U1 , V1 ), (U2 , V2 )] + δF [(Z1 , Z2 ), (U1 , V1 ), (U2 , V2 )]
δ1 δF [(Z3 , Z1 ), (U1 , V1 ), (U2 , V2 )]
,
δF [(Z2 , Z3 ), (U1 , V1 ), (U2 , V2 )] + δF [(Z3 , Z1 ), (U1 , V1 ), (U2 , V2 )] + δF [(Z1 , Z2 ), (U1 , V1 ), (U2 , V2 )]
δ1 δF [(Z1 , Z2 ), (U1 , V1 ), (U2 , V2 )]
.
δF [(Z2 , Z3 ), (U1 , V1 ), (U2 , V2 )] + δF [(Z3 , Z1 ), (U1 , V1 ), (U2 , V2 )] + δF [(Z1 , Z2 ), (U1 , V1 ), (U2 , V2 )]
We can seek to use these expressions as we did the ones in 8.2. The point W0 = (α0 , β0 , γ0 )
has not got with respect to the conic
α2 + β 2 + γ 2 = 0,
a polar
α0 α + β0 β + γ0 γ = 0
415
13.7. EXTENDED REAL CONICS
which is a line, when α0 = β0 = γ0 . Then W0 is the centroid of the triangle [Z1 , Z2 , Z3 ]. Thus the
centroid plays here the role that O did in 8.2.
(i) Under Plücker reciprocation we make correspond to the point W ≡ (u, v) the line with
equation ux + vy + 1 = 0. This is the polar of W with respect to the circle x2 + y 2 + 1 = 0, which
is non-real.
However the polar of −W ≡ (−u, −v) with respect to the unit circle x2 + y 2 − 1 = 0 is
−ux − vy − 1 = 0 which is equivalent to ux + vy + 1 = 0. Thus Plücker reciprocation can be
expressed as the composition of reciprocation with respect to the unit circle over central symmetry
in the origin. The latter can also be expressed as rotation about O through 180F . We can in fact
handle Plücker reciprocation as in 7.7.1 on putting a2 = −1.
EXERCISES
11.1 Prove that
13.7.8
ξ
∂[(ξ1 , η1 , ζ1 ), (ξ2 , η2 , ζ2 ), (ξ3 , η3 , ζ3 )](α + β + γ) = ξ2 − ξ1
ξ3 − ξ1
Equation of a line
η
η2 − η1
η3 − η1
ζ
ζ2 − ζ1
ζ3 − ζ1
.
If we rewrite the identity (3.2.1) as
δF (Z1 , U, V )δF (Z, Z2 , Z3 ) δF (Z2 , U, V )δF (Z, Z3 , Z1 ) δF (Z3 , U, V )δF (Z, Z1 , Z2 )
+
+
,
δF (Z, U, V )δF (Z1 , Z2 , Z3 ) δF (Z, U, V )δF (Z1 , Z2 , Z3 ) δF (Z, U, V )δF (Z1 , Z2 , Z3 )
(13.7.1)
the expression on the right is projectively invariant. It does not make sense when Z, U, V are
collinear. But we can handle the image of a line under a projective transformation by then reverting
to the form (3.2.1) for a line and its image, using a continuity argument.
1=
b
b
b
b
b
b
b
b
Figure 10.6. See Exercise 10.1.
b
416
CHAPTER 13. EXTENSION OF EUCLIDEAN TO PROJECTIVE PLANES
Chapter 14
Further projective methods and
results
14.1
EXTENDED CONICS, THE IDEAL CIRCULAR POINTS
14.1.1
Projective transformations in terms of homogeneous coordinates
As we noted in 2.4.3 a projective transformation f can be expressed in terms of Cartesian coordinates in the form
d1 x + e1 y + f1 ′ d2 x + e2 y + f2
x′ =
, y =
,
d3 x + e3 y + f3
d3 x + e3 y + f3
where

d1
M =  d2
d3
is non-singular. Now on taking
x=
e1
e2
e3

f1
f2 
f3
η′
η
ξ′
ξ
, y = , x′ = ′ , y ′ = ′ ,
ζ
ζ
ζ
ζ
we see that any transformation of the form
ξ′
=
d1 ξ + e1 η + f1 ζ,
′
=
=
d2 ξ + e2 η + f2 ζ,
d3 ξ + e3 η + f3 ζ,
η
ζ′
where M is as above, represents a projective transformation in homogeneous coordinates. It is a
function on Πe onto Πe , and this represents one of the compelling reasons for working in Πe , as
non-affine projective transformations on Π have domains Π less a line.
The pre-image of the ideal line ζ ′ = 0 is the line with equation
d3 ξ + e3 η + f3 ζ = 0,
and this is also the ideal line if and only if d3 = e3 in which case f reduces to an affine transformation. Thus an affine transformation is a projective transformation which maps the ideal line to
the ideal line.
In Chapter 8 we concentrated on the whole on affine invariants as the simplest to handle. In
effect, in tacitly considering only affine transformations, the ideal line always corresponded to
itself. This kept the ideal line at the edge of affairs. But the ideal line can be mapped onto any
other line by means of an appropriate projective transformation, and provided we now work with
417
418
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
projective invariants, we can transfer to any line relationships which seemed special to the ideal
line. This process gives greatly increased flexibility and provides an opportunity for considerable
generalization.
14.1.2
Duo-sectors in projective terms
In 8.1.5 with the parametric form
ξ = x1 + λx2 , η = y1 + λy2 , ζ = 1 + λ,
we used the sensed ratios
λ=
Z1 Z
<> 0
ZZ2
as specifying the open extended segments with end-points Z1 , Z2 . A projectively invariant extension of this is
cr(0, ∞, λ, −1) = cr(Z1 , Z2 , Z, Z1 Z2∞ ) = −λ
corresponding to
−
We also used
Z1 Z
.
ZZ2
δF (Z, Z1 , Z2 )
<> 0
δF (Z, Z3 , Z1 )
in specifying duo-sectors. A projectively invariant extension of the latter is
∞
δF (Z, Z1 , Z2 ) δF (Z4 , Z3 , Z23
)
∞ , Z ) <> 0.
δF (Z, Z3 , Z1 ) δF (Z4 , Z23
2
∞
Here Z23
is the point where Z2 Z3 meets the ideal line.
From it, what does
δF (Z, Z1 , Z2 ) δF (Z4 , Z3 , Z5 )
<> 0
δF (Z, Z3 , Z1 ) δF (Z4 , Z5 , Z2 )
indicate? It is just that (Z1 ; Z2 , Z3 ; Z) has extended orientation the same as, or the opposite to,
that of (Z4 ; Z2 , Z3 ; Z5 ).
In this now let us take Z1 ≡ (ξ1 , η1 , 0), Z2 ≡ (ξ2 , η2 , 0), and then we obtain
δF (Z4 , Z3 , Z5 )
ξ1 η2 − ξ2 η1
.
−η1 (x − x3 ) + ξ1 (y − y3 ) −η2 (x4 − x5 ) + ξ2 (y4 − y5 )
This being positive or negative corresponds to a half-plane.
14.1.3
The ideal circular points
A striking first specific feature of the use of non-real homogeneous coordinates is as follows. An
extended complex conic is the set of all extended points satisfying an equation
aξ 2 + bη 2 + cζ 2 + 2f ηζ + 2gζξ + 2hξη = 0,
where the coefficients are complex numbers and not all zero. Corresponding to this we can have
the Euclidean conic with equation
ax2 + by 2 + c + 2f y + 2gx + 2hxy = 0,
when the coefficients are real. Following Poncelet in 1822, if we look at when the Euclidean conic
is a circle we find that it is when the corresponding extended conic meets the ideal line in the
14.1. EXTENDED CONICS, THE IDEAL CIRCULAR POINTS
419
points I = (1, ı, 0), J = (1, −ı, 0). These are then very distinctive points and are called the ideal
circular points. To quote Poncelet’s historic statement (p.48):
‘Des cercles placés arbitrairement sur un plan ne sont donc pas tout à fait indépendants entre
eux, comme on pourrait le croire au premier abord; ils ont idéalement deux points imaginaire
communs à l’infini, et, sous ce rapport, ils doivent jouir de certaines propriétés appartenant à la
fois à tout leur systême, et analogues à celles dont ils jouissent quand ils ont une sécante commune
ordinaire: . . . .’
This is a powerful reason for working in Πce .
14.1.4
Ratio of sensed areas interpreted
For a triple (Zr , Zs , Zt ) of non-collinear points in our non-extended complex plane, as well as
δF (Zr , Zs , Zt ) we use


ξr ηr ζr
1
δh (Zr , Zs , Zt ) = det  ξs ηs ζs  ,
2
ξt ηt ζt
where (ξj , ηj , ζj ) are homogeneous coordinates for Zj . We use the subscript h to indicate that
general homogeneous coordinates are being used, and we clearly have the relationship
δF (Zr , Zs , Zt ) =
1
δh (Zr , Zs , Zt ).
ζr ζs ζt
We use these terms δh (Zr , Zs , Zt ) only in equipoised fractions, that is rational expressions in
which each of the denominator and numerator is a finite product of such homogeneous sensedareas and in which the set of vertices, including multiplicity, in the denominator is a permutation
of the set of vertices in the numerator. We use the same homogeneous coordinates for any vertex
at all its occurrences. We combine this with noting that, on taking as usual the coordinates
(1, ı, 0), (1, −ı, 0), for I and J , the circular ideal points, we have


1
ı
0
1
δh (I, Zj , J ) = det  ξj ηj ζj  = ıζj .
2
1 −ı 0
This yields
δF (Zr , Zs , Zt ) = −ı
δh (Zr , Zs , Zt )
.
δh (I, Zr , J )δh (I, Zs , J )δh (I, Zt , J )
This is not in projectively invariant form, and for that we take a ratio
δh (Z1 , Z2 , Z3 )δh (I, Z4 , J )δh (I, Z5 , J )δh (I, Z6 , J )
δF (Z1 , Z2 , Z3 )
=
.
δF (Z4 , Z5 , Z6 )
δh (Z4 , Z5 , Z6 )δh (I, Z1 , J )δh (I, Z2 , J )δh (I, Z3 , J )
This is an equipoised fraction, but unfortunately it has eight terms.
14.1.5
Ratio of distances interpreted
To obtain an approach to dealing with distance in
that

ξj ηj
1
ı
δh (Zj , I, Zk ) = det  1
2
ξ
η
=
a projectively invariant way, we further note

ζj
0 
ζk
k
k
1
1
[−(ηj ζk − ηk ζj ) + ı(ξj ζk − ξk ζj )] = ıζj ζk [xj − xk + ı(yj − yk )].
2
2
420
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
On combining these we obtain that
δh (Z1 , I, Z2 )δh (I, Z3 , J )δh (I, Z4 , J )
Z1 Z2
=
,
δh (Z3 , I, Z4 )δh (I, Z1 , J )δh (I, Z2 , J )
Z3 Z4
a ratio of complex-valued distances.
In particular
Z1 Z2
δh (Z1 , I, Z2 )δh (I, Z3 , J )
=
.
δh (Z1 , I, Z3 )δh (I, Z2 , J )
Z1 Z3
If we multiply the first of these by its complex conjugate, we obtain
δh (Z1 , I, Z2 )δh (Z1 , J , Z2 )δh (I, Z3 , J )2 δh (I, Z4 , J )2
|Z1 , Z2 |2
=
,
2
|Z3 , Z4 |
δh (Z3 , I, Z4 )δh (Z3 , J , Z4 )δh (I, Z1 , J )2 δh (I, Z2 , J )2
and in particular
δh (Z1 , I, Z2 )δh (Z1 , J , Z2 )δh (I, Z3 , J )2
|Z1 , Z2 |2
=
.
|Z1 , Z3 |2
δh (Z1 , I, Z3 )δh (Z1 , J , Z3 )δh (I, Z2 , J )2
These are the basis of our dealing with ratios of complex-valued distances and with the ratio
of squares of distances. We follow Cayley (1859) in utilizing I and J for square of distance, but
our details are different.
14.1.6
Angle-measure interpreted
Following Laguerre (1853), to prepare for angle-measure we note similarly that
=
=
=
δh (Z1 , Z3 , I)δh (Z1 , Z2 , J )
δh (Z1 , Z3 , J )δh (Z1 , Z2 , I)
ı[x3 − x1 + ı(y3 − y1 )](−ı[x2 − x1 − ı(y2 − y1 )])
−ı[x3 − x1 − ı(y3 − y1 )](ı[x2 − x1 + ı(y2 − y1 )])
(z3 − z1 )z2 − z1
z3 − z1 (z2 − z1 )
e2ı(θ3 −θ2 ) ,
where θ2 , θ3 are the polar angles for [Z1 , Z2 , [Z1 , Z3 , respectively, and so this is
e2ı∡F Z2 Z1 Z3 .
This expression is complex-valued and has absolute value 1. We note that it can also be expressed
as
[x3 − x1 + ı(y3 − y1 )]2 [x2 − x1 + ı(y2 − y1 )]2
/
.
(x3 − x1 )2 + (y3 − y1 )2 (x2 − x1 )2 + (y2 − y1 )2
14.2
UTILIZATION OF ANALOGUES OF RATIOS OF
SENSED-AREAS
14.2.1
Basic expression
In the foregoing the ideal line clearly had a special role in the complex extended plane, but this is
due to how we introduced it. There is in fact no specially distinguished line, and given any two
lines either can be mapped on to the other by a projective transformation, with an assigned pair
of points on the first line corresponding to an assigned pair; then any projective invariants related
to the first line will convert to analogues related to the second line.
14.2. UTILIZATION OF ANALOGUES OF RATIOS OF SENSED-AREAS
421
To utilize the material above in this way, we take two distinct fixed points W1 , W2 and will take
these as distinguished points on the distinguished line W1 W2 . The other points that we deal with
will generally be off this line.
From
δF (Z1 , Z2 , Z3 )
δh (Z1 , Z2 , Z3 )δh (I, Z4 , J )δh (I, Z5 , J )δh (I, Z6 , J )
=
,
δF (Z4 , Z5 , Z6 )
δh (Z4 , Z5 , Z6 )δh (I, Z1 , J )δh (I, Z2 , J )δh (I, Z3 , J )
we take
δh (Z1 , Z2 , Z3 )δh (W1 , Z4 , W2 )δh (W1 , Z5 , W2 )δh (W1 , Z6 , W2 )
δF (Z1 , Z2 , Z3 )
=
,
δF (Z4 , Z5 , Z6 )
δh (Z4 , Z5 , Z6 )δh (W1 , Z1 , W2 )δh (W1 , Z2 , W2 )δh (W1 , Z3 , W2 )
as a projectively invariant generalization of a ratio of two sensed areas.
We can apply this to generalize results such as those in 5.7 or 5.8. For example, we rewrite the
form of equation of a conic
f βγ + gγα + hαβ = 0,
used in 4.1.4, as
f
δF (Z, Z1 , Z2 )
δF (Z, Z1 , Z2 )
+g
+h=0
δF (Z, Z2 , Z3 )
δF (Z, Z3 , Z1 )
and now can infer from this the projectively invariant form
f
14.2.2
δF (Z, Z1 , Z2 ) δF (W1 , Z3 , W2 )
δF (Z, Z1 , Z2 ) δF (W1 , Z3 , W2 )
+g
+ h = 0.
δF (Z, Z2 , Z3 ) δF (W1 , Z1 , W2 )
δF (Z, Z3 , Z1 ) δF (W1 , Z2 , W2 )
Sine and cosine
For a very substantial application, from 5.7.1 we now let C be a proper conic and O1 an interior
point for it, not the centre in the case of an ellipse or hyperbola. Let W1 and W2 be any points on
the polar of O1 . Let Q be a point on the conic and let H5 be one of the closed half-planes with
edge O1 Q. Let O1 Q meet W1 W2 at U0 ; then the polar of U0 is a line through O1 . We let R be the
point in H5 in which this polar cuts the conic. For any point Z on the conic, let O1 Z meet W1 W2
at U and let Z ∗ be the point in which the polar of U meets the conic and for which (O1 , Q, R) and
(O1 , Z, Z ∗ ) are similarly oriented. Let QZ meet W1 W2 at L, and let M be the harmonic conjugate
of L with respect to {Q, Z}. We use [O1 , M as a modified indicator of the angles with support
|QO1 Z, taking the wedge-angle if [O1 , M is in the interior region and taking the reflex-angle if
[O1 , M is in the exterior region. Among the angles with arm [O1 , Q we confine our attention to
those for which the modified indicator lies in the half-plane H5 , that is [O1 , M ⊂ H5 . We denote
the angle in this class which has the support |QO1 Z by the somewhat under-detailed notation
∡F QO1 Z.
422
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
b
Z
b
b
b
W2
W1
b
U
b
M
O1
b
b
b
b
U0
Q
b
Z∗
b
R
Figure 9.1
b
L
b
Then we define
sin ∡F QO1 Z
=
cos ∡F QO1 Z
=
δF (O1 , Q, Z) δF (W1 , R, W2 )
,
δF (O1 , Q, R) δF (W1 , Z, W2 )
δF (O1 , Q, Z ∗ ) δF (W1 , R, W2 )
.
δF (O1 , Q, R) δF (W1 , Z ∗ , W2 )
We can interpret this as a generalized sine and cosine. Of course, we must check what familiar
properties, if any, are preserved in this generalization. In connection with it we also generalize the
addition of angles. For points Z1 and Z2 on the conic, let Z1 Z2 meet W1 W2 in the point V , and
let QV meet the conic again at Z3 . Then the angle ∡F QO1 Z3 specified in the above way for Z3
is defined to be the sum of ∡F QO1 Z1 and ∡F QO1 Z2 in that order.
b
b
O1
b
b
U0
b
V
Q
b
W1
W2
b
b
b
Z1
Z2
b
R
b
b
Z3
Figure 9.2
14.2. UTILIZATION OF ANALOGUES OF RATIOS OF SENSED-AREAS
423
To keep the calculations from being too unwieldy, we check this definition in the case of the
parabola with equation y 2 = 4x. We recall that the polar of Z0 ≡ (x0 , y0 ) is the line with equation
yy0 = 2(x + x0 ). We take O1 ≡ (a, 0) where a is an arbitrary but fixed positive number. The polar
of O1 is the line with equation x = −a, and we take W1 ≡ (−a, v1 ), W2 ≡ (−a, v2 ). Initially we
take Q ≡ (t20 , 2t0 ). The line O1 Q has equation 2t0 (x − a) + (a − t20)y = 0 and we take H5 to satisfy
2t0 (x − a) + (a − t20 )y ≤ 0. The line O1 Q meets W1 W2 at the point U0 ≡ (−a, −4at0 /(t20 − a)).
This has polar
4at0
− 2
y = 2(x − a),
t0 − a
and this polar meets the parabola in the points
(a +
8a2 t20 − 4a3/2 t0 (t20 + a) 4at0 − 2a1/2 (t20 + a)
,
),
(t20 − a)2
a − t20
(a +
8a2 t20 + 4a3/2 t0 (t20 + a) 4at0 + 2a1/2 (t20 + a)
,
).
(t20 − a)2
a − t20
The first of these is the one in H5 and so it is R.
For Z ≡ (t2 , 2t) we have similarly that
Z ∗ ≡ (a +
8a2 t2 − 4a3/2 t(t2 + a) 4at − 2a1/2 (t2 + a)
,
).
(t2 − a)2
a − t2
Now we can calculate that
δF (O1 , Q, Z)
t0 t2 + (a − t20 )t − at0
2
=
,
δF (W1 , Z, W2 )
v1 − v2
t2 + a
and that
(v1 − v2 )a1/2
δF (W1 , R, W2 )
=−
,
δF (O1 , Q, R)
t20 + a
and so
sin θ = −2a1/2
t0 t2 + (a − t20 )t − at0
.
(t2 + a)(t20 + a)
Similarly we can find that
1
(t20 − a)(t2 − a) + 4at0 t
δF (O1 , Q, Z ∗ )
=
−
,
δF (W1 , Z ∗ , W2 )
t2 + a
(v1 − v2 )a1/2
and with
δF (W1 , R, W2 )
δF (O1 , Q, R)
as before, we find that
cos θ =
(t20 − a)(t2 − a) + 4at0 t
.
(t20 + a)(t2 + a)
Also
Q ≡ (t20 , 2t0 ), V ≡
and so for Z3 we have
t1 t2 − a
−a, 2
,
t1 + t2
t0 t1 t2 + a(t1 + t2 − t0 )
1
y=
,
2
t0 (t1 + t2 ) − t1 t2 + a
and the latter is the parameter of Z3 .
424
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
We now specialize by taking t0 = 0 and so Q = Q0 ≡ (0, 0) henceforth. With this understanding,
we now have
√
at
t2 − a
sin ∡F Q0 O1 Z = −2 2
, cos ∡F Q0 O1 Z = − 2
,
t +a
t +a
and the parameter corresponding to Z3 is
a(t1 + t2 )
.
a − t1 t2
From these we find that
√ (a + t1 t2 )(t2 − t1 )
(t21 − a)(t22 − a) + 4at1 t2
sin ∡F Q0 O1 Z3 = −2 a 2
,
cos
∡
Q
O
Z
=
.
F
0
1
3
(t1 + a)(t22 + a)
(t21 + a)(t22 + a)
We also have
√
at1
t2 − a
, cos ∡F Q0 O1 Z1 = − 12
,
+a
t1 + a
sin ∡F Q0 O1 Z1 = −2
t21
sin ∡F Q0 O1 Z2 = −2
t22 − a
at2
,
cos
∡
Q
O
Z
=
−
.
F
0
1
2
t22 + a
t22 + a
√
From these it is straightforward to verify the addition formulae
sin ∡F Q0 O1 Z3
cos ∡F Q0 O1 Z3
=
=
sin ∡F Q0 O1 Z1 cos ∡F Q0 O1 Z2 + cos ∡F Q0 O1 Z1 sin ∡F Q0 O1 Z2 ,
cos ∡F Q0 O1 Z1 cos ∡F Q0 O1 Z2 − sin ∡F Q0 O1 Z1 sin ∡F Q0 O1 Z2 ,
and the identity
(sin ∡F Q0 O1 Z)2 + (cos ∡F Q0 O1 Z)2 = 1.
If we replace Z1 by the point with parameter −t0 and Z2 by Z we find that for the new Z3 we
have parameter a(t − t0 )/(a + tt0 ) and so for it
√ (a − t0 t)(t2 + t0 )
(t20 − a)(t2 − a) − 4at0 t
sin ∡F QO1 Z3 = −2 a 2
,
cos
∡
QO
Z
=
.
F
1
3
(t0 + a)(t2 + a)
(t20 + a)(t2 + a)
But these have the form of our original sine and cosine of ∡F QOZ and so be should regard
∡F QO1 Z as ∡F Q0 O1 Z − ∡F Q0 O1 Q.
14.2.3
Analogue of rotors
From 3.12.2? we consider a generalization of rotors. To the condition
δF (Z, Z8 , Z9 )
= 1, for all Z,
δF (Z, Z10 , Z11 )
we now make correspond
δF (Z, Z8 , Z9 )δF (W1 , Z10 , W2 )δF (W1 , Z11 , W2 )
= 1,
δF (Z, Z10 , Z11 )δF (W1 , Z8 , W2 )δF (W1 , Z9 , W2 )
for all Z. It follows easily from this that Z10 and Z11 are on the line Z8 Z9 . If we let
Z10 =
λ
1
µ
1
Z8 +
Z9 , Z11 =
Z8 +
Z9 ,
1+λ
1+λ
1+µ
1+µ
then we find that
=
[δF (W1 , Z8 , W2 ) + λδF (W1 , Z9 , W2 )][δF (W1 , Z8 , W2 ) + µδF (W1 , Z9 , W2 )]
(µ − λ)δF (W1 , Z8 , W2 )δF (W1 , Z9 , W2 ).
This yields a related range.
14.3. ANALOGUES OF RATIOS OF COMPLEX-VALUED DISTANCES
425
14.3
ANALOGUES OF RATIOS OF COMPLEX-VALUED
DISTANCES
14.3.1
Basic expression
From
δh (Z1 , I, Z2 )δh (I, Z3 , J )δh (I, Z4 , J )
δh (Z3 , I, Z4 )δh (I, Z1 , J )δh (I, Z2 , J )
we start with
δF (Z1 , W1 , Z2 )δF (W1 , Z3 , W2 )δF (W1 , Z4 , W2 )
δF (Z3 , W1 , Z4 )δF (W1 , Z1 , W2 )δF (W1 , Z2 , W2 )
as a generalization of the ratio of complex-valued distances
Z1 Z2
.
