Translation of the Kimberling’s Glossary into barycentrics
(v72-pdf2)
March 30, 2017
.
Acknowledgement. This document began its life as a private copy of the Glossary accompanying
the Encyclopedia of Clark Kimberling ETC. This Glossary –as its name suggests– is organized
alphabetically. As a never satisfied newcomer, I would have preferred a progression from the
easiest to the hardest topics and I reordered this document in my own way. It is unclear whether
this new ordering will be useful to someone else ! In any case a detailed index is provided.
Second point, the Kimberling’s Glossary is written using trilinear coordinates. From an advanced point of view, these coordinates are neither better nor worse than the barycentric coordinates. Nevertheless, having some practice of the barycentrics, and none of the trilinears, I
undertook to translate everything, from one system to another. In any case, this was a formative exercise, and this also puts the focus on the covariance/contra-variance properties that were
subsequently systematized.
Drawings are the third point. Everybody knows –or should know– that geometry is not possible
at all without drawings. Having no intention to pay royalties for using rulers and compasses, I
turned to an open source software (kseg) in order to produce my own drawings. Thereafter, I have
used Geogebra, together with pstricks. What a battle, but no progress without practice !
Subsequently, other elements have been incorporated from other sources, including materials
about cubics, from the Gibert web site. Finally, original elements were also added. As it will
appear at first sight, "pldx" is addicted to a tradition that requires a precisely specified universal
space for each object to live in.
A second massive addiction of the author is computer algebra. Having at your fingertips a
tool that gives the right answer to each and every expansion or factorization, and never lost the
small paper sheet where the computation of the week was summarized is really great. Moreover,
being constrained to explain everything to a computer helps to specify all the required details.
For example, the "equivalence up to a proportionality factor" doesn’t apply in the same way to
a matrix whether this matrix describes a collineation, a triangle, a trigon or a set of incidence
relations.
In this document, "beautiful geometrical proofs" are avoided as most as possible, since they
are the most error prone. A safe proof of "the triangle of contacts of the inscribed circle and the
triangle of the mid-arcs on the circumcircle admit the similicenter of the two circles as perspector
(and therefore as center of similitude)" is :
ency(persp(matcev(pX(7)),matucev(pX(1)))) 7→ 56
where the crucial point is ency, i.e. a safe implementation of the Kimberling’s search key method
to explore the database.
To summarize, the present document is rather a "derived work", where the elements presented
are not intended to be genuine, apart perhaps from the way to assemble the pieces and cook them
all together.
Index
=well-known, 2
circles, 122
conics, 100
cubics, 192
lines, 28
matrices, 67
surds, 121
triangles, 41
Brocard
Brocard
Brocard
Brocard
line, 28
points, 68
second circle, 129
triangles, 41
C
center of a circle, 120
center of a conic, 93
central line, 33
central point, 33
central triangle, 35
ceva conjugacy, 49
cevadiv, 49, 176, 177, 219
cevamul, 48, 176, 219
cevapoint, 49
cevian collineation, 176
cevian conics, 101
cevian conjugacies, 177
cevian lines, 40
cevian nest, 46
cevian triangle, 40
circle, 119
circlekit, 120
list of well-known circles, 122
circum-anticevian, 83
circumcevian, 83
circumcircle, 100, 123
standard eq, 64
circumconic, 94
asymptote, 110
circum-hyperbola, 110
circumRH, 111
gudulic point, 102
parameterization, 95
circum-hyperbola, 110
circumpolar, 89, 161
circumRH, 111
cK(F,U), 204
cocevian triangle, 43
collinear, 22
collineation, 173
collineation algorithm, 173
Collings transform, 222
combos, 39
complement, 37, 176
complementary conjugate, 221
completed line, 138
complex affix, 155
concurrent, 23
conic, 89
A
affix, 155
complex, 155
Morley, 154
affix, Morley, 157
alephdivision, 177
Al-Kashi, 54, 62
angle between lines, 66
anticevian triangle, 42
anticomplement, 37, 176
anticomplementary conjugate, 221
anticomplementary triangle, 37, 41
antigonal conjugacy, 188
antimedial triangle, 37
antiorthic axis, 28
Apollonius circle, 126
Arbelos, 131
area, 62
complex affixes, 155
Heron, 62
matrix, 62
auxiliary line (of an inconic), 96
B
backward matrix, 158
backward substitutions, 158
backward-2 matrix, 163
backward-2 substitutions, 163
barycentric division, 26
barycentric multiplication, 25, 175
barycentrics, 22, 25
barycentrics (standard), 23
barymul, 220
base points (pencil of circles), 141, 185
Bevan circle, 125
Bevan line, 28
bicentric points, 33
Brocard
angle, 68
Brocard 3-6 circle, 128, 214
Brocard ellipse, 107
3
4
center, 93
cross-ratio, 99
perspector, 93
proper, 92
tangential conic, 96
conico-pivotal isocubic, 204
conjugate circle, 125
Conway symbols, 35
cosinus of projection, 71
cot versus Conway, 69
Cremona, 181, 183
exceptional curves, 183
indeterminacy points, 183
cross conjugacy, 47
crossdiff, 220
crossdifference, 27
crossdiv, 47, 176, 219
crossmul, 47, 176, 219
crosspoint, 47
cross-ratio, 38
conic, 99
Laguerre formula, 157
of parameters, 38
Riemann sphere, 38
crosssum, 203
crosstriangle, 44
cubic, 191
cK(F,U), 204
list of well-known cubics, 192
nK(P,U,k), 203
nK0(P,U), 203
pK(P,U), 192
pK: 22 points property, 193
PKA, 196
triangular, 196
cycle, 137
cyclocevian conjugate, 134
cyclopedal conjugate, 76
D
dalethdivision, 179
Danneels (first) perspector, 50
Darboux cubic, 199
DC point, 85
de Longchamps axis, 28
Desargues
theorem, 44
triangle, 44
trigone, 44
desmic mate, 85
directrix, 108
distance point-line, 68
distance point-point, 63
div formula, 219
dual of a conic, 93
dual triangle, 85
duality, 26
March 30, 2017 14:09
INDEX
E
eigencenter, 180
eigentransform, 202
embedded vector, 62
Euler line, 28
excentral triangle, 41
exceptional curves (Cremona), 183
Exceter point, 83
extouch triangle, 41
F
focus, 102
folium, 90
forward matrix, 158
forward substitutions, 158
forward-2 matrix, 163
forward-2 substitutions, 163
fourth harmonic, 38, 39
Fuhrmann 4-8 circle, 130
Fuhrmann triangle, 41, 215
G
geogebra, 24
Gergonne axis, 28
Gibert-Simson transform, 205
gimeldivision, 178
Gt mapping, 118
gudulic point, 102
H
harmonic division, 39
helethdivision, 179
Hexyl triangle, 41
Hirst inverse, 221
hirstpoint, 221
homography, 181
multiplier, 182
homothety, 73
horizon, 138
hyperbola, 110
circum-hyperbola, 110
circumRH, 111
hyperbola (inhyperbola), 111
I
incentral triangle, 41
incircle, 100, 101, 124
incircle transform, 124
inconic, 96
auxiliary line, 96
center, 96
parametrization, 97
perspector, 96
inconic (in-hyperbola), 111
indeterminacy points (Cremona), 183
infinity line, 28
in-hyperbola, 111
intouch triangle, 41
inversion in a circle, 122, 142
published under the GNU Free Documentation License
INDEX
5
involutory collineation, 174
isoconjugacy, 185
isodynamic points, 71, 129, 146
isogonal conjugacy, 76
Morley, 186, 196
isopivot of a cubic, 193
isoptic pencil, 141
isosceles triangles, 130
isotomic conjugacy, 41
isotomic pencil, 141
isotropic lines, 66
J
Jerabek RH, 100
K
K001 Neuberg, 197
K002 Thomson, 198
K003 McKay, 197
K004 Darboux, 199
K007 Lucas, 199
K010 Simson, 204
K060 cubic, 201
K155 EAC2, 200
K162 cubic, 205
K170 cubic, 193
K219, 112
K-ellipse, 107
Kiepert parabola, 100, 161
Kiepert RH, 100, 130
K-matrix, 62
kseg, 24
L
Laguerre formula, 157
Lemoine axis, 28, 100
Lemoine first circle, 126
Lemoine inconic, 100
Lemoine second circle, 127
Lemoine transform, 163, 184
LFIT, 235, 236
area center, 238
circle of similarity, 243
fixed points, 244
graphs (Catalan), 241
graphs (hexagonal), 237
hexagonal conic, 236
observer, 251
parabolas, 246
parametrization, 239
pilar conic, 240
pilar point, 240
slowness center, 237
temporal graphs, 240
velocities, 237
Vertex-Miquel circle, 243
limit points (pencil of circles), 141, 185
line, 22
at infinity, 23, 155
completed, 138
list of well-known lines, 28
Linear Families of Inscribed Triangles, 235
Longchamps circle, 125
Lubin parametrization, 157
Lubin-2 parameterization, 163
Lucas cubic, 199
M
MacBeath circumconic, 100
MacBeath-inconic, 100
major center, 33
matrix
K, 62
M, 66, 156
N, 257
OrtO, 65, 156
Pyth, 63, 156
Q, 138, 168
Q^-1, 138
W, 61, 156
McKay cubic K003, 197
medial triangle, 37, 41
Menelaus theorem, 43
Miguel theorem, 43
mimosa transform, 178
Miquel circle (quadrilateral), 256
Morley affix, 154, 157
Morley space, 154
mul formula, 219
multiplier of an homography, 182
N
Nagel line, 28
Neuberg circles, 69
Neuberg K001 cubic, 197
Newton axis, 255
Newton line, 43, 116
nine-points circle, 101, 124
NK transform, 203
nK(P,U,k), 203
nK0(P,U), 203
non pivotal isocubic, 191
O
orthic axis, 28, 78
orthic triangle, 41
orthoassociate, 81
orthocentroidal 2-4 circle, 129
orthocorrespondent, 80
orthodir, 66, 156
orthogonal cycle, 140
orthogonal projector, 71, 175
orthogonal reflection, 71, 175
orthologic (triangles), 80
orthopoint, 65
orthopole, 257
OrtO-matrix, 65
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
6
P
parabola, 93, 108
parallelogic (triangles), 80
pedal triangle, 75
Pelle à Tarte, 130
pencil, 141
pencil, bitangent, 113
pencil, Miquel, 115
perspectivity, 44
perspectivity kit, 45
perspector, 44
perspector of a conic, 93
perspectrix, 44
pivot of a cubic, 192
pivotal conic, 204
pivotal umbilical pK, 197
PK# transform, 193
pK(P,U), 192
pK: 22 points property, 193
PKA cubics, 196
Plucker
formulas, 91
representation, 57
point, 22
polar wrt a conic, 93
polar wrt a curve, 89
polardiv, 221
polarmul, 95
polarmul of lines, 221
polarmul of points, 221
polars-of-points triangle, 179
pole wrt a conic, 93
poles-of-lines triangle, 179
polynomial center, 33
Poulbot, first point, 168
Poulbot, second point, 168
power, 63
barycentric formula, 63, 119
Veronese formula, 140
projector (orthogonal), 71
psi-Kimberling, 177
Pythagoras theorem, 63
Pyth-matrix, 63
Q
quadrilateral, 255
conjugate circles, 256
embedded triangles, 255
Miquel circle, 256
Miquel point, 256
Newton axis, 255
Newton pencil, 255
reciprocal lines, 255
R
radical center, 141
radical trace, 121
radius of a circle, 120, 140
rational center, 34
March 30, 2017 14:09
INDEX
reciprocal conjugacy, 186
reciprocal of a line, 41
rectangular hyperbola, 110
reflection, 37
representative, 137
of a point-circle, 138
residual triangles, 212
RH, 110
Rigby point, 259
rotation, 175
RS, 259
S
S=area(ABC), 35
S_a, 35
Saragossa points, 85
search key
Kimberling, 23
Morley, 24, 159
patched, 24
second Danneels perspector, 51
shortest cubic, 198
sideline triangle, 44
sideline trigone, 44
similarity, 174
similitude (centers of), 121
Simson cubic, 204
Simson line, 258
Simson-Moses point, 260
sine-triple-angle circle, 128
Six-circles transform, 111
slowness center, 237
SM, 260
Soddy (but not so) circles, 148
Soddy circles, 147
Soddy line, 28
Sondat, 227
Spieker circle, 126
sqrtdiv, 26, 184
sqrtmul, 26
squarediv, 50
squaremul, 50
SR, 259
star triangle, 41, 215
Steinbart transform, 83
Steiner circumellipse, 51, 100
Steiner in-ellipse, 100, 105
Steiner line, 77
symbolic substitution, 35
symmetric functions, 157
symmetry (orthogonal), 71
T
tangent, 66
tangent of two lines, 67, 156
tangent to a curve, 89
tangential conic, 96
tangential triangle, 41
Tarry point, 69
published under the GNU Free Documentation License
INDEX
7
TCC-perspector, 83
Tg mapping, 118
Thomson K002 cubic, 198
transcendental center, 33
translation, 174
transpose, 27
triangle, 22
list of well-known triangles, 41
may be degenerate, 22
triangle center, 33
trigon, 23
trilinear pole (deprecated), 27
trilinears, 25
tripolar, 27, 28
tripolar centroid, 43
tripolar line, 43
tripole, 27, 28
type-crossing, 25
type-keeping, 25
U
umbilics, 136
unary cofactor triangle, 180
V
vandermonde, 157
Veronese map
barycentric, 137
Morley space, 168
vertex associate, 86
vertex triangle, 44
vertex trigon, 44
virtual circle, 143
visible point, 154
W
wedge operator, 26
W-matrix, 61
Y
Yff parabola, 100
Z
Z(U) cubic, 195
Z+cubic, 203
zosma, 179
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
8
March 30, 2017 14:09
INDEX
published under the GNU Free Documentation License
List of Figures
1.1
Point P and line A”B”C” are the tripolars of each other. . . . . . . . . . . . . . . .
28
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
Obtain Q from A,P,C and auxiliary D . . . . . . . . .
Ceva’s Theorem . . . . . . . . . . . . . . . . . . . . . .
Cevian, anticevian and cocevian triangles . . . . . . .
Triangles ABC and UVW that admit P as perspector.
P is cevamul(U,X) . . . . . . . . . . . . . . . . . . . .
crossmul, crossdiv . . . . . . . . . . . . . . . . . . . . .
P is crossmul of U and X . . . . . . . . . . . . . . . .
cevamul, cevadiv . . . . . . . . . . . . . . . . . . . . .
Lamoen’s construction of X=cevamul(P,U) . . . . . .
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39
40
42
46
46
48
48
49
50
7.1
Level curves of the Brocard angle of the pedal triangle of point M . . . . . . . . .
72
8.1
Cyclopedal congugates are isogonal conjugates . . . . . . . . . . . . . . . . . . . .
76
9.1
9.2
The orthopoint transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Reflection triangle and its perspector. . . . . . . . . . . . . . . . . . . . . . . . . .
78
82
10.1 Saragossa points of point P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
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11.1 Folium of Descartes . . . . . . . . . . . . . . . . . . . .
11.2 Dual curve of the folium . . . . . . . . . . . . . . . . . .
11.3 An inconic and a circumconic with the same perspector
11.4 Conjugacy between a line and a circumconic . . . . . . .
11.5 How to generate an inscribed conic from its perspector.
11.6 Stereographic projection . . . . . . . . . . . . . . . . . .
11.7 Focus of an inconic . . . . . . . . . . . . . . . . . . . . .
11.8 The Steiner in ellipse . . . . . . . . . . . . . . . . . . . .
11.9 The K-ellipse . . . . . . . . . . . . . . . . . . . . . . . .
11.10Two transformations of the focal cubic. . . . . . . . . . .
11.11The Miguel pencil and its focal cubic . . . . . . . . . . .
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90
92
94
96
97
99
104
106
107
114
117
12.1 Sin triple angle circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
12.2 Cake server (Pelle a Tarte) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
12.3 Arbelos configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
13.1
13.2
13.3
13.4
No point-shadow fall outside of the IC(X76) inconic
Euler pencil and incircle . . . . . . . . . . . . . . . .
Lemoine and Brocard pencils . . . . . . . . . . . . .
Apollonius circles of the not so Soddy configuration.
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142
144
145
149
14.1 Poulbot’s points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
14.2 The Euler pencil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
16.1 Antigonal conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
17.1 pK(2,4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
9
10
LIST OF FIGURES
17.2
17.3
17.4
17.5
17.6
The Darboux and Lucas cubics . . . . . . . . .
The Simson cubic (as depicted in Gibert-CTP)
The Simson diagram . . . . . . . . . . . . . . .
Angels . . . . . . . . . . . . . . . . . . . . . . .
Cartesian ovals by inversion . . . . . . . . . . .
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199
205
206
209
209
18.1 Special Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
18.2 The star triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
19.1 The complementary and anticomplementary conjugates . . . . . . . . . . . . . . . 221
19.2 Collings configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
19.3 Collings locus is a ten points rectangular hyperbola . . . . . . . . . . . . . . . . . . 224
20.1 Orthology and perspective. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
21.1
21.2
21.3
21.4
21.5
Fixed triangle abc is inscribed into the fixed trigon ABC . . .
Variable triangle abc is inscribed into fixed triangle ABC . .
Using the inscribed hexagon to construct the variable triangle
Miquel circles . . . . . . . . . . . . . . . . . . . . . . . . . . .
Two parabola . . . . . . . . . . . . . . . . . . . . . . . . . . .
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235
237
237
244
247
22.1 Point U is the orthopole of line A0 B 0 C 0 . . . . . . . . . . . . . . . . . . . . . . . . . 257
22.2 Revisiting the Sister Mary Cordia Karl’s correspondence . . . . . . . . . . . . . . . 261
March 30, 2017 14:09
published under the GNU Free Documentation License
List of Tables
1.1
Some well-known lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.1
Major centers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
4.1
4.2
Some well-known triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Three cases of cevian nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
47
7.1
All these matrices
67
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.1 Some Inconics and Circumconics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
11.2 Perspector and focuses of some in-conics . . . . . . . . . . . . . . . . . . . . . . . . 105
12.1
12.2
12.3
12.4
Some usual square roots (circle kit 1) .
Some usual notations (circle kit 2) . .
Some circles . . . . . . . . . . . . . . .
Similicenters on Kiepert RH . . . . . .
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133
14.1 Action of the Klein group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
16.1 Reduction of a Cremona transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
17.1 Some well-known cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
17.2 Some cubic shadows on EAC2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
24.1 Structure of the database . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
26.1 The various species of points in the Kimberling’s database . . . . . . . . . . . . . . 289
11
12
March 30, 2017 14:09
LIST OF TABLES
published under the GNU Free Documentation License
Contents
Acknowledgement
2
Index
3
List of Figures
8
List of Tables
11
Table of Contents
13
1 Introduction
1.1 Special remarks for French natives . . . .
1.2 Basic objects : points and lines . . . . . .
1.3 Search key . . . . . . . . . . . . . . . . . .
1.4 Figures . . . . . . . . . . . . . . . . . . .
1.4.1 Using kseg . . . . . . . . . . . . .
1.4.2 Using Geogebra . . . . . . . . . . .
1.5 Type-keeping and type-crossing functions
1.6 Duality between point and lines . . . . . .
1.7 Isoconjugacy has moved . . . . . . . . . .
21
22
22
23
24
24
24
25
26
29
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2 Teaching Geometry to a computer
31
3 Central objects
3.1 Triangle centers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Central triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Symbolic substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
33
34
35
4 Cevian stuff
4.1 Centroid stuff . . . . . . . . . . . . . . . . . . . .
4.2 Cross-ratio and fourth harmonic . . . . . . . . .
4.3 About combos . . . . . . . . . . . . . . . . . . .
4.4 Cevian, anticevian, cocevian triangles . . . . . .
4.5 Transversal lines, Menelaus and Miquel theorems
4.6 Perspectivity . . . . . . . . . . . . . . . . . . . .
4.7 Cevian nests . . . . . . . . . . . . . . . . . . . .
4.8 The cross case (aka case I, cev of cev) . . . . . .
4.9 The ceva case (aka case II, cev and acev) . . . .
4.10 The square case (aka case III, acev of acev) . . .
4.11 Danneels perspectors . . . . . . . . . . . . . . . .
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37
37
38
39
40
42
44
46
47
48
50
50
French touch
The random observer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
An involved observer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
From an involved observer to another one . . . . . . . . . . . . . . . . . . . . . . .
53
53
54
55
5 The
5.1
5.2
5.3
13
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14
6 Brief extension to 3D
6.1 Basic results . . . .
6.2 Electric notations .
6.3 Examples . . . . .
CONTENTS
spaces
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Euclidean structure using barycentrics
7.1 Lengths and areas . . . . . . . . . . . . . . . . .
7.2 Embedded Euclidean vector space . . . . . . . .
7.3 About circumcircle and infinity line . . . . . . . .
7.4 Orthogonality . . . . . . . . . . . . . . . . . . .
7.5 Angles between straight lines . . . . . . . . . . .
7.6 Distance from a point to a line . . . . . . . . . .
7.7 Brocard points and the sequel . . . . . . . . . . .
7.7.1 Some results . . . . . . . . . . . . . . . .
7.7.2 Results related to the Kiepert RH . . . .
7.7.3 Spoiler: study of the Neuberg pencil . . .
7.7.4 Spoiler: Brocard angle of a pedal triangle
7.8 Orthogonal projector onto a line . . . . . . . . .
57
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58
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61
61
62
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68
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70
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70
71
8 Pedal stuff
8.1 Pedal triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Isogonal conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3 Cyclopedal conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
75
76
76
9 Orthogonal stuff
9.1 Steiner line . . . . . . . . . . . . . . . . . . .
9.2 Spoiler: Moebius-Steiner-Cremona transform
9.3 Orthopole . . . . . . . . . . . . . . . . . . . .
9.4 Orthology . . . . . . . . . . . . . . . . . . . .
9.5 Parallelogy . . . . . . . . . . . . . . . . . . .
9.6 Orthocorrespondents . . . . . . . . . . . . . .
9.7 Isoscelizer . . . . . . . . . . . . . . . . . . . .
9.8 Orion Transform . . . . . . . . . . . . . . . .
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77
77
78
79
80
80
80
81
81
10 Circumcevian stuff
10.1 Circum-cevians, circum-anticevians
10.2 Steinbart transform . . . . . . . . .
10.3 Circum-eigentransform . . . . . . .
10.4 Dual triangles, DC and CD Points
10.5 Saragossa points . . . . . . . . . .
10.6 Vertex associates . . . . . . . . . .
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83
83
83
84
85
85
86
11 About conics
11.1 Tangent to a curve . . . . . . . . . . .
11.2 Folium of Descartes . . . . . . . . . .
11.3 General facts about conics . . . . . . .
11.4 Circumconics . . . . . . . . . . . . . .
11.5 Tangential conics . . . . . . . . . . . .
11.6 Inconics . . . . . . . . . . . . . . . . .
11.7 Conic cross-ratios . . . . . . . . . . . .
11.8 Some in- and circum- conics . . . . . .
11.9 Cevian conics . . . . . . . . . . . . . .
11.10Direction of axes . . . . . . . . . . . .
11.11Focuses of a conic . . . . . . . . . . . .
11.12Focuses of an inconic . . . . . . . . . .
11.13Focuses of the Steiner inconic . . . . .
11.13.1 Using barycentrics . . . . . . .
11.13.2 Using Morley affixes . . . . . .
11.14The Brocard ellipse, aka the K-ellipse
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89
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101
101
102
103
105
105
107
107
March 30, 2017 14:09
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published under the GNU Free Documentation License
CONTENTS
11.15Parabola . . . . . . . . . . . . . . . . .
11.15.1 Inscribed parabola . . . . . . .
11.15.2 Circum-parabola . . . . . . . .
11.16Hyperbola . . . . . . . . . . . . . . . .
11.16.1 Circum-hyperbolas . . . . . . .
11.16.2 Circum-rectangular-hyperbolas
11.16.3 Inscribed hyperbolas . . . . . .
11.17Diagonal conics . . . . . . . . . . . . .
11.17.1 Pencils of diagonal conics . . .
11.18The bitangent pencil . . . . . . . . . .
11.19The Miquel pencil of conics . . . . . .
11.20Tg and Gt mappings . . . . . . . . . .
15
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108
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12 More about circles
12.1 General results . . . . . . . . . . . . . . . .
12.2 Antipodal Pairs on Circles . . . . . . . . . .
12.3 Inversion in a circle . . . . . . . . . . . . . .
12.4 Circumcircle . . . . . . . . . . . . . . . . . .
12.5 Incircle . . . . . . . . . . . . . . . . . . . . .
12.6 Nine-points circle . . . . . . . . . . . . . . .
12.7 Conjugate circle . . . . . . . . . . . . . . . .
12.8 Longchamps circle . . . . . . . . . . . . . .
12.9 Bevan circle . . . . . . . . . . . . . . . . . .
12.10Spieker circle . . . . . . . . . . . . . . . . .
12.11Apollonius circle . . . . . . . . . . . . . . .
12.12First Lemoine circle . . . . . . . . . . . . .
12.13Second Lemoine circle . . . . . . . . . . . .
12.14Sine-triple-angle circle . . . . . . . . . . . .
12.15Brocard 3-6 circle and second Brocard circle
12.16Orthocentroidal 2-4 circle . . . . . . . . . .
12.17Fuhrmann 4-8 circle . . . . . . . . . . . . .
12.18Kiepert RH and isosceles adjunctions . . . .
12.19Cyclocevian conjugate . . . . . . . . . . . .
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119
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134
13 Pencils of Cycles in the Triangle Plane
13.1 Introductory remarks . . . . . . . . . . . . . . . . . .
13.1.1 How many points at infinity should be used ?
13.1.2 Umbilics . . . . . . . . . . . . . . . . . . . . .
13.1.3 Notations . . . . . . . . . . . . . . . . . . . .
13.2 Cycles and representatives . . . . . . . . . . . . . . .
13.3 Fundamental quadric and orthogonality . . . . . . .
13.4 Pencils of cycles . . . . . . . . . . . . . . . . . . . . .
13.5 Classification of pencils . . . . . . . . . . . . . . . .
13.6 Euler pencil and incircle . . . . . . . . . . . . . . . .
13.7 The Brocard-Lemoine pencils . . . . . . . . . . . . .
13.8 The Apollonius configuration . . . . . . . . . . . . .
13.8.1 Tangent cycles in the representative space . .
13.8.2 An example: the Soddy circles . . . . . . . .
13.8.3 An example: the not so Soddy circles . . . .
13.8.4 Yet another example : the three excircles . .
13.8.5 The special case . . . . . . . . . . . . . . . .
13.9 Elementary properties of the Triangle Lie Sphere . .
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135
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—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
16
CONTENTS
14 Morley spaces, or how to use complex numbers
14.1 How to deal with complex conjugacy ? . . . . . . . . .
14.2 The usual operators of the Morley space . . . . . . . .
14.3 Lubin representation of first degree . . . . . . . . . . .
14.4 Comparing barycentric and Morley spaces . . . . . . .
14.5 Some examples of first degree . . . . . . . . . . . . . .
14.6 Lubin representation of second degree . . . . . . . . .
14.7 Poncelet representation . . . . . . . . . . . . . . . . .
14.8 Poulbot’s points (using the Lubin-4 parameterization)
14.9 Pencil of cycles . . . . . . . . . . . . . . . . . . . . . .
14.9.1 Veronese map . . . . . . . . . . . . . . . . . . .
14.9.2 Back to the Euler pencil . . . . . . . . . . . .
14.9.3 Back to the Brocard-Lemoine pencil . . . . . .
14.10Comparison with Cartesian and Artinian metrics . . .
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153
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15 Collineations
15.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . .
15.2 Involutory collineations . . . . . . . . . . . . . . . . .
15.3 Usual affine transforms as collineations . . . . . . . . .
15.4 Barycentric multiplication as a collineation . . . . . .
15.5 Complement and anticomplement as collineations . . .
15.6 Collineations and cevamul, cevadiv, crossmul, crossdiv
15.7 Cevian conjugacies . . . . . . . . . . . . . . . . . . . .
15.8 Miscellany . . . . . . . . . . . . . . . . . . . . . . . . .
15.8.1 Poles-of-lines and polar-of-points triangles . . .
15.8.2 Unary cofactor triangle, eigencenter . . . . . .
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16 Cremona group and isoconjugacies
16.1 Introducing the Cremona transforms
16.2 Working out some examples . . . . .
16.3 Isoconjugacy and sqrtdiv operator .
16.3.1 Morley point of view . . . . .
16.3.2 The isogonal Morley formula
16.3.3 Isoconjkim . . . . . . . . . .
16.4 Antigonal conjugacy . . . . . . . . .
16.5 Isogonality and perspectivity . . . .
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17 About cubics
17.1 Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . .
17.2 Pivotal isocubics pK(P,U) . . . . . . . . . . . . . . . .
17.3 Another description . . . . . . . . . . . . . . . . . . .
17.4 pK isocubics relative to pole X(6) . . . . . . . . . . . .
17.4.1 ABCIJKL cubics: the Lubin(2) point of view
17.4.2 A more handy basis . . . . . . . . . . . . . . .
17.5 Darboux and Lucas cubics aka K004 and K007 . . . .
17.6 Equal areas (second) cevian cubic aka K155 . . . . . .
17.7 The cubic K060 . . . . . . . . . . . . . . . . . . . . . .
17.8 Eigentransform . . . . . . . . . . . . . . . . . . . . . .
17.9 Non pivotal isocubics nK(P,U,k) and nK0(P,U) . . . .
17.10Conicopivotal isocubics cK(#F,U) . . . . . . . . . . .
17.11Simson cubic, aka K010 . . . . . . . . . . . . . . . . .
17.12Brocard second cubic aka K018 . . . . . . . . . . . . .
17.13A family of circular cubics . . . . . . . . . . . . . . . .
March 30, 2017 14:09
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published under the GNU Free Documentation License
CONTENTS
17
18 Special Triangles
18.1 Changing coordinates, functions and equations
18.2 Residual triangles . . . . . . . . . . . . . . . . .
18.3 Incentral triangle . . . . . . . . . . . . . . . . .
18.4 Excentral triangle . . . . . . . . . . . . . . . . .
18.5 Medial triangle . . . . . . . . . . . . . . . . . .
18.6 Antimedial triangle . . . . . . . . . . . . . . . .
18.7 Orthic triangle . . . . . . . . . . . . . . . . . .
18.8 Tangential triangle . . . . . . . . . . . . . . . .
18.9 Brocard triangle (first) . . . . . . . . . . . . . .
18.10Brocard triangle (second) . . . . . . . . . . . .
18.11Brocard triangle (third) . . . . . . . . . . . . .
18.12Intouch triangle (contact triangle) . . . . . . .
18.13Extouch triangle . . . . . . . . . . . . . . . . .
18.14Hexyl triangle . . . . . . . . . . . . . . . . . . .
18.15Fuhrmann triangle . . . . . . . . . . . . . . . .
18.16Star triangle . . . . . . . . . . . . . . . . . . . .
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211
211
212
212
212
212
213
213
213
214
214
214
214
215
215
215
215
19 Binary operations
19.1 cevamul, cevadiv, crossmul, crossdiv . . . . . . . .
19.2 Formal operators and conics . . . . . . . . . . . . .
19.3 crossdiff, crosssum, polarmul, polardiv . . . . . . .
19.4 Complementary and anticomplementary conjugates
19.5 Hirstpoint aka Hirst inverse . . . . . . . . . . . . .
19.6 Line conjugate . . . . . . . . . . . . . . . . . . . .
19.7 Collings transform . . . . . . . . . . . . . . . . . .
19.8 Brisse Transform . . . . . . . . . . . . . . . . . . .
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219
219
220
220
221
221
222
222
224
20 Sondat theorems
20.1 Perspective and directly similar . . . . . . .
20.2 Perspective and inversely similar . . . . . .
20.3 Orthology, general frame . . . . . . . . . . .
20.4 Simply orthologic and perspective triangles
20.5 Bilogic triangles . . . . . . . . . . . . . . . .
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227
227
229
231
232
234
21 Linear Families of Inscribed Triangles
21.1 Inscribed triangle . . . . . . . . . . . .
21.1.1 Induced hexagon . . . . . . . .
21.2 Linear families . . . . . . . . . . . . .
21.2.1 Slowness center . . . . . . . . .
21.2.2 Equicenter . . . . . . . . . . .
21.2.3 Parametrization of a LFIT . .
21.3 Other hexagons, other graphs . . . . .
21.3.1 Hexagonal graphs . . . . . . .
21.3.2 Temporal and Catalan graphs .
21.3.3 Graphs, the general case . . . .
21.4 Temporal point of view . . . . . . . .
21.5 Miquel circles . . . . . . . . . . . . . .
21.6 Constructions . . . . . . . . . . . . .
21.7 Similarities and Cremona transforms .
21.8 Parabolas . . . . . . . . . . . . . . . .
21.9 Equilateral triangles . . . . . . . . . .
21.9.1 General formula . . . . . . . .
21.9.2 Some remarks . . . . . . . . . .
21.9.3 Equilateral pedal triangles . . .
21.10Families of constant area . . . . . . . .
21.10.1 Three non concurrent lines . .
21.10.2 Three concurrent lines . . . . .
21.11When the graphs are given . . . . . .
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235
235
235
236
236
238
238
239
239
240
241
242
243
244
245
246
246
246
246
247
247
247
248
249
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—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
18
CONTENTS
21.11.1 Catalan graphs . . . . . . . . . . . .
21.11.2 Hexagonal graphs . . . . . . . . . .
21.11.3 The marvelous formula . . . . . . .
21.12Observers (about perspectivities) . . . . . .
21.12.1 Metric observer . . . . . . . . . . . .
21.12.1.1 Forcing the orthocenter . .
21.12.1.2 Forcing the isogonal center
21.12.2 Cevenol graphs . . . . . . . . . . . .
21.12.3 Poulbot observers . . . . . . . . . .
21.12.4 Degenerate observers . . . . . . . . .
21.12.5 Counting in brute force . . . . . . .
22 Quadrilateral
22.1 Newton stuff . . . . . . . . . . . . . .
22.2 Steiner stuff . . . . . . . . . . . . . . .
22.3 Pedal LFIT and orthopole . . . . . . .
22.4 Some properties of the Simson lines .
22.5 Rigby points . . . . . . . . . . . . . .
22.6 Sister Marie Cordia Karl . . . . . . . .
22.7 The four pedal LFIT of a quadrilateral
22.8 Exercises . . . . . . . . . . . . . . . .
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249
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255
255
256
257
258
259
260
261
262
23 Combos
265
23.0.1 More-more combos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
23.0.2 More-more-more combos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
23.0.3 more-more-more-more combos . . . . . . . . . . . . . . . . . . . . . . . . . 267
24 Import the 4995-7000 part of ETC
24.1 Requirements . . . . . . . . . . . . . . . . . . . . . . .
24.2 Recreating the database . . . . . . . . . . . . . . . . .
24.2.1 The context . . . . . . . . . . . . . . . . . . . .
24.2.2 Encapsulating everything in *.php commands .
24.2.3 Main view of the structure . . . . . . . . . . .
24.2.4 The main fields . . . . . . . . . . . . . . . . . .
24.2.5 Ancilary fields . . . . . . . . . . . . . . . . . .
24.2.6 Information fields . . . . . . . . . . . . . . . . .
24.3 Obtaining compliant bar1 and hc . . . . . . . . . . . .
24.3.1 Creating data records from the ETC Web Site
24.3.2 Searchkeys . . . . . . . . . . . . . . . . . . . .
24.3.3 Towards a Maple compliant syntax . . . . . . .
24.4 SQL requests . . . . . . . . . . . . . . . . . . . . . . .
24.5 Killed points (tagged as special) . . . . . . . . . . . .
24.5.1 special . . . . . . . . . . . . . . . . . . . . . . .
24.5.2 pending to shorten . . . . . . . . . . . . . . . .
24.6 Some errors in the web page . . . . . . . . . . . . . . .
24.7 Further work . . . . . . . . . . . . . . . . . . . . . . .
24.7.1 Coherence . . . . . . . . . . . . . . . . . . . . .
24.7.2 Trigonometric lines . . . . . . . . . . . . . . . .
24.7.3 Sqrt and other dramas . . . . . . . . . . . . . .
24.8 Some items to revisit . . . . . . . . . . . . . . . . . . .
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269
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283
283
283
283
283
25 Import the 3588-4994 part of ETC
285
25.1 Some errors in the web page . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
30 mars 2017 14:09
published under the GNU Free Documentation License
CONTENTS
26 Describing ETC
26.1 Various kinds of points . . . . . . . . . . . . . . . .
26.2 Special points . . . . . . . . . . . . . . . . . . . . .
26.2.1 X(368) . . . . . . . . . . . . . . . . . . . . .
26.2.2 X(369), X(3232) trisected perimeter points
26.2.3 X(370) . . . . . . . . . . . . . . . . . . . . .
26.2.4 X(1144) . . . . . . . . . . . . . . . . . . . .
26.2.5 X(2061) . . . . . . . . . . . . . . . . . . . .
26.3 orthopoint, points on circles . . . . . . . . . . . . .
26.4 points on lines . . . . . . . . . . . . . . . . . . . .
19
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March 30, 2017 14:09
CONTENTS
published under the GNU Free Documentation License
Chapter 1
Introduction
Many changes have occurred in the way we are doing geometry, from the old ancient times of
Euclid and Apollonius. Most of them are related to yet another way of performing "automated"
computing of properties, rather than relying on intuition to find "beautiful geometric proofs".
Many individuals have contributed to this long process, and attributing a given discovery to a
given individual is not an easy task (Coolidge, 1940).
The most eminent milestones along this long road are the individuals that have summarized
the discoveries of their time into an efficient way of writing down the questions to solve. Each
time, the new way of writing was appearing as doing the job by itself and providing the required
answers through something like a least action trajectory. Heroes are no more required, replaced
by computing power.
Writing numbers and calculations in a tractable manner is associated with Al-Khwarizmi and
his Algebra (825). Using coordinates (x, y) to describe points and compute their geometrical
properties (as well as the exponent notation for polynomials) is associated with Descartes and
his Géometrie (1637). Using homogeneous coordinates x : y : z to implement the principle of
continuity when dealing with objects that escape to infinity is associated with Moebius and his
barizentrische Calcul (1827).
More recently, the very idea of stamping an hash-code on each noteworthy point involved
in Triangle Geometry and then practice some kind of computer aided inventory management
(Kimberling ETC, 1998-2014) has changed the practice of geometers. This idea has emerged from
a more general trend, where
barycentrics are understood to define points, lines, circles, triangle centers, etc., and
zero determinants are understood to define collinearity and concurrence. Doing that
way, triangle geometry, formally speaking, is much more general than the study of
a single Euclidean triangle. In the formal treatment, sometimes called transfigured
triangle geometry, the symbols a, b, c are regarded as algebraic unknowns, so that
points, defined as functions of a, b, c, are not the usual points of a two-dimensional
plane. When (a, b, c) are real numbers restricted by the "triangle inequalities" for
sidelengths, the resulting geometry is traditional triangle geometry (Kimberling, 1998).
When possible, computed proofs are given that use formal computing tools. This kind of proof
is deprecated by several authors. Nevertheless, these proofs are the easiest since all the messy
job is done by a computer and are also the safest. A construction that sounds like a "beautiful
geometrical proof" is too often invalid due to some hidden exception. During a computerized proof,
exceptions are appearing as multiplicative factors, according to the polynomial model :
conclusion × exceptions = hypothesis
To quote the Knuth’s foreword to Petkovsek et al. (1996) :
Science is what we understand well enough to explain to a computer. Art is everything
else we do. During the past several years important parts of mathematics has been
transformed from an Art to a Science.
21
22
1.1
1.1. Special remarks for French natives
Special remarks for French natives
Commençons par deux faux amis. Le terme US "line" désigne une "droite" (opposée à une courbe),
tandis qu’une "ligne" (opposée à une colonne) est désignée par le terme US "row". Ensuite, une
remarque sur les différences de "feeling" vis à vis des livres scientifiques. Lorsque vous écrivez pour
un public américain, mieux vaut commencer par montrer que le sujet est suffisamment intéressant
pour mériter du temps et de la peine. Lorsque vous écrivez pour un public français, mieux vaut
commencer par montrer que l’auteur dispose d’une hauteur de vue suffisante. Si vous percevez les
choses de cette façon, le Chapitre 5 est le bon endroit par où commencer.
1.2
Basic objects : points and lines
Abbreviation etc. will stand for "et cyclically". After an "A-object" has been defined as F (A, B, C),
then the B-object is F (B, C, A), obtained by the cyclic permutation ABC 7→ BCA, and the Cobject is, likewise, F (C, A, B). Example: If A0 is the point where lines AP and BC meet, and B 0
and C 0 are defined cyclically, then B 0 is where lines BP and CA meet, and C 0 is where lines CP
and AB meet.
Definition 1.2.1. A point is an element of PR R3 . To tell the same thing more simply, a point
is represented by a column of three numbers (the barycentrics of the point), not all of them being
zero, and such a column is dealt "in a projective manner", i.e. up to a proportionality factor. An
efficient way to write such a projective column is the colon notation :

 

p
kp

 

P =p:q:r
meaning that
P '  q  '  k q  ∀k ∈ R \ {0}
r
kr
Definition 1.2.2. A line is an element of the dual of the point space. To tell the same thing
more simply, a line is represented by a row of three numbers (the barycentrics of the line), not all
of them being zero, and such a row is dealt "in a projective manner", i.e. up to a proportionality
factor. A line will be described as :
∆ ' ρ σ τ ' λρ λσ λτ
∀λ ∈ R \ {0}
Notation 1.2.3. In all these definitions, property p2 + q 2 + r2 6= 0 and ρ2 + σ 2 + τ 2 6= 0 are ever
intended. Colon notation will *ever* be restricted to inline equations describing columns, and
*never* be used for rows. Anyway, such a notation would be hopeless when dealing with matrices.
We will use instead the ' sign to denote proportionality. Therefore, except from expressions
written in colons, the equal sign will ever indicate a strong, component-wise equality.
Definition 1.2.4. Incidence relations. We will say that a line ∆ ' (p, q, r) contains a point
U = u : v : w, or that ∆ goes through U , or that U belongs to ∆ when their dot product vanishes,
i.e. :
U ∈ ∆ ⇐⇒ pu + qv + rw = 0
Remark 1.2.5. It is clear that incidence relation is projective, i.e. holds for any choice of the
proportionality factors.
Definition 1.2.6. The collinearity of three points P = p : q : r, U = u : v : w, X = x : y : z is
defined by the following determinant equation :
x p u (1.1)
y q v =0
z r w When P 6= U , the set of all the X that satisfies (1.1) is what is usually called the line P U .
Definition 1.2.7. A triangle is an ordered set of three non collinear points. Its natural representation is an invertible square "matrix of columns", where each column is defined up to a
proportionality factor (right action of a diagonal matrix). On the contrary, a "may be degenerate
triangle" is a matrix such that (i) columns are not proportional to each other and (ii) rank is at
least two. When at least two points are equal, the triangle is "totally degenerate".
March 30, 2017 14:09
published under the GNU Free Documentation License
1. Introduction
23
Remark 1.2.8. Without explicit permission, a triangle is not allowed to be degenerate, while totally
degenerate triangles are (quite ever) to be avoided.
Definition 1.2.9. The concurrence of three lines ∆1 ' (d, e, f ), ∆2 ' (p, q, r) and ∆3 ' (u, v, w)
is defined by the following determinant equation :
d e f (1.2)
r s t =0
u v w When ∆1 6= ∆2 , the set of all the ∆3 that satisfies (1.2) is usually called the pencil generated by
the two lines.
Definition 1.2.10. A trigon is a set of ordered three non concurrent lines. Its natural representation is an invertible square "matrix of rows", where each row is defined up to a proportionality
factor (left action of a diagonal matrix).
Proposition 1.2.11. The reciprocal matrix of a triangle is a trigon and conversely. Adjoint
matrices can be used instead of inverses due to proportionality. Relation T −1 . T = Id is nothing
but the incidence relations : A0 ∈
/ B 0 C 0 , A0 ∈ A0 B 0 , A0 ∈ A0 C 0 and cyclically.
Remark 1.2.12. It should be noticed that a tetra-angle defines an hexa-gone, while an tetra-gone
(quadrilateral) defines an hexa-angle : n = n (n − 1) /2 holds only when n = 3.
Definition 1.2.13. The line at infinity L∞ is the locus of points x : y : z such that x+y +z = 0,
so that any point out of L∞ can be described by a triple such that x + y + z = 1. Using only this
representation would discard L∞ and is nothing but the usual affine geometry.
Definition 1.2.14. Barycentric basis. Special points A = 1 : 0 : 0, B = 0 : 1 : 0, C = 0 : 0 : 1
are usually identified with the vertices of a triangle in the Euclidean1 plane, so that all (barycentric)
points can be mapped onto the Euclidean plane (completed with the appropriate line at infinity).
The side lengths of this triangle are denoted BC = a, CA = b and AB = c. Since a triangle is not
a two-angle, none of the a, b, c are allowed to vanish.
1.3
Search key
Definition 1.3.1. Standardized barycentrics are defined as follows :
x:y:z∈
/ L∞
x : y : z ∈ L∞
1
x+y+z
1
1 1
7→ (x, y, z) ×
+ +
x y z
7→ (x, y, z) ×
Exercise 1.3.2. Both normalizing factor don’t vanish at the same time, except from points :
√
√
−1 ± i 3 : −1 ∓ i 3 : 2
Determine objects that contains these points.
Definition 1.3.3. The Kimberling’s search key is defined as follows. Triangle a = 6, b =
9, c = 13 is used and then barycentrics are converted according to :
x:y:z∈
/ L∞ , key
=
x : y : z ∈ L∞ , key
=
x
bc
×
x+y+z
2R
x
a
b
c
×
+ +
a
x y z
(1.3)
1 This kind of plane would be better called euclidian rather than Euclidean, since the very idea to compute figures
instead of drawing computations is as far as possible from the thoughts and practices of the historical figure who
wrote the celebrated Elements.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
24
1.4. Figures
Definition 1.3.4. The patched search key has been introduced to deal with points like X(5000)
that would otherwise have a non real searchkey. When x/ (x + y + z) ∈
/ R, then
x
.
key = arg
∈ [−π, +π]
x+y+z
If one uses Maple, arg(z) is obtained by arctan (=z, <z). Using arctan (=z/<z) would be wrong
when <z < 0.
Remark 1.3.5. When X ∈ L∞ , key (X) is the first normalized trilinear. Otherwise key (X) is
simply the directed distance
√ between X and sideline BC of the reference triangle. The actual
value of factor bc ÷ 2R is 4 35/3. Reference values of this search key are provided by the ETC
(Kimberling ETC, 1998-2014).
Definition 1.3.6. The Morley’s search key is the search key associated with a point defined
by its Morley affix. Details and formula are provided in the corresponding chapter. See Proposition 14.3.10.
Example 1.3.7. Key conflicts can occur. For example, using a = 6, b = 9, c = 13 leads to :
X667
isc (X1319 )
= −288 : 1701 : −2197
=
252 : 219 : 215
where isc is the inverse in Spieker circle. Both points have the same search value, namely 18/49.
Remark 1.3.8. In 2009/08/27, the minimal distance between two search keys was 4.8 E − 7. In
2017/01/11, this minimal distance is 2.95 E − 8. This can only become smaller as more points are
added to the database. Therefore, one has to be careful when computing a search key, in order to
face the possibility of huge canceling terms. Using Maple Digits:=20 seems to be safe.
1.4
1.4.1
Figures
Using kseg
In this document, figures were initially drawn using kseg (KSEG, 1999-2006). Indeed, the very idea
to pay something for using Pythagoras theorem seems terrific. The only two problems encountered :
1. While figures are fine in the Linux version, constructions are badly saved. Use Linux version
for figures, and win$ version (through wine) for the macros.
2. In kseg, postscript figures are intended to be landscape... while your favorite lyx+tex+xdvi
configuration hate rotating. You can play with the portrait option (in kseg) but this will not
work. Use a script to :
• replace "90 rotate" by "0 794 translate" and "Landscape" by "Portrait" in the *.ps file
• convert *.ps to *.eps... using ps2eps -f
• change the bounding box g+40, d+310
• anyway, you have to act on the bounding box
1.4.2
Using Geogebra
Due to an increasing community using Geogebra, some figures are drawn using this tool.
1. The best graphical output is obtained as a *.pdf file. The result is a page, that must be
reduced to its bounding box by a tool like pdfcrop.
2. Obtaining the second intersection of two objects requires some cryptic incantations :
LListe1={Intersect[object1, object2]}"
Liste1=RemoveUndefined[ "Sequence[If[Element[LListe1, i] = M,
(NaN, NaN), Element[LListe1, i]], i, 1.0, 2.0]
xx=Element[Liste1,1]
March 30, 2017 14:09
published under the GNU Free Documentation License
1. Introduction
25
3. Dealing with macros is not easy. Most of the time, you have to unzip the *.ggb file and
modify it with a text editor.
4. Nevertheless, a command like
X186 = TriangleCenter[A, B, C, 186]
can be used when n < 3053 (Geogebra 5.0.309.0-3D, tested 2016-12-12).
1.5
Type-keeping and type-crossing functions
Remark 1.5.1. The type-keeping/type-crossing properties are better understood when they are
described in terms of collineations. Chapter 15 will be devoted to this topic. The aim of the
current Section is only to provide some useful tools as soon as possible.
Definition 1.5.2. Trilinears and barycentrics. Triangle people splits into a barycentric tribe
and a trilinear tribe. The trilinear tribe thinks that trilinears, i.e. p : q : r2 are better looking than
barycentrics and redefine everything according to their preferences. The barycentric tribe thinks
that barycentrics, i.e. p : q : r3 are better looking than trilinears and redefine everything according
to their preferences.
Remark 1.5.3. Trilinears can be measured directly on the figure, since they are the directed
distances to the sidelines. When compasses were actual compasses and not a button to click
over, using trilinears was a must. Nowadays, the existence –and the persistence– of both systems can be used for an interesting renewal of the Capulet against Montague story, as in http:
//mathforum.org/kb/message.jspa?messageID=1091956. But this could also be used to gain a
better insight over many point-transforms used in the Triangle Geometry.
Definition 1.5.4. Vectors are covariant, while forms are contravariant. Therefore, coordinates
that measure a vector are forms and are contravariant. At the same time, coordinates that measure
a form are covariant. In other words, p : q : r is contravariant, while [p, q, r] is covariant.
Definition 1.5.5. We will say that a function P 7→ f (P ) : p : q : r 7→ u : v : w is type-keeping or
type-crossing or type scrambling according to :


when f (αp : βq : γr) = αu : βv : γw
type keeping
type crossing
when f (αp : βq : γr) = αu : βv : w
γ


type scrambling otherwise
For a function of several variables, global type-keeping means :
f (αp : βq : γr, αu : βv : γw) = αx : βy : γz
when
f (p : q : r, u : v : w) = x : y : z
Remark 1.5.6. An object that is intended to describe a point has to be contravariant. An object
that is intended to describe a line has to be covariant, while relationships like collinearity (1.1) and
concurrence(1.2) have to be invariant. Therefore a function whose input and output are points
has to be type-keeping. In the same way, a function whose input and output are lines has to be
type-keeping. On the contrary, a function whose entries are points and output are lines has to be
type-crossing. In the same way, a function whose entries are lines and output are points has to
be type-crossing. These facts are the reasons why both tribes, using barycentrics or trilinears, are
proceeding to the same geometry.
Definition 1.5.7. Barycentric multiplication is the multiplication component by component
of the barycentrics of two points. This operation is denoted :
P ∗ X = px : qy : rz
b
(1.4)
Component-wise multiplication of trilinears would be another possibility. Trilinear people acts like
that.
2 someone from the barycentric tribe would write them p/a : q/b : r/c, since she would use p : q : r for the
barycentrics
3 someone from the trilinear tribe would write them ap : bq : cr, since she would use p : q : r for the trilinears
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
26
1.6. Duality between point and lines
Definition 1.5.8. Barycentric division is the division component by component of the barycentrics
of two points. This operation is denoted :
P ÷X =
b
p q r
: :
x y z
(1.5)
Component-wise division of trilinears would be another possibility. Trilinear people acts like that.
Remark 1.5.9. These transforms are introduced here to provide an easy description of some other
transforms. The study of their geometrical meaning is postponed to Chapter 16. For our present
needs, we only need to remark that both :
X 7→ X ∗ P ÷ U
b
b
and
X 7→ P ∗ U ÷ X
b
b
are globally type-keeping transforms, that can be used to obtain points from points, or lines from
lines.
Definition 1.5.10. Sqrtdiv. Let F = f : g : h be a fixed point, and U a moving point, restricted
to avoid the sidelines of ABC. The mapping defined by :
.
. f 2 g 2 h2
:
:
sqrtdivF (U ) = UF# =
u
v w
(1.6)
is globally type-keeping and describes a pointwise transform, whose fixed points are the four ±f :
.
±g : ±h. This map U 7→ UF# is exactly the same as U 7→ UP∗ defined by UP∗ = P ÷ U and P = F ∗ F .
The second form is often used, introducing a fictitious point P = f 2 : g 2 : h2 .
b
b
Remark 1.5.11. Using # instead of ∗ in this context is yet been done by cubics’ people (Ehrmann
and Gibert, 2005) : fixed points of the transform (F and its relatives) have a clearer geometrical
meaning than P . On the other hand, when P crosses the borders of ABC, the coordinates of point
F become imaginary, and the configuration is less visual.
√
Remark
√ 1.5.12. The converse operation of sqrtdiv would be sqrtmul defined as (U, X) 7→ ux :
√
vy : wz but this map is multivalued. When U, X are triangle centers and ux, vy, wz are perfect
squares, it makes sense to fix signs so that F is also a triangle center.
1.6
Duality between point and lines
Does equation pu + qv + rw = 0 means P ∈ ∆U or U ∈ ∆P ? Without any further indication, one
cannot decide which is the point and which is the line. This is called duality. If you want to be
specific, you have to say :




p
u
h
i
h
i




u v w .  q  = 0 or
p q r . v =0
r
w
and remember how points/lines are mapped into columns/rows. In any case, points aren’t lines and
columns aren’t rows. An efficient formulation of incidence axioms must recognize this elementary
fact.
Definition 1.6.1. The wedge operator is the universal factorization of the determinant. This
means that wedge of two columns is a row, while wedge of two rows is a column. One has :

 

p
u

 

 q  ∧  v  ' (qw − rv, ru − pw, pv − qu)
r
w


qw − rv


(p, q, r) ∧ (u, v, w) '  ru − pw 
pv − qu
March 30, 2017 14:09
published under the GNU Free Documentation License
1. Introduction
27
Proposition 1.6.2. When P 6= U , the barycentrics of the line P U are provided by operation P ∧U .
As it should be, this operation is commutative and is type-crossing.
Proof. The wedge of two points cannot be 0 : 0 : 0 when the points are different, therefore ∆ = P ∧U
defines a line. By definition, we have :
  
 

p u x x
u
p
  
 

 q  ∧  v  .  y  = q v y r w z z
w
r
and the conclusion comes from the fact that inclusion of a line into another implies equality. Typecrossing is obvious from stratospheric reasons... but can also be checked on the components (up
to a global αβγ factor). When dealing with lines, the same argument shows that "line wedge line"
is a point.
Proposition 1.6.3. When line ∆12 is given by points P1 and P2 (with P1 6= P2 ) and line ∆3 is
given by its barycentrics then either both lines are equal or their intersection M is given by :
∆12 ∩ ∆3 ' (∆3 . P1 ) P2 − (∆3 . P2 ) P1
Proof. Call M this object. It is clear that M ∈ P1 P2 . And we can check that ∆3 . M = 0. Another
proof is that ' is in fact a component-wise identity.
Proposition 1.6.4. Suppose that lines ∆12 and ∆34 are respectively defined by points P1 , P2 and
points P3 , P4 . Then either both lines are equal or their intersection M is given by :
.
M = (P1 ∧ P2 ) ∧ (P3 ∧ P4 ) ' P2 det [P1 P3 P4 ] − P1 det [P2 P3 P4 ]
Proof. Obvious from the previous proposition and the definition det [P1 P3 P4 ] = (P3 ∧ P4 ) . P1 .
Another proof is that ' is in fact a component-wise identity.
Definition 1.6.5. The wedge point X∧ of a line is what is obtained by a simple transposition
the barycentrics. This way of doing is based on a misperception of the wedge operation since
P U = (P ∧ U ) is a line (row) and not a point (column). When written in trilinears, this object
don’t look good. Not without reason.
Definition 1.6.6. The Weisstein point XW of a line is what is obtained when applying the
same misperception to trilinears. Applied to line P U , this leads to the crossdifference of P and
U . When written in barycentrics :
XW = crossdiff (P, U ) = a2 (qw − rv) : b2 (ru − pw) : c2 (pv − qu)
this object don’t look good. Not without reason.
A better founded concept must lead to a type-crossing transform.
Definition 1.6.7. The tripoleof a line and the tripolar of a point is what is obtained by "transpose and reciprocate". Clearly, the one-to-one correspondence between pole and polar is lost when
a coordinate vanishes (line through a vertex, or point on a sideline).
Remark 1.6.8. A less stratospheric definition of the tripolar is given in Definition 4.5.2.
Remark 1.6.9. When applying "transpose and reciprocate", both tribes are thinking they are
acting "their way", and are talking about "trilinear pole" and "barycentric pole". But the result
is the same since reciprocation of barycentrics (aka isotomic conjugacy, Section 4.4) acts over X∧
while reciprocation of trilinears (aka isogonal conjugacy, Section 8.2) acts over XW . To summarize
(using later introduced concepts) :
XΛ = t ∆ ; XW = t ∆ ∗ X6 ; tripole = isotom (XΛ ) = isogon (XK )
b
(1.7)
Remark 1.6.10. When tripole is at infinity, the line is tangent to the Steiner in-ellipse (cf Example 11.8.1).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
28
1.6. Duality between point and lines
name
line
tripo
Euler
Infinity
Brocard
Nagel
Bevan
Soddy
?
antiorthic
?
X2 -X3
X30 -X511
X3 -X6
X1 -X2
X1 -X3
X1 -X7
X2 -X6
X44 -X513
X1 -X6
X1 -X4
X4 -X6
X187 -X237
X230 -X231
X325 -X523
X3 -X8
X241 -X514
X1 -X1
X648
X2
X110
X190
X651
X658
X99
X1
X100
X653
X107
X6
X4
X76
???
X7
X1
Lemoine
orthic
Longchamps
Gergonne
model
t
∆
L
X525
X2
X850
X514
???
X3239
X523
X75
X693
???
X3265
X76
X69
X6
???
X8
X1
L647
L6
L523
L649
L650
X657
L512
L1
L513
L652
L520
L2
L3
L32
L3310
L55
X1
∞
∞⊥
⊥po
X30
X523
X125
X511
X519
X517
X516
X524
X513
X518
X515
X1503
X512
X523
X523
X952
X514
X1
X512
???
X513
X514
X1499
X517
X3309
X522
X525
X511
X30
X30
X900
X516
X1
X115
???
X11
X1565
???
X1512
???
???
X1562
X1513
X1514
X1531
X3259
X1541
X1
#
404
269
229
142
128
97
92
92
64
40
37
29
30
14
21
Table 1.1: Some well-known lines
Example 1.6.11. Table 1.1 describes some well-known lines. P
For example, the Euler line goes
through X2 (centroid) and X3 (circumcenter) . Its equation is
x (b2 − c2 )(b2 + c2 − a2 ) = 0.
t
2
2
2
2
2
Formally, center (b − c )(b + c − a ) is X525 = ∆, while center a2 (b2 − c2 )(b2 + c2 − a2 ) is
X647 = XW . This center has been used sometimes to describe lines, leading to Euler = L647 . The
next column (∞) gives the infinity point while the remaining two columns give the later defined
orthopoint (∞⊥ ) and orthopole (⊥pole).
Proposition 1.6.12. Tripole and tripolar, being correctly typed, are constructible (Figure 1.1).
Start from P . Draw AP and obtain A0 = AP ∩ BC. Construct A00 ∈ BC so that division BCA0 A00
is harmonic (Section 4.2). Act cyclically and obtain B 00 and C 00 . Then A00 B 00 C 00 are collinear, and
the line they define is nothing but the tripolar of P . (and are named tripo in Table 1.1).
A
B''
C''
A''
B
C'
A'
P
C
B'
Figure 1.1: Point P and line A”B”C” are the tripolars of each other.
March 30, 2017 14:09
published under the GNU Free Documentation License
1. Introduction
29
Remark 1.6.13. One of the most important consequence of all these duality formulas is the rocksolid equality giving the intersection of two lines each of them defined by two points :
(P Q ∩ RS) = (P ∧ Q) ∧ (R ∧ S)
Proposition 1.6.14. A symmetric parametrization in ρ, σ, τ of the points U = u : v : w that lie
on line ∆ ' [p, q, r] is :
(1.8)
U = qτ − rσ : rρ − pτ : pσ − qρ
A symmetric parametrization in ρ, σ, τ of the points U = u : v : w that lie on line ∆ = tripolar (P )
where P = p : q : r is :
(1.9)
U = p (τ − σ) : q (ρ − τ ) : r (σ − ρ)
Proof. The first formula is U = ∆ ∧ [ρ, σ, τ ], defining a point on a line as the (projective) intersection of this line with another one. The second one is obvious.
Proposition 1.6.15. Line tripolar (P ) is the locus of the tripoles U of the tangents to the inconic
IC (P ). Moreover, the contact point of tripolar (U ) is Q = U ∗ U ∗ P .
b
b
Proof. This is the right place for the assertion, but not for its proof... Postponed to Proposition 11.6.11.
Remark 1.6.16. Caveat : A triangle is not a conic. When U lies on the polar of P wrt a
conic, then P lies on the polar of U wrt the same conic. When U lies on tripolar (P ) (the polar of
P wrt triangle ABC) we have :
u v w
+ + =0
p
q
r
and this isn’t a commutative relation.
1.7
Isoconjugacy has moved
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
30
March 30, 2017 14:09
1.7. Isoconjugacy has moved
published under the GNU Free Documentation License
Chapter 2
Teaching Geometry to a computer
To be written
31
32
March 30, 2017 14:09
2. Teaching Geometry to a computer
published under the GNU Free Documentation License
Chapter 3
Central objects
Centrality is a key notion. Emphasis on this concept was put by the founding paper of Kimberling
(1998). The corresponding definitions have been tailored so that a central point is something like
I = X (1) or G = X (2) or O = X (3) etc., while a central triangle is something like ABC itself or
JKL (the triangle of the excenters).
3.1
Triangle centers
Three points defines six triangles when taking the order into account. But there exists only one
orthocenter. A central point is something that behaves like that. Moreover, geometrical theorems
are not supposed to change when the king’s foot shortens (not to speak of what happens when
the king itself is shortened) : barycentric functions have to be homogeneous. This leads to the
following definition.
Definition 3.1.1. A triangle center (or a central line) is a point (or a line) of the form


f (a, b, c)


 f (b, c, a) 
f (c, a, b)
where f is a nonzero function satisfying two conditions:
1. f is homogeneous in a, b, c; i.e., there is a real number h such that f (λa, λb, λc) = λh f (a, b, c)
for all (a, b, c) in the domain of f ;
2. f is symmetric in b and c; i.e., f (a, c, b) = f (a, b, c).
Definition 3.1.2. Bicentric points. When condition (2) is relaxed, we have to consider the two
different points

 

f (a, b, c)
f (a, c, b)

 

 f (b, c, a)  ,  f (b, a, c) 
f (c, a, b)
f (c, b, a)
They are called bicentric, together comprising a bicentric pair. Example: the Brocard points,
ω + = a2 b2 : b2 c2 : c2 a2 and ω − = c2 a2 : a2 b2 : b2 c2 (cf Proposition 7.7.1).
Definition 3.1.3. A polynomial center is a triangle center X that can be defined by a polynomial
function (homogeneous and symmetric). Examples: X(1), X(2), · · · , X(12), but not X(13).
Definition 3.1.4. A transcendental center is a triangle center X that cannot be defined as X
= f (a, b, c) : f (b, c, a) : f (c, a, b) using an algebraic function f . Examples: X(359) and X(360).
Definition 3.1.5. A major center is a triangle center X for which there exists a function f (A)
such that X = f (A) : f (B) : f (C). Examples: X(1), X(2), X(3), X(4), X(6). Major centers solve
certain problems in functional equations (Kimberling, 1993, 1997).
33
34
1
16
36
63
92
202
264
300
323
359
471
507
562
608
1085
1119
1135
1253
1327
1442
1586
1820
1994
2160
2323
2964
3200
3.2. Central triangle
2
17
41
68
93
203
265
301
326
360
472
508
563
728
1088
1123
1136
1259
1328
1443
1593
1847
2003
2161
2351
2965
3201
3
18
47
69
94
212
266
302
328
365
473
509
571
738
1092
1124
1137
1260
1335
1488
1597
1857
2006
2165
2361
3043
3205
4
19
48
75
158
215
269
303
331
366
479
554
577
847
1093
1127
1139
1264
1336
1489
1598
1870
2052
2166
2362
3076
3206
6
24
49
76
173
219
273
304
340
371
480
555
601
999
1094
1128
1140
1265
1395
1496
1599
1917
2066
2174
2477
3077
3218
7
25
50
77
174
220
278
305
341
372
483
556
602
1000
1095
1129
1143
1267
1397
1497
1600
1928
2067
2175
2671
3082
3219
8
31
55
78
179
222
279
312
345
378
485
557
603
1028
1096
1130
1147
1270
1398
1501
1659
1969
2089
2207
2672
3083
9
32
56
79
184
236
281
317
346
393
486
558
604
1049
1102
1131
1151
1271
1399
1502
1748
1973
2151
2212
2673
3084
13
33
57
80
186
255
289
318
348
394
491
559
605
1077
1106
1132
1152
1274
1407
1583
1802
1974
2152
2289
2674
3092
14
34
61
85
188
258
298
319
357
400
492
560
606
1081
1115
1133
1250
1321
1411
1584
1804
1989
2153
2306
2962
3093
15
35
62
91
200
259
299
320
358
470
506
561
607
1082
1118
1134
1251
1322
1435
1585
1807
1993
2154
2307
2963
3179
Table 3.1: Major centers
Consider two examples, X(9) and X(37), of which first trilinears are b + c − a and b + c,
respectively. It is not clear from these trilinears that X(9) is a major center, whereas X(37) is not.
Indeed, X(9) also has first trilinear cot(A/2), so that X(9) is a major center, but there remains
this problem: how to establish that X(37) and others are not major. In April, 2008, Manol Iliev
found a criterion for a triangle center to be not a major center (**reference missing**). He applied
his test to the first 3236 triangle centers in ETC and found that exactly 292 of them are major, as
listed Table 3.1.
Definition 3.1.6. A rational center is a triangle center X that can be defined
by X = f (a, b, c) :
f (b, c, a) : f (c, a, b) where f is a polynomial that belongs to C a2 , b2 , c2 , S where S is the surface
of ABC. When coordinates of the vertices are rational, the coordinates of the center are made of
rational quantities and fixed reals.
Example 3.1.7. X(13) is not polynomial but nevertheless rational since
1
1
√
√ ::
:: '
abc + 3 (b2 + c2 − a2 ) R
4 S + (b2 + c2 − a2 ) 3
On the contrary, X(1) is polynomial, but not rational.
3.2
Central triangle
Definitions of this Section are tailored so that triangle ABC itself as well as later defined (Section 4.4) cevian and anticevian triangles are central objects. The corresponding matrices are,
March 30, 2017 14:09
published under the GNU Free Documentation License
3. Central objects
35
columnwise, as follows :

0

CP '  q
r
p
0
r


−p
p


q  , AP '  q
r
0

p
p

−q
q 
r −r
Definition 3.2.1. Suppose that f, g are two homogeneous functions having the same degree of
homogeneity. One of them (but not both) can be the zero function. Then the (f, g)central triangle
is defined as :


f (a, b, c) g (a, c, b) g (a, b, c)


(3.1)
[A0 , B 0 , C 0 ] '  g (b, c, a) f (b, c, a) g (b, a, c) 
g (c, b, a) g (c, a, b) f (c, a, b)
Example 3.2.2. Triangle ABC is (1, 0) while CP is (0, f ) and AP is (−f, f ) when point P is
defined by center function f .
Proof. We have P = f (a, b, c) : f (b, c, a) : f (c, a, b), together with f (a, b, c) = f (a, b, c).
Example 3.2.3. As a summary, we have:

bc
 b2 + bc + c2

c2

T bc, c2 = 
2
 b + bc + c2

b2
b2 + bc + c2
c2
a2 + ca + c2
ca
a2 + ca + c2
a2
a2 + ca + c2
b2
a2 + ab + b2
a2
2
a + ab + b2
ab
2
a + ab + b2
Notation 3.2.4. Among the functions often used, we have:
cot ω
area (ABC). Not twice the area, as used in ETC.
the Conway symbols. Sa = b2 + +c2 − a2 /2
s
(a + b + c) /2, so that 2s is the perimeter.
S
Sa , Sb , Sc







where ω is the Brocard angle, cot ω = a2 + b2 + c2 /2S
Usage of R = abc/2S or r = S/2s is deprecated since r, R are often perceived as positive, while
S is oriented.
3.3
Symbolic substitution
Definition 3.3.1. Symbolic substitution. Suppose p(a, b, c), q(a, b, c), r(a, b, c) are functions
of a, b, c, all of the same degree of homogeneity. As the transfigured plane consists of all functions
of the form X = x(a, b, c) : y(a, b, c) : z(a, b, c), the substitution indicated by
a 7→ p(a, b, c), b 7→ q(a, b, c), c 7→ r(a, b, c)
maps the transfigured plane into itself.
Remark 3.3.2. Such a substitution may have no clear geometric meaning, as suggested by the
name, symbolic substitution. On the other hand, symbolic substitutions are of geometric interest
because they map lines to lines, conics to conics, cubics to cubics, and they preserve incidence.
Example 3.3.3. The symbolic substitution (a, b, c) 7→ (1/a, 1/b, 1/c) maps every triangle center
to a triangle center, every pair of bicentric points to a pair of bicentric points, every circumconic
to a circumconic, etc. However, when (a, b, c) = (3, 4, 5), for example, then a, b, c are sidelengths
of a Euclidean triangle, but 1/a, 1/b, 1/c are not.
Symbolic substitutions were introduced in Kimberling (2007).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
36
March 30, 2017 14:09
3.3. Symbolic substitution
published under the GNU Free Documentation License
Chapter 4
Cevian stuff
4.1
Centroid stuff
Definition 4.1.1. The reflection of point U = u : v : w in point P = p : q : r (not at infinity) is
the point X such that :


(p − q − r)u + 2p(v + w)


(4.1)
X '  (q − p − r)v + 2q(u + w) 
(r − p − q)w + 2r(u + v)
Proof. We want to obtain X = 2P − U when P, U, X are finite and in normalized form. When P
is finite and U ∈ L∞ then X = U (OK). Taking P ∈ L∞ would result into X = P for any value
of U , not an acceptable result.
Stratospheric proof. X is obtained under the action described by :

 

 

1 0
p
p

 

 

2  q  . L∞  − L∞ .  q  .  0 1
0 0
r
r

0

0 
1
Definition 4.1.2. Complement and anticomplement are inverse transforms defined so that
the complement of a vertex of triangle ABC is the middle of the opposite side. In other words
(using barycentrics),
complement (U )
=
anticomplement (Q)
=
(3 G − U ) /2 = q + r : p + r : p + q
(3 G − 2 Q) = −p + q + r : p − q + r : p + q − r
According to Court, p. 297, the term complementary point dates from 1885, and the term anticomplementary point dates from 1886.
Definition 4.1.3. The medial triangleC2 is the complement of triangle ABC. The A-vertex of
C2 is the middle of segment BC, and cyclically. In other words :


0 1 1


C2 =  1 0 1 
1 1 0
Definition 4.1.4. The antimedial triangleA2 is the anticomplement of triangle ABC (and is
also called the anticomplementary triangle). Each sideline of A2 contains a vertex of ABC and
corresponding sidelines of both triangles are parallel. In other words :


−1 1
1


A2 =  1 −1 1 
1
1 −1
These triangles are the model used to define the next coming cevian and anticevian triangles (they
are the triangles related to the centroid X2 ).
37
38
4.2
4.2. Cross-ratio and fourth harmonic
Cross-ratio and fourth harmonic
Definition 4.2.1. The cross-ratio of four (different) parameters zj ∈ C is defined as
. (z4 − z2 ) (z3 − z1 )
cross-ratio C (z1 , z2 , z3 , z4 ) =
(z4 − z1 ) (z3 − z2 )
z2 −z3
3
Remark 4.2.2. This quantity is introduced as zz11 −z
−z4 ÷ z2 −z4 , i.e. as (z4 , z3 , z2 , z1 ) by (Pedoe, 1970,
p 212) and (Schwerdtfeger, 1962, p. 35), the intent being to emphasize the definition as a ratio
of ratios. Our intent in using a product form is to emphasize the choice made among the six
possibilities described just below.
Definition 4.2.3. The cross-ratio of four elements of the Riemann sphere Z : T is defined as :
z1 z2 z3 z4
.
, , ,
cross-ratio (z1 : t1 ; z2 : t2 ; z3 : t3 ; z4 : t4 ) = cross-ratio C
t1 t2 t3 t4
with the usual rules to resolve indeterminacies.
Proposition 4.2.4. Cross-ratio is a projective quantity
on the Riemann sphere and is invariant
under the homographies, i.e. the group PGLC C2 .
Proof. well-known result. The general case is an identity about polynomials and the special cases
come from continuity.
Proposition 4.2.5. Group of the cross-ratio. The cross ratio remains unchanged under the action
of the bi-transpositions like [a, b, c, z] 7→ [b, a, z, c]. Therefore the action of S4 generates only 6
values for the cross-ratio. We have :
a
b
c
z
a
a
z
c
b
b
z
b
a
c
z
c
c
b
a
c
b
a
z
z
1
k
k−1
1
k
1−k
k
k−1
k
1−k
Special cases are {exp (+iπ/3) , exp (−iπ/3)} (equilateral triangle and one of its centers) {1, 0, ∞}
(two points are equal) and {−1, 2, 1/2} (harmonicity) .
Proof. Direct examination.
Definition 4.2.6. Consider the linear projective family defined from two fixed rows (or columns)
X, Y (where X 6' Y is assumed)
∆ = { zX + tY | (z, t) 6= (0, 0)}
The cross-ratio of four elements of ∆ is defined as the cross-ratio of their parameters, i.e.
cross-ratio (zj X + tj Y ) = cross-ratio C (zj /tj )
Theorem 4.2.7. The cross-ratio of four elements of a linear projective family (assuming at least
three different elements) is intrinsic, i.e. doesn’t depend of the columns (or rows) chosen to describe
the family. When four different collinear objects P, Q, R, S are given, then non zero multipliers
p, q, r, s and a constant λ can be found such that :
rR
= pP + qQ
sS
= λpP + qQ
Moreover quantity λ depends only on the four objects and their order. This quantity is the former
defined cross-ratio of the 4-uple.
Proof. To prove the existence, we write P = z1 X + t1 Y, etc and solve the resulting set of 4
equations.The value obtained for λ is the former defined cross-ratio "using X, Y ". The uniqueness
of λ comes from the uniqueness, up to a proportionality factor, of r : p : q and s : λp : q since P, Q
is yet another generating family.
March 30, 2017 14:09
published under the GNU Free Documentation License
4. Cevian stuff
39
Corollary 4.2.8. When the family is parametrized as Pj = kj X+(1 − kj ) Y , then λ = cross-ratio C (kj ).
Proof. By definition, λ = cross-ratio (kj / (1 − kj )). But cross-ratio C () is invariant under z 7→
z/ (1 − z).
Corollary 4.2.9. Let the Pj be given by their barycentrics pj : qj : rj . Assuming that two of the
pj are not 0, then λ = cross-ratio (pj / (pj + qj + rj )).
Proof. This holds only if the points are on the same line, but not the infinity line !
Definition 4.2.10. Four points A, B, J, K on an ordinary straight line form an harmonic division when :
AJ
AK
÷
= −1
BJ
BK
Since cross-ratio is a projective invariant, this relationship is carried along collineations.
Proposition 4.2.11. Let be given three members P, Q, R of a linear projective family with, at least
P 6= Q. Then it exists exactly one member S of the family such that cross-ratio (P, Q, R, S) = −1.
This object is called the fourth harmonic of the first three.
Proof. Cross-ratio is an homographic function of parameter kS , and therefore bijective between λ
and kS .
Proposition 4.2.12. The fourth of an harmonic division can be constructed using a point out of
the line of the first three. In Figure 4.1, points A, P, C and external D are given. Then ∆ is the
parallel to AD through C, E is DP ∩ ∆ and F the reflection of E into C. Point Q result from the
fact that division E, F, C, ∞∆ is clearly harmonic.
Distance A P Distance P C = 2.3101
Distance A Q Distance Q C = 2.3101
D
F
C
P
A
Q
E
Figure 4.1: Obtain Q from A,P,C and auxiliary D
Remark 4.2.13. In the Cartesian plane, the fourth harmonic is also the reflection of the third point
into the circle having the first two as diameter.
Remark 4.2.14. Conic cross-ratio is described in Section 11.7.
4.3
About combos
Definition 4.3.1. combos. When P ' p : q : r and U ' u : v : w are points at finite distance
and f = f (a, b, c), g = g(a, b, c) are nonzero homogeneous functions having the same degree of
homogeneity, then the (f, g) combo of P and U , denoted as f × P + g × U , is :
f ×P +g×U '
U
P
f (a, b, c) +
g (a, b, c)
L∞ · P
L∞ · U
Remark 4.3.2. Written that way, one has not to discuss if barycentrics or trilinears are used.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
40
4.4. Cevian, anticevian, cocevian triangles
A
C'
B'
P
B
C
A'
Cevian Triangle of P
Distance A B' Distance B C' Distance C A' = 9028475.5223
Distance A C' Distance B A' Distance C B' = 9028475.5223
Figure 4.2: Ceva’s Theorem
Proposition 4.3.3. With the same hypotheses, points P, U, f P + gU, f P + hU are collinear and
cross-ratio (P, U, f P + gU, f P + hU ) = h/g
As a special case, f P ± g U are harmonic conjugate wrt P, U .
Proposition 4.3.4. When f, g, h are homogeneous symmetric functions all of the same degree
of homogeneity, and X, X 0 , X”are triangle centers, then f X + g X 0 + h X”is a triangle center.
Conversely, given three non collinear triangle centers, any fourth triangle center is a combo of the
first three, using symmetric functions as coefficients.
Proof. This amounts to say that, in R3 , any invertible matrix defines a basis of the space.
Remark 4.3.5. Part of the time, normalization is useless. Knowing that X(482) ' s × nX(1) +
(r + 4R) × nX(7) can be required, but using X(482) ' vX(1) + 4S vX(7) (where vX is what is
given in the table) can be sufficient.
Proposition 4.3.6. The columns of a matrix T describing the vertices of a central triangle
have to be normalized in such a way that L∞ · T ' L∞ . Such matrices form a group under
multiplication. And then T · P is (another) triangle center when P is a triangle center.
4.4
Cevian, anticevian, cocevian triangles
Theorem 4.4.1 (Ceva). Let A0 ∈ BC, B 0 ∈ CA and C 0 ∈ AB be three points on the sidelines
of triangle ABC, but different from the vertices. Then lines AA0 , BB 0 , CC 0 are concurrent if and
only if :
AB 0 BC 0 CA0
= −1
CB 0 AC 0 BA0
Proof. The usual proof uses Menelaus theorem. Another proof, using determinants, is given below.
Definition 4.4.2. Cevian triangle. Let P be a point not on a sideline of ABC. The lines
AP, BP, CP are the cevians of P . Let Ap = AP ∩ BC. Define Bp and Cp cyclically. Triangle
Ap Bp Cp is the cevian triangle of triangle ABC.



p
0



P '  q  =⇒ cevian(P ) =  q
r
r
March 30, 2017 14:09
p
0
r

p

q 
0
(4.2)
published under the GNU Free Documentation License
4. Cevian stuff
cevian
incentral
medial
orthic
intouch
extouch
anticevian
excentral
antimedial
tangential
other
Fuhrmann
Brocard triangles
Hexyl
star
41
18.3
18.5
18.7
18.12
18.13
18.4
18.6
18.8
incenter
centroid
H-center
Gergonne
Nagel
incenter
centroid
Lemoine
X(1)
X(2)
X(4)
X(7)
X(8)
X(1)
X(2)
X(6)
bary (p)
a
1
tan A
−1
(b + c − a)
b+c−a
bary
a
1
a2
G
O
I
X(2)
X(51)
X(5)
X(5)
X(10)
X(4)
X(210)
G
X(165)
X(2)
X(154)
X(1158)
O
X(40)
X(4)
X(26)
I
X(164)
X(8)
X(3)
X(3576)
X(3817)
X(1)
X(946)
?
18.15
18.9
18.14
18.16
Table 4.1: Some well-known triangles
Example 4.4.3. Examples of cevian triangles are given in Table 4.1.
Proposition 4.4.4. Equation4.3 gives the condition for an inscribed triangle to be the Cevian
triangle of some point P .


0 p2 p3


(4.3)
0 q3  is Cevian ⇐⇒ p2 q3 r1 − p3 q1 r2 = 0
 q1
r1 r2
0
and then p : q : r = r1 p2 : q1 r2 : r1 r2
An absolutely hopeless formula, but nevertheless more "obviously symmetrical" is :
q
q
q
p : q : r = 3 p22 p23 q1 r1 : 3 q32 q12 p2 r2 : 3 r12 r22 p3 q3
Proof. Line AP1 is (0, r1 , −q1 ) and cyclically. These lines are concurrent when their determinant
vanishes. Their common point is then given by any column of the adjoint matrix, or the cubic root
of their product.
Proposition 4.4.5. Isotomic conjugacy. Suppose U = u : v : w is a point not on a sideline of
ABC. Let AU = AU ∩ BC and define BU , CU cyclically. Reflect AU BU CU about the midpoints
of sides BC, CA, AB, respectively, to obtain points A0 , B 0 , C 0 . Then lines AA0 , BB 0 , CC 0 are
concurrent. Their common intersection is called the isotomic conjugate of U . The corresponding
barycentrics are :
1 1 1
isot (u : v : w) = : :
(4.4)
u v w
Proof. Immediate computation.
Remark 4.4.6. The fixed points of this transform are the gravity center and its relatives, so that
#
isot (U ) = UG
while U ∗ isot (U ) =X(2). It should be noticed that X (2) plays together the role of
b
points F and P of Definition 1.5.10.
Proposition 4.4.7. Reciprocal conjugacy. Suppose ∆ = [ρ, σ, τ ] is a line not through a vertex of
ABC. Let A∆ = BC ∩∆, etc the traces of ∆ on the sidelines. Reflect them about the corresponding
midpoint and obtain TA = B + C − A∆ , etc. The points TA , TB , TC are aligned, defining a line
called the reciprocal of the given one since we have:
1 1 1
recip ([ρ, σ, τ ]) '
, ,
ρ σ δ
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
42
4.5. Transversal lines, Menelaus and Miquel theorems
Proof. Immediate computation.
Definition 4.4.8. Anticevian triangle. Let P be a point not on a sideline of ABC. Let
Ap = AP ∩ BC. (The lines AP , BP , CP are the cevians of P , and triangle Ap Bp Cp is the cevian
triangle of P ). Let PA be the harmonic conjugate of P with respect to A and Ap . Define PB and
PC cyclically. Triangle PA PB PC is the anticevian triangle of triangle ABC.


−p
p



P '  q  =⇒ anticevian(P ) =  q
r
r


p
p

−q
q 
r −r
(4.5)
Example 4.4.9. Examples of anticevian triangles are given in Table 4.1.
Proposition 4.4.10. (1) Triangle ABC is inscribed in triangle PA PB PC . (2) ABC is the cevian
triangle of P wrt the anticevian triangle. (3) Anticevian triangle of point PA wrt ABC is P PC PB
(the two remaining points are permuted).
Figure 4.3: Cevian, anticevian and cocevian triangles
4.5
Transversal lines, Menelaus and Miquel theorems
Proposition 4.5.1 (Genuine Menelaus theorem). Let A0 ∈ BC, B 0 ∈ CA and C 0 ∈ AB be
three points on the sidelines of triangle ABC, but different from the vertices. Then A0 , B 0 , C 0 are
collinear if and only if :
AB 0 BC 0 CA0
= +1
CB 0 AC 0 BA0
Proof. Let us parametrize the situation by A0 = ka B + (1 − ka )C, etc. Alignment is described by :
0
1 − kb
kc 0
1 − kc = 0
ka
1 − ka
kb
0
March 30, 2017 14:09
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4. Cevian stuff
43
while A0 − B = (1 − ka ) (C − B) and A0 − C = ka (B − C).
Proposition 4.5.2. Cocevian triangle. Let P = p : q : r be a point not on a sideline of ABC,
and let AP BP CP , PA PB PC be respectively the cevian and anticevian vertices of P . Let TA be the
harmonic conjugate of AP wrt B and C. This is also the harmonic conjugate of A wrt PB and PC .
Define TB and TC cyclically. The cocevian triangle of P is the triangle TA TB TC . This triangle is
degenerate and its vertices belong to the tripolar line of P : x/p + y/q + z/r = 0.