Z3 Z4
b
Z2
b
b
W2
b
Z1
b
W1
b
Z4
b
Z3
Figure 9.3.
b
By keeping Z3 and Z4 fixed, we can interpret this as a generalization of complex-valued distance
Z1 Z2 .
Of course, we must check what familiar properties are preserved in this generalization. First of
all we note that if we interchange Z1 and Z2 , or Z3 and Z4 we pass to its additive inverse, while
if we interchange (Z1 , Z2 ) and (Z3 , Z4 ) we pass to its multiplicative inverse.
14.3.2
Distance 0
If its value is 0, does that imply that Z1 = Z2 ? In this we are given that
δF (Z1 , W1 , Z2 )δF (W1 , Z3 , W2 )δF (W1 , Z4 , W2 ) = 0.
Now we are taking the points Z1 , Z2 , Z3 , Z4 not to be on the distinguished line W1 W2 , so
δF (W1 , Z3 , W2 ) 6= 0, δF (W1 , Z4 , W2 ) 6= 0,
and thus we have as a condition that
δF (Z1 , W1 , Z2 ) = 0, i.e. W1 ∈ Z1 Z2 .
Thus our ratio is equal to 0 if Z1 = Z2 as expected but also if Z1 Z2 passes through W1 . This latter
is a complicating feature and, because of it, we call any line which passes through W1 an isotropic
line. For this ratio to exist we need that Z3 , Z4 are distinct and Z3 Z4 is not an isotropic line.
426
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
14.3.3
Addition
We next check if
Z1 Z2
Z2 Z5
Z1 Z5
+
=
.
Z3 Z4
Z3 Z4
Z3 Z4
This would require
=
δF (Z1 , W1 , Z2 )
δF (Z2 , W1 , Z5 )
+
δF (W1 , Z1 , W2 )δF (W1 , Z2 , W2 ) δF (W1 , Z2 , W2 )δF (W1 , Z5 , W2 )
δF (Z1 , W1 , Z5 )
δF (W1 , Z1 , W2 )δF (W1 , Z5 , W2 )
and so
1
δF (Z2 , W1 , Z5 )
δF (Z1 , W1 , Z2 )
δF (Z1 , W1 , Z5 )
=−
−
.
δF (W1 , Z1 , W2 ) δF (W1 , Z2 , W2 ) δF (W1 , Z5 , W2 )
δF (W1 , Z2 , W2 )δF (W1 , Z5 , W2 )
This can be deduced from our identities.
14.3.4
Complex-valued ratio constant
b
Z2
b
b
W2
b
Z1
b
W1
b
Z3
b
Figure 9.4.
As a particular case of the above, from
δh (Z1 , I, Z2 )δh (I, Z3 , J )
Z1 Z2
,
=
δh (Z1 , I, Z3 )δh (I, Z2 , J )
Z1 Z3
we regard
δF (Z1 , W1 , Z2 )δF (W1 , Z3 , W2 )
δF (Z1 , W1 , Z3 )δF (W1 , Z2 , W2 )
as a generalization of a ratio of complex-valued distances
Z1 Z2
.
Z1 Z3
For it, if for distinct points Z2 and Z3
ZZ2
= k,
ZZ3
what can we conclude? In this we are given that
δF (Z, W1 , Z2 )δF (W1 , Z3 , W2 ) − kδF (Z, W1 , Z3 )δF (W1 , Z2 , W2 ) = 0.
14.3. ANALOGUES OF RATIOS OF COMPLEX-VALUED DISTANCES
427
If k is such that
then we have that
δF (W1 , Z3 , W2 ) − kδF (W1 , Z2 , W2 ) 6= 0,
δF (W1 , Z3 , W2 )Z2 − kδF (W1 , Z2 , W2 )Z3
= 0.
δF Z, W1 ,
δF (W1 , Z3 , W2 ) − kδF (W1 , Z2 , W2 )
This implies that Z is collinear with W1 and the point
δF (W1 , Z3 , W2 )Z2 − kδF (W1 , Z2 , W2 )Z3
.
δF (W1 , Z3 , W2 ) − kδF (W1 , Z2 , W2 )
(14.3.1)
Thus Z need not be on Z2 Z3 . When Z is on Z2 Z3 and W1 6∈ Z2 Z3 , Z must be the point (9.3.1);
here Z2 Z3 is not to be isotropic.
What happens in the excluded case above when
δF (W1 , Z3 , W2 ) − kδF (W1 , Z2 , W2 ) = 0?
Now we have
δF (Z, W1 , Z2 ) − δF (Z, W1 , Z3 ) = 0.
This gives δF (Z − W1 , Z2 − Z3 , O) = 0, so that W1 Z k Z2 Z3 . Here we cannot have Z ∈ Z2 Z3
unless the latter is isotropic.
To continue with the possible cases, take Z2 Z3 to be isotropic and so let
W1 =
λ
1
Z2 +
Z3 .
1+λ
1+λ
Then we have
−δF (Z, Z2 , Z3 )[λδF (W1 , Z3 , W2 ) + kδF (W1 , Z2 , W2 )] = 0.
The term in the square bracket here is equal to
λ
1
δF (W2 , Z3 , Z2 ) + k
δF (W2 , Z2 , Z3 )
1+λ
1+λ
λ
=
(1 − k)δF (W2 , Z2 , Z3 ).
1+λ
Thus when k 6= 1 we must have that Z is in Z2 Z3 but is not otherwise restricted.
Finally, continuing with Z2 Z3 isotropic, when k = 1 we have
λ
δF (Z, W1 , Z2 )δF (W1 , Z3 , W2 ) − δF (Z, W1 , Z3 )δF (W1 , Z2 , W2 ) = 0.
Thus
and so
δF [(Z, W2 ), ((W1 , Z2 ), (W1 , Z3 )] = 0,
δF (W1 , Z, W2 )δF (Z3 , W1 , Z2 ) = 0.
This is identically true as W1 ∈ Z2 Z3 .
Looking to when Z1 is the mid-point of Z2 and Z3 , we take k = −1 and Z1 ∈ Z2 Z3 . Then we
have that
δF (W1 , Z3 , W2 )Z2 + δF (W1 , Z2 , W2 )Z3
Z1 =
.
δF (W1 , Z3 , W2 ) + δF (W1 , Z2 , W2 )
If W3 is the point where Z2 Z3 meets W1 W2 , we have
W3 =
λ
1
W1 +
W2 ,
1+λ
1+λ
where δF (W1 , W2 , W3 ) = 0, and so
λ=−
δF (W1 , W2 , Z2 )
.
δF (W1 , W2 , Z3 )
This shows that (Z2 , Z3 , Z1 , W3 ) is a harmonic range.
428
14.3.5
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Ratio of square of distance
From the formula
|Z1 , Z2 |2
δh (Z1 , I, Z2 )δh (Z1 , J , Z2 )δh (I, Z3 , J )2 δh (I, Z4 , J )2
=
,
2
|Z3 , Z4 |
δh (Z3 , I, Z4 )δh (Z3 , J , Z4 )δh (I, Z1 , J )2 δh (I, Z2 , J )2
we take
δh (Z1 , W1 , Z2 )δh (Z1 , W2 , Z2 )δh (W1 , Z3 , W2 )2 δh (W1 , Z4 , W2 )2
|Z1 , Z2 |2
=
,
2
|Z3 , Z4 |
δh (Z3 , W1 , Z4 )δh (Z3 , W2 , Z4 )δh (W1 , Z1 , W2 )2 δh (W1 , Z2 , W2 )2
as a generalization of ratio of square of distance.
In particular, we take the generalization
|Z1 , Z2 |2
δh (Z1 , W1 , Z2 )δh (Z1 , W2 , Z2 )δh (W1 , Z3 , W2 )2
=
.
2
|Z1 , Z3 |
δh (Z1 , W1 , Z3 )δh (Z1 , W2 , Z3 )δh (W1 , Z2 , W2 )2
This latter gives
δh (Z1 , W1 , Z)δh (Z1 , W2 , Z)δh (W1 , Z3 , W2 )2
= 1,
δh (Z1 , W1 , Z3 )δh (Z1 , W2 , Z3 )δh (W1 , Z, W2 )2
as the generalization of the circle with centre Z1 which passes through Z3 . The equation γ 2 = 4kαβ
used in 4.2.1 is thus of this form.
14.4
UTILIZATION OF ANALOGUE OF ANGLE-MEASURE
14.4.1
Basic expression
To turn now to angle-measure, from
δh (Z1 , Z3 , I)δh (Z1 , Z2 , J )
,
δh (Z1 , Z3 , J )δh (Z1 , Z2 , I)
with fixed (W1 , W2 ) and vertex Z1 6∈ W1 W2 , we consider
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
,
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
as a generalization of
e2ı∡F Z2 Z1 Z3 .
On keeping Z2 and Z3 fixed, the locus of points Z such that an exponential of the generalized
measure of ∡F Z2 ZZ3 is constant, has the form
δF (Z, Z2 , W2 )δF (Z, Z3 , W1 )
= k,
δF (Z, Z2 , W1 )δF (Z, Z3 , W2 )
and this has the form in 1.6.2? and 6.1.1. However because this exponential expression can be
negative, it is more difficult to handle the actual measure of an angle.
With
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
presenting naturally as a way of measuring the sensed duo-angle ∢F Z2 Z1 Z3 , should we not work
with this as it is and would it not be rather perverse to insist on taking logarithms? We look at
properties of this rather as we did of ratio of complex-valued distances in 9.3.
(a)
If we interchange Z2 and Z3 we pass to its multiplicative inverse.
14.4. UTILIZATION OF ANALOGUE OF ANGLE-MEASURE
429
(b)
If W1 ∈ Z1 Z3 or W2 ∈ Z1 Z2 it is equal to 0. If W2 ∈ Z1 Z3 or W1 ∈ Z1 Z2 it becomes infinite.
(c)
It has a multiplicative property that the measure of ∢F Z2 Z1 Z3 by the measure of ∢F Z3 Z1 Z4
is equal to the measure of ∢F Z2 Z1 Z4 . For
=
14.4.2
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z3 ) δF (W2 , Z1 , Z3 )δF (W1 , Z1 , Z4 )
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z3 ) δF (W1 , Z1 , Z3 )δF (W2 , Z1 , Z4 )
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z4 )
.
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z4 )
Angles of equal measures
We ask when the measures of ∢F Z2 Z1 Z3 and ∢F Z2 Z1 Z4 are equal, that is
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z4 )
=
?
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z4 )
This happens trivially if W2 ∈ Z1 Z2 , and both sides are potentially infinite when W1 ∈ Z1 Z2 .
On excluding these cases, the condition for equality is
δF (W1 , Z1 , Z3 )
δF (W1 , Z1 , Z4 )
=
.
δF (W2 , Z1 , Z3 )
δF (W2 , Z1 , Z4 )
On excluding too the cases when W2 ∈ Z1 Z3 or W2 ∈ Z1 Z4 , the condition becomes
δF [((W1 , W2 ), (Z1 , Z3 ), (Z1 , Z4 )] = 0,
and this can be manipulated to
δF (Z1 , W1 , W2 )δF (Z4 , Z1 , Z3 ) = 0.
Now Z1 6∈ W1 W2 so the condition is that Z4 ∈ Z1 Z3 .
14.4.3
Analogue of null angles
This also answers the question of when the measure of ∢F Z4 Z1 Z3 is equal to 1, i.e.
δF (W2 , Z1 , Z4 )δF (W1 , Z1 , Z3 )
= 1.
δF (W1 , Z1 , Z4 )δF (W2 , Z1 , Z3 )
When Z1 Z3 and Z1 Z4 are both non-isotropic, it is when Z1 , Z3 , Z4 are collinear.
14.4.4
Analogue of perpendicularity
What corresponds to Z1 Z2 and Z1 Z3 being perpendicular? It is
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
= −1,
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
which can be rewritten as
δF (Z1 , Z2 , W2 )δF (Z1 , Z3 , W1 )
= −1.
δF (Z1 , Z2 , W1 )δF (Z1 , Z3 , W2 )
This is the condition that Z1 (Z2 , Z3 , W2 , W1 ) be a harmonic pencil.
430
14.4.5
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Alternate angles
What corresponds to ∢F Z2 Z1 Z3 and ∢F Z4 Z3 Z1 being alternate duo-angles for parallel lines Z1 Z2
and Z3 Z4 ? It is that
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
δF (W2 , Z3 , Z4 )δF (W1 , Z3 , Z1 )
=
.
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
δF (W1 , Z3 , Z4 )δF (W2 , Z3 , Z1 )
On performing cancellations corresponding to isotropic lines, this reduces to
δF [(W2 , W1 ), (Z1 , Z2 ), (Z3 , Z4 )] = 0,
which is the condition that Z1 Z2 , Z3 Z4 meet on W1 W2 .
14.4.6
Mid-lines
We now consider when Z1 Z4 is the mid-line of the pair of lines Z1 Z2 , Z1 Z3 . For this we need
δF (W2 , Z1 , Z4 )δF (W1 , Z1 , Z3 )
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z4 )
=
.
δF (W1 , Z1 , Z2 )δF (W2 , Z1 , Z4 )
δF (W1 , Z1 , Z4 )δF (W2 , Z1 , Z3 )
Then
δF (W2 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
δF (W2 , Z1 , Z4 )2
=
.
2
δF (W1 , Z1 , Z4 )
δF (W1 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
Let Z1 Z4 meet W1 W2 at W4 , with
W4 =
1
λ
W1 +
W2 .
1+λ
1+λ
As δF (Z1 , Z4 , W4 ) = 0 we have
λ=−
and so
λ2 =
δF (W1 , Z1 , Z4 )
,
δF (W2 , Z1 , Z4 )
δF (W1 , Z1 , Z2 )δF (W1 , Z1 , Z3 )
.
δF (W2 , Z1 , Z2 )δF (W2 , Z1 , Z3 )
Thus Z1 W1 , Z1 W2 and the two possible positions of Z1 Z4 form a harmonic pencil.
14.4.7
Suppose now that in 9.4.6 we have that Z4 is the mid-point of Z2 and Z3 . We let W3 be the point
in which Z2 Z3 meets W1 W2 . Then we have that
W3 =
1
λ
1
λ
Z2 +
Z3 , Z4 =
Z2 −
Z3 ,
1+λ
1+λ
1−λ
1−λ
where
λ=−
δF (W1 , W2 , Z2 )
.
δF (W1 , W2 , Z3 )
We now suppose additionally that Z1 Z4 is perpendicular to Z1 W3 . Then, by 9.4.4, we have
δF (W2 , Z1 , Z4 )δF (W1 , Z1 , W3 )
= −1.
δF (W1 , Z1 , Z4 )δF (W2 , Z1 , W3 )
If we now substitute for Z4 and W3 in this in terms of λ, the resulting expression reduces to
λ2 =
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z2 )
,
δF (W2 , Z1 , Z3 )δF (W1 , Z1 , Z3 )
14.4. UTILIZATION OF ANALOGUE OF ANGLE-MEASURE
431
and so from the earlier expression for λ we have
δF (W2 , Z1 , Z2 )δF (W1 , Z1 , Z2 )
δF (W1 , W2 , Z2 )2
.
=
δF (W2 , Z1 , Z3 )δF (W1 , Z1 , Z3 )
δF (W1 , W2 , Z3 )2
If we now look at the ratio
=
|Z1 , Z2 |2
|Z1 , Z3 |2
δF (Z1 , W1 , Z2 )δF (Z1 , W2 , Z2 )δF (W1 , Z3 , W2 )2
,
δF (Z1 , W1 , Z3 )δF (Z1 , W2 , Z3 )δF (W1 , Z2 , W2 )2
the last equation is just the condition that this ratio be equal to 1.
14.4.8
Analogue of isosceles triangle
We now note that the condition that the duo-angles ∢F Z3 Z2 Z1 and ∢F Z1 Z3 Z2 are equal in
magnitude is
δF (W2 , Z3 , Z1 )δF (W1 , Z3 , Z2 )
δF (W2 , Z2 , Z3 )δF (W1 , Z2 , Z1 )
=
,
δF (W1 , Z2 , Z3 )δF (W2 , Z2 , Z1 )
δF (W1 , Z3 , Z1 )δF (W2 , Z3 , Z2 )
and so
δF (Z1 , W1 , Z2 )δF (Z1 , W1 , Z3 )δF (W2 , Z2 , Z3 )2
= 1.
δF (Z1 , W2 , Z2 )δF (Z1 , W2 , Z3 )δF (W1 , Z2 , Z3 )2
If we replace Z1 by Z in this, we obtain the equation
δF (Z1 , W1 , Z2 )δF (Z1 , W1 , Z3 )δF (W2 , Z2 , Z3 )2
= 1,
δF (Z1 , W2 , Z2 )δF (Z1 , W2 , Z3 )δF (W1 , Z2 , Z3 )2
which is that of a quadratic locus. If we take
Z=
1
λ
Z2 +
Z3 ,
1+λ
1+λ
and substitute it in, we find that every point of the line Z2 Z3 lies on the locus. As the locus
contains as well the point U which is the intersection of W1 Z2 and W2 Z3 , and the point V which
is the intersection of W1 Z3 and W2 Z2 , the locus is the union of the lines Z2 Z3 .
If we now take from 9.3.5 the condition that the ratio
|Z, Z2 |2
|Z, Z3 |2
is equal to 1, we obtain
δF (Z, W1 , Z2 )δF (Z, W2 , Z2 )δF (W1 , Z3 , W2 )2
= 1.
δF (Z, W1 , Z3 )δF (Z, W2 , Z3 )δF (W1 , Z2 , W2 )2
We find similarly that every point of W1 W2 is on this quadratic locus, and so are U and V , so it
is the union of the lines W1 W2 and U V . Thus these two conditions hold for Z on U V .
We thus have an analogue of the result that the base angles in an isosceles triangle are equal
in magnitude.
432
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
14.5
ANALOGUES OF APPOLONIUS’, PYTHAGORAS’
AND PTOLEMY’S THEOREMS
14.5.1
Appolonius’ characterisation of a circle
If P1 6= P2 , then by Appolonius’ characterization of a circle, the locus of
|P, P1 |2 /|P, P2 |2 = k, where k > 0, is a circle. This converts to
δF (Z, W1 , Z1 )δF (Z, W2 , Z1 )δF (W1 , Z2 , W2 )2
= k,
δF (Z, W1 , Z2 )δF (Z, W2 , Z2 )δF (W1 , Z1 , W2 )2
which is a conic through W1 and W2 .
b
W1
b
b
Z2
b
b
b
b
W2
b
b
b
Z1
Figure 9.6
In this Z1 and Z2 come from points inverse with respect to the circle, and each therefore should
lie on the polar of the other with respect to the conic. Now from the scratch equation of the
equation above, the polar of the point Z3 is
[δF (Z, W1 , Z1 )δF (Z3 , W2 , Z1 ) + δF (Z3 , W1 , Z1 )δF (Z, W2 , Z1 )]δF (W1 , Z2 , W2 )2
− k[δF (Z, W1 , Z2 )δF (Z3 , W2 , Z2 ) + δF (Z3 , W1 , Z2 )δF (Z, W2 , Z2 )]δF (W1 , Z1 , W2 )2
= 0.
On inserting Z1 for Z3 , this becomes
δF (Z, W1 , Z2 )δF (Z1 , W2 , Z2 ) + δF (Z1 , W1 , Z2 )δF (Z, W2 , Z2 ) = 0.
This is clearly satisfied by Z2 . By symmetry, Z1 lies on the polar of Z2 .
In the source P1 P2 passes through the centre of the circle, so Z1 Z2 should pass through the
pole of W1 W2 . In the above take
Z3 =
µ
1
Z1 +
Z2 .
1+µ
1+µ
Then the equation of the polar of Z3 becomes
µ[δF (Z, W1 , Z1 )δF (Z2 , W2 , Z1 ) + δF (Z2 , W1 , Z1 )δF (Z, W2 , Z1 )]δF (W1 , Z2 , W2 )2
− k[δF (Z, W1 , Z2 )δF (Z1 , W2 , Z2 ) + δF (Z1 , W1 , Z2 )δF (Z, W2 , Z2 )]δF (W1 , Z1 , W2 )2
= 0.
14.5. ANALOGUES OF APPOLONIUS’, PYTHAGORAS’ AND PTOLEMY’S THEOREMS433
We wish this to be the line W1 W2 , and on inserting W1 and W2 in turn for Z, we find that this is
so if we choose µ to satisfy the equation
µ = −k
δF (W1 , W2 , Z1 )
.
δF (W1 , W2 , Z2 )
As Z1 , Z2 and the pole of W1 W2 are collinear, the polars of W1 and W2 , and W1 W2 are
concurrent.
14.5.2
Analogue of Pythagoras’ theorem
We next consider an analogue of Pythagoras’ theorem as follows. We start with a circle with
diameter [P2 , P3 ]. Then for any other point P on the circle, P2 P ⊥ P3 P and so
|P2 , P |2
|P3 , P |2
+
= 1.
|P2 , P3 |2
|P3 , P2 |2
b
Z
Z3
b
b
W2
b
Z2
b
b
W1
Z1
Figure 9.7
Now consider a proper conic through the points W1 , W2 , Z2 , Z3 , so that it has an equation of
the form
δF (Z, Z2 , W1 )δF (Z, Z3 , W2 ) − k1 δF (Z, Z3 , W1 )δF (Z, Z2 , W2 ) = 0,
for some k1 6= 0. We wish Z2 Z3 to contain the pole Z1 of W1 W2 ; as Z1 corresponds to the centre,
Z2 Z3 will then correspond to a diametral line. Now the polar of a general point Z4 with respect
to this conic has equation
δF (Z, Z2 , W1 )δF (Z4 , Z3 , W2 ) + δF (Z4 , Z2 , W1 )δF (Z, Z3 , W2 )
− k1 [δF (Z, Z3 , W1 )δF (Z4 , Z2 , W2 ) + δF (Z4 , Z3 , W1 )δF (Z, Z2 , W2 )] = 0.
On taking
Z4 =
1
µ
Z2 +
Z3 ,
1+µ
1+µ
this becomes
δF (Z, Z2 , W1 )δF (Z2 , Z3 , W2 ) + µδF (Z3 , Z2 , W1 )δF (Z, Z3 , W2 )
− k1 [µδF (Z, Z3 , W1 )δF (Z3 , Z2 , W2 ) + δF (Z2 , Z3 , W1 )δF (Z, Z2 , W2 )] = 0.
434
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
This polar contains W1 and W2 if
µδF (Z3 , Z2 , W1 )δF (W1 , Z3 , W2 ) − k1 δF (Z2 , Z3 , W1 )δF (W1 , Z2 , W2 )] = 0,
δF (W2 , Z2 , W1 )δF (Z2 , Z3 , W2 ) − k1 µδF (W2 , Z3 , W1 )δF (Z3 , Z2 , W2 ) = 0.
These equations lead to k12 = 1 and we take k1 = −1 as we have a proper conic. Thus our conic
has equation
δF (Z, Z2 , W1 )δF (Z, Z3 , W2 ) + δF (Z, Z3 , W1 )δF (Z, Z2 , W2 ) = 0,
and we have that
µ=−
δF (W1 , Z2 , W2 )
.
δF (W1 , Z3 , W2 )
Now, by 4.2.2, this conic also has equation
δF (Z, W1 , W2 )2 − k2 δF (Z, W1 , Z1 )δF (Z, W2 , Z1 ) = 0,
for some k2 6= 0, and since it passes through Z2 this equation must yield
δF (Z2 , W1 , W2 )2 δF (Z, W1 , Z1 )δF (Z, W2 , Z1 ) = δF (Z, W1 , W2 )2 δF (Z2 , W1 , Z1 )δF (Z2 , W2 , Z1 ).
To eliminate Z1 in this we recall that
Z1 =
µ
1
Z2 +
Z3 ,
1+µ
1+µ
where
µ=−
and then
δF (W1 , Z2 , W2 )
,
δF (W1 , Z3 , W2 )
δF (Z, W1 , Z1 ) =
µ
1
δF (Z, W1 , Z2 ) +
δF (Z, W1 , Z3 ),
1+µ
1+µ
δF (Z, W2 , Z1 ) =
1
µ
δF (Z, W2 , Z2 ) +
δF (Z, W2 , Z3 ).
1+µ
1+µ
On inserting these in the equation of the conic we obtain
µ2 [δF (Z2 , W1 , W2 )2 δF (Z, W1 , Z3 )δF (Z, W2 , Z3 ) − δF (Z, W1 , W2 )2 δF (Z2 , W1 , Z3 )δF (Z2 , W2 , Z3 )]
+ µ[δF (Z, W1 , Z2 )δF (Z, W2 , Z3 ) + δF (Z, W1 , Z3 )δF (Z, W2 , Z2 )]
+ δF (Z2 , W1 , W2 )2 δF (Z, W1 , Z2 )δF (Z, W2 , Z2 ) = 0.
The coefficient of µ here vanishes, and on inserting the value of µ and simplifying we obtain
δF (Z2 , W1 , W2 )2 δF (Z, W1 , Z3 )δF (Z, W2 , Z3 ) − δF (Z, W1 , W2 )2 δF (Z2 , W1 , Z3 )δF (Z2 , W2 , Z3 )
+ δF (Z3 , W1 , W2 )2 δF (Z, W1 , Z2 )δF (Z, W2 , Z2 ) = 0.
This is our analogue of Pythagoras’ theorem.
14.5.3
Ptolemy’s theorem
We now turn to Ptolemy’s theorem. A significant feature for this is that it involves not a quadratic
expression equated to 0, but the square of a quadratic expression equated to 0. Starting with
points P1 , P2 , P3 , P4 on a circle, with [P1 , P3 ] ∩ [P2 , P4 ] 6= ∅, we recall the result
s
s
|P1 , P2 |2 |P4 , P3 |2
|P1 , P4 |2 |P2 , P3 |2
+
= 1.
|P1 , P3 |2 |P4 , P2 |2
|P1 , P3 |2 |P2 , P4 |2
14.5. ANALOGUES OF APPOLONIUS’, PYTHAGORAS’ AND PTOLEMY’S THEOREMS435
As an analogue of this, we take a proper conic containing the distinct points W1 , W2 , Z1 , Z2 , Z3 , Z4 ,
and then an analogy of the above is
s
δF (Z1 , W1 , Z2 )δF (Z4 , W1 , Z3 ) δF (Z1 , W2 , Z2 )δF (Z4 , W2 , Z3 )
δF (Z1 , W1 , Z3 )δF (Z4 , W1 , Z2 ) δF (Z1 , W2 , Z3 )δF (Z4 , W2 , Z2 )
s
δF (Z1 , W1 , Z4 )δF (Z2 , W1 , Z3 ) δF (Z1 , W2 , Z4 )δF (Z2 , W2 , Z3 )
+
δF (Z1 , W1 , Z3 )δF (Z2 , W1 , Z4 ) δF (Z1 , W2 , Z3 )δF (Z2 , W2 , Z4 )
= 1.
Now the conic has an equation of the form
δF (Z1 , Z, Z2 )δF (Z4 , Z, Z3 )
= j,
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
for some constant j. Then we have
1
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
= ,
δF (Z1 , Z, Z2 )δF (Z4 , Z, Z3 )
j
and
δF (Z1 , Z, Z2 )δF (Z4 , Z, Z3 )
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 ) − δF (Z1 , Z, Z2 )δF (Z4 , Z, Z3 )
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
δF [(Z1 , Z4 ), (Z, Z3 ), (Z, Z2 )]
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
δF [(Z, Z2 ), (Z1 , Z4 ), (Z, Z3 )]
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
δF (Z, Z1 , Z4 )δF (Z2 , Z, Z3 )
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
δF (Z1 , Z, Z4 )δF (Z2 , Z, Z3 )
,
δF (Z1 , Z, Z3 )δF (Z2 , Z, Z4 )
1−
=
=
=
=
=
so that
δF (Z1 , Z, Z4 )δF (Z2 , Z, Z3 )
= 1 − j.
δF (Z1 , Z, Z3 )δF (Z2 , Z, Z4 )
On combining these operations, we can see that on keeping Z1 fixed and varying Z2 , Z3 and Z4 we
can obtain the values
1
j
1 j−1
,
,
,
,
j, 1 − j,
j
j
1−j j−1
for the above expression. We note that when 0 < j < 1, j and 1 − j lie between 0 and 1; when
j > 1, 1/j and 1 − 1/j lie between 0 and 1; when j < 0, 1/(1 − j) and 1 − 1/(1 − j) lie between
0 and 1. On omitting the cases when j = 0 and j = 1 we see that by relabelling the points
Z2 , Z3 , Z4 appropriately among themselves, without loss of generality we can take the conic to
have an equation of the form
δF (Z1 , Z, Z2 )δF (Z4 , Z, Z3 )
= k,
δF (Z1 , Z, Z3 )δF (Z4 , Z, Z2 )
where 0 < k < 1.
Now this is satisfied by Z = W1 and by Z = W2 , so the product of the two terms under the
first radical is k 2 . Similarly the product of the two terms under the second radical is (1 − k)2 .
436
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
The stated result is then equivalent to
and this is true as 0 < k < 1.
√
p
k 2 + (1 − k)2 = 1,
Figure 9.8.
In the diagram we have taken
Z1 ≡ (1, 1), Z2 ≡ (−1, −1), Z3 ≡ (1, −1), Z4 ≡ (−1, 1).
The equation is x2 − 1 − k(y 2 − 1) = 0, and the loci for which 0 < k < 1 are hyperbolas of the
form shown.
14.6
THE NEWTON AND CLASSICAL GREEK PROPERTIES
14.6.1
Newton’s property
As a projective analogue of Newton’s theorem in 4.6.1 the following is suggested by the method of
6.10. Let W1 , W2 , W3 , W4 be distinct collinear points and C a proper conic. For a variable point
W 6∈ W1 W2 let W W3 meet C at Z4 , Z5 , and W W4 meet C at Z6 , Z7 . We wish to show that
E=
δF (W, W1 , Z4 )δF (W1 , Z6 , W2 )δF (W, W1 , Z5 )δF (W1 , Z7 , W2 )
δF (W, W1 , Z6 )δF (W1 , Z4 , W2 )δF (W, W1 , Z7 )δF (W1 , Z5 , W2 )
is constant as W varies on the plane.
437
14.6. THE NEWTON AND CLASSICAL GREEK PROPERTIES
W
b
Z7
b
Z5
b
b
Z6
b
b
Z4
W4
b
W2
b
W1
Figure 9.9
b
W3
Let the conic have equation
f βγ + gγα + hαβ = 0,
and
W3
=
Z4
=
Z6
=
1
λ
1
µ
W1 +
W2 , W4 =
W1 +
W2 ,
1+λ
1+λ
1+µ
1+µ
1
ρ1
1
ρ2
W+
W3 , Z5 =
W+
W3 ,
1 + ρ1
1 + ρ1
1 + ρ2
1 + ρ2
σ1
1
σ2
1
W+
W4 , Z7 =
W+
W4 .
1 + σ1
1 + σ1
1 + σ2
1 + σ2
With
Z=
1
ρ
W+
W3 ,
1+ρ
1+ρ
we have
δF (W, W1 , Z) =
ρ
δF (W, W1 , W3 ),
1+ρ
and so
δF (W, W1 , Z4 )δF (W, W1 , Z5 ) =
ρ1 ρ2
δF (W, W1 , W3 )2 .
(1 + ρ1 )(1 + ρ2 )
Similarly
δF (W1 , Z4 , W2 )δF (W1 , Z5 , W2 ) =
Hence
Similarly
It follows that
1
δF (W1 , W, W2 )2 .
(1 + ρ1 )(1 + ρ2 )
δF (W, W1 , W3 )2
δF (W, W1 , Z4 )δF (W, W1 , Z5 )
.
= ρ1 ρ2
δF (W1 , Z4 , W2 )δF (W1 , Z5 , W2 )
δF (W1 , W, W2 )2
δF (W, W1 , Z6 )δF (W, W1 , Z7 )
δF (W, W1 , W4 )2
.
= σ1 σ2
δF (W1 , Z6 , W2 )δF (W1 , Z7 , W2 )
δF (W1 , W, W2 )2
E=
Now
Z=
ρ1 ρ2 δF (W, W1 , W3 )2
.
σ1 σ2 δF (W, W1 , W4 )2
1
ρ
W+
W3 ,
1+ρ
1+ρ
438
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
so
α =
β
=
γ
=
ρ
1
α′ +
α′ ,
1+ρ
1+ρ 3
1
ρ
δF (Z, Z3 , Z1 ) =
β′ +
β′ ,
1+ρ
1+ρ 3
1
ρ
δF (Z, Z1 , Z2 ) =
γ′ +
γ′ ,
1+ρ
1+ρ 3
δF (Z, Z2 , Z3 ) =
where Z, W, W3 have areal coordinates (α, β, γ), (α′ , β ′ , γ ′ ), (α′3 , β3′ , γ3′ ), respectively. Then the
intersection equation for Z to be on the conic is
f (β ′ + ρβ3′ )(γ ′ + ργ3′ ) + g(γ ′ + ργ3′ )(α′ + ρα′3 ) + h(α′ + ρα′3 )(β ′ + ρβ3′ ) = 0,
and so
ρ1 ρ2 =
f β ′ γ ′ + gγ ′ α′ + hα′ β ′
.
f β3′ γ3′ + gγ3′ α′3 + hα′3 β3′
σ1 σ2 =
f β ′ γ ′ + gγ ′ α′ + hα′ β ′
,
f β4′ γ4′ + gγ4′ α′4 + hα′4 β4′
Similarly
and so
E=
f β4′ γ4′ + gγ4′ α′4 + hα′4 β4′ δF (W, W1 , W3 )2
.
f β3′ γ3′ + gγ3′ α′3 + hα′3 β3′ δF (W, W1 , W4 )2
Now let
W3 =
τ1
1
τ2
1
W1 +
W2 , W4 =
W1 +
W2 .
1 + τ1
1 + τ1
1 + τ2
1 + τ2
Then
δF (W, W1 , W3 ) =
τ1
τ2
δF (W, W1 , W3 ), δF (W, W1 , W4 ) =
δF (W, W1 , W3 ),
1 + τ1
1 + τ2
and the quotient of these is constant.
14.6.2
Classical Greek property
We now seek from 6.9 a projective characterization of a proper conic which is an analogue of the
classical Greek one. For this let W1 , W2 , W4 be distinct collinear points and Z4 , Z5 distinct points
not on W1 W2 . For some fixed k 6= 0, consider the locus of points Z 6∈ W1 W2 such that if W4 Z
meets Z4 Z5 in W , then
δF (W, W1 , Z)2 = kδF (W1 , Z, W2 )2 δF (W, W1 , Z3 )δF (W, W1 , Z4 ).
439
14.6. THE NEWTON AND CLASSICAL GREEK PROPERTIES
b
Z5
b
W3
b
W
W2
b
b
Z4
b
b
Z
W1
Figure 9.10
Now
W =
1
λ
Z4 +
Z5 ,
1+λ
1+λ
and δF (W4 , Z, W ) = 0, so
λ=−
Also
δF (W4 , Z, Z4 )
.
δF (W4 , Z, Z5 )
1
λ
δF (Z4 , W1 , Z) +
δF (Z5 , W1 , Z),
1+λ
1+λ
λ
δF (W, W1 , Z4 ) =
δF (Z5 , W1 , Z4 ),
1+λ
1
δF (W, W1 , Z5 ) =
δF (Z4 , W1 , Z5 ).
1+λ
On inserting these in the equation above, we find that
δF (W, W1 , Z) =
[δF (Z4 , W1 , Z)δF (Z5 , W4 , Z) − δF (Z4 , W4 , Z)δF (Z5 , W1 , Z)]2
= kδF (Z4 , W1 , Z5 )2 δF (W1 , Z, W2 )2 δF (W4 , Z, Z4 )δF (W4 , Z, Z5 ).
On the left we have
δF [(Z4 , Z5 ), (W1 , Z), (W4 , Z)] =
=
Moreover
W2 =
δF [(W4 , Z), (Z4 , Z5 ), (W1 , Z)]
−δF (W4 , W1 , Z)δF (Z, Z4 , Z5 ).
µ
1
W1 +
W4 ,
1+µ
1+µ
for some µ so that
µ
δF (W1 , Z, W4 ).
1+µ
On inserting these and cancelling across by δF (W1 , Z, W4 )2 , we obtain
δF (W1 , Z, W2 ) =
δF (Z, Z4 , Z5 )2 = k1 δF (Z, Z4 , W4 )δF (Z, Z5 , W4 ),
where
µ2
δF (Z4 , W1 , Z5 )2 .
(1 + µ)2
This is the equation of a conic, and from its form Z4 Z5 is the polar of W4 with respect to it.
k1 = k
440
14.7
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
GENERALIZATIONS FROM THE HYPERBOLA
14.7.1
Consider a point Z4 on the hyperbola with equation
y2
x2
−
= 1.
a2
b2
The line through Z4 which is parallel to the asymptote with equation y = (b/a)x will meet the
other asymptote with equation y = −(b/a)x at the point
a x4
y4
b x4
y4
W4 ≡ ( ( − ), − ( − )).
2 a
b
2 a
b
Then δF (O, W4 , Z4 ) =
ab
4 ,
and this is constant as Z4 varies on the hyperbola.
b
b
b
Z4
b
O
b
W4
b
b
Figure 9.11.
We seek a projective generalization of this. Let Z2 Z3 be a line and W1 , W2 distinct points
of it. Let Z1 6∈ Z2 Z3 and consider points Z4 , Z such that if Z2 Z4 , Z2 Z meet Z1 Z3 in W4 , W ,
respectively, then
δF (Z1 , Z, W )δF (Z4 , W1 , W2 )δF (W4 , W1 , W2 )
= 1.
δF (Z1 , Z4 , W4 )δF (Z, W1 , W2 )δF (W, W1 , W2 )
Then with a variable Z we have that
δF (Z1 , Z, W )
= k.
δF (Z, W1 , W2 )δF (Z, W1 , W2 )
Let
W =
1
λ
Z1 +
Z3 ,
1+λ
1+λ
and from this
δF (Z2 , Z, Z1 ) + λδF (Z2 , Z, Z1 ) = 0,
λ
δF (Z1 , Z, Z3 ),
1+λ
1
δF (W, W1 , W2 ) =
δF (Z1 , W1 , W2 ).
1+λ
On inserting these we have that
δF (Z1 , Z, W ) =
δF (Z, Z1 , Z2 )δF (Z, Z3 , Z1 ) = kδF (Z1 , W1 , W2 )δF (Z, W1 , W2 )δF (Z, Z2 , Z3 ).
But
W1 =
ρ
1
σ
1
Z2 +
Z3 , W2 =
Z2 +
Z3 ,
1+ρ
1+ρ
1+σ
1+σ
441
14.7. GENERALIZATIONS FROM THE HYPERBOLA
and from these
δF (Z, W1 , W2 ) =
σ−ρ
δF (Z, Z2 , Z3 ).
(1 + ρ)(1 + σ)
Hence
δF (Z, Z3 , Z1 )δF (Z, Z1 , Z2 ) = k1 δF (Z, Z2 , Z3 )2 ,
for some constant k1 . From its form, this locus is a conic through Z2 and Z3 , with respect to which
Z2 Z3 is the polar of Z1 .
b
W
Z
b
b
W2
W1
b
Z3
b
b
b
W4
Z4
b
b
Z2
Z1
Figure 9.12
14.7.2
For the hyperbola with equation
y2
x2
−
= 1,
a2
b2
we take the point W4 ≡ (ar, br) on one asymptote and the distinct point W5 ≡ (as, −bs) on the
other asymptote. Points Z ≡ (x, y) on W4 W5 have coordinates of the form
x = ar + a(s − r)t, y = br + b(−s − r)t.
If the line W4 W5 meets the hyperbola it is in points Z4 , Z5 with parameters t4 , t5 satisfying the
equation
−4rst2 + 4rst − 1 = 0.
Then t4 + t5 = 1 so that t5 = 1 − t4 . It follows that
W4 Z4
= 1,
Z5 W5
and
W4 Z5
= 1.
Z4 W5
442
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
W4
b
b
Z4
Z5
b
b
W5
Figure 9.14?.
As a projectively invariant generalization of this, we have the following. Let Z2 , Z3 be points on
a proper conic and Z1 the pole of Z2 Z3 . Let W1 , W2 ∈ Z2 Z3 , and for points W4 ∈ Z1 Z2 , W5 ∈ Z1 Z3
let W4 W5 meet the conic in the points Z4 , Z5 . Then
δF (W4 , W2 , Z4 )δF (W1 , Z5 , W2 )δF (W1 , W5 , W2 )
= 1,
δF (Z5 , W2 , W5 )δF (W1 , W4 , W2 )δF (W1 , Z4 , W2 )
and
δF (W4 , W2 , Z5 )δF (W1 , Z4 , W2 )δF (W1 , W5 , W2 )
= 1.
δF (Z4 , W2 , W5 )δF (W1 , W4 , W2 )δF (W1 , Z5 , W2 )
b
W5
Z5
b
b
W2
W1
b
b
Z3
Z4
b
b
W4
b
Z2
b
Z1
Figure 9.15
14.8
NEWTON’S ORGANIC METHOD OF GENERATION
14.8.1
Let W5 W6 be a fixed line and W1 , W2 , W3 , W4 fixed points not on it. For a point W varying on
W5 W6 consider the locus of the point Z such that
δF (W4 , W1 , W )δF (W4 , W2 , Z)
δF (W3 , W1 , W )δF (W3 , W2 , Z)
= k1 ,
= k2 ,
δF (W3 , W1 , Z)δF (W3 , W2 , W )
δF (W4 , W1 , Z)δF (W4 , W2 , W )
for non-zero constants k1 , k2 . Then
W =
1
λ
W5 +
W6 ,
1+λ
1+λ
443
14.9. ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS
for a variable λ. From this we have that
δF (W3 , W1 , W5 ) + λδF (W3 , W1 , W6 )
δF (W3 , W2 , W5 ) + λδF (W3 , W2 , W6 )
δF (W4 , W1 , W5 ) + λδF (W4 , W1 , W6 )
δF (W4 , W2 , W5 ) + λδF (W4 , W2 , W6 )
δF (W3 , W1 , Z)
,
δF (W3 , W2 , Z)
δF (W4 , W1 , Z)
= k2
.
δF (W4 , W2 , Z)
= k1
On eliminating λ between these, we obtain
=
δF (W3 , W1 , W5 )δF (W3 , W2 , Z) − k1 δF (W3 , W2 , W5 )δF (W3 , W1 , Z)
−δF (W3 , W1 , W6 )δF (W3 , W2 , Z) + k1 δF (W3 , W2 , W6 )δF (W3 , W1 , Z)
δF (W4 , W1 , W5 )δF (W4 , W2 , Z) − k2 δF (W4 , W2 , W5 )δF (W4 , W1 , Z)
.
−δF (W4 , W1 , W6 )δF (W4 , W2 , Z) + k2 δF (W4 , W2 , W6 )δF (W4 , W1 , Z)
Thus the locus of Z is a conic.
W4
b
b
W1
b
W6
b
W
Z
b
b
b
W5
b
W2
W3
Figure 9.13?.
This is a projective analogue of Newton’s organic method of generating a conic in 1686. The
diagram corresponds to
W1 ≡ (1, 1), W2 ≡ (−1, −1), W3 ≡ (1, −1), W4 ≡ (−1, 1), W5 ≡ (−5, −1.25), W6 ≡ (3, 0.75)
and k1 = −1/2, k2 = 2.
14.9
ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS
We now give a second extension to some material considered in 5.6.
14.9.1
Analogue of centroid
Let C be a proper conic, and Z0 a point not on it and not the centre in the case of a central conic.
Let the line l be the polar of Z0 with respect to C, and Z1 , Z2 and Z3 be non- collinear points on
the conic. With (Z1 , Z2 , Z3 ) as the basis of our areal coordinates, C will have an equation of the
form
f βγ + gγα + hαβ = 0,
and the polar l will have an equation
(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ = 0.
444
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
We suppose that the line Z2 Z3 meets l in a point Z4 ; taking points Z on Z2 Z3 , of the form
Z=
λ
1
Z2 +
Z3 ,
1+λ
1+λ
for Z4 we have
λ=−
f γ0 + hα0
,
gα0 + f β0
and then for the point Z5 which is the harmonic conjugate of Z4 with respect to {Z2 , Z3 } we have
λ=
f γ0 + hα0
,
gα0 + f β0
and so Z5 has areal coordinates
1
1+
f γ0 +hα0
gα0 +f β0
f γ0 + hα0
δ1 ,
0, δ1 ,
gα0 + f β0
where δ1 = δF (Z1 , Z2 , Z3 ). From these we find that the line Z1 Z5 has equation
−(f γ0 + hα0 )β + (gα0 + f β0 )γ = 0.
Z3
b
b
Z5
b
b
b
Z2
b
b
b
Z4
Z0
b
b
Z1
b
b
Figure 9.16
We also suppose that the line Z3 Z1 meets l in a point Z6 , and that the point Z7 is the harmonic
conjugate of Z6 with respect to {Z3 , Z1 }, and that the line Z1 Z2 meets l in a point Z8 , and that
the point Z9 is the harmonic conjugate of Z8 with respect to {Z1 , Z2 }. Then as above, the lines
Z2 Z7 and Z3 Z9 have equations
(gγ0 + hβ0 )α − (f β0 + gα0 )γ
(−(gγ0 + hβ0 )α + (hα0 + f γ0 )β
= 0,
= 0.
On forming the determinant of the coefficients of the equations of the lines Z1 Z5 , Z2 Z7 and Z3 Z9 ,
we find that these lines are concurrent or all parallel. In the former case, we regard the point of
concurrency as an analogue of the centroid of the triangle [Z1 , Z2 , Z3 ] in 5.6.
445
14.9. ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS
14.9.2
Analogue of orthocentre
Continuing with the notation of 9.9.1, the polar of Z4 has equation
[−g(hα0 + f γ0 ) + h(f β0 + gα0 )]α − f (hα0 + f γ0 )β + f (f β0 + gα0 )γ0 = 0,
and we suppose that it meets l at a point Z10 . Now any line through Z10 has an equation of the
form
[−g(hα0 + f γ0 ) + h(f β0 + gα0 )]α − f (hα0 + f γ0 )β + f (f β0 + gα0 )γ0
+ k[(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0.
We choose k so that this passes through Z1 and in this way find that Z1 Z10 has equation
(hα0 + f γ0 )[−f (gγ0 + hβ0 ) + g(hα0 + f γ0 ) − h(f β0 + gα0 )]β
(f β0 + gα0 )[f (gγ0 + hβ0 ) + g(hα0 + f γ0 ) − f (f β0 + gα0 )]γ = 0.
Similarly, we suppose that the polar of Z6 meets l at a point Z11 and we find that Z2 Z11 has
an equation
(gγ0 + hβ0 )[g(hα0 + f γ0 ) + h(f β0 + gα0 ) − f (gγ0 + hβ0 )]α
(f β0 + gα0 )[−g(hα0 + f γ0 ) + h(f β0 + gα0 ) − f (gγ0 + hβ0 )]γ = 0.
Similarly, we suppose that the polar of Z8 meets l at a point Z12 and we find that Z3 Z12 has
an equation
(gγ0 + hβ0 )[−h(f β0 + gα0 ) + f (gγ0 + hβ0 ) − g(hα0 + f γ0 )]α
(f β0 + gα0 )[h(f β0 + gα0 ) + f (gγ0 + hβ0 ) − g(hα0 + f γ0 )]β = 0.
Z3
b
b
b
b
b
Z2
b
b
b
Z4
Z0
b
b
b
Z1
b
b
b
Z10
Figure 9.17
By forming the determinant of the coefficients in the equations of the lines Z1 Z10 , Z2 Z11 and
Z3 Z12 , we find that these lines are either concurrent or all parallel. In the former case, we regard
the point of concurrency as an analogue of the generalized orthocentre in the context of 5.6.3.
446
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
14.9.3
Analogue of the Euler line
The analogue of centroid in 9.9.1 has areal coordinates proportional to
(f β0 + gα0 )(hα0 + f γ0 ), (gγ0 + hβ0 )(f β0 + gα0 ), (hα0 + f γ0 )(gγ0 + hβ0 )(f β0 + gα0 )).
The line joining this point to Z0 has equation
α0 (hβ0 − gγ0 )(hβ0 + gγ0 )α + β0 (f γ0 − hα0 )(f γ0 + hα0 )β + γ0 (gα0 − f β0 )(gα0 + f β0 )γ = 0,
and if we insert the coordinates of this line in the last determinant instead of those for Z3 Z12
we obtain that Z0 , the analogue of centroid and the analogue of orthocentre are collinear, thus
yielding a further extension of the Euler line 5.6.5.
14.9.4
Analogue of the Wallace-Simson line
Still with the situation in 9.9.1, we consider a variable point Z ′ with areal coordinates (α′ , β ′ , γ ′ ).