0
p



P '  q  =⇒ cocevian(P ) =  −q
r
r

p
0
−r

−p

q 
0
(4.6)
.
Proof. We have AP ' 0 : q : r so that TP = 0 : −q : r is the fourth harmonic of B, C, AP .
We have PB ' p : −q : r, PC ' p : q : −r so that PB + PC ' A while PB − PC ' TA and
(PB , PC , A, TA ) = −1 is proved. Alignment is obvious from the determinant. Menelaus condition
is pqr − prq = 0. Relation with the tripolar, i.e. the "transpose and reciprocate" line is clear.
Proposition 4.5.3 (Miguel theorem). . Let A0 ∈ BC, B 0 ∈ CA and C 0 ∈ AB be three points on
the sidelines of triangle ABC, but different from the vertices. Then circles AB 0 C 0 , A0 BC 0 , A0 B 0 C
are passing through a same point M , the Miquel point of A0 B 0 C 0 wrt ABC.
Proof. Equation of circle AB 0 C 0 is :
a2 y z + b2 z x + c2 x y − (x + y + z) y kc c2 + z (1 − kb ) b2
Therefore their last common point is :


a2 ka (ka − 1) a2 + (ka − 1) (kb − 1) b2 + c2 kc ka


M '  b2 ka kb a2 + kb (kb − 1) b2 + (kb − 1) (kc − 1) c2 
c2 (kc − 1) (ka − 1) a2 + b2 kb kc + kc (kc − 1) c2
Since this expression is symmetric, this expression is on the third circle.
Theorem 4.5.4 (Extended Menelaus theorem). Let A0 ∈ BC, B 0 ∈ CA and C 0 ∈ AB be three
points on the sidelines of triangle ABC, but different from the vertices. All the following are
necessary and sufficient conditions for A0 , B 0 , C 0 to be collinear :
(i) the Menelaus condition : ((1 − ka ) (1 − kb ) (1 − kc ) + ka kb kc ) = 0
(ii) the midpoints Ma = (A + A0 )/2, Mb = (B + B 0 )/2, Mc = (C + C 0 )/2 are on the same line (the
so-called Newton line of the quadrilateral)
(iii) the Miquel point of A0 B 0 C 0 is on the circumcircle
of ABC.
(iv) the homographic application ϕ defined in PC C2 by A 7→ A0 , B 7→ B 0 , C 7→ C 0 is involutory.
Proof. (i) obvious ; (ii) determinant ; (iii) condition for M ∈ is the Menelaus condition times an
ugly factor that can be written as :
2 a2 ka + 2 b2 kb + 2 c2 kc − c2 − b2 − a2
2
+ 16 S 2
(iv) Condition for ϕ to be involutory is det3 [1, α + α0 , α α0 ] = 0. We obtain S4 times the Menelaus
condition (see Section 14.3 for notations and more details).
Proposition 4.5.5. When P is neither a vertex nor X(2), the centroid of the cocevian triangle is
well defined (perhaps on L∞ ), and called the tripolar centroid of P (Stothers, 2003b).
T G (P ) = p(q − r)(2p − q − r) : q(p − r)(p − 2q + r) : r(q − p)(p + q − 2r)
Remark 4.5.6. When q = r then T G (P ) = 0 : 1 : −1 (at infinity on BC). When all p, q, r are
different, T G (P ) is a finite point.
Proposition 4.5.7. When all p, q, r are different, then it exists exactly another point that shares
the same T G (P ), namely :
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
44
4.6. Perspectivity
other (P ) =
q + r − 2p
r + p − 2q
p + q − 2r
:
:
rp + qp − 2 qr qp + qr − 2 rp qr + rp − 2 qp
Proof. Direct computation. When eliminating k, u, v in T G (P ) = k T G (U ), special cases are
p, q − r, u, v − w, qw − rv, 2p − q − r et cyclically. For all named points X, it happens that
other (X) is not named.
Example 4.5.8. Points X(1635) to X(1651) are defined that way. Examples include :
1
3
4
6
7
8
4.6
1635
1636
1637
351
1638
1639
98
99
100
105
190
262
1640
1641
1642
1643
1644
3569
263
512
513
514
523
524
2491
1645
1646
1647
1648
1649
525
648
957
1002
1022
2394
1650
1651
3310
665
244
125
Perspectivity
Definition 4.6.1. Vertex trigon, vertex triangle. Let T1 = A1 B1 C1 and T2 = A2 B2 C2 be
two triangles. Their vertex trigon T3∗ is the set of three lines La = A1 A2 , Lb = B1 B2 , Lc = C1 C2 ,
while their vertex triangle is the set of points A3 = Lb ∩ Lc , etc. The vertex triangle is the dual of
the vertex trigone.
Remark 4.6.2. Exclude the case where all points are on the same line, the rank of the vertex trigone
is either 3 or 2, while the rank of the vertex triangle is either 3 or 1 (adjoint matrix).
Definition 4.6.3. Perspector. Let T1 = A1 B1 C1 and T2 = A2 B2 C2 be two triangles. When
the vertex trigon degenerates, i.e. when the lines La , Lb , Lc , concur at some point P , this point is
called the perspector (replacing center of perspective) of the (ordered) triangles.
Definition 4.6.4. Sideline triangle, sideline trigon. Let T1 = A1 B1 C1 and T2 = A2 B2 C2
be two triangles. Their sideline triangle T4 is the set of three points A4 = B1 C1 ∩ B2 C2 , B4 =
C1 A1 ∩ C2 A2 , C4 = A1 B1 ∩ A2 B2 , while their sideline trigon is the set of lines B4 C4 , C4 A4 , A4 B4
(i.e. the dual of the sideline triangle).
Remark 4.6.5. Excluding the case where all lines are through the same point, the rank of the
sideline triangle is either 3 or 2, while the rank of the sideline trigon is either 3 or 1 (adjoint
matrix).
Definition 4.6.6. Perspectrix . Let T1 and T2 be two triangles. When the sideline triangle
degenerates, i.e. when points A4 , B4 , C4 are on the same line, this line is called the perspectrix
(replacing axis of perspective) of the triangles.
Theorem 4.6.7. [Desargues] . When none of the triangles T1 and T2 are degenerate, the existence of a perspector is equivalent to the existence of a perspectrix.
Proof. In this context, trigone T3∗ is called the Desargues trigon and triangle T4 is called the
Desargues triangle. The result comes from
det T4 = det T1 det T2 det T3∗
Definition 4.6.8. Crosstriangle. The crosstriangle of two given triangles T1 = A1 B1 C1 and
T2 = A2 B2 C2 is defined as the triangle T4 = B1 C2 ∩ B2 C1 , C1 A2 ∩ C2 A1 , A1 B2 ∩ A2 B1 . Its name
comes from the fact we are crossing the vertices of the homologue sidelines.
Example 4.6.9. Let T1 be the reference triangle ABC and T2 the cevian triangle Ap Bp Cp of a
point P . Then
1. P is the perspector of both triangles,
2. the Desargues’ triangle is nothing but the (degenerate) cocevian triangle of P
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4. Cevian stuff
45
3. the perspectrix is the trilinear polar of P . Additionally, we have:
Ap = 0 : q : r, Bp = p : 0 : r, Cp = p : q : 0
A0 = 0 : −q : r,
B 0 = p : 0 : −r, C 0 = −p : q : 0
(x : y : z) ∈ A0 B 0 C 0 is q r x + r p y + p q z = 0
Example 4.6.10. Reference triangle and the anticevian triangle PA PB PC of a point P is another
example. Let us recall that :
PA = −p : q : r, PB = p : −q : r, PC = p : q : −r
Proposition 4.6.11. Perspectivity kit. We define such an object as p : q : r : u : v : w, i.e.
six numbers up to a common proportionality factor, none of them being 0. This amounts to give
points P ' p : q : r, U ' u : v : w (none on an ABC- sideline) together with a synchronization
factor k = (p + q + r) / (u + v + w). Exchanging P and U changes k in 1/k. We define triangle
T0 as ABC and triangle T1 and T2 by :




p u u
u p p




T1 '  q v q  ; T2 '  v q v 
w w r
r r w
1. T0 and T1 are perspective wrt to P , T0 and T2 are perspective wrt U , while T2 , T3 admit P + U
as perspector
2. Each triangle T1 , T2 T3 is the cross-triangle of the other two.
3. All pair of triangles have the same Desargues triangle, defining the same perspectrix:
1
1
1
∆'
;
;
p−u q−v r−w
4. Reciprocally, any triangle in perspective with ABC can be parametrized that way.
Proposition 4.6.12. Suppose additionally
A, B, C, define

(qw − rv) (p − u) p

Q '  (ru − pw) (q − v) q
(pv − qu) (r − w) r
that line P U doesn’t go through any of the vertices

(qw − rv) (p − u) u



 ; V '  (ru − pw) (q − v) v 
(pv − qu) (r − w) w


and consider the triangles T0 , T1 , T2 , the cevian triangles of P : T0P , T1P , T2P and the cevian
triangles of U : T0U , T1U , T2U .
Then: (1) P is the perspector of each pair : (T0 , T1 ), (T0P , T1P ), (T0 , T0P ), (T0 , T1P ), (T1 , T0P ),
(T1 , T1P );
(2) Perspector of (T1 , T2 ) is p + u : q + v : r + w.
(3) Q 6= V and line ∆ = QV is the perspectrix of each pair : (T0 , T1 ), (T0P , T1P ), (T1 , T2 ),
(T0 , T2 ), (T0U , T2U ) ;
(4) Moreover, the perspectrices of (T0 , T0P ), (T0 , T1P ), (T1 , T0P ), (T1 , T1P ) are going through Q
(and symmetrically for U ).
h
i
Proof. The perspectrix of (T0 , T0P ) is ∆P = qr pr pq , the tripolar of P and we have :
T0
T1
T1
T1P
T0P
T1P
pqr ∆1 − det T2 ∆P
2 pqr ∆1 − det T2 ∆P
3 pqr ∆1 − det T2 ∆P
Remark 4.6.13. Behavior wrt isogonal transform is examined in Proposition 16.5.2.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
46
4.7. Cevian nests
Figure 4.4: Triangles ABC and UVW that admit P as perspector.
4.7
Cevian nests
Definition 4.7.1. Cevian nest. Suppose T1 , T2 , T3 are triangles and that T1 is inscribed in T2
and T2 is inscribed in T3 . If any two of the tree triangles are perspective, it is well-known that
each is perspective to the third : T1 is a cevian triangle of T2 for some point P , T3 is an anticevian
triangle of T2 for some point U and T1 , T3 are perspective wrt some point X. Such configuration
is called a cevian nest.
B'u
A
C'u
P = cevamul U,X
Cp
U
P
B
Bp
Ap
X
C
A'u
Figure 4.5: P is cevamul(U,X)
March 30, 2017 14:09
published under the GNU Free Documentation License
4. Cevian stuff
47
Proposition 4.7.2. Map P = P (U, X) giving the perspector of T1 , T2 from the other two perspectors of a cevian nest is symmetric, while –for a given P – map X = X (U ) is involutory.
Proof. Given the vertices Si (i = 1, 2, 3) of triangle T3 and perspector U , the vertices Si (i = 4, 5, 6)
of triangle T2 are obtained by S4 = (S1 ∧ U ) ∧ (S2 ∧ S3 ) and cyclically. Process can be iterated,
obtaining vertices Si (i = 7, 8, 9) of T1 . Then X is obtained as X = (S1 ∧ S7 ) ∧ (S2 ∧ S8 ). It can
be checked that substituting U by X (and keeping everything else unchanged) leads back to U ,
proving the second part. The first part follows immediately.
When triangle ABC belongs to such a nest, three possibilities can occur. The corresponding
operations are summarized in Table 4.2, where "mul" stands for multiplication (giving P ) and
"div" for the converse operation. The Kimberling’s name is also given.
X, P
Kimberling
X = crossdiv (P, U )
cross − conj
CP wrt T2
CU
ABC
P = crossmul (U, X)
cross − point
X = cevadiv (P, U )
ceva − conj
CP
ABC
AU
P = cevamul (U, X)
ceva − point
X = sqrtdiv (P, U )
ABC
AP
AU wrt T2
P = sqrtmul (U, X)
All these operations are (globally) type-keeping, since they transform points into constructible
points.
T2
T1
T3
Table 4.2: Three cases of cevian nets
4.8
The cross case (aka case I, cev of cev)
Definition 4.8.1. Crossmul(U,X). As in Table 4.2 (I), let T3 (the biggest triangle) be ABC,
T2 = AU BU CU the (usual) cevian triangle of U and T1 = A0 B 0 C 0 the triangle inscribed in T2
obtained by A0 = AX ∩ BU CU and cyclically. Then T1 and T2 have a perspector (P ), and mapping
(U, X) 7→ P is called cross-multiplication.
Definition 4.8.2. Crossdiv(P,U). As in Table 4.2 (I), let T3 (the biggest triangle) be ABC,
T2 = AU BU CU the (usual) cevian triangle of U and T1 = A0 B 0 C 0 the cevian triangle of P wrt
T2 , obtained by A0 = AU P ∩ BU CU and cyclically. Then T1 and T3 have a perspector (X), and
mapping (P, U ) 7→ X is called cross-division.
Proposition 4.8.3. Computing rules of crossmul and crossdiv are given (using barycentrics) in
Figure 4.6. Map (U, X) 7→ P is commutative and behaves like ordinary multiplication (eponymous
property). Map (P, U ) 7→ X behaves wrt crossmul like division behaves wrt ordinary multiplication
(eponymous property).
Proof. Barycentrics p : q : r are defining point P 0 with respect to triangle T3 . Call P : Q : R its
barycentrics with respect to triangle T2 , so that (p : q : r) = T2 (P : Q : R). Then :

0

T1 '  v
w
u
0
w
 
u
0 P
 
·
v Q 0
0
R R
 
P
P
 
·
Q  0
0
0



0 0

 
Q 0 '


0 R
u (Q + R)
QR
v
Q
w
R
u
P
v (P + R)
PR
w
R
u
P
v
Q
w (P + Q)
PQ







where the diagonal matrix had been chosen to "synchronize" the columns of triangle T2 . Then
Proposition 4.6.11 shows that T3 and T1 are perspective wrt point u/P : v/Q : w/R.
Remark 4.8.4. The cross-multiplication and cross-division were introduced in Kimberling (1998),
using the names crosspoint and "cross conjugacy", together with the notation X = C(P, U )
and also X = P cU . In our opinion, a name that emphasizes the properties is more efficient.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
48
4.9. The ceva case (aka case II, cev and acev)
complem ◦ isotom
X/U ✛∗
P/U ✛
bU−
1
∗b U
✛
✛
.......................... crossmul ............................
(U, X)
P ✛
1
∗b X
✛ X−
U/X ✛
P/X ∗ b
complem ◦ isotom
crossmul (u : v : w, x : y : z) = (v z + w y) u x : (u z + w x) v y : (u y + v x) w z
∗b U
X ✛
(4.7)
isotom ◦ anticomplem
∗b U −1
X/U ✛
P/U ✛
(P, U )
✛
....................................................................................................................
crossdiv
crossdiv (p : q : r, u : v : w) =
u
v
w
:
:
quw + ruv − pvw pvw + ruv − quw pvw + quw − ruv
(4.8)
Figure 4.6: crossmul, crossdiv
The factorization given can be interpreted in terms of isoconjugacies (see Chapter 16). First
compute what happens when U =X(2) and obtain complem ◦ isotom. Then transmute this map
by the transform ∗U that fixes A, B, C and sends G =X(2) onto U .
b
Proposition 4.8.5. The crossmul P of U, X is also the point of concurrence of (1) the line through
points AX ∩ BU and AU ∩ BX, (2) the line through points BX ∩ CU and BU ∩ CX, (3) the line
through points CX ∩ AU and CU ∩ AX.
A
Bu
A"
Bp
P
C'
Ap
Cp
A'
Cu
B'
U
Au
B
C
Figure 4.7: P is crossmul of U and X
4.9
The ceva case (aka case II, cev and acev)
Definition 4.9.1. Cevamul(U,X). As in Table 4.2 (II), let T2 (the middle triangle) be ABC,
T3 = UA UB UC the anticevian triangle of U and T1 = A0 B 0 C 0 be the triangle inscribed in T2 = ABC
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4. Cevian stuff
49
obtained by A0 = UA X ∩ BC and cyclically. Then T1 and T2 have a perspector (i.e. T1 is the
cevian triangle of some point P ), and mapping (U, X) 7→ P is called ceva-multiplication.
Definition 4.9.2. Cevadiv(P,U). As in Table 4.2 (II), let T2 (the middle triangle) be ABC,
T3 = UA UB UC the anticevian triangle of U and T1 = AP BP CP be the cevian triangle of P ,
obtained (as usual) by AP = AP ∩ BC and cyclically. Then T1 and T3 have a perspector (X), and
mapping (P, U ) 7→ X is called ceva-division.
Proposition 4.9.3. Computing rules of cevamul and cevadiv are given (using barycentrics) in
Figure 4.8. Map (U, X) 7→ P is commutative i.e. :
(
X is the perspector of CP and AU
U is the perspector of CP and AX
and behaves like ordinary multiplication (eponymous property). Map (P, U ) 7→ X behaves wrt
crossmul like division behaves wrt ordinary multiplication (eponymous property).
isotom ◦ complem
X/U ✛∗
P/U ✛
bU−
1
∗b U
✛
✛
........................... cevamul .............................
(U, X)
P ✛
1
∗b X
✛ X−
U/X ✛
P/X ∗ b
isotom ◦ complem
cevamul (u : v : w, x : y : z) =
(uz + wx)(uy + vx) : (vz + wy) (uy + vx) : (vz + wy) (uz + wx) (4.9)
cevadiv (p : q : r, u : v : w) =
u(−q r u + r p v + p q w) : v(q r u − r p v + p q w) : w(q r u + r p v − p q w) (4.10)
Figure 4.8: cevamul, cevadiv
Remark 4.9.4. The ceva-multiplication and ceva-division were introduced in Kimberling (1998),
using the names cevapoint and "ceva conjugacy", together with the notation X = P ©U . In
our opinion, a name that emphasizes the properties is more efficient.
Proposition 4.9.5. Suppose U = u : v : w and X = x : y : z are distinct points, neither lying
on a sideline of ABC. Let Ax = XA XB XC and Au = UA UB UC be the anticevian triangles of X
and U (wrt ABC). Define A0 as UA X ∩ XA U and B 0 , C 0 cyclically (Figure 4.5). Then triangle
A0 B 0 C 0 is inscribed in ABC and is, in fact, the cevian triangle of P = cevamul (U, V ).
Proof. Direct computation.
Proposition 4.9.6 (Floor Van Lamoen (2003/10/17)). Point cevamul(U, X) can be constructed
from the cevian triangles : let AU BU CU be the cevian triangle of U , and AX BX CX the cevian
triangle of X. Define :
Aux = Au Cx ∩ Ax Bu
Bux = Bu Ax ∩ Bx Cu
Cux = Cu Bx ∩ Cx Au
Axu = Au Bx ∩ Ax Cu
Bxu = Bu Cx ∩ Bx Au
Cxu = Cu Ax ∩ Cx Bu
Then, as seen in Figure 4.9, triangle ABC is perspective to both triangles Aux , Bux , Cux and
Axu , Bxu , Cxu , and the perspector in both cases is the cevamul (P, U ).
Proof. Direct computation.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
50
4.10. The square case (aka case III, acev of acev)
Apu
A
Bup
Bp
Cu
Cup
Cp
Cpu
P
X
Aup
B
Bpu
U
Bu
Au
Ap
C
Figure 4.9: Lamoen’s construction of X=cevamul(P,U)
4.10
The square case (aka case III, acev of acev)
Definition 4.10.1. sqrtdiv(P,U). As in Table 4.2 (III), let T1 (the smallest triangle) be ABC,
T2 = PA PB PC the anticevian triangle of P and T3 be the anticevian triangle of U wrt T2 . Then
T1 and T3 have a perspector X and mapping (P, U ) 7→ X is called sqrtdiv. We have formula :
x:y:z=
p2 q 2 r2
:
:
u v w
Definition 4.10.2. sqrtmul(U,X). The inverse mapping of sqrtdiv should be the mapping
sqrtmul (U, X) 7→ P "defined" by :
√
√
√
p : q : r = ± ux : ± vy : ± wz
... but (1) the solution is not unique and (2) in fact, the problem cannot be stated clearly.
4.11
Danneels perspectors
Definition 4.11.1. First Danneels perspector. Let T1 = AU BU CU be the cevian triangle of
a point U = u : v : w. Let LA be the line through A parallel to BU CU , and define LB and LC
cyclically. The lines LA , LB , LC determine a triangle T2 perspective to T1 (and in fact homothetic,
with factor 2). The corresponding perspector is DP1 (U ), the first Danneels perspector (#11037)
of U . Using barycentrics :
DP1 (U ) = u2 (v + w) : v 2 (w + u) : w2 (u + v)
Proof. Compute (from left to right) the row BU ∧ CU ∧ L∞ ∧ A and cyclically. Obtain a matrix
describing a trigon and takes the adjoint to obtain T2 . Then compute the perspector. One can
also remark that T2 is the anticevian triangle of :
X = u(v + w) : v(u + w) : w(u + v)
and obtain DP1 (U ) as cevadiv (U/X) (homothetic property is obvious... and useless to compute
the perspector).
Remark 4.11.2. This point is named D (U ) in ETC. No name of only one letter! Moreover this
conflicts with the Maple’s derivation operator.
Proposition 4.11.3. Point G = X (2) is invariant under DP1 . Moreover, G, U and DP1 (U ) are
ever collinear. For example, Euler line is globally invariant.
Proof. Check that det (X2 , U, DP1 (X2 + λU )) = 0.
March 30, 2017 14:09
published under the GNU Free Documentation License
4. Cevian stuff
51
Proposition 4.11.4. When DP1 (X) = G then either X = G or X lies on the Steiner circumellipse.
Proof. Write DP1 (X) = k G and solve. Except from X = G, xy + yz + zx = 0 is obtained.
Proposition 4.11.5. When DP1 (U ) 6= G, i.e. when U 6= G and U not on the circumSteiner, it
exists two other points that verify DP1 (X) = DP1 (U ). Using barycentrics, we have :
X
W2

v + w − 2u − W

'  w + u − 2v − W
u + v − 2w − W
=
1
(u+w)(u+v)
1
(u+v)(v+w)
1
(u+w)(v+w)
2



where
(u + v) (v + w) (w + u) u (v + w) + v 2 (w + u) + w2 (u + v) − 6 uvw
Proof. Direct computation, assuming xy + yz + zx 6= 0. The main difficulty is to re-obtain a
symmetrical expression after elimination of k, z and resolution on y.
Example 4.11.6. Here is a list of pairs (I, J) of named points such that DP1 (X (I)) = X(J):
1
2
3
4
5
6
7
8
42
2
418
25
3078
3051
57
200
20 3079
25 3080
30 3081
69
394
75
321
99
2
100
55
110
184
189
190
264
290
366
648
651
653
1422
2
324
2
367
2
222
196
664
666
668
670
671
886
889
892
2
2
2
2
2
2
2
2
903
2
1113 25
1114 25
1121
2
1370 455
1494
2
2479
2
2480
2
2481
2966
3225
3226
3227
3228
2
2
2
2
2
2
Definition 4.11.7. Second Danneels perspector. Suppose T1 = AU BU CU is the cevian
triangle of a point U . Let LAB be the line through B parallel to AU BU , and let LAC be the
line through C parallel to AU CU . Define A0 = LAB ∩ LAC and B 0 , C 0 cyclically. Finally obtain
A” = BB 0 ∩ CC 0 and B”, C” cyclically. It happens that triangle T2 = A”B”C” is perspective to
T1 = AU BU CU . The corresponding perspector is DP2 (U ), the second Danneels perspector of U
(Danneels, 2006). Using barycentrics :
DP2 (U ) = u(v − w)2 : v(w − u)2 : w(u − v)2
Proof. Compute (left to right) LAB = AU ∧ BU ∧ L∞ ∧ B, LAC accordingly, then A0 = LAB ∧ LAC
and B 0 , C 0 cyclically. Obtain A” = (B ∧ B 0 ) ∧ (C ∧ C 0 ) and B”, C”cyclically. See that T2 =
A”B”C” is the anticevian triangle of point :
.
X = u(v − w) : v(w − u) : w(u − v)
and obtain DP2 (U ) as cevadiv (U, X).
Proposition 4.11.8. The circumconic through A, B, C, U, isot (U ) admits DP2 (U ) as center and
u v 2 − w2 : v w2 − u2 : w u2 − v 2
as perspector. And therefore isotomic conjugates have the same second Danneels’ perspector.
Proof. Immediate computation.
Example 4.11.9. List of (U, U ∗ , DP2 (U )) :
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
52
4.11. Danneels perspectors
U
1
3
4
6
7
9
10
20
30
U ∗ DP
75
244
264 2972
69
125
76 3124
8
11
85 3119
86 3120
253
122
1494 1650
U
37
42
57
81
94
98
99
100
200
U∗
274
310
312
321
323
325
523
693
1088
DP
3121
3122
2170
3125
2088
868
1649
3126
2310
U
394
519
524
536
538
2394
2395
2397
2398
U∗
2052
903
671
3227
3228
2407
2396
2401
2400
DP
3269
1647
1648
1646
1645
1637
2491
3310
676
and of points without named isotomic :
[43, 3123], [88, 2087], [694, 2086], [1022, 1635], [1026, 2254], [2421, 3569]
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 5
The French touch
5.1
The random observer
When dealing with triangle geometry, we have to organize the coexistence of three kinds of objects.
We have vectors, we have points and we have 3-tuples. A vector describes a translation of the
"true plane". A vector has a direction but also a length and therefore is not defined "up to a
proportionality factor". The set of all these vectors is a 2-dimensional vector space V (more about
it in what follows). A point is either an element of the "true plane", i.e. an ordinary point at finite
distance, or a point at infinity describing the direction of some line. Such points have therefore to
be described "up to a proportionality factor" by a column that belongs to a given copy of PR R3 .
In the same vein, lines
are described "up to a proportionality factor" by a row that belongs to
another copy PR R3 .
And we need to talk with our computer in order to let it compute all the required results.
These computations are done using 3-tuples and tools
acting over 3-tuples so that computations
must be described using R3 rather than V or PR R3 . How to organize the coexistence of these
three points of view is the object of a well-known theorem.
b in such a way that E
Theorem 5.1.1. Any affine space E can be embedded into a vector space E
b
becomes an affine hyperplane of E.
Proof. By definition, E is not empty. Choose ☼ ∈ E (the random observer) and write E = ☼ + V
−−→
where V is the vector space associated with E. Choose , ∈
/ E and write E = V ⊕ R,☼. Define
ζ as the last coordinate in vector space E. Then E = {m ∈ E | ζ (m) = 1}. Moreover the vector
hyperplane {m ∈ E | ζ (m) = 0} is an isomorphic copy of V.
Notation 5.1.2. An affine description of a point at finite distance P is a 3-tuple (ξ, η, ζ) where
ζ = 1 is assumed. The semantic of these coordinates is the pre-existence of some random observer,
that uses a Cartesian frame (ξ, η) to describe what is occurring before her eyes.
Let us take an example and begin with an informal approach (cf http://www.les-mathematiques.
net/phorum/read.php?8,585414,586289#msg-586289). We have a point P , given by a column,
and two triangles T1 , T2 given by the columns of their vertices :




√
√
252 + 130 35
1284 −744 40 35
√
√
1 
1 


P =
 105 − 312 35  , T1 =
 535 −310 −96 35 
1183
169
1183
169
169
169


√
√
√
−9000 + 325 35 14400 + 520 35 5040 − 2600 35
√
√
√
1 

T2 =
 −3750 − 780 35 6000 − 1248 35 2100 + 6240 35 
1690
1690
1690
1690
As it should be, each column verifies ζ = 1, which is the equation of the affine plane E when seen
as a subspace of R3 .
We define W as the matrix
that transforms the matrix T of a triangle (Pj ) into the matrix of
−−−−−−→
the sideline vectors Pj+1 Pj+2 of this triangle (indices are taken modulo 3 so that P4 = P1 etc).
t
And now, we compute K = W . t T . T . W for both triangles. We have :
53
54
5.2. An involved observer

1 −1

0
1 
−1
0

 
a2 −Sc −Sb
−62

 
b2 −Sa 
−107  =  −Sc
−Sb −Sa
c2
169

 
α2 −Sγ −Sβ
−702/5

 
β 2 −Sα 
−351/4  =  −Sγ
−Sβ −Sα
γ2
4563/20



K1 = 
K2


=
0

W =  −1
1
36
26
26
81
−62 −107
4212/5
−702
−702 3159/4
−702/5 −351/4
Here we use a, b, c for the sidelengths
of the first triangle and α, β, γ for the second one. The
Conway’s symbols Sa = b2 + c2 − a2 /2 etc. are defined accordingly. The letter used to name the
matrix K has been chosen by reference to the Al-Kashi formula. Clearly, row (1, 1, 1) belongs to
the kernel of matrix K . The characteristic polynomial of this matrix is µ3 − a2 + b2 + c2 µ2 +
12 S 2 µ. One eigenvalue is µ = 0 and the other two are real (from symmetry of K ) and positive.
Remark 5.1.3. Matrix M used to compute the orthodir of a line is nothing but M = K /2S.
A more "stratospheric", but nevertheless equivalent, definition for these Al-Kashi matrices
would be :
. t
K = W . t T . P yth3 . T . W
where P yth3 describes any R3 -quadratic form that embeds (ξ, η) 7→ ξ 2 + η 2 , the quadratic form
of the ordinary affine Euclidean plane. This embedding quadratic form depends on three arbitrary
parameters since 6 coefficients are needed for dimension three, while only 3 are needed for dimension
two.
5.2
An involved observer
Now, we will describe how things are looking when the observer is no more a random ☼ but rather
an actor of the play. For example, we can take triangle T1 as a new vector basis inside vector space
R3 and calculate everything again using this new basis. From :




ξ
x




 η  = T1  y 
ζ
z
it is clear that condition ζ = 0 that shows that a 3-tuple belongs to V becomes now x + y + z = 0.
t
Defining L∞ = (1, 1, 1) this can be rewritten as L∞ . (x, y, z) = 0. Seen "up to a proportionality
factor" this will gives L∞ . (x : y : z) = 0, i.e. the condition for x : y : z ∈ PR R3 to belong to the
line at infinity. But this is not our purpose for the moment.
The R3 -metric is now described by matrix t T1 . P yth3 .T1 . This matrix depends in turn on
three arbitrary parameters. In fact, any other matrix that can be written as :
t
t
T1 . P yth3 .T1 + U.L∞ + (U.L∞ )
using an arbitrary column U will be just as well to calculate the Pythagoras of V-vector. A zero
diagonal gives a more nice looking matrix, and is also more efficient for computing. Therefore we
define the matrix P yth by this property and we obtain :


11618 −6465 169
1 

t
T1 . P yth3 .T1 =
4013 169 ,
−6465
169
169
169 2409
March 30, 2017 14:09

0 169
1
P yth = −  169 0
2
81 36

81

36 
0
published under the GNU Free Documentation License
5. The French touch
55
Now, matrix K1 can be computed using :
K1

36
26 −62
t


= W . P yth . W =  26
81 −107 
−62 −107
169

while the coordinates of the other triangle and the extra point are transformed according to :




−30
48
240
6
1 
1 


T2 = T1−1 . T2 =
360  , P[1] = T1−1 . P =
 45 −72
 9 
80
28
65 104 −520
13
And now, matrix K2 is computed using :
K2

4212/5
t

t
= W . T2 . P yth . T2 . W = 
−702
−702/5
−702
3159/4
−351/4

−702/5

−351/4 
4563/20
The quadratic form P yth2 can be obtained directly from P yth , using formula :
t
P yth2 = t T2 . P yth . T2 + +U.L∞ + (U.L∞ )
and thereafter choosing U in order to get a zero diagonal. We obtain :



0
0
−4563/40 −3159/8
1


P yth2 =  −4563/40
0
−2106/5  = −  γ 2
2
β2
−3159/8 −2106/5
0
5.3
γ2
0
α2

β2

α2 
0
From an involved observer to another one
All the preceding computations are standard ones. As long as points have no dependence relations
all together, nothing else can be done. On the contrary, when points are constructed from each
other, another pointof view is more powerful. Taking ABC as reference triangle, and describing
our points in PR R3 , we obtain P = 6 : 9 : 13, A0 = −6 : 9 : 13, B 0 = 6 : −9 : 13, C 0 = 6 : 9 : −13.
If this only a random coincidence, nothing more can be said.
On the contrary, if these points are really defined as P = a : b : c, A0 = −a : b : c, B 0 = a : −b : c,
0
C = a : b : −c, then things get more interesting. We have :
P[2]
K2
so that :
= b+c−a:c+a−b:a+b−c

4 a2 bc
−2 abc
 (a − b + c) (b + a − c)
b
+a−c


−2
abc
4 ab2 c
= 

b+a−c
(b + c − a) (b + a − c)


−2 abc
−2 abc
a−b+c
b+c−a
−2 abc
a−b+c
−2 abc
b+c−a
4 c2 ab
(b + c − a) (a − b + c)








4 a2 bc
α =k
, etc
(a − b + c) (b + a − c)
Since barycentrics are defined only "up to a proportionality factor", the former set of equations
has not to be solved explicitly. Only the ratios between the quantities a, b, c are needed. And this
is simple to obtain. One gets :
(a : b : c) = α2 β 2 + γ 2 − α2 : β 2 γ 2 + α2 − β 2 : γ 2 β 2 + α2 − γ 2
2
so that :
1
1
1
:
:
β 2 + γ 2 − α2 γ 2 + α2 − β 2 α2 + β 2 − γ 2
This results identifies the incenter of the reference triangle ABC as the orthocenter of the
excentral triangle. In the same vein, the circumcenter of ABC can be identified as the nine-points
center of A0 B 0 C 0 . More details on this specific relation will be given in Subsection 18.7.
P[2] =
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
56
March 30, 2017 14:09
5.3. From an involved observer to another one
published under the GNU Free Documentation License
Chapter 6
Brief extension to 3D spaces
Previous chapters were dealing with 2D spaces (planes), represented as the projective of a 3D
vector space. In this chapter, we are using a 4D vector space to describe a 3D geometric space.
6.1
Basic results
Definition 6.1.1. A 3D point is a projective column in PR R4 .
Definition 6.1.2. We will say that four points Mj are coplanar when det1..4 (Mj ) = 0.
Theorem 6.1.3 (Universal factoring). Given four columns, we have :





x2
y2
z2
t2


 
 
∧
 
det (M1 M2 M3 M4 ) = t M1 · (M2 ∧ M3 ) · M4
where
 
x3
0
t3 z2 − t2 z3 t2 y3 − t3 y2 y2 z3 − y3 z2


y3  .  t2 z3 − t3 z2
0
t3 x2 − t2 x3 x3 z2 − x2 z3
=
z3   t3 y2 − t2 y3 t2 x3 − t3 x2
0
x2 y3 − x3 y2
t3
y3 z2 − y2 z3 x2 z3 − x3 z2 x3 y2 − x2 y3
0





Proof. Operator det is multilinear: this ensures the existence of the central matrix. The actual
values of the coefficients are obtained by partial derivatives.
Definition 6.1.4. Matrix ∆ is anti-symmetric, and thus depends of 6 parameters. The Plucker
column representation is defined as
col ∆ = D1,2 : D1,3 : D2,3 : D1,4 : D2,4 : D3,4
where
(6.1)


0
D1,2 −D1,3 D1,4
 −D
0
D2,3 D2,4 


1,2
∆ =

 +D1,3 −D2,3
0
D3,4 
−D1,4
−D2,4
−D3,4
0
Remark 6.1.5. When M2 6= M3 , the points collinear with M2 , M3 are the elements of ker ∆ where
∆ = (M2 ∧ M3 ). Moreover, the characteristic polynomial is :
4
χ∆ (λ) = λ + λ
2
X
2
Djk
6
so that
P
3
+
X
3
Di,j Dk,l
!2
Di,j Dk,l is null.
Proposition 6.1.6. When M ∈
/ ∆, the row characterizing the plane containing both point M and
line ∆ is obtained as Π = t M · ∆ .
57
58
6.2. Electric notations
Proposition 6.1.7. Let us define ∆∗ so that the column characterizing the point common to
plane Π and line ∆ is M = ∆∗ · t Π (obviously, ∆ 6⊂ Π is assumed). Then, using notations of
(6.1),we have:
col ∆∗ = D3,4 : D2,4 : D2,3 : D1,4 : D1,3 : D1,2
→
−
Proof. This result is obtained by solving Π · M = 0, ∆ · M = 0 and then generically expressing
M as ∆∗ · t Π.
Proposition 6.1.8. Given two (different) planes Π1 , Π2 , the matrix ∆ of their diedral line can
be obtained directly as:
∆ ' t Π1 · Π2 − t Π2 · Π1
Given two (different) points P1 , P2 , the dual matrix ∆∗ of line P1 P2 can be obtained directly as:
∆ ∗ ' P1 · t P2 − P2 · t P1
Proof. These objects are clearly projective objects and antisymmetric matrices of rank 2. The first
one satisfies ∆ · X = 0 : 0 : 0 : 0 as soon as Π1 · X = Π2 · X = 0. The second one satisfies
Y · ∆∗ = [0, 0, 0, 0] as soon as Y · P1 = Y · P2 = 0.
Proposition 6.1.9. Let us define form φ by polarization of det ∆ , leading to :
φ
∆, ∆
0
= D1,2 E3,4 + D1,3 E2,4 + D2,3 E1,4 + D1,4 E2,3 + D2,4 E1,3 + D3,4 E1,2
t
=
∗
0
· col ∆
col ∆
where
0
= E1,2 : E1,3 : E2,3 : E1,4 : E2,4 : E3,4
col ∆
0
0
is exactly equal to det (M1 M2 M3 M4 ) when ∆ = M1 ∧ M4 , ∆ = M2 ∧ M3 .
∆, ∆
0
= 0 is the condition for the two lines to be coplanar.
Moreover, φ ∆ , ∆
Then φ
Proposition 6.1.10. Assuming that φ
is the point:





∆, ∆
0
= 0, the intersection of the two lines ∆, ∆0

D2,4 E3,4 − D3,4 E2,4
D3,4 E1,4 − D1,4 E3,4
D1,4 E2,4 − D2,4 E1,4
D1,2 E3,4 + D1,3 E2,4 + D2,3 E1,4




while the plane generated by the two lines is :
t




6.2
D1,2 E1,3 − D1,3 E1,2
D2,3 E1,2 − D1,2 E2,3
D1,3 E2,3 − D2,3 E1,3
D1,2 E3,4 + D1,3 E2,4 + D2,3 E1,4





Electric notations
Definition 6.2.1. Electric notations are as follows:



0
Bz −By
x1
!
→
−
 −B
 x 
0
Bx
X
z
 2 

X=
; ∆ =
=
 By −Bx
 x3 
0
x
x4
−Ex −Ey −Ez
March 30, 2017 14:09
Ex
Ey
Ez
0



=

B
←
−
−E
←
− !
E
0
=
→
−!
B
←
−
E
published under the GNU Free Documentation License
6. Brief extension to 3D spaces
59
→
−
←
− →
−
Mind that D12 = Bz (not Bx ), while B is the antisymmetric matrix such that B · E = B × E .
Proposition 6.2.2. Using these notations, we have
line through two points
−
→
line by a point X and a direction W
plane through a line and a point
→
− ←
− →
− ←
−
incidence of lines B : E , C : F
→
−
→
−!
yX − xY
X ∧ Y =
→
− →
−
X×Y
−
→ !
−
→ xW
X ∧ W :0 = −
→ →
−
W ×X
t →
− →
−
←
−!
B
×
V
−
v
E
t
X· ∆ =
←
− →
−
E·V
←
− ←
−!
E×F
←
− →
−
F ·B
→
− →
−!
C ×B
←
− →
−
F ·B
←
− →
− ←
− →
−
E · C + F · B = 0 =⇒ X =
←
− →
− ←
− →
−
E · C + F · B = 0 =⇒ Π =
t
Exercise 6.2.3. Consider four points P1 , P2 , P3 , P4 and a collineation A. Write them in C4 , i.e.
not to a proportionality factor.



P =

53 −58
0
76
−97 −9 −16 40
−74 32 −43 45
79 −34 58 39






 ; det P = 24803093 ; A = 


1
−1
−1
1
−1
−1
1
1
0
−1
−1
−1
0
0
−1
0
Let Q1 , Q2 , Q3 , Q4 be the images of the Pj by the collineation. We have:
A−1 =
1
4





+2
−2
0
−4
−1
−1
−2
+2
0
0
0
−4
1
1
−2
2




.


 ; Q = A P =


53
−97
−74
79
−58
−9
32
−34
0
−16
−43
58
76
40
45
39



 ; det A = 4




 ; det Q = 99212372

Using the wedge formula, we compute line P23 from P2 , P3 and Q23 from Q2 , Q3 :



P23 = 

0
−394
−1066
−899
394
0
3364
2494
And also the planes P123 =
1066
−3364
0
−928
V
899
−2494
928
0






 ; Q23 = 


0
1692
6786
4094
−1692
0
−261
703
−6786
261
0
3451
−4094
−703
−3451
0





(P1 , P2 , P3 ) and Q123 :
P123 = [46081, −31028, 309494, 220893] ; Q123 = [−729354, −192255, −883572, −162149]
Proposition 6.2.4. Action of a collineation. We have the following properties:
1. t P1 · P23 · P4 = P123 · P4 = det P , t Q1 · Q23 · Q4 = Q123 · Q4 = det Q = det P
2. Q23 = det A
t
det A
A−1 · P23 · A−1
3. Q123 = det A P123 · A−1
4. P123 ∧ P234 = det P dual (P23 ) ; t P123 · P234 − t P234 · P123 = det P P23
→
←
− ←
−
− →
−
5. The same dual B , E = E , B is used for P and Q.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
60
6.3. Examples
Exercise 6.2.5. Loosely speaking and using Cartesian coordinates: a line goes through (−1, 1, 5)
and is incident with line D1 : {x + y = 2, x − y = 2} together with line D2 : (1, 3, 1) + λ(2, −2, 1).
We have two planes Π11 ' [1, 1, 0, −2], Π12 ' [1, −1, 0, −2] and three points P0 ' −1 : 1 : 5 : 1,
P21 ' 1 : 3 : 1 : 1, P22 ' 2 : −2 : 1 : 0. Thus D1 = dual (Π11 ∧ Π12 ), D2 = P21 ∧ P22 , and we have
to solve the set of equations :
∆ · P0 = 0 : 0 : 0 : 0, φ (∆, D1 ) = φ (∆, D2 ) = 0
After expansion, we obtain:
(
)
−6 By − 3 Ez ; −By − Bx + Ez ; Bz + 5 Bx + Ey ; Bz − 5 By + Ex ;
Ex − Ey − 5 Ez ; −8 Bz + By + 5 Bx − 2 Ex + 2 Ey − Ez
←
− →
−
so that B , E = −3 : 1 : 1 : 4 : 14 : −2. The incidence points are: P4 ' 2 : 0 : 4 : 1 (who clearly
belongs to both planes) and P5 ' 5 : −1 : 3 : 1 (obtained with λ = 2). Finally, P0 , P4 , P5 are
aligned, since P4 is the middle of the other two.
One can also obtain the rows that describes the plane through P0 , D1 , the plane through P0 , D2
and obtain ∆ using Proposition 6.1.8.
6.3
Examples
Proposition 6.3.1. Consider a tetrahedron A, B, C, D and six coefficients a, b, c, d, e, f . Define the
four points A+ , B + , C + , D+ by D+ = a A+b B+c C, A+ = f B+e C +a D, B + = f A+d C +b D,
C + = e A + d B + c D. Then lines AA+ , BB + , CC + , DD+ are imbedded in the same hyperboloid.
Proof. A simple computation gives:



H=

0
(ad − be) c (cf − ad) b
(ad − be) c
0
(be − cf ) a
(cf − ad) b (be − cf ) a
0
(be − cf ) d (cf − ad) e (ad − be) f
(be − cf ) d
(cf − ad) e
(ad − be) f
0





When computing D+ and A+ , points A and D are multiplied by the same weight a that can be
attributed to the opposite edge BC, and so on for the other edges.
Corollary 6.3.2. If we take a = |BC| and so on, we obtain a property relative to the four lines
joining a vertex to the incenter of the opposite face in a tetrahedron. And the same occurs for the
Lemoine centers. And for their isotomic X(75) and X(76). When a = b = c = d = e = f (the
gravity centers), the four lines are simply concurrent.
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 7
Euclidean structure using
barycentrics
Barycentric coordinates were intended to describe affine properties, i.e. properties that remains
when points are moved freely. Therefore describing Euclidean properties when using barycentrics
is often presented as contradictory. We will show that, on the contrary, all the required properties
can be described simply. The key fact is that orthogonality only depends on the directions of lines
so that all what is really needed is a bijection that sends each point at infinity onto the point that
characterizes the orthogonal direction.
7.1
Lengths and areas
Notation 7.1.1. In this section, T0 describes the reference triangle ABC in the affine plane, and T
describes a generic triangle Pj (indices are dealt modulo 3, i.e. P4 = P1 , etc. In other words :
ξα ξβ ξγ ξa ξb ξc (7.1)
T0 = ηa ηb ηc , T = ηα ηβ ηγ 1
1 1 1 1
1 We will use |BC| = a, |P2 P3 | = α, etc together with Sa = b2 + c2 − a2 /2, Sα = β 2 + γ 2 − α2 /2,
etc. In other words,
2
2
(ξa − ξb ) + (ηa − ηb ) = c2 , etc
(7.2)
Lemma 7.1.2. Let T be the matrix describing triangle (Pj ) wrt triangle ABC, i.e. the matrix
whose columns are the barycentrics pi : qi : ri of the Pi . Then :


p1 + q1 + r1
0
0


T = T0−1 . T . 
0
p2 + q2 + r2
0

0
0
p3 + q3 + r3
Definition 7.1.3. Matrix W. The matrix W is defined by :

0

W =  −1
1

1 −1

0
1 
−1
0
(7.3)
Proposition 7.1.4. When a matrix T, as defined in (7.1), gives the vertices Pj of a triangle, then
−−−−−−→
T . W gives the sideline vectors Pj+1 Pj+2 of this triangle (using modulo 3 indices). On the other
hand, matrix W can be used to compute the point at infinity U ∈ PR R3 of a line given by its
barycentrics ∆. In this case, we have :
U = ∆ ∧ L∞ = W .
61
t
t
∆ ' ∆. W
62
7.2. Embedded Euclidean vector space
−−−−→
Proof. The first part is obvious from Pj Pj+1 = Pj+1 − Pj that holds when Pj are 3-tuples in the
ζ = 1 plane. The second part is the very definition of the ∧ operator.
Lemma 7.1.5. Matrix K. We have the following Al-Kashi formula :


α2 −Sγ −Sβ
t


K = W . t T . T . W =  −Sγ
β 2 −Sα 
−Sβ −Sα
γ2
D−−→ −−→E
D−−→ −→E
D−−→ −→E
Proof. Diagonal elements are BC | BC , etc and the others are BC | CA = − CB | CA , etc.
Proposition 7.1.6. Let matrix T define a (finite) triangle T with vertices Pi = pi : qi : ri . Then
area of T is given by :
S×Q
det T
(7.4)
(pi + qi + ri )
while the area S of the reference triangle is given by the Heron formula :
S2 =
1
(a + b + c)(b + c − a)(c + a − b)(a + b − c)
16
Proof. The first formula is obvious from Lemma 7.1.2 and the well-known formula
ξa ξb ξc 1 S = ηa ηb ηc 2 1 1 1 (7.5)
(7.6)
that gives the oriented area of a triangle. The denominator of (7.4) enforces the required invariance
wrt multiplicative factors acting on barycentrics, and recognizes the fact that only triangles with
finite vertices have an area.
As it should be, this formula acknowledges that area of ABC is S. The Heron formula can be
proved in many ways, one of them being :
D−−→ −→E2
2
2
4S 2 = |AB| |AC| sin2 A = b2 c2 − AB | AC = b2 c2 − Sa2
Remark 7.1.7. A key point is that formula (7.4) is of first degree in S : once the orientation of the
reference triangle is chosen, all other orientations are fixed.
7.2
Embedded Euclidean vector space
Definition 7.2.1. When points P, Q at finite distance are given
u : v : w, the embedded vector from P to U is defined as :



u
1
1



−
→ (P, U ) =
vec
 v −

u+v+w
p+q+r
w
by their barycentrics p : q : r and
 

p
ρ
 

q = σ 
r
τ
(7.7)
Proposition 7.2.2. Embedded vectors belong to the vector plane V : x + y + z = 0, seen as a
subspace of the Cartesian (non projective) vector space R3 . These vectors obey to the usual Chasles
rule :
−
→ (P , P ) = −
→ (P , P ) + −
→ (P , P )
vec
vec
vec
1
3
1
2
2
3
−
−
−
→
and space V is isomorphic to the usual vector space R2 where P1 P2 = (ξ2 − ξ1 , η2 − η1a ).
Proof. Obvious from definition. Nevertheless, a touch stone for the sequel. Defined this way, vector
space V is how the plane ζ = 0 is perceived when using triangle ABC as reference triangle.
March 30, 2017 14:09
published under the GNU Free Documentation License
7. Euclidean structure using barycentrics
63
metric of the usual Euclidean plane V
Proposition 7.2.3. When using (ξ, η, ζ) coordinates, the
is described by matrix :

1 0 ∗
. 
P yth3 =  0 1 ∗
∗ ∗ ∗



(7.8)
where ∗ are placeholders for three arbitrary parameters. When using barycentrics (wrt triangle
ABC), this matrix is replaced by t T0 . P yth3 .T0 , that depends also on three arbitrary parameters.
In fact, any other matrix that can be written as :
t
t
(7.9)
T0 . P yth3 . T0 + U . L∞ + (U . L∞ )
can be used to define the metric of vector space V.
Proof. This is the usual formula for a change of basis in a bilinear form. Independence from column
U comes from L∞ . V = 0.
Theorem 7.2.4. Pythagoras theorem. Define the Pythagoras bilinear form by : R3 × R3 ,→
t −−→
−−→
−→ | −
−→
R : h−
vec
1 vec2 i = vec1 . P yth . vec2 where :

0
1
P yth =  −c2
2
−b2
−c2
0
−a2

−b2

−a2 
0
(7.10)
−−→
→ (P, Q) into an
When restricted to V, this form extends the transformation R2 ,→ V : P Q →
7 −
vec
isomorphism of Euclidean spaces. Therefore, the squared distance between two (finite) points of
the triangle plane is given by :
t−
→
→ (P, U )
vec (P, U ) . P yth . −
vec
= − a2 στ + b2 τ ρ + c2 ρσ
2
|P U |
−−−−−→2
(ρ, σ, τ )
=
(7.11)
Proof. Matrix P yth is what is obtain when choosing column U in (7.9) in order to obtain a zero
diagonal instead of using K directly. A less "stratospheric" proof is direct inspection. We have :
→ (A, B) | −
→ (A, B)i
h−
vec
vec
=
c2
(etc)
and
→ (A, B) | −
→ (A, C)i
h−
vec
vec
=
Sa (etc)
Since Sa = bc cos A, property holds for a basis. And linearity extends the result to all the other
vectors.
7.3
About circumcircle and infinity line
Definition 7.3.1. The power of a point X = x : y : z (at finite distance) with respect to the
circle Ω centered at P with radius R is :
.
2
power (Ω, X) = |P X| − R2
Theorem 7.3.2. The power formula giving the Ω-power of any point X = x : y : z from the
power at the three vertices of the reference triangle is :
power (Ω, X)
=
a2 yz + b2 xz + c2 xy
ux + vy + wz
−
2
x+y+z
(x + y + z)
(7.12)
where u = power (Ω, A) , etc
−−→
Proof. Use (7.7) to obtain P X and then Theorem 7.2.4 to obtain power (Ω, X). Substitute y =
z = 0 to obtain u, etc. Then a simple subtraction leads to the required result.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
64
7.4. Orthogonality
Definition 7.3.3. The standard equation of the circumcircle is defined as :
. a2 yz + b2 xz + c2 xy
Γstd (x, y, z) = −
x+y+z
(7.13)
Proposition 7.3.4. The equation of any circle can be written as :
.
Ω (x, y, z) = (ux + vy + wz) + Γstd (x, y, r)
=
0
where u = power (Ω, A) , etc
where x : y : z is the arbitrary point and Γstd is the standard equation of the circumcircle, as
defined just above.
Proof. Obvious from (7.12) and power (Γ, A) = 0, etc.
Remark 7.3.5. "Can be written" must be understood as "when required, multiply by x + y + z and
use polynomials". For more details, see Chapter 12.
Corollary 7.3.6 (Heron). Center and radius of the circumcircle are :
X3 = a2 b2 + c2 − a2 : b2 c2 + a2 − b2 : c2 a2 + b2 − c2
R2
=
a2 b2 c2
(a + b + c) (a + b − c) (b + c − a) (c + a − b)
n
o
2
Computed Proof. Direct elimination from |XA| = R2 , etc and (7.7,7.11).
Proposition 7.3.7. For a line ∆ = ρ : σ : τ , not the infinity line, point Q = ∆ ∧ L∞ is on L∞
while U = isogon (Q) is on the circumcircle. Therefore, rational parametrizations of the line at
infinity and the circumcircle are :
L∞
circumcircle
= {σ − τ : τ − ρ : ρ − σ | ρ : σ : τ 6= L∞ }
2
a
b2
c2
=
:
:
| ρ : σ : τ 6= L∞
σ−τ τ −ρ ρ−σ
(7.14)
(7.15)
Proof. Point Q is the point at infinity of the line ρx + σy + τ z = 0, U ∈ Γ is obvious from (7.13)
and bijectivity of Q 7→ U is clear.
Remark 7.3.8. Information conveyed by a 3-tuple like (7.7) is multiple. A first part is the direction
of line P U , described –up to a proportionality factor– by the point ρ : σ : τ ∈ L∞ . Another part is
2
the squared length |P U | given by (7.11). In this formula, circumcircle appears as the conic that
defines how lengths are computed in each direction.
7.4
Orthogonality
Proposition 7.4.1. Given a 3-tuple P = (p, q, r) ∈ R3 , but not on V, it exists a linear transform
ψ such that (i) ψ (P ) = 0, (ii) ψ (V) = V (iii) for all V ∈ V, hψ (V ) | ψ (V )i = hV | V i while
hψ (V ) | V i = 0. Then its characteristic polynomial is µ3 + µ and we have :
Orth (P )
=

qSb − rSc
− ra2 + pSb
1

rSc − pSa
 rb2 + qSa
2 (p + q + r) S
2
− qc + rSa
pc2 + rSb

qa2 + pSc

− pb2 + qSc 
pSa − qSb
The opposite of this matrix is the only other solution to the problem.
Computed Proof. Assertions M . P = 0, L∞ . M = 0 and hψ (V ) | V i = 0 when V = x : y : −x − y
gives nine equations. Elimination leads to the given matrix. Then hψ (V ) | ψ (V )i = hV | V i leads
to the coefficient. Division by p + q + r enforces the fact that P is at finite distance.
March 30, 2017 14:09
published under the GNU Free Documentation License
7. Euclidean structure using barycentrics
65
Corollary 7.4.2. Matrices Orth (P ) and Orth (U ) relative to finite points P, U are related by the
"translation" formula :




p
1




Orth(P ) = Orth (U ) . 1 −
 q  . L∞ 
p+q+r
r
Proof. Acting at infinity, both matrices induces a quarter-turn. At finite distance, it remains only
to move the kernel to the right place.
Theorem 7.4.3. Orthopoint. A point at infinity V ∈ L∞ defines a direction of lines. The
point W ∈ L∞ that defines the orthogonal direction is called the orthopoint of V. We have W '
Orth (P ) . U whatever the choice of P . Nevertheless, the choice O =X(3) is interesting since the
Orth (O) matrix verifies :
. t
OrtO = W . P yth ÷ 2S
t
W . P yth matrix is independent of the U -choice appearing in (7.9), and we have :


c2 − b2
−a2
a2
1 

(7.16)
OrtO =
b2
a2 − c2
−b2 

4S
−c2
c2
b2 − a2
t
Proof. Straightforward computation. We have the formal rule orthopoint = L∞ ∧ U ÷ X4 ÷
This
(−4S) when U ∈ L∞ .
b
Proposition 7.4.4. Matrix OrtO describes the +90◦ rotation in the V space, while matrix
W . P yth ÷ 2S describes the −90◦ rotation.
Proof. The only thing to prove is the orientation. Go back to the ordinary Cartesian coordinates
by(7.1), substitute the squared sidelengths with (7.2), and obtain :


0 −1 ∗
−1


T0 . OrtO . T0
= 1 0 ∗ 
0 0 ∗
Remark 7.4.5. Any Orth (P ) matrix describes a quarter-turn in the Euclidean plane V. In order
2
to provide
a better perception of this result, let O =X(3), H =X(4), put W32 = 16 |OH| S 2 =
P
P
6
4 2
2 2 2
3a −
6 a b +3a b c and consider the linear transform whose matrix is φ = [X (3) , X (30) , X (523)].
−−→
2
When computing |OH| , vector OH is involved and therefore X(30) (the direction of the Euler
line), while X(523) is known to be the direction orthogonal to those of the Euler line. We have :


2
a2 b2 + c2 − a2
2 a4 − b2 − c2 − a2 b2 + c2
b2 − c2
2


φ =  b2 c2 + a2 − b2
2 b4 − c2 − a2 − b2 c2 + a2
c2 − a 2 
2
c2 a2 + b2 − c2
2 c4 − a2 − b2 − c2 a2 + b2
a2 − b2


−16 a2 c2 b2 S 2
0
0


t
φ . P yth . φ = 
0
16 S 2 W32
0 
0
0
W32


0 0
0


−1
φ . OrtO . φ =  0 0 (4S)−1 
0 4S
0


0
0
0
t


t
φ . OrtO . P yth . OrtO . φ =  0 16 S 2 W32
0 
0
0
W32
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
66
7.5. Angles between straight lines
Proposition 7.4.6. The orthodir U of any line ∆ (except from the line at infinity) is defined as
the orthopoint of ∆ ∧ L∞ . It can be computed as U = M . t ∆ where :
1
.
M = OrtO . W =
2S
t

a2
1 
W . P yth . W =
 −Sc
2S
−Sb
−Sc
b2
−Sa

−Sb

−Sa 
c2
(7.17)
Proof. This comes directly from the orthopoint formula. Normalization factor 1/2S is useless here,
but will be required when measuring, i.e. computing angles or distances. One can check that, for
example, the first column gives the direction of the first altitude (orthogonal to sideline BC).
Remark 7.4.7. Characteristic polynomial of M is : µ3 + µ2 Sω /S + 3µ and it can be checked that
its left null space is [1, 1, 1], the row associated with L∞ : for any column X, M . X ∈ L∞ (as it
should be).
Proposition 7.4.8. Isotropic lines. A line that satisfies ∆ · M · t ∆ = 0 is orthogonal to itself,
whence the name ’isotropic’ given to these lines. From
∆ · M · t∆
1
a2 p2 + b2 q 2 + c2 r2 − 2 Sa rq − 2 Sb rp − 2 Sc qp
2S
' (∆ · Ωx ) × (∆ · Ωy )
=
this is equivalent to going through either umbilic.
Remark 7.4.9. Spoiler: all the isotropic lines form a ’degenerate tangential conic’ (see Proposition 11.5.2). Therefore adjoint of matrix M is t L∞ . L∞ (the "all ones" matrix) and describes
the
line Ωx , Ωy , i.e. L∞ . Rank is one (since M has rank n − 1), and product Adjoint
is the null matrix.
Fact 7.4.10. The triangle of the midpoints of

a2

 Sc
Sb
7.5
M
. M
the altitudes has barycentrics :

Sc Sb

b2 Sa 
Sa c2
Angles between straight lines
−−→ →
−
Proposition 7.5.1. Let P, U1 , U2 be three points at finite distance, such that P Ui 6= 0 . Then :
−−→ −−→
|P U1 | . |P U2 | . sin P U1 , P U2
−−→ −−→
|P U1 | . |P U2 | . cos P U1 , P U2
=
2S
det (P, U1 , U2 )
Q
(p + q + r) (ui + vi + wi )
t
=
(P ∧ U1 ) . M . (P ∧ U2 )
2S
2Q
(p + q + r)
(ui + vi + wi )
Proof. The sin formula comes from (7.4), while the cos formula can be obtained by using (7.11)
2
2
2
into |P U1 | + |P U2 | − |U1 U2 | and rearranging.
Theorem 7.5.2. Let P be a point at finite distance, and U1 , U2 two other points (at finite distance
or not). Then the angle between straight lines P U1 , P U2 is characterized by its tangent, according
to :
z }| {
det (P, U1 , U2 )
tan P U1 , P U2
= (p + q + r)
(7.18)
t
(P ∧ U1 ) . M . (P ∧ U2 )
When U1 , U2 are at infinity, the angle between all the lines having the given directions can be
computed as :
z }| {
2S (v1 w2 − w1 v2 )
tan∞ U1 , U2 =
(v1 w2 + w1 v2 ) Sa + w1 w2 b2 + v1 v2 c2
March 30, 2017 14:09
published under the GNU Free Documentation License
7. Euclidean structure using barycentrics
namex
ψ
L∞
2
∆ ∧ L∞
W
pythagoras
P yth
orthopoint
OrtO
orthodir
M
#
3
(7.3)
5
(7.10)
4
(7.16)
3
(7.17)
tangent of
z }| {
∆1 , ∆2
(13.8)
(14.21)
Q
OrtO =
barycentrics
#
Morley
[1; 1; 1]

0
1 −1


0
1 
 −1
1 −1
0


0
−c2 −b2
1

0
−a2 
 −c2
2
−b2 −a2
0


2
2
2
c −b
−a
a2
1  2

a2 − c2 −b2 
 b
4S
−c2
c2
b2 − a2


a2 −Sc −Sb
1 

b2
−Sa 
 −Sc
2S
−Sb −Sa
c2
(14.3)
[0; 1; 0]

0 0 −1


 0 0 0 
+1 0 0


0 0 1
2
R 

 0 −2 0 
2
1 0 0


1 0 0


i 0 0 0 
0 0 −1


0 0 1


−i  0 0 0 
1 0 0

∆ 1 . W . t ∆2
(7.19)
circles
1
2S
t
67

1
16S 2
W . P yth
;
∆1 . M . t ∆2
(14.4)
(14.5)
(14.6)
(14.7)
∆1 . Wz . t ∆2
(14.8)
a2
−Sc
−Sb −a2 Sa
 −S
b2
−Sa −b2 Sb

c

 −Sb
−Sa
c2
−c2 Sc
−a2 Sa −b2 Sb −c2 Sc a2 b2 c2
t






OrtO . P yth . OrtO = P yth
;
∆1 . Mz . t ∆2


0 0 1 0

1
 0 0 0 −1 


2 1 0 0 0 
0 −1 0 2
M = OrtO . W
transform bar 7→ Lub ψ : (1) point: X 7→ Lu X ; (2) line: X 7→ X Lu−1
t
(3) line to point Lu X Lu ; (4) point to point Lu X Lu−1 ; (5) quad form
t
Lu−1 X Lu−1
Table 7.1: All these matrices
Proof. The key point here is that formula (7.18) is square-root free. Extension to Ui at infinity
is obtained by continuity after cancellation of the (ui + vi + wi ). Formula tan∞ is not formally
symmetrical (P = A has been used). But the ui are nevertheless present since ui = −vi − wi .
z }| {
Theorem 7.5.3. Tangent of two lines. If the triangle plane is oriented according to BC, BA =
+A, then oriented angle from line ∆1 to line ∆2 is characterized by :
z }| {
∆1 . W . t ∆ 2
tan ∆1 , ∆2 =
∆1 . M . t ∆2
(7.19)
where W and M are exactly as given in (7.3) and (7.17) (i.e. not up to a proportionality
factor).
Proof. Simple use of ∆1 ∧ ∆2 , L∞ ∧ ∆1 , L∞ ∧ ∆2 in (7.18). Among other things, this formula tells
us that ϑ = 0 when each line contain the point at infinity of the other, while |ϑ| = π/2 when each
line contains the orthopoint of the other (formula is anti-symmetric).
Stratospheric proof. Start from the affine space E. Equations of both lines are aj ξ + bj η + cj = 0
and their angle is given by
z }| {
det (L3 , D1 , D2 )
a1 b2 − a2 b1
tan ∆1 , ∆2 =
=
a1 a2 + b1 b2
D1 . Orth3 . t D2
where L3 = [0, 0, 1] and Orth3 is the matrix of quadratic form ξ 2 + η 2 –precisely this one, without
any of the extra terms used in (7.8). Taking now ABC for basis, we have ∆j = Dj . T0−1 , inducing
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
68
7.6. Distance from a point to a line
a factor 1/2S in the numerator. Let us now compare the following two expressions :


ηb − ηc
−ηa + ηc
ηa − ηb
1 
t −1

T0 =
ξa − ξc
−ξa + ξb 
 −ξb + ξc
2S
ξb ηc − ξc ηb −ξa ηc + ξc ηa ξa ηb − ξb ηa


−ξb + ξc ξa − ξc −ξa + ξb


T0 . W =  −ηb + ηc ηa − ηc −ηa + ηb 
0
0
0
t
Due to the specific value of Orth3 , we can replace T0−1 by T0 . W /2S in the change of basis
2
formulas and obtain Orth3 7→ K / (2S) . Using the orthodir matrix instead of the Al-Kashi one
leads to a formula without remaining factors.
Remark 7.5.4. Rotations are examined at 14.2
7.6
Distance from a point to a line
Definition 7.6.1. The distance of point P = p : q : r to line ∆ = (ρ, σ, τ ) is the lower bound
of the distance of P to a point U that belongs to ∆. By continuity, this bound is attained and is
equal to the distance of P to its orthogonal projection P0 on ∆.
Theorem 7.6.2. Distance from point P = p : q : r to line ∆ = (ρ, σ, τ ) is given by :
dist (P, ∆) =
√
2S
∆.P
q
(L∞ . P ) ∆ . M . t ∆
(7.20)
where L∞ = [1, 1, 1] and M is as given in (7.17) (not up to a proportionality factor).
Proof. One obtains easily :
4S 2 (ρ p + σ q + τ r)
2
|P P0 | =
2
2
(p + q + r) (a2 ρ2 + b2 σ 2 + c2 τ 2 − 2 σ τ Sa − 2 ρ τ Sb − 2 ρ σ Sc )
where each factor is recognizable (cf Theorem 7.5.3).
Remark 7.6.3. Formula (7.20) is invariant when barycentrics of P or ∆ are modified by a proportionality factor. Denominators are enforcing the fact that P is supposed to be finite, and ∆
is not supposed to be the infinity line. The square
√ root is the operator norm of the application
P 7→(ρ p + σ q + τ r) / (x + y + z). And finally 2S is the square root of the standard area, i.e.
the standard unit of length.
7.7
7.7.1
Brocard points and the sequel
Some results
Proposition 7.7.1. Brocard points. It exists exactly one point ω + and one point ω − such that :
∠ (Aω + , AC)
∠ (AB, Aω − )
= ∠ (Bω + , BA) = ∠ (Cω + , CB)
= ∠ (BC, Bω − ) = ∠ (CA, Cω − )
They are given by ω + = a2 b2 : b2 c2 : c2 a2 and ω − = c2 a2 : a2 b2 : b2 c2 . Moreover, when defined
exactly that way, both angles are equal. This quantity is called the Brocard angle and one has :
cot ω =
a2 + b2 + c2
4S
(7.21)
Proof. Equating the tangents of the angles and eliminating,
one obtains a third degree equation
√
with one simple real root and tho others that involves −S 2 . As it should be Brocard points are
isogonal conjugates of each other.
March 30, 2017 14:09
published under the GNU Free Documentation License
7. Euclidean structure using barycentrics
69
Proposition 7.7.2. cot versus Conway. We have the following equalities between Conway
symbols and some cotangents:
[Sa , Sb , Sc , Sω ] = [2 S cot A, 2 S cot B, 2 S cot C, 2 S cot ω]
cot
A
bc + Sa
B
ac + Sb
C
ab + Sc
=
; cot
=
; cot
=
2
2S
2
2S
2
2S
Proof. First three are obvious, the ω one is given just above. This proves the Volenec (2005)
formula :
cot ω = cot A + cot B + cot C
.
Remark 7.7.3. A subsection about the so-called Brocard triangles is located at Subsection 18.9.
Remark 7.7.4. The ETC points on the Brocard line are 9 (Mittenpunkt), 512 (at infinity), 881,
882, 2524, 2531.
Lemma 7.7.5. Since tan ω > 0 and |ω| ≤ π/6, we have :
2S
c 2 + a 2 + b2
, sin (ω) = √
cos (ω) = √
2 a2 b2 + a2 c2 + b2 c2
a2 b2 + a2 c2 + b2 c2
(7.22)
Fact 7.7.6. Each Brocard point is at the intersection of three isogonal circles, according to :
∠ ω + B, ω + C
= ∠ (BA, BC)
−
−
∠ ω B, ω C
= ∠ (CB, CA)
Proposition 7.7.7. Neuberg circles. The locus of A when B, C, ω are given is a circle (Neuberg
circle of vertex A). Barycentric equation, center and radius are :
a2 yz + b2 xz + c2 xy − a2 (x + y + z)(y + z) = 0


 
a2 c2 + a2 + b2
−a cos (ω)

 

Na '  a2 + b2 c2 − b4 − a4  '  b cos (C + ω) 
a2 + c2 b2 − c4 − a4
c cos (B + ω)
p
a
ρA =
cot2 ω − 3
2
2
Proof. Straightforward computation, replacing b2 by |A0 C| etc. The form given proves the circular
shape.
Proposition 7.7.8. Tarry point. Triangle Na Nb Nc is perspective with ABC and perspector is
the Tarry point X(98)=
1
a2 b2
+
a2 c2
−
b4
−
c4
:
1
b2 c2
+
b2 a2
−
c4
−
a4
:
1
c2 a2
+
c2 b2
− a4 + b4
Conversely, Neuberg center Na is common point of the A cevian of X98 and the perpendicular
bisector of side BC, etc.
Proof. Direct computation.
Proposition 7.7.9. The Steiner angles ω1 > ω2 are defined as follows. 2ω1 is the maximal value
of A when ω is given and 2ω2 is the minimal value. We have the following relations :
cot 2ωj + 2/ cot ωj = cot ω
p
cot ω1 = cot ω − cot2 ω − 3
p
cot ω2 = cot ω + cot2 ω − 3
sin (2ωj + ω) = 2 sin ω
ω + ω1 + ω2 = π/2
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
70
7.7. Brocard points and the sequel
Proof. First formula comes from (7.21) (at extremum, triangle ABC is isosceles). This gives a
second degree equation whose discriminant cot2 ω − 3 is non negative, and last formula comes from
cot (ω1 + ω2 ) depends on sum and product of the cot ωj .
Proposition 7.7.10. Any Neuberg circle is viewed from another vertex under angle 2ϑ where :
cos ϑ = 2 sin ω = sin (2ωj + ω)
Proof. The polar of B cuts circle Na in two points T1 , T2 (equation of second degree, ∆ = a4 + b4 +
c4 −a2 b2 −a2 c2 −b2 c2 ). And we have 2ϑ = ∠ (BT1 , BT2 ). A better choice is T0 = midpoint(T1 , T2 )
and ϑ = ∠ (BT1 , BT0 ) ... taking orientation into account !
7.7.2
Results related to the Kiepert RH
This section has moved to Proposition 12.18.1.
7.7.3
Spoiler: study of the Neuberg pencil
Let us consider the level curves of the Brocard angle ω (M, B, C) when vertex M moves in the
ABC plane. Barycentric coordinates are surely not the best system here, and using the Veronese
map is probably not required. Therefore, this subsection should rather be considered as a spoiler
related to the chapter devoted to pencils of cycles.
1. Let us note M ' x : y : z, |M B| = γ, |M C| = β and u/v = tan ω. Then
a2 + β 2 + γ 2
u
=
v
4 SM
where β 2 =
z (x + z) a2 + x (x + z) c2 − b2 zx
(x + y + z)
2
, etc. ; SM =
xS
x+y+z
2. It can be seen that numerator of u/v − · · · collects nicely, leading to equation
v × C (x, y, z) − 4 uS × (x (x + y + z)) = 0
3. In the second term, one recognizes that x (x + y + z) describes the line BC seen as a cycle, i.e.
completed by the line at infinity. On the other hand, the conic C0 (x, y, z) can be recognized
as the circle whose equation is the column
: V ' Sω : a2 : a2 : 1. And therefore is centered
q
√
t
at Q · V ' 0 : 1 : 1 : and has radius
V · Q · V = ia 3/2.
4. Thus the set of the level curves is the pencil of cycles generated by the two special cases, i.e.
the Neuberg pencil. Since C0 , the cycle centered on the radical axis, is virtual, the pencil
is an isotomicpencil, whose limit points P± are on the real circle associated with C0 : these
points are the vertices of the equilateral triangles whose basis is BC.
5. From u2 /(u2 + v 2 ) = cos(ωM )2 , we have u = v cot (ωM ) so that the level curves are the
circles:
V
q
t
'
Q ·V
'
V · Q ·V
=
Sω − 2 S cot (ω) : a2 : a2 : 1
a cos (ω) : −b cos (ω + C) : −c cos (ω + B) : −abc cos (A + ω)
1 p 2
a cot ω − 3
2
6. And therefore, C0 is the (virtual) locus of vertices M such that ωM = 90°.
7.7.4
Spoiler: Brocard angle of a pedal triangle
2
1. Consider M ' x : y : z and its pedal triangle. Apart from a common factor 1÷(2R (x + y + z)) ,
area and squared sidelengths are:
SM = S a2 yz + b2 zx + c2 yx ; a2M = a2 z (y + z) b2 + y (y + z) c2 − a2 yz ; etc.
Thus the level curves of the Brocard angle are given by
u
a2 + b2M + c2M
= cot (ωM ) = M
v
4SM
March 30, 2017 14:09
published under the GNU Free Documentation License
7. Euclidean structure using barycentrics
71
2. As before, the numerator collects nicely and the locus is given by:
v C (x, y, z) − 4u S a2 yz + b2 zx + c2 yx = 0
3. This locus is a circle, whose Veronese image V , center E and radiusρ are:



 

b2 c 2 v
p
a cos (A − ωM )
a2 (2 Sv + Sa u)


2 2
c a v
R cot2 (ωM ) − 3

 
 2


V '
 ; E '  b (2 Sv + Sb u)  '  b cos (B − ωM )  ; ρ =
cot (ωM ) + cot (ω)


a2 b2 v
c cos (C − ωM )
c2 (2 Sv + Sc u)
2 uS + Sω v
4. One sees thatv = 0 describes the circumcircle. Projections of M are aligned on the Simson
line, so that ωM = 0.
√
5. One sees that ωM = ±30°, i.e. v = cot (ωM ) = ± 3, u = 1 characterizes two points (ρ = 0).
They are X(15), X(16), the isodynamic points. No other pedal triangle is equilateral.
6. The locus characterized by ωM = ωABC is centered at X(182). One identifies the 3-6-Brocard
circle. The locus characterized by ωM = −ωABC is a straight line (ρ = ∞). Therefore, this
line is the perpendicular bisector of segment X(15),X(16): the line X(187),X(512).
7.
7.8
Orthogonal projector onto a line
Proposition 7.8.1. The matrix π∆ of the orthogonal projector onto line ∆ ' [p, q, r] –not the
line at infinity – is given by :
π∆ = ∆ . M . t ∆ − M . t ∆ . ∆
(7.23)
where matrix M is defined by (7.17).
Proof. Compute in R3 , and define W = M . t ∆ (the projective point associated to W is the
.
orthodir of ∆). In this context, rank one matrix M = W . ∆ = M . t ∆ . ∆ is interesting: its action can be interpreted as redirecting the output of the standard projector 1/ p2 + q 2 + r2 t ∆ . ∆.
Multiplying M by any point of ∆ gives 0, and multiplying by W gives λW where λ = ∆ . W 6= 0
(since ∆ 6= L∞ ). Therefore, (1/λ) M is a rank one projector of R3 and λ − M is a rank 2
projector of the barycentric plane, the required πP .
Proposition 7.8.2. The matrix σ∆ of the orthogonal reflection wrt line ∆ ' [p, q, r] –not the
line at infinity – is given by :
σ∆ = ∆ . M . t ∆ − 2 M . t ∆ . ∆
(7.24)
where matrix M is defined by (7.17).
Proof. Obvious from the preceding proof.
−
Proposition 7.8.3. Cosine of a projection. Consider line ∆1 ' [p, q, r] and use →
e1 =
(q − r, r − p, p − q) as unit vector for this direction. Consider also line ∆2 ' [u, v, w] and use
→
−
e2 = (v − w, w − u, u − v) as unit vector for that other direction. Then orthogonal projection π
onto ∆1 transforms ∆2 -vectors into ∆1 -vectors according to :
∆ 1 . M . t ∆2
−
−
π (→
e2 ) = →
e1
∆ 1 . M . t ∆1
−
Proof. Formula is homogeneous, as it should be. Vectors →
ei are not normalized, that the reason
t
why this formula is square-root free. Formula 2S M = W . P yth . W indicates that scaling
factor can be interpreted in terms of a "cosine of projection".
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
72
7.8. Orthogonal projector onto a line
ωM ≈ +30°
ωM = −ωABC
ωM ≈ −30°
ωM = +ωABC
Figure 7.1: Level curves of the Brocard angle of the pedal triangle of point M
March 30, 2017 14:09
published under the GNU Free Documentation License
7. Euclidean structure using barycentrics
73
Proposition 7.8.4. Consider the homothety of center P = p : q : r (not on the infinity line) and
ratio k (not 0 !). Then points U are transformed as U 7→ h (P, k) . U while lines ∆ are transformed
as ∆ 7→ ∆ . h (P, 1/k) where :




1 0 0
p p p
1−k
P · L∞




h (P, k) =  0 1 0  +
(7.25)
 q q q  ' k + (1 − k)
k (p + q + r)
L∞ · P
0 0 1
r r r
→
−
→
−
Proof. Applied to column P , the last formula gives P . Applied to a vector V , this gives k V since
→
−
a vector is defined by L∞ · V = 0. Another method: we want U 7→ X such that, given U , we have
(X − P ) = k (U − P ). Expressed in barycentrics, this leads to :



 

u
p
x
 (1 − k) (u + v + w) 

 

 q 
 y ' v +
k (p + q + r)
w
r
z
Property about lines comes from the fact that action over lines and action over points are inverse
of each other.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
74
March 30, 2017 14:09
7.8. Orthogonal projector onto a line
published under the GNU Free Documentation License
Chapter 8
Pedal stuff
In a previous life, this Section was intended as foreword to Chapter 7 (orthogonality). Now, this
Section is rather a the symmetric aisle of the former Chapter 4 (cevian stuff).
8.1
Pedal triangle
Definition 8.1.1. The pedal triangle of point P is the triangle whose vertices are the orthogonal
projections of P on the sides of the triangle.
Remark 8.1.2. Crossover the Channel, the pedal triangle is called "triangle podaire", while "pédal
triangle" is used to denote the Cevian triangle. Plaisante vérité qu’une rivière borne (Pascal, 1852,
p. 41).
Proposition 8.1.3. We have the following barycentrics (each point is a column) :
 
0
p
 

pedal  q  =  Sc p + a2 q
Sb p + a2 r
r


Sc q + b2 p Sb r + c2 p

0
Sa r + c2 q 
Sa q + b2 r
0
2
2
2
where Sa = b + c − a /2, etc.
(8.1)
Proof. Use (7.23) and obtain directly the result.
Proposition 8.1.4. Condition for an inscribed triangle P1 P2 P3 to be the pedal triangle of some
P is :
q1 − r1 2 r2 − p2 2 p3 − q3 2
a +
b +
c
(8.2)
q1 + r1
p2 + r2
p3 + q3
In such a case, point P is the perspector between P1 P2 P3 and triangle M and is given by either
following expressions :


b2 c2 p2 p3 − (Sc r2 − Sa p2 ) (Sb q3 − Sa p3 )


b2 c2 p2 q3 + (Sb q3 − Sa p3 ) b2 r2


2 2
2
b c p3 r2 + (Sc r2 − Sa p2 ) c q3


a2 c2 p3 q1 + (Sa p3 − Sb q3 ) a2 r1
 2 2

 a c q1 q3 − (Sa p3 − Sb q3 ) (Sc r1 − Sb q1 ) 
a2 c2 q3 r1 + (Sc r1 − Sb q1 ) c2 p3


a2 b2 p2 r1 + (Sa p2 − Sc r2 ) a2 q1


a2 b2 q1 r2 + (Sb q1 − Sc r1 ) b2 p2


2 2
a b r1 r2 − (Sb q1 − Sc r1 ) (Sa p2 − Sc r2 )
Proof. P is on the line through P1 and orthopoint of BC etc. The required condition is the
determinant of these three lines. The various ways of writing P are the wedge product two at a
time. A more symmetrical formula would be great...
75
76
8.2
8.2. Isogonal conjugacy
Isogonal conjugacy
Definition 8.2.1. isogonal conjugate. Suppose P = p : q : r is a point not on a sideline of
ABC. Let L(A) be the line issued from A and such that (AB, AP ) = (L (A) , AC) –the equal
inclination property. Define L(B) and L(C) cyclically. The lines L(A), L(B), L(C) are concurrent,
and their common point is called the isogonal conjugate of P . Using barycentrics, we have :
isog (x : y : z) =
a2 b2 c2
:
:
x y z
(8.3)
Remark 8.2.2. The isogonal conjugate of a point P is often noted P −1 in ETC since its trilinears
are 1/p : 1/q : 1/r when those of P are p : q : r.
8.3
Cyclopedal conjugate
Proposition 8.3.1 (Matthieu). When P and U are isogonal conjugates, their pedal triangles are
cocyclic and the center of this circle is the middle of P and U (cf Figure 8.1)
Proof. Straightforward computation.
Figure 8.1: Cyclopedal congugates are isogonal conjugates
Remark 8.3.2. The definition of "cyclopedal conjugacy" has been coined to enforce symmetry with
the cyclocevian conjugacy, cf. Section 12.19. Some examples are :
point
code
bary
cycp
circumcenter
incenter
centroid
Lemoine
circumcenter
orthocenter
X (1)
X (2)
X (6)
X (3)
X (4)
a
1
a2
a2 −a2 + b2 + c2
1/ −a2 + b2 + c2
X (1)
X (6)
X (2)
X (4)
X (3)
X (1)
X (597)
X (597)
X (5)
X (5)
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 9
Orthogonal stuff
9.1
Steiner line
Proposition 9.1.1 (Steiner line). Let U be a point in the barycentric plane, define A0 as its
symmetric relative to side BC and define B 0 , C 0 cyclically. Then A0 , B 0 , C 0 are aligned if and
only if point U is either on the circumcircle or on the line at infinity.
When U ∈ L∞ , A0 = B 0 = C 0 = U and Steiner (U ) = L∞ .
When U = u : v : w ∈ Γ, so that Q = isogon (U ) ∈ L∞ , then Steiner (U ) goes through the
orthocenter X4 and the coefficients of Steiner (U ) are given by :
t
t
Steiner (U ) ' isogon U ∗ X4 '
Q ÷ X4
b
b
' [Sa (τ − σ) , Sb (ρ − τ ) , Sc (σ − ρ)]
2
a Sa b2 Sb c2 Sc
,
,
'
u
v
w
(9.1)
Proof. The symmetrics are aligned if and only if the projections are aligned, while the determinant
of (8.1) factors into :
32 (p + q + r) S 2 a2 qr + b2 rp + c2 pq
The case p + q + r = 0 is the line at infinity and is not to be discarded, since the orthogonal
projection of a point at infinity onto a line at finite distance is the point at infinity of this line.
When U ∈ Γ, computing (9.1)is straightforward.
Remark 9.1.2. The study of the properties of the so called Simson lines is delayed until Section 22.4.
Indeed, their envelop is a third degree curve, and not a simple point as for the Steiner lines.
Proposition 9.1.3. When U1 and U2 are on the circumcircle, then :
z
}|
{
1 −−−→ −−−→
X3 U1 , X3 U2
Steiner (U1 ) , Steiner (U2 ) = −
2
where the "overbrace" denotes the oriented angle between two straight lines. Therefore, it exists a
one-to-one correspondence between lines through X4 and points U on the circumcircle. Moreover,
Steiner lines relative to diametrically opposed points on the circumcircle are orthogonal to each
other.
Proof. From elementary Euclidean geometry... or using tan (2ϑ) = 2 tan ϑ/ 1 − tan2 ϑ and (7.19).
Corollary 9.1.4. For each point on the circumcircle, the isogonal conjugate is the orthopoint of
the Steiner line (cf. Figure 9.1, and also –far below– Figure 17.4).
Proof. Let Q1 be a point at infinity. Take the isogonal conjugate of Q1 and obtain U1 ∈ Γ. Take
the Steiner line of U1 and obtain S1 . Take the point at infinity of S1 and obtain Q2 . Take the
isogonal conjugate of Q2 and obtain U2 ∈ Γ. Take the Steiner line of U2 and obtain S2 . Now Q1
is the point at infinity of S2 while Q1 , Q2 are orthopoints of each other and U1 , U2 are antipodes
of each other on the circumcircle.
77
78
9.2. Spoiler: Moebius-Steiner-Cremona transform
ref lection in X3
U1 ∈ Γ ✲ S1 = t (Q1 ÷b X4 )
U2 ∈ Γ ✛
.........
.
.
✲
.
.
.
.
.
.
.........
.........
........................
.
.
.
.
.
.
.
.
..is
....
∞
∞
og ........
... isog
.
.
.
.
.........
.
.
.
.
.
.
.
.
.
.
.
.
.
.........
.....
.........
❄
❄ ...............
.
✲
Q2 = L∞ ∩ S1
Q1 = L∞ ∩ S2
orthopoint
S2 = t (Q2 ÷b X4 ) ✛
orthopoint (Q)
Steiner (P )
= L∞ ∧
t
Q ÷ X4
b
(9.2)
= X4 ∧ orthopoint (isogon (P ))
(9.3)
Figure 9.1: The orthopoint transform
Proposition 9.1.5. When Q1 , Q2 are orthopoints, their barycentric product lies on the orthic
axis, i.e. the tripolar of X4 .
Proof. Use parameterization (7.14), eliminate k, σ in k X = Q1 ∗ Q2 and obtain an equation of
b
first degree.
Example 9.1.6. The following list gives the triples (I, J, K) where X (I) and X (J) are named
orthopoints of each other and X (K) is their (named) barycentric product :
30
511
513
514
2574
3307
9.2
523
512
517
516
2575
3308
1637
2491
3310
676
647
3310
Spoiler: Moebius-Steiner-Cremona transform
This section should be skiped by a reader not familar with Morley spaces (Chapter 14) and Cremona
transforms (Chapter 16).
Definition 9.2.1. Let α : 1 : 1/α, etc be the reference triangle ABC, and z1 : t1 : ζ1 the Morley
affix of point U1 . Then affix of point A’ (symmetric of U1 wrt line BC) is β + γ − β γ ζ1 /t1 . The
triangle A0 B 0 C 0 is called the Steiner triangle of U1 and we have:


t1 (β + γ) − βγ ζ1 t1 (γ + α) − γα ζ1 t1 (α + β) − αβ ζ1


t1
t1
t1

T1 ' 
 t (β + γ) − z
t1 (γ + α) − z1
t1 (α + β) − z1 
1
1
βγ
γα
αβ
Proposition 9.2.2. The linear transform characterized by ψ1 : A 7→ A0 , B 7→ B 0 , C 7→ C 0 sends
O on H and satisfies:
ζ1 s3
ζM
z (ψ (M )) = −zM + s1 −
t1
Proof. The matrix of ψ1 is:



T1 · Lu ' 

March 30, 2017 14:09
1
−s1
0
z1
s3 t1
−1
s2
−
s3

ζ1 s3
t1 

0 

1
published under the GNU Free Documentation License
9. Orthogonal stuff
79
Proposition 9.2.3. Let us consider two points and their Steiner triangles. Define the collineation
φ by T1 7→ T2 ; Lz 7→ Lz . Its matrix is :
φ
−1
'
−1
ψ2 · ψ1
' T2 · T1

(ζ2 t1 − ζ1 t2 ) (z1 σ1 − σ2 t1 )
t1 t2 − ζ2 z1

t1 
' 
0
t21 − z1 ζ1 t2 /t1

 1
(z2 t1 − z1 t2 ) (ζ1 σ2 − σ1 t1 )
(z2 t1 − z1 t2 )
σ3
t1 σ3

σ3 (ζ2 t1 − ζ1 t2 ) 


0


t1 t2 − z2 ζ1
its characteristic polynomial is
!
t22 − z2 ζ2 t21
t1 (2 t1 t2 − ζ2 z1 − z2 ζ1 )
(µ − 1) µ −
µ+ 2
t2 (t21 − z1 ζ1 )
(t1 − z1 ζ1 ) t22
2
and we have φ (H)=H.
Remark 9.2.4. Considering some special cases:
1. For φ to be parallelogic, condition is φ11 + φ33 = 0, i.e: z2 ζ1 + ζ2 z1 − 2 t1 t2 = 0. Each point
is on the circle-polar of the other.
2. For φ to be orthologic, condition is φ11 − φ33 = 0, i.e. z2 ζ1 − ζ2 z1 = 0. Both points are
aligned with the circumcenter X(3).
3. For φ to be an indirect similitude, condition is φ11 = φ33 = 0. Both points are inverse in the
circumcircle.
4. For triangles T1 , T2 to be in perspective (homologic), condition is
σ4 (z2 t1 − t2 z1 ) (ζ2 t1 − ζ1 t2 ) ((z1 t2 − t1 z2 ) σ2 + (ζ1 z2 − z1 ζ2 ) σ3 + (t1 ζ2 − ζ1 t2 ) σ3 σ1 ) = 0
In other words, either the points are equal in one of the spheric maps, or they are aligned
with X(4). In this case, the perspector is a point S on the circumcircle, and line U1 HU2 is
nothing but the Steiner line of S.
Proof. Compute the perspector of T (M1 ) and T (k M1 + (1 − k) H).
Proposition 9.2.5. In the map Z : T, the homography µ defined by A 7→ A0 , B 7→ B 0 , C 7→ C 0
is called the Moebius-Steiner transform related to U . This transform is involutive and we have:
t2 Z − t2 σ1 − tζ σ2 + ζ 2 σ3 T
µ (Z : T) =
ζ tZ − t2 T
Image µ (U ) is isogon (U ), i.e. the other focus of the inscribed conic whose U is the first focus.
Moreover, µ (∞) = t/ζ, i.e. the symmetric of U wrt the circumcircle. Finally, A0 , B 0 , C 0 , H are
cocyclic when either U is on the circumcircle (the Steiner property) or U is on the orthoptic of the
conjugate circle (centered at H).
Proof. Write the reality of the cross-ratio and obtain t2 − zζ s1 s3 tζ + s2 tz − s3 zζ − s3 t2 =
0.
9.3
Orthopole
The section has moved to "Pedal LFIT and orthopole" (Section 22.3), while "orthojoin" has
disappeared.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
80
9.4
9.4. Orthology
Orthology
Definition 9.4.1. Two triangles T1 and T2 are orthologic when perpendiculars from the vertices
of T1 to the corresponding sides of T2 are concurrent.
Proposition 9.4.2. Orthology is a symmetric relation.
Proof. Using computer, symmetry is straightforward : condition of concurrence is the product of
det T2 by a polynomial that is invariant by exchange of the two triangles.
Example 9.4.3. Cevian triangles of X(2) and X(7) are orthologic. Orthology centers are X(10)
and X(1).
Exercise 9.4.4. (spoiler). Morley spaces: when φ is a collineation that maps Lz to Lz , its matrix
verifies φ21 = φ23 = 0. Suppose that a pair T1 , φ (T1 ) is orthologic. Then prove that φ11 = φ33 .
Conclude.
Exercise 9.4.5. Barycentrics: when φ is a collineation that maps L∞ to L∞ , its matrix verifies φ11 + φ21 + φ31 = 1, etc. Suppose that a pair T1 , φ (T1 ) is orthologic. Then prove that
trace φ · OrtO = 0. Conclude.
9.5
Parallelogy
Definition 9.5.1. Two triangles T1 and T2 are parallelogic when parallels from the vertices of T1
to the corresponding sides of T2 are concurrent.
Proposition 9.5.2. Parallelogy is a symmetric relation.
Proof. Using computer, symmetry is straightforward : condition of concurrence is the product of
det T2 by a polynomial that is invariant by exchange of the two triangles.
Proposition 9.5.3. A collineation φ such that L∞ 7→ L∞ is parallelogic if and only if L∞ · φ =
trace (φ) L∞ .
Proof. Direct computation of the determinant that describes parallelogy.
Exercise 9.5.4. (spoiler). Morley spaces: conditions are φ21 = φ23 = 0 to enforce Lz 7→ Lz and
φ11 + φ33 = 0 to have the right trace.
9.6
Orthocorrespondents
Definition 9.6.1. Suppose P is a point in the plane of triangle ABC. The perpendiculars through
P to the lines AP, BP, CP meet the lines BC, CA, AB, respectively, in collinear points. Let L
denote their line. The trilinear pole of L is P ⊥ , the orthocorrespondent of P. This definition is
introduced in Gibert (2003). If P = p : q : r is given in barycentrics, then P ⊥ = u : v : w is given
by :
− b2 + c2 − a2 p2 + a2 − b2 + c2 pq + a2 + b2 − c2 pr + 2qra2
Remark 9.6.2. If follows that if P = x : y : z is given in trilinears, then P ⊥ has trilinears given
cyclically by :
yz + (−x cos A + y cos B + z cos C)x
March 30, 2017 14:09
published under the GNU Free Documentation License
9. Orthogonal stuff
81
Example 9.6.3. Pairs (I,J) for which the orthocorrespondent of X(I) is X(J) include the following:
1
2
3
4
5
6
7
8
9
10
57
1992
1993
2
1994
1995
1996
1997
1998
1999
11
13
14
15
16
19
32
33
36
61
651
13
14
62
61
2000
2001
2002
2003
2004
62
80
98
100
101
103
105
106
107
108
2005
2006
287
1332
1331
1815
1814
1797
648
651
109
111
112
113
114
115
117
118
119
120
1813
895
110
2986
2987
110
2988
2989
2990
2991
125
132
186
242
403
468
915
917
1300
1560
648
287
1994
1999
1993
1992
2990
2989
2986
895
1566 677
1785
57
1845 2006
1878 1997
3563 2987
Proposition 9.6.4. The orthocorrespondent of every point on the line at infinity is the centroid.
Conversely, given a finite point U , different from the centroid, it exists exactly two orthoassociate
points P1 and P2 (real or not, distinct or not) that share the same orthocorrespondent U . When
P1 is given, then :
a2 − b2 + c2 2 a2 + b2 − c2 2
q +
r
p2 ' (q1 + r1 )p1 + 2
a − b2 − c2 1 a2 − b2 − c2 1
When U is given, the condition of reality is ∆ ≥ 0 where :
.
∆ = S 2 (u + v + w)2 − u(w + v)Sc Sb − v(u + w)Sa Sc − w(u + v)Sb Sa
S = area and Sa = b2 + c2 − a2 /2. Then, cyclically, we have :
p1 , p2 '
S ((u − w)(u + v − w)Sb + (u − v)(u − v + w)Sc )
√
± ((u − w)Sb + (u − v)Sc ) ∆
!
Proof. Write orthocorr (P ) = kU and eliminate (rationally) k, p. Obtain a second degree equation,
2
whose discriminant is S 2 (v − w) ∆. In order to obtain a symmetrical form for the barycentrics
√ 2
p : q : r, all these expressions must be simplified using
∆ = ∆, then rationalized and simplified
again, and finally normalized using p + q + r = 1.
9.7
Isoscelizer
An isoscelizer is a line perpendicular to an angle bisector. If P is a point, then the A-isoscelizer of
P is the line L(P, A) through P perpendicular to the line that bisects vertex angle A; the B- and
C- isoscelizers are defined cyclically. Let D and E be the points where L(P, A) meets sidelines AB
and AC. Unless D = E = A, the triangle ADE is isosceles.
In ETC, there are several triangle centers defined in terms of isoscelizers. These were discovered
or invented by Peter Yff, in whose notebooks the word isoscelizer dates back to 1963.
9.8
Orion Transform
Definition 9.8.1. Reflect P through the sidelines of its cevian triangle AP BP CP . The obtained
triangle is perspective with triangle ABC, and the perspector is called the Orion transform of P.
In barycentrics, we have :
2
a
b2
c2
Sa
OT 1 (P ) = p − 2 + 2 + 2 + 2
:: etc
p
q
r
qr
Proposition 9.8.2. (Ehrmann, 2003) For any point P , the reflection triangle and the anticevian
triangle of the same point P are perspective, and the perspector is the isogonal conjugate of P wrt
the orthic triangle.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
82
9.8. Orion Transform
Pb
A
Pc
Rc
Ch
Cp
P
B
Bh
H Bp
C
Ah
Ap
Rb
Ra
Pa
Figure 9.2: Reflection triangle and its perspector.
Proof. Let PA PB PC be the anticevian triangle of P , AP BP CP its cevian triangle and AH BH CH
the orthic triangle. Division (A, AP , P, PA ) is harmonic; so is pencil (AH A, AH AP , AH P, AH PA )
where AH A ⊥ AH AP ( don’t forget that AH AP = BC !). Hence :
1. line BC is a bisector of ∠ (AH P, AH PA ). Thus RA , the reflection of P , is on AH PA . Proving
concurrence of lines PV RV is now proving concurrence of lines PV VH for V = A, B, C.
2. line AH A is a bisector of ∠ (AH P, AH PA ). But AH A is also a bisector of ∠ (AH BH , AH CH ) ;
it follows that AH P and AH PA are isogonal wrt AH AB and AH CH . Since the VH P obviously
concur in P, the PV VH concur wrt the isogonal conjugate of P wrt the orthic triangle.
Computer-aided proof. . The barycentrics of the reflection triangle are :


−a2 p
2 Sc q + b2 p 2 Sb r + c2 p


TR =  2 Sc p + a2 q
−b2 q
2 Sa r + c2 q 
2 Sb p + a2 r 2 Sa q + b2 r
−c2 r
Perspectivity is obvious and the perspector is :
OT 2 (P ) = p (−Sa p + Sb q + Sc r) : q (Sa p − Sb q + Sc r) : r (Sa p + Sb q − Sc r)
The link with the orthic triangle is given in Figure 18.1.
Example 9.8.3. Among pairs (I, J) for which OT (X(I)) = X(J) are these:
1
2
3
35
69
2055
March 30, 2017 14:09
4
6
7
24
2056
57
8
15
16
2057
2058
2059
20
40
63
2060
2061
2062
69
75
98
2063
2064
2065
99
100
110
249
59
250
published under the GNU Free Documentation License
Chapter 10
Circumcevian stuff
10.1
Circum-cevians, circum-anticevians
Definition 10.1.1. Circumcevian. Let P be a point, not on Γ (the circumcircle of ABC). Let
A0 be the other intersection of line AP with Γ and define B 0 , C 0 cyclically. Then A0 B 0 C 0 is the
circumcevian triangle of P . Using barycentric columns,


−qra2
p
p
 c2 q + b2 r



2


−rpb

(10.1)
circumcevian(P ) = 
q
q


a2 r + c2 p



−pqc2 
r
r
2
b p + a2 q
Proposition 10.1.2. A circumcevian triangle is a central triangle. When P is on Γ, the corresponding triangle is totally degenerated. Three points on the circumcircle form a circumcevian
triangle when the triangle obtained by killing the diagonal of the matrix is a cevian triangle.
Proof. Centrality follows directly from (10.1) (that’s a reason to keep denominators), while killing
Q
3
the diagonal gives the intersections with sidelines. Determinant is Γ (P ) over a2 q + b2 p.
Definition 10.1.3. Circum-anticevian. Consider the anticevian triangle PA PB PC of a point
P that is not a vertex of triangle ABC. Line PB PC cuts circumcircle Γ at A. Let A0 be the
other intersection and define B 0 , C 0 cyclically. Then A0 B 0 C 0 is the circum-anticevian triangle of
P . Using barycentric columns,


qra2
−p
p
 c2 q − b2 r



2


rpb

circumanticevian(P ) = 
(10.2)
q
−q


2
2
a r−c p




pqc2
−r
r
2
2
b p−a q
Proposition 10.1.4. This triangle should be a central triangle (does the definition allows that
?). When one of the three other points ±p : ±q : ±r is on Γ, the circum-anticevian triangle
degenerates.
10.2
Steinbart transform
Definition 10.2.1. Exceter point. The circumcevian triangle of the centroid, X2 , is perspective
to the tangential triangle A6 . The perspector, X22 , is named Exceter point, for Phillips Exceter
Academy in Exceter, New Hampshire, USA, where X22 was detected in 1986 using a computer.
Definition 10.2.2. Steinbart transform. The circumcevian triangle of a point P is ever perspective to the tangential triangle A6 . The corresponding perspector has been called Steinbart
point by Funck (2003) and was called in TCCT (p. 201). This transformation carries triangle
centers to triangle centers. Using barycentrics :
83
84
10.3. Circum-eigentransform
Steinbart (P ) = a2
c4
a4
b4
+ 2− 2
2
q
r
p
: b2
b4
c4
a4
− 2+ 2
2
p
q
r
: c2
b4
c4
a4
+ 2− 2
2
p
q
r
Example 10.2.3. On the circumcircle, Steinbart transform is the identity. Here is a list of other
(I,J) such that Steinbart(X(I)) = X(J) :
1
3
2
22
3 1498
4
24
5 1601
6
6
7 1602
8 1603
9 1604
13 1605
14 1606
17 1607
18 1608
19 1609
21 1610
25 1611
28 1612
31 1613
54 1614
55 1615
56 1616
57 1617
58 595
59 1618
63 1619
64 1620
81 1621
83 1078
84 1622
88 1623
162 1624
163 1625
174 1626
188 2933
251 1627
254 1628
259 198
266
56
275 1629
284 1630
365
55
366 1631
509 1486
648 1632
651 1633
662 1634
Points X(1601)-X(1634) have been contributed in ETC by Jean-Pierre Ehrmann (August 2003).
Remark 10.2.4. See Grinberg (2003d) and his Extended Steinbart Theorem in Hyacinthos #7984,
2003/09/23.
10.3
Circum-eigentransform
Definition 10.3.1. The circum-eigentransform of point U = u : v : w, different from X6 , is the
eigencenter of the circumcevian triangle of point U and is denoted by CET (U ). In trilinears, we
have :
avw
bwu
cuv
:
: 2
2
2
2
+
− buv − cuw bw + bu − cvw − avu cu + cv − awu − bwv
and in barycentrics (cyclically) :
av 2
aw2
a2 vw
a2 c2 v 2 + a2 b2 w2 − b2 c2 uv − bc2 uw
Proposition 10.3.2. Point CET (U ) lies on the circumcircle, and we have CET (U ) = isog (U )
if and only if U ∈ L∞ .
Remark 10.3.3. My own computations are leading to :
a2 vw
−a2 c2 v 2 + a2 b2 w2 − b2 c2 uv + bc2 uw
This point is also on the circumcircle, but property CET (U ) = isog (U ) if and only if U ∈ L∞ is
lost. Some signs have changed : why ?
Example 10.3.4. Apart from points on the infinity line, pairs (I, J) such that X(J) = CET (X(I))
include :
1
106
41 767
74 1294
238 741
2
729
42 2368
75 701
265 1300
3 1300
43 106
81 2375
670 3222
4 1294
44 106
110
99
694
98
9 1477
55 2369
125 827
895 2374
19 2365
56 2370
184 2367
1084 689
25 2366
57 2371
194 729
1279 1477
31
767
58 2372
213 2368
1634 689
32 2367
67 2373
219 2376
37
741
69 2374
220 2377
March 30, 2017 14:09
published under the GNU Free Documentation License
10. Circumcevian stuff
85
Exercise 10.3.5. For a given point P on the circumcircle, which points U satisfy CET (U ) = P
? For example, CET carries each of the points X(1), X(43), X(44), X(519) to X(106).
10.4
Dual triangles, DC and CD Points
Definition 10.4.1. Dual triangle. Suppose DEF is a triangle (at finite distance) in the plane of
triangle ABC. Let D0 be the isogonal conjugate of the point at infinity of line EF . Define E 0 and
F 0 cyclically. The triangle D0 E 0 F 0 is called the dual of DEF . Its vertices lie on the circumcircle.
Proposition 10.4.2. The dual triangle D0 E 0 F 0 characterizes the class of all triangles homothetic
to DEF . Moreover, this triangle is similar to the original one.
Proof. Vertices of D0 E 0 F 0 depends only on direction of sidelines DE, EF, F D and conversely. Similarity between DEF and D0 E 0 F 0 can be proved in many ways. Brute force method : Pythagoras
Theorem 7.2.4 applied to both triangles leads to proportional sidelengths.
Remark 10.4.3. (Proof of similarity follows from Theorem 6E in TCCT, as the "gamma triangle"
there is the dual of a triangle whose sidelines are respectively perpendicular to those of DEF .)
Definition 10.4.4. DC point. Suppose U = u : v : w is a point having cevian triangle DEF
and dual triangle D0 E 0 F 0 . It happens that the later triangle is also the circum-anticevian triangle
of some point. This point will be described as DC (U ). Using barycentrics :
DC (U ) =
b2
c2
a2
:
:
u (v + w) v (w + u) w (u + v)
Remark 10.4.5. The barycentrics of triangle D0 E 0 F 0 are :
a2
wu − vu
b2
vu + wv
−c2
wv + wu
−a2
wu + vu
b2
vu − wv
c2
wv + wu
a2
wu + vu
−b2
vu + wv
c2
wv − wu
Proposition 10.4.6. To construct DC(U ) from U and D0 E 0 F 0 , let A0 = AD0 ∩ BC and let A”
be the harmonic conjugate of A0 with respect to B and C. Define B” and C” cyclically. The lines
AA”, BB” and CC” concur in DC(U ). We have also the formula :
DC (U ) = cevamul (isog (U ) , X (6))
Example 10.4.7. There are 115 pairs (I,J) such that I<2980 and X(J) = DC(X(I)). Among them
(1,81), (2,6), (3,275), (4,2), (5,288), (6,83), (7,1), (8,57), (9,1170), (10,1171). The longest chain for
this relation is : 69 7→ 4 7→ 2 7→ 6 7→ 83 7→ 3108.
Proposition 10.4.8. Inversely, the circum-anticevian triangle of a point P is the dual of the
cevian triangle of a point CD(P), given for P = p : q : r by the inverse of the DC-mapping; that is:
−a2 qr
1
1
1
: 2
: 2
2
2
2
2
2
+ b pr + c pq a qr − b pr + c pq a qr + b pr − c2 pq
In other words, we have :
CD (P ) = isog (cevadiv(P, X(6)))
10.5
Saragossa points
Definition 10.5.1. Saragossa points. Let P be a point not on the circumcircle of ABC. Let
T 0 = A0 B 0 C 0 be the cevian triangle of P and T 00 = A00 , B 00 , C 00 the circumcevian triangle of P.
Consider triangle T that is the desmic mate of T 0 and T 00 , i.e. the triangle whose vertices are
U = B 0 C 00 ∩ B 00 C 0 , V = C 0 A00 ∩ C 00 A0 and W = A0 B 00 ∩ A00 B 0 . Then (Figure 10.1) triangles ABC,
T 0 , T 00 and T are pairwise perspective ((Grinberg, 2003c). The first, second and third Saragossa
points of P are the perspectors of T with, respectively, ABC, T 0 and T 00 . The name Saragossa
refers to the king who proved Ceva’s theorem before Ceva did (Hogendijk, 1995).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
86
10.6. Vertex associates
A
C"
Q1
U
Q3
Q2
C' V
B"
B'
W
P
C
A'
B
A"
Figure 10.1: Saragossa points of point P
Proposition 10.5.2. The barycentrics of the Saragossa points of P = p : q : r are (cyclically) :
g1 (a, b, c)
=
g2 (a, b, c)
=
g3 (a, b, c)
=
a2
a2 qr − (a2 qr + b2 pr + c2 pq)
1
1
p−
+ 2
a2 qr + b2 pr + c2 pq
2
c q rb
1
1
2p −
+ 2
a2 qr + b2 pr + c2 pq
2
c q rb
(10.3)
Points P, Q2 , Q3 are clearly collinear. When one of the Saragossa points is equal to P then P is
X(6) or lies on the circumcircle.
Proof. Computations are straightforward.
Example 10.5.3. The following table give the Saragossa points of the X (I) whose number is
given in the first line.
1
2
58
251
386 1180
1193 1194
10.6
3
4
5 6
19
4
54 1166 6 284
1181
?
? 6 1182
185 389
? 6 1195
21 24
25 28
31
961 847
2 943
81
1183
? 1184
? 1185
?
? 1196
? 1197
Vertex associates
Definition 10.6.1. Vertex associate. Consider the circumcevian triangles Ap Bp Cp , Au Bu Cu
of points P, U (not both on the circumcircle) and draw their may be degenerate vertex triangle T
i.e. the triangle whose sidelines are Ap Au , Bp Bu , Cp Cu . It happens that T is perspective to ABC
: the corresponding perspector X is called the vertex associate of P and U .
Proposition 10.6.2. When P ∈ Γ but U ∈
/ Γ, Ap Bp Cp and T are totally degenerate at P , so that
X = P (regardless of U ). Otherwise, the barycentrics of the vertex associate of p : q : r and u:v:w
are (cyclically) :
vertexthird (P, U ) =
wra4 qv
−
p (wb2
a2
+ vc2 ) u (rb2 + c2 q)
Proof. When both P, U are on Γ, both circumcevians are totally degenerate and X is not defined.
March 30, 2017 14:09
published under the GNU Free Documentation License
10. Circumcevian stuff
87
Remark 10.6.3. The definition of vertex conjugate allows X = U. To extend the geometric interpretation to the case that X = U, as X approaches U, the vertex triangle approaches a limiting
triangle which we call the tangential triangle of U, a triangle perspective to ABC with perspector
U-vertex conjugate of U.
Proposition 10.6.4. When P is not on Γ, but U is on PΓ∗ (the Γ-polar of P ), then T is totally
degenerate to a point X that is the Γ-pole of line P U . Finally, triangle P U X is autopolar wrt Γ.
Proof. Concurrence of Ap Au , Bp Bu , Cp Cu in a point X is straightforward, and X ∈ PΓ∗ too.
Proposition 10.6.5. Operation vertexthird is commutative and "formally involutory" i.e. :
vertexthird (P, vertexthird (P, U )) ' U
unless P lies on the circumcircle (where vertexthird (P, U ) = P , regardless of U ).
Proof. Commutativity is from the very definition, and the formal involutory property is from
straightforward computations. It remains only to track degeneracies. Determinant of the vertex
triangle T is the square of determinant of the corresponding trigon (T is either a genuine triangle,
or totally degenerate). The factors are the conditions for U ∈ Γ, P ∈ Γ and the condition for one
point to be on the Γ-polar of the other.
Exercise 10.6.6. In the general case, what can be said about the way the three circumcevian
triangles are lying on Γ ?
Example 10.6.7. Here are some vertex conjugates X(I), X(J), X(K) :
1
2
3
4
5
6
7
8
9
1
2
3
4
56
3415
84
3417
3415
25
3424
3425
84
3424
64
4
3417
3425
4
3
5
6
7
8
2163 3418
1383
3426 3427
3431
9
3420
3432
2163
3418
1383
3426
3427
3431
6
3433
3435
3420
1436
Proposition 10.6.8. For a given U, not on the circumcircle, the associated first Saragossa point
(10.3) is the sole and only point X such that vertexthird (U, X) = X.
Proof. We want U = V T (X, V T (X, U )) = V T (X, X) while sarag1 (V T (P, P )) = P is straightforward.
Proposition 10.6.9. Vertex association wrt X3 maps the Darboux cubic to the Darboux cubic
(X3 is the reflection center of this cubic, whose pole is X6 and pivot X20 ). The appearance of (I,J)
in the following list means that X(I),X(J) are on the Darboux cubic and that X(3),X(I),X(J) are
vertex associates :
1
84
3 4
64 4
20
3346
40 1490 1498 2131 3182
3345 3347 3348 3183 3354
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
88
March 30, 2017 14:09
10.6. Vertex associates
published under the GNU Free Documentation License
Chapter 11
About conics
Notation 11.0.10. C is a conic, C is the matrix of the punctual equation while C ∗ is the matrix
of the tangential equation. Point P is (often) the perspector, U is (often) the center.
11.1
Tangent to a curve
Definition 11.1.1. An algebraic curve C is the set of all the points x : y : z that satisfy a
polynomial equation C (x, y, z) = 0. In order to be a projective property, the polynomial C (x, y, z)
is required to be homogeneous (this is ever assumed in what follows).
Proposition 11.1.2. Consider an algebraic curve C (not necessarily a conic). The line tangent
to C at point P = p : q : r is given by :
−−→
grad (C)p:q:r =
∂C
∂x
,
X=P
∂C
∂y
,
X=P
∂C
∂z
X=P
(11.1)
Proof. Let P ∈ C be the contact point and P + kU be a point in the vicinity. If we require
−−→
P + kU ∈ C, we must have C (P + kU ) − C (P ) = O k 2 and this is grad (C)p:q:r . U = 0. But
−−→
polynomial C is homogeneous and we have grad (C)p:q:r . P = dg (C) C and the result follows.
Exercise 11.1.3. Use parametrization (7.15) to describe the points P of the circumcircle. Obtain
the tangent at P . Take the orthodir and obtain the normal. Differentiate and wedge to catch the
contact point of the envelope of all the normals... and obtain X(3).
Definition 11.1.4. Pole and polar. The polar line of point X with respect to an algebraic
(homogeneous) curve C is the line whose affix is the gradient of C evaluated at point X. Point X
is called a pole of its polar.
Remark 11.1.5. When point X is a simple point on an algebraic curve, its polar is nothing but the
line tangent at X to the curve. Finding all the points whose polar is a given line is not an easy
task in the general case.
Proof. well-known result. In fact, this is the rationale for the concept of polarity.
Remark 11.1.6. Triangle ABC can be seen as the curve xyz. Gradient evaluated at point P '
p : q : r is [qr, rp, pq] i.e. the already defined tripolar. And we can see why the tripolar transform
is not "as nice as" the usual circumpolar transform: the degree of the underlying curve is not 2:
there are no more "reciprocity".
Definition 11.1.7. To avoid misunderstanding, it is often useful to specify the curve used to
polarize. So we will use circumpolar to describe polarity wrt the circumcircle, and conipolar to
describe polarity wrt a given specified conic.
89
90
11.2
11.2. Folium of Descartes
Folium of Descartes
The curve know as "the folium of Descartes" is not a conic ! But, in our opinion, it could be useful
to see how some general methods are working in the general case, before to use them in the rather
specific situation of the algebraic curves of degree two.
Definition 11.2.1. The folium F is the curve used by Descartes to check his methods regarding the coordinate system. The Cartesian equation of this curve is x3 + y 3 − 6xy = 0, and its
homogeneous equation is
X3 + Y 3 − 6 X Y T = 0
Proposition 11.2.2. The folium presents a double point at 0 : 0 : 1. If we cut by the line Y = pX,
we obtain the parametrization : M ' 6p : 6p2 : 1 + p3 . A better parametrization is :
2
2
M ' 3 (1 + q) (1 − q) : 3 (1 + q) (1 − q) : 3 q 2 + 1
Then the tangent ∆M at M ∈ F is given by
N = x : y : z ∈ ∆M
when


x
2


3X − 6T, 3Y 2 − 6T, −6XY ·  y  = 0
z
Proof. Homography q = (p − 1) / (p + 1) has been used to move point p = −1 at q = ∞ in order
−−→ −−→
to have a "one piece" curve. Tangency condition is gradφ · M N = 0. But, due to the Darboux
−−→
property, we already have gradφ · M = 0.
Example 11.2.3. When q1 =
h 1/3 and q2 i= h−3, we obtainipoints M1 = 4 : −5 : 8 and M2 =
24 : −12 : 7. Tangents are 4 −5 8 , 17 20 24 while their common point is (see
Figure 11.1) given by :
h
4
−5
8
i
∧
h
17
20
24
i
= 56 : −8 : −33 ≈ −1.70 : +0.24 : 1
Visible asymptote is the tangent at the visible T = 0 point, i.e. at +1 : −1 : 0. Using the gradient
at that point, we see that asymptote is [3, 3, −6] ' [1, 1, −2].
(a) Two intersecting tangents
(b) Dividing the plane in three regions
Figure 11.1: Folium of Descartes
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
91
Proposition 11.2.4. Tangential equation of the folium (i.e. the condition for a line ∆ ' [u, v, w]
to be tangent to the curve is :
.
F ∗ (u, v, w) = 48 u2 v 2 − 32 w u3 + v 3 + 24 uvw2 − w4 = 0
Proof. Tangency requires a contact point, so that :
n
o
2
2
(q − 1) 3 q 3 + 9 q + 3 q 2 + 1 = Kv, (q + 1) 3 q 3 + 9 q − 3 q 2 − 1 = Ku, 6 (q + 1) (q − 1) = Kw
is required. Apart from w = 0 or q = ±1, last equation gives a K value, that can be substituted
into the other equations. Writing that the remaining two polynomials have the same roots, we
obtain the resolvant :
−w − 6 v −9 w − 6 v −3 w + 6 v −3 w + 6 v
0
0
0
−w − 6 v −9 w − 6 v −3 w + 6 v −3 w + 6 v
0
0
0
−w − 6 v −9 w − 6 v −3 w + 6 v −3 w + 6 v w + 6 u −9 w − 6 u 3 w − 6 u −3 w + 6 u
0
0
0
w + 6u
−9 w − 6 u 3 w − 6 u −3 w + 6 u
0
0
0
w + 6u
−9 w − 6 u 3 w − 6 u −3 w + 6 u Suppressing the non vanishing factors leads to the given result.
Example 11.2.5. Start from a point N ' x : y : z and search the u, v, w such that {ux + vy + wz = 0, Ψ = 0}.
We have three different possibilities, that are exemplified by :
N ' 1 : 1 : 1,
u
1.0
1.0
1.0
1.0
v
−0.51903 − 1.16372 i
−0.31967 + 0.71674 i
−0.51903 + 1.16372 i
−0.31967 − 0.71674 i
w
−0.48097 + 1.16372 i
−0.68033 − 0.71674 i
−0.48097 − 1.16372 i
−0.68033 + 0.71674 i
N ' 4 : 2 : 1,
u
1.0
1.0
1.0
1.0
v
1.0
−1.52334
−0.73833 − 1.09791 i
−0.73833 + 1.09791 i
w
−6.0
−0.95333
−2.52334 + 2.19582 i
−2.52334 − 2.19582 i
N =−
1
1
: − : 1,
2
2
u
1.0
1.0
1.0
1.0
v
1.35307
0.73906
−0.43584
−2.29442
w
1.17653
0.86953
0.28208
−0.64721
In other words, the curve and its asymptote are dividing the plane into three zones. From a
magenta point (see Figure 11.1b) no tangents can be drawn to the curve, but two from a green
point and four from a cyan point.
Proposition 11.2.6 (Plucker formulas). Let d, δ, κ be respectively the degree of a curve C, its
number of nodes (double points with two tangents), its number of cups (double point with a single
tangent) and d0 , δ 0 , κ0 be the corresponding numbers for the dual curve C 0 then we have the following
relations :
d0
0
κ
= d (d − 1) − 2δ − 3κ
=
3d (d − 2) − 6δ − 8κ
d = d0 (d0 − 1) − 2δ 0 − 3κ0
κ = 3d0 (d0 − 2) − 6δ 0 − 8κ0
1
. 1
(d − 1) (d − 2) − δ − κ = (d0 − 1) (d0 − 2) − δ 0 − κ0
g =
2
2
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
92
11.3. General facts about conics
Figure 11.2: Dual curve of the folium
Proof. Proof is not obvious since the formula turns weird when points with multiplicity greater
than two are occurring. There is no simpler formula giving δ than g = g 0 . A remark : we have
3d − κ = 3d0 − κ0 . Application : see Figure 11.2. Cups are occurring at x = y = 1/2 and at
x = j/2, y = j 2 /2 and conjugate.
11.3
General facts about conics
Definition 11.3.1. A conic C is a curve whose barycentric equation is an homogeneous polynomial
of second degree. This can be written using the usual matrix apparatus :

 
m11 m12 m13
x

 
t
X ∈ C ⇐⇒ X . C . X = (x, y, z)  m12 m22 m23   y  = 0
m13 m23 m33
z
Lemma 11.3.2. Let M be a n × n matrix and M ∗ the matrix of the cofactors (at the right
place, such that cofactors of a row form a column). Then :
M . M ∗ = M ∗ . M = det M 1n
Moreover rank M ∗ = n when rank M = n, rank M ∗ = 1 when rank M = n−1, and otherwise
M ∗ is the 0 matrix.
Proof. When det M 6= 0, both matrices can be inverted. In the second case a row of M ∗
describes the wedge of the hyperplane of the columns of M . In the last one, all minors are 0 since
rank is less than n − 1.
Definition 11.3.3. Proper conic. A quadratic form can be written as the sum of as many
squares of independent linear forms as its rank, leading to the following classification :
1. When rank is one, C is a straight line, whose points are each counted twice (strange object).
2. When rank is 2, C is the union of two intersecting lines. When these lines are complex
conjugate of each other (not real), only their intersection is real and this point appears as an
isolated point. When one of these lines is the line at infinity, C is considered as some kind of
extended circle (see Chapter 12).
3. When det C 6= 0, intersection of C and any straight line contains exactly two points (real or
not, may be a double point). A such conic is called a proper conic.
March 30, 2017 14:09
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11. About conics
93
Definition 11.3.4. Pole and polar. According to the general definitions, the polar of point X
wrt proper conic C is the line t X C , i.e. the locus of the points Y such that t X C Y = 0, while
the pole of line {Y | ∆ Y = 0} is the point X = adj C t ∆.
Remark 11.3.5. This polarity is not to be confused with polarity wrt the main triangle. Therefore,
t
it can be useful to describe line t X C as the conipolar of X and line isotom (X) as the tripolar
of X.
Remark 11.3.6. The relation "point U belongs to the polar of P " is symmetric. When point P
belongs to C, its polar wrt the conic is nothing but the line tangent at X to the conic.
Definition 11.3.7. The dual of a conic C1 is the conic C2 such that point x : y : z belongs to C2
when line (x, y, z) is tangent to C1 . When dealing with proper conics, we have C2 = adj C1
and conversely. When rank is 2, the dual is rank 1 : all tangents have to pass through the common
point.
Definition 11.3.8. The center U of a conic C is the pole of the line at infinity L∞ with respect
to the conic. Its barycentrics are :
−m223 + (m13 + m12 ) m23 + m22 m33 − m13 m22 − m12 m33
−m213 + (m12 + m23 ) m13 + m33 m11 − m12 m33 − m23 m11
−m212 + (m23 + m13 ) m12 + m11 m22 − m23 m11 − m13 m22
Definition 11.3.9. A parabola is a conic whose center is at infinity (more about parabolas in
Section 11.15). Two parallel lines make a non proper parabola. The union of the line at infinity
and another line is ... some kind of circle rather than a "special special" parabola.
Fact 11.3.10. When a conic goes through its center and this center is not at infinity, the conic
is the union of two different straight lines. When a conic is a single line whose points are counted
twice, det C vanishes and center has no meaning.
Proposition 11.3.11. When C is not a parabola, its center is the symmetry center of the conic.
Computed Proof. Substitute (4.1) into the equation and obtain C (x, y, z) times the square of the
condition to be a parabola.
Proposition 11.3.12. The polar line of a point P wrt a conic C is the locus of the points Y 0 Z ∩Y Z 0
where Y Y 0 and ZZ 0 are chords of C that meet in P .
Proof. Parametrize the line at infinity. Cut the conic by line from P to U (µ), obtain two points
depending of a radical W (µ). Repeat for another value τ of the parameter. Then check that
determinant of lines Y 0 Z, ZY 0 and t P C is zero.
Remark 11.3.13. Since Z 0 Z is also a chord, the required conipolar is the line Y 0 Z ∩ Y Z 0 ; Y Z ∩ Y 0 Z 0
Lemma 11.3.14. Let A, B be some columns and M an invertible matrix (dimension 3 is assumed).
Then
t
t
A . M ∧ t B . M = M ∗ . (A ∧ B)
Proof. We have :
X.
t
A.M ∧
t
B .M
det X . M −1 . M, t A . M, t B . M
= det M × det X . M −1 , t A, t B
t
= X . det M × M −1 . (A ∧ B)
=
Proposition 11.3.15. Perspector of a conic. Let triangle A0 B 0 C 0 be described by the columns
of matrix T . Then trigon A0 B 0 C 0 is described by the lines of T ∗ and the triangle of the conipolars
t
of the sidelines are described by the columns of C ∗ . T ∗ (remember, a∗ denotes the adjoint
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
94
11.4. Circumconics
matrix). Consider specially the case of triangle ABC itself. Either triangle ABC is "autopolar"
or both triangles are perspective, with perspector P given by :
P =
1
1
1
:
:
m11 m23 − m13 m12 m22 m13 − m23 m12 m12 m33 − m13 m23
Proof. First part is from previous lemma, and the simplicity of the result from the special choice
of the triangle.
B'p
A
C'p
Cp
B
Bp
P
Ap
C
A'p
Figure 11.3: An inconic and a circumconic with the same perspector
Proposition 11.3.16. Let P be a point not on the sidelines of ABC. Six points are obtained by
intersecting a sideline with a parallel through P to another sideline. These points are on the same
conic C. Equation, perspector T and center U are :
X
X
C =
(q + r)qrx2 −
(p2 + pq + pr + 2qr)pyz
p
q
r
T =
:
:
2 pr + 2 pq + qr 2 pq + pr + 2 qr pq + 2 pr + 2 qr
U = p 2 qr + pr + pq − p2 : q 2 pr + pq + qr − q 2 : r 2 pq + pr + qr − r2
Center U is at infinity (and C is a parabola) when P is at infinity or on the Steiner inconic. Points
P and Q = p (q + r − p) : q (r + p − q) , r (p + q − r) are leading to the same center U . Point Q is
at infinity when P is on the Steiner inconic.
Proof. Equation in Q is of third degree. The discriminant factors into minus a product of squares.
Other computations are straightforward. Examples are [P,U], [115, 523], [1015, 513], [1084, 512],
[1086, 514], [1146, 522], [2482, 524], [3163, 30] where P is on the Steiner inconic and [P,U], [2, 2],
[6, 182], [3, 182], [9, 1001], [1, 1001], [190, 1016], [664, 1275] for other points.
11.4
Circumconics
Definition 11.4.1. A circumconic is a conic that contains the vertices A, B, C of the reference
triangle. Its equation can be written as :


0 r q


CC (P ) = t X Cc X = 0
where Cc =  r 0 p 
(11.2)
q p 0
Construction 11.4.2. Graphical tools can construct any conic from five points. Given the perspector P , the other points on the cevian lines are −p : 2q : 2r, etc i.e points 2qB + 2rC − p A.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
95
Proposition 11.4.3. When p : q : r is the perspector, a parametrization of CC(P) is
M (t) '
p q
−r
: :
1 t 1+t
(11.3)
Proof. Direct inspection.
Theorem 11.4.4 (circumconics). We have the following four properties :
(i) Point P is the perspector of the conic, and the polar triangle of ABC wrt CC (P ) is the
anticevian triangle of P wrt ABC (in other words, CC (P ) is tangent at A to PB PC etc.
(ii) When U is the center of CC (P ) then P is the center of CC (U ). Both are related by :
U
= cevadiv(X2 , P )
=
P ∗ anticomplem (P )
b
(iii) Circumconic CC (P ) is the P isoconjugate of L∞ . Inter alia,
X ∈ circumcircle ⇐⇒ isog (X) ∈ L∞
(iv) The polar line of U wrt CC (P ) is '[qw + rv, ru + pw, pv + qu] ... aka the polarmul of (P, U ).
Proposition 11.4.5. Points at infinity. A circumscribed conic is an ellipse, a parabola or an
hyperbola when its perspector is inside, on or outside the Steiner in-ellipse. Moreover, its points
at infinity, expressed from the perspector Q = p : q : r have the following barycentrics :


−2 p


where W 2 = p2 + q 2 + r2 − 2 pq − 2 qr − 2 rp
M∞ '  p + q − r − W 
p+r−q+W
When the point at infinity of a circumparabola is u : v : w, its perspector is P = u2 : v 2 : w2 .
Proof. Immediate computation. Mind the fact that W 2 = −3 when P is at X(2). For the second
2
part, start from T = u : v : −u − v, and compute the circumconic relative to u2 : v 2 : (u + v) .
Remark 11.4.6. Properties of the CircumRH, aka the circumscribed rectangular hyperbola, are
collected at Subsection 11.16.2
Proposition 11.4.7. Two (non equal) circumconics CC (P ) and CC (U ) have exactly four points
in common, namely the three vertices and the tripolar of line P U . Formally, this point is isotom (P ∧ U ).
Proof. Conics that share five distinct points are equal. The value of X follows by direct inspection.
Proposition 11.4.8. The perspector P of the circumconic through additional points U1 , U2 is the
intersection of the tripolars of U1 and U2 (caveat: this is not the tripole of line U1 U2 ). In other
words, :
P = U1 ∗ U2 ∗ (U1 ∧ U2 )
b
b
Proof. Direct inspection.
Remark 11.4.9. Collineations can be used to transform any circumconic into the circumcircle or
the Steiner out-ellipse, so that many proofs can be done assuming such special cases. More details
in Proposition 15.4.3.
Proposition 11.4.10. Let be given ∆ and K where ∆ is a line (tripole G) that cuts the sidelines
BC, CA, AB in A0 , B 0 , C 0 and K is a circumconic (perspector P) . Let M = x : y : z be a point
on ∆, and A” the other intersection of M A with K and cyclically B” ∈ M B ∩ K, C” ∈ M C ∩ K.
Then lines A0 A”, B 0 B”, C 0 C” are concurrent on a point Q ∈ K. Moreover, the conjugacy that
exchanges P and G exchanges also M and Q (M ).
Proof. Use ∆ ' [ρ, σ, τ ], P = p : q : r, M = 1/ρ : t/σ : − (1 + t) /τ (where t is a parameter) and
obtain
p
q
r
Q=
:
:
ρx σy τ z
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
96
11.5. Tangential conics
Figure 11.4: Conjugacy between a line and a circumconic
11.5
Tangential conics
Definition 11.5.1. A point-conic is a ’conic as usual’, i.e. a locus of points, whose equation is
M · C · M = 0. In the general case, a line is tangent to C when ∆ · C ∗ · t ∆ = 0. A line-conic or
t
tangential conic is what is obtained when seeing C in equation ∆ · C · t ∆ = 0 as the primitive
object, and seeing the punctual conic t M · C∗ · M = 0 as a derivative object.
Proposition 11.5.2. Degenerate line-conic. When det C = 0, the equation splits in two first
degree factors, and a line belongs to the conic when it goes through either of the fixed points (the
centers) defined by each factor (supposed distinct).
Example 11.5.3. The degenerate line-conic of the isotropic lines (see Proposition 7.4.8).
11.6
Inconics
Definition 11.6.1. An inconic is a conic that is tangent to the three sides of the reference triangle.
Theorem 11.6.2 (inconics). The punctual and tangential equations of an inconic Ci
as :



f 2 −fg −fh
0 h



t
∗ t
∗
2
X . Ci . X, ∆ . Ci . ∆
where Ci '  −fg g
−hg  , Ci '  h 0
−fh −hg h2
g f
can be written

g

f 
0
(11.4)
.
where Q = [f, g, h] is the so-called auxiliary line of the conic. Let us note P = tripolar (Q) ' p :
q : r. The dual conic of Ci is precisely CC (Q) (that acts over lines). The contact points of Ci
are 0 : h : g ' 0 : q : r, etc (mind the order !). They are the cevians of point P . This point P
is the perspector between triangle ABC and its polar triangle with respect to the conic. Formally,
the center of Ci is the complement of the auxiliary line. In fact, the center is the complement of
isotom (P ).
Proof. By definition, Ci∗ must have a zero diagonal. Then Ci is its adjoint. Perspectivity is
obvious, while center is the pole of the line at infinity.
Corollary 11.6.3. Direct relations between center and perspector are as follows :
U
=
complem (isot (P ))
P
=
isot (anticomplem (U ))
= crossmul(X2 , P )
=
= P ∗ complem (P )
b
crossdiv (U, X2 )
Construction 11.6.4. Graphical tools can construct any conic from five points. Given the perspector P ' p : q : r, the points on the cevian lines are A1 = qB + rC, A2 = qB + rC + 4pA.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
97
Proposition 11.6.5. When p : q : r is the perspector, a parametrization of IC(P) is
2
M (t) ' p : q t2 : r (1 + t)
(11.5)
Proof. Direct inspection.
Proposition 11.6.6. The ’complem formula’ is as follows:
Mcc (t)
Mic (t)
P
1
+2
=3
=P
tripolar (P ) · Mcc (t)
tripolar (P ) · Mic (t)
tripolar (P ) · P
where Mcc (t) is the standard parametrization of CC(P) and Mic (t) the standard parametrization
of IC(P). This amounts to use tripolar (P ) as ’line at infinity’ to normalize the columns in the
assertion Mcc + 2Mic ' P .
Proof. From (11.3), (11.5) and the obvious :

 




t (1 + t) p
p
p/3
1
 1/2
1 






1× 2
qt2
 = 3 ×   q/3 
 (1 + t) q  + 2 ×  2

t +t+1
t +t+1
3
2
−tr
r (1 + t)
r/3


Construction 11.6.7. An inscribed conic can be generated as follows. Given the perspector P '
p : q : r, draw the cevians, obtain AP BP CP and draw the cevian triangle. Draw an arbitrary line
∆ through B. Define Ba = (BP AP ) ∩ ∆ and Bc = (BP CP ) ∩ ∆. Then M = Ba CP ∩ Bc AP is on
the conic. Thereafter, you can define ∆ by a point X on the circumcircle, and generate Ci as the
locus of M (parametrization by a turn).
Proof. Use X = x : y : z so that ∆ ' [−z, 0, x]. Thereafter :






prx
prx
pr2 x2






Ba '  pqz − qrx  , Bc '  qrx − pqz  , M '  q (pz − rx)2 
zpr
zpr
z 2 p2 r
One can check that M is on the conic and that t =
pz
− 1.
rx
Figure 11.5: How to generate an inscribed conic from its perspector.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
98
11.7. Conic cross-ratios
Proposition 11.6.8. Points at infinity. An inscribed conic is an ellipse, a parabola or an
hyperbola when its perspector is inside, on or outside the Steiner circum-ellipse. Moreover, its
points at infinity, expressed from the auxiliary point Q = p : q : r have the following barycentrics :
M∞

2
(r + q)


'  pq − r (p + q + r) + 2 W 
pr − q (p + q + r) − 2 W

where W 2 = −pqr (p + q + r)
Proof. Immediate computation. Steiner property comes from W 2 = −p2 q 2 r2
1 1
q r
+
1 1
r p
+
1 1
p q
.
Proposition 11.6.9. An inscribed conic is a rectangular hyperbola if, and only if, its auxiliary
point is on the Longchamps circle. Equivalently, its center is on the conjugate circle. See Section 12.7 and Section 12.8.
Proof. Write that trace M C = 0, see Proposition 11.16.4.
Proposition 11.6.10. When center U is given, the perspectors Pi , Pc of the corresponding inand circum-conics are related by :
Pi ∗ Pc = U
;
b
Pi = anticomplem (Pc )
When perspector P is given, the centers Ui , Uc are aligned with P = p : q : r together with
p2 : q 2 : r 2 .
Proof. Direct inspection.
Proposition 11.6.11. Let P be a fixed point, not on the sidelines, and U be a point moving point
on tripolar (P ). The envelope of all the lines ∆ = tripolar (U ) is the inconic IC (P ). Moreover,
the contact point of ∆ is T = U ∗ U ÷ P .
b
b
Proof. Yet given result. But now, this is the right place to prove it. Write :
P
= p:q:r
U
= p (σ − τ ) : q (τ − ρ) : r (ρ − σ)
' (1 ÷ p(σ − τ ) : 1 ÷ q(τ − ρ) : 1 ÷ r(ρ − σ))
∆
2
T
and check that : ∆ . adj
(11.4)
11.7
2
= p (σ − τ ) : q (τ − ρ) : r (ρ − σ)
Ci
2
. t ∆ = 0, ∆ . T = 0 and t T . Ci . T = 0 where Ci is given in
Conic cross-ratios
Proposition 11.7.1. All the lines λ through a given point P form a linear projective family F .
Consider a transversal line D (i.e. a line that doesn’t go through P ). Then cross ratio remains
unchanged by application F ,→ D, λ 7→ λ ∩ D.
Proof. Obvious since the wedge operator is a linear transform λ 7→ λ ∧ D: the parametrization is
preserved.
Proposition 11.7.2. Consider four fixed points A, B, C, U with no alignments and define the
moving cross-ratio of a point M in the plane as the cross-ratio of lines M A, M B, M C, M U . The
level lines of this function are the conics passing through A, B, C, U . Using barycentrics with respect
to ABC, we have the more precise statement: the level line of a given µ is the circumscribed conic
whose perspector is P ' −u : vµ : w (1 − µ), on the tripolar of U = u : v : w.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
99
Definition 11.7.3. Conic-cross-ratio. Consider a fixed, non degenerated conic C, and four
points A, B, C, U lying on this conic. Then the cross-ratio of lines M A, M B, M C, M U does not
depends upon the choice of the auxiliary point M as long as this point M remains on the conic. If
we consider C as a circumconic with perspector P = p : q : r wrt triangle ABC, we have :


µ2 − µ p


cross-ratio (A, B, C, U ) = µ when U '  (1 − µ) q 
µr
Remark 11.7.4. In the complex plane, the quantity defined in Theorem 4.2.7 is nothing but the
usual cross-ratio, as computed from complex affixes. In order to be sure that, given four points
on a circle, the conic cross-ratio and the usual cross-ratio are the same quantity, let us consider
the stereographic projection. Start from P (c, s). Define M (C, S) by doubling the rotation and
T (0, t) by intersection of SM with the y-axis. Apply Thales to similar triangles SOT and SKM ,
and Pythagoras to OHP . This gives :
t
S
c2 + s2 = 1, (C, S) = c2 − s2 , 2cs , =
1
1+C
and leads to t = s/c, proving that (SO, ST ) = (ON, OP ) and therefore that
cos 2ϑ =
1 − t2
2t
, sin 2ϑ =
1 + t2
1 + t2
Moreover, circular cross-ratio between Mj points is equal to linear cross-ratio between Tj points
(this is the definition), while complex cross-ratio between Mj points is equal to complex cross-ratio
between Tj points due to the homography :
z = (1 + i t)/(1 − i t) ; t = i (1 − z)/(1 + z)
Figure 11.6: Stereographic projection
11.8
Some in- and circum- conics
A list of specific in- and circum- conics is given in Table 11.1. Kiepert RH is studied at Brocard
Section.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
100
11.8. Some in- and circum- conics
P
center
IC
X1
X2
X3
X4
X5
X6
X7
X8
X99
X190
X264
∞X523
X598
X647
X650
inSteiner 11.13
orthic
Brocard 11.14
incircle 12.5
Mandart
Kiepert parabola 14.5.3
Yff parabola
inMacBeath 11.8.2
inLemoine11.2
X37
X2
X216
X6
X233
X39
X1
X9
IX523
IX514
X5
X115
X597
???
???
CC
circumSteiner
outMacBeath 11.8.3
circumcircle 12.4
Kiepert RH 12.18.1
Jerabek RH
Feuerbach RH
center
X9
X2
X6
X1249
X216
X3
X3160
X3161
???
???
???
X115
???
X125
X11
RH is rectangular hyperbola
The Feuerbach RH is the isogonal conjugate of the line (circumcenter,incenter).
The Jerabek RH is the isogonal conjugate of the Euler line.
Table 11.1: Some Inconics and Circumconics
Example 11.8.1. The Steiner ellipses (centers=perspector=X2 ) are what happen to both the
circum- and in-circle of an equilateral triangle when this triangle is transformed into an ordinary
triangle by an affinity. The Steiner circumellipse (S) is the isotomic conjugate of L∞ and the
isogonal conjugate of the Lemoine axis. Since isog (L∞ ) is the circumcircle (C), the mapping
X 7→ X ∗ X6 sends (S) onto (C). Steiner in-ellipse is the envelope of the line whose tripole is at
b
infinity (more about Steiner in-ellipse in Section 11.13).
Example 11.8.2. The MacBeath-inconic was introduced as follows:
Lemma 1: Let O, H be the common points of a coaxal system of circles. Let a
variable circle of the system cut the line of centers at C. Let T be a point on the
circumference such that T C = k ∗ OC, where k is a fixed ratio. Then the locus of T is
a conic with foci at O, H (Macbeath, 1949)
Its perspector is X(264), center X(5) and foci X(3) and X(4) . Its barycentric equation is :
X
X b2 c2 yz
a4 −a2 + b2 + c2 x2
−2
2
2
2
2
2
2
(c + a − b ) (a + b − c )
b2 + c2 − a2
and this conic goes through X(I) for I =
339, 1312, 1313, 2967, 2968, 2969, 2970, 2971, 2972, 2973, 2974
Example 11.8.3. The
MacBeath-circumconic, is the dual to the MacBeath-inconic. Its perspector is X(3) and its
center X(6). Its barycentric equation is :
a2 (b2 + c2 − a2 )yz + b2 (c2 + a2 − b2 )zx + c2 (a2 + b2 − c2 )xy = 0
and it goes through X(I) for I =
110, 287, 648, 651, 677, 895, 1331, 1332, 1797, 1813, 1814, 1815
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
11.9
101
Cevian conics
Proposition 11.9.1. For any points P = p : q : r and U = u : v : w, not on a sideline of ABC,
the cevians of P and U are on a same conic, whose equation in x : y : z is conicev (P, U ) given
by :
1 2
1 2
1
1
1
1
1
1
1 2
x +
y +
z −
+
yx −
+
zx −
+
zy = 0
up
qv
rw
qu vp
ru wp
rv qw
Proof. Apply (x, y, z) → x2 , xy, y 2 , yz, z 2 , zx to the six points and check that rank isn’t 6.
Example 11.9.2. Any inconic is a cevian conic : IC (P ) = conicev (P, P ). For example, the
incircle is IC (X7 ) = conicev (X7 , X7 ).
A non trivial example is the nine-points circle, aka conicev (X2 , X4 ).
Proposition 11.9.3. Assume that a conic C encounters the sidelines of ABC in six (real) points,
none of them being a vertex A, B, C. Three of these points are the cevians of some point P if and
only if :
m33 m22 m11 − m11 m223 − m22 m213 − m33 m212 − 2 m13 m23 m12
= det M − 4 m13 m23 m12 = 0
In this case, the remaining three intersections are the cevians of some point U
given by :
 p
+m13 m12 + m11 m23 + m13 m212 − m22 m11 /m11
 p
2 −m m
P, U ' 
m
−m
m
−
m
m
+
m
/m22
22
11
22
13
23
12
23

12
p
2
− m12 − m22 m11
and both P, U are




Proof. Hypotheses are implying m11 6= 0 (A ∈
/ C) and m212 − m22 m11 ≥ 0 (existence of intersections).
Proposition 11.9.4. The fourth common point F between cevian conics conicev (P, U1 ) and
conicev (P, U2 ) can be obtained from the tripolars of U1 , U2 . We have :
X
'
F
'
tripolar (U1 ) ∩ tripolar (U2 ) = (U1 ∧ U2 ) ∗ U1 ∗ U2
b
b
anticomplem X ÷ P ∗ X
b
b
Proof. Direct inspection.
11.10
Direction of axes
Proposition 11.10.1. Let C be a conic, but not a circle, and γ an auxiliary circle. Consider, in
any order, the common
X1 , X
2 , X3 , X4 of C and γ. Then axes of C have the same directions
z points
}|
{
as bisectors of angle X1 X2 , X3 X4 .
2
Proof. Use rectangular Cartesian coordinates. Then C is y 2 = 2px + qx2 and γ is (x − a) +
2
(y − b) = r2 . Substitution y 2 = Y gives :
2 by = (1 + q) x2 + (2 p − 2 a) x + b2 + a2 − r2
;
Y = 2px + qx2
By substitution and reorganization :
y2 − y1
y4 − y3
2 p − 2 a (1 + q)
+
=
+
(x1 + x2 + x3 + x4 )
x2 − x1
x4 − x3
b
2b
P
2
Value of xi is obtained from fourth degree equation 0 = Y −y 2 = (1 + q) x4 −4 (1 + q) (a − p) x3 · · · .
This proves that X1 X2 and X3 X4 are symmetrical wrt the axes and conclusion follows. By the
way, it has been proven that points Xi can be sorted in any order without changing the result.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
102
11.11. Focuses of a conic
Definition 11.10.2. Gudulic point. The gudulic point of a circumconic is defined as its fourth
intersection with the circumcircle. Its barycentrics are rational wrt the barycentrics of the perspector.
Gu = isotom b2 r − c2 q : c2 p − a2 r : a2 q − b2 p
Definition 11.10.3. Gudulic point (general method). The former proposition shows that any
pair of orthogonal directions
be specified by giving a point M on the circumcircle Γ of ABC
z can
}| {
such that the bisectors of BC, AM have the required directions. This method was firstly used
by Lemoine (1900) who called it "the point M method". In order to have a more specific name,
the expression "gudulic point" was coined in a discussion at www.les-mathematiques.net. May
be in honor to St Gudula of Brussels.
Proposition 11.10.4. When
but not the circumcircle itself, directions of axes
zC is}|a circumconic,
{
are given by the bisectors of BC, AM where M is the fourth common point of C and Γ. When
C is not a circumconic, it exists nevertheless
z }| {an unique point M ∈ Γ so that axes of C have the
same directions as the bisectors of BC, AM . We have the formulas :

l

C= h
g
h
m
f




1/ (l + m − 2 h) b2 − (n + l − 2 g) c2
g



7  1/ (m + n − 2 f ) c2 − (l + m − 2 h) a2 
f →
1/ (n + l − 2 g) a2 − (m + n − 2 f ) b2
n

1/ rb2 − qc2


CC (P ) 7→  1/ pc2 − ra2 
1/ qa2 − pb2
;


1/ w2 b2 − v 2 c2


IC (U ) 7→  1/ u2 c2 − w2 a2 
1/ v 2 a2 − u2 b2
Proof. First part is preceding proposition. For the second part, we have :
z }| {
z }| {
tan BC, Bδ + tan AM, Bδ
= 0
z }| {
z }| {
tan B∆1 , Bδ + tan B∆2 , Bδ
= 0
where ∆j are the points at infinity of the conic, δ is an unknown direction and M is the unknown
point. The second equation fixes δ, and this is sufficient to fix the direction of AM , i.e. the y, z
barycentrics of M . The x barycentric is obtained from M ∈ Γ.
Some technical details. We note δ = [u, v, w] and M = p : q : r. Points ∆j depend on a
square-root :
2
W 2 = (n + h − g − f ) − (n + l − 2 g) (m + n − 2 f )
It happens the odd powers of W are canceling in the second equation. Then we eliminate q and
obtain :
p w2 b2 − v 2 c2
q= 2 2
w c − w2 a2 + 2 wvc2 + v 2 c2
We extract v 2 as a function of v and the other variables and substitute into the q- expression. This
left
first degree factors in v, and they cancel, leading to the result. Another path could be
z two
}| { . z }| { z }| {
BC, AM = BC, ∆1 + BC, ∆2 , avoiding the unknown δ, but the Maple-length of the
starting expression would be 544846 !
11.11
Focuses of a conic
Definition 11.11.1. A point F is a focus for a curve when the isotropic lines from this point are
tangent to the curve.
Proposition 11.11.2. The geometric focuses of a conic are examples of the preceding definition.
But a conic has, in the general case, four analytical focuses : the two geometrical ones, and two
extra, imaginary, focuses that stay on the other axis.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
103
Proof. Let us consider ellipse x2 /a2 + y 2 /b2 = 1 and isotropic line (x − x0 ) + i (y − y0 ) = 0. Their
2
intersection is given by a second degree equation whose discriminant is (x0 + i y0 ) − a2 − b2 = 0.
It vanishes when y0 = 0, x0 = ±f (as usual) but also when x0 = 0, y0 = ±i f .
Proposition 11.11.3. (spoiler) When using the Morley space to compute the focuses, one equation
applies to map Z : T and the other one applies to map Z : T. When going back to the Morley
space, there are four ways of pairing the two focuses of the first map with the two focuses of the
second map.
√
√
Proof. When, for a visible conic, √
the first map√
says z = √
u±v w, the second map says ζ = u±v w
(with w ∈ R). But conjugate of w is either w or − w, depending of the sign of real w.
Proposition 11.11.4. Let be given two points F1 ' f1 : g1 h1 and F2 ' f2 : g2 h2 . All the conics
that admit F1 , F2 as foci form a tangential pencil. This pencil is generated by (1) the tangential
conic {F1 , F2 } of all the lines ∆ ' [u, v, w] through F1 or F2 , whose equation is :
(f1 u + g1 v + h1 w) (f2 u + g2 v + h2 w)
and (2) the tangential conic {Ω+ , Ω− } of all the isotropic lines (through one or another umbilic),
whose equation is : ∆ · M · t ∆.
Remark 11.11.5. Focus of inscribed parabola and circumscribed parabola are considered at Section 11.15.
11.12
Focuses of an inconic
Proposition 11.12.1. Let F = p : q : r be a point not lying on the sidelines. As in Figure 11.7,
we note Fa , Fb , Fc and Fa0 , Fb0 , Fc0 the projections and the reflections of F about the sidelines ; G
the isogonal conjugate of F , ω = (F + G) /2 and Pa = F G0a ∩ Fa0 G, etc. Then points Pa , Pb , Pc
.
are the cevians of P = (isotom ◦ anticomplem) (ω), and are the contact points of C = inconic (P ).
0 0 0
This conic admits ω as center and F, G as geometrical focuses. Moreover circle Fa Fb Fc , centered
at G is the circular directrix of this conic wrt focus F while circle Fa Fb Fc is the principal circle
(tangent at major axis).
When either F or G is at infinity, the other is on the circumcircle, P is on the Steiner circumconic, and C is a parabola.
Proof. One obtains easily :
G0a = −
a4 Sc a2
a2 b2 Sb a2
a2 c2
:
+
:
+
p
p
q
p
r
leading to the symmetrical expression :
2
|Fa0 G|
=
c2 q 2 + 2Sa qr + b2 r2
a2 r2 + 2Sb rp + c2 p2
2
b2 p2 + 2Sc pq + a2 q 2
2
(r + q + p) (a2 qr + b2 pr + c2 pq)
.
proving that γ = Fa0 Fb0 Fc0 is centered at G. When P is inside ABC, we can define an ellipse C
by focus F and circular directrix γ. We have Pa ∈ C since |F Pa | = |Fa0 Pa |. Moreover BC is
the bisector of (Pa F, Pa Fa0 ), again by symmetry. Therefore C is the inscribed conic tangent at
Pa , Pb , Pc and we have :


rq

 q 2 c2 + 2 Sa rq + r2 b2

rb2 + qc2 p2 + (2 p + q + r) qra2

rp


ω '  pc2 + ra2 q 2 + (p + 2 q + r) rpb2  ; P ' 
2
2
2 2

 c p + 2 Sb rp + a r
qa2 + pb2 r2 + (p + q + 2 r) pqc2

qp

b2 p2 + 2 Sc qp + q 2 a2







When P is outside ABC, the simplest method is to revert the process and define P , and therefore
C, using the given formula and, thereafter, check that lines F Ω± are tangents to C.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
104
11.12. Focuses of an inconic
Figure 11.7: Focus of an inconic
Remark 11.12.2. When substituting F = p : q : r by an umbilic Ω± , the ω formula gives 0 : 0 : 0.
Therefore, umbilics are expected to appear as artifacts when trying to revert the ω formula to
obtain the foci.
Remark 11.12.3. The Moebius-Steiner-Cremona transform (Section 9.2) provides another point of
view... but computations aren’t easier (nor worse).
Proposition 11.12.4. The focus of an inconic can be obtained from the perspector by two successive second degree equations (ruler and compass construction). Let P = p : q : r be the perspector.
Then the four focuses can be written as :


√
p (r + q) + K
√


Fi ' 
q (r + p) + t K

√
r (q + p) − (1 + t) K
Then t is homographic in K, while K is solution of a second degree equation. The converse is true :
K is homographic in t, while t is solution of a second degree equation (with same discriminant).
Obviously, the two solutions in t lead to orthogonal directions (orthopoints at infinity)
Proof. Straightforward elimination.
Example 11.12.5. Table 11.2 gives some examples of perspectors and focuses. In this table,
expressions like X5 ± X30 are not "up to a proportionality factor", but are addressing the usual
simplified values. All expressions are "centered" at the center, the ± term being at infinity.
Imaginary focuses associated with perspector X(80) have a very simple expression, namely a : bj :
cj 2 and a : bj 2 : cj where j is the third of a full turn.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
105
P
7
ω
1
F
1
G
1
FG
X1
598
597
2
6
X597 ± X524
264
5
4
3
X5 ± X30
6
39
ω+
ω−
X39 ± X512
X44 ±
HK
X1
i
X597 ±
X1499
4S
X5 ± i4S X523
i
X39 ±
X511
4S
√
nom
inscrit
Lemoine
M acBeath
Brocard
√
X44 ± i 3 X513
3
X517
4S
80
44
673
3008
X3008 ± W X514
X3008 ±
i
W X516
4S
694
3229
X3229 ± W X512
X3229 ±
i
W X511
4S
Table 11.2: Perspector and focuses of some in-conics
W673
W694
p
a2 + c2 + b2 − 2 bc − 2 ac − 2 ba
s
1 1 1
1 1 1
1 1 1
1 1 1
2 2 2
= a b c
+ +
+ −
− +
− + +
a b
c
a b
c
a b
c
a b
c
=
11.13
Focuses of the Steiner inconic
11.13.1
Using barycentrics
Definition 11.13.1. The Steiner in-ellipse is what happen to the incircle of an equilateral triangle when this triangle is transformed into an ordinary triangle by an affinity. One has center
=perspector =X2 . Moreover, the Steiner inconic is the envelope of the lines whose tripoles are at
infinity. Equation of this conic is :

+1 −1

x2 + y 2 + z 2 − 2xy − 2yz − 2zx = (x, y, z)  −1 +1
−1 −1


x
−1


−1   y  = 0
+1
z
Notation 11.13.2. In this section, the following radicals will be used :
p
a4 + b4 + c4 − a2 b2 − b2 c2 − c2 a2
q
q
q
2
2
2
2
2
2
2
2
2
Wa = (W + a ) − b c ; Wb = (W + b ) − c a ; Wc = (W + c2 ) − a2 b2
W =
Proposition 11.13.3. All these radicals ares real. Moreover, assuming c > a, c > b, we have :
c −b
Proof. Let Wa =
q
2
2
Wc = Wa + Wb
Wb = W Wa + b2 − a2 Wa
c2 − b2 Wc = W Wa + c2 − a2 Wa
2
(−W + a2 ) − b2 c2 . Then Wa Wa
2
= −16S 2 b2 − c2
2
< 0, while Wa2 , Wa2
are real and Wa2 > Wa2 . Therefore each Wi is real (and positive). The expression
(Wa + Wb + Wc ) (−Wa + Wb + Wc ) (Wa − Wb + Wc ) (Wa + Wb − Wc )
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
106
11.13. Focuses of the Steiner inconic
depends only on the Wi2 and is therefore rational in W . Using the value of W 2 , this expression
simplifies to 0 : one of the Wi is the sum of the other two. If c is the greatest side, this leads to
Wc = Wa + Wb . In the same manner, the product :
c2 − b2 Wc + c2 − a2 + W Wa
c2 − b2 Wc − c2 − a2 + W Wa
simplifies to 0. If c is the greatest side, the first factor cannot vanish, leading to the Wc formula.
The Wb formula results by subtraction.
Proposition 11.13.4. The foci of the Steiner inconic are given by F± = X3413 ± Q X2 where :

 

b2 − c2 a4 − b2 c2 − a2 W
1


 
X2 =  1  ; X3413 =  c2 − a2 b4 − a2 c2 − b2 W 
1
a2 − b2 c4 − a2 b2 − c2 W
p
Q = 2a2 b2 c2 W 3 − 16 (a4 b4 + b4 c4 + c4 a4 ) S 2 + a2 b2 c2 (a2 b2 + b2 c2 + c2 a2 ) (a2 + b2 + c2 )
Proof. Isotropic lines through a focus are tangent to the curve. Write that F Ω+ is tangent to C
and separate real and imaginary parts. Eliminate one of the coordinates of F from this system. It
remains a fourth degree equation (E) giving the two real and two imaginary foci. The discriminant
of this equation contains W 4 in factor. Using this indication, we factorize (E) over R (W ) and
obtain :
c2 v 2 − 2 wa2 + 2 wW v + b2 w2 c2 v 2 − 2 wa2 − 2 wW v + b2 w2 = 0
2
The discriminants of these second degree factors are Wa2 and Wa . And we obtain the non
symmetric expression :


W + b2 b2 − c2 + W + b2 − a2 Wa


F+ ' 
b2 − c2 Wa + W + a2

2
2
2
b −c c
In order to obtain a more symmetrical expression, one can compute U = L∞ ∧ (F+ ∧ F− ), i.e. the
point at infinity of the focal line. This point happens to be X3413 , the first Kiepert infinity point.
The existence of Q is obvious since X2 is the middle of the foci. A straightforward computation
leads to the given formula.
Figure 11.8: The Steiner in ellipse
Figure 11.8 summarizes these properties. The hyperbola is Kiepert RH, the Tarry point X(98)
is the gudulic point of the KRH axes, while it’s circumcircle antipode, the Steiner point X(99), is
the gudulic point of both the KRH asymptotes and the Steiner axes.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
11.13.2
107
Using Morley affixes
Proposition 11.13.5. Using Morley affixes, the tangential equation

2 σ2 σ3
2 σ1 σ3 σ2 σ1 − 3 σ3

∗
Cz ' 
2 σ1 σ3
6 σ3
2 σ2
σ2 σ1 − 3 σ3
2 σ2
2 σ1
of the Steiner inconic is :



Proof. Start from barycentric equation and transmute. Ordinary equation is not so handy, and we
know that the adjoint matrix will look better.
Proposition 11.13.6. The four focuses of the inSteiner ellipse can be written as :


s
s
σ1 ± W0 Wf
2


√
σ
−
3
σ
σ22 − 3 σ1 σ3
2
1
3

Fj ' 
where W0 = σ3 , Wf =
, Wg =
 σ

σ3
σ3
Wg
2
±
σ3
W0
The conjugate of W0 is 1/W0 , while Wf , Wg are the conjugate of each other. From all the four
possibilities for the ±, two of them lead to visible points (the real focuses), the other two lead to
non visible points (the analytical focuses).
Proof. Write that isotropic lines Ω± Fj are tangent to the ellipse. The only difficulty is a sound
managment of the conjugacies.
11.14
The Brocard ellipse, aka the K-ellipse
Remark 11.14.1. Since the K-circumconic, i.e. CC(X(6)), is nothing but the circumcircle, the name
"K-ellipse" applies only to the K-inconic.
Figure 11.9: The K-ellipse
1. Equation of the K-ellipse is :
x2
y2
z2
xy
yz
zx
+ 4 + 4 −2 2 2 −2 2 2 −2 2 2 =0
4
a
b
c
a b
b c
a c
perspector is X(6)= a2 : b2 : c2 , center U is X(39) = a2 b2 + c2 , etc.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
108
11.15. Parabola
2. Draw the circumcircle of the contact points AK BK CK and obtain :
P 2 2 4
xb c b + c4 + a2 b2 + b2 c2 + a2 c2 − a4
a2 yz + b2 xz + xyc2 − (x + y + z)
=0
2 (b2 + c2 ) (a2 + c2 ) (a2 + b2 )
3. Compute the fourth intersection of this circle with the conic and obtain :
Q = a2 b2 − c2
2
a4 + a2 b2 + a2 c2 − b4 − b2 c2 − c4
2
, etc
z
}|
{
4. The axes are the lines through the center that are parallel to the bisectors of AK CK , BK Q .
Therefore, compute :
T
t
z
}|
{
= tan AK CK , AK BK
z
}|
{
= tan AK CK , U V
where V ' ρ : 1 : −1 − ρ is an unknown point at infinity, substitute
into T = 2t/ 1 − t2
and solve. Solutions are rational, leading to V1 = a2 b2 − c2 , etc =X(512) and V2 =
a2 a2 b2 + a2 c2 − b4 − c4 , etc =X(511).
5. Compute the axes as U ∧ V1 = a4 − b2 c2 ÷ a2 , etc and U ∧ V2 = b2 − c2 ÷ a2 , etc : the
Brocard axis X(3)X(6).
Having the perspector on an axis is special.
6. The sideline AC and the perpendicular to AC through BK cut the first axis in P1 , Q1 and
the second in P2 , Q2 . The idea is to draw circle having diameter [P1 , Q1 ], then the circle
centered at U orthogonal to the former and obtain the focuses by intersection with the axis.
7. More simpler, write Fi = µP1 +(1 − µ) Q1 and find µ such that U Fi /U P1 ÷ U Fi /U Q1 = 1.
These ratios involve vectors that all have the same direction, and no radicals are appearing.
In our special case, the equation factors, leading to a well-known result (the Brocard points) :
F1 = a2 b2 : b2 c2 : c2 a2
;
F2 = c2 a2 : a2 b2 : b2 c2
8. Proceed the same way with the other axis. Obtain an equation that doesn’t factors directly,
but whose discriminant splits nevertheless when using the Heron formula (7.5). Finally,


4 a2 b2 + a2 c2 S + i b4 + c4 − a2 c2 − a2 b2 a2


F3 , F4 =  4 b2 c2 + b2 a2 S + i a4 + c4 − a2 b2 − b2 c2 b2 
4 c2 a2 + c2 b2 S + i a4 + b4 − a2 c2 − b2 c2 c2
9. To summarize, F1 , F2 =X(39)±X(511), F3 , F4 =X(39)±i X(512)/4S. As it should be, the
focal distance (from center to a focus) is the same since X(511) and X(512)/4S are obtained
by a rotation (in space V).
11.15
Parabola
For the sake of completeness, let us recall the definition.
Definition 11.15.1. A parabola is a conic tangent to the infinity line. Two parallel lines make a
non proper parabola. The union of line at infinity and another line is ... some kind of circle rather
than a "special special" parabola.
Corollary 11.15.2. The conic defined by matrix C is a parabola when L∞ . adj C . t L∞ = 0.
Definition 11.15.3. The directrix of a parabola is the polar line of the focus.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
11.15.1
109
Inscribed parabola
Remark 11.15.4. (Steiner theorems). From Proposition 11.12.1, the focus F of an in-parabola is
the isogonal conjugate of its point at infinity U (and is therefore on the circumcircle), while the
perspector P is the isotomic conjugate of U (and is therefore on the Steiner circumconic).
11.15.2
Circum-parabola
Remark 11.15.5. Let T0 = u : v : w be the barycentrics of the point at infinity of a circumparabola.
Then, from Proposition 11.16.7, its perspector is P = u2 : v 2 : w2 and lies on the inSteiner ellipse.
Proposition 11.15.6. Using Morley affixes, let κ : 0 : 1 be the point at
circumparabola. Then equation, perspector and focus are:

2 σ3
−κ2 − σ1 σ3
−2 σ3 κ

2
2
Cz '  −κ − σ1 σ3 2 σ1 κ + 4 σ3 κ + 2 σ2 σ3 −σ2 κ2 − σ32
−2 σ3 κ
−σ2 κ2 − σ32
2 σ3 κ2
 3 σ2 − 4 σ12
σ1 σ22
4 σ22
+
κ
+
− 12 σ1 + (σ2 σ1 − 9 σ3 )

σ32
σ3

σ3 2

1
2 σ2
σ1
σ2 σ1
P '
−6
κ+2
− 18 + 2 σ12 − 6 σ2

2
σ3 σ3
σ3
κ
  σ2 σ1
9
4 σ12 − 12 σ2
σ12 σ2 − 4 σ22
−
κ+
+ 3 σ1 +
σ32
σ3
σ3
σ3
 
2
2
4 σ1
σ3
σ2
2
κ
+
8
κ
+
2
σ
−
−
2

2 
σ
σ32
3

κ 


1 2 σ2
1

F '
4
κ +
κ + σ1 + σ3


σ3
σ3
κ


 −1 3 2 σ1

1
2
κ +
κ + 8 − 4 σ2 − σ1
2
σ3
σ3
κ
infinity of a given




1
κ 





1 
κ
Proof. Use point κ + i κ h with h → 0 as the fifth point of the conic and compute the determinant.
Thereafter, compute the polar triangle and its perspector. Finalize by writing that Ω ± F are
tangent to the conic.
Proposition 11.15.7. The locus of the foci of all the circumscribed parabola is a circular quintic.
Singular focus (not on the curve) is X(143), the nine points center of the orthic triangle. Other
asymptotes are through points whose barycentrics are respectively, 2 : 1 : 1, 1 : 2 : 1, 1 : 1 : 2. Its
equation is :
1024 σ33 ZZ Z + βγ Z Z + γα Z Z + αβ Z


4 σ2 − σ12 σ32 Z4 + 4 σ22 + 9 σ1 σ3 − σ12 σ2 σ32 Z3 Z
2


−256 
+ 3 σ32 + 13 σ1 σ2 σ3 − σ13 σ3 − σ23 σ32 Z2 Z
T
3
4
2
2
3
2
 + 4 σ1 σ3+ 9 σ2 σ3 − σ1 σ2 σ3 ZZ + σ3 4 σ1 σ3 − σ2 Z 3 
512 σ32 σ3 + 4 σ2 σ1 − σ13 Z3 + σ33 + 4 σ1 σ2 σ32 − σ23 σ3 Z


2  2
+64 
 + σ13 σ22 − 4 σ14 σ3 − 8 σ1 σ23 + 30 σ12 σ2 σ3 + 28 σ22 σ3 + 33 σ1 σ32 σ32 ZZ  T
+ σ12 σ23 − 4 σ24 − 8 σ13 σ2 σ3 + 30 σ1 σ22 σ3 + 28 σ12 σ32 + 33 σ2 σ32 σ3 Z2 Z


2 −σ13 σ22 + 4 σ1 σ23 + 12 σ14 σ3 − 42 σ12 σ2 σ3 − 24 σ22 σ3 − 29!σ1 σ32 Z2



 3
12 σ1 σ24 − 3 σ13 σ23 + 12 σ14 σ2 − 31 σ12 σ22 − 28 σ23 σ3

T
2
ZZ
+16 σ3 
3
3

− 28 σ1 + 177 σ1 σ2 σ3 − 77 σ3


2
2
3
4
2
2 2
2 3
2 −σ1 σ2 + 4 σ1 σ2 σ3 + 12 σ2 − 42 σ1 σ2 σ3 − 24 σ1 σ3 − 29 σ2 σ3 σ3 Z
!
σ12 σ24 − 5 σ13 σ22 σ3 + 4 σ14 σ32 − 4 σ25 + 14 σ1 σ23 σ3 + 6 σ12 σ2 σ32 + 13 σ22 σ32 + 35 σ1 σ33 Z
+32 σ3
T4
+ σ14 σ22 − 5 σ12 σ23 + 4 σ24 − 4 σ15 σ3 + 14 σ13 σ2 σ3 + 6 σ1 σ22 σ3 + 13 σ12 σ32 + 35 σ2 σ32 Z
!
σ14 σ24 − 8 σ12 σ22 σ23 + σ13 σ3 + 16 σ26 + σ16 σ32 − 80 σ14 σ2 σ3 + σ1 σ24 σ3
+
T5
+52 σ13 σ23 + 2 σ12 σ22 σ3 − 4 σ1 σ2 σ32 σ3 − 104 σ23 + σ13 σ3 σ32 − 343 σ34
Proof. Elimination is straightforward. The real asymptotes are parallel to the sidelines.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
110
11.16
11.16. Hyperbola
Hyperbola
Definition 11.16.1. An hyperbola is a conic that intersects the line at infinity in two different
points. An ellipse is a special hyperbola (the intersection points are not visible) and a parabola is
not an hyperbola.
Proposition 11.16.2. Let ∆1 ' (ρ, σ, τ ) and ∆2 ' (u, v, w) be the asymptotes of an hyperbola C.
Then equation of C can be written as :
2
k (x + y + z) + (ρx + σy + τ z) (ux + vy + wz) = 0
(11.6)
Proof. Consider the line ∆1 ' (ρ, σ, τ ) and its point at infinity T1 = σ − τ : τ − ρ : ρ − σ. The
matrix of the quadratic form is :
C =
1
2
t
∆1 . ∆ 2 + t ∆2 . ∆1 + k
t
L∞ . L∞
(11.7)
It can be seen that t T1 . C . T1 = 0 (T1 belongs to conic) while t T1 . C = ∆1 (the tangent to the
conic at T1 is line ∆1 ). Another method is ∆1 . adj C . t ∆1 = 0 (line ∆1 is tangent to the conic)
while C . t ∆1 = T1 (the contact point of ∆ is T ).
Corollary 11.16.3. Equation (11.6) is the parametrization in k of the pencil of hyperbola that
share a given pair of asymptotes.
Proposition 11.16.4. A rectangular hyperbola is an hyperbola with orthogonal asymptotes. Such
an RH is characterized among all the conics by :
trace C . M = 0
Proof. We havetrace t ∆1 . ∆2 . Q = ∆2 . Q . t ∆1 as soon as matrix Q is symmetric. Therefore
trace C . M equals ∆2 . M . t ∆1 and the result follows.
11.16.1
Circum-hyperbolas
Proposition 11.16.5. A circumconic C can be characterized by one asymptote ∆1 ' (ρ, σ, τ ).
Then the second asymptote ∆2 is k/ρ : k/σ : k/τ where k is the constant appearing in (11.6).
Perspector P , center U , contacts at infinity T1 , T2 are given by :
P
C
T1
T2
2
= ρ (σ − τ )
= ρ σ2 − τ 2
=
σ−τ
=
ρ (σ − τ )
2
: σ (τ − ρ)
: σ τ 2 − ρ2
:
τ −ρ
:
σ (τ − ρ)
2
: τ (ρ − σ)
: τ ρ2 − σ 2
:
ρ−σ
:
τ (ρ − σ)
while the equation of the conic can be rewritten into T1 ∗ T2 ÷ X ∈ L∞ , and asymptotes as T1 ∗
b
b
b
X ÷ T2 ∈ L∞ and T2 ∗ X ÷ T1 ∈ L∞ . Moreover, C = T1 ∗ T2 ∗ polarmul (T1 , T2 ).
b
b
b
b
b
Proof. Direct examination.
Proposition 11.16.6. Consider a circumconic and its perspector
given by :

q 2 + r2 − pq − pr p + p (q − r) IST
 2
 r + p2 − qr − qp q + q (r − p) IST
p2 + q 2 − rp − qr r + r (p − q) IST
P . The points at infinity are



where IST 2 = p2 + q 2 + r2 − 2pq − 2qr − 2rp is the equation of the Steiner in-ellipse.
Proof. Direct inspection.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
11.16.2
111
Circum-rectangular-hyperbolas
Proposition 11.16.7. When a circumscribed conic is a rectangular hyperbola, its perspector is on
the tripolar of H =X(4) –the so-called orthic axis–, while its center C = cevadiv (X2 , P ) is on the
Euler circle.
Proof. Write that trace M C = 0 and obtain a first degree equation for the perspector, then
substitute the perspector from the center. (2) We have U ∈ L∞ and the U = cevadiv(X2 , P )
formula.
Proposition 11.16.8. Let P, Q, R be three distinct points on a RH. Then the Euler circle of P QR
goes through the center of the RH.
Proof. Obvious from preceding proposition.
Fact 11.16.9. The perspector of a circumRH can be written as:
1 −µ µ − 1
:
:
Sa Sb
Sc
on the tripolar of X(4), while the RH itself is :
−Sa x (Sb y − zSc ) µ + Sb y (Sa x − zSc ) = 0
Kiepert RH is µ = a2 − c2 Sb ÷ b2 − c2 Sa .
Proof. A direct proof is that H belongs to the pencil generated by BC ∪ AH and AC ∪ BH.
Proposition 11.16.10. [Six-circles transform] Given a fourth point M ' p : q : r, the perspector
P of the RH that goes through A, B, C, P is :
P ' (Sb q − Sc r) p : (Sc r − Sa p) q : (Sa p − Sb q) r
while the center of the RH is:
U ' p c2 pq − b2 pr + c2 − b2 qr (Sb q − Sc r) ::
The map M 7→ U is sometimes called "six-circles transform" since U is the common point to any
Euler circles of an inscribed triangle.
Proof. Comes from P ∈ tripolar (H) and U ∈ conicir (P ).
11.16.3
Inscribed hyperbolas
Proposition 11.16.11. An inconic C can be characterized by one asymptote ∆1 ' (ρ, σ, τ ). Then :
∆1
T1
N1
∆2
T2
N2
P
C
'
=
'
'
=
'
=
=
ρ
σ−τ
ρσ + ρτ − στ
ρ/f
(σ − τ ) f 2
−1
(ρ σ + ρ τ − σ τ )
1 ÷ (σ − τ ) ρ2
(σ − τ ) f
,
σ
:
τ −ρ
:
ρσ + στ − ρτ
;
σ/g
:
(τ − ρ) g 2
−1
; (ρ σ + σ τ − ρ τ )
:
1 ÷ (τ − ρ) σ 2
:
(τ − ρ) g
,
τ
:
ρ−σ
:
ρτ + στ − ρσ
;
τ /h
:
(ρ − σ) h2
−1
; (σ τ + ρ τ − ρ σ)
:
1 ÷ (ρ − σ) τ 2
:
(ρ − σ) h
where (f, g, h) ' N1 ' anticomplem (isot (∆)) is the Newton line associated with line ∆1 (cf
Proposition 11.19.5). Therefore :
∆2 = ∆1 ÷ N1 ; T2 = T1 ∗ N1 ∗ N1 ; N2 = G ÷ N1
b
b
b
b
C = T1 ∗ N1 = T2 ∗ N2 = crossmul (G, P ) ; P = crossdiv (G, C)
b
b
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
112
11.17. Diagonal conics
Proof. Let M be the matrix IC (p : q : r) of the general inconic withperspector p : q : r. Then
formula giving P from ∆ is obtained by elimination from ∆ . adj M . t ∆ = 0 (tangency) and
L∞ . adj M . t ∆ = 0 (T ∈ L∞ ). Thereafter, all formulas are proven by direct computing from
matrix C = IC (P ) where P is as given (cf. Stothers, 2003a).
Proposition 11.16.12. Consider an inconic and its center U . The points at infinity are given
by :




u2 v 4 + w4 − u2 v 2 − u2 w2
v 2 − w2 u2
 2 4



 v u + w4 − v 2 w2 − u2 v 2  ± OST (v + w − u) (w + u − v) (u + v − w)  w2 − u2 v 2 
w2 u4 + v 4 − u2 w2 − v 2 w2
u2 − v 2 w2
where OST 2 = −qr − rp − pq is the equation of the Steiner out-ellipse.
Proof. Direct inspection.
Proposition 11.16.13. Consider point U = v − w : w − u : u − v ∈ L∞ and its tripolar
∆0 ' [1/ (v − w) , 1/ (w − u) , 1/ (u − v)]. This line is tangent to Steiner in-ellipse and the contact
2
2
2
point is T0 = (v − w) : (w − u) : (u − v) . Define (index i=inscribed, c=circumscribed) lines
∆i = tripole (T0 ), ∆c ' [v − w, w − u, u − v] and point
2
2
C = T G (U ) = (v − w)2 (v + w − 2u) : (w − u) (u + w − 2v) : (u − v) (u + v − 2w)
Then C is the common center of a circum-hyperbola with asymptotes ∆0 , ∆c and an in-hyperbola
with asymptotes ∆0 , ∆c . The locus of C (and also of the circum-perspector) is K219 :
X
X
x3 −
x2 y + 3xyz = 0
3
6
P 2
while the locus of the in-perspector
P 3is : P6 x2 y − 6xyz = 0. All lines ∆c contain G = X2 while
envelope of the ∆i is cubic :
3x +3
6 x y − 21xyz = 0.
Proof. In all these cubics, G is an isolated point (and don’t belong to the locus). Otherwise,
computing as usual.
11.17
Diagonal conics
Triangle ABC is autopolar wrt conic C if, and only if, the non-diagonal coefficients vanish. Such
conic is called either autopolar or diagonal.
Remark 11.17.1. The only autopolar circle is the conjugate circle (see Section 12.7).
Remark 11.17.2. An autopolar parabola is tangent to [1, 1, 1] (the line at infinity) and therefore to
[±1, ±1, ±1] the sidelines of the medial triangle A0 B 0 C 0 . If U is its point at infinity, its focus F is
the A0 B 0 C 0 -isogonal of U (on the Euler circle) while the A0 B 0 C 0 -perspector is the A0 B 0 C 0 -isotomic
of U .
11.17.1
Pencils of diagonal conics
Proposition 11.17.3. Let γ (µ) be a pencil of diagonal conics :
.
γ (µ) = (1 − µ) (α1 x2 + β1 y 2 + γ1 z 2 ) + µ ∗ (α2 x2 + β2 y 2 + γ2 z 2 ) = 0
and U = u : v : w a point, not a vertex of ABC. Then polar lines of U wrt all the conics of the
pencil are concurring at a point U ∗ that will be called the isoconjugate of U wrt the pencil. In fact,
U ∗ = UP∗ –cf (16.2)– where :
P = β1 γ2 − γ1 β2 : γ1 α2 − α1 γ2 : α1 β2 − β1 α2
√
√
√
The four fixed points (real or not) of the conjugacy, i.e. the points ± p : ± q : ± r, are the
points common to all conics of the pencil.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
113
When a pair of isoconjugates U1 and U2 is known, P is known and therefore the isoconjugacy.
The pencil contains the conic γ1 through U1 , cevadiv (U2 , U1 ) and the vertices of their respective
anti-cevian triangles. Conic γ2 is defined cyclically. Both conics are tangent to U1 U2 .
Circumconic CC (P ) is together the P -isoconjugate of L∞ , the locus of centers of the conics
of the pencil and the conic that contains the six midpoints of the quadrangle formed by the four
fixed points.
11.18
The bitangent pencil
Definition 11.18.1. All of the conics that are tangent to two fixed lines at two given points
form a pencil, called the bitangent pencil F. We define B, C as the contact points and A as the
intersection of the tangents. Two points of interest are called M and A0 . Point M = (B + C) /2
defines the median AM which is the line of centers. Point A0 is the second intersection of the ABC
circumcircle and the A-symmedian of this triangle.
Theorem 11.18.2. The punctual and tangential equation of these conics are :




λ 0 0
1 0 0
t
−1
−1
t




Cb '  0 0 1  ; Cb∗ =  0 0 λ  ; Cz = Lu · Cb · Lu ; Cz∗ = Lu · Cb∗ · Lu
0 1 0
0 λ 0
The Morley affixes Z : T : Z of a focus of the conic C (λ) ∈ F are bound to the parameter λ by :
Z − T/β Z − T/γ
(Z − βT) (Z − γT)
−1
=
=
(11.8)
2
2
2λ
(Z − αT)
Z − T/α
and the focus is located on the "focal cubic" K :
2
2
1
1
1
1
2
− −
− 2 Z2 T + α2 − β γ Z T+
Z2 Z + (γ + β − 2 α) ZZ +
α β
γ
βγ
α
2
α α β+γ
β+γ
2α
2 βγ
α2
α2
α
βγ
2
2
+ −
−
−
−
−
2
ZZT+
ZT
+
ZT
+
T3
γ
β
α
α2
βγ
α
γ
β
βγ
α2
This curves goes through both umbilics, vertices B, C and two times through A. The asymptote
∆∞ is the parallel to the median AM through point Ω0 = (3A − A0 ) /2
t
t
Proof. The Plucker definition gives (F ∧ Ωx ) · Cz∗ · (F ∧ Ωx ) = 0 = (F ∧ Ωy ) · Cz∗ · (F ∧ Ωy ). A
separation of the variables occurs, giving one equation in the upper view Z : T and another in the
lower view Z : T. Umbilical property is obvious. The real asymptote is easily obtained from the
gradient.
Proposition 11.18.3. The focal cubic of the bitangent pencil can be parametrized/constructed as
follows. A variable point K = κ : 1 : 1/κ on the circumcircle defines a variable line AK. Reflect
B, C into AK and obtain Bk , Ck . Then point F = BCk ∩ CBk is on the cubic.
Proof. From (11.8), line F A is a bisector of angle F B, F C. We obtain :


α (α − β) (α − γ) κ3 − α2 (αβ + αγ − 2 βγ) κ2 + αβγ α2 − β γ κ


F (κ) ' 
−α (αβ + αγ − 2 β γ) κ2 + αβγ (2 α − β − γ) κ

2
2
β γ − α κ + β γ (2α − β − γ) κ + β γ (α − β) (α − γ)
Proposition 11.18.4. In this construction, the other visible focus F 0 of the same conic is obtained
by using the line AK 0 such that lines AK and AK 0 are equally inclined on lines AB, AC. Moreover,
in the upper view Z : T, the focuses are exchanged by the involutory homography whose fixed points
are A and A0 , the second intersection of the A-symmedian with the circumcircle.
Proof. For the second part, write and factor λ Z : T : Z − λ (z : t : ζ) from (11.8). This gives
(zT − tZ) together with another first degree factor with respect to Z, T and also with respect to
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
114
11.18. The bitangent pencil
Figure 11.10: Two transformations of the focal cubic.
z, t. In the upper view Z : T, this induces the identity together with another homography, whose
matrix is
!
β γ − α2
α2 β + γ α2 − 2 α β γ
ψ '
β + γ − 2α
α2 − β γ
.
A fixed point is A = α : 1. The other is A0 = αβ + αγ − 2 β γ : 2α − β − γ on the circumcircle.
One can check that :
ωAA0 × ωAM =
α (γ α + α β − 2 β γ) α β γ (β + γ − 2 α)
×
= (−αβ) × (−αγ) = ωAB × ωAC
β + γ − 2α
γ α + αβ − 2β γ
For the first part, write that f (ν) ' ψ (f (κ)) where f (κ) is the upper part, i.e. Z : T of F (κ).
This leads to a second degree equation. One solution verifies νκ = βγ as required. The other is
not a turn (and leads to a non visible focus). Another method is a direct substitution of F (κ) into
(11.8). This leads to a fourth degree polynomial whose roots are κ, βγ/κ and other two that are
not turns (and lead to both invisible focuses).
Proposition 11.18.5. When center U is chosen on the median AM , the focuses F, F 0 can be
constructed as follows. Draw the bisectors ∆1 , ∆2 of U A, U A0 . Cut them at H1 , H2 by the
perpendicular bisector of [A, A0 ]. Draw circle γ1 (H1 , A) and cut ∆2 . Additionally, draw circle
γ2 (H2 , A0 ) and cut ∆1 . This gives the four focuses.
Proof. This comes from the involutory homography ψ.
Proposition 11.18.6 (Cissoidal property). Consider circle γ through A and centered at Ω0 =
(3A − A0 ) /2. Define P as the second intersection of γ and line AF . Define P 0 from F 0 . Then P 0
is the reflection of P into the asymptote ∆∞ of the cubic K. Define Q as the intersection of P F
−−→ −→
with ∆∞ , and also Q0 as P 0 F 0 ∩∆∞ . Then we have the cissoidal property : P Q = AF . By the
way,
−−−
−−0→ −−
→
→2
intersection Q0 of ∆∞ and external bisector of AB, AC belongs to γ. Moreover Ω Q· Ω0 Q0 =Ω0 Q0 : points Q, Q0 are the reflection of each other into circle γ.
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
115
Proof. Direct computation. In order to deal with with Q0 , we have to use the second degree Lubin
frame.
∗
Proposition 11.18.7. The isogonal conjugates F ∗ , F 0 of the focuses of the C ∈ F belong to a
∗
circle. A diameter of this circle is provided by the feet J1 , J2 of the A-bisectors. Points F ∗ , F 0
∗
are symmetrical with respect to this diameter. Moreover, the point F is the image of point Q
(see preceding proposition) by a skew similitude whose fixed directions are the bisectors of the angle
AM, BC.
Proof. The κ-degrees of the affixes of F are 3, 2, 2. The κ-degrees of the affixes of F ∗ are 1, 1, 2:
these points belong at least to a conic. Umbilics belong to the cubic. Their isogonal transforms
belong to the conic: we have a circle. Using the second degree Lubin frame, the matrix of the
locus is :


0
2 α2 − β 2 − γ 2
β 2 γ 2 − α4


ϕ '  2 α2 − β 2 − γ 2
2 β 2 γ 2 − 2 α4
α2 α2 β 2 + α2 γ 2 − 2 β 2 γ 2 
2
β 2 γ 2 − α4
α2 α2 β 2 + α2 γ 2 − 2 β 2 γ 2
0
Its intersections with line BC are


α2 β γ + α2 β 2 + α2 γ 2 − β 2 γ 2 γ β


J1 ' 
α2 + β γ γ β

2
β 2 + β γ + γ 2 − α2
and J2 obtained by βγ 7→ −βγ. Its center is BC ∩ OΩ, where O =X(3).
We search for a matrix Q such that F ∗ ' Q . P . In order to success into identifying with
respect to powers of κ, we have to normalize both P and F ∗ . We obtain :


(β + γ − 2 α) αβγ
0
Q
o


α2 − β γ




Q ' 
where
0
1
0

 αβ + αγ − 2 βγ

Qo
0
β γ (α2 − β γ)
2
Qo
=
3βγ (β + γ − 2 α)
βγ (β − γ)
β2 + γ2 + 4 β γ
+
+
2
β+γ
α −βγ
(2 βγ − αβ − αγ) (β + γ)
umbilics are exchanged and the conclusion follows.
11.19
The Miquel pencil of conics
Definition 11.19.1. All of the conics that are tangent to four fixed lines form a pencil, called the
Miquel pencil F of these four lines.
Lemma 11.19.2. Let us use describe our four lines as the three sidelines of ABC and the transversal whose tripole is p : q : r. See Theorem 4.5.4 for more details. In the Lubin frame, the transversal, the Newton line, the Miquel point Ω of the quadrilateral and the homography ψ are :

 
 

βq−γr
αp − γ r
βq − αp

 
 

q − r ,  p − r ,  q − p 
A0z , Bz0 , Cz0 ' 


 q


r
r
p 
p
q
−
−
−
β
γ
α γ
β
α

t
α (γ − β) qr + β (α − γ) rp + γ (β − α) pq


N ewton '  α β 2 − γ 2 qr + β γ 2 − α2 rp + γ α2 − β 2 pq 
((γ − β) qr + (α − γ) rp + (β − α) pq) αβγ




c1 /c0
c1 −c2 0




Ω '  1  , ψ =  c0 −c1 0 
c0 /c1
0
0
0
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
116
11.19. The Miquel pencil of conics

c2



c
1
where

c
0



c−1
= α2 (γ − β) p + β 2 (α − γ) q + γ 2 (β − α) r
= α (γ − β) p + β (α − γ) q + γ (β − α) r
= (γ − β) p + (α − γ) q + (β − α) r
= α1 (γ − β) p + β1 (α − γ) q + γ1 (β − α) r
Remark 11.19.3. Quantity c−1 is only for cosmetic purposes. In fact, we have the bounding
relation :
c2 − c1 σ1 + c0 σ2 − c−1 σ3 = 0
Proposition 11.19.4. Fixed points of the isoconjugacy Ψ generated by ψ are given by :


c1 /c0 + Wu /c0


Φ'
1

c0 /c1 + αβγ Wd /c1
where the up and down radicals are given by :
Wu2
Wd2
(α − β) (β − γ) (γ − α) × (qr (γ − β) + rp (α − γ) + pq (β − α))
γ−β
(α − β) (β − γ) (γ − α)
α−γ
β−α
×
qr
=
+
rp
+
pq
α2 β 2 γ 2
βγ
αγ
αβ
=
They can be constructed as follows. Lines ∆1 , ∆2 are the common bisectors of ΩA, ΩA0 , ΩB, ΩB 0 ,
ΩC, ΩC 0 and δA is the perpendicular bisector of [A, A0 ]. Then H1 = ∆1 ∩ δA (resp. H2 = ∆2 ∩ δA )
is the center of a circle by A, A0 that cuts ∆2 (resp ∆1 ) at the four Φj points.
Proof. These points are characterized by
c0 Z2 − 2 c1 ZT + c2 T2 = 0
;
2
c1 Z − 2 c0 ZT + c−1 T2 = 0
Proposition 11.19.5 (Newton). All the conics that are tangent to four given lines have their
centers on another same line, that goes through the midpoints of the diagonal pairs AA0 , BB 0 , CC 0
(the Newton line). When U ∈ N ew is defined as K Mb + (1 − K) Mc the conic can be written as :
Cb∗