Any line through Z10 (where the polar of Z4 meets l) has an equation of the form
[−g(hα0 + f γ0 ) + h(f β0 + gα0 )]α − f (hα0 + f γ0 )β + f (f β0 + gα0 )γ
+ k1 [(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0,
and we choose k1 so that this passes through Z ′ . This line Z10 Z ′ meets the line Z2 Z3 in a point
Z13 with coordinates proportional to
(0, (f + k1 )(f β0 + gα0 ), (f − k1 )(hα0 + f γ0 )),
with the above choice of k1 .
Similarly any line through Z11 (where the polar of Z6 meets l) has an equation of the form
g(gγ0 + hβ0 )α + [f (gγ0 + hβ0 ) − h(f β0 + gα0 )]β − g(f β0 + gα0 )γ
+ k2 [(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0,
and we choose k2 so that this passes through Z ′ . This line Z11 Z ′ meets the line Z3 Z1 in a point
Z14 with coordinates proportional to
((g − k2 )(f β0 + gα0 ), 0, (g + k2 )(gγ0 + hβ0 )),
with the above choice of k2 .
Similarly any line through Z12 (where the polar of Z8 meets l) has an equation of the form
−h(gγ0 + hβ0 )α + h(hα0 + f γ0 )β + [−f (gγ0 + hβ0 ) + g(hα0 + f γ0 )]γ
+ k3 [(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0,
and we choose k3 so that this passes through Z ′ . This line Z12 Z ′ meets the line Z1 Z2 in a point
Z15 with coordinates proportional to
((h + k3 )(hα0 + f γ0 ), (h − k3 )(gγ0 + hβ0 )),
with the above choice of k3 .
447
14.10. ANALOGUE OF POWER PROPERTY OF CIRCLES
Z′
Z3
b
b
b
Z13
b
b
b
b
Z2
b
b
b
Z4
Z0
b
b
b
Z1
b
b
b
Z10
Figure 9.18
The points Z13 , Z14 and Z15 are collinear if and only if
2(f β0 + gα0 )(gγ0 + hβ0 )(hα0 + f γ0 )(f gh + f k2 k3 + gk3 k1 + hk1 k2 ) = 0.
On assuming that the polar of Z0 contains none of the points Z1 , Z2 , Z3 , we can discard the first
three factors after 2, and then on inserting for the values of k1 , k2 and k3 we find that the above
points are collinear if and only if
f β ′ γ ′ + gγ ′ α′ + hα′ β ′ = 0,
that is, if and only if Z ′ lies on the conic. This is an analogue of the extension of the Wallace-Simson
property noted in 5.6.6.
14.10
ANALOGUE OF POWER PROPERTY OF CIRCLES
14.10.1
Can we obtain a projectively invariant version of the material of 7.6? The extension of concepts
to the extended plane suggests the following.
448
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Z2
b
b
Z0
b
b
Z3
b
Figure 9.19.
b
Z1
Let Ck be a proper conic, Z0 a point not on it, and W1 , W2 distinct points on the polar l0 of
Z0 . Let C be a proper conic which meets Ck at two points on the polar l0 . For a variable point Z1
on Ck let the line Z0 Z1 meet the conic C in the points Z2 , Z3 . Then as Z1 varies,
E=
δF (Z0 , W2 , Z2 ) δF (W1 , Z1 , W2 ) δF (Z0 , W2 , Z3 ) δF (W1 , Z1 , W2 )
δF (Z0 , W2 , Z1 ) δF (W1 , Z2 , W2 ) δF (Z0 , W2 , Z1 ) δF (W1 , Z3 , W2 )
is constant.
We take areal coordinates (δ1 , 0, 0), (0, δ1 , 0) for the points in which Ck meets the polar l0 , and
(0, 0, δ1 ) for Z0 . Then by 4.2.1, Ck has an equation of the form
γ 2 = 4jαβ,
and so has parametric equations with (α, β, γ) proportional to (θ2 , j, 2jθ). The points W1 , W2 on
the polar l0 with equation γ = 0 have coordinates
((1 − s1 )δ1 , s1 δ1 , 0), ((1 − s2 )δ1 , s2 δ1 , 0).
As the conic C passes through the points with coordinates (δ1 , 0, 0), (0, δ1 , 0), it has an equation of
the form
cγ 2 + 2f βγ + 2gγα + 2hαβ = 0.
Points on the line Z0 Z1 have coordinates of the form
(tθ2 , tj, δ1 + t(2jθ − δ1 )).
This is on the conic C when t satisfies the quadratic equation
t2 [c(2jθ − δ1 )2 + 2f j(2jθ − δ1 ) + 2g(2jθ − δ1 )θ2 + 2hjθ2 ]
+ 2t[cδ1 (2jθ − δ1 ) + f jδ1 + gδ1 θ2 ] + cδ12 = 0.
Let the roots of this be t2 and t3 , so that Z2 and Z3 have coordinates
(t2 θ2 , t2 j, δ1 + t2 (2jθ − δ1 ), (t3 θ2 , t3 j, δ1 + t3 (2jθ − δ1 ).
449
14.11. ANALOGUE OF PROTO-FOCI
Then we find that
2δF (Z0 , W2 , Z2 )
2δF (W1 , Z1 , W2 )
= δ12 t2 [(1 − s2 )j − s2 θ2 ],
= −2jδ12 θ(s2 − s1 ),
2δF (W1 , Z2 , W2 )
= −(s2 − s1 )δ12 [δ1 + t2 (2jθ − δ1 )],
= δ12 t3 [(1 − s2 )j − s2 θ2 ],
= δ12 [(1 − s2 )j − s2 θ2 ],
2δF (Z0 , W2 , Z3 )
2δF (Z0 , W2 , Z1 )
= −(s2 − s1 )δ12 [δ1 + t3 (2jθ − δ1 )].
2δF (W1 , Z3 , W2 )
On inserting these and performing cancellations, we find that the equipoised quotient E is equal
to
4j 2 θ2 t2 t3
E= 2
.
δ1 + δ1 (2jθ − δ1 )(t2 + t3 ) + (2jθ − δ1 )2 t2 t3
But from the quadratic equation in t,
t2 t3 =
cδ12
,
c(2jθ − δ1 )2 + 2f j(2jθ − δ1 ) + 2g(2jθ − δ1 )θ2 + 2hjθ2
cδ1 (2jθ − δ1 ) + f jδ1 + gδ1 θ2
.
t2 [c(2jθ − δ1 )2 + 2f j(2jθ − δ1 ) + 2g(2jθ − δ1 )θ2 + 2hjθ2
On inserting these, we find that
2jc
.
E=
h
Thus we add an assumption that h 6= 0 to make our proof valid.
t2 + t3 = −2
14.11
ANALOGUE OF PROTO-FOCI
14.11.1
We now consider a projective version of the property of the proto-focus of a central conic considered
in 7.4. Let W1 , W2 , W4 be distinct collinear points and W5 , W6 distinct points not on W1 W2 .
Consider the locus of points Z 6∈ W1 W2 such that if W4 Z meets W5 W6 at W , then
δF (W, W1 , Z)δF (W, W2 , Z)δF (W1 , W6 , W2 )2
δF (W, W1 , W5 )δF (W, W2 , W5 )δF (W1 , W6 , W2 )2
+
δF (W, W1 , W6 )δF (W, W2 , W6 )δF (W1 , Z, W2 )2
δF (W, W1 , W6 )δF (W, W2 , W6 )δF (W1 , W5 , W2 )2
= k3 .
b
Z
b
b
W5
W
b
b
b
W4 W1
W6
b
b
W2
Figure 9.20.
450
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Let
W =
1
λ
W5 +
W6 ,
1+λ
1+λ
and then δF (W, Z, W4 ) = 0 gives
δF (W5 , Z, W4 ) + λδF (W6 , Z, W4 ) = 0.
Now
δF (W, W1 , Z) =
δF (W, W2 , Z) =
δF (W, W1 , W6 ) =
δF (W, W1 , W5 ) =
1
[δF (W5 , W1 , Z) + λδF (W6 , W1 , Z)],
1+λ
1
[δF (W5 , W2 , Z) + λδF (W6 , W2 , Z)],
1+λ
1
1
δF (W5 , W1 , W6 ), δF (W, W2 , W6 ) =
δF (W5 , W2 , W6 ),
1+λ
1+λ
λ
λ
δF (W6 , W1 , W5 ), δF (W, W2 , W5 ) =
δF (W6 , W2 , W5 ).
1+λ
1+λ
On inserting these we obtain
[δF (W5 , W1 , Z)δF (W6 , Z, W4 ) − δF (W6 , W1 , Z)δF (W5 , Z, W4 )].
[δF (W5 , W2 , Z)δF (W6 , Z, W4 ) − δF (W6 , W2 , Z)δF (W5 , Z, W4 )].
δF (W1 , W6 , W2 )2 δF (W1 , W5 , W2 )2
+ δF (W5 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W6 , W1 , W5 )δF (W6 , W2 , W5 )δF (W1 , W6 , W2 )2
= k3 δF (W6 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W5 , W1 , W6 )δF (W5 , W2 , W6 )δF (W1 , W5 , W2 )2 .
The first bracket here equals
δF [(W5 , W6 ), (W1 , Z), (Z, W4 )] = δF (Z, W5 , W6 )δF (W4 , W1 , Z),
and the second equals
δF [(W5 , W6 ), (W2 , Z), (Z, W4 )] = δF (Z, W5 , W6 )δF (W4 , W2 , Z).
On inserting these we obtain
δF (Z, W5 , W6 )2 δF (Z, W4 , W1 )δF (Z, W4 , W2 )δF (W1 , W6 , W2 )2 δF (W1 , W5 , W2 )2
+ δF (W5 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W6 , W1 , W5 )δF (W6 , W2 , W5 )δF (W1 , W6 , W2 )2
= k3 δF (W6 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W5 , W1 , W6 )δF (W5 , W2 , W6 )δF (W1 , W5 , W2 )2 .
But for some constant µ,
W4 =
µ
1
W1 +
W2 ,
1+µ
1+µ
from which
µ
1
δF (W2 , W1 , Z), δF (W4 , W2 , Z) =
δF (W1 , W2 , Z),
1+µ
1+µ
1
[δF (W5 , Z, W1 ) + µδF (W5 , Z, W2 )],
δF (W5 , Z, W4 ) =
1+µ
1
δF (W6 , Z, W4 ) =
[δF (W6 , Z, W1 ) + µδF (W6 , Z, W2 )].
1+µ
δF (W4 , W1 , Z) =
On inserting these and taking out a common factor δF (Z, W1 , W2 )2 we obtain
(−µ)δF (Z, W5 , W6 )2 δF (W1 , W6 , W2 )2 δF (W1 , W5 , W2 )2
=
+ [δF (W5 , Z, W1 ) + µδF (W5 , Z, W2 )]2 δF (W6 , W1 , W5 )δF (W6 , W2 , W5 )δF (W1 , W6 , W2 )2
k3 [δF (W6 , Z, W1 ) + µδF (W6 , Z, W2 )]2 δF (W5 , W1 , W6 )δF (W5 , W2 , W6 ).
This is clearly the equation of a conic.
451
14.12. ANALOGUE OF OBLIQUE AXIAL SYMMETRIES
14.12
ANALOGUE OF OBLIQUE AXIAL SYMMETRIES
14.12.1
With the ellipse
y2
x2
+ 2 = 1,
2
a
b
take a point Z0 =
6 (0, 0). Let l1 be a line through Z0 with equation
y − y0 = µ(x − x0 ),
and let Z1 be the pole of l1 so that
(x1 , y1 ) =
−b2
µa2
,
µx0 − y0 µx0 − y0
.
For a variable point Z, points on the line Z1 Z have the form
1
λ
Z1 +
Z,
1+λ
1+λ
and this meets the line l1 at a point Z2 ≡ (x2 , y2 ) given by
where
Then
x2 − x0
=
y2 − y0
=
1
λ
(x1 − x0 ) +
(x − x0 ),
1+λ
1+λ
λ
1
(y1 − y0 ) +
(y − y0 ),
1+λ
1+λ
1
λ
λ
1
(y1 − y0 ) +
(y − y0 ) = µ
(x1 − x0 ) +
(x − x0 ) .
1+λ
1+λ
1+λ
1+λ
λ=
With this λ we have
−µ(x1 − x0 ) + y1 − y0
.
µ(x − x0 ) − (y − y0 )
Z2 =
so that
Z=
1
λ
Z1 +
Z,
1+λ
1+λ
λ+1
1
Z1 −
Z2 .
1 − (λ + 1)
1 − (λ + 1)
Then for (Z1 , Z2 , Z, Z ′ ) to be a harmonic range we take
Z′ =
λ+1
1
Z1 +
Z2 .
1 + (λ + 1)
1 + (λ + 1)
But
Z2 − Z0 =
1
λ
(Z1 − Z0 ) +
(Z − Z0 ),
1+λ
1+λ
Z ′ − Z0 =
2
λ
(Z1 − Z0 ) +
(Z − Z0 ).
2+λ
2+λ
and so
Hence
2[µ(x − x0 ) − (y − y0 )](x1 − x0 ) + [−µ(x1 − x0 ) + y1 − y0 ](x − x0 )
,
2µ(x − x0 ) − 2(y − y0 ) − µ(x1 − x0 ) + y1 − y0
2[µ(x − x0 ) − (y − y0 )](y1 − y0 ) + [−µ(x1 − x0 ) + y1 − y0 ](y − y0 )
.
y ′ − y0 =
2µ(x − x0 ) − 2(y − y0 ) − µ(x1 − x0 ) + y1 − y0
x′ − x0 =
This is a projective analogue of the oblique axial symmetries we considered in 5.10.
452
14.13
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
GENERALIZATIONS FROM THE HYPERBOLA
14.13.1
Consider a point Z4 on the hyperbola with equation
y2
x2
−
= 1.
a2
b2
The line through Z4 which is parallel to the asymptote with equation y = (b/a)x will meet the
other asymptote with equation y = −(b/a)x at the point
a x4
y4
b x4
y4
W4 ≡ ( ( − ), − ( − )).
2 a
b
2 a
b
Then δF (O, W4 , Z4 ) =
ab
4 ,
and this is constant as Z4 varies on the hyperbola.
b
b
b
Z4
b
O
b
W4
b
b
Figure 9.11.
We seek a projective generalization of this. Let Z2 Z3 be a line and W1 , W2 distinct points
of it. Let Z1 6∈ Z2 Z3 and consider points Z4 , Z such that if Z2 Z4 , Z2 Z meet Z1 Z3 in W4 , W ,
respectively, then
δF (Z1 , Z, W )δF (Z4 , W1 , W2 )δF (W4 , W1 , W2 )
= 1.
δF (Z1 , Z4 , W4 )δF (Z, W1 , W2 )δF (W, W1 , W2 )
Then with a variable Z we have that
δF (Z1 , Z, W )
= k.
δF (Z, W1 , W2 )δF (Z, W1 , W2 )
Let
W =
1
λ
Z1 +
Z3 ,
1+λ
1+λ
and from this
δF (Z2 , Z, Z1 ) + λδF (Z2 , Z, Z1 ) = 0,
λ
δF (Z1 , Z, Z3 ),
1+λ
1
δF (W, W1 , W2 ) =
δF (Z1 , W1 , W2 ).
1+λ
On inserting these we have that
δF (Z1 , Z, W ) =
δF (Z, Z1 , Z2 )δF (Z, Z3 , Z1 ) = kδF (Z1 , W1 , W2 )δF (Z, W1 , W2 )δF (Z, Z2 , Z3 ).
But
W1 =
ρ
1
σ
1
Z2 +
Z3 , W2 =
Z2 +
Z3 ,
1+ρ
1+ρ
1+σ
1+σ
453
14.13. GENERALIZATIONS FROM THE HYPERBOLA
and from these
δF (Z, W1 , W2 ) =
σ−ρ
δF (Z, Z2 , Z3 ).
(1 + ρ)(1 + σ)
Hence
δF (Z, Z3 , Z1 )δF (Z, Z1 , Z2 ) = k1 δF (Z, Z2 , Z3 )2 ,
for some constant k1 . From its form, this locus is a conic through Z2 and Z3 , with respect to which
Z2 Z3 is the polar of Z1 .
b
W
Z
b
b
W2
W1
b
Z3
b
b
b
W4
Z4
b
b
Z2
Z1
Figure 9.12
14.13.2
For the hyperbola with equation
y2
x2
−
= 1,
a2
b2
we take the point W4 ≡ (ar, br) on one asymptote and the distinct point W5 ≡ (as, −bs) on the
other asymptote. Points Z ≡ (x, y) on W4 W5 have coordinates of the form
x = ar + a(s − r)t, y = br + b(−s − r)t.
If the line W4 W5 meets the hyperbola it is in points Z4 , Z5 with parameters t4 , t5 satisfying the
equation
−4rst2 + 4rst − 1 = 0.
Then t4 + t5 = 1 so that t5 = 1 − t4 . It follows that
W4 Z4
= 1,
Z5 W5
and
W4 Z5
= 1.
Z4 W5
454
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
W4
b
b
Z4
Z5
b
b
W5
Figure 9.14?.
As a projectively invariant generalization of this, we have the following. Let Z2 , Z3 be points on
a proper conic and Z1 the pole of Z2 Z3 . Let W1 , W2 ∈ Z2 Z3 , and for points W4 ∈ Z1 Z2 , W5 ∈ Z1 Z3
let W4 W5 meet the conic in the points Z4 , Z5 . Then
δF (W4 , W2 , Z4 )δF (W1 , Z5 , W2 )δF (W1 , W5 , W2 )
= 1,
δF (Z5 , W2 , W5 )δF (W1 , W4 , W2 )δF (W1 , Z4 , W2 )
and
δF (W4 , W2 , Z5 )δF (W1 , Z4 , W2 )δF (W1 , W5 , W2 )
= 1.
δF (Z4 , W2 , W5 )δF (W1 , W4 , W2 )δF (W1 , Z5 , W2 )
b
W5
Z5
b
b
W2
W1
b
b
Z3
Z4
b
b
W4
b
Z2
b
Z1
Figure 9.15
14.14
NEWTON’S ORGANIC METHOD OF GENERATION
14.14.1
Let W5 W6 be a fixed line and W1 , W2 , W3 , W4 fixed points not on it. For a point W varying on
W5 W6 consider the locus of the point Z such that
δF (W4 , W1 , W )δF (W4 , W2 , Z)
δF (W3 , W1 , W )δF (W3 , W2 , Z)
= k1 ,
= k2 ,
δF (W3 , W1 , Z)δF (W3 , W2 , W )
δF (W4 , W1 , Z)δF (W4 , W2 , W )
for non-zero constants k1 , k2 . Then
W =
1
λ
W5 +
W6 ,
1+λ
1+λ
455
14.15. ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS
for a variable λ. From this we have that
δF (W3 , W1 , W5 ) + λδF (W3 , W1 , W6 )
δF (W3 , W2 , W5 ) + λδF (W3 , W2 , W6 )
δF (W4 , W1 , W5 ) + λδF (W4 , W1 , W6 )
δF (W4 , W2 , W5 ) + λδF (W4 , W2 , W6 )
δF (W3 , W1 , Z)
,
δF (W3 , W2 , Z)
δF (W4 , W1 , Z)
= k2
.
δF (W4 , W2 , Z)
= k1
On eliminating λ between these, we obtain
=
δF (W3 , W1 , W5 )δF (W3 , W2 , Z) − k1 δF (W3 , W2 , W5 )δF (W3 , W1 , Z)
−δF (W3 , W1 , W6 )δF (W3 , W2 , Z) + k1 δF (W3 , W2 , W6 )δF (W3 , W1 , Z)
δF (W4 , W1 , W5 )δF (W4 , W2 , Z) − k2 δF (W4 , W2 , W5 )δF (W4 , W1 , Z)
.
−δF (W4 , W1 , W6 )δF (W4 , W2 , Z) + k2 δF (W4 , W2 , W6 )δF (W4 , W1 , Z)
Thus the locus of Z is a conic.
W4
b
b
W1
b
W6
b
W
Z
b
b
b
W5
b
W2
W3
Figure 9.13?.
This is a projective analogue of Newton’s organic method of generating a conic in 1686. The
diagram corresponds to
W1 ≡ (1, 1), W2 ≡ (−1, −1), W3 ≡ (1, −1), W4 ≡ (−1, 1), W5 ≡ (−5, −1.25), W6 ≡ (3, 0.75)
and k1 = −1/2, k2 = 2.
14.15
ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS
We now give a second extension to some material considered in 5.6.
14.15.1
Analogue of centroid
Let C be a proper conic, and Z0 a point not on it and not the centre in the case of a central conic.
Let the line l be the polar of Z0 with respect to C, and Z1 , Z2 and Z3 be non- collinear points on
the conic. With (Z1 , Z2 , Z3 ) as the basis of our areal coordinates, C will have an equation of the
form
f βγ + gγα + hαβ = 0,
and the polar l will have an equation
(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ = 0.
456
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
We suppose that the line Z2 Z3 meets l in a point Z4 ; taking points Z on Z2 Z3 , of the form
Z=
λ
1
Z2 +
Z3 ,
1+λ
1+λ
for Z4 we have
λ=−
f γ0 + hα0
,
gα0 + f β0
and then for the point Z5 which is the harmonic conjugate of Z4 with respect to {Z2 , Z3 } we have
λ=
f γ0 + hα0
,
gα0 + f β0
and so Z5 has areal coordinates
1
1+
f γ0 +hα0
gα0 +f β0
f γ0 + hα0
δ1 ,
0, δ1 ,
gα0 + f β0
where δ1 = δF (Z1 , Z2 , Z3 ). From these we find that the line Z1 Z5 has equation
−(f γ0 + hα0 )β + (gα0 + f β0 )γ = 0.
Z3
b
b
Z5
b
b
b
Z2
b
b
b
Z4
Z0
b
b
Z1
b
b
Figure 9.16
We also suppose that the line Z3 Z1 meets l in a point Z6 , and that the point Z7 is the harmonic
conjugate of Z6 with respect to {Z3 , Z1 }, and that the line Z1 Z2 meets l in a point Z8 , and that
the point Z9 is the harmonic conjugate of Z8 with respect to {Z1 , Z2 }. Then as above, the lines
Z2 Z7 and Z3 Z9 have equations
(gγ0 + hβ0 )α − (f β0 + gα0 )γ
(−(gγ0 + hβ0 )α + (hα0 + f γ0 )β
= 0,
= 0.
On forming the determinant of the coefficients of the equations of the lines Z1 Z5 , Z2 Z7 and Z3 Z9 ,
we find that these lines are concurrent or all parallel. In the former case, we regard the point of
concurrency as an analogue of the centroid of the triangle [Z1 , Z2 , Z3 ] in 5.6.
457
14.15. ANALOGUES OF SOME AFFINE AND SIMILARITY RESULTS
14.15.2
Analogue of orthocentre
Continuing with the notation of 9.9.1, the polar of Z4 has equation
[−g(hα0 + f γ0 ) + h(f β0 + gα0 )]α − f (hα0 + f γ0 )β + f (f β0 + gα0 )γ0 = 0,
and we suppose that it meets l at a point Z10 . Now any line through Z10 has an equation of the
form
[−g(hα0 + f γ0 ) + h(f β0 + gα0 )]α − f (hα0 + f γ0 )β + f (f β0 + gα0 )γ0
+ k[(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0.
We choose k so that this passes through Z1 and in this way find that Z1 Z10 has equation
(hα0 + f γ0 )[−f (gγ0 + hβ0 ) + g(hα0 + f γ0 ) − h(f β0 + gα0 )]β
(f β0 + gα0 )[f (gγ0 + hβ0 ) + g(hα0 + f γ0 ) − f (f β0 + gα0 )]γ = 0.
Similarly, we suppose that the polar of Z6 meets l at a point Z11 and we find that Z2 Z11 has
an equation
(gγ0 + hβ0 )[g(hα0 + f γ0 ) + h(f β0 + gα0 ) − f (gγ0 + hβ0 )]α
(f β0 + gα0 )[−g(hα0 + f γ0 ) + h(f β0 + gα0 ) − f (gγ0 + hβ0 )]γ = 0.
Similarly, we suppose that the polar of Z8 meets l at a point Z12 and we find that Z3 Z12 has
an equation
(gγ0 + hβ0 )[−h(f β0 + gα0 ) + f (gγ0 + hβ0 ) − g(hα0 + f γ0 )]α
(f β0 + gα0 )[h(f β0 + gα0 ) + f (gγ0 + hβ0 ) − g(hα0 + f γ0 )]β = 0.