0 0

' (r − p)  0 0
−p q


0
−p


q  + K p p − q
0
r−p
p−q
0
q−r

r−p

q−r 
0
The first part is the tangential conic "all lines through C 0 or C", the second part is a non degenerated
parabola (with same direction as the Newton line).
Proof. Let ρ : σ : τ be the "service point" of a conic C ∈ F. We have Cb∗ = [0, τ, σ; τ, 0, ρ; σ, ρ, 0]
and therefore pρ + qσ + rτ = 0. Since U = σ + τ : τ + ρ : ρ + σ, the Newton line is :
[q + r − p, r + p − q, p + q − r]
and the conclusions follow (remember that p : q : r is the tripole of the transversal).
Theorem 11.19.6. The Morley affixes Z : T : Z of a focus of the conic C (K) ∈ F are bound to
the parameter K by :
K=
(q − p) (Z − β T) ((r − p) Z − (γ r − α p) T)
p (Zc0 − Tc1 ) T
(11.9)
and the focus is located on the "focal cubic" K :
c1 2
c0
c2 + c0 σ2
c0 σ 1 2
c1 σ 2 2
c2 − c1 s1 3
2
2
Z Z + c0 ZZ −
TZ2 −
TZZ − c1 TZ +
T Z+
T Z+
T
σ3
σ3
σ3
σ3
σ3
s3
March 30, 2017 14:09
published under the GNU Free Documentation License
11. About conics
117
(a) The general case
(b) The unicursal case
Figure 11.11: The Miguel pencil and its focal cubic
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
118
11.20. Tg and Gt mappings
t
Proof. Matrix Cz∗ is obtained as Lu . Cb∗ . Lu and then Plucker method is used. Some factors
(q − r) are appearing during the elimination process, but not all the (p − q) (q − r) (r − p). Nothing
special occurs when P is on a median (but not at the centroid).
Theorem 11.19.7. The focuses Fj of a given conic C (K) ∈ F are exchanged by homographies
ψ, ψ. They can be constructed as follows. Call N ew⊥ the perpendicular to the Newton line at Ω.
Draw the bisectors ∆1 , ∆2 of U Φ1 , U Φ2 . Cut them at H1 , H2 by N ew⊥ . Draw circle γ1 (H1 , Φ1 )
and cut ∆2 . Additionally, draw circle γ2 (H2 , Φ2 ) and cut ∆1 .
Proof. Write and factor K Z : T : Z − K (z : t : ζ) from (11.9). This gives (p − q) (p − r) but not
(q − r), being smooth except at the centroid (i.e. when the fourth line at infinity). Otherwise, this
gives (zT − tZ) together with another first degree factor with respect to Z, T and also with respect
to z, t. In the upper view Z : T, this induces the identity together with another homography. Since
the later has to provide A ←→ A0 , B ←→ B 0 , C ←→ C 0 , it has to be ψ. The same occurs in the
lower view Z : T, and the conclusion follows.
Remark 11.19.8. In the Geogebra Figure 11.11, the orange conic is drawn as follows. Reflect
focuses F, F 0 into sideline AC and obtain Fb , Fb0 . Then point Eb = F Fb0 ∩ F 0 Fb on sideline AC
belongs to the conic. In the same way, obtain the point Ec on AB. We draw both conics, an ellipse
and an hyperbola, with focuses F, F 0 that go through Eb , but only the one that goes also through
Ec is displayed.
Proposition 11.19.9. When the transversal is tangent to one of the inexcircles of triangle ABC,
the pencil contains one circle and the focal conic has a double point. This point is the center of
the circle (see Figure 11.11b). When the transversal touches two of the inexcircles, it degenerates
into the Newton line and a circle having the corresponding inexcenters as antipodal points.
Proof. Only circles have equal focuses.
11.20
Tg and Gt mappings
Definition 11.20.1. Tg and Gt mappings. Suppose U is a point not on a sideline of ABC.
Let :
gU = isogonal conjugate of U , tU = isotomic conjugate of U
tgU = isotomic conjugate of gX, gtU = isogonal conjugate of tU
GtU = intersection of lines U − tU and gU − gtU ,
T gU = intersection of lines U − gU and tU − tgU .
If U = u : v : w (barycentrics), then :
a2 b2 − c2
b2 c2 − a2
c2 a2 − b2
GtU = 2
:
:
(v − w2 ) u (w2 − u2 ) v (u2 − v 2 ) w
T gU =
b2 − c 2
c2 − a2
a 2 − b2
:
:
(w2 b2 − v 2 c2 ) u (u2 c2 − w2 a2 ) v (v 2 a2 − u2 b2 ) w
Proposition 11.20.2. For any point U , not on a sideline of ABC, points A, B, C, gU, tU, T gU, GtU
are on a same conic (Tuan, 2006). The perspector of this conic is X512 ÷ U . This conic is the
b
isogonal image of line U − gtU and also the isotomic image of line U − tgU .
Proof. Straightforward computation.
Example 11.20.3. Points X(3112) to X(3118) are related to Gt and T g functions.
The X(31)-conic passes through X(I) for
I = 75, 92, 313, 321, 561, 1441, 1821, 1934, 2995, 2997, 3112, 3113
The X(32)-conic passes through X(I) for
I = 76, 264, 276, 290, 300, 301, 308, 313, 327, 349, 1502, 2367, 3114, 3115
The X(76)-conic passes through X(I) for
I = 6, 32, 83, 213, 729, 981, 1918, 1974, 2207, 2281, 2422, 3114, 3224, 3225
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 12
More about circles
12.1
General results
Let us start by recalling two key results.
Theorem 12.1.1 (Already stated in Section 7.3 as Theorem 7.3.2). Let Ω be the circle centered
at P with radius ω. The power formula giving the Ω-power of any point X = x : y : z from the
power at the three vertices of the reference triangle is :
.
2
power (Ω, X) = |P X| − ω 2 =
where u
=
ux + vy + wz
a2 yz + b2 xz + c2 xy
−
2
x+y+z
(x + y + z)
(12.1)
power (Ω, A) , etc
Definition 12.1.2 (Already stated in Section 7.3 as Definition 7.3.3). From power (Γ, A) = 0, etc,
we have defined the standard equation of the circumcircle as :
. a2 yz + b2 xz + c2 xy
Γstd (x, y, z) = −
=0
x+y+z
(12.2)
Proposition 12.1.3. Let C be a conic, with matrix C = (mjk ) (notations of Definition 11.3.1) .
Then C is a cycle if and only if, for a suitable factor k, we have :




2 m11
m11 + m22 m33 + m11
0 c2 b2
1



C −  m11 + m22
2 m22
m22 + m33  = k  c2 0 a2 
2
m33 + m11 m22 + m33
2 m33
b2 a 2 0
Proof. Obvious from (12.1).
Proposition 12.1.4. Four points at finite distance belong to the same cycle (aka circle or straight
line) when their barycentrics pi : qi : ri are such that :
i=4
det [pi , qi , ri , Γstd (pi , qi , ri )] = 0
i=1
(12.3)
Proof. Obvious from (12.1). Don’t forget how Γstd was defined in (12.2) !
Computed Proof. Denominators are a reminder of the fact that circles don’t escape to infinity.
Write the Cartesian equation of the circle as :
. i=4 ∆cart = det ξi2 + ηi2 , ξi , ηi , 1 = 0
i=1
where ξ, η are the Cartesian coordinates of the points. Substitute these coordinates by :
ξ=
xξa + yξb + zξc
xηa + yηb + zηc
,η=
x+y+z
x+y+z
119
120
12.1. General results
and obtain another determinant ∆0 (x, y, z). Then compute F . ∆0 . T −1 . G where F is the diagonal
matrix diag (pi + qi + ri ) and



T =

1
0
0
0
0
ξa
ξb
ξc
0
ηa
ηb
ηc

0
1
1
1


,




G=

−1
2
ξa + ηa2
ξb2 + ηb2
ξc2 + ηc2

0 0 0
1 0 0 


0 1 0 
0 0 1
Matrix F acts on rows and kills quite all denominators, T acts on the last three columns and
goes back to barycentrics while G acts on the first column to kill all square terms. After what
everything simplifies nicely and leads to (12.3)
Proposition 12.1.5. The barycentric equation of circle with center P = p : q : r and radius ω is :
2
(u0 x + v.0 y + w0 z) (x + y + z) − ω 2 (x + y + z) − a2 yz + b2 zx + c2 xy = 0
(12.4)
where quantities u0 , v0 , w0 are defined as :
u0
v0
w0
.
2
= |P A| =
.
2
= |P B| =
.
2
= |P C| =
2
c2 q 2 + b2 r2 + b2 + c2 − a2 qr ÷ (p + q + r)
2
a2 r2 + c2 p2 + c2 + a2 − b2 rp ÷ (p + q + r)
2
b2 p2 + a2 q 2 + b2 + a2 − c2 pq ÷ (p + q + r)
(12.5)
Proof. Obvious from (12.1). The added value here is the emphasis on center and ω 2 . It must be
noticed that u : v : w is not a point nor a line. Quantities u, v, w are strongly defined objects and
are not defined up to a proportionality factor. They are to be considered exactly as ω 2 , i.e. are of
the same nature as a surface. It can be observed that u (or v or w) is zero-homogeneous wrt the
barycentrics of point P = p : q : r. More details are given in Chapter 13
Proposition 12.1.6. Center. The center of a circle defined by

 
a2
a2 Sa
1



vX(3) − K . t U =  b2 Sb  −  −Sc
center '
2
−Sb
c2 Sc
1
'
a2 Sa : b2 Sb : c2 Sc − M · t U
2S
its equation (12.1) is
 
u
−Sc −Sb
 
2
b
−Sa  ·  v
w
−Sa
c2
given by :



(12.6)
(12.7)
Proof. As stated in Definition 11.3.8, the center of a conic is the pole of the line at infinity wrt
the conic. Computations are straightforward. It can be noticed that product K . t U gives the
orthodir of line U (i.e. the point at infinity in the orthogonal direction). Since the line of centers is
orthogonal to the radical axis of Γ and Ω, this formula describes the coefficients
to be used when
the circumcenter X(3) is described as in ETC by vX (3) = a2 b2 + c2 − a2 ::. Let us recall the
Al-Kashi formula K = 2S M .
Remark 12.1.7. In both the center and the radius formulas, everything must be used exactly as
written, and not to a proportionality factor. A more efficient formulation will be given later, with
formula (13.13)
Proposition 12.1.8. Radius. The radius of a circle defined by its equation (12.1) is given by :
ω2 =
1 U · K ·t U − U.vX (3) + a2 b2 c2
2
16 S
(12.8)
2
Proof. Subtract formula (12.1) from |P X| obtained from (12.6) and Pythagoras formula.
Definition 12.1.9. circlekit. Some usual square roots are given in Table 12.1, and some other
notations in Table 12.2.
Example 12.1.10. Table 12.3 describes some of the usual circles in triangle geometry. For further
information on many circles, refer to
http://mathworld.wolfram.com/Circle.html
March 30, 2017 14:09
published under the GNU Free Documentation License
12. More about circles
name
#
Lemoine
W1
Brocard
W2
Euler
W3
Furhmann
W4
121
value
√
√
where
a4 + b4 + c4 − a2 b2 − a2 c2 − b2 c2
pP
3
a6 −
pP
3
P
6
a3 −
a4 b2 + 3 a2 b2 c2
P
6
a2 + b2 + c2 + 4 iS
2W1
2 W2 R
|OK| = 2
a + b2 + c 2
W3
R
|OH| =
=
W3
4S
abc
r
1
|HN | = 2 W4 R
abc
eiω =
a2 b2 + a2 c2 + b2 c2
a2 b + 3 abc
(12.12)
(12.13)
(12.14)
(12.15)
Table 12.1: Some usual square roots (circle kit 1)
#
value
name
s
(a + b + c) /2
half-perimeter
abc
p
s (s − a) (s − b) (s − c)
abc
4R
a2 + b2 + c2
2S
exp (i ω) =
+i
2W1
W1
q
4
4
4
a +b +c
b2 c2 +c2 a2 +a2 b2 − 1
circumradius
R
S
ω
e
r
S
abc
=
s
2R (a + b + c)
area of triangle
Brocard angle
p
1 − 4 sin2 ω
inradius
Table 12.2: Some usual notations (circle kit 2)
Definition 12.1.11. Pencil of cycles. When C1 and C2 are two circles, then λC1 + µC2 is also a
cycle. The family generated from two given circles is called a pencil. Then all centers are on the
same line, which is orthogonal to the only line contained in the pencil (the so called radical axis).
More details in Chapter 13
Definition 12.1.12. The radical trace of two non-concentric circles is the point of intersection
of the radical axis of the circles and the line of the centers of the circles. (For examples, see X(I)
for I = 6, 187, 1570, 2021-2025, 2030-2032.)
12.2
Antipodal Pairs on Circles
Proposition 12.2.1. Centers of similitude. Here, all barycentrics are supposed to be in their
normalized form. Let (O1 , r1 ), (O2 , r2 ) be two circles. Points U, V defined by :
U=
r1 O2 − r2 O1
r1 O2 + r2 O1
, V =
r1 + r2
r1 − r2
are respectively the internal and the external centers of similitude of these two circles. When
X ∈ (O1 ) then :
(r1 + r2 ) U − r2 X ∈ (O2 )
and
(r1 − r2 ) V + r2 X ∈ (O2 )
Proof. When r1 = r2 , point V defines a translation, not an homothety. Otherwise, all steps are
obvious.
Proposition 12.2.2. Suppose (O1 ) and (O2 ) are circles and that P, P 0 are antipodes on (O1 ).
Let U = insim(O1 , O2 ) and V = exsim (O1 , O2 ) be the respective internal and external center of
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
122
12.3. Inversion in a circle
Name
Center
Radius
circumcircle
incircle
nine-point circle
conjugate circle
Longchamps circle
Bevan circle
Spieker circle
Apollonius circle
1st Lemoine
2nd Lemoine
Sin-triple-angle
Brocard circle
Brocard second
Orthocentroidal
Fuhrmann
X(3)
X(1)
X(5)
X(4)
X(20)
X(40)
X(10)
X(970)
X(182)
X(6)
X(49)
X(182)
X(3)
X(381)
X(355)
R
r
1
2R
√
−Sa Sb Sc ÷ 2S
√
−Sa Sb Sc ÷ S
2R
r/2
r2 + s2 ÷ 4 r
R ÷ 2 cos ω
abc/ a2 + b2 + c2
Rsta
eR ÷ 2 cos ω
eR
|OH| /2
|HN | /2
12.4
12.5
12.6
12.7
12.8
12.9
12.10
12.11
12.12
12.13
12.14
12.15
12.15
12.16
12.17
Table 12.3: Some circles
similitude of circles (O1 ) and (O2 ). Define Q = P U ∩ P 0 V and Q0 = P V ∩ P 0 U . Then Q, Q0 are
antipodes on (O2 ). Moreover, the lines P P 0 and QQ0 are parallel.
Proof. The result is quite obvious, but giving a non-circular proof is not so obvious. To construct
the centers of similitude, you can draw an arbitrary diameter P0 O1 P00 of the first circle and then
the parallel diameter Q0 O2 Q00 of the second (arbitrary means : don’t choose the center line !).
Intersections P0 Q0 ∩ P00 Q00 and P00 Q0 ∩ P0 Q00 are the required points (in hand drawing, point V is
better defined that way than using the center-line as in Coxeter (1989, page 70). Define a point
P ∈ (O1 ) as "near" when it lies in the same half-plane wrt diameter P0 P00 than O2 , and define it
as "far" otherwise. Act conversely on (O2 ). Then U -similitude converts far into far, near into near
while V -similitude converts far into near and near into far.
Suppose that P ∈ (O1 ) is far. Then its V -image on (O2 ) is the near point of P V ∩ (O2 ) (call
it R). The antipode P 0 is near, and its V image on (O2 ) is the far point of P 0 V ∩ (O2 ) (call it
R0 ). By similitude, RR0 is the diameter of (O2 ) that is parallel to the diameter P P 0 of (O1 ) and
P, R0 are far while P 0 , R are near. The same thing can be done using the insimilitude. But now,
the image of a far point is a far one: the U -image of P is the far point S of P U ∩ (O2 ), while the
U -image of P 0 is the near point S 0 of P 0 U ∩ (O2 ). Here again, SS 0 is the diameter of (O2 ) that is
parallel to the diameter P P 0 of (O1 ) and we must have S = R0 and S 0 = R.
In the following examples, suppose P = p : q : r on the first circle.
12.3
Inversion in a circle
Definition 12.3.1. Two points X1 , X2 Dare inverses Ein a given circle with center P and radius ω
−−→ −−→
when P, X1 , X2 are in straight line and P U1 | P U2 = ω 2 .
Proposition 12.3.2. The inverse of point X = x : y : z in circle Ω having center P = p : q : r
and radius ω is given by :
inv (X) =
Γstd (P )
Γstd (X)
2
+
− ω nor (P )
x+y+z
p+q+r
+
March 30, 2017 14:09
ω2 −
2
(p + q + r)
2
P . t P . P yth
!
. nor (X) (12.9)
published under the GNU Free Documentation License
12. More about circles
123
This formula is to be compared with formula (13.17) given in Theorem 13.5.5.
Proof. By definition, the inverse of point X = u : v : w is given by :
ω2
pytha (X, P )
Multiplying by pytha (X, P ), we obtain inv (X) ' pytha (X, P ) − ω 2 nor (P ) + ω 2 nor (X).
Then pytha is expanded using its definition, and we conclude using t P . X P = P . t P . X.
nor (inv (X)) = nor (P ) + (nor (X) − nor (P ))
Remark 12.3.3. When circle Ω is given by its equation (12.1) and P, ω are obtained from (12.6)
and (12.8), the following identity can be useful :
Γstd (P )
2
(p + q + r)
− ω2 =
U . (vX (3) /2) − a2 b2 c2
8 S2
(remember: U and vX (3) are "as is" and not "up to a proportionality factor", while P is projective
and ω is a number).
12.4
Circumcircle
Definition 12.4.1. The circumcircle is the circle through A, B, C. Perspector is X6 and center
X3 . Equation is :
a2 yz + b2 zx + c2 xy = 0
Its standard parametrization is (7.15), i.e. :
b2
c2
a2
:
:
| ρ : σ : τ 6= L∞
σ−τ τ −ρ ρ−σ
Lemma 12.4.2. The distance |P X3 | to center from any point is given by :
2
|P X3 | = R2 −
a2 qr + b2 rp + c2 pq
(p + q + r)2
Proof. Direct inspection using Theorem 7.2.4. As it should be, |X3 X3 | = 0 while the equation of
2
the circumcircle is |P X3 | = R2 .
Proposition 12.4.3. For any finite point, other than the circumcenter X3 , the inverse-in-circumcircle
of P = p : q : r has barycentrics u : v : w obtained cyclically from :
a2 b2 + c2 − a2
b2 − a2
c2 − a2
2
pq +
pr +
qr
u = −p +
b2
c2
b2 c2
1
36
2
23
4
186
5 2070
6
187
10 1324
15
16
20 2071
21 1325
22
858
24
25
26
27
28
29
32
35
39
40
403
468
2072
2073
2074
2075
1691
484
2076
2077
54
55
56
57
58
67
115
131
182
237
1157
1155
1319
2078
1326
3455
2079
2931
2080
1316
352
371
372
399
667
859
1054
1145
1384
1687
353
2459
2460
1511
1083
3109
1283
2932
2030
1688
1692 3053
2482 2930
2935 3184
3110 3286
3438 3480
3439 3479
3513 3514
Proof. Points X(3), P, U are on the same line, and distance from X(3) to U is R2 /|P X3 |. Therefore,
in normalized barycentrics, we have : u = x3 + (p − x3 ) R2 /|P X3 |2 .
Remark 12.4.4. On ETC n ≤ 3587, there are :
• 258 named points that belongs to Γ
• 47 pairs of "true" inverses that both are named
• 220 named points of Γ that have a named isogonal conjugate (among the 229 points of L∞ )
• 62 pairs of named antipodal points
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
124
12.5. Incircle
12.5
Incircle
Definition 12.5.1. The incircle is one of the four circles that are tangent to the sidelines. This
circle is inside the triangle, and also inside the nine point circle (these circles are tangent). Center
X(1), perspector X(7), radius r = S/s = abc ÷ 2R (a + b + c), equation :
Γstd +
Proof. well-known properties.
1X
2
x (b − a + c)
4
(12.10)
Remark 12.5.2. On ETC n ≤ 3587, there are :
• 39 named points on the incircle
• 7 pairs of "true" inverses that both are named
• 10 pairs of named antipodal points
Proposition 12.5.3. Centers of similitude with the circumcircle are in=X(55), and ex=X(56).
When p : q : r ∈ Γ, then Q, Q0 are antipodal points on the incircle :
2
Q = (b − c) p + a2 q + a2 r (b + c − a) , etc
2
Q0 =
Proof. Proposition 12.2.2.
(b + c) p + a2 q + a2 r
, etc
b+c−a
Proposition 12.5.4. Incircle transform. Let U = u : v : w be a point other than the symmedian
point, X6 . Then reflection of C7 (the intouch triangle) in the line U X1 is perspective with triangle
ABC. The isogonal conjugate of the corresponding perspector is called the incircle transform of U .
Its barycentrics are :
2
IT (u : v : w) =
2
2
a2 (bw − cv)
b2 (cu − aw)
c2 (av − bu)
:
:
b+c−a
a+c−b
b+a−c
and this point is on the incircle.
Proof. Line U X1 is a diameter of the incircle and the reflected triangle T is also inscribed in the
incircle. The barycentrics of U X1 are [bw − cv, cu − aw, av − ub]. Reflection in this line is obtained
using (7.24), and barycentrics of T are obtained. Perspectivity and perspector are easily computed
and conclusion follows by substituting in the incircle equation.
Remark 12.5.5. In ETC another formula is given... and the point is also on the incircle :
2
2
2
a2 b2 w − c2 v
b2 c2 u − a2 w
c2 a2 v − b2 u
IT 2 (U ) = 2 2
:
:
b c (b + c − a) a2 c2 (c + a − b) a2 b2 (a + b − c)
One has IT (X) = IT (U ) when U, X aligned with X1 while IT 2 (X) = IT (U ) when U, X aligned
with X6 .
12.6
Nine-points circle
Definition 12.6.1. The nine point circle is the circumcircle of the orthic triangle. It goes also
through the six midpoints of the orthocentric quadrangle ABCH. Center X(5), radius R/2 (half
the ABC circumradius), perspector :
1
1
1
:
:
a4 − a2 b2 − a2 c2 − b2 c2 b4 − a2 c2 − b2 c2 − a2 b2 c4 − a2 c2 − b2 c2 − a2 b2
is not named, equation :
Γstd +
March 30, 2017 14:09
1
(xSa + ySb + zSc )
2
(12.11)
published under the GNU Free Documentation License
12. More about circles
125
Proposition 12.6.2. Centers of similitude with Γ are in=X(2), and ex=X(4). Ω intersects Γ when
ABC is not acute. Radical trace X(468), direction of center axis (Euler line) X(30), direction of
radical axis X(523). Standard parametrization (similitude from circumcircle) :


(σ − τ ) c2 τ − b2 σ + b2 ρ − c2 ρ


U '  (τ − ρ) a2 ρ − c2 τ + c2 σ − a2 σ 
(ρ − σ) b2 σ − a2 ρ + a2 τ − b2 τ
Remark 12.6.3. On ETC n ≤ 3587, there are :
• 37 named points on the nine points circle
• 9 pairs of "true" inverses that both are named
• 12 pairs of named antipodal points
Proposition 12.6.4 (Feuerbach). The nine-point circle is tangent to the incircle and the three
excircles.
Proof. Use (12.10) and (12.11) to obtain the radical axis.
12.7
Conjugate circle
Definition 12.7.1. The conjugate circle is the only circle√whose matrix is diagonal (and triangle
ABC is autopolar). Center X(4), the orthocenter, radius −Sa Sb Sc ÷ 2S, equation :
X
1
Sa x2 = Γstd + (x Sa + y Sb + z Sc ) = 0
x+y+z
Proposition 12.7.2. This circle belongs to the same pencil as the circum- and the nine-points
circles. This circle is real only when triangle ABC is not acute. Therefore, no named points can
belong to this circle.
Proposition 12.7.3. The conjugate circle is the locus of the centers of the inscribed rectangular
hyperbolas (cf Section 11.6).
12.8
Longchamps circle
Definition 12.8.1. The Longchamps circle of ABC√is the conjugate circle of the antimedial
triangle. Center X(20), the Longchamps point, radius −Sa Sb Sc /S, equation :
Γstd + x a2 + y b2 + z c2 = 0
Proposition 12.8.2. The Longchamps circle is the locus of the auxiliary points of the inscribed
rectangular hyperbolas (cf Section 11.6).
12.9
Bevan circle
Definition 12.9.1. The Bevan circle is the circumcircle of the excentral triangle. Perspector
X(57), center X(40), radius 2R, equation :
Γstd + (−bcx − acy − abz) = 0
Proposition 12.9.2. Centers of similitude with Γ are in=X(165), and ex=X(1). Radical trace
X(1155), center axis X(517), radical axis X(513). Moses parametrization leads to Q (bad looking)
and Q0 = − (a + 2 b + 2 c) p + aq + ar.
Remark 12.9.3. On ETC n ≤ 3587, there are :
• 9 named points on the Bevan circle, namely : 1054, 1282, 1768, 2100, 2101, 2448, 2449, 2948,
3464
• 5 pairs of "true" inverses that both are named
• 2 pairs of named antipodal points
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
126
12.10. Spieker circle
12.10
Spieker circle
Definition 12.10.1. Spieker circle is the incircle of the medial triangle. Perspector X(2), center
X(10), radius r/2 (half the ABC inradius), equation :
Γstd +
1 X
x 5 c2 + 5 b2 − 3 a2 + 2 ac + 2 ab − 6 bc
16
Proposition 12.10.2. Centers of similitude with Γ are in=X(958), and ex=X(1376). Radical
trace not named, center axis X(515), radical axis X(522). Moses parametrization leads to :
Q = (b + c − a)
ab2 + ac2 + b3 − b2 c − c2 b + c3 p + a (q + r) a2 + ab + ac + 2 bc
Q0 = ab2 + c2 a + b2 c + c2 b − c3 − b3 p + a (q + r) +a2 − ab − ac + 2 bc , etc
Centers of similitude with the incircle are in=X(8) and ex=X(2).
Remark 12.10.3. On ETC n ≤ 3587, there are :
• 8 named points on the Spieker circle, namely : 3035, 3036, 3037, 3038, 3039, 3040, 3041,
3042
• no pairs of "true" inverses that both are named
• 2 pairs of named antipodal points [3035,3036], [3042,3042]
12.11
Apollonius circle
Definition 12.11.1. The Apollonius
circle is tangent to the three excircles and encloses them.
P
Center X(970), radius (abc + 6 a2 b) ÷ 8S, perspector not named, equation :
Γstd −
a + b + c X a2 + ab + ac + 2 bc
x=0
4
a
Proposition 12.11.2. Centers of similitude with Γ are in=X(573), and ex=X(386). Radical trace
not named, center axis X(511), radical axis X(512). Moses parametrization leads to bad looking Q
and
2
Q0 = − (b + c) (a + c) (a + b) p + ab + ac + bc + b2 + c2 a2 q + r ab + ac + bc + b2 + c2 a2 , etc
Centers of similitude with the nine-points circle are in=X(10) and ex=X(2051).
Remark 12.11.3. On ETC n ≤ 3587, there are :
• 8 named points on the Apollonius circle, namely : 2037, 2038, 3029, 3030, 3031, 3032, 3033,
3034
• no pairs of "true" inverses that both are named
• 1 pairs of named antipodal points [2037, 2038].
12.12
First Lemoine circle
Definition 12.12.1. The first Lemoine circle of ABC is obtained as follows. Draw parallels to
the sidelines of ABC through Lemoine point X6 . The six intersections of these lines with sidelines
are concyclic on the required circle. The following surd is useful :
W1 =
March 30, 2017 14:09
p
a2 b2 + a2 c2 + b2 c2
(12.12)
published under the GNU Free Documentation License
12. More about circles
127
Proposition 12.12.2. Center is X(182) (i.e. [O, K] midpoint), radius R÷2 cos ω = W1 R/ a2 + b2 + c2 ,
perspector :
2 a2 b2
b2
c2
a2
, 2 2
, 2 2
2
2
2
2
2
2
2
2
+ 2 a c + b c a c + 2 a b + 2 b c a b + 2 a2 c2 + 2 b2 c2
is not named, equation :
Γstd +
1
(a2 + b2 +
2
c2 )
X
x b2 + c2 b2 c2
This circle is concentric with and external to the first Brocard circle.
2
Proof. Difference of squared radiuses factors into abc/ a2 + b2 + c2 .
Proposition 12.12.3. Centers of similitude with Γ are in=X(1342), and ex=X(1343). Radical
trace X(1691), center axis X(511), radical axis X(512). Moses parameterization leads to bad looking
Q and Q0 . Poncelet centers of the pencil : X(1687) (inside) and X(1688) outside.
Proof. In order to see that X(1687) is inside, compute Ω (X(182)) × Ω (X (1687)) and obtain a
quantity that is clearly positive.
Remark 12.12.4. On ETC n ≤ 3587, there are :
• 2 named points on the first Lemoine circle, namely : 1662, 1663 (intersection with the Brocard
axis, X(3)X(6)).
• 7 pairs of "true" inverses that both are named :
"
3
6
32
39
2456 1691 1692 2458
371
2461
372
2462
1687
1688
#
• 1 pairs of named antipodal points [1662, 1663].
12.13
Second Lemoine circle
Definition 12.13.1. The second Lemoine circle of ABC is obtained as follows. Draw parallels
to the sidelines of orthic triangle through Lemoine point X6 . The six intersections of these lines
with sidelines are concyclic on the required circle.
Proposition 12.13.2. Center is X(6) itself, radius abc/ a2 + b2 + c2 , perspector X(3527), equation :
X
4
Γstd +
x Sa b2 c2 = 0
2
2
2
2
(a + b + c )
Centers of similitude with Γ are in=X(371), and ex=X(372). Radical trace X(1692), center axis
X(511), radical axis X(512). Moses parameterization
leads
to bad looking Q and Q0 . Poncelet
p P
P
centers of the pencil are involving radical 6
a2 b2 − 5 a4 and are not named. Moreover, the
second Lemoine circle is bitangent to the Brocard ellipse.
Remark 12.13.3. On ETC n ≤ 3587, there are :
• 2 named points on the second Lemoine circle, namely : 1666, 1667 (intersection with the
Brocard axis, X(3)X(6)).
• 5 pairs of "true" inverses that both are named :
"
#
3
576 1316 1351 2452
1570 1691 2451 1692 3049
• 1 pairs of named antipodal points [1666, 1667].
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
128
12.14. Sine-triple-angle circle
12.14
Sine-triple-angle circle
Definition 12.14.1. Define inscribed triangles T1 and T2 by the properties :
∠ (AB, AC) = ∠ (B1 A, B1 C1 ) = ∠ (C2 B2 , C2 A) , etc
the idea being to obtain isosceles "remainders". Then all the six vertices are on the same circle.
Figure 12.1: Sin triple angle circle
Proof. Using the tangent formula, the six points are easily obtained. T1 is a central triangle, and
each vertex of T2 is obtained by a transposition.


0


A1 '  a2 − ac − b2 a2 + ac − b2 a2 + b2 − c2 
a2 − bc − c2 a2 + bc − c2 c2
Proposition
12.14.2.
Center X(49), perspector not named, equation horrific, radius Rtsa =
2
3
2
R / |OH| − 2 R . Centers of similitude with Γ are in=X(1147), and ex=X(184), direction of
radical axis X(924). Moses parameterization leads to bad looking Q and :
Q0 = a2 a2 − c2 a2 − b2 p − a4 b2 + c2 − a2 (q + r) , etc
Remark 12.14.3. On ETC n ≤ 3587, there are :
• 6 named points on the Sine Triple Angle circle, namely : 3043, 3044, 3045, 3046, 3047, 3048
• 0 pairs of "true" inverses that are both named
• 1 pairs of named antipodal points [3043, 3047].
12.15
Brocard 3-6 circle and second Brocard circle
Definition 12.15.1. Brocard
circle has [X3 , X6 ] for diameter. Center X(182). Radius e R ÷
2 cos ω = W2 R/ a2 + b2 + c2 where :
p
W2 = a4 + b4 + c4 − (b2 c2 + a2 b2 + a2 c2 )
(12.13)
while the perspector :
a2
b2
c2
:
:
2a4 + b2 c2 2b4 + a2 c2 2c4 + a2 b2
is not named. Equation :
Γstd +
March 30, 2017 14:09
(a2
1
x b2 c2 + y c2 a2 + z a2 b2 = 0
2
2
+b +c )
published under the GNU Free Documentation License
12. More about circles
129
Proposition 12.15.2. Centers of similitude with Γ are in=X(1340), and ex=X(1341). Radical
trace X(187), center axis X(511), radical axis X(512). Moses parameterization leads to :
!
a2 a4 − a2 c2 − a2 b2 − 2 b2 c2 (p + q + r)
Q=
, etc
±W a2 b2 + a2 c2 + 2 b2 c2 − b4 − c4 p − a2 b2 + c2 − a2 (q + r)
Poncelet centers of the pencil are X(15) and X(16), the isodynamic points (see Section 13.7).
Remark 12.15.3. On ETC n ≤ 3587, there are :
• 4 named points on the first Brocard circle, namely : 3, 6, 1083, 1316. Moreover, this circle
goes through the Brocard points (cf 7.7.1).
• 47 pairs of "true" inverses that are both named
• 1 pairs of named antipodal points [3, 6].
Definition 12.15.4. First anti-Brocard circle: circumcircle of the first anti-Brocard triangle.
Equation:
X a2 b2 + bc + c2 b2 − bc + c2
Γstd −
x=0
W22
3
Definition 12.15.5. Second√Brocard circle is centered on X(3) and goes through the Brocard’s
centers. Radius eR = W2 R/ a2 b2 + b2 c2 + c2 a2 , while the perspector :
2a4
is not named. Equation :
−
Γstd +
a2 b2
a2 b2
a2
, etc
− a2 c2 + b2 c2
a2 b2 c2
=0
+ a2 c2 + b2 c2
Remark 12.15.6. On ETC n ≤ 3587, there are :
• 6 named points on the second Brocard circle, namely : 1670, 1671, 2554, 2555, 2556, 2557.
Moreover, this circle goes through the Brocard points (cf 7.7.1).
• 17 pairs of "true" inverses that are both named
6
15
16
32
61
39
3105
3104
3094
3107
62 3106
76
99
182 3095
371 3103
372 3102
1340 3558
1341 3557
1666 2563
1667 2562
1689 1690
2026 2561
2027 2560
• 3 pairs of named antipodal points [1670, 1671], [2554, 2555], [2556, 2557].
12.16
Orthocentroidal 2-4 circle
Definition 12.16.1. Orthocentroidal circle has [X2 , X4 ] for diameter. Center is X(381) and radius
R W3 ÷ 3abc where :
p
W3 = a6 + b6 + c6 − (a4 b2 + a4 c2 + a2 b4 + a2 c4 + b4 c2 + b2 c4 ) + 3a2 b2 c2
(12.14)
while the perspector :
b2 c 2
is not named. Equation :
Γstd +
+
2 a2
1
(b2 + c2 − a2 )
2
(x Sa + y Sb + z Sc ) = 0
3
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
130
12.17. Fuhrmann 4-8 circle
Proposition 12.16.2. Centers of similitude with Γ are in=X(1344), and ex=X(1345). Radical
trace X(468), center axis X(30), radical axis X(523). Moses parameterization leads to :
!
abc a4 + a2 b2 + a2 c2 − 2 b4 + 4 b2 c2 − 2 c4 (p + q + r)
Q=
, etc
±W a2 b2 + a2 c2 + 2 b2 c2 − b4 − c4 p − a2 b2 + c2 − a2 (q + r)
Poncelet points are real when triangle is acute.
Remark 12.16.3. On ETC n ≤ 3587, there are :
• 2 named points on the orthocentroidal circle, namely (antipodal) 2, 4
• 45 pairs of "true" inverses that are both named
Proposition 12.16.4. Let AH BH CH be the feet of the altitudes, and A0 = (A + 2AH ) /3, etc.
This circle goes through points A0 , B 0 , C 0 .
12.17
Fuhrmann 4-8 circle
Definition
√ 12.17.1. Fuhrmann circle has [X4 , X8 ] for diameter (see also ). Center X(355), radius
R W4 ÷ abc where :
p
W4 = a3 + b3 + c3 − (a2 b + a2 c + ab2 + ac2 + b2 c + bc2 ) + 3 abc
(12.15)
perspector not named (and not handy), equation :
Γstd +
2
(x aSa + y bSb + z cSc ) = 0
a+b+c
Remark 12.17.2. The only named points of this circle are X(2) and X(4). Five inverse pairs are :
1
11
72 2475 3434
80 1837 3419 3448 3436
12.18
Kiepert RH and isosceles adjunctions
Proposition 12.18.1. Kiepert RH. Chose angle φ and construct isosceles triangles BA0 C,
CB 0 A, AC 0 B with basis angle ∠ (BC, BA0 ) = φ (φ < 0 when A0 is outside). Then triangles
ABC and A0 B 0 C 0 are perspective wrt a point N (φ) :
N (φ) '
a
1
, etc '
, etc
2S cot φ − Sa
sin (A − φ)
(12.16)
and Kiepert RH is the locus of such points. Perspector of this conic is X(523), DeLongchamp point
at infinity, and center is X(115).
Proof. One has : A0 = a2 tan φ : 2 S − Sc tan φ : 2 S − Sb tan φ.
Remark 12.18.2. Triangle A0 B 0 C 0 is perspective from X(3) to the tangential triangle of ABC...
Proposition 12.18.3. For a given K, the points A0 B 0 C 0 are on the same cubic as the vertices,
the inexcenters, orthocenter, circumcenter and the points A0 B 0 C 0 relative to the opposite value of
K. This cubic can be written as K 2 + 1 K001 + 3 − K 2 K003 where K001 and K003 are the
standardized equations of, respectively, the Neuberg and the McKay cubics, as given in Proposi2
tion 17.4.11
and Proposition
17.4.10. The pole is X(6), while pivot is 3 vX(2) − K vX(20) i.e. :
2
2
2
1 + K s1 s3 : 3 − K s3 : 1 + K s2 .
Proof. Details are given in Proposition 17.4.8.
Remark 12.18.4. Points at infinity of Kiepert RH are parametrized by :
−1
2 W2
−1 2
cot φ =
1+ 2
cot (ω) =
a + b2 + c2 + 2 W2
2
2
3
a +b +c
12S
p
W2 =
a4 − a2 b2 − a2 c2 + b4 − b2 c2 + c4
is the Brocard radical
March 30, 2017 14:09
published under the GNU Free Documentation License
12. More about circles
131
Example 12.18.5. Fixed values of angle φ can be obtained by adding some regular shape to each
side of the reference triangle. In Figure 12.2, a "cake server" (Pelle à Tarte) likeAF U GC, made
z }| {
of a square AF GC and an equilateral triangle F U G defines angle φ = AU, AC = −75◦ . When
the triangle
the square, we obtain a Pelle Pliée (folded shovel) like AF W GC that defines
zis inside
}| {
angle φ = AU, AC = −15◦ .
Figure 12.2: Cake server (Pelle a Tarte)
Example 12.18.6. Arbelos. Another idea to obtain some adding object is as follows. Divide
[AB] in a given ratio, obtaining D. Use the same ratio to obtain E, F . Draw the half circles
and perpendicular DM , then the√tangent
circles. The triangle of the centers is perspective with
5 − 1 /2. Perspector is X(2672). And belongs to Kiepert RH.
ABC if and only if the ratio is
√
Therefore, we have isosceles triangle. Value is ± tan φ = 3 − 5 (plus is inward, minus is outward).
Figure 12.3: Arbelos configuration
Example 12.18.7. When using the center of a regular triangle, square, pentagon, we have φ = 30◦ ,
φ = 45◦ , φ = 54◦ ; when using the farthest vertex (or the midpoint of the farthest side), φ = 60◦ ,
.
φ = at2 = arctan 2, φ = 72◦ . We even have points relative to at3 = arctan 3. Points are collected
into 2 × 2 blocks corresponding to −φ, φ − 90◦ , 90◦ − φ, φ.
−90◦
4
4
90◦
H
−at2
1131
1132
at2
0
2
2
0
G
| −75◦
| 3391
| 3367
|
75◦
| P aT
at2 − 90◦
3316
3317
◦
90 − at2
−15◦
3366
3392
15◦
PPl
| −72◦ −18◦
| 1139 3370
| 1140 3397
|
72◦
18◦
|
P enta
| −60◦ −30◦
|
13
17
|
14
18
◦
| 60
30◦
|
F ermat
| −54◦
| 3381
|
?
|
54◦
|
| −at3
| 1327 ?
| 1328 ?
| at3
|
−36◦
?
3382
36◦
| −67.5◦
|
3387
|
3374
|
67.5◦
|
|
−arb
| ? 2671
| ? 2672
|
arb
| Arbelos
−22.5◦
3373
3388
22.5◦
| −45◦ −45◦
|
485
485
|
486
486
◦
|
45
45◦
|
V ecten
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
132
12.18. Kiepert RH and isosceles adjunctions
Example 12.18.8. Brocard angle is often involved since :
Pφ
'
1
, etc
(b2 + a2 + c2 ) cot (φ) − Sa cot (ω)
This leads to four new blocks :
−2 ω 2 ω − π2
3407
3399
1916
3406
π
2ω
2 − 2ω
Gibert
|
|
|
|
|
−ω ω − π2
83
262
76
98
π
ω
2 −ω
Brocard
| − 12 ω 12 ω − π2
| 1676
?
|
?
1677
1
π
1
|
2 ω
2 − 2 ω
| Lemoine circle
Proposition 12.18.9. Let σ be a constant
through point T (σ) where :

a2 Sa
 2
T (σ) '  b Sb
c2 Sc
|
|
|
|
|
− π4 − 12 ω 12 ω − π4
?
2009
2010
?
π
1
π
1
4 + 2 ω
4 − 2 ω
Galatly circle
value. Then all lines N (φ) N (σ − φ) are passing

a2



 + 2S cot σ  b2 
c2


Therefore all the N (φ) N (−φ)
lines (rows in a block) are passing through X(6), the Lemoine point.
And all the N (φ) N π2 − φ lines (columns in a block) are passing through X(3) the circumcenter.
All the T (σ) are on the Brocard axis X(3)X(6).
Proof. Obvious from (12.16).
Proposition 12.18.10. Let δ be a constant value. Then all lines N (φ) N (φ + δ) are tangent to
a conic D (δ). When δ = 0 this conic is the Kiepert RH itself. When δ = π/2 (diagonals in a
block),Qthe conic reduces to a real point : the Euler center X(5). In the general case, D (δ) =
16S 2
a2 − b2 cot2 (δ) D (0) + D (π/2).
Proof. This can be proved using the usual techniques : differentiate and wedge. When δ is rational
wrt π, conics D (δ) and D (0) are in a Steiner configuration.
Lemma 12.18.11. Consider circle C0 : x2 + y 2 = 1 and points U = (u, 0), V = (v, 0) in the
Cartesian plane. When u + v 6= 0, these points are the centers of similitude of circles C0 and
C (P, ρ) where
u − v
2 uv
P =
, 0 , ρ = u+v
u + v
Proof. Consider M = (0, +1) and N = (0, −1). Their counterparts in circle C are obtained by the
intersections J = U M ∩V N and K = U N ∩V M . We have P = (J + K) /2 and ρ = |J − K| /2.
Proposition 12.18.12. A circle C (P, ρ) can be found such that points U = N (φ), V = N (φ + π/2)
are the centers of similitude between C and the nine-points circle. We have :




a2 Sa
a2




P ' cot (2 φ)  b2 Sb  − 2 S  b2  ∼
= cot (2 φ) X3 − 2 S X6
2
2
c
c Sc
p
abc 1 + cot2 (2 φ)
ρ =
2
a + b2 + c2 − 4 S cot (2 φ)
Proof. These two points are aligned with E=X(5). Define W = (U + V ) /2. From the above
lemma, we have :
P = E + k (U − E) ; ρ = (1 − k)
R
2
where k =
V −E
W −E
Computations are greatly simplified when remarking that the results depends not really from cot φ
itself, but rather from cot φ − 1/ cot φ. That is the reason why all these formulas are involving
cot (2φ).
March 30, 2017 14:09
published under the GNU Free Documentation License
12. More about circles
133
φ
0
U
2
V
4
P
3
π/12
3392
3391
16
−π/12
3366
3367
15
π/10
−π/10
3397 1139 3393
3370 1140 3379
arctan (3)
1328
?
?
− arctan (3)
1327
?
?
π/8
3388
3387
372
−π/8
3373
3374
371
arctan (2)
1132
3316
?
− arctan (2)
1131
3317
?
π/6
18
13
62
−π/6
17
14
61
π/5
−π/5
3382
?
3381
?
3395
3368
2672
?
?
2671
?
?
486
485
6
10
2051
970
?
?
?
2ω
1916
3399
?
−2 ω
ω
3407
76
3406
?
262 3095
−ω
83
98
3398
ω/2
?
?
511
−ω/2
1676
1677
182
π/4 + ω/2
2010
2009
39
−π/4 − ω/2
?
?
32
arctan 3 −
√ 5
− arctan 3 −
π/4
√ 5
2
(a + b + c)
4S
2
(a + b + c)
−arccot
4S
arccot
ρ
R
√
2 3abc
√
12 S − 3√
(a2 + b2 + c2 )
2 3abc
√
12 S + 3 (a2 + b2 + c2 )
5 abc
16 S + 3 (a2 + b2 + c2 )
5 abc
16 S − 3√
(a2 + b2 + c2 )
2abc
2
2
2
4 S − (a
√ +b +c )
2abc
4 S + (a2 + b2 + c2 )
5 abc
12 S + 4 (a2 + b2 + c2 )
5 abc
12 S − 4 (a2 + b2 + c2 )
2 abc
√
4 S − 3 (a2 + b2 + c2 )
2 abc
√
4 S + 3 (a2 + b2 + c2 )
−−
−−
√
3 4 5 + 5 abc
√ 2
2
2
44 S − 2 9 +√
5 5 (a
+b +c )
3 4 5 + 5 abc
√ 44 S + 2 9 + 5 5 (a2 + b2 + c2 )
abc
a2 + b2 + c2
r2 + p2
4r
name
circum
2◦ Lemoine
Apollonius
a2 b2 + a2 c2 + b2 c2 R
a4 + b4 + c4 + a2 c2 + a2 b2 + b2 c2
R
a2 b2 + a2 c2 + b2 c2 R
a4 + b4 + c4 + a2 c2 + a2 b2 + b2 c2
∞
R
2 cos (ω)
R sin (ω) abc b2 + a2 + c2
2 (b4 + c4 + a4 ) cos (ω)
Linf ty
1◦ Lemoine
Gallatly
Table 12.4: Similicenters on Kiepert RH
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
134
12.19
12.19. Cyclocevian conjugate
Cyclocevian conjugate
Definition 12.19.1. Two points are called cyclocevian conjugates when their cevian triangles
share the same circumcircle. This definition has to be compared with cyclopedal conjugacy, see
Section 8.3.
Proposition 12.19.2 (Terquem). Each point not on the sidelines has a cyclocevian conjugate. Its
barycentrics are given by (Grinberg, 2003a) :
U = cyclocevian (P ) = (isot ◦ anticomplement ◦ isog ◦ complement ◦ isot) (P )
Proof. This proposition asserts that the other intersections of the circumcircle of AP BP CP with
the sidelines are the vertices of another cevian triangle. As it should be, the formula is involutory.
The center of the common circumcircle is the middle of [P U ].
Example 12.19.3. Some examples :
point
code
bary
cycc
circumcenter
Gergonne
centroid
orthocenter
Nagel
X (7))
X (2)
X (4)
X (8)
1/ (−a + b + c)
1
1/ −a2 + b2 + c2
−a + b + c
X (7)
X (4)
X (2)
X (189)
X (1)
X (5)
X (5)
X (1158)
Point X7 is the only center that is invariant by cyclocevian. Three other points share this
property, obtained by changing one of the a, b, c into its opposite in the barycentrics of X7 .
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 13
Pencils of Cycles in the Triangle Plane
13.1
13.1.1
Introductory remarks
How many points at infinity should be used ?
In Chapter 7, orthogonality of lines has been reduced to a polarity wrt operator M . Here, the
same treatment will be applied to cycles, i.e. the family of all curves that are either a circle or a
line. This leads to a fundamental quadric Q in space PR R4 . Finite points on the quadric can
be interpreted as representatives of point-circles, quite the same thing as an ordinary point in the
triangle plane. Points outside this quadric are representatives of cycles, while points inside are
assigned to virtual circles.
When dealing with tangency of cycles, the best description is given by a Lie sphere, embedded
into a 5D space and obtained by a double coating of the former 4D space (oriented cycles). A
special attention is devoted to objects at infinity –especially the umbilics– since all these projective
spaces are tailored to implement the Poncelet’s continuity principle.
Efficient notations are powerful, poor notations can be confusing. In the opinion of the author,
this kind of problem arises when studying pencils and bundles of circles in the context of the
Triangle Geometry. On the one hand, the usual context for studying pencils and bundles of circles
is Riemann sphere PC C2 where (z1 , z2 ) ' (λz1 , λz2 ) for any
non-zero λ ∈ C (Poncelet, 1822,
1865). On the other hand, a Triangle point lives in PR R3 i.e. is described as x : y : z in a
projective space where x : y : z = kx : ky : kz for any non-zero k ∈ R.
Obviously, both points of view are reducing to the same elementary Cartesian coordinates when
restricted to the finite domain. But they are conflicting where they are the most useful, i.e. where
they are implementing the continuity principle for objects at infinity. An ordinary line must be
completed in a way or another to become a "circle with infinite radius" and having a clear definition
of this completion is required in order to unify the three concepts of circle (0 < ρ < ∞), point
(ρ = 0) and line (ρ = ∞) into a single
concept of cycle.
In the Riemann sphere PC C2 , there exists only one point at infinity (noted ∞). In this context,
a "circle with infinite radius" is an ordinary line ∆ completed by point ∞, i.e. ∆ = ∆ ∪ {∞},
while point-circles are either circles
with radius 0 around an ordinary point or {∞}.
In the Triangle Plane PR R3 , there exists a whole line L∞ of points at infinity, verifying
x + y + z = 0. In this document, barycentrics are used. Using trilinears would only change some
formulas, but not the very nature of the underlying space. In this context, the barycentric equation
of an ordinary circle leads to define a cycle as a second degree curve (a conic) that goes through the
umbilics Ω± . Therefore, the equation of a completed line becomes (x
+ y + z) (ux + vy + wz) = 0
so that we must define ∆ as ∆ ∪ L∞ , while the role of {∞} in PC C2 is played now by the horizon
2
circle C∞ defined by (x + y + z) = 0, i.e. as an object having the same points as L∞ but each of
them counted twice.
13.1.2
Umbilics
Lemma 13.1.1. We have OrtO . OrtO W = − W . Therefore, restricted to V, OrtO
nothing but an half turn. Multiplied by P yth , this leads to OrtO
135
3
2
is
+ OrtO = 0, so that
136
13.1. Introductory remarks
eigenvalues of OrtO are 0, +i, −i.
Definition 13.1.2. Umbilics. Let U ∈ L∞ be a point at infinity, .i.e a point such that u+v +w =
0. The associated umbilics are the complex points :
Ω+ ' 1 − i OrtO . U ; Ω− ' 1 + i OrtO . U
Caveat : signs are crossed ! A possible choice is :
Ω+
'
'
4SX512 − i X511




a2 b2 − c2
a2 a2 b2 + c2 − b4 − c4




4S  b2 c2 − a2  ± i  b2 b2 c2 + a2 − c4 − a4 
c2 a2 − b2
c2 c2 a2 + b2 − a4 − b4
where i is the imaginary unit and S the area of the triangle. Another choice is :


Sb − 2 iS


Ω+ '  Sa + 2 iS 
−c2
(13.1)
(13.2)
This is not a symmetrical expression, but computations are easier.
Proposition 13.1.3. When seen as elements of VC , all Ω± are eigenvectors of operator OrtO ,
with eigenvalues
(respectively) ±i and therefore belong to the light cone. When seen as elements
of PC C3 , points Ω+ and Ω− are now independent of the choice of U , and are the fixed points
of the orthopoint transform. They both belong to the circumcircle, and are isogonal conjugates of
each other.
Proof. See Postnikov (1982, 1986) for better insights on real-complex spaces. Property L∞ . Ω = 0
is obvious. Umbilics are eigenvectors because of
OrtO . Ω+ = OrtO . 1 − i OrtO . U = OrtO + i . U = +iΩ+
Since eigenvalues
of OrtO are simple, the C-dimension of eigenspaces is one, and uniqueness
in PC C3 follows. For this reason, points Ω± are also called the "circular points at infinity".
Finally, intersection of circumcircle and the infinity line must be invariant by isogonal conjugacy,
while Ω+ ∗ Ω+ cannot be real, even up to a complex proportionality factor.
b
Proposition 13.1.4. A circle is a conic that goes through the umbilics. Above all, choosing the
umbilics is deciding which of the circum- ellipses is the circumcircle.
Proof. By definition, umbilics are the (non real) points where line at infinity intersects the circumcircle. By (12.3), these points belong to any circle. For the converse, consider the values taken by
x2 , y 2 , z 2 , xy, yz, zx at A, B, C together with both umbilics. This gives a 5 × 6 matrix whose rank
is 5 : the first three lines are 1, 0, 0, 0, 0, 0 etc. and it remains only to show that rank of submatrix
4..5,4..6 is two. A direct inspection shows that a critical factors are a2 + b2 − c2 (straight angle,
that can occur only once) and a4 + b4 + c4 − b2 c2 − a2 b2 − a2 c2 (condition of equilaterality). In
such a case, the property remains when umbilics are written as 1 : j : j 2 and 1 : j 2 : j.
Remark 13.1.5. When the umbilics are given, the Euclidean structure of the Triangle Plane is
known. From Ω+ ∗ Ω− = a2 : b2 : c2 , the P yth matrix is known (up to the value of R), while
b
the orthopoint transform, and its matrix OrtO is characterized by its diagonal shape, namely
(0, +i, −i), when using the triple X(3), Ω+ , Ω− as barycentric basis.
13.1.3
Notations
We have done our best effort to use unified notations. In this whole chapter,
Notation 13.1.6. Conventions about letters.
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published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
137
P, Q
denote some flat true points in the Triangle Plane and Xn a Kimberling ETC-named
triangle center, all of them being 3-columns.
Γ, Ω
Γ denotes the circumcircle of the fundamental triangle ABC (and nothing else) while
Ω denotes a cycle, both of them being curves, i.e. a set of points. The 3 × 3-matrices
describing their equations as a quadratic form are noted using a box.
U, A
U denotes the representative of a circle-point, while A denotes the representative of
any cycle, all of them being columns in PR R4 .
denotes the representative of an oriented cycle in PR R5
Y
G
denotes a Gram matrix. Its diagonal elements are noted w2 along the diagonal and W
for the non-diagonal elements.
Notation 13.1.7. To design a circle known by a pair center/radius, parentheses will be used, exemplified by Γ = (X3 , R). Using parentheses around a single Roman letter –e.g. (P )– will be
reserved to denote the circle (P, 0) i.e. the circle whose unique real point is P itself.
13.2
Cycles and representatives
Definition 13.2.1. The (barycentric) Veronese map is the correspondence that maps a PR R3
column into a PR R4 row proportional to:
t




x (x + y + z)
y (x + y + z)
z (x + y + z)
2
−a yz − b2 xz − c2 xy





Proposition 13.2.2. The Veronese map is obviously homogeneous. Both umbilics are send to
[0, 0, 0, 0] (points of indeterminacy). The other points at infinity are send to [0, 0, 0, 1]. Otherwise,
the map is injective.
Definition 13.2.3. Veronese map. For points at finite distance, we will use (7.13) and define
V er (x, y, z) by the simpler formula:
V er (x : y : z) = [x, y, z, Γstd (x, y, z)]
(13.3)
Remark 13.2.4. Requiring that four points are on the same circle leads to Proposition 12.1.4, i.e.
to equation :
i=4
det [pi , qi , ri , Γstd (pi , qi , ri )] = 0
i=1
But, conversely, this equation only implies that our four points are on the same circle or on the
same straight line. To summarize both situations under a single concept, we introduce :.
4
Definition 13.2.5. The cycle Ω associated
with
the
representative
A
'
u
:
v
:
w
:
t
∈
P
R
Ω
R
is the locus of the points X ∈ PR R3 that satisfies equation :
V er (X) · (u : v : w : t) = 0
(13.4)
For example, the representative of circumcircle Γ is AΓ ' 0 : 0 : 0 : 1.
Remark 13.2.6. When t 6= 0, cycle Ω is the (ordinary) circle whose standardized equation is :
v
w
u
x + y + z + Γstd (x, y, z)
t
t
t
(13.5)
Remark 13.2.7. The representative of a circle is seen as a 3D point (described by a column). The
Veronese of a 2D point is an action over the 3D points: i.e. a plane, described by a row.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
138
13.2. Cycles and representatives
Definition 13.2.8. Cycle C∞ represented by A∞ ' 1 : 1 : 1 : 0 has to be understood as the
line at infinity L∞ described twice, and will be called the horizon circle. This object has to be
perceived as a circle "whose center is everywhere and circumference nowhere" (Empedocles).
The representative itself, i.e. the point A∞ ' 1 : 1 : 1 : 0, will be called Sirius, following
Kimberling ETC (1998-2014) in using stars to coin the name given to a point. While using that
specific star for a very distant point is from (Voltaire, 1752).
Definition 13.2.9. Otherwise, the cycle represented by u : v : w : 0 is the union of an ordinary
line and L∞ , and will be called a completed line.
Theorem 13.2.10. The representative of the point-circle (P ) associated with a point at finite
distance P ' p : q : r is given by:


0 c2 b2 1
 c2 0 a2 1 


t
(13.6)
where Q−1 =  2
UP ' Q−1 · V er (P )

 b a2 0 1 
1
1
1
0
2
UP = c2 q 2 + b2 r2 + 2 Sa qr : c2 p2 + a2 r2 + 2 Sb pr : b2 p2 + a2 q 2 + 2 Sc pq : (p + q + r)
(13.7)
Conversely, we have:
V er (P ) ' t UP · Q
where Q =
1
16S 2





a2
−Sc
−Sb
−a2 Sa
−Sc
b2
−Sa
−b2 Sb
−Sb
−Sa
c2
−c2 Sc
−a2 Sa
−b2 Sb
−c2 Sc
a2 b2 c2





(13.8)
Proof. Direct computation. Mind the fact that, written that way, Q · Q−1 = −1/2 !
Remark 13.2.11. It must be taken into account that radiuses that are not "up to a proportionality
factor". The values of matrix Q and Q−1 were chosen to obtain the best looking formulas at
(13.9) and (13.13). But this −2 tries to reappear at (13.16): be prudent !
Remark 13.2.12. Sometimes, "the representative of the point-circle P " will be shortened into "the
representative of P ". This object (a column) is not to be confused with the Veronese image of P
(a row) !
Example 13.2.13. Here are some point representatives :
P \x
u
1:0:0
0
0:1:1
2 b2 + 2 c2 − a2
X(1)
bc (b + c − a)
X(2)
2 b2 + 2 c2 − a2
R2
X(3)
2
X(4) R2 a2 b2 + c2 − a2
X(6)
b2 c2 2 b2 +2 c2 −a2
umb
0
∞
1
v
c2
a2
ac (c + a − b)
2 a2 + 2 c2 − b2
R2
2
R2 b2 c2 + a2 − b2
a2 c2 2 c2 +2 a2 −b2
0
1
w
b2
a2
ab (b + a − c)
2 b2 + 2 a2 − c2
R2
2
R2 c2 a2 + b2 − c2
a2 b2 2 a2 +2 b2 −c2
0
1
t
1
4
a+b+c
9
1
a2 b2 c2
2
a2 +b2 +c2
0
0
The fact that formula (13.7) *would* give Sirius ' 1 : 1 : 1 : 0 for each point on L∞ is the
reason of their exclusion from the definition of the point representatives.
Corollary 13.2.14. The representative of the circle (P, ω) where P ' p : q : r is a point at
finite distance is obtained by :
P
Q−1 · t V er
− ω 2 Sirius
(13.9)
p+q+r
t
where Sirius is exactly [1, 1, 1, 0] and Q−1 is exactly as in (13.6).
March 30, 2017 14:09
published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
139
Proof. Obvious from the above theorem.
Example 13.2.15. Compute the representative of the incircle. Two equivalent methods are:
mQQI.(Tr@FActor@Ver@norb@vX)(1)-ri^2*Sirius:
method1:= (FActor@subs)(kitcircleS, valS6, %);
method2:= (nor4@wedge3)(seq(Ver(j), j=Column(matcev(vX(7)), 1..3)));
1 2
2
2
(a − b − c) : (b − c − a) : (c − a − b) : 4
4
Remark 13.2.16. Assuming that representatives are living in PR R4 has many advantages. The
top one could be to enforce the fact that a representative is not a point in the Triangle Plane. It
is a key fact that the triple [b
u, vb, w]
b appearing in the standardized equation (13.5) is ***not***
defined up to a proportionality factor. The same remark applies to the so-called "circle function"
[b
u ÷ bc, vb ÷ ca, w
b ÷ ab] ∈ R3 that appears when using trilinears as in Weisstein (1999-2009).
incircle =
13.3
Fundamental quadric and orthogonality
Theorem 13.3.1. Any point representative U = u : v : w : t belongs to the quadric Q :
t
U · Q ·U =0
=
0
(13.10)
where Q is as given in (13.8).
t
Proof. One has V er (P ) . Q−1 · V er (P ) = 0 by the very definition of V er (P ).
Proposition 13.3.2. Object Sirius ' 1 : 1 : 1 : 0 is the only (real) point at infinity of the quadric
Q. Therefore, Q is a paraboloid.
Proof. Substitute t = 0, then compute the discriminant with respect to w and obtain −(u −
v)2 a2 b2 c2 /R2 . This requires u = v, etc.
Remark 13.3.3. In the usual PC C2 model, L∞ is "in the South plane" while the horizon circle
C∞ is nothing but the point-circle {∞}.
Proposition 13.3.4. An element A = u : v : w : t of PR R4 is the representative of a (real) cycle
t
if only if A is outside of Q (i.e. on the same side as 0 : 0 : 0 : 1 characterized by (u : v : w : t) ·
Q · (u : v : w : t) ≥ 0 when (13.10) is used.
Proof. Obvious from (13.9), that states that representative of (P, ω) is "below"
the representative
of (P ) , while representatives of completed lines are at infinity in PR R4 and therefore outside of
paraboloid Q.
Theorem 13.3.5. Orthogonal cycles. Consider two cycles Ω1 , Ω2 with representatives A1 ,A2 .
When A2 belongs to the polar plane of point A1 wrt the fundamental quadric then cycles Ω1 and
Ω2 are orthogonal –and conversely.
Computed Proof. Begin with two circles. Write representative Aj as in (13.9) from representative
Uj of point-circle (Pj ).This implies that Aj [4] = 1. Compute t A1 · Q · A2 and –using (7.11)–
obtain :
1 2
2
t
A1 · Q · A2 =
ω1 + ω22 − |P1 P2 |
(13.11)
2
Compute now t A1 · Q · A3 where A3 = u3 : v3 : w3 : 0 and obtain :
t
A1 · Q · A3 = −
1
(p1 u3 + q1 v3 + r1 w3 ) ÷ (p1 + q1 + r1 )
2
(13.12)
In both cases, the result is the orthogonality condition times a non vanishing factor. Finally, when
the representatives of two lines are involved, the conclusion follows directly from the properties of
the orthopoint transform.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
140
13.3. Fundamental quadric and orthogonality
Corollary 13.3.6. The locus of the representatives of the points of a given cycle Ω is the intersection between Q and the polar plane –wrt Q– of the representative of Ω.
Proof. By definition, point P belongs to cycle Ω if and only if Ω and (P ) are orthogonal.
Theorem 13.3.7. Back to barycentrics. Let A = u : v : w : t ∈ PR R4 be a representative.
When t = 0 then either A ' 1 : 1 : 1 : 0 and A is Sirius, the representative of the horizon circle or
A 6' 1 : 1 : 1 : 0 and A represents a line. When t 6= 0, then A represents a circle (may be reduced
to a point). The associated squared radius is given by :
t
ω2 =
A· Q ·A
(13.13)
t2
The representative of the center is obtained as U = A/t + ω 2 Sirius ∈ Q while the center itself is
P = p : q : r where :
(13.14)
[p, q, r, θ] ' t A · Q
Proof. The radius formula is a corollary of the preceding theorem. Let A = x : y : z : τ be any
cycle representative and U ∈ Q be the representative of point P = p : q : r. Then t W · Q · U = 0
implies
xp + yq + zr + τ Γstd (p, q, r) = 0
where Γstd (p, q, r) = − a2 qr + b2 rp + c2 pq ÷ (p + q + r), so that equation (13.14) must hold for
rank reason (and can be checked directly). Conversely, starting from any A and applying (13.14)
and then (13.7) leads back to A.
Example 13.3.8. Re obtain the radius of the incircle.
subs(kitcircleS, valS6, %/ri^2); 7→
factor(Tr(incircle).mQQ.incircle);
1
Proposition 13.3.9. Power of a point wrt a circle. When V is the Veronese of a point M and A
is the representative of circle (P, ω), then
.
2
power (M, (P )) = |P M | − ω 2 =
V
A
·
V1 + V2 + V3 A4
(13.15)
Proof. Obvious from definitions. Mind the normalizations !
Theorem 13.3.10. Common orthogonal cycle. Let be given three cycles Ω1 , Ω2 , Ω3 . If they
don’t belong to the same pencil, the bundle they generate is exactly the set of all the cycles orthogonal
to a fixed cycle Ω⊥ . We have the formulas (see (13.8), (13.6) for the exact values):
W
A⊥
center
.
= A1 ∧ A2 ∧ A3
=
(a row)
(13.16)
t
Q−1 · W
= W1 : W2 : W 3
t
A⊥ · Q · A⊥
−1 W · A⊥
2 (A⊥ [4])2
−1
Proof. Let j = 1, 2, 3. By definition, W · Aj = 0, so that t Aj · Q · Q
· t W vanishes. Then
the center follows by (13.14) and the radius as well. Mind the normalizing factor (see 13.2.11 )
and don’t simplify anything in the formula giving the radius !.
squared radius =
(A⊥ [4])
2
=
deprecated
Example 13.3.11. Compute the radius of the Euler circle (useless factor K, means whatever).
tmp1:= K*(wedge3)(seq(mQQI.Tr(Ver(j)), j= Column(matcev(vX(2)),1..3) )):;
tmp2:= FActor(mQQI.Tr(tmp1)): # (Tr@reduce)(tmp1[1..3]); ency(%);
methode1:= (-1/2)* tmp1.tmp2/tmp2[4]^2: subs(rapbpc, factor(%)); 7→ R2 /4
methode2:= Tr(tmp2).mQQ.tmp2/tmp2[4]^2: subs(kitRcircle, factor(%));
March 30, 2017 14:09
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13. Pencils of Cycles in the Triangle Plane
141
Definition 13.3.12. Radical center. The ever visible point W1 : W2 : W3 in the former theorem
is called the radical center of the three cycles. Having the same power wrt all the cycles of the
bundle is a characteristic property of this point.
Remark 13.3.13. As emphasized later, the nature of the radius of Ω4 , i.e. real, zero or imaginary
fixes the nature of the bundle defined by Ω1 , Ω2 , Ω3 .
Example 13.3.14. The example of the three excircles is examined in Subsection 13.8.4. Center
is X(10) and radius is
s
b2 c + ab2 + bc2 + a2 b + ac2 + a2 c + acb
ω4 =
4 (a + b + c)
13.4
Pencils of cycles
Definition 13.4.1. Pencil. When Ω1 , Ω2 are distinct cycles (with non proportional representatives), all curves λ1 Ω1 + λ2 Ω2 = 0, where (λ1 , λ2 ) 6= (0, 0), are cycles and the set of all these cycles
is called the pencil generated by Ω1 , Ω2 . It is clear that representatives of the cycles of a given
pencil are on the same projective line in PR R4 –called the representative of the pencil.
Example 13.4.2. Formula (13.5) describes circle Ω as a member of the pencil generated by the
circumcircle and a completed line. Therefore, the ordinary line ux + vy + wz = 0 is the radical
axis ∆ of both circles Ω and Γ. That’s another way to see that knowing u : v : w is not enough to
determine a circle.
−1
· t V er (P ) − ω 2 C∞ , describes the circle
Example 13.4.3. Formula (13.9), i.e. Ω (P, ω) = Q
(P, ω) as a member of the pencil generated by the point-circle (P ) and the horizon circle, i.e. the
pencil of all circles concentric with (P, 0).
Remark 13.4.4. Here again, the triple u : v : w is not sufficient to specify P , and [u; v; w; t] must
be used. It can be checked that representative is well specified, i.e. doesn’t depends on whichever
triple (kp, kq, kr) is chosen as barycentrics of point P .
13.5
Classification of pencils
Theorem 13.5.1 (Classification). Pencils of cycles
fall in three classes, depending on the way
their representative line P intersects –in PR R4 – the fundamental quadric Q.
Q, P tangent : P is the tangent pencil of all the cycles containing a given point ω0 and tangent
at ω0 to a line ∆1 containing ω0 . Archetype : ω0 = ∞ and P is "all the lines parallel
to a given line ∆1 ".
Q, P secant : P is the isotomic pencil generated by two different point-circles {ω1 } and {ω2 }
(ωi are the limit points of P). Archetype : ω2 = ∞ and P is, apart from {∞}, "all
the circles centered at a finite point ω1 ".
Q, P disjoint : P is the isoptic pencil of all the cycles going through two different points ω1 and
ω2 (the base points). Archetype : ω2 = ∞ and P is "all the lines through a finite point
ω1 ".
When P is a tangent pencil, so is P ⊥ (using ω0 and ∆⊥
1 orthogonal to ∆1 at ω0 ). When P is
isoptic (ω1 , ω2 ) then P ⊥ is isotomic (ω1 , ω2 ) and conversely. In all cases, representative of P and
P ⊥ are conjugate lines wrt Q.
Proof. Everything goes as in (Pedoe, 1970) –using another paraboloid– or (Douillet, 2009) –using
a sphere. The
complex plane, namely
only striking thing is that the usual point at infinity of the
2
∞ ∈ PC C2 , has to be replaced by the horizon circle C∞ : (x + y + z) = 0.
Proposition 13.5.2. A pencil of cycles that contains two lines is a pencil of lines. A concentric
pencil contains the horizon cycle. All other pencils (i.e. all the non archetypal pencils) contain
exactly one straight line (the radical axis of the pencil).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
142
13.5. Classification of pencils
Proof. The representative of P ever intersects the plane at infinity of PR R4 .
Proposition 13.5.3. Let P = p : q : r be a point in the Triangle Plane. Define its shadow in the
Triangle Plane as point S = u : v : w where u, v, w are defined in (13.7). Then S is not outside the
inconic IC(X76 ). Any point on the border of IC (X76 ) is the shadow of exactly one point on the
circumcircle, while a point inside IC (X76 ) –except from X2 – is the shadow of exactly two points.
Moreover, these points are inverse in the circumcircle.
Remark 13.5.4. Figure 13.1 shows the shadows of all the named points in ETC (Kimberling ETC,
1998-2014), using the standard values a = 6, b = 9, c = 13. One can see two lines of points :
L (X2 , X6 ) and L (X2 , X39 ) containing the shadows of points from L (X3 , X2 ) –Euler line– and
L (X3 , X6 ) –Brocard axis– respectively.
Figure 13.1: No point-shadow fall outside of the IC(X76) inconic
Proof of Proposition 13.5.3 . The locus of representatives
of the points P0 that belongs to Γ is the
Q
intersection of quadric Q and the polar plane
of 0 : 0 : 0 : 1, namely :
ua2 b2 + c2 − a2 + vb2 c2 + a2 − b2 + wc2 a2 + b2 − c2 − 2 ta2 b2 c2 = 0
Extracting t and substituting in Q leads (apart from a constant non-zero factor) to :
u2 a4 + b4 v 2 + c4 w2 − 2 uva2 b2 − 2 vwb2 c2 − 2 wuc2 a2 = 0
i.e. the equation of IC (X76 ).
When two points P1 , P2 in the Triangle
Plane share the same shadow, then points U1 , U2
and 0 : 0 : 0 : 1 are collinear in PR R4 so that cycles (P1 ), (P2 ) and Γ belongs to
Q the same
.
pencil. Therefore P1 , P2 are inverse in the circumcircle. Moreover U0 = U1 U2 ∩
is inside
IC (X76 ) –otherwise U0 would be the representative of a real circle belonging to pencil (P1 ) , (P2 )
and orthogonal to Γ.
Theorem 13.5.5. Inversion in a cycle. Let Ω0 be a fixed cycle with representative A0 and Ω1
c1 the interanother cycle with representative A1 . Assume that Ω0 is not a point-circle and call A
c1 , A1 , A3
section of line A0 A1 with the polar plane of A0 . Construct A3 such that division A0 , A
is harmonic. Then A3 represents the cycle Ω3 inverse of Ω1 in cycle Ω0 (inversion in a straight
line is the ordinary reflection in this line) and is given by :
t
A1 · Q · A0
.
A3 ' σ (A1 ) = A1 − 2 t
A0
A0 · Q · A0
(13.17)
Moreover, when A2 is yet another cycle representative, we have the conservation law :
t
March 30, 2017 14:09
σ (A1 ) · Q · σ(A2 ) = t A1 · Q · A2
(13.18)
published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
143
c1 as α1 A0 +A1 in t A0 · Q · A
c1 = 0 and then obtain A3 as 2α1 A0 +A1 since division
Proof. Write A
(∞, 1, 0, 2) is harmonic. Equation (13.18) is obvious from (12.9), and shows that σ preserves
orthogonality. Moreover, (12.9) shows that cycles orthogonal to Ω0 are invariant while cycles
concentric with Ω0 are transformed into cycles concentric with Ω0 : all together, this proves that
σ is the inversion in cycle Ω0 .
Proposition 13.5.6. Points inside of Q are representative of virtual circles (real center, imaginary
radius). The reason to imagine such circles is that inversion in such a circle is a real transform.
Moreover a real cycle Ω is orthogonal to (X, iω) when Ω cuts (X, ω) along a diameter.
Proof. Straightforward computation.
Proposition 13.5.7. Formally, the isoptic (ω1 , ω2 ) pencil is also the isotomic (ω3 , ω4 ) pencil where
⊥
ω3 + ω4 = ω1 + ω2 and ω3 − ω4 = i (ω1 − ω2 ) (ω1 , ω2 are supposed to be at finite distance and
normalized , while orthopoint is obtained using OrtO ).
2
Proof. Denote the common middle by ω0 and use Pythagoras theorem to compute |ω1 ω3 | . One
2
2
2
has |ω1 ω3 | = |ω0 ω1 | + |ω0 ω3 | = 0 and thus both {ω3 },{ω4 } are orthogonal to {ω1 },{ω2 }. Using
OrtO , i.e. a rotation acting over V.
13.6
Euler pencil and incircle
Consider C1 = (X1 , r), C3 = (X3 , R), C5 = (X5 , R/2) and Cz = (Xz = X381 , |GH| /2) i.e., respectively, the in-, circum- nine points and orthocentroidal circles (Figure 13.2a). Let Uj , Aj , xj , cj
be the respective representatives of centers and circles, together with their respective shadows
(Figure 13.2b). Then :
1. Circles (X1 ) , C1 , C∞ are concentric so that U1 , A1 , Sirius are aligned and therefore x1 , c1 , G
are aligned too. The same happens for j = 5 and j = z.
2. Cycles C3 , C5 , Cz belong to the same (Euler) pencil, together with their radical axis AR3,5 , so
that representatives A3 , A5 , Az , A3,5 are aligned and therefore c3 , c5 , cz , ar3,5 are aligned
too. Since c3 is "far below the paper sheet", we have c5 = cz = ar3,5 . For the same reason,
c1 = ar3,1 .
.
3. Circles C1 and C5 are tangent at F = X11 , the Feuerbach point and cycles (F ) , C1 , C5
belong to the same pencil, together with their common tangent AR1,5 . Representatives
Uf , A1 , A5 , A1,5 are aligned and so are xf , c1 , c5 , ar1,5 .
4. Cycles AR1,3 , AR1,5 , AR3,5 are on the same pencil (they concur in the radical center Xk )
and their shadows ar1,3 , ar1,5 , ar3,5 are aligned.
5. In fact line c1 c5 is not representative of a specific pencil, but rather of the bundle generated
by C1 , C3 , C5 . We have :
t


A1 ' 

and therefore :
t


Ak ' 

2
(b + c − a)
2
(c + a − b)
2
(a + b − c)
4

t




, A5 ' 


b2 + c2 − a2
c 2 + a 2 − b2
a2 + b2 − c2
4

t




, A3 ' 


0
0
0
1
(c − b) (b + c − a) b2 + c2 − a2 b2 + c2 − ab − ac
(a − c) (c + a − b) c2 + a2 − b2 c2 + a2 − bc − ba
(b − a) (a + b − c) a2 + b2 − c2 a2 + b2 − ca − cb
4 (a − b) (b − c) (c − a) (a + b + c)










From Ak , the well-known result Xk = X676 and the obvious rk = |Xk Xf | can be re-obtained.
6. As it should be, xk , ck , G are aligned (small insert, at the bottom of Figure 13.2b ).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
144
13.6. Euler pencil and incircle
(a) lines and circles
(b) shadows
Figure 13.2: Euler pencil and incircle
March 30, 2017 14:09
published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
145
Figure 13.3: Lemoine and Brocard pencils
7. Consider Wk at intersection between line (Vk Sirius) and plane (V1 , V3 , V5 ). This points represents a circle that is both concentric and orthogonal to Ck . This circle is therefore (Xk , i rk )
and its shadow co belongs to both Gxk and c5 c1 . Moreover, division G, xk , co, ck is harmonic.
13.7
The Brocard-Lemoine pencils
Proposition 13.7.1. Let Ia and Ja be the points on sideline BC met by the interior and exterior bisectors of angle A. The circle Ωa having diameter [Ia , Ja ] goes through A and is called
A-Apollonian circle. The B- and C- Apollonian circles are similarly constructed. These circles
belong to a same pencil whose base points are the isodynamic points X(15), X(16). The orthogonal
pencil contains the circumcircle and the Brocard 3-6 circle.
Proof. These assertions depend of the following Lemmas.
Lemma 13.7.2. Lemoine Pencil. Point Ia is the A cevian of X(1). Thus Ia ' 0 : b : c and
Ja ' 0 : b : −c. Now, we take the wedge of the Veronese of these points and we obtain:


0
 −a2 c2 


Aa ' [1, 0, 0, 0] ∧ [0, b (b + c) , (b + c) c, −a2 bc] ∧ [0, b (c − b) , c (b − c) , −a2 bc] ' 

2
2
 a b

b2 − c2
Let us remember that Aa is a column that describe a point, while each Veronese is a plane incident
to this point. Here, it is obvious that the three Aj are not independent from each other. Using the
usual formalism to describe the corresponding pencil, one obtains :


0
a2 b2 b2 − a2
a2 c2 c2 − a2
a4 b2 c2
 a2 b2 a2 − b2
^
0
b2 c2 c2 − b2
a2 b4 c2 
.


Lemoinepoint = Aa Ab '  2 2 2
2
2
2 2 4 
2
2 2


b
−
c
0
a
b
c
a
c
a
−
c
b
c
4
4 2 2
2 4 2
2 2 4
−a b c
−a b c
−a b c
0
where the index ’point’ is used to remember that this object has to be multiplied by a point (i.e. a
column) to determine if the point belongs to the pencil.
Lemma 13.7.3. In order to determine the point-circles that belong to the pencil, we solve
V er(x : y : z) · Q
−1
· Lemoinepoint = 0. We obtain both umbilics, together with

√   2
√

a2 ∓ ia2 3
a 2 a2 − b2 − c2 ± i 3a2 b2 − c2
√


  2
E± ' 
−2 b2
 '  b 2 b2 − a2 − c2 ± i 3b2 c2 − a2 
√
√
c2 ± ic2 3
c2 2 c2 − b2 − a2 ± i 3c2 a2 − b2
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
146
13.8. The Apollonius configuration
√
The fact that E± ' X (187) ± i 3 X (512) is not real indicates that the pencil is an isogonal one.
From these values, one sees the the line of centers is the Lemoine axis (eponymous property).
.
Lemma 13.7.4. The polar planes Πj of the Aj are obtained by Πj = t Aj · Q . Their pencil is
described by


^
.

Brocardplane = Πa Πb ' 

4
0
0
0
−b2 c2
0
0
0
−a2 c2
b2 c 2
a2 c2
a2 b2
0
0
0
0
−a2 b2





where the index plane is to remember that this object has to be multiplied by a plane (i.e. a row)
to determine if the plane belongs to the pencil of planes. To obtain the matrix acting over points
(here: the representatives of cycles), one must take the dual. This leads to:



Brocardpoint ' 

0
−a2 b2
a2 c2
0
a2 b2
0
−b2 c2
0
−a2 c2
b2 c2
0
0
0
0
0
0





Lemma 13.7.5. Using the same algorithm as above, we obtain the point-circles that belong to the
pencil. We obtain both umbilics, together with points X(15),X(16) the isodynamic points.
Proof. The expression obtained isn’t symmetrical... and rather complicated. The best thing to do
is using the Kimberling’s search keys. Thereafter, one can check that:

√

a2 2 a2 − b2 − c2 + 4S3 a2 b2 + c2 a2 − b4 − c4
√


F± =  b2 2 b2 − c2 − a2 + 4S3 b2 c2 + a2 b2 − c4 − a4 
√
c2 2 c2 − a2 − b2 + 4S3 c2 a2 + b2 c2 − a4 − b4
√
i.e. F± ' X (187) ± 3 OrtO · X (512) (see Proposition 13.5.7). From these values, one sees the
the line of centers is the Brocard axis (eponymous property).
Lemma 13.7.6. One sees that the circumcircle Γ and the 3-6 Brocard circle belong to the second
pencil.
13.8
The Apollonius configuration
In the general case, it exists eight cycles Ω tangent to three given cycles Ω1 , Ω2 , Ω3 (not from the
same pencil). A survey of this question is Gisch and Ribando (2004), while the usual disjunction
into ten cases is Wikipedia: WillowW et al. (2006). The best space where this Apollonius problem
can be discussed
is PR R5 (cf Section 13.9). Nevertheless, most of the results can be formulated
4
in PR R ... and it will appear that only one situation is really special (cycles through the same
point), all the other belonging to the same general case.
13.8.1
Tangent cycles in the representative space
Proposition 13.8.1. Two cycles are tangent when their pencil line is tangent to the fundamental
quadric. Therefore, the locus of the representatives of all cycles tangent to a given (real) cycle Ω
represented by V (not inside Q) is the cone whose vertex is V and that goes through Q ∩ polar (V ).
Proof. Two tangent cycles are defining a tangent pencil !
Remark 13.8.2. When Ω is a point circle, its representative U belongs to the fundamental quadric,
and the cone of the tangent cycles degenerates into a doubly coated plane.
March 30, 2017 14:09
published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
147
Definition 13.8.3. The Gram matrix Gp,q,··· ,r of Xp , Xq , · · · , Xr ∈ R4 is the matrix of all the
products t Xp Q Xq . In this context, notation Wpq = t Xp · Q · Xq and wp2 = t Xp · Q · Xp will be
used, leading to
!
wp2 Wpq
(13.19)
Gpq =
Wpq wq2
Proposition 13.8.4. Two cycles Ω1 , Ω2 are secant, tangent or external when signum det G12 is
(respectively) +1, 0 or −1.
Theorem 13.8.5. Special cases of the Apollonius problem are (1) cycles from the same pencil
and (2) cycles through the same point (tangent bundle). Otherwise, representatives Vj of the three
given cycles and their common orthogonal cycle Ω4 formPa basis that splits the problem into four
pairs of solutions. One of the solutions is given by V0 =
kj Vj where :
k1
=
k2
=
k3
=
k4
(w2 w3 − W23 ) (−w1 w2 w3 − w1 W23 + w2 W13 + w3 W12 )
(w1 w3 − W13 ) (−w1 w2 w3 + w1 W23 − w2 W13 + w3 W12 )
(w1 w2 − W12 ) (−w1 w2 w3 + w1 W23 + w2 W13 − w3 W12 )
p
=
−2 (w2 w3 − W23 ) (w1 w3 − W13 ) (w1 w2 − W12 ) G123 /w4
(13.20)
and the others are obtained by changing k4 into −k4 (inversion through Ω4 ) or changing the signs
of w1 , w2 , w3 . A solution is real/imaginary or "unimaginable" (object that would have a non real
center) according to theQsign of k42 . Globally, the number of "imaginable" solutions changes when
the tangency condition Gjk vanishes.
Proof. When Ωj , j = 1, 2, 3 is a basis of a non tangent bundle, then Ωj , j = 1, 2, 3, 4 is a basis
of the whole representative space. The fundamental quadratic form is described, in this basis, by
matrix G1234 where Wj4 = 0 for j = 1, 2, 3. Computing, in thispbasis, the tangency condition of
Ω0 and any of the Ωj leads to 0. Since the wj are defined as Wjj we have 4 choices of signs
leading, due to the possibility of a global proportionality factor, to eight different values.
13.8.2
An example: the Soddy circles
Proposition 13.8.6. Soddy circles are three mutually, externally, tangent circles. Let A, B, C
be their centers. Then the common orthogonal circle of the Soddy’s is the incircle of ABC.
Proof. Let x be the radius of circle (A) , etc. We have a = y + z, b = z + x, c = x + y. Therefore
x = b + c − a, etc, while the contact point of the (B) , (C) circles is Ga ' 0 : y : z. This is also the
contact point of (I) with line BC and the conclusion follows.
Proposition 13.8.7. The Apollonius circles of the Soddy circles are twice each of them, a small
circle (inside the intouch triangle) and an outer circle. The center of the smaller circle is called
X(176), the other is called X(175). Let Ha be the branch of hyperbola that goes through A, Ga and
has B, C as focuses. Then the three branches concur at X(176), while the other three branches of
the whole hyperbolas concur at X(175).
Proof. This is clear from a = y + z, etc.
Proposition 13.8.8. Centers and radiuses of the Soddy circles are given by :
nX(175)
=
nX(176)
=
4R + ρ
−2s
nX(7) +
nX(1)
4R + ρ − 2s
4R + ρ − 2s
4R + ρ
2s
nX(7) +
nX(1)
4R + ρ + 2s
4R + ρ + 2s
1
1
4 1
1
8R + 2ρ
−
= ;
+
=
ρ6
ρ5
ρ ρ6
ρ5
sρ
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
148
13.8. The Apollonius configuration
Proof. Using (13.9), the representatives of circles Ω1 · · · Ω4 are :





2
(b + c − a)
(b − 3 c − a) (b + c − a)
(c − 3 b − a) (b + c − a)
2
(b + c − a)
(a − 3c − b) (c + a − b) (a − 3b − c) (a + b − c)
2
(c + a − b)
(b − 3 a − c) (a + b − c)
2
(c − 3 a − b) (c + a − b)
(a + b − c)
2
2
(c + a − b)
(a + b − c)
Their Gram matrix is :

2
(b + c − a)
− (b + c − a) (c + a − b) − (a + b − c) (b + c − a)
2
 − (b + c − a) (c + a − b)
(c + a − b)
− (c + a − b) (a + b − c)


2
 − (a + b − c) (b + c − a) − (c + a − b) (a + b − c)
(a + b − c)


0
0
0
−4
−4
−4
+4






0
0
0
16 S 2
2
(b + a + c)
Then (13.20) gives the decomposition of the Soddy’s circles on the Ω basis. We have :
1
1
1
b+a+c
t
K'
;
;
;
b+c−a c+a−b a+b−c
2S






and the conclusion follows (here a + b + c = 2s).
13.8.3
An example: the not so Soddy circles
Proposition 13.8.9. The not so Soddy circles of a triangle are the circles centered at A with
radius a = BC, etc. Their common orthogonal circle is the Longchamps circle λ (i.e. the conjugate
circle of the antimedial triangle).
Proof. The circle γa , centered at A with radius a is described by :
yza2 + b2 zx + c2 yx + (x + y + z) a2 x + y a2 − c2 + z a2 − b2 = 0
Its intersections with the circumcircle are :
Pb ' a2 − c2 : −b2 : c2 − a2
and
Pc ' a2 − b2 : b2 − a2 : −c2
while its intersections with γb (resp. γc ) are Pc and C 0 ' 1 : 1 : −1 (resp. Pb and B 0 ' 1 : −1 : 1).
Points A0 , B 0 , C 0 are on circle (H, 2R) and form the antimedial triangle. Lines Pa A0 are the altitudes
.
of A0 B 0 C 0 and concur at L = H 0 =X(20). But they are also the radical axes of our circles. The
radius rL of the orthogonal circle is obtained by :
2
2
rL
= |AL| − a2 , etc =
−Sa Sb Sc
= 4 (2R + ρ + s) (2R + ρ − s)
S2
Proposition 13.8.10. The Apollonius circles of the not so Soddy circles are obtained by extraversions (i.e a 7→ −a) from their central versions. These central circles are centered at the Soddy
points X(175) and X(176).
Proof. The representatives and the Gram

−a2
c2 − a2 b2 − a2
 c2 − b2
−b2
a2 − b2

 2
 b − c2 a2 − c2
−c2
2
2
+a
+b
+c2
matrix of
 
1

1 
 
, 
1  
1
γa , γb , γc , λ are :
a2
Sc
Sb
0
Sc
b2
Sa
0
Sb
Sa
c2
0
0
0
0
−Sa Sb Sc ÷ S 2





Changing w1 7→ −w1 is a 7→ −a, proving the extraversions. Formula(13.20) gives the coefficients :


(bc − Sa ) (−bca − aSa + bSb + cSc )
 (ca − S ) (−bca − bS + cS + aS ) 

b
b
c
a 
K'

 (ab − Sc ) (−bca − cSc + aSa + bSb ) 
16 S 3 ÷ (a + b + c)
March 30, 2017 14:09
published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
Figure 13.4: Apollonius circles of the not so Soddy configuration.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
149
150
13.8. The Apollonius configuration
Organizing the obtained equations, we have :
γ6
=
γ5
=
4s (2R + ρ + s)
4R + ρ + 2s
4s (2R + ρ − s)
λ + (x + y + z) (ax + by + cz) ×
4R + ρ − 2s
λ + (x + y + z) (ax + by + cz) ×
This leads to the centers. Additionally, this proves that ax + by + cz = 0 is the radical axis of
the three circles. Moreover, the Soddy conic (through A, B, C, with focuses X(175), X(176) and
perspector X(7)) is tangent to the Longchamps circle at the common points of λ, γ6 , γ5 since we
have :
2
conic = λ + (ax + by + cz)
13.8.4
Yet another example : the three excircles
Taking the three excircles as Ω1 , Ω2 , Ω3 leads to a well-known situation (Stevanovic, 2003).
1. Representative of point X1 is :
U0 = [bc (b + c − a) ; ca (c + a − b) ; ab (a + b − c) ; a + b + c]
while radius of the incircle is :
r=
s
(a + b − c) (c + a − b) (b + c − a)
4 (a + b + c)
2. Representative of the incircle, given by (13.9), is :
h
i
2
2
2
V0 = (a − b − c) ; (a − b + c) ; (a + b − c) ; 4
3. Centers, radiuses and representatives Va , Vb , Vc of the excircles are obtained by changing one
of the sidelengths into its opposite in the respective formulas for the incircle.
4. The representative of the common orthogonal circle, as computed from (13.16), is :
V4 = [(c + a − b) (a + b − c) ; (a + b − c) (b + c − a) ; (b + c − a) (c + a − b) ; −4]
5. The radius of this circle, as computed from (13.13), is :
s
b2 c + ab2 + bc2 + a2 b + ac2 + a2 c + acb
ω4 =
4 (a + b + c)
while the representative of the center is :
t


U4 = 

2 a b2 + c2 − acb + b3 + c3 − a3
2 b c2 + a2 − acb + c3 + a3 − b3
2 c a2 + b2 − acb + a3 + b3 − c3
4 (a + b + c)
and the center itself is :





b + c : a + c : a + b = X10
6. The pairs of solutions of the Apollonius problem, as given by (13.20), are :


!
b2 + c2 − a2
c2 + a2 − b2
a2 + b2 − c2
4
S1

= 
2 bc
2 ca
2 ab
−4
S5
a+b+c+
a+b+c+
a+b+c+
a
b
c
a+b+c
March 30, 2017 14:09
published under the GNU Free Documentation License
13. Pencils of Cycles in the Triangle Plane
t


S2 = 

1
0
0
0
t





, S6 = 


151
(a + b + c) b2 + ab + ac + c2
(b + c) (a − b − c) (a + b − c)
(b + c) (a − b − c) (a − b + c)
4 (b + c)





Point S1 is the representative of the nine-points circle, centered at X5 while S5 is related
to the Apollonius circle, centered ad X970 . Points S2 , S3 , S4 are the representatives of lines
BC, CA, AB while S6 and S7 , S8 (obtained cyclically) are the representatives of the last
three solutions.
13.8.5
The special case
Proposition 13.8.11. Let Ω1 , Ω2, Ω3 be three cycles generating a bundle whose common orthogonal
cycle is a point-cycle (ω5 ), and ω4 be any other point. The representative of one of the cycles tangent
P3
to Ω1 , Ω2 , Ω3 is given by V0 = 1 kj Vj + 4U4 where :
w2 w3 − W2,3
1,2,3
k1 =
G1,2,3,4 −2 G2,3,4 ÷ ∆2,3,4
(13.21)
(w1 w3 − W1,3 ) (w1 w2 − W1,2 )
∆1,2,3
2,3,4 is the minor obtained by deleting row 1 and column 4 in G1,2,3,4 , while k2 , k3 are obtained
cyclically. Three other cycles are obtained by changing one of the w1 , w2 , w3 into its opposite. The
other solutions are four times the point cycle ω5 .
Proof. In this special case, G1,2,3 = 0 and Ω4 is chosen so that w4 = 0. When assuming that
Ω1 , Ω2, Ω3 aren’t pairwise tangent, a direct substitution shows that Ω0 is tangent to any of the
given cycles.
Example 13.8.12. Using Ω1 = 1 : 0 : 0 : 0 (representative of line BC) etc, leads to ω5 = Sirius.
An efficient choice for ω4 is any vertex. Using, for example, U4 = 0 : c2 : b2 : 1, one re-obtains
easily the in/excircles.
Example 13.8.13. The Apollonius circles relative to the three circles (ABH), (BCH), (CAH)
are (H, 0) four times, (H, 2R) once and three other circles, T a, T b, T c.
Circles T a, T b, T c are ever external to each other, and their common orthogonal circle T o is
real. Condition of (external) tangency is :
2
a2 − b2 − a2 + b2 c2 = 0
(etc) or ABC rectangular. The Apollonius circles of T a, T b, T c are (ABH) etc, their inverses in
T o and two others. Center X = x : y : z and radius ω of the first one are :
x = b2 + c2 − a2 ×
2
4 a8 − 2 b2 + c2 a6 + 2 b4 − b2 c2 + c4 a4 − 2 b2 + c2 b2 − c2 a2 + b2 − c2
a2 b2 c2 R
a6 + b6 + c6 − a2 b4 − a4 b2 − c2 b4 − a4 c2 − b2 c4 − c4 a2 + 4 a2 b2 c2
while the second is less simple.
ω=2
13.9
Elementary properties of the Triangle Lie Sphere
Definition 13.9.1. The Triangle Lie Sphere is the locus of points X = x : y : z : t : τ ∈ PR R5
such that t X . Q5 . X = 0 where fundamental matrix is defined by :
Q5 =
"
Q
0
0
−4S 2
#




=



a2
−Sc
−Sb
−a2 Sa
0
−Sc
b2
−Sa
−b2 Sb
0
−Sb
−Sa
c2
−c2 Sc
0
−a2 Sa
−b2 Sb
−c2 Sc
a2 b2 c2
0
0
0
0
0
−4 S 2








—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
152
13.9. Elementary properties of the Triangle Lie Sphere
Proposition 13.9.2. Using notations of Theorem 13.3.5, an element of the Lie sphere represents,
when t 6= 0 and τ 6= 0, an oriented circle :

t
2
b2 r12 + c2 q12 + 2 Sa q1 r1 − ω12 (p1 + q1 + r1 )

 2 2
 c p1 + a2 r12 + 2 Sb r1 p1 − ω12 (p1 + q1 + r1 )2 


2 
2 2
2 2
2
Y1 = 
 a q1 + b p1 + 2 Sc p1 q1 − ω1 (p1 + q1 + r1 ) 


2
(p1 + q1 + r1 )


2
2 ω1 (p1 + q1 + r1 )
(where signum of ω1 defines the orientation) or, when t = 0 and τ 6= 0, an oriented line :
1 p
t
Y3 = u3 ; v3 ; w3 ; 0; ±
∆.M. ∆
2S
–cf (9.2) for definition of M– or, when τ = 0, a non-oriented point.
Proof. All these results follow directly from Theorem 13.3.5.
t
Proposition 13.9.3. Two oriented cycles are tangent if and only if
Yj · Q5 · Yk vanishes.
Proof. This result is the rationale behind the former definitions. Using notations of Theorem 13.3.5,
we obtain for two circles :
2
2
2
2
t
Y1 · Q5 · Y2 = − |P1 P2 | − (ω1 − ω2 ) × 8S 2 (p1 + q1 + r1 ) (p2 + q2 + r2 )
(13.22)
By continuity, the Proposition holds also when lines are involved. For the sake of completeness,
one can nevertheless compute :
q
p1 u3 + q1 v3 + r1 w3
1
2
t
Y1 · Q5 · Y3 = −
∆3 . M . t ∆3 × 8S 2 (p1 + q1 + r1 ) (13.23)
+ ω1
p1 + q1 + r1
2S
for a circle and a line, and (after multiplication by conjugate quantity) :
2
− ((v4 − w4 ) u3 + (w4 − u4 ) v3 + (−v4 + u4 ) w3 ) × 8 S 2
for two lines Y3 , Y4 . In the three cases, this is the condition of tangency times a non vanishing
factor.
Remark 13.9.4. When using the Lie representation, objects that don’t belong
to Q5 are meaningless, while Q5 itself is obtained by double coating the outside of Q in PR R4 . Therefore, imaginary
circles are lost : when the radius decreases to 0 in an isotomic pencil, the differentiable continuation is going back to positive radiuses (with the other orientation) and not escaping to imaginary
values.
Proposition 13.9.5. Let Ω0 be a cycle, but not a point-circle, σ the inversion in cycle Ω0 as
described in Theorem 13.5.5, V1 ∈ PR R4 a representative of cycle Ω1 and Y1 = (V1 , τ1 ) ∈ PR R5
a representative of one of the corresponding oriented cycles. Then applications :
σ+
: σ + (V1 , τ1 ) = (σ (V1 ) , +τ1 )
σ−
: σ − (V1 , τ1 ) = (σ (V1 ) , −τ1 )
are describing inversions wrt each of the oriented cycles corresponding to Ω0 . The conservation
law (13.18) can be rewritten in order to describe a projective invariant by inversion, namely :
t
−
Y1 · Q5 · Y2
4S 2 τ1 τ2
2
=
2
|P1 P2 | − (ω1 − ω2 )
2ω1 ω2
or, for a circle and a line,
t
−
Y1 · Q5 · Y3
4S 2 τ1 τ3
=1+
p1 u3 + q1 v3 + r1 w3
ω1 (p1 + q1 + r1 ) τ3
and for two lines :
t
−
Y3 · Q5 · Y4
= 1 ± cos (∆3 , ∆4 )
4S 2 τ3 τ4
Remark 13.9.6. Searby (2009) illustrates how this invariant (named jk ) can be used to summarize
tangencies in various situations.
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 14
Morley spaces, or how to use
complex numbers
Our aim in this chapter is to describe how to translate into complex numbers all of the methods
we have described in the previous chapters. This is equivalent to give the complex version of all
the operators we have already described.
In fact, some of these operators have a very simple form when using complex numbers and can
be introduced directly. Just after that, we will prove the equivalence between "the Morley version"
and "the barycentric version" of these operators.
14.1
How to deal with complex conjugacy ?
When writing points M as (ξ, η) and describing curves C by polynomials P so that M ∈ C
when P (ξ, η) = 0, it is efficient to use homogeneous coordinates (X, Y, T ) and describe curves by
homogeneous polynomials
X Y
n
,
Pn (X, Y, T ) = T P
T T
where n = dg (P ) is the degree of the curve. Indeed, we have a theorem stating that curves C1 and
C2 have exactly m n common points when polynomials Pn and Pm have no non-constant common
factor. To obtain this result, all the points have to be taken into account, including points with non
real coordinates as well as points at infinity, and also considering the multiplicities of the solutions.
When we want to reformulate this result using complex affixes, we need the following proposition.
Proposition 14.1.1. Define the complex polynomial Qn of an algebraic curve by
Z + iZ Z − iZ
Qn Z, T, Z = Pn
,
,T
2
2i
where Z (read it as "big z"), Z (read it as "big ζ") and T are algebraic variables. Define conj (Qn )
as the complex polynomial associated to polynomial Pn (obtained by complex conjugation of the
coefficients). Then
!
X
X
p
p q r
conj
cp, q, r Z Z T
=
cp, q, r Zq Z Tr
p+q+r
p+q+r
in other words, conjugate the coefficients and exchange Z with Z.
Remark 14.1.2. An algebraical variable is a placeholder used to write polynomials. What could be
the C-conjugate of a placeholder ? Therefore, a notation was to be found in order to satisfy the
following constraints :
1. Have capital letters, since polynomials are usually written P (X1 , X2 , · · · , Xk )
2. Avoid indices, and deal with the fact that letter "big ζ" has the same shape as letter "big z"
(nevertheless, read Z as "big zeta").
153
154
14.1. How to deal with complex conjugacy ?
3. Have a robust cursive version, to facilitate hand computations: a Z is clearly a Z, while a Z
is clearly a Z !
4. Enforce nevertheless the fact that Z is not the C-conjugate of Z.
This leads to the following definition.
Definition 14.1.3. The Morley space is PC C3 : a point is a complex triple defined up to a
non-zero complex factor. A point of the Morley space that can be written as :


zP


ζP '  1 
zP
for some zP ∈ R2 is referred as the Morley-affix of the ordinary point P (shortened as : ζP is a
finite point). A point of the Morley space that can be written as


exp (+i ϑ)


ζϑ ' 
0

exp (−i ϑ)
for some ϑ ∈ R is referred as the Morley-affix of the angle whose measure is ϑ + k π (shortened as :
ζϑ is a direction). Taken together, finite points and directions are referred as the visible points.
All other points of the Morley space are described as being invisible.
Remark 14.1.4. The invisible points of the Morley space correspond to the points that cannot be
written as (x : y : t) with x, y, t ∈ R in the ξ, η, t representation. In this context they are referred
as "non real". In the context of the Morley-space, this designation is no more suitable, and another
word must be used.
Remark 14.1.5. From an abstract point of view, it could be tempting to define the Morley space
as PR (C × C × R), i.e. to describe points by triples (z, ζ, t) such that z, ζ ∈ C and t ∈ R, each
triple being defined up to a real proportionality factor. But in real life, enforcing t ∈ R can only
be done by carrying annoying factors. We better have the following definition.
Proposition 14.1.6. An algebraic curve is said to be "reduced to a point" when it contains only
one visible point, and "visible" when it contains an infinite number of visible points. The polynomial
of an irreducible visible curve must be proportional to its conjugate.
Remark 14.1.7. From an abstract point of view, using a constant factor to enforce conj(P ) = P is
always possible. But in real life, this can only be done by carrying annoying factors and has to be
avoided. Before any simplification, a polynomial obtained from a determinant, as the equation of
a line or a circle, verifies conj(P ) = −P .
Proposition 14.1.8. In the Morley space, the equation of a visible line is a Z + a Z + bT = 0
(with b ∈ R). Its point at infinity is :
a : 0 : −a = cos ϑ + i sin ϑ : 0 : cos ϑ − i sin ϑ = ω 2 : 0 : 1
where ϑ is the oriented angle from the real axis to the line. This is an angle between straight lines
and not an angle between vectors. We have the formula :
ω 2 = − coeff ∆, Z ÷ coeff (∆, Z)
Proof. Straight line y = p x + q can be rewritten as (p + i)Z + (p − i)Z + 2q T = 0. Point at infinity
is [p + i, 2 q, p − i] ∧ [0, 1, 0].
Proposition 14.1.9. In the Morley space, the equation of a visible cycle is p ZZ + qZ + qZ T +
r T2 = 0 (with p, r in R). Each circle (p 6= 0 ) goes through the umbilics :
 
 
0
1
.  
 