Z3
b
b
b
b
b
Z2
b
b
b
Z4
Z0
b
b
b
Z1
b
b
b
Z10
Figure 9.17
By forming the determinant of the coefficients in the equations of the lines Z1 Z10 , Z2 Z11 and
Z3 Z12 , we find that these lines are either concurrent or all parallel. In the former case, we regard
the point of concurrency as an analogue of the generalized orthocentre in the context of 5.6.3.
458
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
14.15.3
Analogue of the Euler line
The analogue of centroid in 9.9.1 has areal coordinates proportional to
(f β0 + gα0 )(hα0 + f γ0 ), (gγ0 + hβ0 )(f β0 + gα0 ), (hα0 + f γ0 )(gγ0 + hβ0 )(f β0 + gα0 )).
The line joining this point to Z0 has equation
α0 (hβ0 − gγ0 )(hβ0 + gγ0 )α + β0 (f γ0 − hα0 )(f γ0 + hα0 )β + γ0 (gα0 − f β0 )(gα0 + f β0 )γ = 0,
and if we insert the coordinates of this line in the last determinant instead of those for Z3 Z12
we obtain that Z0 , the analogue of centroid and the analogue of orthocentre are collinear, thus
yielding a further extension of the Euler line 5.6.5.
14.15.4
Analogue of the Wallace-Simson line
Still with the situation in 9.9.1, we consider a variable point Z ′ with areal coordinates (α′ , β ′ , γ ′ ).
Any line through Z10 (where the polar of Z4 meets l) has an equation of the form
[−g(hα0 + f γ0 ) + h(f β0 + gα0 )]α − f (hα0 + f γ0 )β + f (f β0 + gα0 )γ
+ k1 [(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0,
and we choose k1 so that this passes through Z ′ . This line Z10 Z ′ meets the line Z2 Z3 in a point
Z13 with coordinates proportional to
(0, (f + k1 )(f β0 + gα0 ), (f − k1 )(hα0 + f γ0 )),
with the above choice of k1 .
Similarly any line through Z11 (where the polar of Z6 meets l) has an equation of the form
g(gγ0 + hβ0 )α + [f (gγ0 + hβ0 ) − h(f β0 + gα0 )]β − g(f β0 + gα0 )γ
+ k2 [(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0,
and we choose k2 so that this passes through Z ′ . This line Z11 Z ′ meets the line Z3 Z1 in a point
Z14 with coordinates proportional to
((g − k2 )(f β0 + gα0 ), 0, (g + k2 )(gγ0 + hβ0 )),
with the above choice of k2 .
Similarly any line through Z12 (where the polar of Z8 meets l) has an equation of the form
−h(gγ0 + hβ0 )α + h(hα0 + f γ0 )β + [−f (gγ0 + hβ0 ) + g(hα0 + f γ0 )]γ
+ k3 [(gγ0 + hβ0 )α + (hα0 + f γ0 )β + (f β0 + gα0 )γ] = 0,
and we choose k3 so that this passes through Z ′ . This line Z12 Z ′ meets the line Z1 Z2 in a point
Z15 with coordinates proportional to
((h + k3 )(hα0 + f γ0 ), (h − k3 )(gγ0 + hβ0 )),
with the above choice of k3 .
459
14.16. ANALOGUE OF POWER PROPERTY OF CIRCLES
Z′
Z3
b
b
b
Z13
b
b
b
b
Z2
b
b
b
Z4
Z0
b
b
b
Z1
b
b
b
Z10
Figure 9.18
The points Z13 , Z14 and Z15 are collinear if and only if
2(f β0 + gα0 )(gγ0 + hβ0 )(hα0 + f γ0 )(f gh + f k2 k3 + gk3 k1 + hk1 k2 ) = 0.
On assuming that the polar of Z0 contains none of the points Z1 , Z2 , Z3 , we can discard the first
three factors after 2, and then on inserting for the values of k1 , k2 and k3 we find that the above
points are collinear if and only if
f β ′ γ ′ + gγ ′ α′ + hα′ β ′ = 0,
that is, if and only if Z ′ lies on the conic. This is an analogue of the extension of the Wallace-Simson
property noted in 5.6.6.
14.16
ANALOGUE OF POWER PROPERTY OF CIRCLES
14.16.1
Can we obtain a projectively invariant version of the material of 7.6? The extension of concepts
to the extended plane suggests the following.
460
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Z2
b
b
Z0
b
b
Z3
b
Figure 9.19.
b
Z1
Let Ck be a proper conic, Z0 a point not on it, and W1 , W2 distinct points on the polar l0 of
Z0 . Let C be a proper conic which meets Ck at two points on the polar l0 . For a variable point Z1
on Ck let the line Z0 Z1 meet the conic C in the points Z2 , Z3 . Then as Z1 varies,
E=
δF (Z0 , W2 , Z2 ) δF (W1 , Z1 , W2 ) δF (Z0 , W2 , Z3 ) δF (W1 , Z1 , W2 )
δF (Z0 , W2 , Z1 ) δF (W1 , Z2 , W2 ) δF (Z0 , W2 , Z1 ) δF (W1 , Z3 , W2 )
is constant.
We take areal coordinates (δ1 , 0, 0), (0, δ1 , 0) for the points in which Ck meets the polar l0 , and
(0, 0, δ1 ) for Z0 . Then by 4.2.1, Ck has an equation of the form
γ 2 = 4jαβ,
and so has parametric equations with (α, β, γ) proportional to (θ2 , j, 2jθ). The points W1 , W2 on
the polar l0 with equation γ = 0 have coordinates
((1 − s1 )δ1 , s1 δ1 , 0), ((1 − s2 )δ1 , s2 δ1 , 0).
As the conic C passes through the points with coordinates (δ1 , 0, 0), (0, δ1 , 0), it has an equation of
the form
cγ 2 + 2f βγ + 2gγα + 2hαβ = 0.
Points on the line Z0 Z1 have coordinates of the form
(tθ2 , tj, δ1 + t(2jθ − δ1 )).
This is on the conic C when t satisfies the quadratic equation
t2 [c(2jθ − δ1 )2 + 2f j(2jθ − δ1 ) + 2g(2jθ − δ1 )θ2 + 2hjθ2 ]
+ 2t[cδ1 (2jθ − δ1 ) + f jδ1 + gδ1 θ2 ] + cδ12 = 0.
Let the roots of this be t2 and t3 , so that Z2 and Z3 have coordinates
(t2 θ2 , t2 j, δ1 + t2 (2jθ − δ1 ), (t3 θ2 , t3 j, δ1 + t3 (2jθ − δ1 ).
461
14.17. ANALOGUE OF PROTO-FOCI
Then we find that
2δF (Z0 , W2 , Z2 )
2δF (W1 , Z1 , W2 )
= δ12 t2 [(1 − s2 )j − s2 θ2 ],
= −2jδ12 θ(s2 − s1 ),
2δF (W1 , Z2 , W2 )
= −(s2 − s1 )δ12 [δ1 + t2 (2jθ − δ1 )],
= δ12 t3 [(1 − s2 )j − s2 θ2 ],
= δ12 [(1 − s2 )j − s2 θ2 ],
2δF (Z0 , W2 , Z3 )
2δF (Z0 , W2 , Z1 )
= −(s2 − s1 )δ12 [δ1 + t3 (2jθ − δ1 )].
2δF (W1 , Z3 , W2 )
On inserting these and performing cancellations, we find that the equipoised quotient E is equal
to
4j 2 θ2 t2 t3
E= 2
.
δ1 + δ1 (2jθ − δ1 )(t2 + t3 ) + (2jθ − δ1 )2 t2 t3
But from the quadratic equation in t,
t2 t3 =
cδ12
,
c(2jθ − δ1 )2 + 2f j(2jθ − δ1 ) + 2g(2jθ − δ1 )θ2 + 2hjθ2
cδ1 (2jθ − δ1 ) + f jδ1 + gδ1 θ2
.
t2 [c(2jθ − δ1 )2 + 2f j(2jθ − δ1 ) + 2g(2jθ − δ1 )θ2 + 2hjθ2
On inserting these, we find that
2jc
.
E=
h
Thus we add an assumption that h 6= 0 to make our proof valid.
t2 + t3 = −2
14.17
ANALOGUE OF PROTO-FOCI
14.17.1
We now consider a projective version of the property of the proto-focus of a central conic considered
in 7.4. Let W1 , W2 , W4 be distinct collinear points and W5 , W6 distinct points not on W1 W2 .
Consider the locus of points Z 6∈ W1 W2 such that if W4 Z meets W5 W6 at W , then
δF (W, W1 , Z)δF (W, W2 , Z)δF (W1 , W6 , W2 )2
δF (W, W1 , W5 )δF (W, W2 , W5 )δF (W1 , W6 , W2 )2
+
δF (W, W1 , W6 )δF (W, W2 , W6 )δF (W1 , Z, W2 )2
δF (W, W1 , W6 )δF (W, W2 , W6 )δF (W1 , W5 , W2 )2
= k3 .
b
Z
b
b
W5
W
b
b
b
W4 W1
W6
b
b
W2
Figure 9.20.
462
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Let
W =
1
λ
W5 +
W6 ,
1+λ
1+λ
and then δF (W, Z, W4 ) = 0 gives
δF (W5 , Z, W4 ) + λδF (W6 , Z, W4 ) = 0.
Now
δF (W, W1 , Z) =
δF (W, W2 , Z) =
δF (W, W1 , W6 ) =
δF (W, W1 , W5 ) =
1
[δF (W5 , W1 , Z) + λδF (W6 , W1 , Z)],
1+λ
1
[δF (W5 , W2 , Z) + λδF (W6 , W2 , Z)],
1+λ
1
1
δF (W5 , W1 , W6 ), δF (W, W2 , W6 ) =
δF (W5 , W2 , W6 ),
1+λ
1+λ
λ
λ
δF (W6 , W1 , W5 ), δF (W, W2 , W5 ) =
δF (W6 , W2 , W5 ).
1+λ
1+λ
On inserting these we obtain
[δF (W5 , W1 , Z)δF (W6 , Z, W4 ) − δF (W6 , W1 , Z)δF (W5 , Z, W4 )].
[δF (W5 , W2 , Z)δF (W6 , Z, W4 ) − δF (W6 , W2 , Z)δF (W5 , Z, W4 )].
δF (W1 , W6 , W2 )2 δF (W1 , W5 , W2 )2
+ δF (W5 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W6 , W1 , W5 )δF (W6 , W2 , W5 )δF (W1 , W6 , W2 )2
= k3 δF (W6 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W5 , W1 , W6 )δF (W5 , W2 , W6 )δF (W1 , W5 , W2 )2 .
The first bracket here equals
δF [(W5 , W6 ), (W1 , Z), (Z, W4 )] = δF (Z, W5 , W6 )δF (W4 , W1 , Z),
and the second equals
δF [(W5 , W6 ), (W2 , Z), (Z, W4 )] = δF (Z, W5 , W6 )δF (W4 , W2 , Z).
On inserting these we obtain
δF (Z, W5 , W6 )2 δF (Z, W4 , W1 )δF (Z, W4 , W2 )δF (W1 , W6 , W2 )2 δF (W1 , W5 , W2 )2
+ δF (W5 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W6 , W1 , W5 )δF (W6 , W2 , W5 )δF (W1 , W6 , W2 )2
= k3 δF (W6 , Z, W4 )2 δF (W1 , Z, W2 )2 δF (W5 , W1 , W6 )δF (W5 , W2 , W6 )δF (W1 , W5 , W2 )2 .
But for some constant µ,
W4 =
µ
1
W1 +
W2 ,
1+µ
1+µ
from which
µ
1
δF (W2 , W1 , Z), δF (W4 , W2 , Z) =
δF (W1 , W2 , Z),
1+µ
1+µ
1
[δF (W5 , Z, W1 ) + µδF (W5 , Z, W2 )],
δF (W5 , Z, W4 ) =
1+µ
1
δF (W6 , Z, W4 ) =
[δF (W6 , Z, W1 ) + µδF (W6 , Z, W2 )].
1+µ
δF (W4 , W1 , Z) =
On inserting these and taking out a common factor δF (Z, W1 , W2 )2 we obtain
(−µ)δF (Z, W5 , W6 )2 δF (W1 , W6 , W2 )2 δF (W1 , W5 , W2 )2
=
+ [δF (W5 , Z, W1 ) + µδF (W5 , Z, W2 )]2 δF (W6 , W1 , W5 )δF (W6 , W2 , W5 )δF (W1 , W6 , W2 )2
k3 [δF (W6 , Z, W1 ) + µδF (W6 , Z, W2 )]2 δF (W5 , W1 , W6 )δF (W5 , W2 , W6 ).
This is clearly the equation of a conic.
463
14.18. ANALOGUE OF OBLIQUE AXIAL SYMMETRIES
14.18
ANALOGUE OF OBLIQUE AXIAL SYMMETRIES
14.18.1
With the ellipse
y2
x2
+ 2 = 1,
2
a
b
take a point Z0 =
6 (0, 0). Let l1 be a line through Z0 with equation
y − y0 = µ(x − x0 ),
and let Z1 be the pole of l1 so that
(x1 , y1 ) =
−b2
µa2
,
µx0 − y0 µx0 − y0
.
For a variable point Z, points on the line Z1 Z have the form
1
λ
Z1 +
Z,
1+λ
1+λ
and this meets the line l1 at a point Z2 ≡ (x2 , y2 ) given by
where
Then
x2 − x0
=
y2 − y0
=
1
λ
(x1 − x0 ) +
(x − x0 ),
1+λ
1+λ
λ
1
(y1 − y0 ) +
(y − y0 ),
1+λ
1+λ
1
λ
λ
1
(y1 − y0 ) +
(y − y0 ) = µ
(x1 − x0 ) +
(x − x0 ) .
1+λ
1+λ
1+λ
1+λ
λ=
With this λ we have
−µ(x1 − x0 ) + y1 − y0
.
µ(x − x0 ) − (y − y0 )
Z2 =
so that
Z=
1
λ
Z1 +
Z,
1+λ
1+λ
λ+1
1
Z1 −
Z2 .
1 − (λ + 1)
1 − (λ + 1)
Then for (Z1 , Z2 , Z, Z ′ ) to be a harmonic range we take
Z′ =
λ+1
1
Z1 +
Z2 .
1 + (λ + 1)
1 + (λ + 1)
But
Z2 − Z0 =
1
λ
(Z1 − Z0 ) +
(Z − Z0 ),
1+λ
1+λ
Z ′ − Z0 =
2
λ
(Z1 − Z0 ) +
(Z − Z0 ).
2+λ
2+λ
and so
Hence
2[µ(x − x0 ) − (y − y0 )](x1 − x0 ) + [−µ(x1 − x0 ) + y1 − y0 ](x − x0 )
,
2µ(x − x0 ) − 2(y − y0 ) − µ(x1 − x0 ) + y1 − y0
2[µ(x − x0 ) − (y − y0 )](y1 − y0 ) + [−µ(x1 − x0 ) + y1 − y0 ](y − y0 )
.
y ′ − y0 =
2µ(x − x0 ) − 2(y − y0 ) − µ(x1 − x0 ) + y1 − y0
x′ − x0 =
This is a projective analogue of the oblique axial symmetries we considered in 5.10.
464
CHAPTER 14. FURTHER PROJECTIVE METHODS AND RESULTS
Chapter 15
Inversive Methods and Results
15.1
GENERAL INVERSION
15.1.1
Definition of inversion
Let C1 be a proper conic with equation
a1 x2 + 2h1 xy + b1 y 2 = 1,
so that C1 is an ellipse/circle or hyperbola according as a1 b1 − h21 is positive or negative.
The polar of a point W ≡ (u, v) with respect to C1 has equation
a1 ux + h1 (vx + uy) + b1 vy = 1,
and this meets the line OW in the point Z ≡ (x, y) where
x=
v
u
, y=
.
a1 u2 + 2h1 uv + b1 v 2
a1 u2 + 2h1 uv + b1 v 2
(15.1.1)
Because of 4.? we expect this to be symmetrical in Z and W . In fact it is so as by (10.1.1),
a1 x2 + 2h1 xy + b1 y 2 =
a1 u2 + 2h1 uv + b1 v 2
1
=
,
(a1 u2 + 2h1 uv + b1 v 2 )2
a1 u2 + 2h1 uv + b1 v 2
and so from (20.1.1) we also have
u=
a1
x2
x
y
, v=
.
2
2
+ 2h1 xy + b1 y
a1 x + 2h1 xy + b1 y 2
(15.1.2)
With this introduction, we now let
E1 = {Z ≡ (x, y) : a1 x2 + 2h1 xy + b1 y 2 = 0},
so that E1 consists of O in the case of a circle/ellipse and is the pair of asymptotes in the case of a
hyperbola. Then (10.1.1) defines a bijection from Π \ E1 , and its inverse is given by (10.1.2). We
call it an inversion with respect to C1 .
15.1.2
Lineo-loci
Given any line in Π with equation in W ,
lu + mv + n − 0,
465
(15.1.3)
466
CHAPTER 15. INVERSIVE METHODS AND RESULTS
the corresponding points Z satisfy the equation
lx + my + n(a1 x2 + 2h1 xy + b1 y 2 ) = 0.
(15.1.4)
The solution set of (10.2.2) in Π \ E1 , when non-empty, we call a lineo-locus, and then its solution
set in Π we call an extended lineo-locus. Note that each extended lineo- locus passes through
O, and when n 6= 0 is homothetic to C1 , that is when not straight. The straight extended lineo-loci
are the lines that pass through O, apart from the asymptotes in the case of a hyperbola.
We note that the inversion (10.1.1) maps the restriction of the line (10.2.1) to Π \ E1 to the
lineo-locus defined by (10.2.2).
15.1.3
Parallel lineo-loci
We say that lineo-loci
l1 x + m1 y + n1 (a1 x2 + 2h1 xy + b1 y 2 ) = 0, l2 x + m2 y + n2 (a1 x2 + 2h1 xy + b1 y 2 ) = 0,
are parallel when l1 m2 − l2 m1 = 0. Then two lineo-loci are parallel when their extensions have a
common tangent at O, and in fact distinct straight lineo-loci are not parallel.
15.2
PARAMETRIC EQUATIONS FOR A LINEO-LOCUS
15.2.1
Parametric equations
If W1 , W2 are distinct points on the restriction of the line (10.2.1) to Π \ E1 , then that restriction
will have parametric equations
u = u1 + t(u2 − u1 ), v = v1 + t(v2 − v1 ),
where t ∈ R, but we exclude any values of t for which W ∈ E1 .As any such line meets E1 in at
most two points, at most two parametric values of t are to be excluded.
Then
x =
y
=
u1 + t(u2 − u1 )
,
a1 [u1 + t(u2 − u1 + 2h1 [v1 + t(v2 − v1 )] + b1 [v1 + t(v2 − v1 )]2
v1 + t(v2 − v1 )
,
a1 [u1 + t(u2 − u1 )]2 + 2h1 [v1 + t(v2 − v1 )] + b1 [v1 + t(v2 − v1 )]2
)]2
(15.2.1)
(15.2.2)
with the same parametric domain, gives parametric equations of the unique lineo-locus through
Z1 and Z2 .
15.2.2
Lineo-segments and lineo-half-loci
Increasing t, or decreasing t, on the parametric domain, will induce a natural order on the lineolocus, and we can define lineo-segments and lineo-half-loci by inequalities on t. We have to
exercise care if the parametric domain excludes some points of R, and in any case it is t = ∞ that
yields O. When n = 0, (10.2.1) has parametric equations
u = mt, v = −lt,
and then under (10.1.1)
x =
y
=
m
1
,
2
a1
− 2h1 lm + b1 l t
1
−l
.
a1 m2 − 2h1 lm + b1 l2 t
m2
15.3. DUO-ANGLES BETWEEN TANGENTS
467
Now cross-ratio is invariant under a transformation s = k/t, so we can use cross-ratio to identify
straight lineo-segments and lineo-half-loci.
More generally, from (10.2.1) we have
y
v1 + t(v2 − v1 )
=
,
x
u1 + t(u2 − u1 )
so we can parametrize the general lineo-locus in terms of the slope m = y/x. We can then use
cross-ratios to specify the analogues of segments and half- lines.
15.2.3
Analogue of mid-points
We take t = 21 for the analogue of the mid-point Z0 of Z1 and Z2 . In the case where Z1 Z2 contains
O, we have that (Z1 , Z2 , O, Z0 is a half-range. For O 6∈ Z1 Z2 , Z0 is on the unique lineo- locus
through Z1 and Z2 , and O(Z1 , Z2 , J, Z0 ) is a harmonic pencil, OJ being the y-axis.
15.3
Duo-angles between tangents
From calculus, we know that the tangent to a lineo-locus at a point Z ≡ (x, y) on it, is the line
through Z with slope the derivative m = dy/dx. Then by the chain rule
m=
(a1 u2 − b1 v 2 )µ − 2v(a1 u + h1 v)
,
−a1 u2 + b1 v 2 − 2u(h1 u + b1 v)µ
where µ = dv/du.
For conjugate directions with respect to C1 we have
a1 + h1 (m1 + m2 ) + b1 m1 m2 = 0,
for the slopes m1 and m2 and on inserting into this
m1
=
m2
=
(a1 u2 − b1 v 2 )µ1 − 2v(a1 u + h1 v)
,
−a1 u2 + b1 v 2 − 2u(h1 u + b1 v)µ1
(a1 u2 − b1 v 2 )µ2 − 2v(a1 u + h1 v)
,
−a1 u2 + b1 v 2 − 2u(h1 u + b1 v)µ2
we obtain
(a1
u2
− b1
v2
(a1 u2 + 2h1 uv + b1 v 2 )2
[a1 + h1 (µ1 + µ2 ) + b1 µ1 µ2 ].
+ 2u(h1 u + b1 v)µ1 )(a1 u2 − b1 v 2 + 2u(h1 u + b1 v)µ2 )
Thus we must have that
a1 + h1 (µ1 + µ2 ) + b1 µ1 µ2 = 0,
implies
a1 + h1 (m1 + m2 ) + b1 m1 m2 = 0.
Thus lines which are in conjugate directions with respect to C1 map to lineo-loci, the tangents to
which at the point of intersection, are also in conjugate directions with respect to C1 .
15.4
CIRCULAR INVERSION
15.4.1
Circular inversion
As the general expressions need extensions of our basic concepts to be identified, from now on we
take a1 = b1 = 1, h1 = 0 so that C1 is the unit circle. Then we have
u
v
x= 2
, y= 2
,
u + v2
u + v2
468
CHAPTER 15. INVERSIVE METHODS AND RESULTS
so that
x + ıy =
u + ıv
1
=
.
u2 + v 2
u − ıv
We can write this as
1
,
w∗
z=
where we are now using
15.4.2
∗
to denote complex conjugate.
Complex-valued distance
Now
w2 − w1 = −
z2∗ − z1∗
,
z1∗ z2∗
and so our analogue of complex-valued distance from Z1 to Z2 is
−
Z1 Z2
∗
∗
OZ1 OZ2
∗.
Thus for any points Z1 , Z2 , Z3 in Π \ {O}, we have
−
Z1 Z2
∗
∗
OZ1 OZ2
∗
−
Z2 Z3
∗
∗
OZ2 OZ3
∗
=−
Z1 Z3
∗
∗
OZ1 OZ3
∗,
and if Z3 is between Z1 and Z2 on a circular lineo-locus,
|Z2 , Z3 |
|Z1 , Z3 |
|Z1 , Z2 |
+
=
.
|O, Z1 ||O, Z2 | |O, Z2 ||O, Z3 |
|O, Z1 ||O, Z3 |
We note that this last is Ptolemy’s theorem.
15.4.3
Angle-measure
Now
w3 − w1
z∗
= 2∗
w2 − w1
z3
z3 − z1
z2 − z1
∗
,
and so
=
|W1 , W3 |
cis ∡F W2 W1 W3
|W1 , W2 |
|O, Z2 ||Z1 , Z3 |
cis (∡F Z2 OZ3 − ∡F Z2 Z1 Z3 ).
|O, Z3 ||Z1 , Z2 |
It follows that
∡F W2 W1 W3 = ∡F Z2 OZ3 − ∡F Z2 Z1 Z3 .
As an application, we note that if W3 ∈ W1 W2 then the left-hand side is 0F or 180F , according
as W1 is not, or is, between W2 and W3 . Then for concyclic points O, Z1 , Z2 , Z3 ,
∡F Z2 OZ3 − ∡F Z2 Z1 Z3
is 0F or 180F , according as Z1 is not, or is, between Z2 and Z3 on the circle less O.
15.4. CIRCULAR INVERSION
15.4.4
469
Sensed-area
We have
δF (W1 , W2 , W3 )
x1
y1
2 2
x +y
x21 +y12
1 1x2 1
y2
=
2 22 x22 +y22
2 x2x+y
y3
3
x2 +y
2
x2 +y 2
3
=
=
3
3
1
1
1
3
x y1 x21 + y12 1 1
1
x2 y2 x22 + y22 2
2
2
2
2
2
(x1 + y1 )(x2 + y2 )(x3 + y3 ) 2 x3 y3 x23 + y32 1
|O, Z1 |2 δF (O, Z2 , Z3 ) + |O, Z2 |2 δF (O, Z3 , Z1 ) + |O, Z3 |2 δF (O, Z1 , Z2 ) .
2
2
2
|O, Z1 | |O, Z2 | |O, Z3 |
As a first application, we note that Z is on the circular lineo-locus through Z1 and Z2 if and
only if
|O, Z1 |2 δF (O, Z2 , Z) + |O, Z2 |2 δF (O, Z, Z1 ) + |O, Z|2 δF (O, Z1 , Z2 ) = 0.
As a second application, Z is on a lineo-locus parallel to the circular lineo-locus through Z1
and Z2 if and only if
1
|O, Z1
|2 |O, Z
2
|2 |O, Z|2
for some constant k.
|O, Z1 |2 δF (O, Z2 , Z) + |O, Z2 |2 δF (O, Z, Z1 ) + |O, Z|2 δF (O, Z1 , Z2 ) = k,
Exercises
10.1 For circular inversion, in the case of a parallelogram make out the analogue of equal senseddistances associated with opposite sides, and of the coincidence of mid-points of diagonals.
470
CHAPTER 15. INVERSIVE METHODS AND RESULTS
Appendix A
More Complicated Geometric
Algebra
Special Notation
Because the expressions which we encounter are complicated and take up a lot of space, we
introduce the following notation :
δF Z1 , Z2 , Z3 = Z(1,2,3) , δF Z, Z2 , Z3 = Z(∼,2,3) ,
δF Z 1 , Z 2 , Z 3 δF Z 4 , Z 5 , Z 6
(1,2,3)(4,5,6) δF Z, Z1 , Z3 δF Z, Z2 , Z4
(∼,1,3)(∼,2,4)
= Z(1,2,4)(3,5,6) ,
= Z(∼,1,4)(∼,2,3)
,
δF Z 1 , Z 2 , Z 4 δF Z 3 , Z 5 , Z 6
δF Z, Z1 , Z4 δF Z, Z2 , Z3
and will use obvious extensions of these.
A.1
Sensed-area, points given indirectly
A.1.1
Sensed-area, two points given indirectly
As preparation for the next subsection, we suppose that W1 W2 and W3 W4 are non-parallel lines
and denote by W ′ their point of intersection; similarly that W5 W6 and W7 W8 are non-parallel lines
and denote by W ′′ their point of intersection. For any point W we wish to consider δF (W ′ , W ′′ , W ).
Now we have that
W ′ = (1 − r)W1 + rW2 ,
W ′′ = (1 − s)W5 + sW6 ,
where by (1.1.17)
1−r =−
δF (W3 , W4 , W2 )
,
δF (W2 − W1 , W4 − W3 , O)
r=
δF (W3 , W4 , W1 )
,
δF (W2 − W1 , W4 − W3 , O)
s=
δF (W7 , W8 , W5 )
.
δF (W6 − W5 , W8 − W7 , O)
and by this applied to the second pair
1−s=−
δF (W7 , W8 , W6 )
,
δF (W6 − W5 , W8 − W7 , O)
Then
δF (W ′ , W ′′ , W )
=δF (1 − r)W1 + rW2 , (1 − s)W5 + sW6 , W )
=(1 − r)(1 − s)δF (W1 , W5 , W ) + (1 − r)sδF (W1 , W6 , W )
+ r(1 − s)δF (W2 , W5 , W ) + rsδF (W2 , W6 , W )
471
472
APPENDIX A. MORE COMPLICATED GEOMETRIC ALGEBRA
and so
δF (W ′ , W ′′ , W )δF (W2 − W1 , W4 − W3 , O)δF (W6 − W5 , W8 − W7 , O)
=δF (W3 , W4 , W2 )δF (W7 , W8 , W6 )δF (W1 , W5 , W ) − δF (W3 , W4 , W2 )δF (W7 , W8 , W5 )δF (W1 , W6 , W )
−δF (W3 , W4 , W1 )δF (W7 , W8 , W6 )δF (W2 , W5 , W ) + δF (W3 , W4 , W1 )δF (W7 , W8 , W5 )δF (W2 , W6 , W ).
This expression can be written in two ways
δF (W2 , W3 , W4 )δF [(W5 , W6 ), (W, W1 ), (W7 , W8 )] − δF (W1 , W3 , W4 )δF [(W5 , W6 ), (W, W2 ), (W7 , W8 )]
=δF (W6 , W7 , W8 )δF [(W1 , W2 ), (W5 , W ), (W3 , W4 )] − δF (W5 , W7 , W8 )δF [(W1 , W2 ), (W6 , W ), (W3 , W4 )].
(A.1.1)
We can express this as a determinant
δF (W2 , W3 , W4 ) δF [(W5 , W6 ), (W, W2 ), (W7 , W8 )]
δF (W1 , W3 , W4 ) δF [(W5 , W6 ), (W, W1 ), (W7 , W8 )]
.
Having (1.2.1) equal to 0 is then be a necessary and sufficient condition for W ′ , W ′′ and W to be
collinear.
Continuing with W1 , W2 , W3 , W4 , W5 , W6 , W7 , W8 , W ′ , W ′′ as given, for distinct points W9 , W10
we note that W ′ W ′′ k W9 W10 if and only if δF (W ′ , W ′′ , W9 ) − δF (W ′ , W ′′ , W10 ) = 0. Now this
last expression is equal to
W(2,3,4) W[(5,6),(9,1),(7,8)] − W(1,3,4) W[(5,6),(9,2),(7,8)] − W(2,3,4) W[(5,6),(10,1),(7,8)] + W(1,3,4) W[(5,6),(10,2),(7,8)]
=W(2,3,4) W[(9,1),(7,8),(5,6)] − W[(10,1),(7,8),(5,6)] − W(1,3,4) W[(9,2),(7,8),(5,6)] − W[(10,2),(7,8),(5,6)]
=W(2,3,4) W(9,7,8) W(1,5,6) − W(1,7,8) W(9,5,6) − W(10,7,8) W(1,5,6) + W(1,7,8) W(10,5,6)
− W(1,3,4) W(9,7,8) W(2,5,6) − W(2,7,8) W(9,5,6) − W(10,7,8) W(2,5,6) + W(2,7,8) W(10,5,6)
=W(2,3,4) W(9,7,8) − W(10,7,8) W(1,5,6) − W(2,3,4) W(9,5,6) − W(10,5,6) W(1,7,8)
− W(1,3,4) W(9,7,8) − W(10,7,8) W(2,5,6) − W(1,3,4) W(10,5,6) − W(9,5,6) W(2,7,8)
=W(2,3,4) W(1,5,6) δF (W9 − W10 , W7 − W8 , O) − W(1,3,4) W(1,7,8) δF (W9 − W10 , W5 − W6 , O)
(A.1.2)
Thus W ′ W ′′ k W9 W10 if and only if (1.2.2) is equal to 0.
A.2
Sensed-area, three points given indirectly
Suppose that W1 W2 and W3 W4 are non-parallel lines and denote by W ′ their point of intersection;
similarly suppose that W5 W6 and W7 W8 are non-parallel lines and denote by W ′′ their point of
intersection ; finally suppose that W9 W10 and W11 W12 are non-parallel lines and denote by W ′′′
their point of intersection. Then by (1.1.17) we have
W ′ = (1 − r)W1 + rW2 , W ′′ = (1 − s)W5 + sW6 , W ′′′ = (1 − t)W9 + tW10 ,
from which by (1.1.16) we have
W(1,3,4)
W(2,3,4)
W1 +
W2 ,
δF (W1 − W2 , W3 − W4 , O)
δF (W1 − W2 , W3 − W4 , O)
W(5,7,8)
W(6,7,8)
W5 +
W6 ,
W ′′ = −
δF (W5 − W6 , W7 − W8 , O)
δF (W5 − W6 , W7 − W8 , O)
W(10,11,12)
W(9,11,12)
W ′′′ = −
W9 +
W10 .
δF (W9 − W10 , W11 − W12 , O)
δF (W9 − W10 , W11 − W12 , O)
W′ = −
(A.2.3)
A.2. SENSED-AREA, THREE POINTS GIVEN INDIRECTLY
473
Then we can express δF (W ′ , W ′′ , W ′′′ ) by the following
δF (W1 − W2 , W3 − W4 , O)δF (W5 − W6 , W7 − W8 , O)δF (W9 − W10 , W11 − W12 , O)•
W(2,3,4)
W(1,3,4)
δF −
W1 +
W2 ,
δF (W1 − W2 , W3 − W4 , O)
δF (W1 − W2 , W3 − W4 , O)
W(5,7,8)
W(6,7,8)
W5 +
W6 ,
−
δF (W5 − W6 , W7 − W8 , O)
δF (W5 − W6 , W7 − W8 , O)
W(10,11,12)
W(9,11,12)
−
W9 +
W10
δF (W9 − W10 , W11 − W12 , O)
δF (W9 − W10 , W11 − W12 , O)
=j3 δF (W2 , W3 , W4 ) (1 − t)δF [(W5 , W6 ), (W9 , W1 ), (W7 , W8 )] + tδF [(W5 , W6 ), (W10 , W1 ), (W7 , W8 )]
−j3 δF (W1 , W3 , W4 ) (1 − t)δF [(W5 , W6 ), (W9 , W2 ), (W7 , W8 )] + tδF [(W5 , W6 ), (W10 , W2 ), (W7 , W8 )] .
= − δF (W2 , W3 , W4 ) δF (W11 , W12 , W10 )δF [(W5 , W6 ), (W9 , W1 ), (W7 , W8 )]
− δF (W11 , W12 , W9 )δF [(W5 , W6 ), (W10 , W1 ), (W7 , W8 )]
+ δF (W1 , W3 , W4 ) δF (W11 , W12 , W10 )δF [(W5 , W6 ), (W9 , W2 ), (W7 , W8 )]
− δF (W11 , W12 , W9 )δF [(W5 , W6 ), (W10 , W2 ), (W7 , W8 )] .
(A.2.4)
where j3 = δF (W9 − W10 , W11 − W12 , O). Independently of the conditions under which we encountered the last expression in (1.2.4), we denote it by
(A.2.5)
δF [(W1 , W2 ), (W3 , W4 )], [(W5 , W6 ), (W7 , W8 )], [(W9 , W10 ), (W11 , W12 )] .
Thus having (1.2.5) equal to 0 is necessary and sufficient for W ′ , W ′′ and W ′′′ to be collinear.
Somewhat surprisingly we are able to find an identity for the formidable expression (1.2.4).
In relation to (1.2.5) we prove the following identity. The multiplier of −W(2,3,4) in (1.2.4) is
equal to
W(11,12,10) W[(5,6),(9,1),(7,8)] − W(11,12,9) W[(5,6),(10,1),(7,8)]
=W(11,12,10) W[(9,1),(7,8),(5,6)] − W(11,12,9) W[(10,1),(7,8),(5,6)
=W(11,12,10) [W(9,7,8) W(1,5,6) − W(1,7,8) W(9,5,6) ] − W(11,12,9) [W(10,7,8) W(1,5,6) − W(1,7,8) W(10,5,6) ]
=W(1,5,6) [W(10,11,12) W(9,7,8) − W(9,11,12) W(10,7,8) ] − W(1,7,8) [W(10,11,12) W(9,5,6) − W(9,11,12) W(10,5,6) ]
=W(1,5,6) W[(10,9),(11,12),(7,8)] − W(1,7,8) W[(10,9),(11,12),(5,6) .
Similarly the multiplier of W(1,3,4) in (1.2.4) is equal to
W(11,12,10) W[(5,6),(9,2),(7,8)] − W(11,12,9) W[(5,6),(10,2),(7,8)]
=W(11,12,10) W[(9,2),(7,8),(5,6)] − W(11,12,9) W[(10,2),(7,8),(5,6)]
=W(11,12,10) [W(9,7,8) W(2,5,6) − W(2,7,8) W(9,5,6) ] − W(11,12,9) [W(10,7,8) W(2,5,6) − W(2,7,8) W(10,5,6) ]
=W(2,5,6) [W(10,11,12) W(9,7,8) − W(9,11,12) W(10,7,8) ] − W(2,7,8) [W(10,11,12) W(9,5,6) − W(9,11,12) W(10,5,6) ]
=W(2,5,6) W[(10,9),(11,12),(7,8)] − W(2,7,8) W[(10,9),(11,12),(5,6)] .
Now the sum of the two parts of (1.2.4) is equal to
[−W(2,3,4) W(1,5,6) + W(1,3,4) W(2,5,6) ]W[(10,9),(11,12),(7,8)]
+ [W(2,3,4) W(1,7,8) − W(1,3,4) W(2,7,8) ]W[(10,9),(11,12),(5,6)]
=W[(1,2),(3,4),(7,8)] W[(5,6),(9,10),(11,12)] − W[(1,2),(3,4),(5,6)] W[(7,8),(9,10),(11,12)] ,
which we could also denote as
W{[(7,8),(5,6)],[(1,2),(3,4)],[(9,10),(11,12)]}.
474
APPENDIX A. MORE COMPLICATED GEOMETRIC ALGEBRA
We can represent it as a determinant as follows
W[(1,2),(3,4),(7,8)]
W[(1,2),(3,4),(5,6)]
W[(7,8),(9,10),(11,12)] W[(5,6),(9,10),(11,12)]
.
Now when W1 W2 , W3 W4 are distinct and parallel and so are W5 W6 , W7 W8 , as δF (W2 , W3 , W4 ) =
δF (W1 , W3 , W4 ) 6= 0 the expression in (1.2.4) is equal to the product of δF (W1 , W3 , W4 ) and
− δF (W11 , W12 , W10 ) δF [(W5 , W6 ), (W9 , W1 ), (W7 , W8 )] − δF [(W5 , W6 ), (W9 , W2 ), (W7 , W8 )]
+δF (W11 , W12 , W9 ) δF [(W5 , W6 ), (W10 , W1 ), (W7 , W8 )] − δF [(W5 , W6 ), (W10 , W2 ), (W7 , W8 )] ,
and as δF (W6 , W7 , W8 ) = δF (W5 , W7 , W8 ) 6= 0 this in turn is equal to the product of δF (W5 , W7 , W8 )
and
− δF (W11 , W12 , W10 ){δF (W5 , W9 , W1 ) − δF (W6 , W9 , W1 ) − [δF (W5 , W9 , W2 ) − δF (W6 , W9 , W2 )]}
+δF (W11 , W12 , W9 ){δF (W5 , W10 , W1 ) − δF (W6 , W10 , W1 ) − [δF (W5 , W10 , W2 ) − δF (W6 , W10 , W2 )]}
= − δF (W11 , W12 , W10 )[δF (W5 − W6 , W9 − W1 , O) − δF (W5 − W6 , W9 − W2 , O)]
+ δF (W11 , W12 , W9 )[δF (W5 − W6 , W10 − W1 , O) − δF (W5 − W6 , W10 − W2 , O)]
= − δF (W11 , W12 , W10 )δF (W5 − W6 , W2 − W1 , O) + δF (W11 , W12 , W9 )δF (W5 − W6 , W2 − W1 , O)
=[δF (W11 , W12 , W9 ) − δF (W11 , W12 , W10 )]δF (W5 − W6 , W2 − W1 , O)
=δF (W9 − W10 , W11 − W12 , O)δF (W5 − W6 , W2 − W1 , O).
(A.2.6)
When δF (W5 − W6 , W2 − W1 , O) 6= 0, by (1.2.6) we have (1.2.5) equal to 0 if and only if
δF (W9 − W10 , W11 − W12 , O) = 0, that is W9 W10 k W11 W12 . Thus we have three pairs of parallel
lines.
On the other hand, when δF (W9 − W10 , W11 − W12 , O) 6= 0, by (1.2.6) we have (1.2.5) equal
to 0 if and only δF (W5 − W6 , W2 − W1 , O) = 0, thus having four parallel lines and an intersecting
pair.
Finally suppose that W1 W2 and W3 W4 are distinct and parallel but the other details are as
originally. Then W(1,3,4) = W(2,3,4) 6= 0, and (1.2.4) becomes W(1,3,4) times
W(11,12,10) W[(9,2),(7,8),(5,6)] − W(11,12,9) W[(10,2),(7,8),(5,6)]
− W(11,12,10) W[(9,1),(7,8),(5,6)] + W(11,12,9) W [(10, 1), (7, 8), (5, 6)].
We note that under the correspondence
W1 ↔ W9 , W2 ↔ W10 , W3 ← W11 , ; W4 ← W12 , W7 ↔ W5 , W8 ↔ W6 ,
this becomes
W(2,3,4) W[(1,10),(5,6),(7,8)] − W(1,3,4) W[(2,10),(5,6),(7,8)] − W(2,3,4) W[(1,9),(5,6),(7,8)] + W(1,3,4) W[(2,9),(5,6),(7,8)]
=W(2,3,4) W[(10,1),(7,8),(5,6)] − W(1,3,4) W[(10,2),(7,8),(5,6)] − W(2,3,4) W[(9,1),(7,8),(5,6)] + W(1,3,4) W[(9,2),(7,8),(5,6)] .
This is −1 times (1.2.2) and so its being equal to 0 is necessary and sufficient for W1 W2 k W3 W4 k
W ′′ W ′′′ .
A summary diagram for this section is the following :W1
b
W2
b
b
W3
b
W
b
b
′
b
W1
b
b
b
b
W11
W6
W10
b
W5
W
W8
b
W9
Case (i)
b
W5
b
b
b
b
W ′′
b
W4
W6
W7
b
b
′′′
b
b
b
b
W10
b
b
W9
b
W7
b
b
b
W8
b
b
W2
b
W11
W12
b
b
W12
Case (ii)(a). W1 W2 6k W5 W6
b
b
W3
W4
b
475
A.3. PAPPUS’ THEOREM AND DESARGUES’ PERSPECTIVE THEOREM
W1
W11
W2
b
b
W10
b
b
b
b
b
b
W5
b
b
b
W6
b
b
W7
b
b
W
′′′
W8
b
W11
b
W12
b
W9
b
b
W ′′
b
b
W10
b
b
b
W5
b
b
b
b
b
b
b
b
W ′′′
W8
b
W3 W4
Case (ii)(b). W1 W2 k W5 W6
A.3.1
W6
W7
b
W3 W4
A.3
b
W1 W2
b
b
b
b
W9
b
W12
Case (iii)
Pappus’ theorem and Desargues’
Figure 1.3 perspective theorem
Pappus’ theorem
Z3
Z3
b
b
Z1
Z2
Z1
b
b
b
b
Z2
b
b
b
b
b
Z4
Figure 1.4.
b
b
b
b
Z5
Z6
b
b
b
b
Z4
Z5
Z6
Z3
b
Z1Z2
b
b
b
b
b
b
Z6
Z5
Z4
Pappus’ theorem (c. 300A.D.). Let (Z1 , Z2 , Z3 ) be distinct collinear points and (Z4 , Z5 , Z6 )
be distinct collinear points, none of them being the point of intersection of these lines if the latter
meet. Then concerning the pairs of lines {Z1 Z5 , Z4 Z2 }, {Z2 Z6 , Z5 Z3 } and {Z3 Z4 , Z6 Z1 } :
i) if
Z1 Z5 ∩ Z4 Z2 = {W ′ }, Z2 Z6 ∩ Z5 Z3 = {W ′′ }, Z3 Z4 ∩ Z6 Z1 = {W ′′′ },
then W ′ , W ′′ , W ′′′ are collinear ;
(ii) if
Z1 Z5 k Z4 Z2 , Z2 Z6 ∩ Z5 Z3 = {W ′′ }, Z3 Z4 ∩ Z6 Z1 = {W ′′′ },
then Z1 Z5 k W ′′ W ′′′ ;
476
APPENDIX A. MORE COMPLICATED GEOMETRIC ALGEBRA
(iii) if
Z1 Z5 k Z4 Z2 , Z2 Z6 k Z5 Z3 ,
then Z1 Z5 6k Z2 Z6 and Z3 Z4 k Z6 Z1 .
Proof.
δF (Z5 , Z2 , Z4 )λδF (Z1 , Z5 , Z2 )δF (Z2 , Z1 , Z4 )µδF (Z4 , Z5 , Z1 )
− µδF (Z4 , Z2 , Z5 )δF (Z2 , Z5 , Z1 )λδF (Z1 , Z2 , Z4 )δF (Z5 , Z4 , Z1 )
= 0.
For case (ii) we note that if
Z1 Z5 k Z4 Z2 , Z2 Z6 k Z5 Z3 , Z1 Z5 k Z2 Z6 ,
then Z4 Z2 k Z2 Z6 and so Z2 ∈ Z4 Z6 which is ruled out.
A.3.2
Desargues’ perspective theorem
b
Z2
b
b
b
Z1
b
b
Z1
b
Z3
Z2
b
Z1
b
b
b
Z6
Z5
Figure 1.5.
b
b
b
b
b
b
b
Z2
b
Z4
b
Z3
b
b
b
Z4
b
b
Z3
Z4
Z6
b
b
Z5
Z5
b
Z6
Desargues’ perspective theorem (1648). Suppose that (Z1 , Z2 , Z3 ) and (Z4 , Z5 , Z6 ) are
triples of non-collinear points and that the lines Z1 Z4 , Z2 Z5 , Z3 Z6 are either concurrent or all
parallel. Then for the pairs of distinct lines {Z1 Z2 , Z4 Z5 }, {Z2 Z3 , Z5 Z6 } and {Z3 Z1 , Z6 Z4 },
(i) if
Z1 Z2 ∩ Z4 Z5 = {W ′ }, Z2 Z3 ∩ Z5 Z6 = {W ′′ }, Z3 Z1 ∩ Z6 Z4 = {W ′′′ },
then W ′ , W ′′ , W ′′′ are collinear ;
(ii) if
Z1 Z2 k Z4 Z5 , Z2 Z3 k Z5 Z6 ,
then Z3 Z1 k Z6 Z4 .
(iii) if
Z1 Z2 k Z4 Z5 , Z2 Z3 ∩ Z5 Z6 = {W ′′ }, Z3 Z1 ∩ Z6 Z4 = {W ′′′ },
then Z1 Z2 k W ′′ W ′′′ ;
Proof.
We apply §1.2.2 with
Z1 = W1 = W10 , Z2 = W2 = W5 , Z3 = W6 = W9 , Z4 = W3 = W12 , Z5 = W4 = W7 , Z6 = W8 = W11 ,
and δF [(Z1 , Z4 ), (Z2 , Z5 ), (Z3 , Z6 )] = 0
A.3. PAPPUS’ THEOREM AND DESARGUES’ PERSPECTIVE THEOREM
By the identity of Appendix A (????) we have that
n
o
δF [(Z1 , Z2 ), (Z4 , Z5 )], [(Z2 , Z3 ), (Z5 , Z6 )], [(Z3 , Z1 ), (Z6 , Z4 )]
=δF [(Z1 , Z2 ), (Z2 , Z3 ), (Z5 , Z6 )]δF [(Z4 , Z5 ), (Z3 , Z1 ), (Z6 , Z4 )]
− δF [(Z4 , Z5 ), (Z2 , Z3 ), (Z5 , Z6 )]δF [(Z1 , Z2 ), (Z3 , Z1 ), (Z6 , Z4 )]
=δF (Z1 , Z2 , Z3 )δF (Z2 , Z5 , Z6 )δF (Z4 , Z3 , Z1 )δF (Z5 , Z6 , Z4 )
− δF (Z5 , Z2 , Z3 )δF (Z4 , Z5 , Z6 )δF (Z2 , Z3 , Z1 )δF (Z1 , Z6 , Z4 )
h
=δF (Z1 , Z2 , Z3 )δF (Z4 , Z5 , Z6 ) δF (Z2 , Z5 , Z6 )δF (Z4 , Z3 , Z1 )
i
− δF (Z5 , Z2 , Z3 )δF (Z1 , Z6 , Z4 )
=δF (Z1 , Z2 , Z3 )δF (Z4 , Z5 , Z6 )δF [(Z3 , Z6 ), (Z1 , Z4 ), (Z2 , Z5 )]
=0.