Ω x =  0  , Ωy =  0 
1
0
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
155
Proof. Circle centered at M0 = (z0 , z0 , 1) and radius ω is given by :
(Z − z0 ) Z − z0 − ω 2
Remark 14.1.10. When all of the terms are present, variable T can be omitted. But when subtract
ing normalized circles, obtaining the radical axis, it makes a difference to obtain bZ + bZT + c T T
instead of a first degree polynomial !
Proposition 14.1.11. As a consequence, collinearity of three points and concyclicity of four points
are (respectively) :
3
det ([zj , zj , 1]) = 0,
j=1
14.2
4
det ([zj zj , zj , zj , 1]) = 0
j=1
(14.1)
The usual operators of the Morley space
Definition 14.2.1. Complex affix of a point. Let ξP , ηP be the Cartesian coordinates of a
point P in the Euclidean plane. The C-affix of point P is defined as
zP = ξP + i ηP
Remark 14.2.2. In this definition, quantity i is a quarter turn. Since two quarter turns performed
one after another is nothing but one half turn, we have i2 = −1. This equation has another
solution, namely −i, the quarter turn in the opposite orientation. Obviously, the very choice of a
frame to obtain Cartesian coordinates like ξP , ηP ensures a choice of orientation of the plane: when
you look at a plane from above, you measure angles by placing your protractor onto the plane,
seeing zP = ξP + i ηP . But the guy that looks at the plane from below will put his protractor on
to the other face of the plane, and will therefore see zP = ξP − i ηP instead of zP .
Remark 14.2.3. The former affine point of view, describing points of R2 by vectors (ξP , ηP , 1) can
be restated as :


 


zP
ξP + i ηP
ξP


 


P :  1 =
1
 = Y .  ηP 
zP
ξP − i ηP
1


1 +i 0
. 

where Y =  0 0 1 
1 −i 0
collecting together the from above and the from below points of view.
Theorem 14.2.4. In the Morley space, the area of triangleABC is given by :
zA zB zC 1
area (ABC) = − i 1
1
1 4
zA zB zC (14.2)
Proof. Despite it’s great consequences, this result has a very short proof: formula (7.6) using
Cartesian coordinates has to be modified by factor −i/2 since :


1 +i 0


det  0 0 1  = 2i
1 −i 0
Proposition 14.2.5. In the Morley space, the line at infinity is described by
Lz '
0
1
0
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
(14.3)
156
14.2. The usual operators of the Morley space
Proof. The value is obvious. And can also seen as [0, 0, 1] · Y −1 .
Proposition 14.2.6. In the Morley space, the matrix Wz of the operator that changes the affix
D of a line, into the affix P of the point at infinity of that line can be written as :


0 0 −1


P = Wz . t D
(14.4)
where Wz =  0 0 0 
+1 0 0
Proof. This is only P = D ∧ Lz .
Proposition 14.2.7. In the Morley space, the matrix P ythz
squared length of a vector in the V space can be written as :

0 0 1
1

P ythz = R2  0 −2 0
2
1 0 0
of the quadratic form that gives the



(14.5)
Proof. Here, equalities are required since a length is not defined up to a proportionality factor.
Remember that an element of V is obtained as the difference of the normalized columns of two finite
points, and therefore looks like z1 /t1 − z2 /t2 , 0, z1 /t1 − z2 /t2 . The obtained formula is nothing but
2
the usual |ζ| = ζ ζ. The strange looking −2 that acts on the t = 0 component is the counterpart
for the better looking form of the P yth matrix. And the R factor that appears in the formula
will only make sense in the Lubin’s context. Otherwise, the radius of the "unit circle" is supposed
to be 1 !
Proposition 14.2.8. In the Morley space,
direction into its orthogonal direction while
by :

1 0 0

OrtOz = i  0 0 0
0 0 −1
the matrix OrtOz of the operator that transforms a
transforming the circumcenter into 0 : 0 : 0, is given


0


 ' Y ·  −1
0
+1
0
0

0

0  · Y −1
0
(14.6)
Proof. The Cartesian formula x : y : 0 7→ +y : −x : 0 is well-known. Then the change of basis
formulas have to be applied. Eigenvectors of OrtOz are both umbilics 1 : 0 : 0, 0 : 0 : 1 and
0 : 1 : 0 (the circumcenter): the first umbilic is rotated by a quarter of turn, and the second umbilic
is rotated by the opposite amount.
Proposition 14.2.9. Orthodir. In the Morley space, the matrix Mz of the orthodir operator
that changes the affix D of a line, into the affix P of the point at infinity in the orthogonal direction,
can be written as :


0 0 1
. 

P = Mz . t D
where Mz = i  0 0 0  = OrtOz . Wz
(14.7)
1 0 0
Proof. Obvious from definitions. Coefficient i is kept here in order to obtain a better looking
tangent formula.
Theorem 14.2.10. Tangent of two lines. In the Morley space, the oriented angle from a visible
line ∆1 to a visible line ∆2 is characterized by :
z }| {
∆1 . Wz . t ∆2
tan ∆1 , ∆2 =
∆1 . Mz . t ∆2
(14.8)
where Wz , Mz are exactly as given in (14.4) and (14.7) (not up to a proportionality factor).
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
157
Proof. Consider two Cartesian lines y = p1 x + m1 and y = p2 w + m2 . Then tan (D1 , D2 ) is
obviously (p2 − p1 ) ÷ (1 + p1 p2 ) since the slope of a line is the tangent of the angle from the x
axis to the line. It remains to see that Dj is written [pj , −1, mj ] in the Cartesian frame, and then
[pj + i, 2 mj , pj − i] in the Morley frame... and the formula gives the right value. Numerator tell
us when lines are parallel, and denominator when they are orthogonal.
Proposition 14.2.11 (Laguerre Formula). Consider two (distinct) lines ∆1 , ∆2 and their common
point M . Then
z }| {
cross-ratio (M Ωy , M Ωx , ∆1 , ∆2 ) = exp 2i ∆1 , ∆2
Proof. Tedious proof: compute and check using (14.8). Better proof:

 
 
 

σ
τ
0
1
 σ 2
 
 
 

cross-ratio  0  ,  0  ,  0  ,  0  = 2
τ
1/σ
1/τ
1
0
Proposition 14.2.12. In the Morley space, distance from point P to line ∆ is given by :
dist (P, ∆) =
where Lz =
0
1
0
∆.P
q
(Lz . P ) 2i ∆ . Mz . t ∆
(14.9)
and Mz is as given in (14.7) (not up to a proportionality factor).
Remark 14.2.13. Formula (14.9) is invariant when barycentrics of P or ∆ are modified by a proportionality factor. Denominators are enforcing the fact that P is supposed to be at finite distance, and
∆ is not supposed to be the infinity line. The square root is the operator norm of the application.
φ : Z : T : Z 7→ z Z + t T + ζ Z / (T)
Replace Z :p
T : Z by Z + r τ : T : √
Z + r/τ . Derive wrt τ and obtain r z − r ζ/τ 2 . Solve in τ and
obtain τ = ζ/z. So that ∆φ = 2 zζ r.
14.3
Lubin representation of first degree
Proposition 14.3.1. The Morley-affix of
with respect to triangle ABC is given by :

zA

ζP '  1
zA
a point P whose barycentrics are p : q : r ∈ PR R3
zB
1
zB
 

zC
p
 

1 . q 
zC
r
(14.10)
Proof. When p + q + r 6= 0, this is nothing but the usual definition of a barycenter, and ζP is
a Morley finite point. When p + q + r = 0, P is at infinity and ζP is a Morley direction. The
condition p, q, r ∈ R ensures that no invisible points in the Morley-space can be generated from a
real point in the Kimberling space.
Remark 14.3.2. Formula (14.10) requires a lot of conjugacies... and it is well-known that conjugacy
is a terrific process that introduces branching points. Therefore a method is needed to transform
conjugacy into an holomorphic process, at least for a sufficiently large subset of points.
Definition 14.3.3. The Lubin parameterization are obtained by assuming that the circumcircle
of triangle ABC is nothing but the unit circle of the complex plane, together with the relations :
zA = αn , zB = β n , zC = γ n
Since our interest is directed toward central objects, we will largely use the so-called symmetric
functions :
.
.
.
s1 = α + β + γ =, s2 = α β + β γ + α γ, s3 = α β γ, s4 = i (α − β) (β − γ) (γ − α)
(14.11)
σ1 = zA + zB + zC , σ2 = zA zB + zB zC + zA zC , σ3 = zA zB zC
σ4 = i (zA − zB ) (zB − zC ) (zC − zA )
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
158
14.3. Lubin representation of first degree
Remark 14.3.4. Quantity s4 (i times the Vandermonde of the three numbers) is skew-symmetric
and verifies :
s24 = −s21 s22 + 4 s3 s31 + 4 s32 − 18 s3 s1 s2 + 27 s23
so that s4 will not appear by a power greater than one. The "small" symmetric functions
s1 , s2 , s3 , s4 depend on the degree of the representation, while the "big" ones σ1 , σ2 , σ3 , σ4 depend
only on the triangle.
Proposition 14.3.5. Forward and backward matrices. Using the Lubin-1 parameterization,
we have :


α
β
γ
iσ4
4i


(14.12)
= 2S
Lu =  1
1
1  ; det Lu =
σ3
R
1/α 1/β 1/γ


α (β − γ) α γ 2 − β 2
α β γ (β − γ)
1 

Lu−1 =
α β γ (γ − α) 
 β (γ − α) β α2 − γ 2
iσ4
γ (α − β) γ β 2 − α2
α β γ (α − β)
Theorem 14.3.6. Forward substitutions. Suppose that barycentrics p : q : r of point P depends
rationally on a2 , b2 , c2 , S. Then Morley-affix of P is obtained by substituting the identities :
S=
i R2 (α − β) (β − γ) (γ − α)
R (β − γ)
R (γ − α)
R (α − β)
, a= p
, b= p
, c= p
4αβ γ
−β γ
−γ α
−αβ
(14.13)
into p : q : r and applying (14.10). The result obtained is a rational fraction in α, β, γ whose
degree is +1. When P is a triangle center, ζP depends only on σ1 , σ2 , σ3 . When P is invariant
by circular permutation, but not by transposition, a σ4 term appears.
Proof. Cancellation of radicals is assured by condition p, q, r ∈ Q a2 , b2 , c2 , S . Elimination of R
comes from homogeneity. Symmetry properties are evident.
Proposition 14.3.7. Backward substitutions. Let zP ∈ C (α, β, γ) be an homogeneous rational fraction, supposed to be the complex affix of a finite point. Then deg (zP ) = 1 is required.
Alternatively, let ω 2 ∈ C (α, β, γ) is an homogeneous rational fraction, supposed to describe the
Morley affix of a direction. Then deg ω 2 = 2 is required. When these conditions are fulfilled, the
ABC-barycentrics p : q : r of these objects can be obtained as follows. Compute the corresponding
vector :








p
p
zp
ω2








−1
.  1  or  q  = Lu−1 .  0 
 q  = Lu
r
r
zp
1
then apply substitutions
β
=
γ
=
!
+2 i a2 + b2 − c2
a4 + b4 + c4 − 2 a2 + b2 c2
S+
α
a2 b2
2 a2 b2
!
−2 i c2 + a2 − b2
a4 + b4 + c4 − 2 a2 + c2 b2
S+
α
a2 c2
2 a2 c2
to this vector and simplify the obtained expression using the Heron formula :
S2 = −
1
(a + b + c) (b + c − a) (c + a − b) (a + b − c)
16
Proof. Transform α, β, γ into αδ, βδ, γδ. Since this transform is a similitude, barycentrics must
remain unchanged and the zP affix is turned by δ. On the other hand, polynomial zP ishomogeneous and zP is multiplied by δ k where k = deg (zP ). Concerning the directions, arg ω 2 is twice
the angle with the real axis, and degree 2 is required.
−1
Alternatively, the degrees of rows of ζP are +k, −k, 0 while the degrees of columns of Lu
are −1, 0, +1. Quantities p, q, r will therefore be a sum of terms whose degrees are respectively
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
159
k − 1, 0, 1 − k. But homogeneity is required in order that a transformation β = B α, γ = C α can
eliminate α, leading to k = 1.
2
To obtain the substitution formulas, compute β from c2 α β = −R2 (α − β) . A choice of
branch (a sign for i) has to be chosen. Exchange b and c (and therefore change S into −S) and
obtain the corresponding γ.
Remark 14.3.8. The substitution formulas can be written as:
2
2
4 Sc
2 Sb
4Sb
2 Sc
+i S 2 2 −1 α ; γ =
−i S 2 2 −1 α
β=
a2 b2
a b
a2 c2
a c
b = (BC, BA)
i.e. β = α exp (2i C), γ = α exp (−2i B). This result is indeed symmetric, since B
b
while C = (CA, CB).
Remark 14.3.9. When starting with a symmetrical Morley-affix, the obtained p : q : r remains symmetrical in α, β, γ. The given substitutions are breaching the symmetry of individual coefficients
p, q, r, that can only be reestablished by cancellation of unsymmetrical common factors between
the p, q, r. Most of the time, its more efficient to proceed by numerical substitution and use the
obtained search key to identify the point (and proceed back to obtain a proof of the result).
Proposition 14.3.10. The Kimberling search key associated to a visible finite point defined by its
Morley affix (short= Morley’s search key) is obtained by substituting :
zA = 1 ; zB = −
104 √
401
248 √
391
−i
35 ; zC =
−i
35
729
729
1521
1521
into Z/T and then computing :
22 Z
Z .
157 √
321 √
35 − i
35
searchkey
=<
+
T
840
3
T
280
Proof. Kimberling’s search keys are associated
with triangle a = 6, b = 9, c = 13. The radius
√
of the circumcircle is R = (351/280) 35. One can see that sidelengths of triangle αβγ are
6/R, 9/R, 13/R. We apply these substitutions to obtain the numerical value of the "retour"
matrix, and then use (1.3)
Remark 14.3.11. When using Lubin-n with n > 1, adequate substitutions have to be used to
calculate Z/T.
Proposition 14.3.12. The Morley’s searchkey of a visible point at infinity (T = 0) is obtained
from Ω = Z/Z by :
√ √ √ 1108809 241 + 16 i 35 Ω2 − 907686 157 + 176 i 35 Ω + −224394311 + 30270800 i 35
√ √
389191959 Ω2 − 106136082 + 118980576 i 35 Ω + 19397664 i 35 − 371888361
Another method is identifying the isogonal conjugate of the given point, which is simply : σ3 /Ω :
−1 : Ω/σ3 .
Proof. The searchkey of an point at infinity is : xa × xa + yb + zc , leading to this tremendous
expression. But after all, this formula is not designed for hand computation but rather to a
floating evaluation by a computer... .
14.4
Comparing barycentric and Morley spaces
Fact 14.4.1. Concerning the line at infinity (14.3), we have
.
Lz =
0
1
0
= L∞ . Lu−1
Fact 14.4.2. Concerning the operator that takes the direction of a line (14.4), we have


0 0 −1
t


Wz =  0 0 0  = Lu . W . Lu ÷ det Lu
+1 0 0
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
160
14.5. Some examples of first degree
Fact 14.4.3. Concerning the quadratic form acting on the line at infinity (14.5), we have


0 0 1
t
. 1


P ythz = R2  0 −2 0  =
Lu−1 · P yth · Lu−1
2
1 0 0
Remark 14.4.4. Here, strong equalities are required, not equalities up to a proportionality factor.
Values of a, b, c that appear in P yth have to be substituted (see (14.13) ).
Fact 14.4.5. Concerning the orthopoint

1

OrtOz = i  0
0
transform (14.6), we have

0 0

0 0  ' Lu . OrtO . Lu−1
0 −1
Remark 14.4.6. Values of a, b, c that appear in OrtO have to be substituted (see (14.13) ). The
exact application of the change of basis formula leads to a global factor i that can be canceled.
Fact 14.4.7. Concerning the orthodir transform (14.7) that changes the affix D of a line, into the
affix P of the point at infinity in the orthogonal direction, we have


0 0 1
t
.


Mz = −i  0 0 0  = OrtOz . Wz = Lu . M . Lu ÷ det Lu
1 0 0
Proof. With the given definition, the proportionality factor (acting over the right member of the
') is exactly the same factor that was acting over Wz .
Fact 14.4.8. Concerning the tangent of two lines, formulas (7.19) and (14.8) have exactly the
same shape, i.e.
z }| {
∆1 . Wz . t ∆2
tan ∆1 , ∆2 =
∆1 . Mz . t ∆2
because the same proportionality factors were used for both Wz and Mz .
14.5
Some examples of first degree
Example 14.5.1. The
tions are giving :

p

 q
r
circumcenter O. By definition, ζO = 0 : 0 : 1. The preceding transforma

 

α (β − γ) (β + γ)
a2 b2 + c2 − a2

 
 
 '  β (γ − α) (α + γ)  '  b2 c2 + a2 − b2 
γ (α − β) (α + β)
c2 a2 + b2 − c2
Example 14.5.2. Symmedian point X(6), aka Lemoine point.
1. Consider the middle A0 of segment [B, C] and define the A symmedian as the line ∆A that
goes through A and verifies ∠ (AB, ∆A ) = ∠ (AA0 , AC). We will use P instead ζP since
this is more readable ... and hand-writable. A remark : symmetry wrt bisectors would be
irrelevant, since bisectors are unreachable in the Lubin-1 representation !
2. We have A0 = B + C, AA0 = A ∧ A0 etc. and our equations are
∆A . A =
0
0
0
tan (AB, ∆A ) + tan (AC, AA )
=
solving this system, then permuting, gives :
∆A '
2 α − β − γ ; (α γ + α β − 2 β γ) α ; −2 α2 + 2 β γ
∆B '
2 β − γ − α ; (α β + β γ − 2 α γ) β ; −2 β 2 + 2 α γ
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
161
3. Intersecting two symmedians gives a symmetrical result. Therefore, the three symmedians
are concurrent at some point. This point is well-known as Lemoine point, and we have :


2 σ22 − 6 σ3 σ1


K = ζ (6) = ∆A ∧ ∆B =  σ2 σ1 − 9 σ3 
2 σ12 − 6 σ2
4. Going back to barycentrics, we obtain the well-known result :

 

2
a2
α (γ − β)

 

X (6) '  β (α − γ)2  '  b2 
2
c2
γ (α − β)
Example 14.5.3. The Kiepert parabola.
1. Morley equation of the circumcircle is Z Z − T2 = 0. The Morley affix ∆P of the polar line
of point K = z : 1 : ζ wrt the circumcircle is therefore given by :


0 0 1
h
i
.


∆P = z 1 ζ .  0 −2 0 
1 0 0
2. The coefficients of the tangential conic determined by five given lines [uj , vj , wj ] are obtained
as :
^ u2j , vj2 , wj2 , uj vj , vj wj , wj uj
j=1..5
by universal factorization of the corresponding 6 × 6 determinant. Let us consider the inconic
tangent to the infinity line (parabola) and to the circumpolar of point K. Using (BC) '
B ∧ C, etc together with the previous equation, we obtain the symmetric matrix :


2 σ3 z 2 − 2 σ2 σ3 z ζ + 4 σ32 ζ
qsp −σ1 z 2 + σ2 σ3 ζ 2 + 2 σ2 z − 2 σ3 σ1 ζ


C∗ ' 
−σ3 σ1 zζ + σ32 ζ 2 + 2 σ3 z
0
−z 2 + σ2 z ζ − 2 σ3 ζ

2
2
2
−σ1 z + σ2 σ3 ζ + 2 σ2 z − 2 σ3 σ1 ζ qsp
2 σ1 z ζ − 2 σ3 ζ − 4 z
Degrees of all these expressions are :
dg
C∗

5

= 4
3
4
.
2

3

2 
1
3. A focus is a point such that both isotropic lines through that point are tangent to the conic.
Writing that Q = Z : T : Z ∧ (1 : 0 : 0) satisfies Q . C ∗ . t Q = 0 and the similar with the
other umbilic gives two equations whose solution is
 2

z − σ2 zζ + 2 σ3 ζ
 2 z − σ1 zζ + σ3 ζ 2 



F (z) ' 
1


2
 2 z − σ1 zζ + σ3 ζ 
and this point is on the circumcircle.
z 2 − σ2 zζ + 2 σ3 ζ
4. Take K at the Lemoine point X(6). Its circumpolar line is called the Lemoine axis. One
obtains the Kiepert parabola, where :




2 σ2 σ3 σ3 σ1
0
−σ22
2 σ3 σ12
−σ2 σ3 σ1




C ∗ '  σ3 σ1
0
−σ2  , C '  2 σ3 σ12
−4 σ2 σ3 σ1
2 σ22 σ3 
0
−σ2 −2 σ1
−σ2 σ3 σ1
2 σ22 σ3
−σ12 σ32
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
162
14.5. Some examples of first degree
5. As stated in Proposition 11.3.15, the triangle of the circle-polars of the sidelines of triangle
t
T is described by matrix C ∗ . T ∗ . Both triangle are in perspective (lines AA0 , etc are
concurrent). The perspector is obtained as AA0 ∧ BB 0 and concurrence is verified by the
symmetry of the result. It is well-known that P =X(99), the Steiner point.
F =
σ2
σ3 σ12 − 3 σ2 σ3
, P =
σ1
σ22 − 3 σ3 σ1
6. Applying the preceding transformations, the p : q : r associated with z = σ2 /σ1 is obtained
as :


a2
 


 (b + c) (b − c) 


α (β − γ) α β − γ 2 γ α − β 2
p

b2
 
 

2
2


αβ − γ  ' 
 q  '  β (γ − α) β γ − α

(c
+
a)
(c
−
a)


2
γ (α − β) γ α − β 2 β γ − α2
r


c
(a + b) (a − b)
and we can identify X(110), the focus of the Kiepert parabola.
Example 14.5.4. Isogonic, isodynamic and Napoleon.
1. Define j = exp (2 iπ/3). Start from triangle ABC. Construct PA such triangle PA BC is
equilateral. More precisely, the z affix of PA is such that z + j β + j 2 γ = 0, deciding of
the orientation. The three lines APA , BPB , CPC are concurrent leading to the first isogonic
center. Changing j into j 2 leads to the second isogonic point. A simple computation leads
to :
9 σ2 σ3 − 12 σ3 σ12 + 3 σ1 σ22 √
σ4 σ2
± 3 2
2
6 σ2 − 18 σ3 σ1
6 σ2 − 18 σ3 σ1
2. When trying to transform the former expression into barycentrics, the formal computer is
poisoned by the√following fact. Quantity σ4 describes the orientation of the triangle, while
the choice of ± 3 depends of the orientation of the whole plane. We better generalize the
problem using tan (AB, AA0 ) = K. This leads to :


4 K σ22 − 3 σ3 σ1 + σ4 K 2 + 1 σ1



2 K (σ2 σ1 − 9 σ3 ) + σ4 3 + K 2
ζK ' 

σ2 
2
2
4 K σ1 − 3 σ2 + σ4 K + 1
σ3

 

γ−β
1


 (β + γ) K + i (γ − β)   2 S − Sa K 
p

 

α−γ
1

 
 

'
 q '
(α
+
γ)
K
+
i
(α
−
γ)

  2 S − Sb K 



1
r
β−α
2 S − Sc K
(α + β) K + i (β − α)
3. And we obtain a lot of results when changing K, and even more by isogonal conjugacy. In
the following table, line K lists usual values for the tangent of an angle, while the other two
lines give the Kimberling number of the corresponding points. The PK points are on the
Kiepert RH (more details in Proposition 12.18.1).
√
√
−1
−1
+1
+1
√
K
− 3 −1
0 √
1
3 ∞
2
2
3
3
PK
isog (PK )
13
15
485 3316
371 3311
17
61
2
6
18
62
3317 486
3312 372
14
16
4
3
Fact 14.5.5. As of jmax < 3587, there are 367 points whose barycentrics contain R. Among
them :
1. 49 contain other literal radicandes
2. 85 are are in C (a, b, c, S) but aren’t in C a2 , b2 , c2 , S
3. 233 are in fact inside C a2 , b2 , c2 , S . When changing R into −R, three aren’t paired 2981,
3381, 3382: there are 115 pairs. For 46 of them, the isogonal of a known pair is again a
known pair.
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
14.6
163
Lubin representation of second degree
When dealing with half angles, we have to introduce the mid-arcs on the circumcircle of ABC, i.e.
the circumcevians of the in-excenters.
Proposition 14.6.1. Lubin-2 parameterization. When using parameterization zA = α2 , etc,
the mid-arcs are ±αβ, ±βγ, ±γα. But there are only four choices of sign since
product of midarcs = (−1) × product of vertices
must be enforced. When using the symmetric choice, i.e. −αβ, −βγ, −γα then lines AIa , BIb , CIc
concur at −αβ − βγ − γα = −s2 .
Theorem 14.6.2. Lemoine transforms. When using barycentrics, the three transforms a 7→
−a, b 7→ −b and c 7→ −c form a Klein group together with the identity. They are called the Lemoine
transforms (see Lemoine, 1891). They swap the inexcenters by pairs. When using Lubin-2, they
are obtained by α 7→ −α or β 7→ −β or γ 7→ −γ . The Lubin-1 points (aka the strong points) are
unchanged by these transforms.
Proof. The fact that La ◦Lb = Lc comes from the homogeneity required for the formulas of interest.
Remember: a theorem is a proposition with the biggest consequences, not something difficult to
prove.
Proposition 14.6.3. Forward-2 and backward-2 matrices. Using the Lubin-2 parameterization, we have :


α2
β2
γ2
iσ4
i s4 s1 s2 − s3
4i


Lu2 =  1
(14.14)
=
= 2S
1
1  ; det Lu2 =
σ
s
s
R
3
3
3
2
2
2
1/α 1/β
1/γ
Lu−1
2

α2 β 2 − γ 2
1  2 2
=
 β γ − α2
iσ4
γ 2 α2 − β 2
α2 γ 4 − β 4
β 2 α4 − γ 4
γ 2 β 4 − α4

σ3 β 2 − γ 2

σ3 γ 2 − α 2 
σ3 α2 − β 2
Proof. These formulas come from (14.12). Remember that s1 = α + β + γ while σ1 = zA + zB + zC
etc.
Theorem 14.6.4. Forward-2 substitutions. Suppose that barycentrics p : q : r of point P
depends rationally on a, b, c, S. Then Morley-affix of P is obtained by substituting the identities :
i R 2 α2 − β 2 γ 2 − α2 β 2 − γ 2
γ
β
γ
α
α
β
S=
; a = iR
−
; b = iR
−
; c = iR
−
4 α2 β 2 γ 2
β
γ
γ
α
α β
(14.15)
into p : q : r and applying (14.10). The result obtained is a rational fraction in α, β, γ whose
degree is +2. When P is a triangle center, ζP depends only on s1 , s2 , s3 . When P is invariant by
circular permutation, but not by transposition, a s4 term appears.
Proof. Elimination of R comes from homogeneity. Symmetry properties are evident. Sign chosen
for a is irrelevant, but signs of b, c must be chosen accordingly.
Proposition 14.6.5. Backward substitutions. Let zP ∈ C (α, β, γ) be an homogeneous rational fraction, supposed to be the Lubin-2 affix of a finite point. Then deg (zP ) = 2 is required.
Alternatively, let ω 2 ∈ C (α, β, γ) is an homogeneous
rational fraction, supposed to describe the
Lubin-2 affix of a direction. Then deg ω 2 = 4 is required. When these conditions are fulfilled, the
ABC-barycentrics p : q : r of these objects can be obtained as follows. Compute the corresponding
vector :








p
zp
p
ω2








−1
−1
 q  = Lu2 .  1  or  q  = Lu2 .  0 
r
zp
r
1
then apply substitutions
α = −1 ; β =
Sc + 2 iS
Sb − 2 iS
; γ=
ab
ac
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
(14.16)
164
14.7. Poncelet representation
to this vector and simplify the obtained expression using the Heron formula :
S2 = −
1
(a + b + c) (b + c − a) (c + a − b) (a + b − c)
16
Proof. Result about degrees follows Proposition 14.3.7. Substitutions are α = −1, β = exp (+iC) , γ =
b = (BC, BA) while C
b = (CA, CB).
exp (−iB). This result is indeed symmetric, since B
14.7
Poncelet representation
Definition 14.7.1. The parameters of this representations are the contact points of the incircle,
described as ρ, σ, τ in a frame using this circle as unit circle. Thus ρ = 1/ρ, etc.
Proposition 14.7.2. This representation describes the triangle ABC by:


2τ σ
2ρτ
2ρσ


σ+τ
τ +ρ
ρ+σ

P on = 


1
1
1
−1
−1
−1
2 (σ + τ )
2 (τ + ρ)
2 (ρ + σ)
The algebraic direct substitutions are:
a=
2 iρ (σ − τ )
2 iσ (τ − ρ)
2 iτ (ρ − σ)
ri ; b =
ri ; c =
ri
(τ + ρ) (ρ + σ)
(ρ + σ) (σ + τ )
(σ + τ ) (τ + ρ)
2τ ρσ
(ρ − σ) (σ − τ ) (τ − ρ)
S = i ri2
; R = −ri
(τ + ρ) (ρ + σ) (σ + τ )
(τ + ρ) (ρ + σ) (σ + τ )
(14.17)
while the backward substitutions are:
ρ a2 + b2 − c2
ρ a2 − b2 + c2
2 iρ S
−2 iρ S
−
; τ=
−
σ=
ab
2ab
ac
2ac
(14.18)
Proof. Tangent Tρ at ρ to the unit circle is [ρ−1 , −2, ρ], and A = Tσ ∩ Tρ , etc. Signum for a
can be chosen at will, but the other two have to be synchronized. The σ, τ formulas are using
σ = ρ exp i (π + C) , τ = ρ exp i (π − B) i.e. the π − A property, and correct orientations.
Proposition 14.7.3. Going back from the Poncelet affix ζM of a point M to the Lubin-2 affix zM
of this point only requires the similitude:
r 1
i
zM = −s2 + (s2 s1 − s3 ) ζM = −s2 +
(−s3 ) ζM
2
R
Proof. Substituting (14.15) in (14.18), one obtains
(β + γ) (α + γ) (α + β)
ρ = λα ; σ = λβ ; τ = λγ ; ri = −R
2 αβγ
Obviously, we choose λ = 1, and there is no more a change of parameterization. It only remains a
change of the projective basis, that is given by:


1
0
−s2
(s2 s1 − s3 )


2


0
1
0
Lu · P on−1 = 

 s2 s1 − s3

s1
−
0
2
2 s3
s3
Fact 14.7.4. The Poncelet affix of an ordinary point (i.e. ζM ) is an homogeneous fraction whose
total degree is +1. The Lubin-2 affix (i.e. zM ) of the same point has total degree 2... and this is
verified in zM = −s2 + 21 (s2 s1 − s3 ) ζM since the degree of ζ is -1.
Remark 14.7.5. In fact, formula zM = −s2 + rRi (−s3 ) ζM is rather obvious.
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
165
• The translation −s2 is required to move I0 from point ζ = 0 to its new place in the Lubin-2
frame, i.e. z = −βγ − γα − αβ.
• Homothety ri /R is required to acknowledge that the radius of the "unit" circle has changed.
• Exchanging the umbilics is required by the orientations, i.e. the values of :
det Lu /S = 4i/R2 ; det P on /S = −4i/ri2
• And lastly, the −s3 is required by the property used in the Acknowledgment of this book:
the intouch triangle is perspective with the circumcevian triangle of the incenter, while the
perspector X(56) is the similicenter of the incircle and the circumcircle: the mid-arcs are
−βγ, etc. By complex conjugacy, they become −1/βγ and multiplying by −αβγ is the
rotation that gives the required α, etc.
Remark 14.7.6. The only difference between Poncelet and Lubin-2 parametrization is their behavior
wrt the Lemoine transform. Since Lubin-2 coordinates are relative to the strong points Ω± and
Ω0 =X(3), all of the strong points are invariant by the Lemoine transforms. When using Poncelet
representation, the origin changes, and therefore the coordinates of the strong points don’t remain
unchanged by the Lemoine transforms, annoying property.
14.8
Poulbot’s points (using the Lubin-4 parameterization)
Proposition 14.8.1. When using half angles, i.e. A/2, etc, we can introduce α, etc such that
zA = α4 , etc. The intermediate points on the circumcircle can be described as:
β 4 ; iβ 3 γ, −β 2 γ 2 , −iβ γ 3 ; γ 4 ; γ 3 α, γ 2 α2 , γα3 ; α4 ; α3 β, α2 β 2 , α β 3 ; β 4
And then we have:
bc + Sa
A
A
; cos =
cot =
2
2S
2
cos
√
√
√
√
b+c+a b+c−a
A
a+b−c a−b+c
√ √
√ √
; sin =
2
2 b c
2 b c
β2 − γ2
A
β2 + γ2
A
=i
; sin = −
2
2 βγ
2
2 βγ
Proof. Everything goes like in Lubin-2: product of points 2,5,8 must be i times the product of
points 1,4,7, etc.
Remark 14.8.2. The following quantity belongs to Lubin-2 :
A
B
C
ri
S3 − S1 S2
sin
sin
sin
=
=
2
2
2
4R
8 S3
Remember that s1 = α + β + γ, S1 = α2 + β 2 + γ 2 , σ1 = α4 + β 4 + γ 4 .
Example 14.8.3. Consider the circles going through the incenter I and tangent to AB and AC .
1. Clearly the centers of these circles, say Aj , must be on the AI line. Thus:

µ α4 + (1 − µ) −α2 β 2 − α2 γ 2 − β 2 γ 2


1
Aj = µA + (1 − µ) I = 

α2 + β 2 + γ 2
µ
− (1 − µ)
α4
α2 β 2 γ 2





2. Equating the distance to I and the distance to AB (14.9)we obtain :
α2 + γ 2 α2 + β 2
β 2 + γ 2 α2 + γ 2 α2 + β 2
= (1 − µ)
±µ
α2 β γ
2 α2 β 2 γ 2
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
166
14.8. Poulbot’s points (using the Lubin-4 parameterization)
α
β
γ
I0
Ia
Ib
Ic
Aj
Bj
Cj
Ajk
Bkn
Cnj
Ajk
Bkn
Cnj
iα
α
−α
+α
α
β
β
+β
−β
β
γ
γ
γ
γ
−γ
Ia
I0
I0
I0
I0
I0
Ia
Ia
Ia
Ia
Ic
Ib
Ib
Ib
Ib
Ib
Ic
Ic
Ic
Ic
A1
A1
A0
A0
B1
B0
B1
B0
C1
C0
C0
C1
A11
A00
A10
A01
B11
B01
B00
B10
C11
C10
C01
C00
A00
A11
A01
A10
B00
B10
B11
B01
C00
C01
C10
C11
Table 14.1: Action of the Klein group
leading to µ = β 2 + γ 2 ÷ β 2 ± 2 β γ + γ 2 , and to:



A1 = 


S2 α3 − α s23 + 2 S2 s3
−α3 + S1 α − 2 s3
1
S1 s3 − α2 s3 + 2 S1 α3
α3 s3 (−α3 + S1 α − 2 s3 )






 ; A0 = 




S2 α3 − α s23 − 2 S2 s3
−α3 + S1 α + 2 s3
1
S1 s3 − α2 s3 − 2 S1 α3
α3 s3 (−α3 + S1 α + 2 s3 )






Here s3 = αβγ while S1 = α2 + β 2 + γ 2 , etc. Point A1 is unchanged by α 7→ −α (since this
leads also to s3 7→ −s3 ). But β 7→ −β changes only s3 and exchanges A1 and A0 .
3. Obviously, circles (A1 ) and (A0 ) are tangent at I.
4. Let us note Xjk the second intersection of circle Yj and circle Zk . For example, B31 is
(C3 ) ∩ (A1 ). We obtain:
z (A11 ) =
!
2 α3 βγ β 2 + γ 2 + 2 αβ 3 γ 3 + β 2 γ 2 (β + γ) β 2 + γ 2
+α2 β 2 + βγ + γ 2 β 2 − βγ + γ 2 (β + γ)
α2 (β + γ) − 2 αβγ
And then, we use β + γ = s1 − α and βγ = s2 − αs1 + α2 . This leads to "huge" polynomials
in α, that can be reduced using the relation α3 − s1 α2 + s2 α − s3 = 0. As a result:
!
!
2 s31 s3 − s21 s22 + s32
−2 s41 s3 + s31 s22 + 4 s21 s2 s3
α+
−2 s1 s2 s3 + s23
−2 s1 s32 − 7 s1 s23 + 3 s22 s3
z (A11 ) =
α s2 − 3 s3
5. And now consider circles (A, Ajk , I), (B, Bkn , I), (C, Cnj , I). On the Figure, we can see that
these circles concur in a second point. Under the action of the Klein group of the α 7→ −α
transforms, the parity of j + k + n is kept, as described in Table 14.1: two situations are
encountered.
6. Construction. The center B4 of circle (B, B00 , I) can be obtained as A0 C0 ∩ Ma Mc where
A0 , C0 are the centers described above, and Ma , Mc are the mid-arcs relative to the incenter
I0 .
7. Spoiler (Veronese map). Use z (A) = α4 , z (I) = 2s1 s3 − s22 and z (A11 ) as above. Take the
Veronese of these points, and then the wedge of these three rows. Reduce this column, and
take the remainder of each element wrt α3 − s1 α2 + s2 α − s3 . Now, the representative of
circle (I, A, A11 ) is written as
Va = V1 + αV2 + α2 V3
were V1 , V2 , V3 are symmetric in α, β, γ. One can see that the family V1 , V2 , V3 is not independent, so that the Va , Vb , Vc belong to a same pencil.
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
(a) Whole figure
(b) Core figure
orange: 11 circles, blue 00 circles, magenta: 01 or 10 circles
Figure 14.1: Poulbot’s points
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
167
168
14.9. Pencil of cycles
8. Spoiler (pencil of circles). Determine the point-circles in this pencil, i.e. determine K so that
t
(V1 + K V2 ) · Qz · (V1 + K V2 ) = 0. This equation factors gently, giving K and therefore
the point-circles. Going back to the equations in Z, T, Z, we obtain factored equations:
!
! !
!
s21 − 2 s2 T
s31 s2 − 2 s1 s22
2 s41 s2 s3 − s31 s32 − 2 s31 s23 − 3 s21 s22 s3
×
T+
Z =0
−3 s21 s3 + 4 s2 s3
+2 s1 s42 + 10 s1 s2 s23 − 4 s32 s3 − 4 s33
+Z s23
and conjugate. This characterizes the base points of an isoptic pencil. Thus the jkn = 111
case leads to Poulbot’s points of first kind (Ayme et al., 2014) :
zU =
−2 s41 s2 s3 + s31 s32 + 2 s31 s23 + 3 s21 s22 s3 − 2 s1 s42 − 10 s1 s2 s23 + 4 s32 s3 + 4 s33
s31 s2 − 2 s1 s22 − 3 s21 s3 + 4 s2 s3
9. In the same vein, the jkn = 000 case leads to Poulbot points of second kind:
zW =
s21 s32 − 2s31 s2 s3 − 2s42 + 3s1 s22 s3 + 2s21 s23 − 2s2 s23
s21 s2 − 2s22 + s1 s3
and it can be seen that this point is on the Bevan circle Ja Jb Jc . So are the other three.
14.9
14.9.1
Pencil of cycles
Veronese map
Definition 14.9.1. The Veronese map used in the triangle plane was defined at (13.3). The
Veronese map used in the Morley space is:
V erz Z : T : Z ' ZT, T2 , ZT, ZZ − T2
(14.19)
Proposition 14.9.2. Both umbilics are mapped to [0, 0, 0, 0] (points of indeterminacy) while
all other points at infinity are mapped to [0, 0, 0, 1], called Sirius. When using Z : T : Z =
Lu . (x : y : z), we have
!
t
t
Lu
0
z
V er Z : T : Z ' V er (x : y : z) · Lv
where Lv =
0
1/R2
t
Proof. The Lu part is obvious. The 1/R2 acknowledges the fact that (X3 , R) is the model of all
circles in the triangle plane.
Remark 14.9.3. The Veronese map amounts to generate the projective space of all the cycles
from four of them, namely the horizon circle T2 (i.e. the line at infinity described twice), the
fundamental isotropic lines (each of them completed by the line at infinity to obtain ZT and ZT)
and the "unit" circle ZZ − T2 . Here "unit" doesn’t mean "radius one", but the fact that this
circle is used to measure the how circular are the other cycles.
Proposition 14.9.4. The points of the Veronese map belong to a 3D quadric:

0 0 1 0

−1 t
−1
 0 −2 0 −1
V erz (P ) · Qz
· V erz (P ) = 0
where Qz
= 2
 1 0 0 0
0 −1 0 0
And the columns representative of point-circles, i.e. the UPz ' Qz
paraboloid:

0 0
 0 0
t
1

U z · Qz · U z = 0
where Qz = 
2 1 0
0 −1
March 30, 2017 14:09
−1





(14.20)
t
· V erz (P ), belong to a 3D

1 0
0 −1 


0 0 
0 2
(14.21)
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
Proof. We have Qz
−1
= Lv −1 · Q
−1
t
169
t
· Lv −1 and Qz = Lv · Q · Lv .
Proposition 14.9.5. Two point-circles are orthogonal wrt the quadric when their centers share
one of their two coordinates .
Proof. We have the more precise result:
t
V erz Z : T : Z · Qz · V erz (z : t : ζ) = −2tT
Z
z
−
T
t
Z
ζ
−
T
t
Remark 14.9.6. And then, business as usual, using the adapted matrices.
14.9.2
Back to the Euler pencil
This pencil has been treated in detail at Section 13.6. Let us examine again this pencil, in the
light of the new formalism.
1. We use the Lubin-1 representation, i.e. Az ' α : 1 : 1/α. The representative of Γ is known
to be Vo ' 0 : 0 : 0 : 1. The Euler circle goes through the midpoints. Its representative is
therefore:


−2 σ2
 σ σ + 3σ 
^
 1 2
3 
Azeul '
V erz (Az + B z ) ' 

 −2 σ1 σ3 
3
4 σ3
2. Applying the formalism, we have
t
Aeul · Qz · Aeul ÷ V42 =
−σ1 σ3 , −2 σ3 , −σ2 ,
5 σ3 − σ1 σ2
2
. (· · · ) =
1
4
i.e. confirmation of center X(5) z = σ1 /2 and radius 1/2.
3. The line representative of the Euler pencil (generated by circum- and Euler circles) is therefore:


0
2 σ1 σ3 σ1 σ2 + 3 σ3 0

^
−2 σ1 σ3
0
2 σ2
0 
.


Eulerzpoint = Ao Aeul ' 



−σ
σ
−
3
σ
−2
σ
0
0
1 2
3
2
4
0
0
0
0
4. Consider the point z : t : ζ. Its Veronese image is zt, t2 , ζt, t2 − zζ and the associated
point circle is represented by VF ' −tζ : t2 + zζ : −tz : t2 . This circle belongs to the Euler
→
−
pencil when t AF · Eulerzpoint = 0 . One obtains
 σ σ + 3σ
σ3 W
1 2
3
±

σ2
σ2

F± ' 
4
 σ σ + 3σ
W
1 2
3
±
σ1 σ3
σ1





.
where W 2 =
9−
σ1 σ2
σ3
σ1 σ2
1−
σ3
5. The midpoint of F+ , F− is X(468). This is recognizable to the fact that z468 = (z4 + 3z186 ) /4
where z4 = σ1 is the affix of the orthocenter X(4) and z186 = σ3 /σ2 is the affix of the inverse
of X(4) in the circumcircle, i.e. X(186).
6. The direction of line F+ F− is σ1 σ3 : 0 : σ2 i.e. X(30)... the direction of the Euler line. This
was obvious from the beginning.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
170
14.9. Pencil of cycles
(a) Acute triangle
(b) Obtuse triangle
Figure 14.2: The Euler pencil
7. Both factors of quantity W 2 are real. This can be seen from σ1 = σ2 /σ3 . Since |σ1 | < 3, the
first factor is ever positive. The second factor can be split into :
+1 −
(β + γ) (α + γ) (β + α)
σ1 σ2
=−
σ3
αβγ
Thus W 2 = 1 for equilateral triangles (σ1 = 0) and vanishes only for right-angled triangles,
so that W 2 > 0 characterizes acute triangles.
8. Points F± are the base points of the Euler pencil, seen as an isotomic pencil. Consider now
the isopticpencil of the cycles that are orthogonal to all of the cycles of the Euler pencil
(i.e. the orthic pencil). When A1 , A2 are the representatives of two orthogonal cycles, then
t
A1 · Qz · A2 = 0. Thus t Aeul · Qz is a plane that describes the bundle of cycles orthogonal
to Ωeul . And therefore
!
^
.
t z
t z
z
Orthicpoint = dual
Ab · Qz
Aa · Qz
'





4
0
2 σ1 σ3
0
−4 σ1 σ3
−2 σ1 σ3
0
−2 σ2
−σ1 σ2 − 3 σ3
0
2 σ2
0
−4 σ2
4 σ1 σ3
σ1 σ2 + 3 σ3
4 σ2
0





9. Using the same method as above, we search the centers of the point-circles that belong to
the Orthic pencil. We obtain:
 σ σ + 3σ
σ3 W 
1 2
3
±

σ2
σ2 


E± ' 
where Wisasabove
4

 σ σ + 3σ

W
1 2
3
∓−
σ1 σ3
σ1
10. The midpoint of E+ , E− is X(468) again, while the direction of E+ E− is σ1 σ3 : 0 : −σ2 , i.e.
X(511).
14.9.3
Back to the Brocard-Lemoine pencil
This pencil has been treated in detail at Section 13.7. Let us examine again this pencil, in the
light of the new formalism.
1. We use the Lubin-2 representation i.e. Az ' α2 : 1 : α−2 and obtain :

 

β γ α2 β 2 + α2 β γ + α2 γ 2 − β 2 γ 2
β γ α2 β 2 − α2 β γ + α2 γ 2 − β 2 γ 2

 

Iaz , Jaz ' 
α2 + β γ β γ
α2 − β γ β γ
,

2
2
2
2
2
2
β −α +βγ+γ
α −β +βγ−γ
March 30, 2017 14:09
published under the GNU Free Documentation License
14. Morley spaces, or how to use complex numbers
171
2. Now, we take the wedge of the Veronese of these points, and obtain


β 2 + γ 2 − 2 α2
 2 α2 − β γ α2 + β γ 


Aza ' Az ∧ Iaz ∧ Jaz '  2
 ' Lv −1 · Aa
 α 2 β 2 γ 2 − α2 β 2 − α2 γ 2 
α2 − β γ α2 + β γ
Thereafter, we obtain the line describing the pencil containing Aza , Azb , Azc . The expression
is obviously symmetric and, moreover, can be expressed using Lubin-1, i.e. zA = α etc. This
leads to:
^
t
.
Lemoinezpoint = Aza Azb ' Lv · Lemoinepoint · Lv
'





4
3 σ1 σ3 − σ22
0
3 σ2 − σ12
9 σ3 − σ1 σ2
0
2
σ2 − 3 σ1 σ3
0
6 σ1 σ3 − 2 σ22
0
2
σ1 − 3 σ2
0
6 σ2 − 2 σ12
3. The conjugate pencil is obtained by:
Brocardzpoint
.
=
dual



' 

t
Aza
· Qz
0
3 σ1 σ3 − σ22
9 σ3 − σ1 σ2
0
4. Solving V erz Z : T : Z · Qz
−1
4
t
Azb
σ22
· Qz
− 3 σ1 σ3
0
3 σ2 − σ12
0
!





t
' Lv · Brocardpoint · Lv
σ1 σ2 − 9 σ3
σ12 − 3 σ2
0
0
0
0
0
0





→
−
· Brocardzpoint = 0 give the point-circles that generate
the Brocard pencil. One obtains :

F±z
^
2 σ22 − 6 σ1 σ3
σ1 σ2 − 9 σ3
2 σ12 − 6 σ2
0



'


√
σ1 σ2 − 9 σ3
i 3V dm
± 2
σ12 − 3 σ2
σ1 − 3 σ2
2
√
9 σ3 − σ1 σ2
i 3V dm
∓
3 σ1 σ3 − σ22
3 σ1 σ3 − σ22

√
i 3V dm
σ1 σ2 − 9 σ3
± 2
σ12 − 3 σ2
σ1 − 3 σ2
2
√
9 σ3 − σ1 σ2
i 3V dm
±
3 σ1 σ3 − σ22
3 σ1 σ3 − σ22







and one can check that F+ is X(15) and F− is X(16): they are the Poncelet points of the
Brocard pencil.
−1
→
−
5. Solving V erz Z : T : Z · Qz
· Lemoinezpoint = 0 give the point-circles that generate
the Lemoine pencil. One obtains :

z
E±



'








6. Obviously, pairs F± and E± share the same midpoint, while their directions are orthogonal.
But there is a huge difference between both pairs: points of the F pair are visible, the other
are not, one is z1 : 1 : z1 , z2 : 1 : z2 ,the other is z1 : 1 : z2 , z2 : 1 : z1 .
14.10
Comparison with Cartesian and Artinian metrics
To be written.
For all metrics, OrtO
∗
=
OrtO
2 ∗
while trace
OrtO
∗
= 1.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
172
March 30, 2017 14:09
14.10. Comparison with Cartesian and Artinian metrics
published under the GNU Free Documentation License
Chapter 15
Collineations
15.1
Definition
Definition 15.1.1. A collineation is a reversible linear transformation of the barycentrics, i.e.
U = φ (P ) determined by :
 
m11
u
 

 v  '  m21
m31
w

m12
m22
m32


p
m13


m23   q 
r
m33
where ' is a reminder of the fact that barycentrics are determined up a proportionality factor.
Proposition 15.1.2. A collineation is determined by two ordered lists four points: Pi , i = 1, 2, 3, 4,
Ui , i = 1, 2, 3, 4 such that no triples of P points are on the same line, and the same for the Q points.
Proof. If φ is reversible, then det M 6= 0 is required and the φ (Pi ) haven’t alignments when the
Pi haven’t. Conversely, we have the following algorithm.
Algorithm 15.1.3. Collineation algorithm. Let be given the two lists of points Pi , i = 1, 2, 3, 4,
Ui , i = 1, 2, 3, 4. With obvious notations, the question is to find the mij (not all being 0) and the
ki (none being 0) in order to ensure :

u1 u2

 v1 v 2
w1 w2




k1 0 0 0
m11 m12 m13
p1 p2 p3 p4
u3 u4 



  0 k2 0 0  
v3 v4  
 =  m21 m22 m23   q1 q2 q3 q4 
 0 0 k3 0 
m31 m32 m33
r1 r2 r3 r4
w3 w4
0 0 0 k4


This system has 13 unknowns and 12 equations, since a global proportionality factor remains
undetermined. The ki are determined (up to a global proportionality factor) by
1
det Q = det P det M
6=i
ki
6=i
where a 3 × 4 matrix subscribed by an 6= i refer to the square matrix obtained by deleting the i-th
column. Thereafter, M is easy to obtain. To summarize :
ki = det Q ÷ det P
6=i
6=i
M = Q .K .
6=4 6=4
P
−1
6=4
With the given hypotheses, transformation φ is clearly reversible.
Remark 15.1.4. An efficient choice of the Pi , Qi is eight centers, or a central triangle and a center
for the P and the corresponding Q. In such a case, any center is transformed into a center, and
homogenous curves into homogeneous curves of the same degree.
173
174
15.2
15.2. Involutory collineations
Involutory collineations
Proposition 15.2.1. . Let M1 , M2 , N1 , N2 be four (different points). The collineation ψ that
swaps the (M1 , M2 ) pair and also the (N1 , N2 ) pair is involutory. The line through the crossed
intersections M1 N1 ∩ M2 N2 and M1 N2 ∩ M2 N1 is a line of fixed points (the axis of ψ). The
paired intersection, i.e. point P = M1 M2 ∩ N1 N2 is an isolated fixed point (the pole of ψ).
Reciprocally, given an axis ∆ and a pole P (outside of the axis), we obtain an involutory transform
U 7→ X = ψ (U ) by requiring P, U, X aligned together with (P, U, P U ∩ ∆, X) = −1 (harmonic
conjugacy).
Proof. Consider ψ defined by A ←→ P and B ←→ C.
general 15.1.3 . Then the cevian triangle of P provides a


1
0
0
 q
q 

0
− 
ψ ' p
;
TP
r 
 r

r
−
0
p
q

−1
−1

TP
. ψ . TP =  0
0
Its matrix ψ can be obtained by the
diagonalization basis and we have :

0 p

= q 0
r r

0 0

1 0 
0 1

p

q 
0
In the general case, we can chose matrix ψ to enforce det ψ = −1. Then minimal polynomial is
2
µ2 − 1 while characteristic polynomial is (µ − 1) (µ + 1).
15.3
Usual affine transforms as collineations
Remark 15.3.1. Umbilics have been defined in Subsection 13.1.2. A possible choice can be described
as Ω± ' abc X512 ± iR X511 . The exact value is given in (13.2).
→
−
→
−
Proposition 15.3.2. Translation. The matrix of the translation U 7→ U + V where V = (p, q, r)
is given by :

 

1+p
p
p
1 0 0
−

 
 →
(15.1)
1+q
q  =  0 1 0  + V . L∞
 q
r
r
1+r
0 0 1
Proof. Use Pi = A, B, Ω+ , Ω− and Qi = C, D, Ω+ , Ω− . Characteristic polynomial is
2
(X − 1) (X − 1 − p − q − r)
For a translation, p + q + r = 0, and the matrix is not diagonalizable.
−
−
−
→
→
→ −
→
Remark 15.3.3. The translation operator is linear, meaning that M V1 +M V2 = M V1 + V2 .
Proposition 15.3.4. Homothety. When p + q + r is different from 0 and −1 then (15.1) characterizes the homothety centered at point P = p : q : r with ratio µ = 1/ (1 + p + q + r).
Proof. The factor is the reciprocal of the eigenvalue λP since a fixed point should be described by
λ = 1 : all the eigenvalues have to be divided by λP . This result can also be obtained by direct
−−−−−−−→
examination of f (P ) f (U ).
t
Remark 15.3.5. When computing P , the column (p, q, r) can be viewed as "defined up to a
proportionality". This does not apply to the computation of µ. In any case, we are re-obtaining
(7.25).
Proposition 15.3.6. Similarity. When A, B, C, D are points at finite distance, with A 6= B,
C 6= D it exists two similarities φ, ψ, one forwards and the other backwards, that sends A →
7 C
and B 7→ D. As collineations, they are characterized by :
φ = collineate A, B, Ω+ , Ω− ; C, D, Ω+ , Ω−
ψ = collineate A, B, Ω+ , Ω− ; C, D, Ω− , Ω+
March 30, 2017 14:09
published under the GNU Free Documentation License
15. Collineations
175
Proof. The group of all the similarities is the stabilizer subgroup of the pair {Ω+ , Ω− } under the
action of the group of all the collineations. This comes from the fact that any similitude transforms
circles into circles, and therefore must preserve the umbilical pair.
Remark 15.3.7. The matrix π∆ of the orthogonal projector onto line ∆ ' [p, q, r] is :
π∆ = ∆ . M . t ∆ − M . t ∆ . ∆
while the matrix σ∆ of the orthogonal reflection wrt line ∆ ' [p, q, r] is :
σ∆ = ∆ . M . t ∆ − 2 M . t ∆ . ∆
These formulas are recalled from Section 7.8, where more details are given.
Proposition 15.3.8. The matrix of the rotation centered at finite point P = p : q : r with angle
φ is :

p

Φ = q
r
p
q
r
 
r+q
p
 
q  +  −q
−r
r
−p
r+p
−r

−p

−q  cos φ + OrtO (L∞ . P − P.L∞ ) sin φ
q+p
Proof. It suffices to check what happens to P, Ω+ , Ω− : they are fixed points, with respective
eigenvalues : 1, exp (+iφ) , exp (−iφ) — up to a global factor p + q + r.
Stratospherical proof. A rotation with angle φ is multiplication by Φ = cos φ+i sin φ in the complex
plane. Therefore, rotation with center P and angle φ can be written as :
−−→
Φ (X) = P + 1 cos φ + i sin φ P X
Since matrix OrtO describes a "project and turn" action, we have OrtO
i = OrtO . Multiplying, we get : OrtO
4
2
= − OrtO
2
and − OrtO
2
3
= − OrtO , so that
is a projector onto space
V. This gives : 1 = − OrtO . Canceling the denominators, we obtain :



p
x




Φ  y  ' (x + y + z)  q  +
r
z

sin φ OrtO − cos φ OrtO
leading to the required matrix :
Φ = (L∞ . P ) Q + (1 − Q) . P . L∞
15.4




x
p




(p + q + r)  y  − (x + y + z)  q 
z
r
2 

2
.
where Q = sin φ OrtO − cos φ OrtO
Barycentric multiplication as a collineation
Proposition 15.4.1. Barycentric multiplication by P = p : q : r is what happens to the plane
when using collineation φ : (A, B, C, X2 ) 7→ (A, B, C, P ). In other words :


p 0 0
. 

X ∗ P = P .X
where P =  0 q 0 
b
0 0 r
Remark 15.4.2. Obviously, trilinear multiplication can be described using collineations involving
X1 .
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
176
15.5. Complement and anticomplement as collineations
Proposition 15.4.3. The collineation whose matrix is diagonal, with elements U ÷ P transforms
b
(A, B, C, P ) into (A, B, C, U ) and circumconic CC (P ) into CC (U ).
Proof. Direct examination. One obtains :
t
u
p
pqr 

 0
uvw 
0
15.5
0
v
q
0
0


0

 
0 .  w

w
v
r
w
0
u
 u
v
 p
 
u . 0

0
0

0
v
q
0


r q

0 p 
p 0

0

 
0 = r

w
q
r
0
Complement and anticomplement as collineations
Proposition 15.5.1. Complement is what happens to the plane when using collineation (A, B, C, X2 ) 7→
(A2 B2 C2 , X2 ) where A2 B2 C2 is the medial triangle. In other words :


0 1 1
. 

where
C = 1 0 1 
complem (X) = C . X
1 1 0


−1 1
1


anticomplem (X) = C −1 . X
where
C −1 =  1 −1 1 
1
1 −1
Proof. Direct computation.
Proposition 15.5.2. The cevian collineation wrt point P is defined as collineation (A, B, C, P ) 7→
(AP BP CP , P ). Its matrix is :


0 p/q p/r


φP = P . C . P −1 =  q/p 0
q/r 
r/p r/q
0
where P −1 is to be understood as the reciprocal of matrix P .
Proof. Composition of the two former collineations.
Proposition 15.5.3. We have the following relation between conics :
complem (X) ∈ conicev (isotom (U ) , X2 ) ⇐⇒ X ∈ conicir (complem (U ))
15.6
Collineations and cevamul, cevadiv, crossmul, crossdiv
In this Section 15.6, the start point will ever be Table 4.2 (II) i.e. T1 = CP (the cevian of P ),
T2 = ABC, T3 = AU (the anticevian of U ).
Proposition 15.6.1. Start as described, and use φU = U . C . U −1 . This collineation is
tailored so that : φ (T3 ) = ABC, φ (T2 ) = CU and φ (T1 ) is the cevian triangle of φ (P ) wrt
CU . Then :
φ . cevadiv (P, U )
cevamul φ−1 .X, φ−1 .U
= crossdiv (φ . P, φ . U )
= φ−1 . crossmul (X, U )
=
=
u2
p
u2
x
:
:
v2
q
v2
y
:
:
w2
r
w2
z
= PU#
= XU#
Proof. Direct computation. The symmetry between U, X is broken by using φU .
Exercise 15.6.2. Explain why one obtains sqrtdiv (U, P ) and sqrtdiv (U, X), reverting the symmetry.
March 30, 2017 14:09
published under the GNU Free Documentation License
15. Collineations
177
Proposition 15.6.3. Start as described and use ψP = P . C −1 . P −1 . This collineation is
tailored so that ψ (T1 ) = ABC, ψ (T2 ) = AP and ψ (T3 ) is the anticevian of ψ (U ) wrt AP .
Triangles ψ (T3 ) and ψ (T2 ) are perspective wrt ψ (U ) while ψ (T3 ) and ψ (T1 ) are perspective wrt
ψ (X) and :
p2 q 2 r 2
ψ . cevadiv ψ −1 . P, ψ −1 . U = sqrtdiv (P, U ) =
:
:
= UP#
u v w
Proposition 15.6.4. Start as before, but use instead collineation (ABC, X2 ) 7→ (ceva (P ) , P )
i.e. ψP = C −1 . P −1 . Then ψ (T1 ) is ABC while ψ (T2 ) is the anticomplementary triangle.
Perspector between ψ (T1 ) and ψ (T2 ) is ψ (U ) = X2 , perspector between ψ (T2 ) and ψ (T3 ) is
ψ (P ) = anticomplem (U/P ) while perspector between ψ (T1 ) and ψ (T3 ) is isotomic conjugate
of the former. In other words :
X = cevadiv (P, U ) = ψ −1 ◦ isotom ◦ ψ (U )
15.7
Cevian conjugacies
Definition 15.7.1. The psi-Kimberling collineation of pole P is the collineation ψP such that
ABC 7→ cevian (P ) and X1 → P . Therefore :
= P ∗ complem (U ÷b X1 )
b
v w u w u v
+
:q
+
:r
+
= p
b
c
a
c
a
b
−1
ψP (U ) = X1 ∗ anticomplem (U ÷b P )
b
u v w
u v w
u v w
= a − + +
:b
− +
:c
+ −
p
q
r
p
q
r
p
q
r
ψP (U )
(15.2)
Remark 15.7.2. It is clear that ψP , ψP−1 are type-keeping when X1 (a : b : c), P (p : q : r) and
U (u : v : w) are transformed. Moreover, ψP (A) = Ap = 0 : q : r, ψP (X1 ) = P (from the very
definition) while ψP (−a : b : c) = A is obvious.
This ψP collineation has been used by Kimberling (2002a) to construct some new functions,
following the patterns :
φ 7→ ψP ◦ φ ◦ ψP−1
or
φ 7→ ψP−1 ◦ φ ◦ ψP
1. cevadivision of P by U can be re-obtained as ψP ◦ isogon ◦ ψP−1 . The result X is the
perspector of cevian (P ) and anticevian (U ). More about this operation in Section 4.9. One
has the formulas :
X = (−uqr + vrp + wqp) u : (uqr − vrp + wqp) v : (uqr + vrp − wqp) w
−1
P = (vz + wy)
−1
: (uz + wx)
: (yu + xv)
−1
(a) cevadivision et ceva-multiplication are both type-keeping with respect to P and U .
Using isogonal conjugacy result in the disappearing of a, b, c from the equations.
(b) fixed points are P = ψP (X1 ) and the three vertices A = ψP (−a : b : c), since ±a : ±b :
±c are the fixed points of isogon. A brute force resolution leads also to the cevians
of P . A Taylor expansion around 1 : 0 : 0 shows that vertices are really fixed points
of cevadivision, while a Taylor expansion around 0 : p : q shows that undetermined
ψP (0 : q : r) = 0 : 0 : 0 must be determined as ψP (0 : q : r) = 0 : −q : r
2. alephdivision of P by U : ψP−1 ◦ isogon ◦ ψP (Hyacinthos #4111, Oct. 11, 2001). Formulas
(cyclically) :
p2 r2 b2 + q 2 p2 c2 − q 2 r2 a2
x ' a p2 r2 v 2 + q 2 p2 w2 − q 2 r2 u2 +
(vaw + ubw + cuv)
bc
1
1
1
p2 : q 2 : r2 =
:
:
(bw + cv) (bz + cy) (cu + aw) (cx + az) (av + bu) (ay + bx)
Therefore, the alephmultiplication gives four result, one inside the triangle ABC and three
outside
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
178
15.7. Cevian conjugacies
3. bethdivision of P by U : ψP ◦ sym3 ◦ ψP−1 (Hyacinthos #4146, Oct. 26, 2001) where
involution sym3 is the reflection in the circumcenter X3 (17.10). This involution sym3 is
related to the Darboux cubic. Barycentrics are :
x
p (c + b) (a + c − b)
p (c + b) (b + a − c)
v−
w
q (−a + b + c)
r (−a + b + c)
p
b
p
b
= −a u + (c + b) v + (c + b)
where pb : qb : rb = P ∗ X7
qb
rb
' au −
(a) This operation is type-keeping with respect to P, U .
(b) The fixed points of U 7→ β (P, U ) are obtained by ψP from the fixed points of sym3.
They are ψP (X3 ) together with ψP (L∞ ), namely the line : u/b
p + v/b
q + w/b
r = 0.
(c) Bethdivision of X21 = a (b + c − a) / (b + c) by the circumcircle gives the circumcircle.
(d) Bethdivision of P by U is P if and only if U = P ∗ X57 .
b
(e) Bethmultiplication is not simple (equation of third degree).
Exercise 15.7.3. What are the situations where the discriminant vanishes ?
4. gimeldivision of P by U : ψP−1 ◦ sym3 ◦ ψP . Using barycentrics, one obtains :
F (U )
16σ 2
α
β
=
16σ 2 U − α β X1 + 2 α
X48 ∗ isot (P )
b
where
(b + c − a) (a + c − b) (b + a − c) (b + a + c)
(q + r) u (r + p) v (p + q) w
+
+
=
a
b
c
c2 + a2 − b2 b2
a2 + b2 − c2 c2
b2 + c2 − a2 a2
+
+
=
p
q
r
=
(a) The fixed points of U 7→ γ (P, U ) are obtained by ψP−1 from the fixed points of sym3.
They are ψP−1 (X3 ) together with ψP (L∞ ), namely the line : u bc (q + r) + v ca (r + p) +
w ab (p + q) = 0.
(b) Gimel multiplication leads to three points on the triangle sides, and three other points.
5. mimosa aka "much ado about nothing" X(1707)-X(1788). As with other names in ETC,
the name Mimosa is that of a star. Define mimosa (P ) as ψP−1 (X3 ). Using barycentrics, one
obtains :
mimosa (p : q : r) = u : v : w
where
!
−a2 + b2 + c2 a2
b2 a2 − b2 + c2
c2 a2 + b2 − c2
u=a −
+
+
p
q
r
and also :
mimosa (P )
−1
mimosa
(U )
=
cevadiv X92 ∗ P, X1
=
cevamul (U, X1 ) ∗ X63
b
b
Then, marvelously, the Mimosa transform M(X) arises in connection with the equation
gimeldiv(P, X) = X. And there are too many such cases of gimel conjugates for all to
be itemized in ETC... Here is a list of pairs (I,J) for which X(J) = M(X(I)).
March 30, 2017 14:09
published under the GNU Free Documentation License
15. Collineations
179
1
2
3
4
6
7
8
9
10
19
46
19
1
920
1707
1708
1158
1709
1710
1711
20 1712
21
4
27 1713
28 1714
29 1715
31 1716
35 1717
36 1718
37 1719
40 1720
48
54
55
56
57
58
59
60
63
69
43
47
1721
1722
1723
1724
109
580
9
63
71
72
73
74
75
77
78
80
81
84
846
191
1046
1725
1726
57
40
1727
579
1728
85
86
88
89
90
95
96
97
98
99
1729
1730
1731
1732
90
92
91
48
1733
1577
6. zosma, yet another star. X(1824)-X(1907). The Zosma transform of a point X is the isogonal
conjugate of the inverse mimosa transform of X.
7. dalethdivision of P by U : ψP ◦ hirst1 ◦ ψP−1 where hirst1 (X) = hirstpoint (X1 , X) and
thus U 6= P . Using barycentrics, one obtains :
x'
w v
−
r
q
2
p−
u v w
+ +
p
q
r
u−3
u2
p
(a) This operation is type-keeping with respect to P, U .
(b) The locus of fixed points of hirst1 is the circumconic cc (X1 ). Therefore, the locus of
fixed points of dalethP is the conic cvc (P, P ) tangent to the sidelines of ABC at the
cevian points of P .
8. hedivision of P by U : ψP−1 ◦φ◦ψP where φ (X) = hirstpoint (X1 , X) and thus U 6= ψP−1 (P ).
Using barycentrics, one obtains :
x ' −p
v
b
+
w 2 qa u w 2 ra u v 2
+
+
+
+
c
b a
c
c a
b
rqa2 u v u w qcp u w v w brp u v v w +
+
+
+
+
+
+
−
−
cbp a b
a c
br a c
b c
qc a b
b c
(a) The locus of fixed points is a conic, but not a conic of cevians.
15.8
15.8.1
Miscellany
Poles-of-lines and polar-of-points triangles
In what follows, indexes are to be taken modulo 3.
Definition 15.8.1. Polars-of-points triangle. Consider the general triangle T with vertices Ti ,
for i = 1, 2, 3. Taking the tripolars, we obtain a trigon. Taking the dual, we obtain the (may be
degenerate) polars-of-points triangle TU = pntpoltri (T). Its vertices are :
Ui = tripolar (Ti+1 ) ∧ tripolar (Ti+2 )
(15.3)
Definition 15.8.2. Poles-of-lines triangle. Consider the general triangle T with vertices Ti ,
for i = 1, 2, 3. Taking the dual trigon and then the tripoles, we obtain the (may be degenerate)
poles-of-lines triangle TP = linpoltri (T). Its vertices are :
Pi = tripole (Ti+1 ∧ Ti+2 )
(15.4)
Remark 15.8.3. An example of flat line-polar triangle is given by triangles sharing the circumcircle
of ABC
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
180
15.8. Miscellany
Lemma 15.8.4. The determinants of these triangles are:
2
(det T) det Tisot
2
det TU = (det Tisot ) ; det TP = Q
9 Adjoint (T)
where det Tisot is either the determinant of the triangle of the isotomics or the determinant of the
trigon of the tripolars, and Π9 is the condition expressing that two vertices of T are aligned with a
vertex of ABC.
Proposition 15.8.5. When isotomic conjugates are collinear, TU is totally degenerated. Otherwise
pntpoltri (T) is a triangle. Point-polarity is type-keeping (and both tribes share the same formula).
Proof. For example, the polar of P1 is the line x/p1 + y/q1 + z/r1 = 0, and the polars of P2 and
P3 are defined cyclically. Then U1 is obtained as the common point of the last two lines.
Proposition 15.8.6. When T is flat (aligned points), then TP is totally degenerate. When isotomic
conjugates are collinear, TP is flat, i.e. simply degenerate. Otherwise, linpoltri (T) is a triangle.
The line-polarity transform is type-keeping (and both tribes share the same formula).
Proof. For example U2 ∧ U3 gives the barycentrics of line U2 U3 , while tripole is transpose and
invert : being the product of two type-crossing transforms, linpoltri is type-keeping.
Proposition 15.8.7. Line-polar and point-polar transforms are converse of each other... in the
generic case. More precisely, pntpoltri (linpoltri (T)) gives T times det T, going back to any non
degenerate triangle. On the contrary, linpoltri (pntpoltri (T)) gives T times 1/ det Tisot : the converse relation holds certainly when isotomic conjugates of points Pi aren’t collinear and points Pi
aren’t collinear either and points Pi aren’t on the sidelines.
15.8.2
Unary cofactor triangle, eigencenter
Definition 15.8.8. The unary cofactor triangle of triangle Ui (i = 1, 2, 3) is the triangle
whose vertices are the isoconjugates of the vertices of the line-polar triangle of the points Ui . This
operator is type-crossing over the Ui , but is nevertheless type-keeping over all involved points when
using any fixed point F instead of P = F 2 . Using barycentrics :
t
Xi = (Ui+1 ∧ Ui+2 ) ∗ P
b
Proposition 15.8.9. When triangle U1 U2 U3 is degenerate (collinear vertices), then triangle X1 X2 X3
is totally degenerate (reduced to a point). Apart this situation, the unary cofactor transform is involutory.
Definition 15.8.10. Eigencenter. Any triangle U1 U2 U3 and its unary cofactor X1 X2 X3 are
perspective. Their perspector is called the eigencenter of these triangles (formula don’t simplify,
and has Maple-length 945).
When the original triangle is the cevian or the anticevian of a point U , formula shorten into :
∗ eigencenter (CU ) = anticomplem
U ∗U
∗ Up∗ = cevadiv (U, U ∗ )
b
b
P
eigencenter (AU ) = anticomplem U ∗ U ÷ P ∗ U
b
b
b
These points are called, respectively, the eigentransform and the antieigentransform of point U
(see Section 17.8).
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 16
Cremona group and isoconjugacies
16.1
Introducing the Cremona transforms
Definition 16.1.1. The upper spherical map of the Morley space PC C3 is the projection Z :
T : Z 7→ Z : T,
while the lower spherical map is the projection Z : T: Z 7→ Z : T. Each of them
sends PC C3 onto (yet another copy) of the Riemann sphere PC C2 .
Definition 16.1.2. An homography ψ is an element of PGLC C2 . Such object can be seen as
.
z 7→ ψ (z) = (az + b) / (cz + d)
acting on the Riemann sphere C.
Remark 16.1.3.
Some French writers are using ’homography’ as a synonym to collineations acting
on PC C3 . This is as stupid as possible. At High Schools, students have acquired (or should have
acquired) some behavioral skills about these ψ functions. When introducing new concepts, here
another set of points at infinity, one has to preserve what is already acquired.
Theorem 16.1.4. Consider a non-degenerated conic C in the projective plane and two fixed tangents ∆1 , ∆2 to that conic. A moving tangent ∆ cuts ∆1 at M and ∆2 at N . If we adopt two
linear parametrization, k on ∆1 and K on ∆2 , the relation M 7→ N induces an homography between parameters k and K. Moreover
correspondence ∆1 ,→ ∆2 : M 7→ N can be extended into a
transform Ψ that acts into PC C3 and looks like a pair of homographies ψ, ψ, each of them acting
onto one of the spherical maps.
Proof. Consider the inscribed conic whose auxiliary line is Q = [u, v, w]. Consider lines AB, AC
and parametrize by M = k A + (1 − k) B, N = K B + (1 − K) C. Assume that M N is tangent to
C and obtain :
kw
K=
(u + v + w) k − v
This proves the first part. Using the Lubin transmutation, we have :


Z
. 

Mz =  T 
Z


Z0
. 

Nz =  T0 
0
Z
'
'

 
α k + β (1 − k)
k


 
1

aller ·  1 − k  ' 


k
1−k
+
0
α
β

 
α kw + γ (ku + vk − v)
K


 
k (u + v + w) − v
aller · 
0
'
kw ku + vk − v
+
1−K
α
γ

Identifying with respect to parameter k, we are conducted to define
ψ =
(γ (u + v) + α w) − (βγ u + γα v + αβ w)
u+v+w
− (α v + β (u + w))
181
!




182
16.1. Introducing the Cremona transforms
in order to have :

1

 0
0
0
1
0
0

 0
0
0
1
0


 
Z0
0
ψ11
 0  
0   T  =  ψ21
0
0
0
Z

 
Z0
0
0

 
0   T0  =  0
0
1
0
Z
ψ12
ψ22
0
0
ψ22
ψ12


Z
0


0  T 
0
Z


0
Z


ψ21   T 
ψ11
Z
Theorem 16.1.5. (Continued).
Finally, the four focuses of C are the fixed points of the transform
Ψ that acts into PC C3 while the projections
of the focuses onto the upper spherical map are the
two ordinary fixed points of ψ ∈ PGLC C2 .
Proof. Fixed points of ψ are the roots of :
(u + v + w) Z2 − (α (v + w) + β (w + u) + γ (u + v)) ZT + (βγ u + γα v + αβ w) T2 = 0
This equation can be rewritten into :
v
w
u
+
+
=0
Z − αT Z − βT Z − γT
and this equation characterizes the focuses of the conic in the upper map.
Lemma 16.1.6. Let ψ be an element of PGLC C2 , i.e. an homography z 7→ (az + b) / (cz + d)
of the Riemann sphere. If we assume that the fixed points f1 , f2 of ψ are not equal, then ψ is
characterized by a number µ ∈ C that is neither 0 nor ∞. We call it the multiplier of ψ and we
have :
c f2 + d
1
.
µ = cross-ratio (f1 , f2 , z, ψ (z)) =
= ψ 0 (f1 ) = 0
c f1 + d
ψ (f2 )
Proof. To see that cross-ratio (f1 , f2 , z1 , ψ (z1 )) = cross-ratio (f1 , f2 , z2 , ψ (z2 )), use the fact that
cross-ratio (f1 , f2 , z1 , z2 )is invariant by an homography.
Lemma 16.1.7. A more symmetrical quantity is :
2
2
1
(µ − 1)
(a − d) + 4 b c
.
σ =µ+ −2=
=
µ
µ
ad − bc
Proof. Direct computation.
Remark 16.1.8. An homography z 7→ (az + b) / (cz + d) of the Riemann sphere is involutory when
a + d = 0.
Proposition 16.1.9. The focal transform :
Z : T : Z 7→
(f1 − µ f2 ) Z + (µ − 1) f1 f2 T
(g1 − ν g2 ) Z + (ν − 1) g1 g2 T
:1:
(1 − µ) Z + (µ f1 − f2 ) T
(1 − ν) Z + (ν g1 − g2 ) T
is transmuted into the following standardized transform :
(1 + µ) Z + (1 − µ) T
(1 + ν) Z + (1 − ν) T
.
Ψ0 = Z : T : Z 7→
:1:
(1 − µ) Z + (1 + µ) T
(1 − ν) Z + (1 + ν) T
by similitude :
while Ψ0 can be factored into :

f1 − f2

0

0
f1 + f2
2
g1 + g2

0

0

g1 − g2
Ψ0 = trans ◦ σ ◦ multi ◦ trans
March 30, 2017 14:09
published under the GNU Free Documentation License
16. Cremona group and isoconjugacies
183
where the Cremona transform σ, the multiplication and the translation (the same translation is
applied once and again, not once and the reverse afterward) are defined by :





1/Z
Z
σµ

 


σ  T  '  1/T  , multi =  0
1/Z
Z
0
0
−4
0
Proof. Direct computation.


1
0



0  , trans =  0

σν
0
µ+1
µ−1
1
ν+1
ν−1
0



0 

1
Definition
16.1.10. The Cremona group is defined as the set of the bi-rational transforms of
PC C3 . Therefore a transformation Ψ ∈ Cremona can be written as :
Ψ Z : T : Z = ψ1 Z, T, Z : ψ2 Z, T, Z : ψ3 Z, T, Z
where the ψj are three homogeneous polynomials of the same degree. And the existence
of another
transform Φ ∈ Cremona is assumed so that, at least formally, (Φ ◦ Ψ) Z : T : Z ' Z : T : Z .
Exercise 16.1.11. What can be said about the degrees when two transforms are inverse of each
other ? See Diller (2011)
Definition 16.1.12. Given a Cremona transform, we define the indeterminacy points and the
exceptional curves by :
Ind (Ψ)
Exc (Ψ)
= { M | ψ1 (M ) = ψ2 (M ) = ψ3 (M ) = 0 }
)
(
∂ (ψ1 , ψ2 , ψ3 )
=0
=
M det
∂ Z, T, Z
Exercise 16.1.13. What do you think about : A quadratic
transformation
f ∈ Cremona acts
+
+
by blowing up three (indeterminacy) points Ind (f ) = p+
and blowing down the (excep1 , p2 , p3
tional) lines joining them. Typically, the points and the lines are distinct,
with
but
they−can−occur
multiplicity. Then f −1 is also a quadratic transformation and Ind f −1 = p−
consists
1 , p2 , p3
of the images of the three exceptional lines ?
Remark 16.1.14. This simple relation between indeterminacy points and exceptional curves does
not hold for higher degree transforms.
Theorem 16.1.15. The Cremona
group is generated
by collineations and the "inverse everything"
transform i.e. σ : Z : T : Z 7→ TZ : ZZ : ZT .
Proof. A detailed proof can be found in Alberich-Carramiñana (2002), and an historical sketch is
given in Déserti (2009a). The idea is to separate infinitely neighbor points in Ind (ψ) if required
and then proceed to a descending recursion over the cardinal of Ind (ψ).
16.2
Working out some examples
Example 16.2.1. Transform ρ is ρ Z : T : Z = ZZ : TZ : T2 . This is an involution. The
set Ind (ρ) contains 1 : 0 : 0 (twice) and 0 : 0 : 1. The exceptional locus is the reunion of the
contraction lines T = 0 and Z = 0. The decomposition of this transform can be conducted as
described in Table 16.1.
2
Example 16.2.2. Transform µ is µ Z : T : Z = ZZ : Z2 − TZ : Z . This is an involution.
+
+
The points of indeterminacy form a sequence of infinitely close neighbor µ+
3 µ2 µ1 = 0 : 1 : 0.
Similarly, the exceptional locus is line Z = 0, counted three times. Due to this specificity, its
Cremona factorization is longer. A description of this process is given in Déserti (2008–2009),
leading to a nine steps process (φ5 σφ4 σφ3 σφ2 σφ1 ).
.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
184
16.3. Isoconjugacy and sqrtdiv operator
1

(y − z) y


 (x + y) (y − z) 
(x + y) z
3

4

5
6

(x + y) z


yz


(y − z) y

2
result


xz


 yz 
y2

(y − z) y


 (x + y) (y − z) 
(x + y) y

x+y


 y 
y−z



x


 y 
z
Ind (ψ)
  
 

0
1
1
  
 

 0  =  0 ,  0 
1
0
0
  
 
0
1
1
  
 

 0  =  0 ,  0 
1
0
0


 
 
+1
0
1

 
  
 0  ,  0  ,  −1 
−1
1
0


 
 
+1
0
1

 
  
 0  ,  0  ,  −1 
−1
1
0

none
transform

1

 0
0
1
1
−1

0

0 
1
σ

1

 0
0
0
1
1

0

0 
1
σ

1

 0
0
−1
1
1

0

0 
−1
none
Table 16.1: Reduction of a Cremona transform
16.3
Isoconjugacy and sqrtdiv operator
Definition 16.3.1. An heuristic definition of the sqrtdiv operator was already given in Definition 1.5.10. Start from triangle ABC and ’fix’ a point F not on the sidelines. If we call f : g : h
it’s ABC-barycentrics, then UF# (the square-div image) is defined by :
.
. f 2 g 2 h2
sqrtdivF (U ) = UF# =
:
:
u
v w
(16.1)
Remark 16.3.2. This operator is globally type-keeping and therefore sqrtdiv is a pointwise transform. Its fixed points are the four ±f : ±g : ±h, i.e. F and its associates under the Lemoine
transforms.
Theorem 16.3.3. Formal definition of the sqrt operator. Start from four independent points
F1 , F2 , F3 , F4 (three of them are never on the same line). Call ABC their diagonal triangle, i.e.
define
A = F1 F4 ∩ F2 F3 ; B = F2 F4 ∩ F1 F3 ; C = F3 F4 ∩ F1 F2
For any point U in the plane, draw both conics :
.
C12 = C (U, F1 , F2 , F1 F3 ∩ F2 F4 , F1 F4 ∩ F2 F3 ) = C (U, F1 , F2 , A, B)
.
C34 = C (U, F3 , F4 , F1 F3 ∩ F2 F4 , F1 F4 ∩ F2 F3 ) = C (U, F3 , F4 , A, B)
Their fourth intersection is independent of the order chosen for the set {Fj } and is the sqrtdivF
image of U that was former defined wrt the diagonal triangle of the Fj , i.e. the triangle ABC
where A = F1 F2 ∩ F3 F4 ,B = F1 F3 ∩ F2 F4 , C = F1 F4 ∩ F2 F3 .
March 30, 2017 14:09
published under the GNU Free Documentation License
16. Cremona group and isoconjugacies
185
Proof. Choose an order over set {Fj }, use the above defined triangle ABC as the reference triangle
and let f : g : h be the barycentrics of F4 in this context. Then F1 , F2 , F3 is the anticevian triangle
of F4 wrt ABC, enforcing F1 = −f : g : h, etc. Compute the conics using the usual 6 × 6
determinant and obtain :


0
h2 w (f v + gu)
−g f gw2 + h2 uv


C12 '  h2 w (f v + gu)
0
−f f gw2 + h2 uv 
−g f gw2 + h2 uv −f f gw2 + h2 uv
2 f gw (f v + gu)
C34

0

2
'  h w (gu − f v)
g f gw2 − h2 uv
h2 w (gu − f v)
0
f h2 uv − f gw2

g f gw2 − h2 uv

f h2 uv − f gw2 
2 f gw (f v − gu)
Computing their re-intersection is straightforward, and the symmetry of the result implies the
independence from the way the set {Fj } was ordered.
Construction 16.3.4. Given generic A, B, C, U, V the fixed points of the ABC-isoconjugacy that
exchanges U, V are obtained as intersection of conics γU and γV where γU is the conic that goes
through U , the anticevians of U and cevadiv (V, U ), etc. Remark: line U V is tangent at U to γU
and at V to γV .
Proof. Equation of γU is g 2 r2 − h2 q 2 x2 + h2 p2 − f 2 r2 y 2 + f 2 q 2 − g 2 p2 z 2 = 0. It is obvious
that the Fj belong to this conic.
Corollary 16.3.5. Any sqrtdiv operator is an involution. When U is on line Fj Fk , so is UF# and
we have :
cross-ratio Fj , Fk , U, UF# = −1
Proof. We have : det F1 , F2 , UF# = (−gh/vw) × det (F1 , F2 , U ).
Construction 16.3.6. The fourth harmonic (harmonic conjugate of point U ∈ (F1 , F2 ) wrt points
F1 , F2 ) can be constructed by choosing two arbitrary points F3 F4 and then using the sqrtdiv
operator having the four Fj as fixed points. If we chose F3 , F4 in order to obtain the middle of
F1 F2 and the two umbilics as triangle ABC, we obtain the usual construction using one circle from
the pencil admitting F1 F2 as base points and one from the pencil admitting F1 F2 as limit points.
Definition 16.3.7. Isoconjugacy. Forget that f 2 : g 2 : h2 are the square of the barycentrics of
four points, and consider them as the barycentrics of a new point, called the pole P = p : q : r of
the transform. Then the isoconjugate of U = u : v : w wrt pole P is obtained by :
q r
∗ . p
UP∗ = (u : v : w)P = : :
u v w
(16.2)
Remark 16.3.8. This transform was introduced in order to unify isotomic conjugacy (Section 4.4)
and isogonal conjugacy into a common frame... and also to deal with the reluctance towards
imaginary focuses. When using barycentrics, isotomic conjugacy is obtained with P = X2 and
isogonal conjugacy with P = X6 . When using trilinears, you have to use (respectively) P = X75
and P = X1 . When X2 is special, its isogonal conjugate X6 is special too. When X1 is special, its
isotomic conjugate X75 is special too.
Remark 16.3.9. Isoconjugacy U 7→ U ∗ , considered as a function of U alone is type-crossing, so that
it is not so clear to say
√ that this mapping has four fixed points (real or not), namely the points
√
. √
Fi = ± p : ± q : ± r.
Construction 16.3.10. Let be given A, B, C, U, V in generic position (without alignments). Then
the ABC-isoconjugacy ψ that exchanges U, V can be constructed as follows. Once for ever, point R
is chosen on line U V and its cevian triangle AR BR CR is obtained, and triangle T1 is constructed
as :
A1 = V A ∩ U AR , B1 = V B ∩ U BR , C1 = V C ∩ U CR
Then consider a point X not on the sidelines, define triangle T2 by A2 = A1 X ∩ BC (etc) and
triangle T3 as cross (T1 , T2 ) i.e. A3 = B1 C2 ∩ C1 B2 (etc). It happens that triangle ABC and
A3 B3 C3 are perspective, and this perspector is the required point ψ (X).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
186
16.3. Isoconjugacy and sqrtdiv operator
Proof. Put V = p : q : r
wq − vr 6= 0, i.e. A, U, V

ρu

T1 =  q
r

and U = u : v : w. Express R as R = V − ρ U . Existence of A1 requires
not collinear. The result is :



0
ρvx − py ρwx − pz
p
p



T2 =  ρuy − qx
0
ρwy − qz 
ρv q  ,
ρuz − rx ρvz − ry
0
r ρw
p (yw + zv) − ρvwx
puz

T3 = 
qvz
q (zu + xw) − ρwuy
rwy
rwx

puy

qxv

r (xv + yu) − ρuvz
and the perspector is : pu/x : qv/y : rw/z as required. In (Dean and van Lamoen, 2001), ψ was
called reciprocal conjugacy.
Construction 16.3.11. Another construction, with same hypotheses (A, B, C, U, V given, without
alignments). Consider the conic γ that goes through A, B, C, U, V , and the traces of U V on the
.
sidelines, i.e. Ta = U V ∩ BC, etc. For a moving point X, call shadows of X the re-intersections
Xa , Xb , Xc of lines AX, BX, CX and the conic γ. Then traces and shadows are collinear, i.e. Ya
is the second intersection of Ta Xa with γ, etc.
Proof. Direct computation. Here again wq − vr 6= 0, etc is required.
Remark 16.3.12. Most of the time, barycentric multiplication appears as the result of two successive
isoconjugacies, according to :
∗
(XU∗ )P = X ∗ UP∗
b
For example, isot (isog (X)) =
∗
(X6∗ )2
∗
= X ∗ isot (X6 ) while isog (isot (X)) = (X2∗ )6 = X ∗ isog (X2 )
b
b
Remark 16.3.13. Maps sqrtmul and sqrtdiv are related with cevian nests (Table 4.2, case III).
This leads to another construction of ψ (X) when P = U = ψ (P ) is given.
16.3.1
Morley point of view
Proposition 16.3.14. Isogonal conjugacy. The Morley affix of the isogonal conjugate of point
P = Z : T : Z is given by :


Z


isog  T 
Z


2
σ3 Z − ZT − σ2 ZT + σ1 T2


T2 − ZZ

' 
 1
σ1
σ2 2 
2
Z −
ZT − ZT +
T
σ3
σ3
σ3
(16.3)
Proof. This formula can be stated using the representation of first degree. Start from point P and
obtain ∆A , the A-isogonal of line AP , by solving :
∆A . A =
tan (AB, ∆A ) + tan (AC, AP )
=
0
0
Compute ∆A ∧ ∆B and obtain a symmetric expression, proving that ∆C goes also through this
point. One can check that isog (Ω+ ) = Ω− and vice versa.
Proposition 16.3.15. The inexcenters are enumerated by the polynomial :
ζ 4 − 2 σ2 ζ 2 + 8 σ3 ζ + σ22 − 4 σ1 σ3
(16.4)
Proof. The fixed pointsof the isogonal transform are obtained by solving the equation
T isog (M )− T2 − ZZ M = 0. This gives two polynomials of degree 3, that are not self-conjugate
but conjugate of each other. The corresponding algebraic curves are not visible. Intersecting these
curves, we obtain nine points. Among them, are both umbilics. The umbilical pair is fixed, but
a given umbilic is not fixed. They are nevertheless appearing since both T and T2 − ZZ are
March 30, 2017 14:09
published under the GNU Free Documentation License
16. Cremona group and isoconjugacies
187
vanishing here. When computing the Z resultant of these polynomials, we obtain an eight degree
polynomial that factors into :
T×
Y
3
(Z − αT) × poly4 (Z, T)
The umbilical pair is represented by T, vertices are appearing and it remains the required
polynomial of degree 4: this gives the 2 + 3 + 7 = 9 intersections of two degree 3 curves. To be
sure of what happens, we can compute the T resultant of both polynomials and obtain :
ZZ ×
Y
3
Z − α2 Z × poly4 Z, Z
where each point is represented by a specific factor.
Remark 16.3.16. When substituting α = α2 , etc (i.e. using the Lubin representation of second
degree), polynomial (16.4) splits, with roots ±βγ ± γα ± αβ as required.
16.3.2
The isogonal Morley formula
Proposition 16.3.17. Points Z : T : Z and z : t : ζ are isogonal conjugate of each other if and
only if :
(
σ3 ζZ + tZ + zT − σ1 tT
zZ + ζT + tZ σ3 − σ2 tT
= 0
= 0
(16.5)
Since they are complex conjugates, only one relation is required for visible points.
Proof. Direct examination. The method to find such an expression ϑ is as follows. We search
coefficients such that :
zZc00 + tZc01 + ζZc02 + zTc10 + tTc11 + ζTc12 + zZc20 + tZc21 + ζZc22 = 0
when conjugacy occurs. This formula must be symmetrical in M and M ∗ , so that c10 = c01 ,
c20 = c02 , c12 = c21 . Writing that in-excenters are fixed points gives us 4 equations, leaving two
indeterminate coefficients. One obtains, for example, c01 ϑ + c12 conj ϑ. As is, the equation ϑ = 0
defines a line. But this equation is not self-conjugate, and only a point of the line is visible. When
cutting by conjϑ, we obtain a point (and a 2nd degree equation, as it should be).
Exercise 16.3.18. Apply the same method to the isotomic conjugacy, obtain the only possible
formula... and conclude.
16.3.3
Isoconjkim
In the old ancient times, Kimberling (1998) introduced another definition of the isoconjugacy, in
an attempt to unify isotomic and isogonal conjugacies into a broader concept. Thereafter, this
definition was changed into the one given above. To avoid confusion with (16.2), we will use the
term isoconjkim to describe the older concept.
Definition 16.3.19. isoconjkim. For points outside the sidelines of ABC, the Kimberley P isoconjkim of U is the point X such the product of the trilinears of P, U, X gives 1 : 1 : 1.
Restated into barycentrics, this gives :
P ∗ U ∗ X = X1 ∗ X1 ∗ X1
b
b
b
b
(16.6)
In this transformation, P and U play the same role so that isoconjkim acts like a multiplication.
On the other hand U and X play also the same role, so that isoconjkimP is involutory. Description
X = isoconjkimP (U ) reflects the fact that second point of view is the more useful.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
188
16.4. Antigonal conjugacy
Example 16.3.20. Here is a list of various isoconjkim transformations. The isogonal conjugation
is X1 -isoconjkim while the isotomic conjugation is X31 - isoconjkim.
P
barycentrics
X (1)
a2 u1
X (2)
a3 u1
X (3)
a2 / cos A u1
X (4)
a2 cos A u1
X (6)
a u1
X (19)
a cos A u1
X (31)
1 u1
X (48)
(a/ cos A) u1
16.4
trilinears
1 u1
a u1
(1/ cos A) u1
cos A u1
(1/a) u1
(cos (A) /a) u1
1/a2 u1
1/ (a cos A) u1
pole bar(pole) fixed
X (6)
a2
X (1)
3
X (31)
a
X (365)
X (19)
a/Sa
X (???)
X (48)
a3 Sa
X (???)
X (1)
a
X (366)
X (3)
a2 Sa
X (???)
X (2)
1
X (2)
X (4)
1/Sa
X (???)
Antigonal conjugacy
Proposition 16.4.1. Antigonal conjugacy. Let A0 , B 0 , C 0 be the reflections of a point M in
the sidelines BC, CA, AB of the reference triangle. Then circles ABC 0 , AB 0 C, A0 BC concur into
a common point N (see Figure 16.1). This point N is called the antigonal conjugate of M . When
point M is given either by Mb ' p : q : r (barycentrics) or by Mz ' Z : T : Z (Lubin affixes),
point N is given by :


p

 (a2 − b2 − c2 ) p2 + (a2 − b2 ) pq + (a2 − c2 ) pr + a2 qr

q
Nb ' 
 (b2 − c2 − a2 ) q 2 + (b2 − c2 ) qr + (b2 − a2 ) qp + b2 rp

r
(c2 − a2 − b2 ) r2 + (c2 − a2 ) rp + (c2 − b2 ) rq + c2 pq





2
2
−σ3 ZZ + σ2 ZZT + σ3 σ1 Z T − σ1 ZT2 + (σ3 − σ1 σ2 ) ZT2 + σ12 − σ2 T3



2


σ3 Z − ZT − σ2 ZT + σ1 T2



Nz ' 
T

3


2
2
2
2
2
 −σ3 Z Z + σ2 TZ + Tσ1 σ3 ZZ − (σ1 σ2 − σ3 ) T Z − σ2 σ3 ZT + σ2 − σ1 σ3 T 
σ3 Z2 − σ1 ZT − σ3 ZT + σ2 T2
Proof. Direct computation for Nb , then usual transform for Nz .
Proposition 16.4.2. The conic that goes through A, B, C, M, N = antigon (M ) is a rectangular
hyperbola (and therefore goes through H =X(4)). Moreover M, N are antipodes on this conic.
Proof. Direct computation. This suggests that antigonal transform is involutory.
Proposition 16.4.3. The antigonal transform can be factored into :
antigon = isogon ◦ invincircum ◦ isogon
This is an involutory Cremona transform. The indeterminacy set contains six points: A, B, C, H
and both umbilics. The exceptional locus is the reunion of six conics. Each of them goes through five
points of Ind (antigon): circumcircle is blown-down into H, circle BCH (centered at B + C − O)
is blown-down into A, idem for the other two vertices. Finally, curve :
γy = Z2 − σ1 ZT − σ3 ZT + σ2 T2
that goes through A, B, C, H, Ωy is blown-down into Ωy (the same umbilic), idem for the other
umbilic.
Proof. Direct computation.
Proposition 16.4.4. Define the ii transform as ii = invincircum◦isogon then point ii (A, B, C, P )
is independent of the ordering of points A, B, C, P (Hyacinthos 20929).
March 30, 2017 14:09
published under the GNU Free Documentation License
16. Cremona group and isoconjugacies
189
Figure 16.1: Antigonal conjugacy
Proof. Transform triangle ABC into ABP , and thus C into P 0 . Then use the usual change of
triangle formula. Since ii only involves even powers of the sidelength, everything goes fine... and
the result follows.
Proposition 16.4.5. Seen as a Cremona map, the ii transform has the same indeterminacy locus
as the antigonal transform. The exceptional locus contains three lines and three conics. A sideline
like BC blows-down into the opposite vertex A. Curve γx (the same as before) now blows down to
Ωy , whileγy blows-down to Ωx . Additionally, the circumcircle blows-down to its center X(3), while
X (3) 7→ X(186) is regular and X(265) is the only regular point that maps onto H =X(4).
ii
Proof. This lack of symmetry (a curve that blows down to a regular point, a point of indeterminacy
that does not blows-up under the reverse transform) is related to the fact that ii is not involutory.
16.5
Isogonality and perspectivity
Lemma 16.5.1. Results from Proposition 4.6.11. Let T2 be a triangle perspective wrt T1 = ABC.
Then triangle T2 , perspector P and perspectrix ∆ can be written as :


u p p
1
1
1


T2 =  q v q  ; P ' p : q : r ; ∆ 2 =
;
;
p−u q−v r−w
r r w
.
Point U ' u : v : w is the perspector of T1 and T3 = cross (T1 , T2 ). But, in these formulas, the same proportionality factor must be applied to (p, q, r) and (u, v, w), so that factor
.
k = (u + v + w) / (p + q + r) is a projective quantity.
Proposition 16.5.2. Assume that vertices of T2 are not on the sidelines of T1 , and perspectivity
as in Lemma 16.5.1. Define T2∗ as the triangle whose vertices are the isogonal images of the T2
vertices. Then triangles T1 and T2∗ are perspective, and P, U are replaced by P ∗ and U ∗ . Moreover,
T2 and T2∗ share the same perspectrix if and only if U = k P ∗ . In this case, the isogon T2∗ and the
crosstri T3 are equal.
Proof. Computations are easy from the lemma. If A1 is on BC then A2 = A and everything
degenerates, etc.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
190
16.5. Isogonality and perspectivity
Remark 16.5.3. This obviously occurs with the 27 Taylor-Marr triangles since, for example, A, B1 , C2
are aligned on the same trisector (and cyclically).
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 17
About cubics
17.1
Cubic
For a catalog with sketches, visit Gibert (2004-2014), especially Ehrmann and Gibert (2005). Some
notations :
I
G
O
H
N
K
L
X(1)
X(2)
X(3)
X(4)
X(5)
X(6)
X(20)
Definition 17.1.1. A cubic is a curve defined by an homogeneous polynomial of degree 3. Using
barycentrics, or trilinears or Morley affixes is irrelevant, the degree is the same. Notation : K.
Proposition 17.1.2. A cubic is defined by nine general points.
Proof. There are ten coefficients, defined up to a global proportionality factor. We found them by
computing
(17.1)
det x3 , x2 y, xy 2 , y 3 , y 2 z, yz 2 , z 3 , z 2 x, zx2 , xyz
j=10
applied to the nine given points and the generic point.
Example 17.1.3. Pivotal isocubics. Let F, U be two points not on the sidelines and define
pK (#F, U ) as the cubic that goes through the nine points: ABC, AU BU CU (the cevians of U )
and FA FB FC (the anticevians of F ). Its equation, as computed using (17.1), is :
.
.
pK (P, U ) = pK (#F, U ) = h2 y 2 − g 2 z 2 ux + f 2 z 2 − h2 x2 vy + g 2 x2 − f 2 y 2 wz
(17.2)
And it can be seen that pK (P, U ) is invariant by the isoconjugacy whose pole is P . More about
this important class will be given at Section 17.2.
Proposition 17.1.4. Define an isocubic K with pole P as a cubic that is invariant wrt the P
isoconjugacy. Then K is either a pivotal cubic as obtained just above or belongs to the "non pivotal
isocubic" family (Section 17.9), with equation :
.
nK (P, U ) = ux r y 2 + qz 2 + vy p z 2 + r x2 + wz q x2 + p y 2 + k xyz
(17.3)
Proof. Direct inspection from K (XP∗ ) = λK (X). It can be seen that terms like xy 2 and xz 2 are
to be paired, and that terms like x3 are to be avoided. As a corollary, such a cubic goes through
the vertices A, B, C.
∗
Proposition 17.1.5. When the pole P is fixed, and
XP is defined by (16.2), an isocubic K is
∗
2 2 2
characterized by K (XP ) /K (X) = ∓ pqr ÷ x y z . Sign "-" characterizes a pK cubic, sign "+"
characterizes a nK cubic. Each class form a projective space, whose dimensions are respectively 3
and 4.
191
192
17.2. Pivotal isocubics pK(P,U)
pK
name
pKA
vertex 17.4.3
ZU(1)
17.4.4
E1 , E3 hidden 17.4.9
K002 Thomson 17.4.17
K003
McKay 17.4.10
K006
K005
K004 Darboux 17.5
K001 Neuberg 17.4.11
K035
K020
K021
K007
K170
K155
K060
nK
∆
circle
K137
???
K040
K018
K010
K162
shortest 17.4.13
Lucas 17.5
17.2.6
EAC2 17.6.2
17.7
17.4.5
17.4.7
Pelletier
Brocar2 17.14
Simson 17.11
17.11.4
F =
√
F
U ,U ∗
1
A
1
X(1)
1
Ω
1
2,6
1
3,4
1
4,3
1
5,54
1
20,64
1
30,74
1
98,511
1
99,512
1 100,513
1 110,523
1 384,695
1
512,99
1 523,110
2
X(69)
2
X(4)
√
31
238
√
1989 265
F
1
1
1
1
1
1
2
6
E,E ∗
D,D∗
some other points on the cubic
3,4
1075,?
155,254
2120,2121
2130,2131
2132,2133
?,?
39,83
1,1
5,54
?,?
2142,2143
9,57
1745,3362
46,90
3460,3461
3182,3347
3464,?
1756,?
1019,1018
513,100
?,?
?,?
?,?
223, 282, 1073, 1249
371, 372, 485, 486, 487, 488
3, 4, 17, 18, 61, 62, 195, 627, 628
3, 4, 40, 84, 1490, 1498
3, 4, 13, 14, 15, 16, 399, 484, ...
1687, 1688, 2009, 2010
1379, 1380
1381, 1382
1113, 1114
3,4,32,39,76,83
root
513 Z + (X1 X6 )
649 Z + (X1 X2 )
650 Z + (X1 X3 )
523 Z + (X3 X6 )
cK (#X2 , X69 )
cK (#X6 , X3 )
1, 44, 88, 239, 241, 292, 294, 1931
1, 238, 291, 899, 2107
1,105,243,296,518,1155,1156, 2651, 2652
2, 6, 13, 14, 15, 16, 111, 368, 524
2, X(2394) upto X(2419)
6, X(2420) upto X(2445)
√
P (central fixed point), U , UP∗ , E = cevadiv (U, U ∗ ), E ∗ , D = cevadiv(U, P )
Table 17.1: Some well-known cubics
17.2
Pivotal isocubics pK(P,U)
Definition 17.2.1. Pivotal isocubics. Let F, U be two points not on the sidelines and define
pK (#F, U ) as the cubic that goes through the nine points: ABC, AU BU CU (the cevians of U )
.
and FA FB FC (the anticevians of F ). Then U and F0 = F are also on the cubic. Its equation is :
.
.
pK (P, U ) = pK (#F, U ) = h2 y 2 − g 2 z 2 ux + f 2 z 2 − h2 x2 vy + g 2 x2 − f 2 y 2 wz
(17.4)
.
Part of the time, the isocubic is defined using P ' p : q : r = f 2 : g 2 : h2 i.e. the pole of the
conjugacy.
Theorem 17.2.2. isocubic property. When point X is on a sideline of triangle ABC, then
XF# is undefined (geometrically), while the formula gives the third vertex. Otherwise, XF# belongs
to pK (#F, U ) if and only if X belongs to pK (#F, U ). And then U, X, XF# are collinear. For
this reason, point U is called the pivot of the cubic. Alternate formulation: pK (#F, U ) is the
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
193
locus of the X such that U, X, XF# are collinear (to be taken ’Cremona more’ , i.e. allowing
indeterminacies)
Proof. Compute the determinant of the 10 rows (17.1) relative to the nine points given in the
definition
and the
variable point X = x : y : z. And remark that this quantity is proportional to
det U, X, XF# .
Theorem 17.2.3. cevadiv property. When X is on the sidelines of the cevian triangle of U ,
then Y = cevadiv (U, X) is undefined geometrically, while the formula gives 0 : 0 : 0. Otherwise,
this Y belongs to pK (#F, U ) if and only if X belongs to pK (#F, U ). And then UF# , X, cevadiv (U, X)
are collinear. This fact is underlined by the name isopivot given to the point UF# .
Proof. Compute pK (#F, U ) (cevadiv (U, X)) and obtain pK (#F, U ) (X) times the incidence relations,
i.e. the rows of adj
(cevian (U )) · X . And remark that this quantity is proportional to
det UF# , X, cevadiv (U, X) .
Proposition 17.2.4. The 22 points property. The pK (#F, U ) cubic goes through
1. U ,AU BU CU and their conjugates UF# , ABC (8)
2. F0 FA FB FC (cf Theorem 17.2.2) (4)
3. cevadiv U, UF# , the four cevadiv (U, Fj ) and their isoconjugates (10)
Proof. Follows directly from the two theorems.
Example 17.2.5. When U is one of the fixed points of the isoconjugacy (i.e. P = U ∗ U ), the
b
pK (P, U ) cubic degenerates into the lines through the remaining three fixed points.
P
Example 17.2.6. K170 is pK (X2 , X4 ). Equation
x y 2 − z 2 /Sa = 0. On Figure 17.1, one
can see the following alignments (general properties, applicable to any pK) :
1. Fixed points : F0 , Fa , A are collinear, and cyclically for the other fixed points and the other
vertices ;
2. From U : U, X, XF# are collinear (e.g. E and E ∗ are aligned with U ). Therefore, each line
from U to a fixed point is tangent to the cubic at this fixed point ; in the same vein, point
#
Ub = U B ∩ AC is on the cubic and viewing B as (Ub )F makes sense, but not viewing Ub as
#
BF (this object would be "quite all points on line AC").
3. From UF# : UF# , X, cevadiv (U, X) are collinear (e.g. F and D are aligned with UF# ).
Therefore, each line from UF# to a vertex or to U is tangent to the cubic at this point.
Definition 17.2.7. The P KF# (X) point is the intersection of the trilinear polars of points X and
XF# . Using barycentrics and F = f : g : h, we have :
P KF# (X) = f 2 x g 2 z 2 − h2 y 2 : g 2 y h2 x2 − f 2 z : h2 z f 2 y 2 − x2 g 2
Proof. Direct computation.
Remark 17.2.8. This transform was introduced by Ehrmann and Gibert (2005) as PKP , i.e. putting
forwards the pole of the conjugacy rather than its fixed points, and turned into Definition 19.3.1.
Example 17.2.9. Using F=X(1), i.e. P=X(6), we have PK(X(I)) = X(J) for these (I,J):
I
J
2
3
4
5
6
9 19
31
44
54 57 63
512 647 647 2081 512 663 810 2084 3251 2081 663 810
Proposition 17.2.10. Points of indeterminacy of P KF# are the fixed points Fj and their diagonal
vertices i.e. A, B, C. The exceptional curves are the six lines
n through
o two of the fixed points.
#
#
Otherwise, each point P = P KF (X) characterizes a pair X, XF (so that P K # is not a
Cremona transform !).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
194
17.2. Pivotal isocubics pK(P,U)
Figure 17.1: pK(2,4)
Proof. Solving P = 0 : 0 : 0 gives the first result, and solving the Jacobian gives the second.
Otherwise, set P ' p : q : r and obtain :


2 h2 qf 2 prg 2


 −rg 2 +g 2 h2 p2 − f 2 g 2 r2 + f 2 h2 q 2 − rg 2 W 
−qh2 +g 2 h2 p2 + f 2 g 2 r2 − f 2 h2 q 2 + qh2 W
where W = 4i f 2 g 2 h2 S
p q r
f , g, h
and S is the usual Heron formula for the area (7.5).
Proposition 17.2.11. When P KF# (X) belongs to the tripolar line of UF# , then point X belongs to
the cubic pK(P, U ). This amounts to say that X is on pK if and only if the tripolars of X, XF# , UF#
are concurrent.
Proof. Direct computation. In Kimberling (1998, p. 240) the corresponding cubic is noted Z(U, Y )
where P = Y ∗ X6 .
b
Proposition 17.2.12. When point U = u : v : w is at infinity, the pK (P, U ) cubic can be rewritten
as :
2
p
q
r
a ρ b2 σ c2 τ
+ +
(xρ + yσ + zτ ) − (x + y + z)
+
+
=0
x y z
x
y
z
where [ρ, σ, τ ] is any line whose direction is U . In other words u = σ − τ, etc.
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
195
Proof. Direct examination. We can check that pK (P, U ) contains the intersections of both conics
—vertices A, B, C and UP∗ —, the points at infinity of Ccir (P ), the intersections of line [ρ, σ, τ ] with
the associated conic, and U itself !
17.3
Another description
1. Consider pK (#F, P ) and use Fj ' ±f : ±g : ±h (isoconjugacy), P ' p : q : r (pivot). The
name U ' u : v : w will be used for the generic point of the cubic. Let CP(F,P) be the
diagonal conic that goes through P and the four Fj . In fact, this conic goes also through the
3 associates of P since


g 2 r2 − h2 q 2
0
0


CP (F, P ) ' 
0
h2 p2 − f 2 r2
0

2 2
2 2
0
0
f q −g p
2. Cut this conic by line P U and obtain point N . Acting that way, we obtain a symmetric
expression. Cut rather by line U UF# and say that P is the other solution.. We obtain :

−f 2 pv 2 + 2 f 2 quv − g 2 pu2


N '
f 2 qv 2 − 2 g 2 puv + g 2 qu2

2
2
2 2
2 2
2 w f qv − g pu + r g u − f v

Now, define N (U ) by this formula. Then N (U ) = N UF# is equivalent to U ∈ pK... except
when P is one of the Fj .
3. Cut the tangent at N to CF(F,P) with line P PF# and obtain the point M



M '


qw − rv
− q 2 h2
ru − pw
p2 h2 − r2 f 2
pv − qu
2
q f 2 − p2 g 2
r2 g2



 ; t = − (pw − ru) q

(qw − rv) p

4. For all U , point M is aligned with P, PF# . When U is on the cubic,
• N = cevadiv (M, P ).
• M is aligned with N, NF# .
• NF# is the intersection of tangents at U and UF# to the cubic.
• MF# is the intersection of the line P, U, UF# and the fixed circumconic through P, PF# .
#
5. Line cevadiv (P, U ) ; cevadiv P, UF# goes through the fixed point Q = cevadiv P, PF#
.
F
Conversely, the locus of such U is the union of pK (#F, P ) et nK #F, Q, −1 ÷ p2 q 2 r2 .
17.4
pK isocubics relative to pole X(6)
Definition 17.4.1. A Kimberling ZU cubic is a pK (X6 , U ), giving a special place to isogonal
conjugacy. Some examples are given in Table 17.1.
Proposition 17.4.2. Only 8 ZU cubics have a reflection center : the Darboux cubic (center=
X3 ), the four degenerate cubics that are union of the three bisectors through an inexcenter, and
three other (Maple length = 135712 using RootOf, [4948, 5345, 4215] using alias).
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
196
17.4. pK isocubics relative to pole X(6)
ABCIJKL cubics: the Lubin(2) point of view
17.4.1
Proposition 17.4.3. Cubics PKA. The set of all cubics that go through points ABCIJKL is
a projective space. Its dimension is 3. A generating family is made of the three cubics pKA =
(BC) ∪ (AI) ∪ (KL) , etc. Pivot of pKA is A. Its factored equation in Lubin(2) is :

β2

1
pKA = det
1
β2
γ2
1
1
γ2
 2

α
Z
1

T  det
1

Z
α2
αβ + βγ + γα
−1
α+β+γ
αβγ


Z
αβ + βγ − γα


T  det
1
 α−β+γ

Z
αβγ
βγ + γ α − αβ
1
α+β−γ
αβγ

Z
T


Z
while its (not factorisable) equation in Lubin(1) is det (X, XP∗ , A) = 0 where XP∗ is given by the
isogonal conjugacy formula.
Proof. Equation of pK (F, U ) is det X, XF# , U = 0, leading to dimension 3, and allowing to
check that pivot of pKA is the vertex A. After that, we can go back to Lubin(1) since isogonal
conjugacy doesn’t require to identify which is the incenter among the inexcenters.
Example 17.4.4. The Kimberling Z(X(1)) cubic, i.e. (IJ) ∪ (IK) ∪ (IL), is obtained as :
(β + γ)
(γ + α)
(α + β)
pKA +
pKB +
pKC
(β − γ)
(γ − α)
(α − β)
Definition 17.4.5. The triangular cubic is the union of the three sidelines: nK0 = (BC) ∪
(CA) ∪ (AB). Its normalized equation is defined by :
nK0 (X) = det (A, B, X) det (A, X, C) det (X, B, C) ÷ det (ABC)
Theorem 17.4.6. When substituting the isogonal formulas (16.3) into the equation of a #X(1)
pivotal cubic, we have the following equalities :
+pK (X) × nK0 (X)
pK (isog (X))
=
nK (isog (X))
= −nK (X) × nK0 (X)
that are exact identities, not up to a proportionality factor. Therefore, the set of the isocubics splits
into two projective subspaces, the pK one (dimension 3) and the nK one (dimension 4).
Definition 17.4.7. Cubic "circumcircle union infinity", i.e. T ZZ − T2 is a nK.
Proposition 17.4.8. The Kiepert RH construction (see Proposition 12.18.1 for more details) can
be summarized as follows. Let K = cot φ be a fixed real and define circularly the points A0 B 0 C 0
by :
z }| {
z }| {
0
cot BC, BA − K = 0 ; cot BC, CA0 + K = 0
These points are cocubics with ABCIJKL. To prove this result, we obtain :
A0 ' β + γ + i (γ − β ) K : 2 : (β + γ + i (γ − β ) K) ÷ βγ
P
Then substitute A0 and B 0 into 3 Kj pKj = 0. This system has non zero solutions, proving the
result. Moreover, the obtained cubic can be rewritten using the Lubin(1) basis and we have :
pK(K)
=
=
pKT homson + K 2 pKDarboux
K 2 + 1 pKN euberg + 3 − K 2 pKM cKay
Equations of the named cubics are given below. We can see that pK(K) contains also the circumcenter X(3) (the T3 coefficient), the orthocenter (isogonality) and the A0 B 0 C 0 points related with
the other orientation (pK(K) depends only on K 2 ).
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
17.4.2
197
A more handy basis
Proposition 17.4.9. The hidden IJKL cubics. The following cubics are conjugate of each
other, but not self conjugate (hidden curves). Their nine visible intersections are ABCIJKL
(stable by isogonal conjugacy) and the two umbilics (stable as a pair) :
E1
E3
2
= Z2 Z + σ3 T Z − 2 T 2 Z − σ2 T 2 Z + T 3 σ1
1
σ2 3
σ1 2
2
= ZZ +
T Z2 −
T Z − 2 T 2Z +
T
σ3
σ3
σ3
Both curves are pK cubics and pivot of E1 is the umbilic 0 : 0 : 1 while umbilic 1 : 0 : 0 is the pivot
of E3 .
Proof. Start from (16.3) that gives isog (X) when X ∈ PC C3 and compute :
E = T2 − ZZ X − T isog (X)
Quantities E1 , E3 are the Z, Z components of E while factors have been chosen to ensure E2 = 0.
Column E vanishes at an umbilic since the factors are zero. At a vertex, the circle vanishes and
isog (X) is 0 : 0 : 0. At a fixed point, this column is proportional to X but E3 is ever 0 and
the X aren’t at infinity. This result has been proved in the Lubin(1) space, without specifying
who is the incenter among the three inexcenters. Obviously, is can be checked in Lubin(2) by
substitution or even by factoring the T resultant of both equations. Thereafter, pivots are obtained
by identification.
Proposition 17.4.10. K003, the McKay Cubic, pK (6, 3). Expression Z E3 − Z E1 /T gives
a visible cubic, whose equation is :
pKM cKay =
1 3
σ1
σ2 2
3
2
Z − σ3 Z −
T Z2 + σ2 T Z +
T Z − σ1 T 2 Z
σ3
σ3
σ3
p
K003 goes through the orthocenter. Its points at infinity are Θ : 1/Θ : 0 where Θ2 = 3 σ32 . These
points are the directions of the Morley triangles. We obtain the asymptotes as the gradient at the
corresponding points. We obtain :
Θ2
σ2
σ 1 Θ2
σ3
3
;
−
;
−3
σ3
Θ2
σ3
Θ2
and the other two are obtained by substituting Θ by j Θ or j 2 Θ. By super-symmetry, the three
asymptotes concur, at the gravity center. The pivot is X(3).
Proof. Everything is straightforward, except from the pivot. It can be obtained as the intersection
of two well chosen lines P P ∗ , for example for two of the points at infinity.
Proposition 17.4.11. K001, the Neuberg cubic, pK (6, 30). Quantity (σ2 /σ3 ) E1 − σ1 E3
provides another visible cubic. Its equation is :
pKN euberg =
σ1
σ 2 − 2 σ2 2
σ2 2
2 σ1 σ3 − σ22 2
2
2
Z Z − σ1 ZZ −
T Z2 + σ 2 T Z + 1
T Z+
T Z
σ3
σ3
σ3
σ3
This is a circular curve. The third point at infinity is σ1 σ3 : 0 : σ2 , i.e. X(30), the direction of the
Euler line. This is also the pivot. The asymptotes are :
i
h
i h
i h
0 −σ1 σ2 , −σ1 σ2 0 , σ22 σ1 ; −σ13 σ3 + σ23 ; −σ2 σ12 σ3
Intersecting the first two asymptotes, we obtain a singular focus at σ22 : σ1 σ2 : σ12 , i.e. point
X(110). Moreover, the cubic goes through X(3) and therefore through X(4). Visible asymptote
goes through X(74), the isogonal of X(30).
Proof. X(30) is the pivot because it is the only visible common point to the curve and line P P ∗
when P is an umbilic. An asymptote is the gradient evaluated at the corresponding point at
infinity.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
198
17.4. pK isocubics relative to pole X(6)
Corollary 17.4.12. The Neuberg cubic is the locus of points X such that the isogonal line X X ∗
is parallel to the Euler line. See (Gibert, 2004-2014, 2005). Its (barycentric) equation is :
X
2 =0
(17.5)
x y 2 c2 − z 2 b2 2 a4 − b2 + c2 a2 − b2 − c2
cyclic
and can be rewritten as :
2
2
a
b2
c2
a ρ b2 σ c2 τ
+
+
(xρ + yσ + zτ ) − (x + y + z)
+
+
=0
x
y
z
x
y
z
where [ρ, σ, τ ] is the Euler line –remember: tripole = X(648). In other words,
K001=circumcircle×Euler - infinity×Jerabek.
Proof. See Proposition 17.2.12. We can check that K001 contains the intersections of Γ and Jerabek
—vertices A, B, C and X(74)—, the points at infinity of the circumcircle —the umbilics—, the
intersections of Euler line and Jerabek hyperbola —X(3) and X(4)— and X(30) itself.
Proposition 17.4.13. Shortest cubic, pK (6, 523). Quantity (1/σ1 ) E1 + (σ3 /σ2 ) E3 provides
another visible cubic. Its equation is :
pKshortest =
1 2
σ3
1 2
σ 2 + 2 σ2 2
σ3
σ 2 + 2 σ1 σ3 2
2
2
Z Z+
ZZ +
Z T+
TZ − 1
T Z− 2
T Z + 2T3
σ1
σ2
σ2
σ1
σ2 σ1
σ2 σ1
The third point at infinity is −σ1 σ3 : 0 : σ2 , i.e. X(523), the orthodir of the Euler line. This is
also the pivot. The umbilical asymptotes concur at X(110), giving a singular focus. Barycentric
equation of this curve is ;
b2 − c2 x b2 z 2 − c2 y 2 + c2 − a2 y c2 x2 − a2 z 2 + a2 − b2 z a2 y 2 − b2 x2
The ETC databasis doesn’t contains other known points. We have the eight points: the 4 inexcenters, X(110), X(523) and both umbilics.
Remark 17.4.14. The "shortest cubic" has been introduced as the circular pK cubic whose expression requires the shortest number of characters. Its main property is to provide an handy basis
when used together with the McKay and the Neuberg cubics. The shortest cubic contains no ETC
points apart X(1), X(523) and X(110).
Proposition 17.4.15. Resulting pencils. We have the following pencils :
1. The pencil generated by pKN euberg and pKshortest is the set of all the circular pK cubics.
Their pivots are at infinity.
2. The pencil generated by pKM cKay et pKN euberg is the set of the pK cubics that goes through
the X(3), X(4) pair. Their pivots are on the Euler line.
3. The pencil generated by pKM cKay et pKshortest is the set of pK-cubics whose pivots are on
the line X(3), X(523), i.e. the line through X(3) and perpendicular to the Euler line. The
common points of these cubics are on ∆ (horrible formula, with a 24th degree radicand).
Exercise 17.4.16. Does it exist a cubic XXX such that :
1. XXX, K001, K003 provide a basis of the ZU cubics space
2. XXX contains "many" ETC points
3. Pencils (XXX,K001) and (XXX,K003) contain "many" known cubics
4. Lubin equation of XXX remains practicable
Proposition 17.4.17. K002, the Thomson cubic, pK (6, 2) is the locus of points X whose
trilinear polar is parallel to their polar line in the circumcircle. It is a pK cubic with pole=X(6)
and pivot=X(2). It is the K = 0 cubic of the Kieper RH construct. Its equation is :
pKT homson =
2
3 3 σ2 2
σ 2 + σ1 σ3 2
2
3 4σ1
2 σ2 + σ1
Z + Z Z−σ1 ZZ −3 σ3 Z −
T Z2 +4 σ2 T Z +
T 2 Z− 2
T Z
σ3
σ3
σ3
σ3
σ3
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
199
Figure 17.2: The Darboux and Lucas cubics
17.5
Darboux and Lucas cubics aka K004 and K007
In this section, P ∈ Darboux and U ∈ Lucas while pole and pivot are noted otherwise.
Definition 17.5.1. The Darboux cubic K004 is the locus of point P such that the pedal triangle
of P is the Cevian triangle of some other point U , while the Lucas cubic K007 is the locus of point
U such that the cevian triangle of U is the pedal triangle of some other point P . .
Proposition 17.5.2. Darboux cubic is a pK cubic, with X(6) as pole and X(20) as pivot. X(20)
is the de Longchamps point. Lucas cubic is a pK cubic, with X(2) as pole and X(69) as pivot.
X(69) is the anticomplement of X(6). Their equations are :
det (X20 , P, isog (P ))
=
0
(17.6)
det (X69 , U, isot (U ))
=
0
(17.7)
Moreover, K004 has a reflection center at X(3), the circumcenter. Using Morley affixes and expanding, we have :
pKDarboux = −
Z3
σ2 2
3 σ1 σ3 − σ22 2
σ 2 − 3 σ2 2
2
3
+
Z Z − σ1 ZZ + σ3 Z + 1
T Z+
T Z
σ3
σ3
σ3
σ3
Proof. Straightforward from (8.1) and (4.3).
Proposition 17.5.3. The correspondence between P ∈ Darboux and U ∈ Lucas is described by :


 
p
(ar + cp cos B) (bp + aq cos C)

 

(17.8)
ψ  q  '  (aq + bp cos C) (br + cq cos A) 
r
(br + cq cos A) (ar + cp cos B)


a3 b cos C
b2 a2
a2 bc cos A
−
+
+


u
v
w


2 2
3
2
b
a
ab
cos
C
ab
c
cos
B
−1


ψ '
−
+

u
v
w


α β
γ
w
4
2 2
2 2
4
2 2
4
+ − −
a − 2b a − 2a c + c − 2b c + b
u
v
w 4uv
where α = a2 bc cos C cos B, β = ab2 c cos C cos A, γ = abc2 cos A cos B
(17.9)
Proof. Transformation ψ as described in (17.8) and (17.9) is certainly not a central transformation
when applied to the whole plane. But this transformation is central when restricted to the Darboux
cubic and nothing else matters. On the other hand, a more symmetrical formula will involve cubic
roots and, for this reason, will not be efficiently used by formal computing tools.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
200
17.6. Equal areas (second) cevian cubic aka K155
In Kimberling (2002a), the following chains are emphasized. When P is on Darboux then
isog (P ) too (the cyclopedal property), so that ψ (P ) and (ψ ◦ isog) (P ) belong to Lucas. In this
figure are involved two inscribed triangles that share the same circumcircle, so that
(ψ ◦ isog) (P ) = (cyclocev ◦ ψ) (P )
When P = X3 , the corresponding circumcenter is X5 ; when P = X40 , this center is X1158 . Now,
from (17.7), isot is an involution of Lucas. Therefore, ψ −1 ◦ isot ◦ ψ is an involution of Darboux.
On the other hand, the symmetry centered at X3 , i.e. the transformation :
(17.10)
sym3 (U ) ' 2 [α, β, γ] /(α + β + χ) − [u, v, w] / (u + v + w)
where X3 = α : β : γ is a natural involution of the Darboux cubic. It is natural to check if they
are equal. This is straightforward when using (17.8) and (17.9).
sym3
isog
sym3
isog
sym3
isot
cycl
isot
cycl
isot
X1 −−−−→ X40 −−−−→ X84 −−−−→ X1490 −−−−→ X3345 −−−−→ X3182 · · ·






ψ
ψ
ψ
ψ
ψ
ψ
y
y
y
y
y
y
X7 −−−−→ X8 −−−−→ X189 −−−−→ X329 −−−−→ X1034 −−−−→ X??? · · ·
X3
X2
isog
−−−−→

ψ
y
cycl
sym3
isog
sym3
isog
isot
cycl
isot
cycl
X4 −−−−→ X20 −−−−→ X64 −−−−→ X1498 −−−−→ X3346 · · ·





ψ
ψ
ψ
ψ
ψ
y
y
y
y
y
−−−−→ X4 −−−−→ X69 −−−−→ X253 −−−−→ X20 −−−−→ X1032 · · ·
Since X1 is fixed by isog and X3 by sym3, these chains are unidirectional.
Claim 17.5.4. Let Q = (sym3 ◦ isog ◦ sym3 ◦ isog ◦ sym3) P . When P ∈ Darboux, then cevamul (P, Q) =
X20 (for all known points, and randomly). When P is on the branch of X3 , so is Q (obvious).
When P is not on Darboux, ??? In any case, a simple division by polynomial (17.6) isn’t sufficient.
17.6
Equal areas (second) cevian cubic aka K155
Definition 17.6.1. Cubic shadow. Triangle centers on a cubic K yield non-central points on the
cubic; e.g., if Q1 and Q2 are on K, then the line Q1 Q2 meets K in a "third" point, L (Q1 , Q2 ),
possibly Q1 or Q2 . If A0 B 0 C 0 is a central triangle (cf Section 3.2), R a triangle center, A00 =
L (R, A0 ) and cyclically, then triangle A00 B 00 C 00 is a central triangle on K.
Definition 17.6.2. Cubic EAC2, the equal areas (second) cevian cubic is K155 in Gibert (20042014). This cubic is pK(X31, X238), i.e self-isoconjugate wrt P = X31 = a3 : b3 : c3 and pivotal
wrt U = X238 = a3 − abc : b3 − abc : c3 − abc.
Proposition 17.6.3. It happens that P ∈ EAC2. When a point Q is on EAC2, its isoconjugate
Q∗P aka X31 ÷ Q is on EAC2 too. In the following table, for each (I,J), the centers X(I) and X(J)
b
are on EAC2 and are an isoconjugate pair. Each pair is collinear with the pivot X(238).
"
"
2106
2107
1
6
2108
2109
2 105 238 365 1423 1931
31 672 292 365 2053 2054
2110
2111
2112
2113
2114
2115
2116
2117
2118
2119
#
2144
2145
2146
2147
#
Table 17.2 gives some cubic shadows on EAC2. Column 1 gives the perspector R ∈ K. Column
2 gives A0 ∈ K, the A vertex of the original triangle. Columns 3 and 4 give A00 ∈ K, the A
vertex of the shadow triangle. When expressions are growing, these coordinates are given in two
rows. For example, in row 1, the perspector is the centroid, the original triangle is ABC itself and
A00 = −a2 : bc : bc, the third coordinate being obtained by swapping b and c. Points X2 , A0 = A
∗
∗
∗
and A00 are collinear. Obviously, (A00 )P , (B 00 )P , (C 00 )P , is another central triangle inscribed in K.
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
201
R
A0
a3/2
A
X(2)
A
X(238) −a2 : bc :
X (31)
A
X (1)
A
X(238) upright
X (6)
A
X (6) −a2 : bc :
X (1) −abc : b3 :
X (31) −a2 : bc :
X (1)
X (2)
X (6)
A00
B 00
−a3/2
b3/2
−a2
bc
−abc
b3
−abc
b3
−a2 (a + b + c)
b (bc + ca + ab)
−a (bc + ca + ab)
b2 (a + b + c)
−a (bc + ca + ab)
b2 (a + b + c)
2
2
a (a + b + c)
b a + b2 − c (a + b)
a2 (bc + ca + ab)
(c − b) a2 b + b3 (c − a)
2 a2 b2 + ca c2 + ab
b b2 + ca a3 + b3 − c3 − abc
−a2 : bc :
a (a + b) (c + a) a2 + b2 + c2 + bc + ca + ab
b2 (c + a) a2 + b2 + ab − c (a + b + c)
3
−abc : b :
2 a3 bc b2 + ca c2 + ab
b2 + ca b4 c3 + a3 bc3 − a3 b4 − b3 a2 c2
−abc : b3 : a (a + b) (c + a) c2 + bc + b2 a2 + bc (b + c) a + b2 c2
b2 (c + a) c2 − bc − b2 a2 + abc (−b + c) + b2 c2
Table 17.2: Some cubic shadows on EAC2
17.7
The cubic K060
Proposition 17.7.1. Let A0 , B 0 , C 0 be the reflections of a point M into the sidelines BC, CA, AB.
When triangle A0 B 0 C 0 is perspective with ABC, point M lies on the Neuberg cubic K001, while
the resulting perspector N lies on another cubic (K060).
Proof. The matrix of the reflection into the sideline BC is :


−a2
0 0


σA '  a2 + b2 − c2 a2 0 
a2 − b2 + c2 0 a2
Start from M = p : q : r. Compute A0 = σA (M ) , etc and obtain :


−pa2
a2 + b2 − c2 q + pb2
a2 − b2 + c2 r + pc2


A0 B 0 C 0 '  a2 + b2 − c2 p + a2 q
−qb2
b2 + c2 − a2 r + qc2 
a2 − b2 + c2 p + a2 r b2 + c2 − a2 q + b2 r
−rc2
Then det (AA0 , BB 0 , CC 0 ) is computed and identified with K001. Now, start
−1
Compute δA = σA (AN ) = (A ∧ N ) · σA , etc and obtain :

 
δA
2 vSb − 2 wSc
−wa2
va2

 
wb2
2 wSc − 2 uSa
−ub2
 δB  ' 
δC
−vc2
uc2
2 uSa − 2 vSb
from N = u : v : w.



Then det (δA , δB , δC ) is computed, and we obtain yet another pK cubic, defined as K060.
Proposition 17.7.2. K060, the pK (1989, 265) cubic. The pole P , the pivot U and the Morley
equations of this cubic are respectively :
σ13 σ22 − σ1 σ23 − 4 σ14 − 9 σ12 σ2 + 9 σ22 σ3
1
etc ; zP =
P = 2 2
b c − 4 Sa2
3 σ12 σ22 − 6 σ23 + (9 σ1 σ2 − 6σ13 ) σ3
U=
b2 c 2
Sb
Sc
Sa
:
:
− 4 Sa2 a2 c2 − 4 Sb2 a2 b2 − 4 Sc2
;
zU =
σ12 − σ2
σ1
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
202
17.8. Eigentransform
s2 2
s2
2s1
2
2
Z Z − s1 ZZ +
− 22 TZ2 + s21 − 2 s2 TZ +
 s3
s
s
3
3

2
3

3 s2 − s21 s2
s22 s1
s3
s1
s2 − 3 s21
2
T Z+
+ 2
− s1 T2 Z +
− 22 T3
s3
s3
s3
s3
s3





Both umbilics belong to the curve. The corresponding asymptotes intersect at X(3448). The real
asymptote :
[σ1 σ22 , 2 σ3 σ13 − 2 σ23 , −σ12 σ2 σ3 ]
is parallel to the Euler line. The sixth intersection with the circumcircle is X(1141).
Proof. Direct inspection.
D on K001, F=isogD on K001
N d = antig(D) = (isg ◦ inv ◦ isg) (D)
Nf
17.8
= antig (F ) = (isg ◦ inv) (D)
Eigentransform
Definition 17.8.1. The mapping U 7→ cevadiv U, UF# is called eigentransform of U wrt pole
P = F 2 . In ETC, F =X(1), i.e. P =X(6), is assumed, and notation ET (U ) is used. Here, the
same notation is used, but isogonal conjugacy isn’t assumed.
Example 17.8.2. Assuming P =X(6), pairs (I,J) such that X(J) = ET(X(I)) include :
1
2
3
4
5
6
7
8
9
10
1
3
1075
155
2120
194
218
2122
2124
2126
13
14
19
20
30
37
57
63
69
75
62
61
2128
2130
2132
2134
2136
1712
2138
2172
81 3293
86 3294
88
1
92
47
94
49
99
39
100
1
101 2140
110
5
162
1
174 266
190
1
512 2142
648 185
651
1
653
1
655
1
658
1
660
1
662
1
664 2082
673
1
694 384
771
1
799
1
811 2083
823
1
897
1
1113
3
1114
3
1156
1
1492
1
1821
1
1942 1941
Proposition 17.8.3. For any point U not on a sideline of triangle ABC, the following properties
of eigentransform are easy to verify :
1. The barycentrics of ET (U ) are (cyclically) :
vwf 2 u2 v 2 h2 + u2 w2 g 2 − v 2 w2 f 2
2. ET (U ) is the eigencenter of the cevian triangle of U as well as the eigencenter of the anticevian triangle of UP∗ .
.
3. ET (U ) = F = f : g : h (fixed point of the isoconjugacy) if and only if U = F or U lies
on the CC (F ) circumellipse. When P =X(6), then F =X(1) and this locus is the Steiner
circumellipse: yz + zx + xy = 0.
#
4. Points U , ET (U ) and (ET (U ))F are collinear points of the cubic pK (P, U ) .
#
5. Points F , ET (U ) and (cevadiv (U, F ))F are collinear. The last point is also on the cubic.
6. ET(U) is the tangential of UF∗ .
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
17.9
203
Non pivotal isocubics nK(P,U,k) and nK0(P,U)
Definition 17.9.1. The non pivotal isocubic with pole P , root U and parameter k is defined by
the equation :
.
nK (P, U, k) = ux y 2 r + qz 2 + vy z 2 p + rx2 + wz qx2 + py 2 + k xyz
(17.11)
When k = 0, the cubic is noted nK0 (P, U ).
Remark 17.9.2. The more efficient method for specifying k is to indicate a point that belongs to
the cubic. This is noted nK (P, U, X).
Proposition 17.9.3. The "third intersections" of a nK (P, U, k) with the sidelines are the cocevians
of the root U . Therefore, they are aligned. This is to be compared with the fact that, for a pK
cubic, these points are the cevians of the pivot.
Proof. Direct inspection.
Remark 17.9.4. In the general case, a nK0 (P, U ) does not contain P nor U nor any of the four
fixed points F of the conjugacy.
Definition 17.9.5. We define the F -crosssum of two points U = u : v : w and X = x : y : z that
aren’t lying on a sideline of ABC as :
crosssumF (U, X) = f 2 (wy + vz) : g 2 (uz + wx) : h2 (vx + uy)
Remark 17.9.6. In ETC, F =X(1) is assumed. Defined as above, the operation (F, U, X) 7→
crosssumF (U, X) is globally type-keeping and provides a point when the entries are points (F is
any of the four fixed point of the conjugacy X 7→ XP∗ ).
Definition 17.9.7. The NKP (X) point is the pole of the line X XP∗ with respect to the circumconic
that passes through X and XP∗ (Bernard Gibert, 2003/10/1). Using barycentrics and P = p : q :
r = f 2 : g 2 : h2 , we have :
NKp (X)
= p x r y 2 + q z 2 : q y p z 2 + r x2 : rz q x2 + p y 2
=
∗
crosssumF (X, XP∗ ) = crossmul (X, XP∗ ) = (cevamul (X, XP∗ ))P
Proof. Direct computation.
Remark 17.9.8. Here, XP∗ is the perspector of the cevian triangle of NKP (X) and the anticevian
triangle of X.
Example 17.9.9. Using F=X(1), i.e. P=X(6), we have NK(X(I)) = X(J) for these (I,J):
I
J
1
1
2
39
3
185
4
185
6
39
9
2082
19
31
57
63
2083 2085 2082 2083
Proposition 17.9.10. When NKP (X) belongs to the tripolar line of UP∗ , then point X belongs to
the cubic nK0(P, U ).
Proof. Direct computation. In Kimberling (1998, p. 240), notation Z + (XY ) is used to denote
the nK0 (#1, U ) cubic where the pole U is the isogonal of the tripole of line XY . Therefore,
Z + (X1 X6 )
= nK0 (#1, 513)
Z + (X3 X6 )
= nK0 (#1, 523)
+
= nK0 (#1, 649)
+
= nK0 (#1, 650)
Z (X1 X2 )
Z (X1 X3 )
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
204
17.10
17.10. Conicopivotal isocubics cK(#F,U)
Conicopivotal isocubics cK(#F,U)
Definition 17.10.1. A conico-pivotal isocubic cK (#F, U ) (Ehrmann and Gibert, 2005) is a non
pivotal isocubic nK (P, U, k) that contains one of the fixed points of the isoconjugacy (F 6= U is
assumed). Using F = f : g : h instead of P = p : q : r = f 2 : g 2 : h2 , k = −2 (ghu + f hv + f gw)
and equation becomes :
x(gz − hy)2 u + y(f z − hx)2 v + z(f y − gx)2 w = 0
(17.12)
Proposition 17.10.2. The pivotal conic is defined as the conic C tangent to the six lines FB FC ,
AA0U and cyclically where FA FB FC is the anticevian of F and A0U BU0 CU0 the cocevian of U . Then
the dual conic of C is conicev (1/F, 1/U ) and C itself has equation :
X
cyclic
2
(gw − hv) x2 − 2 gu2 h + 3 f (gw + vh) u + f 2 vw zy = 0
Proof. We have the equations :

 
f
f
1 1

 

(FB FC ) = FB ∧ FC =  −g  ∧  g  = [0, 2f h, 2f g] = 0, ,
g h
−h
h
1 1
0
AAU = UB UC = 0, ,
v w

The equation of C follows by duality. Barycentrics of the center are 2 f u−(v + w) f −(g + h) u, etc.
Proposition 17.10.3. The contact conic (K) is defined as the circumconic whose perspector is
g
h
f
K' 2 + +
f, etc
u v w
Assuming F 6= U , three of the intersections of the pivotal and contact conics are the three contacts
of cK with C, the fourth point being :
g
h
f
÷ (gw − vh) , etc
T4 ' 2 + +
u v w
Proof. Eliminate z between (K) and C. Obtain P1 (x, y) P3 (x, y), where degrees are respectively
1 and 3. Solving for P1 gives directly T4 . Eliminate z between cK and C. This leads again to P3 ,
proving that each common point is a contact and belongs also to (K).
Example 17.10.4. A special case is obtained when U = F , i.e. when the root is a fixed point
of the isoconjugacy. Then C = (K) is a circumconic. The Tucker cubic K015 is obtained with
F =X(2), while K228 is obtained with F =X(1) and K229 is obtained with F =X(6).
17.11
Simson cubic, aka K010
Notation 17.11.1. In this section, the involved pole is the centroid, so that X ∗ = isot (X). Due
to the nature of the cubic, the key point is F = X2 (involved as fixed point) rather than P = X2
(involved as pole). Therefore, letter P has been used not to describe the pole, but the independent
moving point of various parametrization.
Definition 17.11.2. The Simson cubic is the locus of the tripoles of the Simson lines. Depicted
as K010 (cf. Figure 17.3) in Gibert (2004-2014). Founding paper is Ehrmann and Gibert (2001).
Proposition 17.11.3. The Simson cubic (K010) is cK (#X2 , X69 ). Centroid G = X2 belongs to
K010 (Simson line of Q is L∞ when Q ∈ L∞ ). Apart from this isolated point, a parameterization
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
205
Centroid G = X2 is isolated, but belongs nevertheless to the cubic.
Figure 17.3: The Simson cubic (as depicted in Gibert-CTP)
of K010 is given in (22.4), starting from L∞ . An other parameterization, using the barycentrics of
the involved point on Γ is as follows :




u
b2 wu2 − c2 vu2 + b2 − c2 wvu




 v  ∈ Γ 7→  c2 uv 2 − a2 v 2 w + c2 − a2 wuv  ∈ K010
w
a2 vw2 − w2 b2 u + a2 − b2 wuv
Definition 17.11.4. Cubic K162 is the isogonal transform of the Simson cubic (that can also be
obtained by Q 7→ Q ÷ X6 ). Therefore, K162 is cK (#X6 , X3 ).
b
Definition 17.11.5. The Gibert-Simson transform is another parameterization of the Simson
cubic that also uses an U ∈ Γ :
"
! #
c2 a2 + b2 − c2
b2 c2 + a2 − b2
−
u
GS (U ) = cyclic
va2
a2 w
P
The lack of uniqueness is due to the binding relation a2 vw = 0. Definition introduced in ETC on
2003/10/19, leading to points X(2394) - X(2419) on the Simson cubic and points X(2420)-X(2445)
–their isoconjugates– on K162.
Remark 17.11.6. Regarding triangle centers that do not lie on the circumcircle, GS(X(I)) = X(J)
for these (I,J): (32,669), (48,1459), (187,1649), (248,879), (485,850), (486,850). Of course, other
realizations of U 7→ K give other results. Here again, only parameterization (22.4) ensures uniqueness.
Example 17.11.7. Use t P = t X525 as entry point Figure 17.4 (arrow at the left of the bottom
diagram). Obtain X30 = Q1 ∈ L∞ by (7.14), then X74 = U1 ∈ Γ by isogonal conjugacy. The
Steiner line St1 hasn’t received any name, while the Simson line Si1 of U1 is t X247 . The trilinear
pole of this line, i.e. X2394 = K1 ∈ K010, can be obtained by isot ◦ (t ) from Si1 , by gs from U1
and also directly from P (the dotted line) using parameterization (22.4).
Proposition 17.11.8 (Fools’ Day Theorem). Direct arrows U1 7→ K2 and U2 7→ K1 have a
∗
geometrical meaning : (gs (U )) is the eigencenter of the pedal triangle of U.
Proof. Point K2 has no other choice : he *is* the unary cofactor of the pedal triangle of U , and
therefore the perspector of this triangle with anything else.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
206
17.11. Simson cubic, aka K010
t
isotom
K1 ∈ K010
✛
...
.✲
.
.
.
..
....
t
h U2 , 12 ........
...
....
.
.
.
...
φ
St2 ✛
U2 ∈ Γ
Si2
✻
❄
Q1 ∈ L∞
t
Si1
✻
t
gs
gs
U1 ∈ Γ
σ3
h U1 , 12
φ ✲
St1
isog
isog
∞
K2 ∈ K010
✲
∞
❄
Q2 ∈ L∞
orthopoint
where φ(U ) =t isog (U1 ∗b X4 ) and ∞(L) = L ∧ L∞
isotom
t
t
2394✛
2407✛
2407
.
.
.
.
.
✲
✲
.
.
....
..
✻
....
....
.
.
.
.... t
..
1
t
.
g
.
h
74,
h 110, 21 ........
.
s
....
2
gs
....
...
.
.
.
.
.
....
....
φ
φ
✲ St1 (?)
✲ t 525 ✛
110
σ3
74
t
2394
✻
t
i so
g
g
i so
∞
❄
30
⊥
∞
❄
523
Figure 17.4: The Simson diagram
Proposition 17.11.9. The barycentric equation of the Simson cubic is
X
x(z 2 + y 2 )(b2 + c2 − a2 ) − 2(a2 + b2 + c2 )x y z
(17.13)
cyclic
In other words, the Simson cubic is nK (X2 , X69 , #X2394 ). Moreover, one of the fixed points
(namely G = X2 ) belongs to the cubic and the Simson cubic is in fact cK (#X2 , X69 ).
Proof. Obtained from the parametric representation. The converse property is more easily obtained
from next coming proposition.
Definition 17.11.10. Special points wrt K010. Points on sidelines of triangle ABC or of
antimedial triangle, together with the centroid are said to be special wrt K010 (because quite
every "general" formula turns wrong when dealing with these points).
Proposition 17.11.11. Among the special points, the following are the sole and only elements of
K010 :
(i) then centroid itself (fixed point under isoconjugacy)
(ii) the vertices of triangle ABC, the cocevians of X69 (the root) and points b + c : b − c : c − b or
b − c : b + c : −b − c and cyclically.
Proof. Direct inspection.
Proposition 17.11.12. When point X is on the Simson cubic but not G, A, B, C then :
(i) the trilinear polars of X and X ∗ are perpendicular
(ii) the trilinear polars of X and X ∗ are concurrent on the nine-point circle
Conversely, when X is not special and either property holds, then X is on the cubic.
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
207
Proof. When X is on the Simson cubic, tripolar (X) is a Simson line and conclusion follows from
tripolar = t ◦isot. Conversely, if (i) then points t X ∗ ∧ L∞ and t X ∧ L∞ are the infinity points
of both tripolars. They have to be the orthopoint of each other, and
Q (17.13) is re-obtained by
elimination. If (ii) then dividing nineq (X ∧ X ∗ ) by (17.13), leads to (y + z) /x2 . When X is on
a sideline of ABC, conjugacy is no more defined, and when y + z = 0 (implying X on the sidelines
of the antimedial triangle) then X ∧ X ∗ is ever 0 : 1 : 1. Outside of these six lines, both conditions
are equivalent.
17.12
Brocard second cubic aka K018
Definition 17.12.1. The Brocard second cubic is inventoried as K018 in (Gibert, 2004-2014). This
cubic is nK0 (X6 , X523 ). It is a circular isogonal focal nK cubic with root X(523) and singular
focus X(111). The real asymptote is parallel to GK. It is also the orthopivotal cubic O(X6) and
Z+(L) with L = X(3)X(6) in TCCT p.241. See also Z+(O) = CL025 and CL034.
Proposition 17.12.2. The barycentric equation of K018 is :
X
x(b2 z 2 + y 2 c2 )(b2 − c2 ) = 0
(17.14)
cyclic
17.13
A family of circular cubics
Proposition 17.13.1. Given the fixed points A, B, C, we define the tripolar curve W (u : v : w)
as the locus of points M such that :u AM + v BM + w CM = 0. This curve is a bicircular quartic,
except if ±u ± v ± w = 0.
Proof. This equation can be rationalized into :
AM4 u4 − 2 AM2 BM2 u2 v 2 − 2 AM2 CM2 u2 w2 + BM4 v 4 − 2 BM2 CM2 v 2 w2 + CM4 w4 = 0
When using second degree Morley coordinates, the Pythagorean formulas leads to a not so huge
expression. When making T = 0, it only remains :
(u − v − w) (u − v + w) (u + v − w) (u + v + w) × α4 β 4 γ 4 Z2 Z
2
Proposition 17.13.2. From now on, ±u ± v ± w = 0 is assumed. In the special cases where
w = 0 and u ± v = 0, etc, the curve is a bipolar conic. Otherwise, the curve is a cubic that admits
four focuses that are concyclic: A, B, C (the fixed points) and a variable point D.
Proof. Replace u : v : w by 1 : s : −1 − s. Write F = z : t : ζ and assume that F Ω ∩ W has
a multiple root. The discriminants are 4th degree, but they split and give the four focuses. It
remains only to check zζ − t2 . Since the expression obtained for s 7→ D is quite discouraging, we
have the next Proposition.
Proposition 17.13.3. Let τ be a turn. Then D ' τ 2 : 1 : 1/τ 2 is the fourth focus of W (ap : bq : cr)
where
p : q : r ' ατ − βγ : βτ − αγ : γτ − αβ
Using σ1 = α + β + γ + τ, etc (the so called symmetric functions), we obtain :
W (α, β, γ, τ ) = 4 σ12 σ4 − σ32 Zσ4 − Z ZZ + T2 + T × f actor
where f actor =
2
8 σ1 σ3 σ42 − 4 σ2 σ32 σ4 + σ34 Z + σ14 − 4 σ12 σ2 + 8 σ1 σ3 Z2 +2 2 σ12 σ2 σ4 − σ12 σ32 − 8 σ1 σ3 σ4 + 2 σ2 σ32 ZZ
Proof. When D is given, there are two values of s, and we can exchange them using τ 7→ −τ . A
choice has to be done, and our’s is :
s = α2 − γ 2 (αγ − βτ ) β ÷ β 2 − γ 2 (ατ − βγ) α
When substituting, one
can see the symmetry of the expression, allowing to use the σj . Moreover,
quantity σ12 σ4 − σ32 = p0 q0 r0 appears as common factor, excluding the degeneracy into a conic
and the line at infinity.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
208
17.13. A family of circular cubics
.
.
Proposition 17.13.4. Let δ be a turn, and define W + = W (+τ ) ; W − = W (−τ ). Then (see
Figure 17.5) :
1. Cubics W ± are going through O and Ω± . The third point at infinity of W + is U + ' σ4 : 0 : 1
and the asymptote is
∆ ' −4 σ4 , σ12 σ4 − σ32 , 4 σ42
2. Each curve is invariant by symmetry wrt Γ0 (so that U is the tangential of O).
3. The tangential of U (τ ) is :


−8 σ1 σ3 σ42 + 4 σ2 σ32 σ4 − σ34 + 16 σ43


Ut (τ ) ' 
4 σ12 σ4 − 4 σ2 σ4 + σ32 σ4

−8 σ1 σ3 + 4 σ12 σ2 − σ14 + 16 σ4 σ4
and it happens that Fs (τ ) = Ut (−τ ) where Fs is the singular focus (not on it’s curve).
4. Apart from Ω± , the 8 intersections of W ± and the unit circle are obtained from −σ3 ÷ σ1 by
changing the signs of α, β, γ, τ (odd changes lead to W − , even ones to W + ).
5. The nine common points of W ± are O, Ω± (3), the Vj (3) and the Vj0 (3) where :
z (Va0 ) = α2 τ 2 − β 2 γ 2 ÷ α2 − β 2 − γ 2 + τ 2 , etc
and Va , Va0 are symmetric wrt Γ0 , etc. Moreover, Va0 ∈ Vb Vc while the Vj form the diagonal
triangle of both quadrangles Γ0 ∩ W + and Γ0 ∩ W − .
6. Both curves are orthogonal at each of their intersections. Among them, ∆+ and ∆− are
1 2 1
1
1 2 1
s − s2 + τ 2 (see below).
orthogonal. Inter, at K (D) = σ1 − σ2 =
4
2
4 1 2
4
Proof. Due to the choice zA = α2 , everything factors and computations are easy. The six Vj , Vj0
depend only on the squares α2 , etc.
Proposition 17.13.5. The locus of Ut , Fs is a circular quintic. Its singular focus is X(143).
The real asymptotes are going through (2A + B + C) /4, etc and the directions of sidelines are the
points at infinity. The envelope of ∆ is the deltoid :
1 2 1
s3
1
M (τ ) =
s − s2 − τ 2 −
4 1 2
4
2τ
whose center is X(140)= (A + B + C + O) /4, its internal and external radiuses being 1/4 and
3/4. Cups are given by τ 3 = s3 , i.e. the Morley’s directions. Moreover
1 2 1
1 2 1
1
+
−
∆ ∩∆ =
σ − σ2 =
s − s2 + τ 2
4 1 2
4 1 2
4
describes a circle around X(140).
Proof. Part one: obvious elimination. Part two: transcribe ∆ into a function of τ using :
σ1 = τ + s1 ; σ2 = s1 τ + s2 ; σ3 = s2 τ + s3 ; σ4 = τ s3
And then obtain M (τ ) = ∆ (τ ) ∧ (d∆ (τ ) ÷ dτ ).
Proposition 17.13.6. Equation of W (ap : bq : cr) can be rewitten as :
p |birap (A, C, B, M )| + q |birap (B, C, A, M )| + r = 0
Therefore reflection of W wrt a circle centered at a focus leads to Cartesian ovals whose focuses are
the images of the remaining three original focuses (see Figure 17.6, where ρ = 1 so that O0 = O).
Proof. Original equation can be rewritten as
p BC MA + q CA MB + r AB MC = 0
March 30, 2017 14:09
published under the GNU Free Documentation License
17. About cubics
209
Figure 17.5: Angels
Figure 17.6: Cartesian ovals by inversion
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
210
March 30, 2017 14:09
17.13. A family of circular cubics
published under the GNU Free Documentation License
Chapter 18
Special Triangles
Central triangles have been defined in Section 3.2.
18.1
Changing coordinates, functions and equations
Proposition 18.1.1. Any triangle T can be used as a barycentric basis instead of triangle ABC.
When columns of triangle T are synchronized, the old barycentrics x : y : z (relative to ABC) can
be obtained from the new ones ξ : η : ζ (relative to T) by :




x
ξ




 y = T . η 
z
ζ
while the converse transformation can be done using the adjoint matrix.
Proof. A column of synchronized barycentrics acts on the matrix of rows containing the projective
coordinates of the vertices of the reference triangle by the usual matrix multiplication.
Remark 18.1.2. By definition of synchronized barycentrics, L∞ . T ' L∞ and the line at infinity
is (globally) invariant.
Proposition 18.1.3. Let α, β, γ the side lengths of triangle T (computed using Theorem 7.2.4).
Consider a central punctual transformation Φ that can be written as :
p : q : r 7→ u : v : w = φ (a, b, c, p, q, r)
with all the required properties of symmetry and homogeneity. Consider now the corresponding
punctual transformation Φ0 with respect to triangle T (written in its normalized form) and define
φT as the action of Φ0 on the old barycentrics (the ones related to ABC). Then :






p
p






φT a, b, c,  q  = T . φ α, β, γ, T −1 .  q 
r
r
Example 18.1.4. Applied to the isogonal transform and some usual triangle, this leads to formulas given in Figure 18.1. The term "complementary conjugate" is a synonym for "medial isogonal conjugate", as is "anticomplementary conjugate" for "anticomplementary isogonal conjugate".
Also, "excentral isogonal conjugate" is "X(188)-aleph conjugate" and "orthic isogonal conjugate"
is "X(4)-Ceva conjugate".
Proposition 18.1.5. Let α, β, γ the side lengths of triangle T. Consider a conic Φ whose matrix
can be written as M (a, b, c) with the required properties of symmetry and homogeneity. Consider
now the corresponding conic Φ0 with respect to triangle T (written in its normalized form) and
define MT as the matrix defining Φ0 wrt the old barycentrics. Then :
t
MT (a, b, c) = T −1 . M (α, β, γ) . T −1
Example 18.1.6. Applied to the circumcircle and some usual triangles, this leads to formulas
given in Figure 18.1.
211
212
18.2
18.2. Residual triangles
Residual triangles
Definition 18.2.1. The residual triangles associated with point P are the triangles APb Pc , Pa BPc ,
Pa Pb C where Pa Pb Pc is the cevian triangle of P (mind the order...).
Proposition 18.2.2. The A-residual of the orthic and the intouch triangles have the following
sidelengths :
[α, β, γ]orthic
2 2 2
α , β , γ intouch
Sa
[a, b, c]
bc
2
(b + c − a)
[(a + b − c) (a − b + c) , bc, bc]
4 bc
=
=
The incenters of the orthic residuals are the orthocenters of the intouch residuals.
18.3
Incentral triangle
definition cevian triangle of the incenter X(1)
pythagoras (strong values)
2
α =
abc a3 + a2 b + a2 c − ab2 − ac2 + 3 abc − b3 + b2 c + bc2 − c3
2
barycentrics (normalized)



C1 = 

(f, g)
18.4
2
(a + b) (a + c)
a
a+c
0
c
a+c
0
b
b+c
c
b+c
a
a+b
b
a+b
0





(0; b)
Excentral triangle
definition triangle of the excenters. And thus, anticevian triangle of the incenter X(1)
pythagoras (strong values)
2 2 2 2R
α ,β ,γ =
[a (b + c − a) , b (c + a − b) , c (a + b − c)]
ρ
thus similar with the intouch triangle (center X(57), ratio 2R/ρ)
barycentrics (normalized)
A1
(f, g)
18.5
(−a; b)



=


−a
b+c−a
b
b+c−a
c
b+c−a
a
c+a−b
−b
c+a−b
c
c+a−b
a
a+b−c
b
a+b−c
−c
a+b−c






Medial triangle
definition cevian triangle of the centroid X(2)
side_length (strong values)
[α, β, γ] =
March 30, 2017 14:09
1
[a, b, c]
2
published under the GNU Free Documentation License
18. Special Triangles
213
barycentrics (normalized)

0
1 
=  1
2
1
C2
18.6
1
0
1

1

1 
0
Antimedial triangle
definition anticevian triangle of the centroid X(2)
side_length (strong values)
[α, β, γ] = 2 [a, b, c]
barycentrics (normalized)

−1

= 1
1
A2
18.7
1
−1
1

1

1 
−1
Orthic triangle
definition cevian triangle of the orthocenter X(4)
side_length (strong values)
[α, β, γ] =
barycentrics (synchronized)
C4
angles
18.8
1 2
a Sa , b2 Sb , c2 Sc
abc

0

=  Sc /a2
Sb /a2
Sc /b2
0
Sa /b2

Sb /c2

Sa /c2 
0
π − 2A, π − 2B, π − 2C
Tangential triangle
definition The sidelines are the tangents to the ABC-circumcircle at the vertices.
key_property Anticevian triangle of X(6).
side_length (strong values)
[α, β, γ] =
2
abc
a Sa , b2 Sb , c2 Sc
2 Sa Sb Sc
Therefore, this triangle is similar to the orthic triangle.
barycentrics (synchronized)
A6

a2
−
 Sa

 b2
=
 Sa
 c2
Sa
a2
Sb
b2
−
Sb
c2
Sb
a2
Sc
b2
Sc
c2
−
Sc







—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
214
18.9
18.9. Brocard triangle (first)
Brocard triangle (first)
Remember that Brocard points are defined by ω + = a2 b2 : b2 c2 : c2 a2 and ω − = c2 a2 : a2 b2 : b2 c2
(see Proposition 7.7.1).
.
definition A1 = Bω − ∩ Cω + ' a2 : c2 : b2 etc. This triangle is inscribed in the Brocard 3-6 circle,
see , diameter X(3)X(6).
side_length (strong values)
[α, β, γ] =
W2
[a, b, c]
2 So
and therefore Broc1 is (anti-)homothetic to ABC. Perspector X(76), the third Brocard
point. Moreover, anti-similar to ABC with center X(2).
barycentrics (synchronized)
Broc1
18.10

a2
 2
= c
b2
c2
b2
a2

b2

a2 
c2
Brocard triangle (second)
definition Ua ' b2 + c2 − a2 : b2 : c2 is the projection of X(3) on the A-symmedian, etc. (and
therefore belongs to the 3-6 Brocard circle).
key_property Triangle ABC and triangle Broc2 share the same isodynamic centers X(15), X(16).
Exercise 18.10.1. The center Ea of circle A, X(15), X(16) is the inverse in circumcircle of Ua ,
etc.
18.11
Brocard triangle (third)
.
definition A3 = Cω − ∩ Aω + ' b2 c2 : b4 : c4 = isog (A1 ).
perspector with ABC: X(32)
18.12
Intouch triangle (contact triangle)
definition Cevian triangle of the Gergonne point X(7).
key_property contacts of the incircle and the sidelines.
pythagoras (strong equality)
2 2 2
ρ
[a (b + c − a) , b (c + a − b) , c (a + b − c)]
α ,β ,γ =
2R
thus similar with the excentral triangle (ratio ρ/2R)
barycentrics (normalized)
C7
March 30, 2017 14:09



=


0
a+b−c
a
c+a−b
a
a+b−c
b
0
b−a+c
b
c+a−b
c
b−a+c
c
0






published under the GNU Free Documentation License
18. Special Triangles
18.13
215
Extouch triangle
definition cevian triangle of the Nagel point X(8).
side_length (strong values)
[α, β, γ] =
barycentrics (normalized)
C8
18.14



=


0
c+a−b
a
a+b−c
a
b−a+c
b
0
a+b−c
b
b−a+c
c
c+a−b
c
0






Hexyl triangle
definition symmetric of the excentral triangle Ja Jb Jc wrt the circumcircle. Thus Ha Jb Hc Ja Hb Jc
is an hexagon whose opposite sides are parallel.
key_property Vertex Ha is the point in which the perpendicular to AB through the excenter Jb
meets the perpendicular to AC through the excenter Jc .
pythagoras (strong values)
2 2 2 2R
[a (b + c − a) , b (c + a − b) , c (a + b − c)]
α ,β ,γ =
ρ
thus similar with the intouch triangle (center I, ratio 2R/ρ).
barycentrics (normalized) Ha ' a (aSa + bSb + cSc + abc) : b (aSa + bSb − cSc − abc) : c (aSa − bSb + cSc − abc)
circumcircle centered at X(1), radius 2R.
18.15
Fuhrmann triangle
Definition 18.15.1. Fuhrmann triangle is A00 B 00 C 00 where A0 B 0 C 0 is the circumcevian triangle
of X1 and A00 is the reflection of A0 in sideline BC (and cyclically for B 00 and C 00 ).
Proposition 18.15.2. Side length of Fuhrmann triangle are :
s
!
r
r
a
b
c
,
,
W4 ×
(a + b − c) (a − b + c)
(b − c + a) (b + c − a)
(c + a − b) (c − a + b)
p
where W4 = a3 + b3 + c3 − (a2 b + a2 c + ab2 + ac2 + b2 c + bc2 ) + 3 abc
Quantity W4 is the Fuhrmann square root (12.15). Circumcircle of the Fuhrmann triangle is the
Fuhrmann circle of ABC (whose diameter is [X4 , X8 ]
Remark 18.15.3. As noticed in (Dekov, 2007), OIF uN a and OIHF u are parallelograms, whose
respective centroids are the Spieker center and nine-point center, respectively, where IF u = 2N I
or IF u = R − 2r.
18.16
Star triangle
Definition 18.16.1. Star triangle. Consider the midpoints A0 B 0 C 0 of the sidelines of triangle
ABC. Draw from each midpoint the perpendicular line to the corresponding bisector. These three
lines determine a triangle A∗ B ∗ C ∗ . This is our star.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
216
18.16. Star triangle
Proposition 18.16.2. The synchronized barycentrics and the sidelengths of the star triangle are :
T∗

b+c

' c−a
b−a
 
b+c−a
b−c
 
0
a−c ·
0
b+a
c−b
c+a
a−b
2 2 2
α ,β ,γ =
R
2ρ
2
−1
0
0

c+a−b
0

0
a+b−c
× [a (b + c − a) , b (c + a − b) , c (a + b − c)]
Similar to the intouch triangle (k 2 = R3 ÷ 2ρ3 ) and to the excentral triangle (k 2 = R ÷ 8ρ). We
have the following central correspondences :
T∗
T
2 3817
3
946
4
10
5
5
6
142
20 4301
25 3452
30
517
39 2140
51
2
52
3
53
141
65
178
113
119
114
118
115
116
125
11
128
113
129
114
130
115
131
117
132
120
T∗
133
134
135
136
137
138
139
143
184
185
235
389
403
418
427
428
511
512
520
523
525
526
T
121
122
123
124
125
126
127
140
226
1
1329
1125
3814
2051
2886
3740
516
514
3667
513
3309
900
T∗
542
647
690
804
924
974
1112
1154
1205
1495
1503
1510
1531
1562
1568
1596
1637
1824
1843
1986
1990
2052
T
2801
3835
3887
926
522
1387
3035
30
3254
908
518
523
1512
4904
1532
3820
4928
2090
9
214
3834
3840
T∗
2393
2501
2574
2575
2679
2777
2781
2782
2790
2794
2797
2799
2848
3258
3564
3566
3574
3575
3917
T
527
4885
3308
3307
1566
2802
528
2808
2810
2809
2821
2820
2832
3259
971
3900
442
960
1699
For example, orthocenter X(4, T∗ ) is Spieker center X(10, T ).
Proof. Straightforward computations. In fact, A0 A∗ and B ∗ C ∗ are orthogonal and the orthic
triangle of T∗ is the medial triangle of T .
March 30, 2017 14:09
published under the GNU Free Documentation License
18. Special Triangles
217
medial
C2
antimedial
A2
orthic
C4
tangential
A6
excentral
A1
medial
antimedial
orthic
tangential
excentral
C2
A2
C4
A6
A1
φT when φ is the isogonal conjugacy
(v + w − u) (u + v − w) b2 + (u − v + w) c2
−a2
b2
c2
+
+
v+w u+w u+v
u (−Sa u + Sb v + Sc w)
2
−a
b2
c2
a2
+
+
c2 v + wb2
uc2 + wa2
ub2 + a2 v
−1
1
1
a (b+c−a)(cv+bw) + (a−b+c)(cu+aw) + (b+a−c)(bu+av)
circumcircle of T
P 2
P
b + c2 − a2 x2 − 2 a2 yz
P
P 2 2
yz
a x + a2 + b2 + c2
P 2
P
b + c2 − a2 x2 − 2 a2 yz
P 2
P 4 2 P 6 P 2
a2 b2 c2
b + c2 − a2 x2 +
a yz
6a b −
3a
P
P
bcx2 + (a + b + c) ayz
Figure 18.1: Special Triangles
Figure 18.2: The star triangle
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
218
March 30, 2017 14:09
18.16. Star triangle
published under the GNU Free Documentation License
Chapter 19
Binary operations
Let us examine again some already defined operations, and consider in more details their formal
properties
19.1
cevamul, cevadiv, crossmul, crossdiv
These operations were considered in Kimberling ETC to formalize various properties related to
cevian nests.
Remark 19.1.1. cevadiv has been defined in Section 4.9 by :
.
cevadiv (P ; U ) = persp (Cevian (dividend, ABC) , Anticevian (divisor, ABC))
' u(−q r u + r p v + p q w) : v(q r u − r p v + p q w) : w(q r u + r p v − p q w)
Property X = cevadiv (P, U ) is sometimes stated as "X is the P-cevaconjugate of U". This
operation is clearly a type-keeping (P, U )-map and an involutory U -map (when P is fixed).
Remark 19.1.2. The cevamul operation is the converse of the previous one, and is sometimes
called cevapoint.
cevamul (u : v : w, x : y : z) = (uz + wx)(uy + vx) : (vz + wy) (uy + vx) : (vz + wy) (uz + wx)
This operation is clearly commutative and type-keeping.
Remark 19.1.3. The crossdiv has been defined in Section 4.8 by :
.
crossdiv (P ; U ) = persp (ABC, Cevian (dividend, Cevian (divisor, ABC)))
v
w
u
:
:
'
quw + ruv − pvw pvw + ruv − quw pvw + quw − ruv
Property X = crossdiv (P, U ) s sometimes stated as "X is the P-crossconjugate of U". This
operation is clearly a type-keeping (P, U )-map and an involutory U -map (when P is fixed).
Remark 19.1.4. The crossmul operation is the converse of the previous one, and is sometimes
called crosspoint.
crossmul (u : v : w, x : y : z) = (v z + w y) u x : (u z + w x) v y : (u y + v x) w z
This operation is clearly commutative and type-keeping.
Proposition 19.1.5. ’div’ formula. The isoconjugacy that exchanges P, U also exchanges
P ↔ U ; crossdiv (P, U ) ↔ cevadiv (U, P ) ; crossdiv (U, P ) ↔ cevadiv (P, U )
Proof. We have crossdiv (P, U ) ∗ cevadiv (U, P ) = P ∗ U .
b
b
Proposition 19.1.6. ’mul’ formula. For any isoconjugacy, we have :
#
cevamul (U, X) = crossmul UF# , XF#
F
#
#
#
crossmul (U, X) = cevamul UF , XF
F
Proof. Direct examination is shorter, using the "div" formula just above is more stratospheric.
219
220
19.2. Formal operators and conics
19.2
Formal operators and conics
Since they are all Cremona transforms of second degree, operators wrt a triangle can be transformed
into operators wrt a well chosen conic.
Proposition 19.2.1. barymul. Let ψ be the ABC-isoconjugacy that swaps points P, U (not on
.
.
the sidelines). Then ψ swaps CP U = conic (A, B, C, P, U ) and LP U = line (P, U ). The pole F 2 of
the conjugacy is the barymul of P, U and therefore the "barysquare" of the four fixed points of ψ:
F 2 ' P ∗ U ; f 2 = kp u, g 2 = kq v, h2 = kr w, k 6= 0
b
Proposition 19.2.2. We have the various ’cross’ formulas
formula
1
1
−
qw rv
1
1
+
rv qw
1
qw − rv
1
qw + rv
name
usefulness
inverse
intersections of the tripolars, perspector of CP U
X ∈ CP U
conipole of LP U wrt CP U
crossdiv
tripole of LP U
X ∈ LP U
U = persp (Acev (P ) , cev (X))
cevadiv
crossmul
cevamul
Thus the ’odd ones’1 aren’t Cremona transforms.
19.3
crossdiff, crosssum, polarmul, polardiv
Definition 19.3.1. The F -crossdiff of two points U = u : v : w and X = x : y : z that aren’t
lying on a sideline of ABC is defined by :
crossdiff F (U, X) = f 2 (wy − vz) : g 2 (uz − wx) : h2 (vx − uy)
Remark 19.3.2. In ETC, F =X(1) is assumed. Defined as above, the operation (F, U, X) 7→
crossdiff F (U, X) is globally type-keeping and provides a point when the entries are points (F is
any of the four fixed point of the conjugacy X 7→ X ∗ ).
Proposition 19.3.3. The F -crosssum of U, X that was defined at Definition 17.9.5 is constructible
using crossmul, cevamul and isoconjugacies. Therefore a better definition is the globally typekeeping function :
crosssumF (U, X)
=
=
f 2 (wy + vz) : g 2 (uz + wx) : h2 (vx + uy)
#
(cevamul (U, X))F = crossmul UF# , XF#
where F = f : g : h is either fixed point of the conjugacy.
Remark 19.3.4. Defined that way, crosssumF (U, X) is really different from the polar line of X wrt
the circumconic CC (U ) : u yz + v zx + w xy = 0 since this line is the next coming polarmul.
.
Proposition 19.3.5. Given U and P = crosssumF (U, X) one can find X using
X ' cevadiv PF# , U
Proof. Direct inspection (here P is generic, not the "square" of the fixed point F ).
1 English
joke. Can not be translated.
March 30, 2017 14:09
published under the GNU Free Documentation License
19. Binary operations
221
Proposition 19.3.6. The polarmul of two points U, X is a line ∆. When U = u : v : w and
X = x : y : z are not on the sidelines, then line ∆ is defined as the conipolar of the point X wrt
the circumconic CC (U ). We have polarmul (U, U ) = tripolar (U ) and
polarmul (U, X) = complem X ÷ U ÷ U
b
b
=
crossmul(tripolar(U ), tripolar(X)) = tripolar (cevamul (U, X))
=
[wy + vz ; uz + wx ; vx + uy]
Operation polarmul is commutative and type-crossing (i.e output is a line when entries are points).
Proposition 19.3.7. The polarmul of two lines D, ∆ is the point P obtained by applying
"the same rules" as above, i.e.
.
polarmul ([u, v, w] , [x, y, z]) =
wy + vz : uz + wx : vx + uy
Then polarmul (∆, ∆) is tripolar (∆) and polarmul (D, ∆) is the conipolar of line ∆ wrt the inconic
whose perspector is tripolar (D).
Proof. Direct inspection. Remember that the dual of an inconic "goes through" the sidelines.
Definition 19.3.8. The polardiv of a line ∆ ' [ρ; σ; τ ] and a point U ' u : v : w is the reverse
operation of the previous one. One has the formula :
polardiv(∆, U )
=
cevadiv(tripolar(∆), U )
=
(σv + τ w − ρu) u : (τ w + ρu − σv) v : (ρu + σv − τ w) w
that correctly defines a point when P is a line and U is a point.
19.4
Complementary and anticomplementary conjugates
V ✛
..
..
..
.
anticomcon
..
..
.❄
V′ ✛
anticompl
X
complem
kP
anticompl
xcomcon =
❄
X′
complem
✲ U
..
..
..
.
comcon
..
..
..❄
✲ U
h2
g2
+
u−v+w u+v−w
xanticomcon = −
f2
g2
h2
+
+
v+w u+w u+v
Figure 19.1: The complementary and anticomplementary conjugates
Definition 19.4.1. comcon, anticomcon. For points P = f 2 : g 2 : rh2 and U = u : v : w,
neither lying on a sideline of ABC, the P -complementary and P -anticomplementary conjugates of
U are defined as in Figure 19.1, where k is the P-isoconjugacy. Most of the time, P = X1 and the
conjugacy reduces to isogonal conjugacy.
19.5
Hirstpoint aka Hirst inverse
Definition 19.5.1. Hirstpoint. Suppose P = p : q : r and U = u : v : w are distinct points,
neither lying on a sideline of ABC. The hirstpoint X is the point of intersection of the line P U
and the polar of U with respect to the circumconic CC (P ) conic :
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
222
19.6. Line conjugate
p yz + q zx + r xy = 0.
Proposition 19.5.2. We have the following properties :
(i) H is a type-keeping operation as a "ramified product" of type-crossing transforms and :
hirstpoint (P, U )
=
(P ∧ U ) ∧ polarmul (P, U )
= u2 q r − p2 v w : p v 2 r − u q 2 w : p q w2 − u v r2
(19.1)
(ii) H is commutative from the duality properties of polarization.
(iii) H (P, U ) = 0 : 0 : 0 occurs only when U = P
(iv) H (P, U ) = P if and only if U lies on the polar line of P
(v) H (P, H (P, U )) is either 0 : 0 : 0 or U . Indeterminate form is obtained (a) on the polar line
of P and (b) on a conic containing P and having P has perspector ... i.e. a conic whose only real
point is P .
Proof. Direct inspection for all properties. To be precise, these properties are valid only on the
real part of the world. For example, (i) gives U = p : jq : j 2 r where j 3 = 1, i.e. U = P and two
other "imagined" solutions.
All these properties show that "Hirst inverse" is a poorly chosen term, since we aren’t dividing,
but multiplying. Concerning the designation "Hirst inverse," see the contribution Gunter Weiss:
http://mathforum.org/kb/message.jspa?messageID=1178474.
19.6
Line conjugate
Suppose P = p : q : r and U = u : v : w are distinct points, neither equal to A, B, or C. The
P -line conjugate of U is the point whose trilinears are given by :
p(v 2 + w2 ) − u(qv + rw) : q(w2 + u2 ) − v(rw + pu) : r(u2 + v 2 ) − w(pu + qv)
This is the point of intersection of line P U and the tripolar of the isogonal conjugate of U .
Using the same formula with barycentrics, another point is obtained, that is the intersection of
P U and the dual of U . So what ?
19.7
Collings transform
Lemma 19.7.1. Let Mi , 1 ≤ i ≤ 5, be five (different) points, not four of them on the same line,
such that midpoint (M1 , M2 ) = midpoint (M3 , M4 ) = P . They determine uniquely a conic whose
center is P and contains ref lection(P, M5 ).
Proof. Take
P as origin of the
Euclidean coordinate system and consider determinant γ whose
lines are x2i , xi yi , yi2 , xi , yi , 1 , the last line (i = 6) referring to the generic point of the plane.
Since x2 = −x1 , · · · γ can be factored into (x1 y3 − x3 y1 ) times an expression without terms of
first degree in x6 , y6 .
Lemma 19.7.2. Let Mi , 1 ≤ i ≤ 4, be four different points, not on the same line. The locus :
Θ = {center (γ) | γ is a conic and M1 , M2 , M3 , M4 ∈ γ}
is a conic. It contains the six midpoint(Mi , Mj ) and its center is K =
P
Mi /4.
Proof. Θ is a conic since degree is two. When P = midpoint (M1 , M2 ), the preceding lemma can be
applied to M1 , M2 , M3 , ref lection (P, M3 ) , M4 , defining a conic whose center is P , so that P ∈ Θ.
Now, the lemma can be applied to Θ itself, since K = midpoint (midpoint (M1 , M2 ) , midpoint (M3 , M4 ))
–and cyclically.
March 30, 2017 14:09
published under the GNU Free Documentation License
19. Binary operations
223
A
B'
C'
P
B
C
A'
Q
Figure 19.2: Collings configuration
Proposition 19.7.3. Let P be a point not on a sideline of ABC, and A0 , B 0 , C 0 the reflections of
A, B, C in P .
(i) It exists a conic γ through the six points A, B, C, A0 , B 0 , C 0 , its center is P and its perspector
is U = P ∗ anticomplem (P ), so that γ = CC (U ). This conic intersects the circumcircle at point
b
Q = isotom (X6 ∧ U ), i.e. the tripole of line U X6 . Moreover, the circumcircles of triangles AB 0 C 0 ,
A0 BC 0 and A0 B 0 C are also passing through point Q.
(ii) Conversely, when Q is given on the circumcircle, the locus of P is conicev (X2 , Q). This conic
goes through the three AB ∩CQ points, the three midpoint (A, Q) and through four fixed points : the
vertices of medial triangle and through the circumcenter X3 . Moreover, this conic is a rectangular
hyperbola, its center is K = (A + B + C + Q) /4 and belongs to the nine points circle of the medial
triangle.
(iii) The anticomplement of this RH is the rectangular ABC-circumhyperbola whose center is the
complement of Q.
Proof. For (i), only Q belongs to circumcircle of AB 0 C 0 has to be proved. Barycentric computation.
For (ii), point AB ∩ CQ lead to the degenerate conic AB ∪ CQ (and cyclically) while the six
midpoints come from the lemma. When P = X3 , conic γ is the circumcircle... and passes through
A, B, C, Q and X3 ∈ Θ. But X3 of ABC is X4 of the medial triangle, and Θ contains an orthic
configuration, characteristic property of a rectangular hyperbola.
For the sake of exhaustivity, if barycentrics of P are p : q : r then barycentrics of Q are
Qx : Qy : Qy
where Qx =
1
r (p + q − r) b2 − q (p − q + r) c2
(and cyclically in a, b, c and p, q, r too). The transformation P 7→ Q was described by Collings
(1974) and was further discussed by (Grinberg, 2003b).
Example 19.7.4. Examples are as follows :
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
224
19.8. Brisse Transform
3
B'
C'
A
P
9
G
10 K
7
5
B
A'
8
4
1 C
6 2
Q
Figure 19.3: Collings locus is a ten points rectangular hyperbola
Q
X74
X98
X99
X100
X107
X110
X476
19.8
points on the conic
125
115, 868
2, 39, 114, 618, 619, 629, 630, 641, 642, 1125
1, 9, 10, 119, 142, 214, 442, 1145
4, 133, 800, 1249
5, 6, 113, 141, 206, 942, 960, 1147, 1209
30
wrt medial triangle
Kiepert hyperbola
Feuerbach hyperbola
Jerabek hyperbola
Brisse Transform
From http://en.wikipedia.org/wiki/Poncelet’s_porism :
In geometry, Poncelet’s porism, named after French engineer and mathematician JeanVictor Poncelet, states the following: Let C and D be two plane conics. If it is possible
to find, for a given n > 2, one n-sided polygon which is simultaneously inscribed in C
and circumscribed around D, then it is possible to find infinitely many of them.
Poncelet’s porism can be proved via elliptic curves; geometrically this depends on the
representation of an elliptic curve as the double cover of C with four ramification points.
(Note that C is isomorphic to the projective line.) The relevant ramification is over the
four points of C where the conics intersect. (There are four such points by Bezout’s
theorem). One can also describe the elliptic curve as a double cover of D; in this case,
the ramification is over the contact points of the four bitangents.
Applied to a point P = p : q : r on the circumcircle Γ of triangle ABC, this gives the existence of
points P2 , P3 on the circumcircle such that the incircle of ABC is also the incircle of P P2 P3 . As
given in Brisse (2001), barycentric equation of P2 P3 is :
(−a + b + c)p
(a − b + c)q
(a + b − c)r
x+
y+
z=0
a2
b2
c2
and the contact point of this line with the incircle is given by the (barycentric) Brice transform :
T (P ) =
March 30, 2017 14:09
a4
b4
c4
:
:
.
(b + c − a)p2 (c + a − b)q 2 (a + b − c)r2
published under the GNU Free Documentation License
19. Binary operations
225
The converse transform, applied to a point U = u : v : w on the incircle, is :
a2
b2
c2
T −1 (U ) = √ √
:√ √
:√ √
u −a + b + c
v a−b+c
w a+b−c
where all involved quantities are non-negative.
Examples:X(11) = Feuerbach point = T (X(109)), X(1317) = incircle-antipode of X(11) =
T (X(106)).
Exercise 19.8.1. Still open is the question posed in (Yiu, 2003) : list all polynomial centers on
the incircle having low degree and prove that there are no others. Here, "degree" of X = p(a, b, c) :
p(b, c, a) : p(c, a, b) [barycentrics] refers to the degree of homogeneity of p(a, b, c), and "low" means
less than 6. For example, the Feuerbach point, X(11), has degree 3.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
226
March 30, 2017 14:09
19.8. Brisse Transform
published under the GNU Free Documentation License
Chapter 20
Sondat theorems
20.1
Perspective and directly similar
Remark 20.1.1. When triangle ABC is translated into A0 B 0 C 0 , these triangles are in perspective
and the perspector is the direction of the translation. The converse situation is not clear, so that
translations will be excluded from what follows, implying the existence of a center. When using
composition, we have to examine if nevertheless translations are reappearing.
Lemma 20.1.2. When σ is a direct similitude (but not a translation) with center S = z : 0 : ζ
and ratio kκ (k is real while κ is unimodular, and kκ 6= ±1) then its matrix in the Morley space
is :


z
kκ
(1 − kκ) 0
t



1
0 
σ = 0


ζ
k
k 
0
1−
t
κ
κ
Proof. One can check that umbilics are fixed points of this transform. The characteristic polynomial of matrix σ is (µ − 1) (µ − k κ) (µ − k/κ). Excluding kκ = ±1 ensures the existence of a
center.
Proposition 20.1.3. When a direct central similitude σ (S, kκ) and a perspector P 6= S are given,
the locus C of points M such that P, M, M 0 = σ (M ) are collinear is the circle through S, P, σ −1 P .
Proof. The locus contains certainly the five points such that M = P or M 0 = P or M = M 0 i.e. S
and both umbilics. The general case results from the fact that det [P, M, M 0 ] is a second degree
polynomial in Z, Z, T so that C is a conic.
Proposition 20.1.4. Suppose that triangles T1 and T2 are together in perspective (center P ) and
strictly similar (center S, ratio kκ, κ 6= ±1). Then P and S are the two intersections of their
circumcircles (P = S cannot occur).
Proof. Use Lubin coordinates relative to T1 = ABC, and note S ' z : t : ζ. Then A0 B 0 C 0 is
obtained as :
A0 B 0 C 0 ' σ · ABC
The determinant of lines AA0 , BB 0 , CC 0 factors as :
σ4 k
k
z ζ − t2
1 − κ2 1 −
(1 − k κ) ×
σ3 κ
κ
t2
proving S ∈ Γ. Then perspector is also on this circle (from preceding proposition). We even have
the more precise result :