In Case (ii), if we had
Z1 Z2 k Z4 Z5 , Z2 Z3 k Z5 Z6 , Z1 Z2 k Z2 Z3 ,
we would have had Z1 Z2 k Z2 Z3 and so Z2 ∈ Z1 Z3 which is ruled out.
477
478
APPENDIX A. MORE COMPLICATED GEOMETRIC ALGEBRA
Appendix B
Lines common to two pencils
As back-up for the investigations in §2.1.3, we include the following detail.
In our substantial analysis, we shall divide into four cases. We adopt the following notation.
When Z4 Z5 and Z7 Z8 are non-parallel we denote their unique point of intersection by W1 , and
choose points W2 and W3 so that
Z5 − Z4 = W2 − W1 ,
Z8 − Z7 = W3 − W1 .
(B.0.1)
δF (Z, Z7 , Z8 ) = δF (Z, W1 , W3 ),
(B.0.2)
Then by (2.1.10) we have that
δF (Z, Z4 , Z5 ) = δF (Z, W1 , W2 ),
for all Z.
Similarly when Z10 Z11 and Z13 Z14 are non-parallel we denote their unique point of intersection
by W7 , and choose points W8 and W9 so that
Z11 − Z10 = W8 − W7 ,
Z14 − Z13 = W9 − W7 .
(B.0.3)
δF (Z, Z13 , Z14 ) = δF (Z, W7 , W9 ),
(B.0.4)
Then by (2.1.10) we have that
δF (Z, Z10 , Z11 ) = δF (Z, W7 , W8 ),
for all Z.
CASE A.
We suppose first that Z10 Z11 and Z13 Z14 are non-parallel so that we have (2.1.13) and (2.1.14).
Then we have that
δF [(Z4 , Z5 ), (Z10 , Z11 ), (Z13 , Z14 )]
=δF (Z4 , Z10 , Z11 )δF (Z5 , Z13 , Z14 ) − δF (Z5 , Z10 , Z11 )δF (Z4 , Z13 , Z14 )
=δF (Z4 , W7 , W8 )δF (Z5 , W7 , W9 ) − δF (Z5 , W7 , W8 )δF (Z4 , W7 , W9 ),
=δF [(Z4 , Z5 ), (W7 , W8 ), (W7 , W9 )]
=δF [(W7 , W8 ), (W7 , W9 ), (Z4 , Z5 )]
= − δF (W8 , W7 , W9 )δF (W7 , Z4 , Z5 ).
Hence
δF [(Z7 , Z8 ), (Z10 , Z11 ), (Z13 , Z14 )] = −δF (W8 , W7 , W9 )δF (W7 , Z7 , Z8 ).
From these combined we have that
J1 = δF (W7 , W8 , W9 )[δF (W7 , Z4 , Z5 ) − δF (W7 , Z7 , Z8 )].
479
(B.0.5)
480
APPENDIX B. LINES COMMON TO TWO PENCILS
Similarly we have that
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )]
=δF (Z4 , Z7 , Z8 )δF (Z5 , Z10 , Z11 ) − δF (Z5 , Z7 , Z8 )δF (Z4 , Z10 , Z11 )
=δF (Z4 , Z7 , Z8 )δF (Z5 , W7 , W8 ) − δF (Z5 , Z7 , Z8 )δF (Z4 , W7 , W8 )
=δF [(Z4 , Z5 ), (Z7 , Z8 ), (W7 , W8 )]
=δF [(W7 , W8 ), (Z4 , Z5 ), (Z7 , Z8 )]
=δF (W7 , Z4 , Z5 )δF (W8 , Z7 , Z8 ) − δF (W8 , Z4 , Z5 )δF (W7 , Z7 , Z8 ).
From this
δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z13 , Z14 )]
=δF (W7 , Z4 , Z5 )δF (W9 , Z7 , Z8 ) − δF (W9 , Z4 , Z5 )δF (W7 , Z7 , Z8 ),
and from these combined we have that
J2 =δF (W7 , Z4 , Z5 )[δF (W8 , Z7 , Z8 ) − δF (W9 , Z7 , Z8 )]
− [δF (W8 , Z4 , Z5 )δF (W7 , Z7 , Z8 ) − δF (W9 , Z4 , Z5 )]
=δF (W7 , Z4 , Z5 )δF (W8 − W9 , Z7 − Z8 , O) − δF (W7 , Z7 , Z8 )δF (W8 − W9 , Z4 − Z5 , O).
(B.0.6)
CASE B.
Now instead of (2.1.13) and (2.1.14) we suppose that Z10 Z11 k Z13 Z14 so that
Z14 − Z13 = k2 (Z11 − Z10 ),
(B.0.7)
for some non-zero number k2 . Then by (2.1.8) we have that
δF [(Z4 , Z5 ), (Z10 , Z11 ), (Z13 , Z14 )]
=δF (Z4 , Z10 , Z11 )δF (Z5 , Z13 , Z14 ) − δF (Z5 , Z10 , Z11 )δF (Z4 , Z13 , Z14 )
=k2 δF (Z4 , Z10 , Z11 )[δF (Z5 , Z10 , Z11 ) − δF (Z13 , Z10 , Z11 )
−δF (Z5 , Z10 , Z11 )[δF (Z4 , Z10 , Z11 ) − δF (Z13 , Z10 , Z11 )]
=k2 δF (Z13 , Z10 , Z11 )[δF (Z5 , Z10 , Z11 ) − δF (Z4 , Z10 , Z11 )]
=k2 δF (Z10 , Z11 , Z13 )δF (Z5 − Z4 , Z10 − Z11 , O).
Hence
δF [(Z7 , Z8 ), (Z10 , Z11 ), (Z13 , Z14 )] = k2 δF (Z10 , Z11 , Z13 )δF (Z8 − Z7 , Z10 − Z11 , O).
From these combined we have that
J1 =k2 δF (Z10 , Z11 , Z13 )[δF (Z5 − Z4 , Z10 − Z11 , O) − δF (Z8 − Z7 , Z10 − Z11 , O)]
=k2 δF (Z10 , Z11 , Z13 )δF Z5 − Z4 − (Z8 − Z7 ), Z10 − Z11 , O .
(B.0.8)
481
We also have that
J2 =δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] − δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z13 , Z14 )]
=δF (Z4 , Z7 , Z8 )δF (Z5 , Z10 , Z11 ) − δF (Z5 , Z7 , Z8 )δF (Z4 , Z10 , Z11 )
− [δF (Z4 , Z7 , Z8 )δF (Z5 , Z13 , Z14 ) − δF (Z5 , Z7 , Z8 )δF (Z4 , Z13 , Z14 )
= − δF (Z4 , Z7 , Z8 )[δF (Z5 , Z13 , Z14 ) − δF (Z5 , Z10 , Z11 )]
+ δF (Z5 , Z7 , Z8 )[δF (Z4 , Z13 , Z14 ) − δF (Z4 , Z10 , Z11 )]
= − δF (Z4 , Z7 , Z8 ) k2 [δF (Z5 , Z10 , Z11 ) − δF (Z13 , Z10 , Z11 )] − δF (Z5 , Z10 , Z11 )]
+ δF (Z5 , Z7 , Z8 ) k2 [δF (Z4 , Z10 , Z11 ) − δF (Z13 , Z10 , Z11 )] − δF (Z4 , Z10 , Z11 )]
= − δF (Z4 , Z7 , Z8 ) (k2 − 1)δF (Z5 , Z10 , Z11 ) − k2 δF (Z13 , Z10 , Z11 )
+ δF (Z5 , Z7 , Z8 ) (k2 − 1)δF (Z4 , Z10 , Z11 ) − k2 δF (Z13 , Z10 , Z11 )
=(1 − k2 )δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] + k2 δF (Z13 , Z10 , Z11 )[δF (Z4 , Z7 , Z8 ) − δF (Z5 , Z7 , Z8 )]
=(1 − k2 )δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] + k2 δF (Z13 , Z10 , Z11 )δF (Z4 − Z5 , Z7 − Z8 , O).
(B.0.9)
COMBINED CASE C(i).
We now suppose that (2.1.13) and (2.1.11) hold.
Then from (2.1.13), by (2.1.15) we have that
J1 = δF (W7 , W8 , W9 )[δF (W7 , Z4 , Z5 ) − δF (W7 , Z7 , Z8 )],
(B.0.10)
and by (2.1.16) we have that
J2 = δF (W7 , Z4 , Z5 )δF (W8 − W9 , Z7 − Z8 , O) − δF (W7 , Z7 , Z8 )δF (W8 − W9 , Z4 − Z5 , O). (B.0.11)
Moreover from (2.1.11) by interchange of the two pencils we have from (2.1.20) that
J2 = δF (W1 , W2 , W3 )[δF (W1 , Z10 , Z11 ) − δF (W1 , Z13 , Z14 )],
(B.0.12)
and from (2.1.21)
J1 = δF (W1 , Z10 , Z11 )δF (W2 − W3 , Z13 − Z14 , O) − δF (W1 , Z13 , Z14 )δF (W2 − W3 , Z10 − Z11 , O).
(B.0.13)
Now on applying (2.1.11) in (2.1.20)
J1 =δF (W7 , W8 , W9 )[δF (W7 , W1 , W2 ) − δF (W7 , W1 , W3 )]
=δF (W7 , W8 , W9 )[δF (W2 , W7 , W1 ) − δF (W3 , W7 , W1 )]
=δF (W7 , W8 , W9 )δF (W2 − W3 , W7 − W1 , O),
(B.0.14)
and we would derive this also by applying (2.1.13) in (2.1.23).
Similarly on applying (2.1.13) in (2.1.22) we have that
J2 =δF (W1 , W2 , W3 )[δF (W1 , W7 , W8 ) − δF (W1 , W7 , W9 )]
=δF (W1 , W2 , W3 )[δF (W8 , W1 , W7 ) − δF (W9 , W1 , W7 )]
=δF (W1 , W2 , W3 )δF (W8 − W9 , W1 − W7 , O),
(B.0.15)
and we would derive this also by applying (2.1.11) to (2.1.21).
COMBINED CASE C(ii).
Suppose now that (2.1.13) holds for the first pencil, and that for the second pencil we have
Z8 − Z7 = k1 (Z5 − Z4 ),
(B.0.16)
482
APPENDIX B. LINES COMMON TO TWO PENCILS
for some k1 6= 0 in R, analogous to (2.1.17). Then from (2.1.13) we have (2.1.20) and (2.1.21)
again, and from (2.1.24) we have by interchanging in (2.1.18)
(B.0.17)
J2 = k1 δF (Z4 , Z5 , Z7 )δF Z11 − Z10 − (Z14 − Z13 ), Z4 − Z5 , O ,
and by interchanging in (2.1.19)
J1 = (1 − k1 )δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] + k1 δF (Z7 , Z4 , Z5 )δF (Z10 − Z11 , Z13 − Z14 , O).
(B.0.18)
By applying (2.1.26) to (2.1.20) we have that
J1 = − δF (W7 , W8 , W9 ) k1 [δF (W7 , Z4 , Z5 ) − δF (Z7 , Z4 , Z5 )] − δF (W7 , Z4 , Z5 )
=δF (W7 , W8 , W9 ) (1 − k1 )δF (W7 , Z4 , Z5 ) + k1 δF (Z7 , Z4 , Z5 ) ,
(B.0.19)
and this could also be derived on applying (2.1.13) to (2.1.28).
By applying (2.1.13) to (2.1.27) we have that
J2 =k1 δF (Z4 , Z5 , Z7 )δF (W8 − W7 − (W9 − W7 ), Z4 − Z5 , O)
=k1 δF (Z4 , Z5 , Z7 )δF (W9 − W8 , Z5 − Z4 , O),
(B.0.20)
and this could also be derived on applying (2.1.26) to (2.1.21).
COMBINED CASE C(iii).
In this we interchange the roles of the two pencils in COMBINED CASE C(ii).
COMBINED CASE C(iv).
We now suppose that (2.1.17) and (2.1.26) hold. From (2.1.17) we have that (2.1.18) and
(2.1.19) hold,
J1 = k2 δF (Z10 , Z11 , Z13 )δF Z5 − Z4 − (Z8 − Z7 ), Z10 − Z11 , O ,
and
J2 = (1 − k2 )δF [(Z4 , Z5 ), (Z7 , Z8 ), (Z10 , Z11 )] + k2 δF (Z13 , Z10 , Z11 )δF (Z4 − Z5 , Z7 − Z8 , O).
From (2.1.26), on interchanging the pencils, we have that
and
J2 = k1 δF (Z4 , Z5 , Z6 )δF Z11 − Z10 − (Z14 − Z13 ), Z4 − Z5 , O ,
(B.0.21)
J1 = (1 − k1 )δF [(Z10 , Z11 ), (Z13 , Z14 ), (Z4 , Z5 )] + k1 δF (Z7 , Z4 , Z5 )δF (Z10 − Z11 , Z13 − Z14 , O).
(B.0.22)
Now on applying (2.1.26) to (2.1.18) we obtain
J1 =k2 δF (Z10 , Z11 , Z13 )δF (k1 − 1)(Z5 − Z4 ), Z10 − Z11 , O
(B.0.23)
=k2 (k1 − 1)δF (Z10 , Z11 , Z13 )δF Z5 − Z4 , Z10 − Z11 , O .
Similarly on applying (2.1.17) to (2.1.31) we obtain
J2 =k1 δF (Z4 , Z5 , Z6 )δF (k2 − 1)(Z11 − Z10 ), Z4 − Z5 , O
=k1 (k2 − 1)δF (Z4 , Z5 , Z6 )δF Z11 − Z10 , Z4 − Z5 , O .
(B.0.24)
We could also derive these from (2.1.19) and (2.1.32).
We could now develop material which is dual to that in §1.2.1 and §1.2.2, but rest content with
having indicated the path and pattern of development.
Appendix C
Material on dual sensed-area
In furtherance of §1.1.4 and §1.1.5 we prove some identities involving dual sensed-area.
C.0.3
A determinant for dual sensed-area
By the identity (3.2.1), we have
δF (Z1 , Z2 , Z3 )δF (Z4 , Z6 , Z7 )
=δF (Z1 , Z6 , Z7 )δF (Z4 , Z2 , Z3 ) + δF (Z2 , Z6 , Z7 )δF (Z4 , Z3 , Z1 ) + δF (Z3 , Z6 , Z7 )δF (Z4 , Z1 , Z2 ),
δF (Z1 , Z2 , Z3 )δF (Z5 , Z8 , Z9 )
=δF (Z1 , Z8 , Z9 )δF (Z5 , Z2 , Z3 ) + δF (Z2 , Z8 , Z9 )δF (Z5 , Z3 , Z1 ) + δF (Z3 , Z8 , Z9 )δF (Z5 , Z1 , Z2 ),
(C.0.1)
and similarly
δF (Z1 , Z2 , Z3 )δF (Z5 , Z6 , Z7 )
=δF (Z1 , Z6 , Z7 )δF (Z5 , Z2 , Z3 ) + δF (Z2 , Z6 , Z7 )δF (Z5 , Z3 , Z1 ) + δF (Z3 , Z6 , Z7 )δF (Z5 , Z1 , Z2 ),
δF (Z1 , Z2 , Z3 )δF (Z4 , Z8 , Z9 )
=δF (Z1 , Z8 , Z9 )δF (Z4 , Z2 , Z3 ) + δF (Z2 , Z8 , Z9 )δF (Z4 , Z3 , Z1 ) + δF (Z3 , Z8 , Z9 )δF (Z4 , Z1 , Z2 ).
(C.0.2)
Now
=
δF (Z1 , Z2 , Z3 )2 δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )]
δF (Z1 , Z2 , Z3 )2 [δF (Z4 , Z6 , Z7 )δF (Z5 , Z8 , Z9 ) − δF (Z5 , Z6 , Z7 )δF (Z4 , Z8 , Z9 )]
and from (C.0.1) and (C.0.2), this has the form
=
(AD + BE + CF )(JG + KH + LI) − (AG + BH + CI)(JD + KE + LF )
(BL − CK)(EI − HF ) + (CJ − AL)(F G − ID) + (AK − BJ)(DH − GE),
where
A =
D =
G =
J
=
δF (Z1 , Z6 , Z7 ), B = δF (Z2 , Z6 , Z7 ), C = δF (Z3 , Z6 , Z7 ),
δF (Z4 , Z2 , Z3 ), E = δF (Z4 , Z3 , Z1 ), F = δF (Z4 , Z1 , Z2 ),
δF (Z5 , Z2 , Z3 ), H = δF (Z5 , Z3 , Z1 ), I = δF (Z5 , Z1 , Z2 ),
δF (Z1 , Z8 , Z9 ), K = δF (Z2 , Z8 , Z9 ), L = δF (Z3 , Z8 , Z9 ).
483
484
APPENDIX C. MATERIAL ON DUAL SENSED-AREA
Thus we have
δF (Z1 , Z2 , Z3 )2 δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )]
= [δF (Z2 , Z6 , Z7 )δF (Z3 , Z8 , Z9 ) − δF (Z3 , Z6 , Z7 )δF (Z2 , Z8 , Z9 )]
[δF (Z4 , Z3 , Z1 )δF (Z5 , Z1 , Z2 ) − δF (Z5 , Z3 , Z1 )δF (Z4 , Z1 , Z2 )]
+[δF (Z3 , Z6 , Z7 )δF (Z1 , Z8 , Z9 ) − δF (Z1 , Z6 , Z7 )δF (Z3 , Z8 , Z9 )]
[δF (Z4 , Z1 , Z2 )δF (Z5 , Z2 , Z3 ) − δF (Z5 , Z1 , Z2 )δF (Z4 , Z2 , Z3 )]
+[δF (Z1 , Z6 , Z7 )δF (Z2 , Z8 , Z9 ) − δF (Z2 , Z6 , Z7 )δF (Z1 , Z8 , Z9 )]
[δF (Z4 , Z2 , Z3 )δF (Z5 , Z3 , Z1 ) − δF (Z5 , Z2 , Z3 )δF (Z4 , Z3 , Z1 )]
= δF [(Z2 , Z3 ), (Z6 , Z7 ), (Z8 , Z9 )]
[δF (Z4 , Z3 , Z1 )δF (Z5 , Z1 , Z2 ) − δF (Z5 , Z3 , Z1 )δF (Z4 , Z1 , Z2 )]
+δF [(Z3 , Z1 ), (Z6 , Z7 ), (Z8 , Z9 )]
[δF (Z4 , Z1 , Z2 )δF (Z5 , Z2 , Z3 ) − δF (Z5 , Z1 , Z2 )δF (Z4 , Z2 , Z3 )]
+δF [(Z1 , Z2 ), (Z6 , Z7 ), (Z8 , Z9 )]
[δF (Z4 , Z2 , Z3 )δF (Z5 , Z3 , Z1 ) − δF (Z5 , Z2 , Z3 )δF (Z4 , Z3 , Z1 )]
on applying (3.2.1) systematically,
= δF [(Z2 , Z3 ), (Z6 , Z7 ), (Z8 , Z9 )]δF [(Z4 , Z5 ), (Z3 , Z1 ), (Z1 , Z2 )]
+δF [(Z3 , Z1 ), (Z6 , Z7 ), (Z8 , Z9 )]δF [(Z4 , Z5 ), (Z1 , Z2 ), (Z2 , Z3 )]
+δF [(Z1 , Z2 ), (Z6 , Z7 ), (Z8 , Z9 )]δF [(Z4 , Z5 ), (Z2 , Z3 ), (Z3 , Z1 )]
on applying (3.2.1) systematically again,
= δF [(Z2 , Z3 ), (Z6 , Z7 ), (Z8 , Z9 )]δF (Z1 , Z4 , Z5 )δF (Z2 , Z3 , Z1 )
+δF [(Z3 , Z1 ), (Z6 , Z7 ), (Z8 , Z9 )]δF (Z2 , Z4 , Z5 )δF (Z3 , Z1 , Z2 )
+δF [(Z1 , Z2 ), (Z6 , Z7 ), (Z8 , Z9 )]δF (Z3 , Z4 , Z5 )δF (Z1 , Z2 , Z3 ),
on expanding via (3.2.1). Thus we have proved the identity that


δF (Z1 , Z4 , Z5 ) δF (Z2 , Z4 , Z5 ) δF (Z3 , Z4 , Z5 )
det  δF (Z1 , Z6 , Z7 ) δF (Z2 , Z6 , Z7 ) δF (Z3 , Z6 , Z7 ) 
δF (Z1 , Z8 , Z9 ) δF (Z2 , Z8 , Z9 ) δF (Z3 , Z8 , Z9 )
=
C.0.4
δF (Z1 , Z2 , Z3 )δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )].
(C.0.3)
A mixed identity for sensed-area and dual sensed-area
From (C.0.3) we deduce the identity
δF (Z1 , Z2 , Z3 )δF [(Z4 , Z5 ), (Z6 , Z7 ), (Z8 , Z9 )]
=δF (Z3 , Z4 , Z5 )δF [(Z1 , Z2 ), (Z6 , Z7 ), (Z8 , Z9 )]
+ δF (Z3 , Z6 , Z7 )δF [(Z1 , Z2 ), (Z8 , Z9 ), (Z4 , Z5 )]
+ δF (Z3 , Z8 , Z9 )δF [(Z1 , Z2 ), (Z4 , Z5 ), (Z6 , Z7 )].
C.0.5
An identity for dual sensed-area
Dually to the fundamental identity (3.2.1), we consider whether there is an expansion
δF [(Z, W ), (Z4 , Z5 ), (Z6 , Z7 )] = pδF [(Z, W ), (Z3 , Z1 ), (Z1 , Z2 )]
+ qδF [(Z, W ), (Z1 , Z2 ), (Z2 , Z3 )] + rδF [(Z, W ), (Z2 , Z3 ), (Z3 , Z1 )].
(C.0.4)
485
This would require
δF [(Z2 , Z3 ), (Z4 , Z5 ), (Z6 , Z7 )] = pδF [(Z2 , Z3 ), (Z3 , Z1 ), (Z1 , Z2 )],
δF [(Z3 , Z1 ), (Z4 , Z5 ), (Z6 , Z7 )] = qδF [(Z3 , Z1 ), (Z1 , Z2 ), (Z2 , Z3 )],
δF [(Z1 , Z2 ), (Z4 , Z5 ), (Z6 , Z7 )] = rδF [(Z1 , Z2 ), (Z2 , Z3 ), (Z3 , Z1 )],
and suggests the identity
δF [(Z2 , Z3 ), (Z3 , Z1 ), (Z1 , Z2 )]δF [(Z, W ), (Z4 , Z5 ), (Z6 , Z7 )] =
δF [(Z2 , Z3 ), (Z4 , Z5 ), (Z6 , Z7 )]δF [(Z, W ), (Z3 , Z1 ), (Z1 , Z2 )]
+ δF [(Z3 , Z1 ), (Z4 , Z5 ), (Z6 , Z7 )]δF [(Z, W ), (Z1 , Z2 ), (Z2 , Z3 )]
+ δF [(Z1 , Z2 ), (Z4 , Z5 ), (Z6 , Z7 )]δF [(Z, W ), (Z2 , Z3 ), (Z3 , Z1 )].
(C.0.5)
Now the first term on the left-hand side here is equal to δF (Z1 , Z2 , Z3 )2 while on the right-hand
side
δF [(Z, W ), (Z3 , Z1 ), (Z1 , Z2 )] =
=
δF [(Z, W ), (Z1 , Z2 ), (Z2 , Z3 )] =
=
δF [(Z, W ), (Z2 , Z3 ), (Z3 , Z1 )] =
=
δF [(Z1 , Z2 ), (Z, W ), (Z3 , Z1 )]
δF (Z1 , Z, W )δF (Z2 , Z3 , Z1 ),
δF [(Z2 , Z3 ), (Z, W ), (Z1 , Z2 )]
δF (Z2 , Z, W )δF (Z3 , Z1 , Z2 ),
δF [(Z3 , Z1 ), (Z, W ), (Z2 , Z3 )]
δF (Z3 , Z, W )δF (Z1 , Z2 , Z3 ).
On adding up the terms on the right now, cancelling δF (Z1 , Z2 , Z3 ) on both sides and replacing Z by Z8 and W by Z9 , we are led to an identity which can be obtained by expanding the
determinant in (C.0.3) in terms of the third row. Thus (C.0.5) is established.