ω 0
0
1 − kκ


P = 0 1
0  · S where ω =
k
1−
0 0 ω −1
κ
227
228
20.1. Perspective and directly similar
Proposition 20.1.5. Consider a fixed triangle ABC, and describe the plane using the Lubin
frame. Consider points P = Φ : 1 : 1/Φ and S = Θ : 1 : 1/Θ on the unit circle (P as Phi, and S as
Sigma. But Sigma is sum, use the next Greek letter). Assume P 6= S. Then all triangles A0 B 0 C 0
that are P -perspective and S-similar to triangle ABC are obtained as follows. Let point O0 on the
perpendicular bisector of (P, S) be defined by property (SO, SO0 ) = κ where κ2 is a given turn.
Draw circle γ centered at O0 and going through P and S. Then A0 = γ ∩ SA, etc.
Proof. Point O0 can be written as P + S + x (ΘΦ : 0 : 1). This point is the σ (S, kκ) image of O if
and only if :


κ2 (Φ − Θ) 1 − κ2 Θ Φ
0
Φ−Θ


k=κ
, σ =
0
Φ − κ2 Θ
0

Φ − κ2 Θ
2
0
1−κ
Φ−Θ
As it should be, σ depends only of κ2 , while the sign of k depends on the choice
square roots of κ2 . The matrix of circle γ is :

0
1 − κ2
κ2 Θ − Φ
t

−1
−1
2
2
γ = σ
· Γ · σ
' 1−κ
2 κ Φ−Θ
1 − κ2 Θ Φ
κ2 Θ − Φ 1 − κ2 Θ Φ
0
And it can be checked that P, A, A0 = σ (A) are collinear.
of κ among the



Proposition 20.1.6. With same hypotheses, the perspectrix of triangles ABC and A0 B 0 C 0 is the
line :
XY Z = Φ − κ2 Θ Φ, −ΦΘ Φ − κ2 σ1 + κ2 κ2 σ3 − σ2 Φ , κ2 Φ − κ2 Θ σ3
Points S, C, C 0 , X, Y are concyclic (and circularly). When κ2 reaches Φ/α, then A0 moves to P
and XY Z becomes the sideline BC. When S is not a vertex A, B, C, the envelope of line XY Z is
the Steiner parabola of point S (focus at S, directrix the Steiner line of S). The tangential equation
of this parabola is given by matrix :


2 Θ σ3
σ3 σ2 − σ1 Θ


P∗ ' 
σ3
0
−Θ

σ2 − σ1 Θ −Θ
−2
Proof. Since XY Z is given by second degree polynomials, the envelope is a conic. It can be
obtained by diff and wedge, then eliminate. Parabola comes from the central 0. Focus is obtained
in the usual way, and one recognizes S. Directrix ∆ is the locus of the reflections of the focus in
the tangents. From the special cases, this is the Steiner line of S.
Last point, compute circle (S, X, Y ). Special cases Θ = α, etc, and Φ = κ2 Θ (O0 at infinity)
are appearing in factor. Otherwise, the equation is :
α Φ − κ2 Θ ZZ + κ2 α − Φ Z + α Θ Z T + Φ Θ − α2 κ2 T2
and this circle goes through A and A0 .
Proposition 20.1.7. When A, B, C, S, P are according the former hypotheses, let H, H 0 be the
respective orthocenters of ABC and A0 B 0 C 0 . Then midpoint of H, H 0 belongs to XY Z if and only
if κ2 = −1 (so that κ is a quarter turn) or :
κ2 = −Φ2
(Θ − σ1 )
,
σ2 Θ − σ3
i.e. κ = (BC, OP ) + (SA, SH)
In the second case, XY Z is the perpendicular bisector of (H, H 0 ).
Proof. We have H 0 = σ (H) and equation in κ2 is straightforward. Then we have :
ω 2 (BC) = −βγ, ω 2 (OP ) = Φ2 , ω 2 (SA) = −Θα, ω 2 (SH) = −σ3 Θ
March 30, 2017 14:09
Θ − σ1
σ2 Θ − σ3
published under the GNU Free Documentation License
20. Sondat theorems
229
Corollary 20.1.8. Start from triangle ABC, and assume that XY
Z = [ρ, σ, τ ] while
κ is a
quarter turn. Then X, Y, Z are X = 0 : τ : −σ, etc. Lines XδA = Sc σ + Sb τ, a2 σ, a2 τ , etc are
the perpendicular at X to BC, etc. Finally, point A0 is Y δB ∩ ZδC , etc. In other words :


ρ (ρSb Sc + σSc Sa + τ Sa Sb ) − 4 S 2 στ


A0 ' 
b2 ρ τ c2 − σSa − ρSb
 , etc
c2 ρ σb2 − τ Sa − ρSc
The perspector and the similicenter are :
P = isogon M · t ∆ ; S = isogon ((σ − τ ) ρ : σ (τ − ρ) : τ (ρ − σ))
while the ratio of the similitude is :
a2 + b2 + c2 ρστ − Sa ρ σ 2 + τ 2 − Sb σ ρ2 + τ 2 − Sc τ ρ2 + σ 2
k=
2 S (σ − τ ) (ρ − τ ) (ρ − σ)
Proof. Straightforward computation.
20.2
Perspective and inversely similar
Lemma 20.2.1. A circumscribed rectangular hyperbola goes through A, B, C, H, Gu where H is the
orthocenter and Gu is the gudulic point, the intersection of the RH and the circumcircle. Directions
of axes are given by the bisectors of, for example, AGu and BC. Then directions of asymptotes
are obtained by a 45◦ rotation (or taking again the bisectors).
Lemma 20.2.2. When a rectangular hyperbola H is known by its implicit equation
1
2
κ2 Z − 2 Z2 + W Z + V Z T + Q T2
κ
then points M ∈ H can be parametrized as :






+V κ2
+i κ
κ





1
+X 0 +Y  0 
1
M= 





2
1
1 
1
−W 2
−i
κ
κ
κ
where X, Y are real quantities linked by :
−i
W2
YX =
κ2 V 2 + 4 Q − 2
16
κ
(20.1)
Proof. This way of writing may look weird but, most of the time, hyperbola equations are appearing
that way.
Lemma 20.2.3. Consider four points Mj on a rectangular hyperbola, parametrized by (20.1).
These points form an orthocentric quadrangle if and only if :
2
W2
2 2
256 x1 x2 x3 x4 = κ V + 4 Q − 2
κ
t
Proof. We write that (M1 ∧ M2 )· Mz · (M3 ∧ M4 ) = 0, and obtain this condition. The conclusion
follows from the symmetry of the the result.
Lemma 20.2.4. When ψ is a central inverse similitude (reflections in a line are allowed, but not
the other isometries), then its matrix in the Morley space can be written as :


z
ζ
0
− k κ2
kκ2


t
t


ψ = 0
1
0 
 k

ζ
k z
−
0
κ2
t
κ2 t
Point S = z : 0 : ζ is the center, axes are directed by ±κ2 and ratio is k (κ is unimodular while k is
real and k = ±1 is a dubious case). One can check that umbilics are exchanged by this transform.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
230
20.2. Perspective and inversely similar
Proof. Let z2 : t2 : ζ2 be the image of the origin 0 : 1 : 0. The characteristic polynomial of matrix :

0

 0
 k
κ2
z2 /t2
1
ζ2 /t2

k κ2

0 

0
is (µ − 1) (µ − k) (µ + k). Excluding k = ±1 ensures the existence of a center. Consider the
reflection δ about line through S and κ2 : 0 : 1. We have :
δ = subs k = ±1, ψ
;

k

ψ · δ = δ · ψ =
 0
0
(1 − k)
z
t
(1 − k)
ζ
t
0


0 

k
Remark 20.2.5. The unimodular κ was an intrinsic quantity when we were dealing with direct similarities. Now, κ2 measure the angle between the real axis and one of the axes of the antisimilarity.
Proposition 20.2.6. When a central antisimilarity ψ (S, kκ) and a perspector P 6= S are given,
the locus H of points M such that P, M, M 0 = ψ (M ) are collinear is the conic through S, P, ψ −1 P
and directions of the ψ-axes (and this conic is a rectangular hyperbola).
Proof. The locus contains certainly the five points such that M = P or M 0 = P or M = M 0 i.e.
S and both directions ±κ2 : 0 : 1. The general case results from the fact that det [P, M, M 0 ] is a
second degree polynomial in Z, Z, T so that C is a conic.
Proposition 20.2.7. Suppose that triangles T1 and T2 are together in perspective (center P ) and
strictly antisimilar (center S, ratio k 6= ±1, direction of axes ±κ2 : 0 : 1). Let X be one of the
points S, P , ±κ2 : 0 : 1. Then the other three are obtained as the remaining intersections between
conic H through A, B, C, H, X and conic H0 through A0 , B 0 , C 0 , H 0 , X.
Proof. Use Lubin coordinates relative to T1 = ABC, and note S ' z : t : ζ. Then A0 B 0 C 0 is
obtained as :
A0 B 0 C 0 ' ψ · ABC
The determinant of lines AA0 , BB 0 , CC 0 factors as :
σ4
k k 2 − 1 × conic
t2 σ3
proving that S belongs to the rectangular hyperbola whose implicit and parametric equations are :
−1 2
2
Z + κ2 Z +
κ2
σ1
κ2
−
κ2
σ3
TZ +
σ2 κ 2
σ3
− 2
σ3
κ
TZ +
σ1 κ2
σ2
− 2
σ3
κ
T2
2
2
2 !
κ4 − σ1 σ3
1
κ
σ1
1 σ3
σ2 κ2
2
+
κ κ
+ 2
− 2
+
 xκ −
2 σ3
16 x
σ3
κ
κ
κ2
σ3



M (x) ' 
1
2
2
2 !

4
2
1
σ
κ
−
σ
1
1
κ
σ1
1 σ3
σ2 κ2

2
3
2
x+
−
κ
+ 2
− 2
+
κ
2 κ4 σ3
16 x κ
σ3
κ
κ
κ2
σ3









Then perspector is also on this hyperbola (from preceding proposition). We even have the more
precise result :
xS − xP
k=
xS + xP
March 30, 2017 14:09
published under the GNU Free Documentation License
20. Sondat theorems
231
Proposition 20.2.8. Consider a triangle A, B, C, its orthocenter H and two points S, P such
that the six points A, B, C, H, P, S are on the same conic H. Then it exists exactly one triangle
A0 B 0 C 0 that is together S-antisimilar and P perspective with ABC. Moreover A0 , B 0 , C 0 , H 0 , P, S
are on the same conic H0 , both conics are rectangular hyperbolas and share the same asymptotic
directions. Finally, the gudulic point Gu of conic H sees triangle ABC at right angles with trigon
A0 B 0 C 0 , and the fourth intersection of conic H0 with circle Γ0 sees triangle A0 B 0 C 0 at right angles
with trigon ABC.
Proof. A conic through A, B, C, D is a rectangular hyperbola. Consider one of its asymptotes
and draw a parallel ∆ to this line through point S. Let A00 , B 00 , C 00 , H 00 be the reflections of
A, B, C, H into ∆. Then we have A’=SA”∩PA, etc. Final result comes from
Proposition 20.2.9. When A, B, C, H, P are fixed, the direction of the perspectrix XY Z is also
fixed.
Proposition 20.2.10. When A, B, C, H, S are fixed and P moves onto the A, B, C, H, S hyperbola,
the envelope of the perspectrix XY Z is the parabola inscribed in triangle ABC whose directrix is
line HS.
20.3
Orthology, general frame
Definition 20.3.1. We say that point P sees triangle A0 B 0 C 0 at right angles to trigon ABC when
P is different from A0 , B 0 , C 0 and verifies P A0 ⊥ BC etc.
Remark 20.3.2. Should point P be at infinity, all sidelines of ABC would have the same direction,
and triangle ABC would be degenerated (flat). Some problems have to be expected...
.
Lemma 20.3.3. The orthodir of the BC sideline is :δA = a2 : −Sc : −Sb . This is also the
direction of line HA, where H is the orthocenter of the triangle.
Proposition 20.3.4. Consider two non degenerate finite triangles ABC, A0 B 0 C 0 and suppose that
it exists a finite point P , different from A0 , B 0 , C 0 , that sees triangle A0 B 0 C 0 at right angles to trigon
ABC. Then it exist a point U that sees triangle ABC at right angles to trigon A0 B 0 C 0 .
Proof. From hypothesis, A0 belongs to line P δA . Then it exists a real number kA 6= ∞ such that
A0 = kA P + δA . And the same for B 0 , C 0 . Now compute :


−kB − kC
.


A00 = M . (B 0 ∧ C 0 ) = −2S (p + q + r) 
kB

kC
t
t
This comes from P ∧ P = 0, together with M . (δB ∧ δC ) = M . t L∞ = 0. Since p + q + r 6= 0
and kj 6= 0 is assumed, column A00 really defines a direction. We can therefore simplify and obtain :


−kB − kC
kA
kA


A00 B 00 C 00 = 
kB
−kA − kC
kB

kC
kC
−kA − kB
This triangle is perspective to ABC, with perspector U = kA : kB : kC . Therefore point U sees
triangle ABC at right angles to trigon A0 B 0 C 0 .
Definition 20.3.5. We say that two triangles are orthologic to each other when it exists a point P
that sees triangle A0 B 0 C 0 at right angles to trigon ABC and a point U that sees triangle ABC at
right angles to trigon A0 B 0 C 0 . We also say that P, U are the orthology centers of the triangles (P
looking at A0 B 0 C 0 , and U looking at ABC). In this definition, flat triangles and centers at infinity
are allowed.
Remark 20.3.6. This property cannot be reworded in a shorter form (the so-called symmetry),
since P ∈
/ L∞ is required to be sure of the existence of U , but this is not sufficient to be sure of
U∈
/ L∞ .
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
232
20.4. Simply orthologic and perspective triangles
Proposition 20.3.7. Assume that triangle ABC is not degenerate and let points finite points P, U
by described their barycentrics P = p : q : r and U = u : v : w (here P = U is allowed). Moreover,
assume that U is not on the sidelines of ABC. Then all triangles A0 B 0 C 0 such that point P sees
triangle A0 B 0 C 0 at right angles to trigon ABC and point U sees triangle ABC at right angles to
trigon A0 B 0 C 0 are given by formula :


pu + a2 ϑ pv − Sc ϑ pw − Sb ϑ


(20.2)
A0 B 0 C 0 '  qu − Sc ϑ qv + b2 ϑ qw − Sa ϑ  = P · t U + 2Sϑ M
2
ru − Sb ϑ rv − Sa ϑ rw + c ϑ
where ϑdescribes an homothety centered at P .
Proof. Since u 6= 0, relation A0 ∈ P δA can be written as uP + ϑA δA , and the same holds for the
other points. Now, compute :
h
i
t
det uP + ϑA δA , vP + ϑB δB , M (C ∧ U ) = 2uvS (p + q + r) (ϑA − ϑB )
Due to 0the hypotheses, all the ϑ must have the same value, leading to the formula. Converse is
obvious, when a triangle is as described by the formulas, both orthologies are verified.
Remark 20.3.8. If P were at infinity, P A0 , P B 0 , P C 0 would have the same direction, and also the
sidelines of ABC. In the formula, this would lead to A0 , B 0 , C 0 at infinity. If coordinate u was 0
then U B k U C so that A0 B 0 k A0 C 0 . In the formula, this would lead to A0 at infinity.
20.4
Simply orthologic and perspective triangles
Definition 20.4.1. When triangles are orthologic with P 6= U , we say they are simply orthologic.
When P = U, we say they are bilogic.
Notation 20.4.2. In this chapter, when triangles ABC and A0 B 0 C 0 are in perspective, their perspector will be noted Ω (and never P , nor U ), while the perspectrix will be noted XY Z with
X = BC ∩ B 0 C 0 , etc.
Theorem 20.4.3 (First Sondat Theorem). Assume that triangle ABC is finite and non degenerate
; P is at finite distance ; U is different from P and is not on the sidelines. Then it exists exactly
one triangle A0 B 0 C 0 such that (1) P sees A0 B 0 C 0 at right angles to trigon ABC (2) U sees ABC at
right angle to trigon A0 B 0 C 0 (3) A0 B 0 C 0 is perspective to ABC. When A0 B 0 C 0 is chosen that way,
the corresponding perspector Ω is collinear with points P, U and the perspectrix XY Z is orthogonal
to P U .
Proof. We have assumed that P = p : q : w is different from U = u : v : w. Then ϑ is fixed by the
condition of being perspective and we have :
u (vSb − wSc ) qr + v (wSc − uSa ) pr + w (uSa − vSb ) pq
(vSb − wSc ) pSa + (wSc − uSa ) qSb + (uSa − vSb ) rSc
Sc w − Sb v Sc w − Sa u Sb v − Sa u
:
:
qw − rv
pw − ru
pv − qu


(Sc w − Sb v) p + Sb u + a2 w q − Sc u + a2 v r


Sb q − Sc r


 (Sa u − Sc w) q + b2 u + Sc v r − Sa v + b2 w p 




Sc r − Sa p


 (Sb v − Sa u) r + c2 v + Sa w p − Sb w + c2 u q 
Sa p − Sb q
ϑ
= −
Ω
=
tripolar (∆)
=
First assertion is proved by Ω ' α U − β P where :
! "
#
α
(qw − rv) puSa + (ru − pw) vqSb + rw (pv − qu) rwSc
'
β
(qw − rv) u2 Sa + (ru − pw) v 2 Sb + (pv − qu) w2 Sc
Second assertion is proved by checking that
∆ · OrtO · (normalized (P ) − normalized (U )) = 0
March 30, 2017 14:09
published under the GNU Free Documentation License
20. Sondat theorems
233
Figure 20.1: Orthology and perspective.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
234
20.5
20.5. Bilogic triangles
Bilogic triangles
Proposition 20.5.1. Bilogic triangles are ever in perspective. When orthology center is U = u :
v : w is finite and not on the sidelines, formula (20.2) becomes


u2 + a2 ϑ uv − Sc ϑ uw − Sb ϑ


A0 B 0 C 0 '  uv − Sc ϑ v 2 + b2 ϑ vw − Sa ϑ  = U · t U + 2Sϑ M
(20.3)
2
2
uw − Sb ϑ vw − Sa ϑ r + c ϑ
while the perspector and the tripole of the perspectrix are respectively :
−1
−1
−1
Ω ' (ϑ vw − Sa ) : (ϑ wu − Sb ) : (ϑ uv − Sc )


vwa2 + u (vSb + wSc − uSa ) ϑ − 4 S 2


tripolar (XY Z) '  uwb2 + v (wSc + uSa − vSb ) ϑ − 4 S 2 
uvc2 + w (uSa + vSb − wSc ) ϑ − 4 S 2
Proof. Straightforward computation.
Proposition 20.5.2. When triangle ABC is fixed and the bilogic center U is given, the locus
of the perspector Ω of the bilogic triangles A0 B 0 C 0 is the rectangular hyperbola that goes through
A, B, C, U, H =X(4). The perspector of this circumconic is :
u (vSb − wSc ) : v (wSc − uSa ) : w (uSa − vSb )
while the envelope of the perspectrix is the inconic whose perspector is :
−1
(vSb − wSc )
−1
: (wSc − uSa )
, (uSa − vSb )
−1
Proof. Eliminate ϑ from Ω. Then eliminate ϑ from XY Z and take the adjoint matrix.
Proposition 20.5.3. Let ABC, A0 B 0 C 0 be two bilogic triangles, with orthology center U , perspector Ω and perspectrix (XY Z) where X = BC ∩B 0 C 0 , etc. Then lines U Ω and XY Z are orthogonal.
Moreover we have (U X) ⊥ (AA0 Ω), etc.
Proof. We compute the orthodir δ of line XY Z and find that :


u2 (w + v) Sa − Sc uw2 − Sb v 2 u θ + Sb Sc (v + w) − a2 Sa u


δ =  v 2 (w + u) Sb − Sa vu2 − Sc vw2 θ + Sc Sa (w + u) − b2 Sb v 
w2 (v + u) Sc − Sa wu2 − Sb wv 2 θ + Sa Sb (u + v) − c2 Sc w
It remains to check that normalized (Ω) − normalized (U ) is proportional to δ.
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 21
Linear Families of Inscribed Triangles
This chapter is an extended version of Douillet (2014c).
Important caveat: in all parts of this chapter, the relation f + g + h − u − v − w = 0 will be ever
assumed. This rule is related to the asymmetric parameterization described at .
21.1
21.1.1
Inscribed triangle
Induced hexagon
Figure 21.1: Fixed triangle abc is inscribed into the fixed trigon ABC
Definition 21.1.1. We say that triangle abc is inscribed in
Using barycentrics wrt ABC, we can write:

0
1−q
r

abc =  p
0
1−r
1−p
q
0
triangle ABC when a ∈ BC, etc.



Definition 21.1.2. As described in Figure 21.1, the induced hexagon is aγbαcβ where α =
b + c − A, etc, so that:


r−q
r
1−q


αβγ =  1 − r p − r
p 
q
1−p q−p
235
236
21.2. Linear families
Remark 21.1.3. Both triangles abc and αβγ have the same area, equal to s S where
.
s = pqr + (1 − p)(1 − q)(1 − r) = det abc = det αβγ
Proposition 21.1.4. Let ψ be the collineation ψ : a 7→ α,b 7→ β, c 7→ γ, L∞ 7→ L∞ . Using
triangle abc as reference, the normalized matrix of ψ is:


(1 − r) q
(p − 1) (q + r − 1)
(q + r − 1) p
1 

ψa =
(p + r − 1) q
(1 − p) r
(q − 1) (p + r − 1) 

s
(r − 1) (p + q − 1)
(p + q − 1) r
(1 − q) p
1+W
1−W
1
χψ (X) = (µ − 1) µ −
µ−
where s =
1 − W2
1−W
1+W
4
and its fixed points are:

 



2q + 2r − 2
2q + 2r − 2
(2 p − 1) (q + r − 1)
−1 

 


K1 = 2  (2 q − 1) (r + p − 1)  ; K2 , K3 '  −2 r + 1 − W  ,  −2 r + 1 + W 
W
−2 q + 1 − W
−2 q + 1 + W
(2 r − 1) (p + q − 1)
(21.1)
Proof. We are not using ABCL∞ as a basis, and the matrix is not αβγ · abc
−1
. On the contrary,
−1
we have ψa = abc
· αβγ . Everything else is straightforward. A more symmetrical form can
be found for the two points at infinity, but largely less handy.
Proposition 21.1.5. Hexagonal conic. A conic goes through the six points aγbαcβ. Using
barycentrics relative to abc, this conic can be written as:


0 p+q−1 p+r−1


C = p+q−1
0 q+r−1 
p+r−1 q+r−1
0
t
One has ψ · C · ψ ' C , and the conic is globally invariant by ψ. Moreover K1 is its center,
while K2 , K3 are its points at infinity.
Proof. The opposite sidelines of polygon aγbαcβ are parallel by pairs, and the points are co-conic.
Seen from abc, this conic is a circumconic, and its perspector is the intersection of the tripolar of
the other three points. Invariance is obvious. Since L∞ is also invariant, the center is invariant,
and therefore is K1 . Computational verifications of all the properties are easy.
Corollary 21.1.6. The hexagonal conic is a parabola when area (a, b, c) = area (A, B, C) ÷ 4, an
ellipse when area (abc) is greater and an hyperbola otherwise. The conic degenerates into a pair of
lines when q + r = 1 or circularly (then bγ and βc are not only parallel, but equal).
21.2
Linear families
21.2.1
Slowness center
Definition 21.2.1. LFIT. When the barycentrics of the vertices a, b, c are non-constant linear
functions of a parameter t this situation is described as a Linear Family of Inscribed Triangles.
Using the values relative to t = 0 and t = 1, this can be written as:
.
abc = t a1 b1 c1 + (1 − t) a0 b0 c0

0

= t  p1
1 − p1
March 30, 2017 14:09
1 − q1
0
q1


r1
0


1 − r1  + (1 − t)  p0
0
1 − p0
1 − q0
0
q0

r0

1 − r0 
0
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
237
Figure 21.2: Variable triangle abc is inscribed into fixed triangle ABC
.
−
−
Definition 21.2.2. As in Figure 21.2, the velocities →
va , etc are defined by →
va = (a1 − a0 ) =
→
− .
[0, p1 − p0 , p0 − p1 ]. The so called "speed vector"is defined by V = [p1 − p0 , q1 − q0. , r1 − r0 ].
Remark 21.2.3. The gravity center g (t) of the moving triangle abc is given by g (t) = (1 − t) g0 +t g1
and therefore its locus is a straight line.
Definition 21.2.4. Graphs Gj and hexagonal construction. The locus of α (t) is nothing
but the graph of the correspondence b (t) ←→ c (t). By definition, this graph Ga is a straight
line. And so are also the other two graphs. Using these graphs, a triangle abc can be constructed
stepwise from one of its vertices according to the closed path aγbαcβa, i.e. by constructing the
associated hexagon (see Figure 21.3).
Figure 21.3: Using the inscribed hexagon to construct the variable triangle
Remark 21.2.5. The point at infinity of the graph b 7→ c is Ga ∧ L∞ ' q0 − r0 − q1 + r1 : r0 − r1 :
q1 − q0 . This point can be computed from the speed vector, with the following strange property:
everything becomes simpler when introducing the reciprocal of these quantities (remember that
t 7→ a, etc are non constant). This leads to the following:
Definition 21.2.6. The slowness center of a linear family of triangles is:
.
S=f :g:h=
1
1
1
:
:
p1 − p0 q1 − q0 r1 − r0
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
238
21.2. Linear families
leading to the following description (where indices 0 have been omitted) :




abc (t) = 


0
t
f
t
1−p−
f
p+
1−q−
0
t
q+
g
t
g

t
h 
t 

1−r−

h 

0
r+
(21.2)
Remark 21.2.7. Considering S as a projective quantity is only recognizing that choosing a time
unit or another is unessential.
Proposition 21.2.8. Graphs Ga , etc are parallels to the sidelines of the anti-cevian triangle of S.
Proof. When t → ∞, α (t) → h − g : g : −h, giving the value of Ga ∧ L∞ .
Theorem 21.2.9 (Main result). The slowness center is the area center of the family, i.e. triangles (S, a (t1 ) , a (t2 )) , (S, b (t1 ) , b (t2 )) and (S, c (t1 ) , c (t2 )) are equivalent.
Proof. Suppose that S is at finite distance, compute the area, obtain:
area (S, a (t1 ) , a (t2 )) = (t2 − t1 ) S ÷ (f + g + h)
and conclude from symmetry. Suppose now that S is at infinity (i.e. h = −f − g), compute the
.
.
width of the strip created by the parallel lines ∆1 = S ∧ a (t1 ) and ∆2 = S ∧ a (t2 ), obtain:
∆2 − ∆1 = [t1 − t2 : t1 − t2 : t1 − t2 ]
and here again conclude from symmetry.
21.2.2
Equicenter
Theorem 21.2.10 (Neuberg). If we apply to abc the barycentrics of S wrt ABC, we obtain a
fixed point, the so called equicenter E.
.
Proof. From equation (21.2), it is obvious that E = abc · S doesn’t depend on t. And we have:
E ' u : v : w ' (1 − q) g + r h : (1 − r) h + p f : (1 − p) f + q g
Remark 21.2.11. When one of the centers S, E is at infinity, the other is also at infinity.
21.2.3
Parametrization of a LFIT
Parametrization (21.2) has many interesting properties, among them its genericity and its symmetry. Nevertheless, there are many situations where this parametrization induces too burdensome
computations. There are several other choices, each of them depending on some assumption about
the areas of the variable triangles.
Proposition 21.2.12. The area of triangle at bt ct equals the area of ABC times the quantity:
A
=
=
(f + g + h) 2 f g (p + q − 1) + gh (q + r − 1) + hf (r + p − 1)
t +
t + pqr + (1 − p) (1 − q) (1 − r)
f gh
f gh
(f + g + h) t2
(f u + 2 f v + gv − hw) t v (f + g − w)
−
+
f gh
f gh
gh
Proof. First formula is the characteristic property of the determinant. The second one doesn’t
apply when S = E ∈ L∞ : in this case, area is constant, but not necessarily −2. More details
below.
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
239
Theorem 21.2.13. When S, E are at finite distance, the area is a second degree quantity. We can
choose the origin of time at the extremal value of the area, leading to:
abc =

0


T
f u + 2 f v + gv − hw

 + +

f
2 (u + v + w) f
 T
f u + 2 f w − gv + hw
− +
f
2 (u + v + w) f
−
T
gv + 2 gu − hw + f u
+
g
2 (u + v + w) g
0
+
T
gv + 2 gw + hw − f u
+
g
2 (u + v + w) g
T
hw + 2 hu + f u − gv
+
h
2 (u + v + w) h
T
hw + 2 hv − f u + gv
− +
h
2 (u + v + w) h
+
0
And then we have:







area (t)
(f + g + h) 2
f + h + h f 2 u2 − 2 f guv − 2 f huw + g 2 v 2 − 2 ghvw + h2 w2
=
t −
2
S
f gh
4 f gh
(u + v + w)
Proof. Only the apparent denominators were used to obtain this formula. The A formula becomes
indeterminate if one tries to use it for S ∈ L∞ . In any case, a quite useless parametrization (too
huge coefficients).
Theorem 21.2.14. When S, E are at infinity but aren’t equal, the area is a first degree quantity.
We can choose the origin of time at A = 0, leading to:


T
u (h + w)
T
u (g + v)
0
− +
−

g
gw − vh
h
gw − vh 
 T
(h + w) v
T
v (f + u) 


abc = 
−
0
− +

 f
gw − vh
h
gw − vh 
 T

(g + v) w
T
w (f + u)
− +
−
0
f
gw − vh
g
gw − vh
Here, we have gw − hv = hu − f w = f v − gu and the area is T (hv − gw) ÷ f gh (caveat: replacing
E 7→ kE changes the family and multiplies the area by k !)
Proof. S ∧ E ' L∞ , hence the equality. No hidden denominators are implied.
Remark 21.2.15. As a matter of experiment, the following parametrization is more computationfriendly as any other one, largely enough to justify a lack of symmetry, and the burden of enforcing
binding rule.
Definition 21.2.16. The asymmetric parametrization of a LFIT is defined as:


t
f −w+g t
h−v
0
−
+
+

g
g
h
h 

t
v 
t


0
− +
Tt = 

h h 
 f


t
t
w−f
1−
+
0
f
g
g
where f : g : h is the slowness center S and u : v : w s the equicenter E. This amounts to enforce
a0 = C together with the rule
f +g+h−u−v−w =0
In other words, when S ∈
/ L∞ , we consider S : E or f : g : h : u : v : w as a unique projective object
subject to a binding relation. And when S 6= E ; S, E ∈ L∞ , we have f + g + h = u + v + w = 0
and so that S, E are independent quantities so that replacing E 7→ kE changes the family (and
multiplies the area by k !).
Remark 21.2.17. CAVEAT. The remaining case, i.e. S = E ∈ L∞ , is really special, worth of a
specific section
21.3
21.3.1
Other hexagons, other graphs
Hexagonal graphs
Local version of Chex is conicir(qt + rt − 1 : rt + pt − 1 : pt + qt − 1). At area’s peak, the local point
with coordinates f : g : h belongs to this conic and therefore the "true" Chex goes through E.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
240
21.3.2
21.3. Other hexagons, other graphs
Temporal and Catalan graphs
In the first section, we have constructed the hexagonal graphs Ga , Gb , Gb by drawing lines bα
through b, parallel to AB together with drawing lines cα through c, parallel to AC. As a result,
triangle αβγ (t) is the crosstri (see 4.6) of triangles abc (t) and triangle dirA , dirB , dirC . This
method can be repeated, using other auxiliary directions than those of the sidelinesBC, CA, AB
and obtaining other graphs than the Gj .
Definition 21.3.1. The pilar point Ω of a LFIT is defined as


g+h−u


Ω = (f + g + h) (A + B + C) − S − E '  h + f − v 
f +g−w
while the pilar conic C is the inscribed conic whose perspector is the isotomic conjugate of Ω, and
.
the center is ω = S + E = f + u : g + v : h + w (remember: by definition, S and E are synchronized
in order to enforce f + g + h − u − v − w = 0 !).
Remark 21.3.2. The name can be tracked back to a joke describing a temporal conic Cs (t) as
a major general (conique divisionnaire), distinguished among all the hexagonal conics (conique
brigadière). In this context, the pilar conic would be something like a lieutenant general in command.
Definition 21.3.3. The graphs ∆sA , etc obtained when using the directions of trigon as bs cs as auxiliary directions, are called the temporal graphs of the LFIT. And the conics through at βts ct αts bt γts
are called the temporal conics Cs (t). This process amounts to observe the LFIT from the fixed
triangle that coincides with abc at time t = s.
Proposition 21.3.4. The temporal graph ∆sA is the tangent to the pilar conic C issued from as
(other than the obvious BC).
Proof. Property as ∈ ∆sA is obvious. For the contact: points α, β, γ are obtained as crosstri of
abc (t) , given at (21.2), and
 s

s
s
s
+ + qs + rs − 1
− − rs
− + 1 − qs
 g h

h
g


s
s
s
s

−
+
1
−
r
+
+
p
+
r
−
1
−
−
p
Ds ' 
s
s
s
s


h
h f
f


s
s
s
s
− − qs
− + 1 − ps
+ + ps + qs − 1
g
f
f
g
which gives the auxiliary directions. Thus, the ABC-equation of the A-temporal graph is
g (1 − ps ) (1 − qs ) − h ps rs h ps rs − g (1 − ps ) (1 − qs )
s
∆A ' g qs + h (1 − rs ) ;
;
ps
1 − ps
Use locusconi, and obtain a line-conic circumscribed to trigon ABC, and see that its perspector is
t
Ω.
Theorem 21.3.5. The temporal conic Cs (t) related to at bt ct doesn’t depends on the s chosen to
construct the graphs. The centers of the C (t) belong to line
[(g + h − u) (gh − vw) , (h + f − v) (hf − wu) , (f + g − w) (f g − uv)]
that goes through ω = S + E. Moreover, when observed from the local point of view, i.e. reported
to the mobile triangle at bt ct , the conic C (t) appears as a fixed circumconic, whose perspector has
the same barycentrics wrt at bt ct as point S ∗ Ω wrt ABC.
b
−1
Proof. Compute at bt ct
· αts βts γts and then, according to Proposition 11.4.8, take the wedge of
the tripolars of two columns. The result doesn’t contain s. Moreover, being equal to S ∗ Ω, this
b
result doesn’t depends on t either. Now use the formula giving center from perspector, go back to
ABC- barycentrics and conclude.
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
241
Theorem 21.3.6. The temporal conic C (t) is bi-tangent to the pilar conic C and the line through
the contacts (that are not necessarily visible) is parallel to line SE.
2
Proof. A point on C is P (s) ' 1 ÷ (g + h − u) : s2 ÷ (h + f − v) : (1 + s) ÷ (f + g − w). Thus,
using SS = s1 + s2 , P P = s1 s2 , an equation of the chord P (s1 ) P (s2 ) is
chord ' [(g + h − u) (2 P P + SS) , (h + f − v) (SS + 2) , − (f + g − w) SS]
t
−1
−1
· conicir S ∗ Ω · at bt ct
· P , and substitute u = f + g + h − v − w.
Then compute t P · at bt ct
b
The result factorizes into det Tt times a perfect square. Use the usual formulas for sum and
product, and conclude by obtaining:
chord(t) = [f + v − w, f − h + v, f + g − w] − 2t [1, 1, 1]
Proposition 21.3.7. When S ∈
/ L∞ , the center of C (t) is ω (and the chord of contacts is SE)
exactly when the area of at bt ct is extremal.
f u + 2 f v + gv − hw
Proof. Substitute t =
in the previous results.
2 (f + g + h)
Proposition 21.3.8. Consider the temporal graph ∆sA . When point αts moves along ∆sA , then
line αts at keeps a constant direction (depending on s, but not on t).
Proof. For any kind of graphs, directions of αts bt and αts ct are constant by the very definition. The
αts at property is special even if easily checked.
Exercise 21.3.9. Determine the graphs that share this property with the temporal graphs.
Definition 21.3.10. When using the directions of SA, SB, SC as auxiliary directions, the results
are called Catalan conics and Catalan graphs (when doing that, we assume that point S is at finite
distance). Directions D and point α are:




f +u


−g − h
f
f
 g − u − v + hw − gv + t f + g + h 




D '
g
−h − f
g
 ; α'
f
f


gv − hw
f +g+h 
h
h
−f − g
h+v+
−t
f
f
Proposition 21.3.11. The Catalan graphs ∆a , etc are the symmetrics of the sidelines wrt ω, the
midpoint of S, E. In fact, the Catalan graphs are nothing else than the temporal graphs related to
s = ∞. When area of abc is extremal, the Catalan conic is centered at ω.
Proof. Make s → ∞ in Ds .
21.3.3
Graphs, the general case
Definition 21.3.12. A "delta -triple" is what is obtained by choosing three directions


u
−1 − v
1
. 

δ=
1
v
−1 − w 
−1 − u
1
w
and using them to draw the graph of bt 7→ ct , etc, i.e. the locus ∆a of the αt = bt δc ∩ ct δb , etc.
Quite obviously, these graphs are straight lines.
Proposition 21.3.13. The directions of these graphs (coded the same way as the original ones)
are:
h (1 + v) (1 + w) − g f (1 + w) (1 + u) − h g (1 + u) (1 + v) − f
0 0
0
[u , v , w ] =
,
,
(g − hv) (1 + w)
(h − f w) (1 + u)
(f − gu) (1 + v)
Only the directions of cevian (S) lead to a 1-sized orbit. The 2-sized orbits are characterized by
(uvw + uw + u) gh + (uvw + uv + v) f h + (uvw + vw + w) f g = 0
For a given triple [u0 , v 0 , w0 ] there are at most two triples [u, v, w].
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
242
21.4. Temporal point of view
Proof. Computations are straightforward. Nevertheless, the general formula giving the other antecedent is rather huge.
Example. Starting from [0, 0, 0] (hexagonal conics), we obtain the directions of anticev (S), and
conversely. The other antecedent of [0, 0, 0] is given by the directions of SA, SB, SC (Catalan
conics).
Exercise 21.3.14. Study the cevian and anticevian conics.
21.4
Temporal point of view
Definition 21.4.1. The temporal embedding is the projective map defined by:




x (t)


 

7 
G  y (t)  , t →

z (t)
x
y
z
t (x + y + z)





Proposition 21.4.2. Given a LFIT, the temporal embeddings of the variable vertices belong to a
quadric Q, whose matrix is Q ':





2 hg q (r−1)
h (gqr + f (1−p) (1−r)) g (f pq + h (1−q) (1−r)) gq + h (r−1)
h (gqr + f (1−p) (1−r))
2 hf r (p−1)
f (hpr + g (1−p) (1−q)) hr + f (p−1)
g (f pq + h (1−q) (1−r)) f (hpr + g (1−p) (1−q))
2 f g p (q−1)
f p + g (q−1)
gq + h (r−1)
hr + f (p−1)
f p + g (q−1)
2


2 v (f − w)
−f h + 2 f v + hw − vw f v + gv − vw −f − v + w
 −f h + 2 f v + hw − vw
2 f (v − h)
f 2 + f g − f w −f + h − v 


'


f v + gv − vw
f2 + fg − fw
0
−f − g + w 
−f − v + w
−f + h − v
−f − g + w
2
2
2
2
And we have det Q = (g + h − u) (h + f − v) (f + g − w) (squared coordinates of Ω in the
triangle plane).
Proof. Well known property of hyperbolic paraboloids.
Remark 21.4.3. This matrix is not symmetric. One could enforce symmetry by choosing t = 0
when area is extremal. But the price to pay would be a huge size for the coefficients.
Remark 21.4.4. The directions of sidelines of ABC are embedded at
0 : 1 : −1 : f ; −1 : 0 : 1 : g ; 1 : −1 : 0 : h
Proposition 21.4.5. The temporal embeddings of the s-temporal graphs, i.e. the lines
. G (∆sA ) = G (αts , t) | t ∈ R , etc
are drawn on the Q quartic. Moreover, G (αts , t)t=s = G (as , s). Therefore G (∆sA ) is nothing but
the other line drawn on Q through G (as , s).
Proof. The first assertion is easily computed. The second is 21.3.4. And the conclusion follows.
Proposition 21.4.6. A point x : y : z on the triangle plane has two temporal embeddings on the
quartic Q (counting multiplicity and complex values). The critical points of this double coating of
the plane are the points of the pilar conic (and the points at infinity).
Proof. First assertion is about the degree in t. Second assertion is about the discriminant, which
2
is (x + y + z) times the equation of the key conic.
March 30, 2017 14:09
published under the GNU Free Documentation License





21. Linear Families of Inscribed Triangles
243
Proposition 21.4.7. The embedding G (C) of the pilar conic lies in the plane :
[−f − v + w, −f + h − v, −f − g + w, 2]
The horizontal plane [s, s, s, −1] cuts this curve in two points that are the embeddings of the contact
points of the pilar conic with the temporal conic C (s). One of the horizontal lines of the pilar plane
is drawn through points G (S, t∗ ) and G (E, t∗ ) where t∗ is the parameter related to the extremal
area of triangle at bt ct .
2
Proof. A point on the pilar conic can be written as: 1 ÷ (g + h − u) : τ 2 ÷ (h + f − v) : (1 + τ ) ÷
(f + g − w). It’s only date of embedding is
t = (f + g − w)
f (g + h − u) τ 2 + (g + h − u) (h + f − v) τ + v (h + f − v)
(f + u) (g + h − u) τ 2 + 2 (g + h − u) (h + f − v) τ + (g + v) (h + f − v)
and the conclusion follows. One should remark that the plane equation is the last line of Q .
Definition 21.4.8. A linear motion M (t) on a given line ∆ is said to be incident to a given
LFIT when the temporal parameters at cutting points are the same on the sidelines and on the
transversal.
Proposition 21.4.9. A given line ∆ ' [l, m, n] is the support of an incident motion if and only
if the line is tangent to the pilar conic. And then, the incident motion is embedded along a line
drawn on Q.
t
Proof. Solving in t the equation [l, m, n] · [0, p + t/f, (1 − p) − t/f ] = 0 tells us when point ∆ ∩ BC
is reached. Substituting into at and doing the same for the other two sidelines leads to the three
embeddings:


0
−n
−m

 −n
0
l





 m
l
0
−nf
(l − n) (f − w) + lg
(l − m) v − lh
Asking for the degeneracy of the plane drawn by these three points gives four equations whose gcd
is the tangential equation of C.
21.5
Miquel circles
Definition 21.5.1. Circles (A, bt , ct ) are called the A−Miquel circles of the LFIT family. Et circ.
for B and C. Their Veronese columns are:

 2
 2

0
c f (v − t)
b f (w − f + t)
 c2 g (h − v + t)  
  a2 g (f − t) 
0




cirA, cirB, cirC '  2



 b h (u + v − h − t)  


a2 ht
0
gh
fh
fg
Proposition 21.5.2. The A−Miquel circles are going through A and another fixed point OA , etc.
And we have




a2 gh − b2 hu − c2 gu
a2 h (v − h − f )
a2 g (w − f − g)




OA , OB , OC '  b2 h (u − g − h)   −a2 hv + b2 f h − c2 f v  
b2 f (w − f − g)

2
2
2
2
2
c g (u − g − h)
c f (−h − f + v)
−a gw − b f w + c f g
(21.3)
Proof. The existence is obvious from the linear nature of the Veronese’s (and the computation is
straightforward).
Proposition 21.5.3. Circle of similarity. For a given t, the three Miquel circles concur at a
point Mt . The locus of this point is the circle (FA , FB , FC ). Moreover, M(t=∞) is S ∗ , the isogonal
.
conjugate of S, i.e. S ∗ = a2 /f : b2 /g : c2 /h. Therefore, S ∗ is the perspector of triangles ABC and
FA FB FC .
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
244
21.6. Constructions
Figure 21.4: Miquel circles
Proof. Compute the circle orthogonal to the three Miquel circles, i.e. take the wedge of their
Veronese. Test that this circle is a point-circle, and that M (t) is second degree in parameter t.
Use locusconi to obtain the locus. Check that we have a circle, and obtain its Veronese, namely
 2 2

b c (g + h − u) f
 c2 a2 (h + f − v) g 


 2 2

 a b (f + g − w) h 
a2 gh + b2 f h + c2 f g
What remains is straightforward. The name "circle of similarity" is explained at Section 21.7.
Proposition 21.5.4. Fixed point.
barycentrics are:

a2 gh
 2
Pa , Pb , Pc '  b f h − b2 hw + c2 gv
b2 hw + c2 f g − c2 gv
All the lines at Mt are going through a fixed point PA whose



a2 gh − a2 hw + c2 f u
a2 gh − a2 gv + b2 f u

 2

b2 f h

  a gv + b2 f h − b2 f u  (21.4)
a2 hw + c2 f g − c2 f u
c2 f g
Points E, Oa , Pa are aligned and Oa , Pa belong to the circle of similitude. Moreover S ∗ PA k BC.
Proof. Equation of line at Mt is
2
b h + c2 g f t − f b2 f h − b2 hw + c2 gv , a2 ghf − a2 ght, −a2 ght
This leads to Pa . One can see directly that Pa = f ghS ∗ + b2 hw − c2 gv δBC and that Oa − Pa
(as written here) is ' E.
21.6
Constructions
Suppose now that S ' f : g : h and E ' u : v : w are given (with f + g + h = u + v + w,as ever).
We have, inter alia, the following constructions.
Construction 21.6.1 (Miquel). Compute the Fj from S, E. When they are different, chose a
point M on circle (FA , FB , FC ). Draw circles (M, A, FA ) , etc. Then cirA, cirB concur on AB to
give point c, etc.
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
245
Remark 21.6.2. Condition FB = FC induces E = S ∗ = Fa = Fb = Fc .
Construction 21.6.3 (Catalan). When S, E are at finite distance, the Catalan graphs are obtained
by symmetry around S + E.
Construction 21.6.4. Using the hexagonal graphs. The A-one goes through B + C − a0 and is
parallel to SA.
21.7
Similarities and Cremona transforms
Now, we describe points A, B, C using the Lubin’s parametrization, i.e. A ' α : 1 : 1/α in the
Z : T : Z complex projective plane. It happens that equations using f, g, h, p, q, r are simpler than
equations using f, g, h, u, v, w (the equicenter).
Proposition 21.7.1. Similarities. The correspondence CA 7−→ AB : b 7→ c induces a similarity
σA of the whole plane. Center is FA , the fixed point of the Miquel circles (21.3), while angle and
ratio are, respectively, −A and −cg/bh. In the Morley frame, the matrix of σA is:


(α − β) g
(α − β) (q γ + (1 − q) α) g
−
αr
+
(1
−
r)
β
+
0
 (α − γ) h

(α − γ) h




σA = 
0
1
0


r
r − 1 (α − β) (q α + (1 − q) γ) g
(α − β) γ g 
0
+ −
+
−
α
β
(α − γ) αβ h
(α − γ) β h
The equicentric property holds for the whole plane, since:
−1
f 1 + g σC + h σB
= E · L∞
Proof. The quicker is to describe σA as the collineation: Ωx 7→ Ωx , Ωy 7→ Ωy , b0 7→ c0 , b 7→ c
and apply the general formulas. The obvious result σC σB σA = 1 can be used to check the
obtained matrices.
Proposition 21.7.2. Anti-similarities. The correspondence CA 7−→ AB : b 7→ c induces an
−
anti-similarity σA
of the whole plane. Center is


h (qg − hr) b2 − g (qg − hr − g + h) c2


Qa ' 
(21.5)
− (qg − hr + h) hb2

2
(qg − hr + h) gc
−
−
−
while σC
σB
σA
is the involutory affinity that fixes AC and reverts the direction of BH.
Proposition 21.7.3. Miquel homography. Assume that S ∈
/ L∞ and define σ as the homography: A 7→ FA , B 7→ FB , C 7→ FC . The matrix of σ in the upper spherical map (Z, T) is:
σ =
zE
1
σ3 zS + zE − zH
−zS
!
; σ
−1
=
zS
1
where zE =
σ3 zS + zE − zH
−zE
!
f α + gβ + hγ
uα + vβ + wγ
, zS =
, z H = σ1
u+v+w
f +g+h
This defines a Cremona transform of the whole projective plane, whose indeterminacy points are
the umbilics and S while exceptional lines are L∞ and the isotropic lines of S. By σ −1 , the center
ω of circle FA FB FC is mapped to 1/ζS (symmetrical of S in the circumcircle).
Proof. Direct computation.
Proposition 21.7.4. Homography σ degenerates when S and E are isoconjugates (thus none of
them at infinity). In this case, the Miquel circle is reduced to point E = S ∗ = Fa = Fb = Fc .
Proof. Obvious from (16.5) (when assuming S ∈
/ L∞ ). Assuming S ∈ L∞ , this would require
(u α + v β + w γ) (α f + β g + γ h) = 0, i.e. E at origin, instead of E ∈ L∞ .
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
246
21.8. Parabolas
21.8
Parabolas
Proposition 21.8.1. The envelope of the line bt ct is a parabola PA . Its point at infinity is
g + h : −g : −h, i.e. the direction of line SA. Its tangential equation is
P∗A

2 (q − 1) g − 2 hr
(1 − q) g − (1 − r) h

=  (1 − q) g − (1 − r) h
0
hr − gq
(1 − r) h + gq

hr − gq

(1 − r) h + gq 
0
This parabola is tangent to AB and AC at their intersections with locus (α). When using metric
properties, the focus is the already encountered point FA , while its directrix is
∆A ' [Sa (g + h) , −Sb g, −Sc h] − uSa L∞
where u = E1 = hr − gq + g
Proof. The coefficients of line bc have degree 2 in t. Therefore, the envelope is a conic. Matrix
P∗A is obtained by locusconi. Then L∞ · P∗A = −g − h : g : h and we have a parabola. The
focus comes from the Plucker method, and the directrix is the polar line of the focus.
Et circularly for the other two parabolas.
21.9
Equilateral triangles
21.9.1
General formula
We require now that abc, the inscribed triangle, is equilateral. Chose a ∈ BC and consider the
−1
rotation ρ (a, +60°). Define
√ b as ρ (c). Then abc is equilateral direct. In
√ c as AB ∩ ρ (CA) and
order to avoid horrible 3, we define Sv3 = S/ 3 and we obtain:

−a2 x + Sc + 2 Sv3

ρ =  (Sc + 2 Sv3 ) x − b2
(Sb − 2 Sv3 ) x + Sa + 2 Sv3
−a2 x + a2
(Sc + 2 Sv3 ) x − Sc + 2 Sv3
(Sb − 2 Sv3 ) x − Sb + 2 Sv3
leading to (using the same norm for all the three points) a, b, c '

−a2 x

(Sc + 2 Sv3 ) x

(Sb − 2 Sv3 ) x + 4 Sv3

 
 

0
+4 Sv3 − (Sb + 2 Sv3 ) x
Sa − b2 + 2 Sv3 + (Sc + 2 Sv3 ) x

 
 

(Sa + 2 Sv3 ) 
x ,
0 ,
+b2 − (Sc + 2 Sv3 ) x 
1−x
Sa − 2 Sv3 + (Sb + 2 Sv3 ) x
0
21.9.2
Some remarks
From this expression, we obtain the area center:
S'
1
1
1
1
1
1
:
:
'
:
:
x1 − x0 y1 − y0 z1 − z0
Sa + 2 Sv3 Sb + 2 Sv3 Sc + 2 Sv3
and we recognize S =X(13), the first Fermat center, i.e. the <i>Pelle à Tarte</i> center obtained
using the vertices of the outer equilateral triangles.
Applying these masses to the variable triangle, we obtain the equicenter:
E ' (Sa + 2 Sv3 ) a2 : (Sb + 2 Sv3 ) b2 : (Sc + 2 Sv3 ) c2
and we recognize E =X(15) the first isodynamic center.
Obviously, changing the orientation, i.e. +Sv3 into −Sv3, gives another family.
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
21.9.3
247
Equilateral pedal triangles
Since triangles abc are "turning around" point E, the triangle of minimal area is obtained by
orthogonal projection. Therefore the pedal triangle of X(15) is equilateral and belongs to the
family.
The center of this triangle is the middle of [S, E]. Moreover, the locus of g = (a + b + c) /3 is
directed by:
Sc − Sb : Sa − Sc : Sb − Sa ' b2 − c2 : c2 − a2 : a2 − b2
One recognizes X(531), the orthopoint of X(30): the locus of g is the perpendicular bisector of
segment [S, E], and therefore orthogonal to the Euler line.
21.10
Families of constant area
21.10.1
Three non concurrent lines
Theorem 21.10.1. When S ' E then S is one of the points Ka , K2 , K3 defined by equations
(21.1). When triangle a0 b0 c0 is given, points K2 , K3 are the points at infinity of the hexagonal
conic. They are real only if area (abc) ≤ S/4. In this case, the other triangles of the family can be
constructed using strips of equal width.
When S ' f : g : h and W are given (with S ∈ L∞ ), then:
abc =

0


t
f (g − h) W
1

+
 + −
 f
2 (f g + gh + hf ) 2

f (g − h) W
1
t
+
− +
f
2 (f g + gh + hf ) 2
t
g (h − f ) W
1
− +
+
g 2 (f g + gh + hf ) 2
0
t
g (h − f ) W
1
+ −
+
g 2 (f g + gh + hf ) 2
t
h (f − g) W
1 
−
+
h 2 (f g + gh + hf ) 2 
t
h (f − g) W
1 

− +
+ 
h 2 (f g + gh + hf ) 2 

0
+
Proof. The first part is from the first section, and the rest from a straightforward computation.
Figure 21.5: Two parabola
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
248
21.10. Families of constant area
Let us go back to the families of constant area. This implies S ∈ L∞ and that S is one of the
Ki . Assuming S = K2 (choosing the sign of W ), one obtains:



g−h
f
 2 (f g + gh + hf ) 


K3 ' W  h − f  −
 g t
f gh
f −g
h

while the locus P of the K1 becomes a conic since the t−degrees of the barycentrics of that point
are now 2, due to f + g + h = 0. After computing P using the procedure locusconi, one can check
that K2 ∈ P together with L∞ ∈ P ∗ : the conic is a parabola.
Moreover, the envelope of line K1 K3 is also a conic, since the t−degree of K1 ∧ K3 is two.
Using again locusconi, we obtain the matrix of the tangential equation:


(1 − W ) f
−W h
−W g


Q∗ '  −W h
(1 − W ) g
−W f

−W g
−W f
(1 − W ) h
When searching for the common points of Q and P, one obtains that these curves are bitangent.
A first contact occurs in K2 , so that Q is a parabola with the same direction as P. The second
contact is the center of the hexagonal conic relative to
t0 =
(g − h) (f − h) (f − g)
W
6 (f g + f h + gh)
when using the former given parameterization. This value is the arithmetical mean of the dates of
the degeneracies.
21.10.2
Three concurrent lines
Suppose now that points a, b, c are moving on concurrent lines. We only have to consider this case
as the limit of what happens when K → 0 to the figure obtained by applying an homothety of
center G and ratio K to the previous results (and replacing X by X/K for X = p, q, r, W, t, while
slownesses f, g, h are unchanged).
Then the inscribed triangle and the equicenter become:




1
t 1
t
1
1
−
q
−
+
r
+
hr
−
gq
+
(f
+
g
+
h)

 3
3
g 3
h 


3
 1


t 1
1
t 
1


−r−
abc '  + p +
 , E '  f p − hr + (f + g + h) 

 3
f 3
3
h 
3



 1
1
t 1
t 1
gq − f p + (f + g + h)
−p−
+q+
3
3
f 3
g 3
while the areal center is the limit of:


(2 f − g − h) K + (f + g + h)


 (2g − f − h) K + (f + g + h) 
(2h − f − g) K + (f + g + h)
and thus is G when S ∈
/ L∞ , but is S when S ∈ L∞ .
The hexagonal conic goes through a, b, c and a0 = b + c − G, etc. When t = 0, its points at
infinity are:
2 q + 2 r : −W − 2 r : W − 2 q
where W 2 = −4 (pq + qr + rp) is the limiting value of the relative area. Therefore:
Proposition 21.10.2. Given three lines that are concurrent at G and an inscribed triangle abc it
exists two families of inscribed triangles that share the same area. Their areal center is one of the
points at infinity of the hexagonal conic defined by a, b, c, G. Point S is real if W 2 ≥ 0. Then, the
family is obtained by constructing strips of equal width (in the S direction).
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
21.11
249
When the graphs are given
In the previous sections, the graphs were the result of the pre-existing mappings a ←→ b ←→
c ←→ a. Let us now examine what happens when these graphs are chosen from the beginning.
21.11.1
Catalan graphs
Proposition 21.11.1. When three lines ∆j are parallel to the sidelines, they are the Catalan
graphs of a LFIT if and only if the bisectors of strips (∆a , BC) , etc are concurrent. And then S
can be chosen at will (outside of L∞ ).
Proof. The necessity comes from the required symmetry wrt ω = (S + E) /2. Since ω is at finite
distance, S, E cannot be chosen at infinity.
Proposition 21.11.2. Define the collineation µ∗ by L∞ 7→ L∞ , BC 7→ ∆A = [p0 q0 , q0 ], CA 7→
∆B = [r1 , q1 , r1 ], AB 7→ ∆C = [p2 , p2 , r2 ]. Then its matrix (acting over the lines !) is


p
q
q
0
0
 q0 − p0

r1
'
 r1 − q1

p2
p2 − r2
∗
µ
2
0
q0 − p0
q1
r1 − q1
p2
p2 − r2
q0 − p0
r1
r1 − q1
r2
p2 − r2





We have χ (X) = (X + 1) (X − λ) Then λ = 1, i.e. µ is a central symmetry, when
2 p0 q1 r2 + q0 r1 p2 − p0 q1 p2 − p0 r1 r2 − q0 q1 r2 = 0
.
In any case, ω = ker (µ − 1) '
E = µ (S) '
q0
r1
p2
:
:
. When ω is the center, we have:
p0 − q0 q1 − r1 r2 − p2
f p0 + gq0 + hq0 f r1 + gq1 + hr1 f p2 + gp2 + hr2
:
:
p0 − q0
q1 − r1
r2 − p2
Proof. Computations are straightforward.
21.11.2
Hexagonal graphs
Lemma 21.11.3. When S, E are know, the hexagonal graphs have the following equations:
(21.6)
Ga ' [f u − gv − gw − hv − hw, f u − gv − gw + hu, f u + gu − hv − hw] , etc
Proof. From the very definition, we have
.
a = (b ∧ (B − A)) ∧ (c ∧ (C − A)) '
0
1
1
−
h g
t−q+r : −
t
t
+1−r : +q
h
g
and then pqr are substituted.
Proposition 21.11.4. When three lines ∆j ' [pj , qj , rj ] are in general position, there exist S, E so
that the ∆j are the hexagonal graphs of the corresponding LFIT when the two following relations:


 r −q 
 q −r 
3
3
2
2
1
 r3 − p3 
 r3 − p3 
 q2 − p2 
. 
. 
. 



S1 ' S2 ' S3
where S1 =  r3 − q3  ; S2 = 
1
 ; S3 =  p1 − r1 
 q −p 
 p −q 
 p1 − q1 
2
2
1
1
1
q2 − r2
p1 − r1
 p1 − q1 − r1 r3 − p3 


p1 − q1 − r1
p1 − r1
r3 − q3
q2 − p2 − r2
E1 ' E2
q2 − r2
r3 − p3 − q3
r3 − q3
are fulfilled. In such case, the common values provide

. 
where E1 = 


p1 − q1




q
− p2 − r2
2
.
 ; E2 = 



 r − p q2−−q p2p − r
3
3
3 1
1
r3 − p3
p1 − q1
S and E.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–





250
21.11. When the graphs are given
Proof. The nine equations saying that Sj ∧Sk = 0 are proportional to a single 3rd degree polynomial
P1 in the nine coefficients. The same occurs for the three equations E1 ∧ E2 = 0 and another
polynomial P2 . On the other hand, from ∆c , ∆a we obtain the graph ∆0b of a 7→ b 7→ c. Equations
∆b ∧ ∆0b = 0 are generated by two polynomials P3 , P4 ... and one sees that ideal (P1 , P2 ) =
ideal (P3 , P4 ).
Proposition 21.11.5. Let us now suppose that S is given. This determines the direction of the
graphs, and therefore one of their coefficients, requiring that :


p1 g
p1 g + (r1 − p1 ) h
r1 g


(G) ' 
p2 h
q2 h
q2 h + (p2 − q2 ) f 
r3 f + (q3 − r3 ) g
q3 f
r3 f
And then, the equicenter is given by



E '



r1 g
+h
r1 − p1


hp2
+f 

p2 − q2

f q3
+g
q3 − r3
subjected to the condition that S ∈ D where
q3
r1
p2
D'
,
,
q3 − r3 r1 − p1 p2 − q2
Proof. Computations are straightforward... but the geometric interpretation of this line is not
clear.
21.11.3
The marvelous formula
Another formulation of condition § 8.2. is as follows. The
triangle of S. We have:

∗ 
−f
f
f
. 


TS∗ =  g −g
g  '
h
h −h
directions are those of the anti-cevian
0
h
g

h g

0 f 
f 0
Consider the collineation µ defined by L∞ 7→ L∞ , TS∗ 7→ (G). its matrix is :
µ = K + Q · L∞
where K = (g + h − f ) f g p0 p1 q2 + (h + f − g) gh q1 q2 r0 + (f + g − h) f h p1 r0 r2 + 2 f gh r0 p1 q2 and


(−f gp0 p1 q2 + ghq1 q2 r0 + f hp1 r0 r2 ) f


Q =  (+f gp0 p1 q2 − ghq1 q2 r0 + f hp1 r0 r2 ) g 
(+f gp0 p1 q2 + ghq1 q2 r0 − f hp1 r0 r2 ) h
Therefore, µ is an homothety centered at Q, whose ratio is: 2 f gh r0 p1 q2 /K.
Let us left-multiply by the homothety centered at G, with ratio −1/2. We obtain another
homothety µG whose center is:


2f gh r0 p1 q2 + f g (g + h) p0 p1 q2 + gh (h − g) q1 q2 r0 + hf (g − h) p1 r0 r2


QG '  2f gh r0 p1 q2 + f g (h − f ) p0 p1 q2 + gh (h + f ) q1 q2 r0 + hf (f − h) p1 r0 r2 
2f gh r0 p1 q2 + f g (g − f ) p0 p1 q2 + gh (f − g) q1 q2 r0 + hf (f + g) p1 r0 r2
When writing that line L ' [f ; g; h] goes through QG , we re-obtain the condition §8.2. This
amounts to say that µ (L0 ) = L with the pappus notations. This replaces the requirement that
S ∈ D y the requirement that µ sends the conipolar of S wrt the Steiner ellipse onto the tripolar
of the isotomic of this point S.
And we are back with the same question: a geometric interpretation would be great.
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
21.12
251
Observers (about perspectivities)
Definition 21.12.1. We say that a (fixed) triangle Ta Tb Tc observes the LFIT when the fixed
triangle is in perspective with all of the triangles of the LFIT.
Our intent is to specify some notations and prove the following theorem:
Theorem 21.12.2. There are two kinds of observers, each family being parametrized by a generic
point M in the triangle plane . The first one, we call it the metric observer, is obtained by using M
∗
as SH
and leading to the PaH , PbH , PcH of Proposition 21.5.4. And then the locus of the perspectors
is the conic through the seven M, PjH , OjH . The second one, we call it the Poulbot observer, is
obtained by using M as a center of parallelogy between trigon ABC and a trigon tangent to each
of the parabolas Pj described at Proposition 21.8.1. And then, the locus of the perspectors is a
straight line.
Conversely, there are five observers that share a given vertex (say PaH ), one metric and four
Poulbot. The B-vertices of the Poulbot triangles (say Bj ,j=1..4) are the common points of the
tangents to PB from PaH and to PA from PbH , etc. for the vertices Ck . And then the Bj and the
Ck are paired by the fact that Bj Cj is tangent to PA .
In what follows, the LFIT is given by S
a case, we have
 

f
 

S, E, Ω '  g  , 
h
and the parameters p, q, r of the t = 0 triangle. In such


gq − hr + h
hr − gq + g


f p − hr + h   hr − f p + f 
f p − gq + g
gq − f p + f
so that the three key points have the same normalizing factor at infinity, namely f + g + h.
Proceeding that way, the case S ∈ L∞ can be treated like the other cases.
Proposition 21.12.3. An observer is necessarily parallelogic with the ABC triangle.
Proof. This is obvious since the observer must be in perspective with T∞ , i.e. with (δBC , δCA , δAB ).
Remark 21.12.4. Spoiler! For a Poulbot observer, the generic point M will be the parallelogy
center of ABC 7→ O. But for a metric observer, M will be the parallelogy center of O 7→ ABC !
21.12.1
Metric observer
By itself, the theory of the LFIT is a projective one, and doesn’t necessitate to use a metric or
another. Nevertheless, we have encountered interesting properties when using a metric (circle of
similarity). In order to reuse these properties, we are now investigating what happens when the
metric is taken as a parameter.
21.12.1.1
Forcing the orthocenter
Definition 21.12.5. When an Euclidean structure is given on the ABC plane, then triangle ABC
receives an orthocenter H ' 1/Sa : 1/Sb : 1/Sc . Reverting the process, i.e. saying that a given
point H ' ρ : σ : τ is the orthocenter of ABC, determines a metric structure on the ABC plane.
This process is called "forcing the orthocenter". The resulting objects will be super-scripted with
∗
a "H", like in X H (except from the isogonal conjugacy, noted SH
, in order to avoid S ∗ H ).
Proposition 21.12.6. Define OH as the complement of H, and ΓH (the forced circumcircle) as
.
the circumconic whose perspector is K H = H ∗ OH . Then



b



σ+τ
ρ (σ + τ )
0
τ (ρ + σ) σ (τ + ρ)






OH '  τ + ρ  , K H '  σ (τ + ρ)  ; ΓH '  τ (ρ + σ)
0
ρ (σ + τ ) 
ρ+σ
τ (ρ + σ)
σ (τ + ρ) ρ (σ + τ )
0
.
Moreover ΓH is the center of the conic, which goes through 2HA −H, etc where HA = AH ∩BC, etc.
Proof. Direct application of Theorem 11.4.4.
Definition 21.12.7. The forced isogonal transform is defined as the ABC-isoconjugacy that ex∗
changes H and OH . In other words, (x : y : z)H ' ρ (σ + τ ) yz : σ (τ + ρ) zx : τ (ρ + σ) xy.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
252
21.12. Observers (about perspectivities)
21.12.1.2
Forcing the isogonal center
Definition 21.12.8. Forcing the isogonal center of a LFIT is making an arbitrary choice M ' x :
∗
y : z for the point SH
. This amounts to force the metric to a2 : b2 : c2 ' xf : yg : zh. The values
a, b, c are not intended to be real, only different from 0.
Proposition 21.12.9. The "fixed points" PA , etc(see Proposition 21.5.4) are obtained from S ∗
by collineations that doesn’t depend from a choice of metric. Indeed, we have Pa = φa · S ∗ where:

1
0

gq
 0
p−
φa ' 
f


gq
0 1−p+
f
0
hr − h
p−
f
hr − h
1−p+
f


1
0
 
  0
'
 

0
w
1−η
f
w
η
f
0
v
η
f
v
1−η
f





using η =
f +g+h
u+v+w
hr − gq − h
v+w
= 1−η
.
f
f
Proper point associated to k is the direction δBC , while the proper line is [0, −w, v] i.e. the line AE.
2
This is an affinity whose charpoly is (X − 1) (X − k) where k =
Proof. Obvious from (21.4)
Proposition 21.12.10. For any point M , the triangle φa M, φb M, φc M is an observer of the
∗
LFIT. The locus (on t) of the perspector KtH is a conic, that goes through M = SH
, the three PjH
H
and the three Oj .
Proof. This is only reformulating some already proven properties (see Proposition 21.5.3)
Proposition 21.12.11. The O 7→ ABC para-center of a metric observer is M ' x : y : z itself.
The ABC 7→ O para-center is


f (uz − wx) (uy − vx)


ϕ (M ) '  g (vx − uy) (vz − wy) 
h (wy − vz) (wx − uz)
We have ϕ (E) = 0 : 0 : 0. Two M that share the same ϕ (M ) are aligned with E, while all the
ϕ (M ) belong to the circumconic γ with perspector S ∗ E. Moreover ϕ (M ) belongs to ΓH , the forced
b
circumcircle. Therefore ϕ (M ) is the forced gudulic center of conic γ.
Proof. Existence is Proposition 21.12.3. Computing the value is easy. The result clearly depends
on M ∧ E, hence the alignment.
21.12.2
Cevenol graphs
Definition 21.12.12. Let αt0 be the point where the parallel to BH through bt cuts the parallel to
CH through ct . As time t flows, the point αt0 draws a straight line cevaH . We call it the A-Cevenol
graph (related to the forced orthocenter H ' ρ : σ : τ ).
Proof. Straight line cevaH is obtained as:
((gq − hr) ρ + (gq − hr − g) τ ) ((gq − hr + h − g) ρ + (gq − hr − g) σ)
(gq − hr + h) ,
,
τ +ρ
ρ+σ
Proposition 21.12.13. The line cevaH is nothing but the line that joins FH;A and QH,A , the
centers of the direct and reverse H-similarities that generalizes the correspondence CA 7−→ AB :
b 7→ c.
Proof. Obvious from (21.3) and (21.5).
Proposition 21.12.14. The threeH-Cevenol graphs concur at a point Gc that belongs to the Hconic of similarity.
Proof. Properties of Gc are easily computed.
March 30, 2017 14:09
published under the GNU Free Documentation License
21. Linear Families of Inscribed Triangles
21.12.3
253
Poulbot observers
Definition 21.12.15. The Poulbot observer related to a point M is the only triangle Pj such that
Pb Pc is together parallel to M A and tangent to PA and circ. for B and C.
Proof. There is one and only one tangent to a parabola that contains a given direction.
Proposition 21.12.16. The "Poulbot observer" observes the LFIT. The equations of this observing trigon are:

zy (gq − hr + h)

O∗ '  (−f px + hrx + f x + f z − hx) z
(f px − gqx + f y)
z (gqy − hry + gz)
xz (−f p + hr + f )
(f py − gqy − f y + gx + gy) x

(gqz − hrz − gz + hy + hz) y

x (−f pz + hrz + hx)

yx (f p − gq + g)
while the locus of the perspectors is a straight line, whole tripolar is:

ghx2 + h (f p − gq − f + g) yx − g (f p − hr) xz + f (gq − hr − g) yz


 f hy 2 + h (f p − gq) yx − g (f p − hr + h) xz + f (gq − hr − g + h) yz 
f gz 2 + h (f p − gq − f ) yx − g (f p − hr − f + h) xz + f (gq − hr) yz

Proof. Write that M ∧ A + λL∞ is tangent to PA . Since L∞ itself is a tangent, we obtain a first
degree equation, leading to O∗ . Then we take the adjoint and the perspectivity, for all t, is easy
to check.
Proposition 21.12.17. Assume that triangle Pa Pb Pc is a Poulbot observer and use Pa = PaH to
force the metric, determining PbH and PcH . Then lines Pb PbH and Pc PcH are tangent to PA (or
degenerate !)
.
.
Proof. Note p1 : q1 : r1 the coordinates of Pa and δb = tb : −1 − tb : 1 and δc = 1 : tc : −1 − tc
the directions of Pa Pb and Pa Pc . Write the conditions hb and hc for lines Pa δb and Pa δC to be
tangent to the required parabolas. On the other hand, the parallelogy center is tb : tb tc : 1, and
the direction of the third tangent is −1 − tb tc : tb tc : 1. Draw this tangent, say ∆ and obtain the
expressions in p1 , q1 , r1 , tb , tc of Pb , Pc . Obtaining those of PbH , PcH is obvious. It only remains to
write the contact condition (length ≈120000) and take the Euclidean remainder modulo h1 and
then modulo h2 . This gives 0.
21.12.4
Degenerate observers
Definition 21.12.18. When the A-vertex of an observer O belongs to the sideline BC, we say
that O is an A-degenerate observer, et circ. for the other vertices.
Proposition 21.12.19. Let line ∆a be tangent to PA . Note Pb the intersection of ∆a and the
second tangent to PA from B, et circ. for Pc . Then Pa Pb Pc is an observer for any Pa ∈ BC.
Conversely, any degenerate observer is obtained that way.
Proof. Substitute p1 = 0 in the perspectivity equation and assume that all the four coefficients of
t are 0. The system separates into Pb ' Pc ' 0 : q3 : r3 , to be discarded, and to:




gq − hr − g
gq − hr − g




Pb ' 
gh s
 ; Pc '  hr − gq + g − h 
hr − gq
1/s
21.12.5
Counting in brute force
1. Four equations. We start from Pa ' p1 : q1 : r1 , Pb ' p2 : q2 : r2 , Pc ' p3 : q3 : r3 and
.
we write that O = Pa Pb Pc observes the LFIT family. The corresponding determinant is a 3
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
254
21.12. Observers (about perspectivities)
degree polynomial in t. This gives four equations with the following degrees:
1
t
t2
t3
length
1428
3388
1768
286
p1 , q1 , r1
1
1
1
1
p2 , q2 , r2
1
1
1
1
p3 , q3 , r3
1
1
1
1
p, q, r
3
2
1
0
f, g, h
0
2
1
0
The t3 equation is nothing but the condition of parallelogy:
trace (O) =
p1
q2
r3
+
+
=1
p1 + q1 + r1
p2 + q2 + r2
p3 + q3 + r3
2. Three cubics. Then we solve two of these first degree equations in p3 q3 r3 , and substitute in
the other two. This leads to the following three cubics:
name
K1
K2
K3
solving
t2 , t3
t2 , t3
t0 , t3
length
7271
10654
16725
f, g, h
2
1
2
p, q, r
1
2
3
p1 , q1 , r1
3
3
3
p2 , q2 , r2
3
3
3
3. Excluding. Line AC is asymptote to K1 and K2 , but not to K3 : one occurrence of δAC must
2
be excluded. Moreover, point p1 (gq1 + (g − f ) r1 ) : g (q1 + r1 ) : f p1 q1 is on K1 and K2 but
not on K3 .
4. It remains 7 common points to the first two cubics. Eliminating q2 , the equation in p2 , r2
splits into four factors whose degrees are: 1,1,1,4 . This allows to identify:
(a) ***
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 22
Quadrilateral
22.1
Newton stuff
Definition 22.1.1. A transversal is a moving line L0 that cuts the sidelines of a fixed triangle
ABC. When we consider that the four lines L0 , LA = BC, LB = CA, LC = CA are playing the
same role, we call this situation a quadrilateral. This induces four embedded triangles and
we denote Tj the triangle which is associated with the trigon "all the four lines except from Lj ".
Remark 22.1.2. Without any other notice, barycentrics are relative to triangle ABC.
Proposition 22.1.3. Newton axis. Let A0 = L0 ∩ LA , , etc be the traces of the transversal L0
on the other three lines. Then the middles of segments AA0 , BB 0 , CC 0 are on a same line, called
the Newton axis of the quadrilateral. Coordinates:
γ ' [−pq − pr + rq, −pq + pr − rq, pq − pr − rq]
Proof. Direct computation. Formally, this is cevadiv (L0 , L∞ ).
Proposition 22.1.4. Reciprocal lines. Let Rj be the reciprocal of line Lj wrt triangle Tj . They
are parallel to each other. Moreover, their equibarycenter is the Newton axis.
Proof. Remember that R0 is the line through B + C − A0 , etc. The barycentrics of these lines are:
R0
RA
RB
RC
γ
'
'
'
'
'
rq
−rq
−2 pr + rq
−2 pq + rq
−pq − pr + rq
,
pr
,
pr − 2 rq
,
−pr
,
−2 pq + pr
, −pq + pr − rq
,
,
,
,
,
pq
pq − 2 rq
pq − 2 pr
−pq
pq − pr − rq
The first one comes from the very definition. The other three are obtained by the usual conjugacy
methods, and adjusted so that all the Rj ∧ L∞ are equal (and not simply proportional), so that
computing R0 + RA + RB + RC makes sense.
Proposition 22.1.5. Newton pencil. The three circles whose diameters are the segments
[A, A0 ] , [B, B 0 ] , [C, C 0 ] belong to a same pencil. Its radical axis ν is called the Steiner axis of
the quadrilateral. Their ABC-barycentrics are:




0
−p Sa
+p Sa
 +q S  
  −q S 
0


b 
b 
νa , νb , νc ' 
(22.1)



 −r Sc   +r Sc  

0
q−r
r−p
p−q
.
ν = Steiner_axis ' [Sa p (q − r) , Sb q (r − p) , Sc r (p − q)]
Proof. Obvious from the coordinates.
255
256
22.2
22.2. Steiner stuff
Steiner stuff
Proposition 22.2.1. The four circumcenters of the four embedded triangles Tj are on a same
circle that is called their Miquel circle. The four circumcircles have a common point, called their
Miquel point. Moreover, the Miquel point belongs to the Miquel circle. Their ABC-barycentrics
are:


a2
2 2 

2
(q − r) Sb r + Sc q − pa b c p
 (q − r) p 


 (r − p) S p + S r − b2 q c2 a2 q 


b2


c
a


;
M
'
ΓM ' 

q


2
2
2
 (p − q) Sa q + Sb p − c r a b r 
 (r −2p) q 


c
8 (p − q) (q − r) (r − p) S 2
r (p − q)
Proof. Using wedge, Veronese and wedge3, one obtains the ABC-barycentrics of the four circumcircles and the four centers:

 2
 2



c p
b p
0


0
 c2 q 
 p−q 
 p−r 






 0 


 a2 q 


0


q
−
p






Γ0 ' 
 ; ΓA '  2
 ; ΓB '  a2 r  ; ΓC '  q − r 
 0 
 b r 




 r−p 
 r−q 
 0 
1
1
1
1




a2 Sa
a2 Sa p (p − q − r) − b2 Sb pr − c2 Sc pq + 8 S 2 rq




O0 '  b2 Sb  ; OA ' 
pb2 Sb p + Sa q − c2 r
 , etc
c2 Sc
pc2 Sc p − b2 q + rSa
And, some wedge, Veronese and wedge3 later, the result is obtained. Other proofs are possible
(Ehrmann, 2004).
Proposition 22.2.2. The Miquel point has the same Simson line with respect to all four embedded
triangles Tj . This line is called the pedal line of the quadrilateral. By Proposition 9.1.1, the image
of this line by the homothety h (M q, 2) contains the orthocenters Hj of the triangles Tj . Moreover,
this common Steiner line of M q wrt the four triangles Tj is the Steiner axis of the quadrilateral
(i.e. the radical axis of the Newton pencil).
Proof. A line is defined by two points, and three projections are on each Simson line. One can also
use the ABC-barycentrics of the orthocenters:


S 2 rq
Sb Sc
Sb Sc p − b2 Sb q − c2 Sc r + 4



p
H0 '  Sa Sc  ; Ha ' 

Sa Sc p − b2 q + Sa r
Sa Sb
S S p + S q − c2 r

a
b
a




Proposition 22.2.3. The conjugate circle associated with a triangle was defined at Section 12.7.
The four conjugate circles relative to the four embedded triangles form a pencil with γ, the Newton
axis. This pencil, called the Steiner pencil, is orthogonal to the Newton pencil. The barycentrics
of these cycles are:



γ=

qr − pq − pr
rp − qr − qp
pq − rp − rq
0






 ; γ0 = 


Sa
Sb
Sc
1


Sa

γ, etc
 ; γA = γ0 − 2
p − p (q + r) p + qr

Proof. Forγ, see the Newton section. For γ0 , see Section 12.7. For γA , make a change of algebraic
basis.
March 30, 2017 14:09
published under the GNU Free Documentation License
22. Quadrilateral
22.3
257
Pedal LFIT and orthopole
Definition 22.3.1. The orthopole of a line ∆ 6= L∞ , is the Neuberg center of the FLIT provided
by the pedal triangles of its points. Some orthopoles are given in Table 1.1.
Proposition 22.3.2. The orthopole E (∆) of a line, and its slowness center, are given by:
S (∆) ' isogon
M · t∆
where M is defined by (7.17) and N by :.
N =1−
2
O · L∞
L∞ · O
; E (∆) '

t
M · t∆ ∗ ∆ · N
(22.2)