486
APPENDIX C. MATERIAL ON DUAL SENSED-AREA
Appendix D
Rotors common to two pencils
We standardize our notation as follows. If for first pencil P1 we suppose that Z4 Z5 and Z7 Z8 are
non-parallel thus meeting in a unique point, we denote this point of intersection by W1 and take
points W2 and W3 so that
hZ4 , Z5 i = hW1 , W2 i,
Z8
b
hZ7 , Z8 i = hW1 , W3 i.
b
Z7
b
Z5
b
W3
b
Z4
b
W2
b
W1
Figure D.1
Similarly for the second pencil P2 , if Z10 Z11 and Z13 Z14 are non-parallel, thus meeting in a
unique point, we denote this point of intersection by W7 , and take points W8 and W9 so that
hZ10 , Z11 i = hW7 , W8 i,
hZ13 , Z14 i = hW7 , W9 i.
487
488
APPENDIX D. ROTORS COMMON TO TWO PENCILS
b
b
W9
W7
Z14
Z13
b
b
b
W8
b
Z10
Figure D.2
b
Z11
CASE A.
We suppose that we have the non-parallel cases for both of the pencils and that the points of
intersection W1 and W7 are distinct.
(i) Suppose first that W1 W7 6k W2 W3 and W1 W7 6k W8 W9 as well. Now hUr , Vr i can be a rotor
hUr′ ′ , Vr′′ i = (1 − r′ )hZ10 , Z11 i + r′ hZ13 , Z14 i, in the second pencil only if Vr ∈ W1 W7 and then we
must also have Vr′′ − W7 = Vr − W1 . In the diagram we take Vr , Vr′′ to be the points in which
W1 W7 meets W2 W3 and W8 W9 , respectively.
Z8
Z7
b
b
b
b
b
Z4
W3
b
W2
W7
b
b
Z5
W9
b
Z14
Z13
b
b
Vr′′
b
Vr
b
W1
b
W8
b
Z10
b
Figure D.3. Rotor in common only when
Vr′′ − W7 = Vr − W1 ; then unique.
Z11
489
(ii) Secondly suppose that as above we now have either W1 W7 k W2 W3 or W1 W7 k W8 W9 . We
take the first of these for our diagram. In this subsection we generally draw just skeleton diagrams
dealing with the essentials of the cases. Then for no Vr ∈ W2 W3 is Vr ∈ W1 W7 , so no rotor in the
first pencil is also in the second pencil.
W9
W2
Vr
W7
b
b
b
b
W3
b
b
W1
Figure D.4. No rotor in common.
b
W8
CASE B.
We suppose that the situation is as in Case A, in that Z4 Z5 6k Z7 Z8 and Z10 Z11 6k Z13 Z14 , but
now the points of intersection W1 and W7 are coincident.
(i) First we note that if we had W2 W3 and W8 W9 non-parallel and so meeting in a unique
point W , then hW1 , W i would be a unique rotor common to the two pencils.
(ii) Secondly suppose that W2 W3 and W8 W9 are parallel and distinct. Then the two pencils
have no rotor in common.
Z13
b
Z14
b
Z8
b
b
Z7
W9
b
b
W3
b
W1 = W7
Z5
b
b
Z4
b
W2
b
b
b
W8
Z10
Z11
Figure D.5. No rotors in common.
(iii) Thirdly suppose that W2 W3 = W8 W9 . Then every rotor in each of the two pencils is also
in the other.
490
APPENDIX D. ROTORS COMMON TO TWO PENCILS
W9
b
b
W3
W2
b
b
b
W1 = W7
W8
Figure D.6. All rotors in common.
CASE C.
We suppose that one of the pencils is based on non-parallel lines and the other on distinct
parallel lines, say that Z4 Z5 6k Z7 Z8 and Z10 Z11 k Z13 Z14 but Z10 Z11 6= Z13 Z14 .
(i) First suppose that W2 W3 k Z10 Z11 . Then no carrier of a rotor in the first pencil is parallel
to Z10 Z11 , as no line joining W1 to a point of W2 W3 is parallel to Z10 Z11 . Thus the pencils have
no rotor in common.
Z8
b
b
W3
b
b
Z5
b
b
b
Z7
Z4
W2
W1
b
b
Z10
Z13
b
Z14
b
Figure D.7. No rotors in common.
Z11
(ii) Secondly suppose that W2 W3 6k Z10 Z11 . Now the rotor hW1 , Vr i can be in the second
pencil only if W1 Vr k Z10 Z11 ; when this parallelism holds, let W1 Vr meet Z10 Z13 at Ur′ ′ . If
Z14 − Z13 = k2 (Z11 − Z10 , we need that
Vr′′ − Ur′ ′ = (1 − r′ )(Z10 − Z11 ) + r′ (Z14 − Z13 ) = (1 − r′ + k2 r′ )(Z11 − Z10 ),
and so must have 1 − r′ + k2 r′ 6= 0. When this is satisfied, the line W1 Vr will meet hyperbola H2
which corresponds to the second pencil at Vr′′ . Then for the rotor hW1 , Vr i to be in the second
491
pencil, we must have further that
Vr′′ − Ur′ ′ = Vr − W1 .
W3
b
b
W2
W1
b
Vr′′
Ur′ ′
b
b
b
Z13
Vr
b
Z14
b
b
Z10
b
Z11
Figure D.8.
A common rotor only if Vr′′ − Ur′ ′ = Vr − W1 .
CASE D.
We suppose that the situation is as in Case C, with Z4 Z5 6k Z7 Z8 but now Z10 Z11 = Z13 Z14 ,
and again Z14 − Z13 = k2 (Z11 − Z10 ).
(i) First suppose that W2 W3 k Z10 Z11 . Then no carrier of a rotor in the first pencil is parallel
to Z10 Z11 , as no line joining W1 to a point of W2 W3 is parallel to Z10 Z11 . Thus the pencils have
no rotor in common.
Z8
b
b
b
Z14
W3
Z7
b
Z4
b
b
Z5
b
b
W2
W1
b
Z10
b
Z13
b
Z11
Figure D.9. No rotors in common.
(ii) Secondly suppose that W2 W3 6k Z10 Z11 . Then unless W1 ∈ Z10 Z11 no rotor in the first
pencil can be in the second pencil.
492
APPENDIX D. ROTORS COMMON TO TWO PENCILS
Z8
b
b
Z7
b
W3
b
Z5
b
Z4
b
W2
b
W1
b
b
b
b
b
Z10
Z11
Vr
Z13
Z14
Figure D.10. No rotors in common.
(iii) Thirdly suppose that W2 W3 6k Z10 Z11 and that W1 ∈ Z10 Z11 . Then there a unique rotor
common to the two pencils.
Z8
b
b
W3
Z7
b
b
Z5
b
W2
Z4
b
b
b
b
b
b
b
Z10
Z11
W1
Vr
Z13
Z14
Figure D.11. Unique common rotor.
CASE E.
Suppose that Z4 Z5 and Z7 Z8 are distinct parallel lines, and that so are Z10 Z11 and Z13 Z14 .
(i) First we suppose that Z4 Z5 6k Z10 Z11 . Then the pencils can have no common rotor.
Z8
b
Z7
b
b
Z5
b
Z4
b
b
Z10
Z13
b
b
Figure D.12. No rotors in common.
Z11
Z14
493
(ii) Secondly we suppose that Z4 Z5 k Z10 Z11 .
Z13
b
Z7
Z10
Z4
b
b
Z14
Z8
Z11
b
b
b
b
b
Z5
Figure D.13
We make the assumptions of §4.3.3 and recall that the locus of Ur is the hyperbola
y=
k1 y7 x − x5
.
k1 − 1 x
(D.0.1)
For the second pencil (1 − r′ )hZ10 , Z11 i + r′ hZ13 , Z14 i, as Z10 and Z13 both are on the line Z4 Z7 ,
they have coordinates of the form (0, y10 ) and (0, y13 ), say. We also take
Z11 − Z10 = k2 (Z5 − Z4 ),
Z14 − Z13 = k2 (Z5 − Z4 ),
so that Z11 , Z14 have coordinates (k2 x5 , y10 ), (k3 x5 , y13 ). Now
Ur′ ′ =
r′ k3 /k2
1 − r′
Z10 +
Z13 ,
′
1 − r + rk3 /k2
1 − r′ + r′ k3 /k2
so that Ur′ ′ has coordinates
and Vr′′ has coordinates
Vr′′ − Ur′ ′ = (1 − r′ + r′ k3 /k2 )k2 (Z5 − Z4 ),
(1 − r′ )y10 + r′ y13 k3 /k2
0,
,
1 − r′ + r′ k3 /k2
(1 − r′ )y10 + r′ y13 k3 /k2
[k2 + (k3 − k2 )r′ ]x5 ,
.
1 − r′ + r′ k3 /k2
If we write
x = [k2 + (k3 − k2 )r′ ]x5 ,
y=
(1 − r′ )y10 + r′ y13 k3 /k2
,
1 − r′ + r′ k3 /k2
(D.0.2)
for these coordinates of Vr′′ , and on assuming initially that k3 − k2 6= 0, on eliminating r′ we obtain
y=
2 x5
k2 x5 y10 + (k3 y13 − k2 y10 ) x−k
k3 −k2
x
,
(D.0.3)
as the equation of the locus H2 of Vr′′ . This locus is also a hyperbola and passes through Z11 and
Z13 .
494
APPENDIX D. ROTORS COMMON TO TWO PENCILS
Now for these to provide a rotor common to the two pencils we must have Ur = Ur′ ′ and
Vr = Vr′′ so that the two hyperbolas have the point Vr = Vr′′ in common. Thus the hyperbolas
either have no point in common, in which case the two pencils have no rotor in common, or else
the two hyperbolas coincide in which case the two pencils have identical membership.
We derive a less straightforward but more elementary characterization of dual-parallelism as
follows. On multiplying across by x/x5 in both (4.6.11) and (4.6.13) we see that the hyperbolas
have a point in common if and only if the lines with equation
y=
y=
k1 y7 x − x5
,
k1 − 1 x5
2 x5
k2 x5 y10 + (k3 y13 − k2 y10 ) x−k
k3 −k2
x5
(D.0.4)
,
(D.0.5)
have a point in common for which x 6= 0.
Z13
b
Z7
Z10
Z4
b
b
Z14
Z8
Z11
b
b
b
b
b
Z5
Figure D.14
To identify the line with equation (4.6.14) we first note that it passes through Z5 . If Z21 is the
point such that [Z4 , Z7 , Z8 , Z21 ] is a parallelogram, let the line through Z21 which is parallel to Z5 Z7
meet Z4 Z7 at Z22 , and then let Z23 be the point such that [Z21 , Z4 , Z22 , Z23 ] is a parallelogram.
Now Z21 , Z22 , Z23 have coordinates (k1 x5 , 0), (0, k1 y7 ), (k1 x5 , k1 y7 ), respectively, and it can be
checked that the line (4.6.14) also passes through the point Z23 . This line Z5 Z23 then meets the
ky7
line Z4 Z7 in the point Z24 with coordinates (0, 1−k
). Similarly for the second pencil we take points
1
Z25 , Z26 , Z27 , Z28 so that [Z10 , Z13 , Z14 , Z25 ] is a parallelogram, the line through Z25 which is
parallel to Z11 Z13 meet Z4 Z7 at Z26 , and [Z25 , Z10 , Z26 , Z27 ] is a parallelogram. Then the line
(4.6.15) is Z11 Z27 and this meets Z4 Z7 in Z28 .
For the general argument in this subsection, we note that
c
a
+b= +d
x
x
is equivalent to
a−c
= d − b,
x
495
and this is equivalent to
a−c
, x 6= 0,
d−b
and this has a unique solution unless a = c or b = d.
CASE F.
Suppose that Z4 Z5 and Z7 Z8 are distinct parallel lines, and that Z10 Z11 = Z13 Z14 .
(i) Suppose first that Z4 Z5 6k Z10 Z11 . Then the two pencils have no rotors in common.
Z8
x=
b
Z7
b
b
Z14
b
Z5
b
Z4
b
Z10
b
Z13
b
Figure D.15. No rotors in common.
Z11
(ii) Suppose secondly that Z4 Z5 k Z10 Z11 . Without loss of generality we take Z4 Z7 ⊥ Z4 Z5 ,
and let Ur be where Z10 Z11 meets Z4 Z7 . Then Vr is on the hyperbola H1 , and as we can take
Ur′ ′ = Z10 and
Vr′′ − Z10 = (1 − r′ + k2 r′ )(Z10 − Z11 ) = Vr − Ur ,
the two pencils have a unique rotor in common.
Z7
Ur
Z4
Z8
b
b
b
b
Vr
Z13
b
b
Z5
Figure D.16
CASE G.
Suppose that Z4 Z5 = Z7 Z8 and Z10 Z11 = Z13 Z14 .
(i) First let Z4 Z5 6k Z10 Z11 . Then the two pencils have no rotor in common.
b
b
Z14
496
APPENDIX D. ROTORS COMMON TO TWO PENCILS
b
Z14
b
Z8
b
b
Z7
Z5
b
Z4
b
Z10
b
Z13
b
Figure D.17. No rotors in common.
Z11
(ii) Secondly suppose that the two lines Z4 Z5 and Z10 Z11 are parallel but distinct. Then the
two pencils have no rotor in common.
Z14
b
Z13
b
Z11
Z10
b
Z8
b
b
b
Z7
b
Z5
b
Z4
Figure D.19. No rotors in common.
(iii) Thirdly suppose that Z4 Z5 = Z10 Z11 . Then all the rotors in either of the pencils are also
in the other pencil.
b
Z14
b
b
b
b
b
b
b
Z4
Z10
Z5
Z7
Z11
Figure D.20. All rotors in common.
Z8
Z13
Appendix E
Note on a paper of W.K. Clifford’s
The paper in mind is On the general theory of anharmonics, [4, pp. 110-114].
In §3 of it Clifford states inter alia that, in our notation,
Z(1,2,3) Z(4,5,6) + Z(1,2,4) Z(5,6,3) + Z(1,2,5) Z(6,3,4) + Z(1,2,6) Z(3,4,5)
is identically equal to 0, with five others obtained from this by permutation. Now consider instead
Z(3,1,2) Z(4,5,6) − Z(4,1,2) Z(3,5,6) + Z(5,1,2) Z(6,3,4) − Z(6,1,2) Z(5,3,4)
=Z[(3,4),(1,2),(5,6)] + Z[(5,6),(1,2),(3,4)]
=Z[(5,6),(3,4),(1,2)] + Z[(5,6),(1,2),(3,4)]
=0.
If Clifford’s were also true we would have
Z(3,1,2) Z(4,5,6) + Z(5,1,2) Z(6,3,4) = 0
and this is manifestly untrue if Z1 , Z2 , Z3 , Z4 , Z5 , Z6 , Z1 are in an anti-clockwise sequence on a
circle.
497
498
APPENDIX E. NOTE ON A PAPER OF W.K. CLIFFORD’S
Appendix F
More on trigonometric tangent
function
F.1
F.1.1
From (12.1.7) we note that
cos 2φ
u−1
sin 2φ
1 − v 2 − u2
=u2 + v 2 + 2
u−1
2u
=0.
u2 + v 2 + 2
Thus the circle with equation
x2 + y 2 + 2
cos 2φ
x − 1 = 0,
sin 2φ
(F.1.1)
passes through the point with coordinates (u, v). It also passes through the points with coordinates
(0, 1) and (0, −1).
From (12.1.6) we obtain the equation
u2 + v 2 − 2
e4ψ + 1
v + 1 = 0.
e4ψ − 1
x2 + y 2 − 2
e4ψ + 1
y + 1 = 0,
e4ψ − 1
Thus the circle with equation
(F.1.2)
also passes through the point with coordinates (u, v).
The equations (F.1.1) represent a system of coaxal circles. The equations (F.1.2) also represent
a system of coaxal circles. By Sommerville p. 88, these systems are conjugate.
F.1.2
When u is non-zero and is kept constant, as v ranges from −∞ to ∞, (12.1.9) and (12.1.10) give
us parametric equations of the locus of (φ, ψ) for a fixed value of u kept constant. Two of these
loci are drawn in the following diagram.
499
500
APPENDIX F. MORE ON TRIGONOMETRIC TANGENT FUNCTION
1
0.5
0
0.5
1
1.5
2
2.5
-0.5
-1
Figure F.1.
For each value of u we obtain a figure of eight, as illustrated in Figure F.1., where the outside
one corresponds to u = 12 and the inside one to u = 1. Each curve crosses itself at the point ( π2 , 0).
However the point ( π2 , 0) is not a point of any of the loci.
Bibliography
[1] J.W. Archbold. Introduction to the Algebraic Geometry of a Plane. Edward Arnold, 1948.
[2] P.D. Barry. Geometry with Trigonometry. Horwood Publications, Chichester, West Sussex,
2001.
[3] M. Berger. Geometry I and Geometry II. Springer-Verlag, 1987.
[4] W.K. Clifford. Mathematical Papers. Macmillan, 1882.
[5] H.S.M. Coxeter. Introduction to Geometry, 2nd ed. Wiley, 1969.
[6] G. Darboux. Principes de Géométrie Analytique. Gauthier-Villars, Paris, 1917.
[7] D. Fearnley-Sander. A royal road to geometry. Mathematics Magazine, 53, No.5, 1980.
[8] J.V. Field and J.J. Gray. The Geometrical Work of Girard Desargues. Springer-Verlag, 1987.
[9] G. H. Forder. The Calculus of Extension. Cambridge University Press, 1941.
[10] G.H. Forder. Higher Course Geometry. Cambridge, 1931.
[11] G.H. Forder. The Foundations of Euclidean Geometry. Cambridge, Dover, 1931, 1958.
[12] J. Frenkel. Géométrie pour l’élève-professeur. Hermann, Paris, 1973,1977.
[13] H.G. Grassmann. Die Ausdehnungslehre. Berlin, 1844, 1862.
[14] L. Hoehn. A simple generalisation of ceva’s theorem. Mathematical Gazette, 73, No.464, 1989.
[15] Roger A. Johnson. Advanced Euclidean Geometry (Modern Geometry). Houghton Mifflin,
Dover, 1929, 1960.
[16] W.J. Johnston. An Elementary Treatise on Analytical Geometry. Oxford, 1893.
[17] L.J. Magnus. Aufgaben und Lehrsätze aus der analytischen Geometrie. Berlin, 1833.
[18] Bruce E. Meserve. Fundamental Concepts of Geometry. Addison-Wesley, 1955.
[19] A. F. Möbius. Der barycentrische Calcul. Barth, Leipzig, 1827.
[20] P.S. Modenov and A.S. Parkhomenko. Geometric Transformations, Volume 2. Academic
Press, 1965.
[21] E. H. Neville. Prolegomena to Analytic Geometry. Cambridge University Press, 1922.
[22] Fathers of the Society of Jesus. A page of Irish history; the story of University College Dublin,
1883-1909. Dublin, 1930.
[23] D.K. Picken. The complete angle and geometrical generality. Math. Gazette, 1923.
501
502
BIBLIOGRAPHY
[24] D.K. Picken. Euclidean geometry of angle. Trans. London Math. Soc., 23, 1924.
[25] J.-V. Poncelet. Traité des Propriétés Projectives des Figures. Paris, 1865.
[26] Gilbert de B. Robinson. The Foundations of Geometry. University of Toronto Press, 1940.
[27] A. Seidenberg. Lectures in Projective Geometry. Van Nostrand, 1962.
[28] C.E. Springer. Geometry and Analysis of Projective Spaces. Freeman, 1964.
[29] Annita Tuller. A Modern Introduction to Geometries. Van Nostrand, 1966.
[30] R. Walker. Cartesian and Projective Geometry. Edward Arnold, 1953.
[31] A.N. Whitehead. A treatise on Universal Algebra with applications. Cambridge UP, ?
Index
Z0 -deleted point-pair conic, 220
Z0 -non-aligned, 205
Z0 -non-incident line, 205
δF (Z1 , Z2 , Z3 ), 1
non-central
conic, 152
affine line property, 3
affine transformation, 53
amenable pair, 31
areal pair-coordinates, 75
asymptotes, 158
axes, 296
axis
radical, 302
conjugate hyperbola, 263
conjugate points
conic, 153
contact, 221
coordinates
oblique, 272
cr(Z1 , Z2 , Z3 , Z4 ), 59
cross ratio, 59
cross-ratio of pencil, 60
deficient incidence, 146
Desargues’ perspective theorem, 15
diagonal lines, 36
diagonal triple, 24
diameter
ellipse, 239
diametral
line
branch, 143
conic, 169
diametral lines
central conic, 151
conjugate, 171
central perspectivity, 16
difference
centre, 239
rotors, 95
conic, 151
difference
of double rotors, 125
Ceva’s theorem, 6, 34
dilatation,
290
chord
directrix,
292
ellipse, 238
distance
concurrent pencil, 31
rotors, 108
dual diametral, 224
double
rotor, 122
conic, 134
double
rotors
central, 151
difference,
125
centre, 151
sum, 125
conjugate points, 153
doubly deficient incidence, 146
diametral line, 169
dual diametral concurrent pencil, 224
directrix, 292
dual equipoised quotients, 82
eccentricity, 292
dual mid-line, 215
focus, 292
dual mid-pair, 215
non-central, 152
dual parallelogram, 115
point-pair, 192
dual polar, 199
proper, 135
dual sensed-area, 6
proto-directrix, 297
dual-centre, 223
similar and similarly situated, 298
dual-diametral concurrent pencils
tangent, 147
conjugate, 226
conjugate
dual-elliptic type, 221
diametral lines, 171
conjugate dual-diametral concurrent pencils, 226 dual-hyperbolic type, 221
503
504
dual-parabolic type, 221
dual-parallel points, 210
dual-parallel rotors, 104
eccentric angle
ellipse, 241
eccentricity
conic, 292
ellipse, 140
chord, 238
diameter, 239
eccentric angle, 241
enlargement, 291
equation
incidence, 146
equipoised quotient, 58
Euler line, 254
exterior region, 144
focus
conic, 292
harmonic pencil, 60
harmonic property, 67
harmonic range, 60
homothetic, 298
hyperbola, 140
branch, 143
conjugate, 263
incidence
deficient, 146
doubly deficient, 146
equation, 146
interior region, 144
intrinsic angle, 109, 110
inversion, 463
involution, 278
line
Z0 -non-incident, 205
line- equation
point, 75
lineo-half-loci, 464
lineo-locus, 464
lineo-segments, 464
lines
diagonal, 36
magnification ratio, 234
Magnus
quotient, 58
Menelaus’ theorem, 5
mid-line
INDEX
dual, 215
mid-pair
dual, 215
mid-rotor, 108
normalized areal coordinates
points, 40
normalized areal pair-coordinates, 206
null rotor, 91
oblique axial symmetry, 270
oblique coordinates, 272
opposite vertices of quadrilateral, 36
pair
Z0 non-aligned, 205
amenable, 31
areal coordinates, 75
pair-coordinates
normalized areal, 206
Pappus’ theorem, 14
parabola, 140
parallel pencil, 31
parallel perspectivity, 18
parallelogram
dual, 115
pencil
concurrent, 31
cross-ratio, 60
harmonic, 60
parallel, 31
pencil of rotors, 100
pencil-perspectivity, 38
perspectivity
central, 16
parallel, 18
point
line-equation, 75
point-pair conic, 192
Z0 -deleted, 220
points
dual-parallel, 210
normalized areal coordinates, 40
polar, 150
pole, 155, 222
projective transformation, 53
projectivity
points, 19
proper conic, 135
proto-directrix, 297
proto-focus, 296
Ptolemy’s theorem, 62
quadrangle, 24
505
INDEX
quadrilateral, 36
opposite vertices, 36
quadruple
sensed area, 25
quotient
equipoised, 58
Magnus, 58
quotients
dual equipoised, 82
radical axis, 302
range
harmonic, 60
ratio
magnification, 234
reciprocal sensed-area, 326
region
exterior, 144
interior, 144
rotor, 91
null, 91
scalar multiple, 91
rotors
difference, 95
distance, 108
dual-parallel, 104
pencil, 100
sensed- distance, 107
sum, 93
vertex, 115
scalar multiple
rotor, 91
self-conjugate triple or triangle, 166
sensed area
quadruple, 25
sensed-area, 1
dual, 6
reciprocal, 326
sensed-distance
rotors, 107
similar and similarly situated, 298
similarity transformations, 287
sum
rotors, 93
sum, double rotors, 125
tangent
conic, 147
theorem
Ceva’s, 6, 34
Desargues’ perspective, 15
Menelaus’, 5
Pappus’, 14
Ptolemy’s, 62
transformation
affine, 53
projective, 53
similarity, 287
triple
diagonal, 24
vertex rotors, 115