−a2 Sa

−b2 Sb 
Sa Sb
(22.3)
Sb Sc
1 
2
=
−b
Sb

4S 2
2
−c Sc
b
−a2 Sa
Sc Sa
−c2 Sc
Proof. Denote the line by ∆ ' [p, q, r]. Describe its points as M = (0 : −r : q) + (q − r) t (L∞ ∧ ∆)
and consider the set of the pedal triangles of M (t). Their normalized expression are:


t
t
rSc
Sb q
0
− −
+


g (q − r) b2
h (q − r) c2


2
 t

r
t
−c
r
+
S
q
a


−
0
− +
T (t) = 

2
f
q
−
r
h
(q
−
r)
c




q
−b2 q + Sa r
t
t
− +
0
−
f
q−r g
(q − r) b2
where f, g, h can be identified as the isogonal conjugate of orthodir (∆). The E formula follows.
Remark: normalization of matrices M , N is not required here.
Proposition 22.3.3. The equicenter (orthopole) E lies on the reciprocal to the image of ∆ in the
circumcenter O.
Proof. Matrix N takes that image. And then, point pu : qv : rw belongs to line [1/u, 1/v, 1/w]
when p + q + r = 1. And, here, M · t ∆ ∈ L∞ .
Proposition 22.3.4. Eponymous property. Let A0 be the projection of A onto ∆. Then line
EA0 is orthogonal to BC, etc.
Proof. Straightforward computation (E is named O in Figure 22.1).
A
O
D
C'
A'
B'
B
C
Figure 22.1: Point U is the orthopole of line A0 B 0 C 0
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
258
22.4. Some properties of the Simson lines
Proposition 22.3.5. Point S (∆) belongs to the circumcircle, and its Simson line is parallel to
∆. Point S 0 = 2O − S is the isogonal conjugate of the direction of ∆. The Simson line of S 0 is
perpendicular to ∆ and, moreover, this line goes through E, the orthopole of ∆.
Proof. Everything is obvious, except the last point. And every computation is straightforward.
One can also follow the Steiner movie, given at Corollary 9.1.4.
Exercise 22.3.6. Vertex-Miquel circles. In the pedal-LFIT relative to L0 , the circles (A, bt , ct ) are
the circles having [A, Mt ] as diameter. Therefore the FA , etc points of Proposition 21.5.2 are the
projection of the vertices on L0 . The Vertex-Miquel point is simply Mt ... whose locus is nothing
than line (FA , FB , FC ) = L0 , whose direction is the orthopoint of S ∗ . This doesn’t contradict
the former assumption S ∗ ∈ cycle (FA , FB , FC ) since this cycle is L∞ ∪ L0 ! And S ∗ remains the
perspector of triangles ABC and FA FB FC .
Proposition 22.3.7. When using the Lubin-1 representation, the orthopole transform is:
q
r
s2
q
ps3
+ s1 − : 2 :
+
−
[p, q, r] 7→
r
p
ps3
s3
r
Proof. Direct computation is easy. One can prefer a change of basis applied to (22.2).
22.4
Some properties of the Simson lines
Proposition 22.4.1. Simson line. The pedal vertices of a point U are collinear if and only if
the point is on the circumcircle or on the line at infinity. When it exists, this line is called the
Simson line of U . When U ∈ L∞ , Simson (U ) = L∞ . When U belongs to the circumcircle of
ABC, the barycentrics and the Lubin-1 coordinates of it’s Simson line are
b2 −c2
1
1+t
−t
Ut ' a2 :
:
7→ Simson (U ) ' 2
,
,
(22.4)
t 1+t
a t + Sc Sc t + b2 Sb t − Sa
U ' τ : 1 : τ −1
7→ Simson (U ) ' [2 τ 2 , −τ 2 s1 − τ 3 + τ s2 + s3 , −2 s3 τ ]
Proof. Collinearity condition is the same as for the Steiner line. Everything else is either the result
of straightforward computations, or obtained from the Steiner line equation (9.1), right multiplied
by h (U, 2) of (7.25).
Proposition 22.4.2. The intersections of the U -Simson line and the nine points circle γ are (i)
the midpoint M of [U, X4 ] (ii) the intersection L of Simson (U ) and Simson (U 0 ) where U 0 is the
Γ-antipode of U .
Proof. By definition, Simson (U ) is obtained from Steiner (U ) by the point-transformation h (U, 1/2).
Since X4 ∈ Steiner (U ), point M is on Simson (U ). Using now h (X4 , 1/2), U ∈ Γ becomes M ∈ γ
and (i) is proved. Therefore midpoint M 0 of U 0 X4 is the γ-antipode of M , while LM ⊥ LM 0 .
Proposition 22.4.3. When ∆ ' [p, q, r] is a Simson line, then we have:
p q 2 + r2 Sa + q p2 + r2 Sb + r p2 + q 2 Sc − 2 Sω pqr = 0
(22.5)
In other words, tripolar (∆) belongs to the Simson cubic K010.
Proof. See Section 17.11.
Proposition 22.4.4. The complex affix of the contact point of a Simson line with its envelope
(the so-called Steiner’s deltoid) is
1
s3 −2
s1 + τ +
τ
2
2
Thus centered at X(5), cups at ρ = 3/2, innermost points at ρ = 1/2.
Proof. This comes from Simson (τ ) ∧
March 30, 2017 14:09
∂
∂τ Simson (τ ).
published under the GNU Free Documentation License
22. Quadrilateral
22.5
259
Rigby points
Proposition 22.5.1. Given three distinct points Uj on the circumcircle, the following conditions
are equivalent:
1. The three Simson lines are concurrent in a point K.
2. Simson line of Uj is orthogonal with line Uj+1 Uj−1
3. Two sidelines of U1 U2 U3 have the same orthopole K (and therefore the third too).
In such a case, the three points Uj are said to form a "Rigby triangle" , and K is their
Rigby result. When using Lubin-1 representation, the condition is τ1 τ2 τ3 = αβγ, and z (K) =
1
2 (α + β + γ + τ1 + τ2 + τ3 ).
Proof. All three properties lead to the same condition, and the same value of z (K).
Exercise 22.5.2. Prove that K is the midpoint of [α + β + γ ; τ + τ2 + τ3 ] and conclude (Honsberger, 1995, p. 136).
Remark 22.5.3. In ETC, the third of a Rigby triangle is called the Simson-Rigby point of the
first two, and noted U3 = SR (U1 , U2 ), while their common result is called the Rigby-Simson
point and noted K = RS (U1 , U2 ) .
Corollary 22.5.4. When using barycentrics, and U1 ' p : q : r, U2 ' u : v : w ∈ Γ, we have:
SR (P, U ) = isog ((P ∧ U ) ∧ L∞ )
=
b2
c2
a2
:
:
(q + r) u − (v + w) p (p + r) v − (u + w) q (p + q) w − (u + v) r
(22.6)
One can also use the Peter Moses (2004/10) expression :
U3
'
ru − pw
pv − uq
qw − rv
: 2
: 2
2
2
2
rwb − qvc
c up − a wr a qv − b2 pu
Example 22.5.5. Centers X(2677) to X(2770) are examples of SR and RS points. In the following
table, the first three of a quadruple is a (sorted) triangle U1 U2 U3 and its Rigby point.
74
74
74
74
74
98
98
98
99
99
99
98
99
110
1113
1294
110
843
1379
110
111
1379
691
842
477
1114
1304
842
1296
1380
691
843
1379
?
?
3258
125
?
2682
?
115
?
?
?
99
99
100
100
100
100
100
100
100
101
102
1380
2378
101
104
105
109
110
1381
1382
109
103
1380
?
2379
?
1308
?
953 3259
840
?
2222
?
1290
?
1381
?
1382
?
929 1521
929
?
102 104 2222
?
103 104 1308
?
104 840 1292
?
104 1381 1382
11
107 110 1304
?
110 110 476 1553
110 112 935 1554
110 827 1287
?
110 930 1291
?
110 1113 1113
?
110 1114 1114
?
Proposition 22.5.6. Third point. For each point U on the circumcircle, it exists exactly one
other point U2 such that SR (U, U3 ) = U . This point is the isogonal conjugate of the orthopoint of
line U ,X(3), belongs to line U U ∗ and is given by :
third (U )
=
=
a2
b2
c2
: 2
: 2
(bw + cv) (bw − cv) v (cu + aw) (cu − aw) w (av + bu) (av − bu)
a2
b2
c2
: 2
: 2
2
2
2
2
2
2
(b − c ) u + (v − w) a
(c − a ) v + (w − u) b
(a − b ) w + (u − v) c2
u2
Proof. The limit of line U1 U2 is orthogonal to line U ,X(3). Everything else follows. Remark:
relation SR (U, third (U )) = U is not granted for a random point U in the triangle plane. But this
relation obviously holds when restricting U to the circumcircle.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
260
22.6. Sister Marie Cordia Karl
Proposition 22.5.7. Simson-Moses point. If points U1 , U2 are on the circumcircle then, us∗
ing isoconjugacy wrt pole P = X6 (isogonal conjugacy), the intersection of lines U1 (U2 )P and
∗
U2 (U1 )P is point SR (U1 , U2 ). When another isoconjugacy is used, the intersection remains on the
circumcircle. Its barycentrics are :
SP∗ (U1 , U2 ) =
−w1 v2 + v1 w2 −w1 u2 + u1 w2 −v1 u2 + u1 v2
:
:
qw1 w2 − rv1 v2 pw1 w2 − ru1 u2 pv1 v2 − qu1 u2
Proof. The barycentrics are straightforward, while parameterization (7.15) leads to the other properties.
Definition 22.5.8. In ETC, the Simson-Moses point is computed using P = X2 (isotomic conjugacy) in SP∗ (U1 , U2 ), and noted SM (U1 , U2 ). Centers X(2855) to X(2868) are examples of
Simson-Moses points.
22.6
Sister Marie Cordia Karl
Proposition 22.6.1. Let L = [u, v, w] and ∆ = [p, q, r] be two lines. The condition for the
orthopole of L belongs ∆ can be written as
where
∆ · H · V (L) = 0

2 a2 Sb Sc 2 b2 Sb Sc 2 c2 Sb Sc

H '  2 a2 Sc Sa 2 b2 Sc Sa 2 c2 Sc Sa
2 a2 Sa Sb 2 b2 Sa Sb 2 c2 Sa Sb
Proof. Obvious from (22.2).

b2 Sb2 + c2 Sc2
−Sc a2 c2 + Sb2
−Sb a2 b2 + Sc2

−Sc b2 c2 + Sa2
a2 Sa2 + c2 Sc2
−Sa a2 b2 + Sc2 
−Sb b2 c2 + Sa2
−Sa a2 c2 + Sb2
a2 Sa2 + b2 Sb2
t
V ([u, v, w]) ' u2 , v 2 , w2 , 2vw, 2wu, 2uv
Definition 22.6.2. Map L 7→ V (L) is the full-Veronese map (whose signature is row(3) gives
column(6): we are building conics here, not circles). For a given ∆, equation ∆ · H · V (L) = 0
is the equation of a tangential conic. We call it the associated parabola of ∆ and note it: H (∆),
while we call H the Sister Mary Cordia Karl’s matrix, due to Karl, E. (Sister Mary Cordia), 1932.
.
Proposition 22.6.3. Conic H (∆)is a parabola. Its point at infinity is H∞ = M · t ∆, while its
focus Hω is the isogonal conjugate of H∞ . Moreover, its directrix (polar of the focus) is homothetic
to ∆ from X(4) (ratio=2).
Proof. Compute H (∆) · t L∞ and recognize the orthodir M · t ∆, proving the contact. Then write
that the isotropic lines of the focus belongs to the conic, and obtain the given result. For the last
point, you find it by computing the directrix, obtaining a linear formula, extracting its matrix and
t
recognizing homot (H, 2)... but you prove it simpler by H (∆) · (∆ · h (H, 1/2)) ' Hω , using (7.25)
(and 1/2, we are acting on lines, not on points).
Example 22.6.4. Taking the sidelines and L∞ as examples, one obtains:
H (BC) (u, v, w) = −a2 u + Sc v + Sb w Sb Sc u − b2 Sb v − c2 Sc w
H (CA) (u, v, w) = −b2 v + Sa w + Sc u Sc Sa v − c2 Sc w − a2 Sa u
H (AC) (u, v, w) = −c2 w + Sb u + Sa v Sa Sb w − a2 Sa u − b2 Sb v
H (L∞ ) (u, v, w) ' u (Sb − 2 iS) + v (Sa + 2 iS) − c2 w u (Sb + 2 iS) + v (Sa − 2 iS) − c2 w
the last one being the conic of isotropic lines.
Proposition 22.6.5. The H (∆) conic is degenerate if and only if ∆ is the Simson line of some
point M (on Γ, the circumcircle). In such a case, the pivots are M and its isogonal conjugate (the
orthodir of ∆, on L∞ ).
Proof. Compute the determinant and obtain (22.5). Since this equation is invariant by isotomic
conjugacy, this tells us that tripolar (∆) is on K010, so that ∆ is a Simson line. Conversely,
computing the conic from parameterization (22.4) of Simson (Mt ), leads to Mt and Mt∗ ∈ L∞ .
March 30, 2017 14:09
published under the GNU Free Documentation License
22. Quadrilateral
261
Figure 22.2: Revisiting the Sister Mary Cordia Karl’s correspondence
Remark 22.6.6. When points are opposite on Γ, Simson lines are orthogonal and also reciprocate
(characteristic property).
Proposition 22.6.7. By a generic point, three Simson lines are going through. By the orthopole
of a known line L, we have the Simson line orthogonal to L, given by:
p−q
q−r
r−p
p−r
,
∆1 '
,
; t=−
2
2
2
Sb r + Sc q − a p Sa r + Sc p − qb Sa q + Sb p − c r
q−r
and the Simson lines of the points of Γ ∩ ∆1 .
Proof. Degree of equation is 3. For an orthopole, the equation splits. The first degree factor leads
to ∆1 . The other part is exactly the condition for Mt ∈ L.
22.7
The four pedal LFIT of a quadrilateral
Proposition 22.7.1. Consider the four LFIT created by the pedal triangles of points on line Lj
wrt the trigon Tj of the other three lines. The four slowness centers Sj are on the Miquel circle,
while the four Neuberg centers Ej are on the Steiner axis.
Proof. Computation to obtain S0 ∈ M iquel and E0 ∈ Steiner, symmetry to obtain the rest. One
can also use the fact that S0 is the perspector of ABC and Oa Ob Oc .
Corollary 22.7.2. Point Sj is therefore the second intersection of the Miquel circle (O0 , Oa , Ob , Oc )
and circumscribed circle (Oj ) , while Ej is the intersection of the Simson line of 2Oj − Sj with the
line (H0 , Ha , Hb , Hc ).
Proposition 22.7.3. When point M moves along the line L0 , the orthopole E0 has a constant
power ρ2 with respect to the pedal circle of M (shorten version of the circle circumscribed to the
pedal triangle of M ) wrt T0 = ABC. The value of ρ2 is


 

 

 


a2 Sa
−a2
Sc
Sb


 

 

 


∆ ·  b2 Sb  ∆ ·  Sc  ∆ ·  −b2  ∆ ·  Sa 
Sa
c2 Sc
Sb
−c2
ρ2 =
(22.7)
2
16 S 4 ∆ · M · t ∆
Moreover ρ2 is (-2) times the algebraic product of dist (O, ∆) and dist (E, ∆).
Proof. Start from the columns of triangle T (t) , compute the Veronese rows and take their wedge.
Then obtain ρ2 (t) ... and see that all the t cancel. Moreover, we have the formulas:
√
Sa a2 p + b2 Sb q + c2 Sc r
q
dist (∆, O) =
2S
8 S2 ∆ · M · t∆
Q
√
Sb r + Sc q − a2 p
dist (∆, E) =
2S
q
3
t
3
8S
∆· M · ∆
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
262
22.8. Exercises
And we can check the homogeneous degrees: in a: (4 + 2 + 2 + 2) − 8 = 2 (since ρ is a length); in
p: 4 − 4 (since ∆ is projectively defined).
Corollary 22.7.4. When point M moves along line LA , the orthopole EA has a constant power
ρ2A with respect to the pedal circle of M wrt trigon TA .
pa2 (p − q − r) Sa − Sb b2 pr − Sc c2 pq + 8 S 2 rq p Sb Sc −a2 p + Sb r + Sc q
2
×
ρA =
4a S (p − r) (p − q)
a3 S (p − r) (p − q)
Proof. Simply using the product of distances. One can also recompute everything, and/or using
an algebraic change of parameters, followed by a collineation.
Proposition 22.7.5. When point M moves along Lj , its pedal circumcircle wrt triangleqTj remains
orthogonal to a fixed circle, the orthopedal circle, centered at Ej and whose radius is
circle is virtual when Oj , Ej are on the same side of Lj .
ρ2j . This
Proof. Immediate from the invariance of ρ2j . See also Thebault, 1946.
Proposition 22.7.6. The four orthopedal circles belong to the Steiner pencil (generated by the
conjugate circles).
Proof. This circle is obviously orthogonal to ν (the Steiner axis). It suffices to see that ρ2 is the
power of point E0 wrt one circle of the Newton pencil (this pencil is the orthogonal pencil of the
Steiner pencil).
22.8
Exercises
Exercise 22.8.1. Use the Lubin-1 representation, and write the transversal as the t-locus of
ikτ + tτ : 1 :
−ik
t
+
τ
τ
where k is real and τ is a turn. Recompute everything, especially dist (∆, E)... that only depends
on τ (see Goormaghtigh, 1939).
Exercise 22.8.2. Point O0 is the perspector of SA , SB , SC with ABC.
Exercise 22.8.3. Let P 0, P a, P b, P c be the projections of M q on the sidelines. There is a similarity P a 7→ A0 , P b 7→ B 0 , P c 7→ C 0 . Elaborate further.
Exercise 22.8.4. Let be Q0, Qa, Qb, Qc the reflections of M q about the sidelines.
The similarity σ defined by Qa 7→ A0 , Qb 7→ B 0 is centered at M q, maps the Steiner line onto L0 ,
sends Qc to C 0 , the circle ABC onto the Miquel circle and A 7→ SA , etc.
Exercise 22.8.5. When points U1 and U2 are antipodes on the circumcircle, the orthopole of
Simson (U1 ) is the intersection of Simson (U2 ) with Steiner (U1 ).
Exercise 22.8.6. When a line goes through the circumcenter, its orthopole belongs to the nine
points circle.
Exercise 22.8.7. When a line goes through a fixed point P , its orthopole belongs to the conic
centered at (P + H) /2 and passing through the projections of P on the sidelines.
Proposition 22.8.8. When X moves on a line through the centroid X2 , then orthopole of trilipo (X)
moves on a line through the orthocenter X4 . For example :
X on line
L(2,1)
L(2,3)
L(2,6)
L(2,7)
March 30, 2017 14:09
orthopole (trilipo (X)) on line
L(4,9)
L(4,6)
L(4,3)=L(2,3)
L(1,4)
published under the GNU Free Documentation License
22. Quadrilateral
263
Proof. Suppose X is not X2 and does not lie on a sideline of triangle ABC. Then, using barycentrics,
we have :
orthopole(trilipo(X)) ∧ X4 =
Sa (y − z) : Sb (z − x) : Sc (x − z) = (Sa : Sb : Sc ) ∗ (X ∧ X2 )
b
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
264
March 30, 2017 14:09
22.8. Exercises
published under the GNU Free Documentation License
Chapter 23
Combos
23.0.1
More-more combos
Many triangle centers can be defined (as just above) by the form Xcom(nT), where T denotes a
central triangle. In order to present such centers, it is helpful to introduce the notation T(f(a,b,c),
g(a,b,c)) for central triangles. Following TCCT, pages 53-54, suppose that each of f(a,b,c) and
g(a,b,c) is a center-function or the zero function, and that one of these three conditions holds:
the degree of homogeneity of g equals that of f; f is the zero function and g is not the zero
function; g is the zero function and f is not the zero function.
There are two cases to be considered: If g(a,b,c)=g(a,b,c), then the central triangle T(f(a,b,c),
g(a,b,c)) is defined by the following 3x3 matrix (whose rows give homogeneous coordinates for the
A-, B-, C- vertices, respectively):
f(a,b,c) g(b,c,a) g(c,a,b) g(a,b,c) f(b,c,a) g(c,a,b) g(a,b,c) g(b,c,a) f(c,a,b)
If g(a,b,c) is not equal to g(a,b,c), then the central triangle T(f(a,b,c),g(a,b,c)) is defined by
the following 3x3 matrix:
f(a,b,c) g(b,c,a) g(c,b,a) g(a,c,b) f(b,c,a) g(c,a,b) g(a,b,c) g(b,a,c) f(c,a,b)
If the homogenous coordinates are chosen to be barycentric, as they are for present purposes,
then the new notation for central triangles is typified by these examples:
Medial triangle = T(1, 0); Anticomplementary triangle = T(-1, 1) Incentral triangle = T(a,
0); Excentral triangle = T(-a, b) Euler triangle = T(2(b4 + c4 - b2c2 - c2a2 - a2b2), (b2 + c2 a2)(a2 + b2 - c2)) 2nd Euler triangle = T(2a2(b4 + c4 - b2c2 - c2a2 - a2b2), (b2 - c2 - a2)(a4 +
b4 + c4 - 2a2c2 - 2b2c2)) 3rd Euler triangle = T((b - c)2, c2 - a2 - bc) 4th Euler triangle = T((b
+ c)2, c2 - a2 + bc) 5th Euler triangle = T(2(b2 + c2), b2 + c2 - a2 ) (The vertices of the five
Euler triangles lie on the nine-point circle.)
The notation F(f(a,b,c),g(a,b,c)) can be used to define several more triangles; in each case, two
of the perspectivities can be used to construct the triangle.
T(bc, b2) is perspective to ABC with perspector X(6); other such pairs: incentral, X(192),
excentral, X(1045); tangential, X(6); anticomplementary, X(192)
T(-bc, b2) is perspective to ABC with perspector X(6); incentral, X(2); tangential, X(6);
T(bc,b2), X(6)
T(bc, c2) is perspective to ABC with perspector X(76); anticomplementary, X(8); 4th Brocard,
X(76), Fuhrmann, X(8); cevian(X(350)),X(1)
T(-bc, c2) is perspective to ABC with perspector X(76); cevian(X(321)),X(75); anticevian(X(8)),
X(2)
T(a2, a2 - b2) is perspective to the medial triangle with perspector X(3); tangential , X(3);
symmedial, X(2); cevian(X(287)), X(98)
T(a2, a2 - c2) is perspective to the tangential triangle with perspector X(25); orthic , X(25);
medial, X(6); 1st Lemoine, X(1383); circummedial, X(251)
T(a2, a2 + b2) is perspective to ABC with perspector X(83); medial, X(6); 1st Neuberg, X(182)
T(a2, a2 + c2) is perspective to ABC with perspector X(141); medial, X(3); 1st Neuberg,
X(182); T(a2,a2 - c2), X(2); cevian(X(69)), X(6); cevian(X(76)), X(2); cevian(X(3313)),X(39)
The triangles IT1 and IT2 are triply perspective to ABC (as are T1 and T3). Order-label IT1
as A’B’C’.
T The
T three perspectivitiesTare then
T given by
AA’ BB’ CC’ = X(3862) AB’ BC’
T CA’
T = f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) =
ab(a2 - bc)(c2 - ab)(b2 + bc + c2) AC’ BA’ CB’ = f(a,c,b) : f(b,a,c) : f(c,b,a)
265
266
23. Combos
Peter Moses noted (12/23/2011) that T(bc, b2) is triply perspective to ABC. With T(bc, b2)
order-labeled as A’B’C’, the three perspectivities with perspectors are as follows:
T
T
T
T
T
T
AA’ BB’ CC’ = X(6); AB’ BC’ CA’ = P(6); AC’ BA’ CB’ = U(6). (The notation P(k)
and U(k) refers to a bicentric pair of points; see More at the top of ETC.) Likewise, T(bc, c2) is
triply perspective to ABC, with perspectors: X(76), P(10), and U(10).
(This section was added to ETC on 12/23/2011.)
23.0.2
More-more-more combos
The Euler triangle and the 2nd, 3rd, 4th, and 5th Euler triangles are discussed in the preamble to
X(3758), along with eight other central triangles using the notation T(f(a,b,c), g(a,b,c)). Recall
that the inverse of a normalized central triangle T, denoted by Inverse(nT) or Inverse(n(T)), is
also a normalized central triangle. Barycentrics for the inverse of each of the 13 triangles and
properties of these triangles are given (Peter Moses, December 2011) as follows:
Inverse(n(Euler triangle)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = (b2 - c2)2 + a2(3a2 - 4b2 4c2) g(a,b,c) = b4 - (c2 - a2)2
Inverse(n(2nd Euler triangle)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = a2(a2 - b2 - c2)(a6 +
b6 + c6 - a4b2 - a4c2 - a2b4 - a2c4 - 2a2b2c2 - b4c2 - b2c4) g(a,b,c) = b2(b2 - c2 - a2(a6 + b6 c6 - a4b2 - 3a4c2 - a2b4 + 3a2c4 + 2a2b2c2 - 3b4c2 + 3b2c4)
Inverse(n(3rd Euler triangle)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = a(b2 + c2 - ab - ac +
bc) g(a,b,c) = b(a2 - ab - ac + bc - c2)
Inverse(n(4th Euler triangle)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = a(a - b - c)(b2 + c2 +
ab + ac) g(a,b,c) = b(-a + b + c)(a2 + ab - ac - bc - c2)
Inverse(n(5th Euler triangle)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = (a2 - 2b2 - 2c2)(3a2 +
b2 + c2) g(a,b,c) = -(a2 - b2 - c2)(2a2 - b2 + 2c2)
Inverse(n(T(bc,b2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = -bc(a2 - bc)(b2 + c2 + bc) g(a,b,c)
= b2(ab - c2)(a2 + c2 + ac)
Inverse(n(T(-bc,b2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = bc(a2 - bc)(b2 + c2 - bc) g(a,b,c)
= b2(ab + c2)(a2 + c2 - ac)
Inverse(n(T(bc,c2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = -a(a2 - bc)(b2 + c2 + bc) g(a,b,c)
= (ab - c2)(a2 + c2 + ac)
Inverse(n(T(-bc,c2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = -a(a2 - bc)(b2 + c2 - bc) g(a,b,c)
= (ab + c2)(a2 + c2 - ac)
Inverse(n(T(a2,a2 - b2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = (3a2 - b2 - c2)(b4 + c4 - b2c2)
g(a,b,c) = (a2 - 3b2 + c2)(a2b2 - 2b2c2 + c4)
Inverse(n(T(a2,a2 - c2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = a2(b2 + c2 - 3a2) g(a,b,c)
=(a2 - c2)(a2 - 3b2 + c2)
Inverse(n(T(a2,a2 + b2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = -(3a2 + b2 + c2)(b4 + c4 +
b2c2) g(a,b,c) = (a2 + 3b2 + c2)(a2b2 + c4)
Inverse(n(T(a2,a2 + c2)) = T(f(a,b,c), g(a,b,c)), where f(a,b,c) = -a2(a2 + b2 + c2)(3a2 + b2
+ c2) g(a,b,c) = (a2 + c2)(a2 + b2 - c2)(a2 + 3b2 + c2)
For present purposes, label these last triangles as IT1, IT2, IT3, IT4, IT5, IT6, IT7, IT8; then
all except IT5 and IT6 are perspective to the reference triangle ABC; perspectors are described
at X(3862)-X(3866). Label the corresponding original triangles T1, T2, T3, T4, T5, T6, T7, T8;
then the following pairs are perspective: (T1, IT1), (T2, IT2), (T3,IT3), (T4, IT4), (T7, IT7),
(T8,IT8); coordinates for the perspectors are long and omitted.
The triangles IT1 and IT2 are triply perspective to ABC (as are T1 and T3). Order-label IT1
as A’B’C’. The three perspectivities are then given by
T
T
T
T
AA’ BB’ CC’ = X(3862) AB’ BC’
T CA’
T = f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) =
ab(a2 - bc)(c2 - ab)(b2 + bc + c2) AC’ BA’ CB’ = f(a,c,b) : f(b,a,c) : f(c,b,a)
Note that the second two perspectors are a bicentric pair (not triangle centers; see TCCT, page
47); likewise for the other triple perspectivity, with IT3 order-labeled as A’B’C’:
T
T
T
T
AA’ BB’ CC’ = X(3864) AB’ BC’
T CA’
T = g(a,b,c) : g(b,c,a) : g(c,a,b), where g(a,b,c) =
a(a2 - bc)(c2 - ab)(b2 + bc + c2) AC’ BA’ CB’ = g(a,c,b) : g(b,a,c) : g(c,b,a)
underbar
March 30, 2017 14:09
published under the GNU Free Documentation License
23. Combos
23.0.3
267
more-more-more-more combos
(Pending: will tell constructions for the following triangles and their inverse normalizations) T(a,
c)
T(-a, c)
T(a, b - c)
T(a, c - b)
T(a2, bc)
T(-a2, bc)
T(bc, bc + c2)
T(-bc, bc + c2)
T(bc, b2 + bc)
T(-bc, b2 + bc) )
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
268
March 30, 2017 14:09
23. Combos
published under the GNU Free Documentation License
Chapter 24
Import the 4995-7000 part of ETC
In this chapter, we give the specifications of our implementation of the Kimberling ETC ETC. At
the present moment, this implementation is restricted to a private use. The intent is to participate
to a distributed maintenance of the database, together with providing "random search keys",
specific to each local copy, in order to allow a distributed method of proof.
At the moment, our main result is the text-only file~/etc/bar_igcta.csv that can be read
and parsed by Maple.
24.1
Requirements
1. A web server is required. We are using apache2. Should start chrooted (user= wwwrun,
#=30) on boot.
List of active modules: access_compat, actions, alias, auth_basic, authn_core, authn_file,
authz_core, authz_groupfile, authz_host, authz_user, autoindex, cgi, core, dir, env, expires,
http, include, log_config, mime, mpm_prefork, negotiation, reqtimeout, setenvif, so, ssl,
systemd, unixd, userdir, status, info php5
Module info is not so easy to launch (don’t list its name at the end, but at the beginning of
the list).
Inside httpd.conf: DirectoryIndex index.html index.php index.html.var
2. A php server is required. Here, we have php5. As said before, php5 must be loaded by the
web server. Remember that *.php scripts are run by wwwrun. This user should receive all
the required permissions to write or execute (when required).
3. A database server is required. We are using mysql. Should start chrooted (user= mysql,
#=60) on boot. Activation by the run-levels.
4. A web interface with mysql is required. We are using phpMyAdmin. The config file is
/etc/phpMyAdmin/config.inc.php. Many things are ugly there. Many "__" and "_".
5. Everything must be functional before trying to work with the ETC database. A good test is
to create/work with the amarok database.
6. Dare to create security holes ! Users mysql and wwwrun are chrooted and must remain
chrooted ! Acceptable are links... when wwwrun is really chrooted and cannot use links to
escape outside.
7. Finally, don’t forget the HAL line when using the Listing environment + prettyref + hyperref
in LYX.
\def\arraystretch{1.2}
\makeatletter
\floatstyle{boxed}
\restylefloat{algorithm}
\def\fname@algorithm{{\scshape Listing}}
\def\fnum@algorithm{\fname@algorithm~\thealgorithm\,\!}
269
270
24.2. Recreating the database
\def\theHalgorithm{\thealgorithm}%%{ HELLO, I am HAL }
\makeatother
24.2
24.2.1
Recreating the database
The context
1. The database is called etcetc and created empty.
2. The user is called etcetc. Privileges are: (1) everything on etcetc from localhost ; (2) FILE
privilege from localhost.
3. The old archived version from mirabel is center-2012-02-17.sql.gz. The etcetc user (at
madiran) imports center-2012-02-17.sql.gz. Using a compressed version is mandatory
(in order to avoid a file too long error. maxsize= 2048 Ko). This creates the main table,
called center. The purpose of this table is to be filled by records reformatted from the ETC
web page. Encoding utf-8. Actualsize= 869 Ko. Message: 77 queries executed. 3899 lines.
4. Therefore, etcetc@mirabel create another export-file, using properties
Add custom comment into header (\n splits lines)
Add IF NOT EXISTS
Add AUTO_INCREMENT value
Enclose table and field names with backquotes
Data: Complete inserts, Extended inserts
Maximal length of created query 50000
Use hexadecimal for BLOB
Export type= insert
Dump all rows
Save as file "bzipped"
5. Then etcetc@madiran has to drop the wrong table etcetc (truncate would keep the old structure). Then etcetc@madiran imports center-2015-03-29.sql.bz2.
Encoding utf-8. Actualsize= 836.6 Ko. 97 queries. 4994 lines.
6. Caveat: user etcetc can export to file using the phpMyAdmin frontal, but not directly using
mysql (and the scripts). Unless receiving the global ’FILE’ property.
24.2.2
Encapsulating everything in *.php commands
1. Writing for eternity implies to keep track of each action. Writing explicit programs is the
best way for that.
2. A file named index.php is somewhere in a file tree that is mapped into a web tree ~/public_html/etc.
This file contains a link to each command file. These *.php files are only writable by ipse,
and stored in a specific directory. Database is somewhere else (in the protected mysql space
/var/lib/mysql/etcetc).
3. These scripts create auxiliary tables or files, import files into temporary tables and commit
some changes from auxiliary tables into the main table (center)..
4. An individual command file is as listed in Incl. 24.1. The core of the file is the mysql line of
command (2). When everything is written by *.php, the redirection is in line (4). Otherwise,
the mysql messages are directed to the web page. When a post-treatment is required, it can
be embedded in an external routine (6).
5. a list of items is build that way :
drop table if exists lesqui;
create table lesqui select 1 as xnum ;
insert into lesqui (xnum) values (829),(1105),(1288),...
March 30, 2017 14:09
;
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
1:
2:
3:
4:
271
<html> <head> <title> structure </title></head>
<body> <h1><center> structure </center></h1>
<a href="../index.php">back to index</a> <br> <hr>
<?php $ici="/home/ipse/public_html/etc";
$cmd = <<<EOF
use etcetc; describe center ;
EOF;
$fp = fopen("$ici/tmp_file.cmd", ’w’);
fwrite ($fp, $cmd); fclose($fp);
$now = <<<EOF mysql -uetcetc -p****** -v --show-warnings
< \$ici/tmp_file.cmd > $ici/tmp_structure.txt
5:
EOF;
echo "----------- mysql -------------<br>\n";
echo $now ."<br><br>\n"; exec(\$now, $retour);
$count = count($retour);
for ($i = 0; $i < $count; $i++)
{ echo $retour[$i] . "<br>\n"; }
echo "----------------------------------<br>\n";
6:
system("$ici/.structure.bat", $retour);
7:
echo "-structure is done-$retour---------------<br>\n";
?> </body> </html>
Listing 24.1: The structure.php file
24.2.3
Main view of the structure
The purpose of the center table is to contain the original records from the ETC web page...
together with the clean data obtained after some extra work. This table contains each and every
information.
The evolving structure of this main table is described in Table 24.1.
24.2.4
The main fields
These are the fields that are extracted towards the bar_igtca.csv file, so that these fields are
read each time Maple is launched.
1. field num is identifier, i.e. the n in the X(n) value.
2. Today (2015/03/27), the fields extracted are: num, ff, gg, tt, cc, ac, bary1, dg, lubin, gerade,
xx,yy,zz,zxx,zyy,zyy, hc.
3. bary1, in strict Maple syntax, is the first barycentric coordinate without any spaces. The
main variables are a, b, c. But quantities S (area), Sa , Sb , Sc (the Conway symbols) are also
used while R is deprecated. Some radicals should be specially quoted.
4. lubin, in strict Maple syntax, is the Lubin affix while dg is the corresponding degree. Variables
are s1,s2,s3,s4
5. boolean : ff= 1 for points at infinity (#229)
6. links are: gg (isogon), tt (isotom), cc (complem), ac (anticomplem), ho (orthopoint).
7. gerade is a longtext listing the lines that contain the point. The syntax used is "1_291=2_76=3_6=4_232=5_11
: "=" between two lines and "_" between the two points defining the line. This list is sorted
according to the head point. In ETC, line "1_8" is recorded as "2_8" in the X(1) list.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
272
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
24.2. Recreating the database
Field
num
name
ff
gg
tt
cc
ac
bary1
dg
lubin
sf
mf
ho
bary2
bary3
bary4
baryf
triliasis
baryasis
combo
lg
gerade
curves
orthj
iccir
ibroc
iortc
iflec
ibevc
iincc
islec
inine
Type
int
varchar
tinyint
smallint
smallint
smallint
smallint
varchar
int
varchar
tinyint
tinyint
smallint
varchar
varchar
varchar
varchar
mediumtext
mediumtext
varchar
smallint
varchar
varchar
smallint
smallint
smallint
smallint
smallint
smallint
smallint
smallint
smallint
5
60
1
6
6
6
6
500
11
500
1
1
6
400
100
100
10
100
6
1500
100
6
6
6
6
6
6
6
6
6
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
Field
ifuhr
circa
cirna
cocj
accj
cycc
reflex
crosc
crosp
crosa
crosd
hirst
alph
beth
midp
ceva
cevc
licon
eigct
eigat
lam
psi
refs
residuel
xx
yy
zz
zxx
zyy
zzz
hc
Type
smallint
smallint
smallint
smallint
smallint
smallint
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
varchar
mediumtext
longtext
varchar
varchar
varchar
varchar
varchar
varchar
varchar
6
6
6
6
6
6
500
250
100
300
100
150
250
250
250
250
250
250
250
250
50
50
30
30
30
30
30
30
30
Table 24.1: Structure of the database
8. Quantities xx,yy,zz,zxx,zyy,zzz are the coordinates in the K and the Z systems. They are
given with 20 decimals.
9. Finally, hc is the searchkey.
24.2.5
Ancilary fields
1. booleans: sf= 1 for the 6 special points, mf= 1 for the bare and Morley points (#33).
2. ho is orthopoint
3. bary2, bary3, bary4, baryf (numer of lines), triliasis, baryasis, combo: raw data from the
ETC web pages.
4. lg is the length of gerade
5. curves is a list containing X(n)
March 30, 2017 14:09
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
273
6. reflex is a list of xx_yy so that yy is the reflection of X(n) in xx.
7. refs: references
8. residuel: all that doesn’t go somewhere else.
24.2.6
Information fields
Inverses in various circles : [iccir iincc inine ibroc iortc ibevc iflec islec ifuhr] are circumcircle,
incircle, npc, Brocard, iortc, Bevan, Lemoine1, Lemoine2, Fuhrman.
orthj
orthojoin of
circa
antipode in circumcircle (special case of reflex)
cirna
antipode in nine-points circle (special case of reflex)
cocj
complementary conjugate of
accj
anticomplementary conjugate of
cycc
cyclocevian conjugate of
reflex
reflection of XI in XJ
crosc
XI-cross conjugate of XJ
crosp
crosspoint of XI and XJ
crosa
crosssum of XI and XJ
crosd
crossdifference of any
hirst
XI-Hirst inverse of XJ
alph
XI-aleph conjugate of XJ
beth
XI-beth conjugate of XJ
midp
midpoint of XI and XJ
ceva
cevapoint of XI and XJ
cevc
XI-Ceva conjugate of XJ
licon
XI-line conjugate of XJ
eigct
eigencenter of cevian triangle of XI
eigat
eigencenter of anticevian triangle of XI
lam
&Lambda;
psi
&Psi;
24.3
Obtaining compliant bar1 and hc
The first aim is to obtain certified pairs of searchkeys and Maple parseable barycentrics. Using
etcetc names, we are dealing with triples (num, bary1, hc) until they become compliant.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
274
24.3.1
24.3. Obtaining compliant bar1 and hc
Creating data records from the ETC Web Site
1. In 2012, ETC was made of two parts. Page one covering the 1-2000 part, and page two
covering the 2001-4994 part.
2. Nowadays (as of 03/2015), ETC is made of five parts:
Centers X(0001) - X(1000): PART 1
Centers X(1001) - X(3000): PART 2
Centers X(3001) - X(5000): PART 3
Centers X(5001) - X(7000): PART 4
Centers X(7001) - X(0000): PART 5
3. Icons: leftri.gif, rightri.gif, s1blue.gif, underbar.gif are usefull to read these files.
4. We use create-data-file.bat to split each web file in two parts:
• the data part, selecting lines that start by "<h3 id=". This generates
1000 103757 1195485 ETC-001.cut
2000 114528 1503577 ETC-002.cut
2000 122237 1256778 ETC-003.cut
1805 185687 1679070 ETC-004.cut
0373 026135 0256865 ETC-005.cut
• the comment part, selecting the other lines, that are gathered into dutu-files.html
5. The data part contains a line per recorded point. The script create-fiche.sh transcripts
each line into an sql request, that is send to the database. Since ETC is often hand-patched,
this script is not so straightforward.
6. Nowadays (X=7000), we use a main script create-fiche.sh to cut a record (= a line in
the *.cut file) into pieces. Part of the job is common to all records, summarized by the
tmp_trig filter, only rewritten when needed by create-tmp_trig.sh. Part of the job is
specific to a given number X(nnn) and is summarized into the tmp_nxt filter, controlled by
create-tmp_nxt.sh (called at each iteration).
7. All and everythig that was a comment is copyed into the tmp_remaining file.
24.3.2
Searchkeys
1. The 6_9_13 searchkeys are imported from http://faculty.evansville.edu/ck6/encyclopedia/
Search_6_9_13.html. Only the first column is useful. Use import-search.sh to select this
column into search-data.txt.
2. Obviously, this searchkey is a pilar of the process since improving the syntax of the barycentric
coordinates is an error prone process...
3. From the past, two searchkeys were corrected: 600 and 3020. Bad coordinates...
4. When sure, use the same barz.csv process as below.
24.3.3
Towards a Maple compliant syntax
1. Table and file barz.csv are used to rework the syntax of the barycentrics, until the string
becomes parseable and gives an expression having the right searchkey.
March 30, 2017 14:09
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
275
center
num
ff
gg
tt
if(bary1<>"", bary1, "x")
quote(bary_asis)
quote(trili_asis)
replace(if(lg<>0, left(gerade,1500),"none")," ","=")
if(bary2<>"", bary2, "x")
barz
bnum
bff
bgg
btt
bbary1
basis
btrili
bgerade
bbary2
alter
varchar(1300)
varchar(1300)
2. The request starts with DROP TABLE IF EXISTS barz. Then newlines are changed into
"QQ". And the table is exported into etc/barz.csv (terminated by ";" enclosed by ("").
3. The Maple file gerer07.mw proceeds in several steps, all resulting into a ryba.csv containing
pairs num, bary1.
(a) Keys already known to Maple (enc_dat) are checked with the keys read from the web
(skey). Then enc_dat is filled with skey.
(b) An automated process tries to proceed from bary1, firstline of basis, firstline of btrili
(11 from 21).
(c) An automated process tries to proceed from the gerade field (4 from 21)
(d) Then file ryba.csv is read into ryba table. And, in a second step, this table is imported
into the center table.
ryba
center
rnum
rbary1
num
bary1
4. Qu’èst-ce
(a) pas-create-fiche
24.4
SQL requests
• modify names in order to replace some &xxx; web characters. The html code must be hidden,
otherwise the text of the command istself is parsed and replaying is scrambled.
# SELECT num, name, replace(name,concat(char(38 using utf8), "omega;"),"omega") as qname FROM
# SET name=replace(name,concat(char(38 using utf8), "pi;"),"Pi") WHERE ‘name‘ like "%&%"
• [b^2+c^2 = -a^2+2*So]
[b^2*c^2 = a^4-2*So*a^2+4*S^2+So^2]
24.5
24.5.1
Killed points (tagged as special)
special
368,369,370,1144,3232
X(5373)
= Equiareality Center
X(5394)
= Congruent Incircles Point
X(5626)
= Center of Electrostatic Potential.
X(6768)
= Mixtilinear-incentral-to-abc Trilinear Image of X(1)
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
276
24.5. Killed points (tagged as special)
24.5.2
pending to shorten
5676
len = 736, 598
5677
len = 757, 729
5678
len = 705, 609
5679
len = 2323, 2140
5680
len = 810, 667
5681
len = 912, 823
5682
len = 911, 822
5973
len = 851
6802
len = 601, 232
March 30, 2017 14:09
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
24.6
277
Some errors in the web page
2136
Better provide a value for the barycentrics: "a*(b+c-a)*(a^2+(2*b+2*c)*a+b^2-6*b*c+c^2)"
2137
Better provide a value for the barycentrics: "a/(b-a+c)/(a^2+(2*b+2*c)*a+b^2-6*b*c+c^2)"
This one is collision prone. And is *not* incident to lines [[44, 380], x, x, [1018, 1400],
[1149, 1450], x], but *is* incident to lines [[56, 1743], [57, 145], [1407, 1420]]
4996
"a^2(b+c-a)(b^2+b^2-a^2=bc)^2" should be "a^2(b+c-a)(b^2+c^2-a^2-bc)^2"
5000
"<sup>)1/2</sup>" should be ")<sup>1/2</sup>"
X(5249)
= complement of X(3319) should be 3219
5351
a^2*(7*b^2+7*c^2-7*a^2-(12)^(1/2)*a^2*S) should be a^2*(7*b^2+7*c^2-7*a^2-(12)^(1/2)*S)
5383
is isogon, not isotom, of 6377
5403
barycentrics: "1/(Sa*(4*S^2+So^2)^(1/2)+4*S^2+Sa*So)"
5420
isogonal conjugate of X(5418) should be 5419.
5454
many errors. (1) f(4C)/3 instead of f(4C/3); (2) missing ")". Should be: "((-cos(2/3*A)+cos(4/3*A)+cos(2/
cos(4/3*B))*(csc(1/3*C+1/6*Pi)*(2*cos(1/3*C)*sec(1/3*A)+sec(1/3*B))-(2+cos(1/3*C)*sec(1/3*A)*sec
(-cos(2/3*A)+cos(4/3*A)+cos(2/3*C)-cos(4/3*C))*(csc(1/3*B+1/6*Pi)*(2*cos(1/3*B)*sec(1/3*A)+sec(
(2+cos(1/3*B)*sec(1/3*A)*sec(1/3*C))*csc(1/3*A+1/6*Pi)) )*a"
5460
extra ] at end.
5465
(b^4-c^4)^2*(b^2+c^2)*(b^2-2*c^2)*(2*b^2-c^2) should be (b^4-c^4)^2*(b^2-2*c^2)*(2*b^2c^2). Factor (b^2+c^2) is already imbedded.
5484
...-3a^2bc-3abc^2... should be ...-3ab^2c-3abc^2... (symmetry).
5497
don’t fit with the searchkey
5498
exponent should be <sup>4</sup>
5497
"a^7-a^6(b+c)-a^5(b+c)^2+a^4(2b^3+b^2c+bc^2+2c^3)-a^2(b^4-b^3c-3b^2c^2-bc^3+c^4)+abc(b^2
c^2)^2" should be: " a(a^6-a^5(b+c) - a^4(b+c)^2 + (b^2-c^2)^2(b^2+c^2) +
a^3(2b^3+b^2c+b c^2+2c^3) - a^2(b^4-b^3c-3b^2c^2-b c^3+c^4) - a(b^5+b^4c+b
c^4+c^5))";
5499
gerade should be "2_3=12_79" (at least !!!). Two errors in the barycentrics: "...a^4*(b^3+b^2*c-b*c^2+c^3)" should be: "...-a^4*(b^3+b^2*c+b*c^2+c^3)" while
"...-a^3*(2*b^4+3*b^3*c+3*b*c^3+2*c^4)" should be "+a^3*(2*b^4+3*b^3*c+3*b*c^3+2*c^4)".
5500
some ^ shouldn’t be there. Moreover 52b^4c^2 should be 152b^4c^2 in -a^16(81b^6 +
152b^4c^2 +152b^2c^4 + 81c^6). This one is shorter: "(-458752*a^2+524288*So)*S^10+(336384*a^6+659456*So*a^4-196608*So^2*a^2-131072*So^3)*S^8+64*a^2*(-a^2+So)*(585*a^6270*So*a^4-1504*So^2*a^2+1216*So^3)*S^6-16*a^4*(465*a^4-1274*So*a^2+944*So^2)*(a^2+So)^3*S^4+4*a^6*(-213*a^2+278*So)*(-a^2+So)^5*S^2-29*a^8*(-a^2+So)^7".
5501
at the end, b^2 is wrong. Should be (b^2-c^2)^6(b^4-4b^2c^2+c^4) (homogeneity).
5507
NOT SURE a^2(bc-2S)(abc(b+c-a)+2(b^2+c^2-a^2)S): wrong key
But a^2(bc-2S)(abc(b+c-a)-2(b^2+c^2-a^2)S) is not S->-S in X(600)
5509_to_5520 the name KIRIKAMI SIX CIRCLES IMAGE OF X(xx) is a joke. This point, say
U, is nothing but the center of the RH that goes through A,B,C,H,X(xx). And then,
U belongs to the Euler circle of any triangle inscribed in the RH. You can have 6, 66,
666, and even 6666 of them if you want.
5519
not anticomplem(X(6078)) but complem(X(6078)).
5542
X lies on line... should be X(5542) lies on these lines:
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
278
24.6. Some errors in the web page
X(5562)
= complement of X(5889) X(5562) = anticomplement of X(389)
5582
a^2(b+c-a)(b-c)^2(b^3+c^3-a^3+5a^2b+5a^2c-5ab^2-5ac^2-abc) should be a^2(b+ca)(b-c)^2(b^3+c^3-a^3+5a^2b+5a^2c-5ab^2-5ac^2-abc)^2 (square missing at the end).
5599
(r+4R)^(1/2) should be (r*R+4R^2)^(1/2)... homogeneity !
5600
5601, 5602, idem
5631
Extra parethesis "cos((2A-4Pi)/3)sec(((A-2Pi)/3)" should be "a*cos((2*A-4*Pi)/3)*sec((A2*Pi)/3)"
5633
missing: "a/sin(1/3*A+1/6*Pi)/sec(1/3*Pi+2/3*A)"
5638
Barycentrics: g(B,C,A ) should be g(B,C,A ), hf(C,A,B) should be f(C,A,B)
f(a,b,c)= "(b^2-c^2)*a^2*(-a^2*b^2-a^2*c^2+b^4+c^4 +(-a^2+b^2+c^2)*(a^4-a^2*b^2a^2*c^2+b^4-b^2*c^2+c^4)^(1/2))"
5639
change W into -W in X(5638)
5652
pending: "(b^2-c^2)*(3*a^6-2*a^4*b^2-2*a^4*c^2+a^2*b^4-3*a^2*b^2*c^2+a^2*c^4+b^4*c^2+b^2*
5653
pending: "a^2*(b^2-c^2)*(a^10+a^8*b^2+a^8*c^2-5*a^6*b^4-8*a^6*b^2*c^2-5*a^6*c^4+a^4*b^6+1
14*a^2*b^6*c^2-3*a^2*b^4*c^4-14*a^2*b^2*c^6+4*a^2*c^8-2*b^10+4*b^8*c^2+b^6*c^4+b^4*c^6+
2*c^10)"
5655
pending: "a^10-6*a^8*b^2-6*a^8*c^2+11*a^6*b^4-a^6*b^2*c^2+11*a^6*c^4-7*a^4*b^6+2*a^4*b^4*
7*a^4*c^6+8*a^2*b^6*c^2-16*a^2*b^4*c^4+8*a^2*b^2*c^6+b^10-3*b^8*c^2+2*b^6*c^4+2*b^4*c^6
3*b^2*c^8+c^10"
5656
pending: "-a^10+7*a^8*b^2+7*a^8*c^2-14*a^6*b^4+12*a^6*b^2*c^2-14*a^6*c^4+10*a^4*b^610*a^4*b^4*c^2-10*a^4*b^2*c^4+10*a^4*c^6-a^2*b^8-12*a^2*b^6*c^2+26*a^2*b^4*c^412*a^2*b^2*c^6-a^2*c^8-b^10+3*b^8*c^2-2*b^6*c^4-2*b^4*c^6+3*b^2*c^8-c^10"
5665
1/((s-a)*((s^2-Sa)) should be 1/((s-a)*(s^2-Sa))
5670
pending: "8*(262144*S^12+(138240*a^4+49152*So*a^2-196608*So^2)*S^10+2304*a^2*(a^2+So)*(-63*a^4+33*So*a^2+32*So^2)*S^8-1728*a^2*(151*a^4-151*So*a^2+48*So^2)*(a^2+So)^3*S^6+1296*a^4*(90*a^4-124*So*a^2+43*So^2)*(-a^2+So)^4*S^4-5832*a^6*(3*a^2+2*So)*(-a^2+So)^6*S^2+729*a^8*(-a^2+So)^8)/(64*S^4+12*(-a^2+So)*(-7*a^2+4*So)*S^29*a^2*(-a^2+So)^3)"
5672
pending: "(256*a*(b+a+c)*S^5+48*b*a^3*(3*a*b+2*a*c-3*b^2+b*c-2*c^2)*S^3+9*a^3*b^2*(a^5a^4*b+a^4*c-2*a^3*b^2-a^3*c^2+2*a^2*b^3-2*a^2*b^2*c+3*a^2*b*c^2-a^2*c^3+a*b^43*a*b^2*c^2-2*a*b*c^3-b^5+b^4*c+b^3*c^2-b^2*c^3)*S)*3^(1/2)+192*a*(a^3+5*a^2*ba^2*c-a*b^2+2*a*b*c-a*c^2+b^3+3*b^2*c+3*b*c^2+c^3)*S^4+36*b*a^3*(2*a^4+2*a^3*c4*a^2*b^2+2*a^2*b*c-2*a^2*c^2-2*a*b^2*c+a*b*c^2-2*a*c^3+2*b^4-2*b^3*c-7*b^2*c^23*b*c^3)*S^2"
5673
pending: "(256*a*(b+a+c)*S^5+48*b*a^3*(3*a*b+2*a*c-3*b^2+b*c-2*c^2)*S^3+9*a^3*b^2*(a^5a^4*b+a^4*c-2*a^3*b^2-a^3*c^2+2*a^2*b^3-2*a^2*b^2*c+3*a^2*b*c^2-a^2*c^3+a*b^43*a*b^2*c^2-2*a*b*c^3-b^5+b^4*c+b^3*c^2-b^2*c^3)*S)*3^(1/2)-192*a*(a^3+5*a^2*ba^2*c-a*b^2+2*a*b*c-a*c^2+b^3+3*b^2*c+3*b*c^2+c^3)*S^4-36*b*a^3*(2*a^4+2*a^3*c4*a^2*b^2+2*a^2*b*c-2*a^2*c^2-2*a*b^2*c+a*b*c^2-2*a*c^3+2*b^4-2*b^3*c-7*b^2*c^23*b*c^3)*S^2"
5674
pending: "(14336*S^7*a^2+96*a^2*(141*a^4-235*So*a^2+104*So^2)*S^5+144*a^2*(a^2+So)*(-17*a^6+39*So*a^4-29*So^2*a^2+4*So^3)*S^3-54*a^4*(a^4-2*So*a^2+3*So^2)*(a^2+So)^3*S)*3^(1/2)+192*a^2*(-53*a^2+120*So)*S^6+144*a^2*(-23*a^2+10*So)*(a^2+So)*(-a^2+4*So)*S^4-108*a^4*(3*a^2+13*So)*(-a^2+So)^3*S^2-81*a^6*(-a^2+So)^5"
5675
pending: "(64*a^2*(-53*a^2+120*So)*S^7+48*a^2*(-23*a^2+10*So)*(-a^2+So)*(-a^2+4*So)*S^536*a^4*(3*a^2+13*So)*(-a^2+So)^3*S^3-27*a^6*(-a^2+So)^5*S)*3^(1/2)-14336*S^8*a^296*a^2*(141*a^4-235*So*a^2+104*So^2)*S^6-144*a^2*(-a^2+So)*(-17*a^6+39*So*a^429*So^2*a^2+4*So^3)*S^4+54*a^4*(a^4-2*So*a^2+3*So^2)*(-a^2+So)^3*S^2"
March 30, 2017 14:09
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
279
5683
pending: "(-3145728*S^12+524288*(5*a^2+So)*(-a^2+So)*S^10+65536*(17*a^4-54*So*a^2+21*So^2)*
a^2+So)^2*S^8-32768*a^2*(-3*a^2+4*So)*(-13*a^2+9*So)*(-a^2+So)^3*S^6+8192*a^4*(47*a^2+45*So)*(-a^2+So)^5*S^4-49152*a^6*(-a^2+So)^7*S^2+2304*a^8*(-a^2+So)^8)*(1024*S^6256*(-a^2+So)*(-a^2-4*b^2+3*So)*S^4-128*a^2*b^2*(-a^2+So)^2*S^2+16*a^4*(-a^2+So)^4)*(1024*S
a^2+So)*(-3*a^2-4*b^2+5*So)*S^4-128*a^2*(-a^2+So)^2*(-a^2-b^2+2*So)*S^2+16*a^4*(a^2+So)^4)*a^2*(2*a^2-2*So)"
5684
pending: "(4096*S^6-16*a^2*(-71*a^2+80*So)*S^4+24*a^2*(-11*a^2+8*So)*(-a^2+So)^2*S^29*a^4*(-a^2+So)^4)*(256*S^4-8*a^2*(-a^2+3*b^2+So)*S^2-2*a^2*(-a^2+So)^2*(a^2-b^2+So))*(256*S^4-8*a^2*(-4*a^2-3*b^2+7*So)*S^2+2*a^2*(-b^2+So)*(-a^2+So)^2)"
5685
pending: "(2*a^3-2*a^2*c-a*b*c-2*b^3-2*b^2*c-2*So*a+2*So*b+2*So*c)*a*(2*a^3+2*a^2*ca*b*c+2*b^3+2*b^2*c-2*So*a-2*So*b-2*So*c)*((-128*a-256*b-256*c)*S^4-12*a*(14*a^4+8*a^3*b+6*a
2*a*b^2*c-22*So*a^2-10*So*a*b-6*So*a*c-8*So*b*c+8*So^2)*S^2+9*a^2*(-a^2+So)^2*(2*a^3-2*a^2*c-3*a*b*c+2*b^3-2*b^2*c+2*So*a-2*So*b+2*So*c))"
5687
"a(a^2-ab^2-ac^2-2abc+2b^2c+2bc^2)" should be "a(a^3-ab^2-ac^2-2abc+2b^2c+2bc^2)"
homogeneity !
5879
ends by "-(b^2-c^2)4))". Should be
"a^2/(a^10-4*a^6*(2*b^4-3*b^2*c^2+2*c^4)-a^2*(b^2-c^2)^2*(b^4+10*b^2*c^2+c^4)+(b^2+c^2)*(a
c^2)^2-(b^2-c^2)^4))"
5896
"Trilinears : f(A,B,C)". Should be "Trilinears f(A,B,C)"
5897
ends by "(b^2-c^2)4))". Should be
"a^2/(2*a^2*(a^8-(4*a^4+b^4+6*b^2*c^2+c^4)*(b^2-c^2)^2)-(b^2+c^2)*(a^8-10*a^4*(b^2c^2)^2+(b^2-c^2)^4))"
5910
"(b^2-c^2)8" instead of "(b^2-c^2)^8" and "(b^2-c^2)4" instead of "(b^2-c^2)^4".
Moreover, a shorter version is "-256*S^10*a^2+(-512*a^6+704*So*a^4-192*So^2*a^2)*S^8+(352*a^10+928*So*a^8-800*So^2*a^6+224*So^3*a^4)*S^6+(-128*a^14+528*So*a^12848*So^2*a^10+656*So^3*a^8-240*So^4*a^6+32*So^5*a^4)*S^4+(-20*a^18+112*So*a^16260*So^2*a^14+320*So^3*a^12-220*So^4*a^10+80*So^5*a^8-12*So^6*a^6)*S^2-a^22+7*So*a^2021*So^2*a^18+35*So^3*a^16-35*So^4*a^14+21*So^5*a^12-7*So^6*a^10+So^7*a^8"
5916
contains some lines relative to X(5917), X(2378), X(2379)
5920
pending: "-(a^7*(b+c)-(b+c)^2*a^6-(b+c)*(3*b^2+8*b*c+3*c^2)*a^5+(3*b^4+2*b^3*c+22*b^2*c^2+
(3*b^4+4*b^3*c-6*b^2*c^2+4*b*c^3+3*c^4)*(b-c)^2*a^2-(b+c)*(b^4+10*b^3*c-6*b^2*c^2+10*b*c^
c)^2*a+(b^2+c^2)*(b+c)^2*(b-c)^4)*(a-b-c)*a"
5924
"...-2(b-c)^2(b^2-c^2)4... " should be "-2(b-c)^2(b^2-c^2)^4". This one is shorter:
"(20*a^2-12*a*b-12*a*c+16*b*c-16*So)*S^4+2*a*(-a^2+So)*(-a^3+a^2*b+3*a^2*c4*a*b*c-2*b^3+2*b^2*c+2*So*a+2*So*b-2*So*c)*S^2-a^3*(-a^2+So)^2*(a^2*c-a*b*cb^3+b^2*c+So*b-So*c)"
X(5930)
= isotomic conjugate of X(5731). Should be 5931.
5931
missing: isotomic of 5930.
5933
barycentrics: "(b^3-a*(3*b^2+4*b*c+3*c^2)-(b+c)*(3*a^2-b^2-c^2))/(b+c-a)" should
be "(a^3-a*(3*b^2+4*b*c+3*c^2)-(b+c)*(3*a^2-b^2-c^2))/(b+c-a)"
5939
Remark: X(5026) is X(187) of the 1st Brocard triangle. While X(187) is X(5939) of
the 1st Brocard triangle.
5947
Huge expression... inducing truncation errors. Not on 3614_5949, but on 3614_5953.
5948
OK. But "(1024*a+1024*b+1024*c)*S^6-16*a^2*(-3*a^3+a^2*b-5*a^2*c-8*a*b*c+6*b^36*b^2*c+2*So*a-6*So*b+6*So*c)*S^4+4*a^2*(-a^2+So)*(3*a^5+a^4*b+5*a^4*c+7*a^3*b*c4*a^2*b^3+4*a^2*b^2*c-9*So*a^3+3*So*a^2*b-7*So*a^2*c+2*So*b^3-2*So*b^2*c+6*So^2*a2*So^2*b+2*So^2*c)*S^2+a^4*(-a^2+So)^3*(-2*a^3+2*a^2*c+7*a*b*c-2*b^3+2*b^2*c+2*So*a+2*S
2*So*c)" is shorter.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
280
24.6. Some errors in the web page
5952
the last factor is wrong.
When sorted, we se the repeated terms "(a^4+2*a^3*b+2*a^3*c+a^2*b*c-2*a*b^33*a*b^2*c-3*a*b*c^2-2*a*c^3-b^4+2*b^2*c^2-c^4++++++++++++ +2*a^3*c+2*a^3*c+2*a^3*c+a
2*a*c^3-2*a*c^3-2*a*c^3-b^4-c^4)"
Should be
"(a^4+2*a^3*b+2*a^3*c+a^2*b*c-2*a*b^3-3*a*b^2*c-3*a*b*c^2-2*a*c^3-b^4+2*b^2*c^2c^4)"
5953
Huge expression... inducing truncation errors. But OK.
5960
shorter: "-(b+c)^2*(a+b)*(a-b+c)/(-a^3+b*a^2-c*a^2+b^2*a+3*a*b*c+c^2*a-b^3b^2*c+b*c^2+c^3)^(1/2)-(b+c)^2*(a+b-c)*(a+c)/(-a^3-a^2*b+a^2*c+a*b^2+3*a*b*c+a*c^2+b^3+b
b*c^2-c^3)^(1/2)+(b+c)*(b-c)^2*(b+a+c)/(a^3+b*a^2+c*a^2-b^2*a+3*a*b*c-c^2*ab^3+b^2*c+b*c^2-c^3)^(1/2)"
5965
(and 5966) Should be "2*a^6-4*a^4*b^2-4*a^4*c^2+3*a^2*b^4+3*a^2*c^4-b^6+b^4*c^2+b^2*c^4c^6"
5975
"a^2*(b^2+c^2+a*b+a*c)^2*(a^3+b^3+c^3+a*b*c-2*b*c^2)*(a^3-b^3+c^3+a*b*c2*b^2*c)" should be "a^2*(b^2+c^2+a*b+a*c)^2*(a^3+b^3-c^3+a*b*c-2*b*c^2)*(a^3b^3+c^3+a*b*c-2*b^2*c)"
6011
Last ")" missing. Should be "a/((b-c)*(a^3-a^2*b-a*b^2+b^3-a^2*c-a*b*c-a*c^2+c^3))"
6013
idem. Should be "a/((b-c)*(2*a*b+2*a*c+b*c))"
6014
idem. Should be "a^2/((5*a-b-c)*(-b+c))"
6016
idem. Should be "a^2/((b-c)*(-3*a^2-a*b-a*c+2*b*c))"
6038
pending: barycentrics= "3*a^10*b^2+3*a^10*c^2-a^8*b^4+2*a^8*b^2*c^2-a^8*c^4+a^6*b^6a^6*b^4*c^2-a^6*b^2*c^4+a^6*c^6-3*a^4*b^8-3*a^4*c^8-4*a^2*b^8*c^2+4*a^2*b^6*c^4+4*a^2*b^
4*a^2*b^2*c^8-2*b^8*c^4+4*b^6*c^6-2*b^4*c^8"
6039
"a^4-a^2*b^2-a^2*c^2+b^4-b^2*c^2+c^4" is not "-2*S^2+Sa^2+Sb^2+Sc^2" (Letter version) nor "-8*S^2+Sa^2+Sb^2+Sc^2" (A4 version) but "-4*area(ABC)^2+Sa^2+Sb^2+Sc^2".
Should be: "Sc*So*Sb*(a^4-a^2*b^2-a^2*c^2+b^4-b^2*c^2+c^4)^(1/2)+(So*a^2+4*S^2So^2)*(-So*a^2+4*S^2)"
6055
"4*sin(A)*sec(A+omega)+sin(B)*sec(B+omega)-sin(C)*sec(C+omega)" should be "4*sin(A)*sec(A+omeg
6078
X(6078) = complement of X(5519); should be anticomplement
6082
is not anticomplement of X(6093)
6093
is not 6092 but "(a^8+a^6*b^2-7*a^6*c^2-a^4*b^4-10*a^4*b^2*c^2+20*a^4*c^4a^2*b^6+14*a^2*b^4*c^2-10*a^2*b^2*c^4-7*a^2*c^6-b^6*c^2-b^4*c^4+b^2*c^6+c^8)*(a^87*a^6*b^2+a^6*c^2+20*a^4*b^4-10*a^4*b^2*c^2-a^4*c^4-7*a^2*b^6-10*a^2*b^4*c^2+14*a^2*b^2*
a^2*c^6+b^8+b^6*c^2-b^4*c^4-b^2*c^6)"
6096
is not 6095 but "a^2/(3*a^4*b^2+3*a^4*c^2+2*a^2*b^4-10*a^2*b^2*c^2+2*a^2*c^4b^6+b^4*c^2+b^2*c^4-c^6)"
6116
"&pi/" should be "&pi;/"
6117
idem
6118
no S at the end of barycentrics
6135
"a/[(b - c)(bc + 2S)" is too short ! Should be "a/[(b - c)(bc + 2S)]"
6136
"a/[(b - c)(bc - 2S)" is too short ! Should be "a/[(b - c)(bc - 2S)]"
X(6101)
= complement of X(6255) should be 6143
March 30, 2017 14:09
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
281
6146
coherent. wrong searchkey ?
0.25027451271196897860. Should be
2.99369649092256991433280478504
6155
Trlinears should be Trilinears
6159
"lies on thess lines": should be "lies on these lines:"
6190
f(a,b,c) : g(B,C,A) : g(C,A,B),
6206
not Sin but sin
6207
idem
6214
S See César Lozada should be at a new line
6300
"Sqrt(3)" should be "sqrt(3)"; idem for 6300,6301,6302,6303,6305,6307
6326
<GBR> somewhere in the code should be <br>
6343
Trilinears should be written on a simple line.
6377
is isogon, not isotom, of 5383
X(6386)
is *not* complement of X(6338)
6388
X(6388) = complement of X(4563) repeated
6400
missing some commas in {4, 69}, {24, 1152} {1307, 3563}, {1587, 3567}, {5871,
6220} {6197, 6404}, {6198, 6405} . pending: "8*(a^2-b^2+c^2)*b^2*(a^2+b^2c^2)*c^2*a^2*S+(a^4*b^2+a^4*c^2-2*a^2*b^4-3*a^2*b^2*c^2-2*a^2*c^4+b^6+c^6)*(a^2+b^2c^2)*(a^2-b^2+c^2)*a^2"
6401
pending: "(a^2-b^2-c^2)/(a^4*b^2+a^4*c^2-2*a^2*b^4-3*a^2*b^2*c^2-2*a^2*c^4+b^6+c^68*S*b^2*c^2)"
6402
pending: "1/(8*(a^2-b^2+c^2)*b^2*(a^2+b^2-c^2)*c^2*S+(a^4*b^2+a^4*c^2-2*a^2*b^43*a^2*b^2*c^2-2*a^2*c^4+b^6+c^6)*(a^2+b^2-c^2)*(a^2-b^2+c^2))"
6465
"(at least)" should be "(more pending)"; pending: "+4*(a^6+a^4*b^2+a^4*c^2+3*a^2*b^4+10*a^2*b^2
3*b^4*c^2-3*b^2*c^4+3*c^6)*a^2*(a^2-b^2-c^2)*S+(a^8-2*a^6*b^2-2*a^6*c^2+20*a^4*b^2*c^2+2*
b^8+8*b^6*c^2-14*b^4*c^4+8*b^2*c^6-c^8)*a^2*(a^2-b^2-c^2)"
6466
"(at least)" should be "(more pending)"; pending: "-4*(a^6+a^4*b^2+a^4*c^2+3*a^2*b^4+10*a^2*b^2
3*b^4*c^2-3*b^2*c^4+3*c^6)*a^2*(a^2-b^2-c^2)*S+(a^8-2*a^6*b^2-2*a^6*c^2+20*a^4*b^2*c^2+2*
b^8+8*b^6*c^2-14*b^4*c^4+8*b^2*c^6-c^8)*a^2*(a^2-b^2-c^2)"
6519
repeated line about lines
6586
Barycentrics "a^2*(b-c)*(b^2+c^2-b*c+a*b+a*c)" should be "a^2*(b-c)*(a*b+a*cb^2-b*c-c^2)"
6592
barycentrics: "(2cosA)(-2+cos(2B-2C)+4cos^2A)(1+2cos2B+2cos2C-2cos2A)+(-1+4cos^2Acos(BC))QQ———QQQQ(2cosA)(-2+cos(2B-2C)+4cos^2A)(1+2cos2B+2cos2C-2cos2A)+(1+4cos^2Acos(B-C))"
X(6684)
= reflection of X(i) in X(j) for thesse (i,j): (5,3634), (1125,140)>BR> X(6684) =
complement of X(956) should be "these", "<br>" and "complement of X(546)"
X(6708)
= complement of X(1215) should be X(1214)
—6723—
no more search keys
6736
not on line (55,5793). Everything else fit together.
6743
Trilinears "(b+c-a)*(b-c)*(2*a^3-(b+c)*a^2-2*(b+c)^2*a+(b^2-c^2))/a" don’t match
with barycentrics "2*(s-a)*(4*R+r)-r*a" which are propto "(a-b-c)*(2*a^3-a^2*b-a^2*c2*a*b^2-4*a*b*c-2*a*c^2+b^3-b^2*c-b*c^2+c^3)" and agree with the list of lines.
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
282
24.6. Some errors in the web page
6770
Barycentrics give, are not simplified... and wrong. Should be: "-64*3^(1/2)*S^3+3*(a^2b^2-c^2)*(3*a^4+b^4-2*b^2*c^2+c^4)"
6773
idem. Should be: "64*3^(1/2)*S^3+3*(a^2-b^2-c^2)*(3*a^4+b^4-2*b^2*c^2+c^4)".
6791
{2,5503), {6,5642}, etc. Should be {2,5503}, {6,5642},... Moreover an extra factor has
appeared in the barycentrics. Should be: "(b^2-c^2)^2*(5*a^2-b^2-c^2)"
6792
Barycentrics given don’t fit with the other properties. Should be: "a^6-a^4*b^2a^4*c^2+3*a^2*b^4-5*a^2*b^2*c^2+3*a^2*c^4-b^6+b^4*c^2+b^2*c^4-c^6"
6802
Shorter barycentrics:
March 30, 2017 14:09
published under the GNU Free Documentation License
24. Import the 4995-7000 part of ETC
24.7
24.7.1
283
Further work
Coherence
When quite everything is OK, save the center table (SQL, compressed). Then export to bar_igtca
(the one which is routinely read by Maple). And see what happens.
1. Rows are not sorted according to num. Correction: Go to Operation, then "Alter table order
by num". And now, the result is sorted.
2. An item with num=0 had appeared. Killed.
3. Now "relire" runs to 6802.
4. Special says "368, 369, 370, 1144, 3232, 5373, 5394, 5497, 5626, 5920". But fdat[j][1..4]="len="
is also a special case.
5. The ff test: 6763 and 6768 are detected. "r" remaining in 6763.
6. Points at infinity. Check "%=30-511=%" in gerade or "line at infinity" in curves
7. Isogonal conjugates. Error detected at 2136/2137 (collision prone). Error at 5420. Wrong
parsing at 5640, 5916, 5917 (conjugates wrt another triangle).
8. Isotomic conjugates. Was (pldx) (4859, 4859) ?, [5383, 6377, 0], [5930, 5731, 5931], [6377,
5383, 0]
24.7.2
Trigonometric lines
1. 211 points were using A, B, C as variables
2. Among them, ...
24.7.3
Sqrt and other dramas
√
√
√
√
√
√
1. 13+2+6 points were using 2 12, 27, 2 48 instead of 4 3, 3 3, 8 3: bad automated
translations (simplify is required, not only factor).
24.8
Some items to revisit
3272-3283 related to Morley
5390
Euler-Morley-Zhao
600
past wrong searchkey. Link with the other point.
3020
past wrong searchkey. (b − a + c) (b − c)
3310
Barycentric Products of Perpendicular Directions (on the orthic axix)
5000
5000-5005: To be fully rewritten !
Coordinates u : v : w are not homogeneous tripolar coordinates, but homogeneous
triradial coordinates!
X(5002) is the point whose polar distances in the plane of triangle ABC are proportional to (a, b, c). If the reference triangle ABC is acute, the actual polar distances are
ka, kb, kc. If triangle ABC is obtuse, then the barycentrics of X(5002) are a nonreal
complex numbers. Nevertheless, for all triangles ABC, the midpoint of X(5002) and
X(5003) is X(858).
A method for converting from homogeneous polar distances to homogeneous barycentrics,
found by Peter Moses (March, 2012), depends on finding the point of intersection of
the radical axes of radical circles centered at A, B, C.
Write the polar distances for a point U as u : v : w, and let
2
2
b2 − bc + c2 .
2Sa = b2 + c2 − a2 , 2Sb = c2 + a2 − b2 , 2Sc = a2 + b2 − c2
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
284
24.8. Some items to revisit
Then barycentrics x : y : z for U are given by
x =
y
=
z
=
x+y+z
=
a2 Sa + k 2 (Sc v 2 + Sb w2 − a2 u2 )
b2 Sb + k 2 (Sa w2 + Sc u2 − b2 v 2 )
c2 Sc + k 2 (Sb u2 + Sa v 2 − c2 w2 )
8S 2
where k 2 has two values (as in the quadratic formula): (−f −g)/h or (−f +g)/h, where
f
g
= −a2 u2 2Sa − b2 v 2 2Sb − c2 w2 2Sc
p
= 2S (−au + bv + cw)(au − bv + cw)(au + bv − cw)(au + bv + cw)
h =
2(a2 (u2 − v 2 )(u2 − w2 ) + b2 (v 2 − w2 )(v 2 − u2 ) + c2 (w2 − u2 )(w2 − v 2 ))
√
Sa a2 u2 + b2 Sb v 2 + c2 Sc w2 ± 2 S ...
a2 u4 + b2 v 4 + c2 w4 − 2 (Sa w2 v 2 + Sb u2 w2 + Sc u2 v 2 )
The meaning of k is as follows: starting with polar distances u : v : w, the actual polar
distances are ku, kv, kw. That is,
|U A| = ku, |U B| = kv, |U C| = kw
5099
inverse in polar circle
xxx
isogonal conjugates wrt various triangles
March 30, 2017 14:09
published under the GNU Free Documentation License
Chapter 25
Import the 3588-4994 part of ETC
25.1
Some errors in the web page
3590
barycentrics are wrong (should be 6S, not 3S). Same apply to 3591-3595
3599
ending )
3611
ending )
3632
lines are unreadable and a − 2b − 2c is 3679
3637
over 500 char ==> shortened using Conway symbols.
3638
ending )
3639
ending )
3641
starting (
3642
3646
√
a squared point. Better written as 2a4 − b4 − c4 + 2b2 c2 − a2 b2 − a2 c2 +4 3S(c2 +b2 )
searchkey is wrong. 2.80169029270208151552171148982
3648
should be (4(2 + 3R/r))S 2 − (ac + 2Sb )(ab + 2Sc ). The other is also wrong (a factor
−a is missing).
3654
a parenthesis is missing before the last term. Should be · · · a2 b2 ) − 3k).
3657
trilinears are wrong
3658
should be
a a3 b + a3 c − a2 b2 − a2 c2 − ab3 + ab2 c + abc2 − ac3 + b4 − 2 b2 c2 + c4
b2 − c2
3660
should be a ba2 + ca2 + 2abc − 2ac2 − 2ab2 + b3 + c3 − cb2 − bc2 /(a − b − c)
3677
initial html tag with an extra )
3684
should be a(a2 − bc)(a − b − c)
3697
should be a(b + c)(a2 − b2 − 4bc − c2 )
remark
: isotom(1292) has the same searchkey as 3617
3702
should be bc(2a + b + c)(a − b − c). What is given is 4723
3706
should be (a − b − c)(2bc + ab + ca). What is given is 3705
3724
should be a3 (b + c)(a2 − b2 + bc − c2 ). Sign of bc.
3738
should be (b − c)(a − b − c)(a2 − b2 + bc − c2 )a. A factor is missing.
285
286
25.1. Some errors in the web page
3771
leading ( is missing
3772
leading ( is missing
3791
should be 2a3 + a2 b + a2 c − b2 c − bc2 . What is given is 3790.
3793
should be (3a2 + b2 + c2 )(2a2 − b2 − c2 )
3800
should be (b2 − c2 )(3a2 + b2 + c2 )
3801
should be (b2 − c2 )(b2 − bc + c2 ) html syntax
3813
should be (a3 − a2 b + a2 c − ab2 − 4abc + b3 + b2 c)(a + b − c) some wrong exponents
3814
leading (
3816
should be ab2 − 4abc + ac2 − b3 + b2 c + bc2 − c3 some wrong exponents
3820
should be a2 b2 + a2 c2 − 4ab2 c − 4abc2 − b4 + 2b2 c2 − c4 sign
3825
should be a2 b2 − 2a2 bc + a2 c2 − ab2 c − abc2 − b4 + 2b2 c2 − c4 some wrong exponents
3829
should be 3ab2 − 4abc + 3ac2 − 3b3 + 3b2 c + 3bc2 − 3c3 homogeneity
3839
missing term 5a4 + 2a2 b2 + 2a2 c2 − 7b4 + 14b2 c2 − 7c4
3846
should be a2 b + a2 c + 3ab2 + 8abc + 3ac2 + 4b2 c + 4bc2 . 3847 is given instead
3847
should be 2abc + b3 + c3 . What is given looks like 3846
3853
should be 6a4 − a2 b2 − a2 c2 − 5b4 + 10b2 c2 − 5c4 . Leading 6 is missing
3866
should be (3a2 + b2 + c2 )(a2 b2 + c4 )(a2 c2 + b4 )
3867
should be (3a2 + b2 + c2 )(b2 + c2 )(a2 − b2 + c2 )(a2 + b2 − c2 )
3879
should be 2a2 + ab + ac − b2 − c2
3894
should be (2a2 b + 2a2 c + 3abc − 2b3 − 2c3 )a. Homogeneity
3896
should be (b + c)(2a2 − bc). What is given is 740.
3905
should be ab3 − b3 c + a4 − bc3 + ac3
3949
should be a(b + c)2 (a2 − b2 − c2 )
3953
should be a(b3 + ab2 − 2abc + c3 + ac2 ) symmetry
3960
should be (b − c)(a2 − b2 + bc − c2 )a
3964
should be (a2 − b2 − c2 )3 a2
3966
should be (a − b − c) ∗ (a2 + a ∗ b + a ∗ c + b2 + c2 )
3968
should be (b + c)(a2 − b2 + 7bc − c2 )a . Symmetry
3984
should be (a − 3b − 3c)(a2 − b2 − c2 )a. Extra parenthesis.
4021
2a2 + 3ab + 3ac + b2 − 2bc + c2
4027
(a2 − bc)2 (a2 + bc)2 . What is given is 3801.
4036
bc(b − c)(b + c)2
4044
bc(b + c)(a2 − ab − ac − 2bc). What is given is 4043.
4049
(b − c)(b + c)/(2a − b − c)
4050
(a − b − c)(ab + ac − 3bc)a. Signum
4085
(b + c)(2a2 + b2 − bc + c2 ). Signum
March 30, 2017 14:09
published under the GNU Free Documentation License
25. Import the 3588-4994 part of ETC
287
4108
(b2 − c2 )(2a4 + b2 c2 ). missing exponents
4148
(b − c)(a2 − bc)(a − b − c)2 missing power
4169
(2a − b − c)(b + c)/(b − c) missing coeff 2
4317
missing leading (
4323
iterated //
4377
(b + c)bc(2bc + a2 ) Missing coeff, giving 4374
4382
repeated record
4383
repeated record
4401
(b − c)(2a2 − bc)a Missing coeff, giving 659
4405
4a2 − b2 − 8bc − c2 Missing coeff
4407
extra ending )
4427
(2a + b + c)/(b − c) Missing a
4433
(b + c)(a − b − c)(a2 − bc)a What is given is 4087.
4435
(b − c)(a − b − c)(a2 − bc)a
4460
5a2 + 4ab + 4ac − b2 − 6bc − c2 What is given is 4440.
4466
missing opening (
4469
a (b2 + bc + c2 )2 /(b + c) Missing exponent
4481
a (b − c)(b2 + bc + c2 )/(b + c) What is given is 984
4490
(b − c)(b2 + 3bc + c2 )a what is given is 4489
4499
4a2 + 2ab + 2ac + b2 + 8bc + c2
4516
(b − c)2 (b + c)(a − b − c)a What is given is 4041
4666
(a2 − 2ab − 2ac + b2 − 4bc + c2 )a
4674
a (b + c)/(2a − b − c) Divide, not multiply.
4750
(b − c)(2a2 − b2 − c2 ) Homogeneity
4756
(a + 2b + 2c)/(b − c) Divide, not multiply. What is given is 10
4758
Extra ending )
4803
(a − 2b − 2c)2 /(b + c) Missing power.
4820
(b − c)(a + 2b + 2c)(a − b − c)
4830
(b − c)(3a + b + c)(a2 − bc) Extra power
4850
(ab + ac + b2 − bc + c2 )a What is given is 3681.
4878
(b + c)(a2 − 2ab − 2ac + b2 + c2 )a2 What is given is 4440.
4902
−3a2 + ab + ac + 4b2 − 8bc + 4c2 Provided result is non homogeneous.
4915
(a − b − c)(a2 − b2 + 10bc − c2 )a
4993
(a4 + a2 b2 + a2 c2 − 2b4 + 4b2 c2 − 2c4 )/(a2 b2 + a2 c2 − b4 + 2b2 c2 − c4 ). No searchkey
provided.
4994
(a4 − 3a2 b2 − 3a2 c2 + 2b4 − 4b2 c2 + 2c4 )/((a2 − b2 − c2 )(a2 b2 + a2 c2 − b4 + 2b2 c2 − c4 )).
No searchkey provided
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
288
March 30, 2017 14:09
25.1. Some errors in the web page
published under the GNU Free Documentation License
Chapter 26
Describing ETC
26.1
Various kinds of points
Table 26.1 describe the various species and subspecies that can be found in the Kimberling’s
database.
no_surd=6177
Only_abcS: 6090, only_abc 5691
species
rational
2nd deg
surds
radicals
1
2
3
4
5
6
7
8
more
angles
special
subspecies
a , b2 , c2 only
a2 , b2 , c2 , S
a, b, c only
a, b, c, S
√
2 (all a2 , S)
√
5 (all a2 , S)
p
√
(among them, #12 5 + 5) (14)
√
3 (#125a2 S, #39 aS) (225,86)
p
a (b + c − a)
p
(a + b + c) ÷ abc used in cos A2 (#14)(18)
√
(among them 2 a )
√
√
√
a, b, c only
p
√
a2 b2 + · · · = 4S 2 + So2
p
√
a4 − b2 c2 + · · · = So 2 − 12 S 2
√
√
a6 − b4 c2 · · · + 3a2 b2 c2 = 9 a2 b2 c2 − 32 S 2 So Euler
q
|OI|
a3 −b2 c···+3abc
(mixed #2)
R =
q abc
p
a2 b···+abc
2
2
ρ +s =
Spieker circle
a+b+c
√
2
2bc − a · · · Poncelet In/circum-circle
p
2abc (a + b + c) (2bc − a2 · · · )
√
√
a8 − a6 b2 + a4 b2 c2 + · · · = 2 16 S 4 − 3 Sa Sb Sc So
e.g. dg10, Euler-Incircle
degree (1/3)
trig (A/3) etc
bare A & weird
9 special + 8 toolong
# means among (i.e. not additive)
2
total
784
241
3593
90
#8
#18
1025
3683
#164
#50
311
68
#18
#66
18
68
#62
#52
72
52
#12
12
#8
8
#2
2
6
4
11
2
58
10
17
#4
#2
236
33
10
5
48
Table 26.1: The various species of points in the Kimberling’s database
289
1075
232
4616
169
8
20
1307
4785
339
284
85
290
26.2
26.2. Special points
Special points
The following six points have no explicit barycentrics, but are defined as the solution of some
special equation.
26.2.1
X(368)
1. Situated on the second Brocard cubic, and on the anticomplement of the Kiepert hyperbola
CC (X523 ). Append x + y + z = 1.
2. Substitute ency_ (=local solution). Eliminate x, y. Obtain a six degree equation in z, with
3z − 1 in factor (barycenter G).
3. Four of the roots are complex, and X368 is the remaining one.
26.2.2
X(369), X(3232) trisected perimeter points
There exist points A0 , B 0 , C 0 on segments BC, CA, AB, respectively, such that :
AB 0 + AC 0 = BC 0 + BA0 = CA0 + CB 0 = (a + b + c)/3
Lines AA0 , BB 0 , CC 0 concur in X(369). Yff found that barycentrics p : q : r for X(369) can be
obtained in terms of the unique real root, K, of the cubic polynomial equation :
2 K 3 − 3 (b + c + a) K 2 + a2 + b2 + c2 + 8 (bc + ca + ab) K
− b2 c + c2 a + a2 b − 5 bc2 + ca2 + ab2 − 9 abc = 0
and gives at the geometry conference held at Miami University of Ohio, 2004/10/02, a symmetric
form for these barycentrics, namely :
p = K 2 − (2 c + a) K − a2 + b2 + 2 c2 + 2 bc + 3 ac + 2 ab
and cyclically. In the reference triangle, we have :
X369 = (0.4090242897, 0.3561467951, 0.2348289151)
On the other hand, there exist points A0 , B 0 , C 0 on segments BC, CA, AB, respectively, such that
B 0 C + C 0 B = C 0 A + A0 C = A0 B + B 0 A = (a + b + c)/3, and the lines AA0 , BB 0 , CC 0 concur in
X(3232), the isotomic conjugate of X(369).
26.2.3
X(370)
From http://pagesperso-orange.fr/bernard.gibert/Tables/table10.html, point X370 lies
on conic :
abc
(xy(a2 − b2 + c2 ) + xz(a2 + b2 − c2 ) + 2zya2 ) − x(y + z + 2x) √ = 0
R 3
and on the other two obtained cyclically. Two such conics have in common a vertex, X370 and
other two (may be not real) points. In any case, X370 is inside triangle ABC.
26.2.4
X(1144)
Suppose P is a point inside triangle ABC. Let Sa be the square inscribed in triangle P BC, having
two vertices on segment BC, one on P B, and one on P C. Define Sb and Sc cyclically. Then X1144
is the unique choice of P for which the three squares are congruent. L(a, b, c) is the common length
of the sides of the three squares. The function L(a, b, c) is symmetric, homogeneous of degree 1,
and satisfies 0 < L(a, b, c) < min (a, b, c). L is the smallest root of :
a2
b2
c2
S
+
+
−2 =0
a−L b−L c−L
L
March 30, 2017 14:09
published under the GNU Free Documentation License
26. Describing ETC
291
and point X1144 is obtained as :
leading to :
a2
b2
c2
:
:
a−L b−L c−L
X1144 = [0.2373201571, 0.3224457580, 0.4402340851]
This point lies on the hyperbola {A, B, C, X(1), X(6)} = CC (X649 ). Indeed, X(1144) lies on
the open arc from X(1) to the vertex of ABC opposite the shortest side. (Jean-Pierre Ehrmann,
12/16/01)
26.2.5
X(2061)
Nothing special, but the length... 743 with best efforts. Point with only one relation. Has been
shortened using Conway symbols.
26.3
orthopoint, points on circles
Remark 26.3.1. Orthopoint. Only 6 orthopoint pairs are registered, but 48 can be computed
(involving 96 points on L∞ , that contains 229 named points). Has been abbreviated into hortopoint
since "or" is a reserved word.
Remark 26.3.2. Inverse in a given circle. Points on L∞ are not considered here.
1. circumcircle. There are 352 named points whose inverse in circumcircle is named too. Among
them, 258 are on the circumcircle itself, and there are 47 pairs of "true" inverse pairs. Among
these 94 points, 29 aren’t listed.
On the circumcircle, 220 named points have a named isogonal conjugate (among the 229
points of L∞ ).
62 antipodal pairs are listed.
2. incircle. There are 53 named points whose inverse in incircle is named too. Among them,
39 are on the incircle itself, and there are 7 pairs of "true" inverse pairs. Among these 14
points, 3 aren’t listed.
3. nine points circle. There are 55 named points whose inverse in the nine points circle is
named too. Among them, 37 are on the nine points circle itself, and there are 9 pairs of
"true" inverse pairs. Among these 18 points, 4 aren’t listed.
4. Bevan circle : 19, 9, 5 (pairs).
5. Brocard circle : 98, 4, 47 (pairs). Among 94 points, only 89 are listed.
6. Spieker circle : 8, 8, 0 (pairs)
7. Orthocentroidal circle : 82, 2, 40 (pairs). Among the 80 points, 20 aren’t listed.
8. Fuhrmann circle : 12, 2, 5 (pairs). Among the 10 points, only 4 are listed (and belong to 4
different pairs).
9. First Lemoine : 16, 2, 7 (pairs).
10. Second Lemoine : 12, 2, 5 (pairs).
Remark 26.3.3. Antipode in circumcircle
1. From a total of 258 named points on the circle, 124 have a named antipode.
2. Antipode is mostly given as a member of the list "reflection", that appears as I_3 in the
corresponding field. 10 are missing.
Remark 26.3.4. Antipode in nine points circle
1. Among 37 named points on the nine points circle, 24 have a named antipode (12 pairs).
2. Among 37 named points on γ, 33 have a named anticomplement
—– pldx : Translation of the Kimberling’s Glossary into barycentrics —–
292
26.4. points on lines
26.4
points on lines
Remark 26.4.1. Bad extractions. Due to lack of regularity (may be human modifications), 50 lists
of lines were badly extracted (mostly : HTML tags, reflection, extra spaces).
SELECT num, gerade FROM center WHERE gerade like "%\_ %"
Lack of pattern for these :
505
680 1154 1180 1300 1573 1593 1689 1944 1951 2051
2173 2388 2442 2503 2708 2719 2723 2725 2727 2758 2776
2974 3037 3110 3111 3113 3308 3340 3354 3363 3501
Remark 26.4.2. There are 32322 quotations, involving 15988 lines.
1. Lines are referred by two other points, so that none of the 169 lines quoted as going through
X1 are quoted under their "true" name.
2. Here are the statistics :
1
2
3
4
5
6
7
8
9436
2700
2207
725
375
151
110
60
9
10
11
12
13
14
15
16
43
32
24
19
14
7
4
10
17
18
19
20
21
22
23
24
3
4
5
2
2
5
1
3
25
26
27
29
30
35
36
38
2
2
1
2
1
2
2
1
43
47
62
75
76
78
81
100
1
1
1
1
1
1
1
1
184
229
312
1
1
1
This has to be read as : there are 9446 lines that are quoted only once, and so on.
3. In fact, most of the lines involving only three points are quoted only at their third point
(except from 120 of them that are not quoted at the third point). For example, X1 is
involved in 294 quoted lines whereas only 169 of them are quoted in the X1 page. When
taking this fact into account, we obtain :
3
4
5
6
7
8
12998
1773
580
220
119
69
9
10
11
12
13
14
15
16
44
34
25
20
14
7
4
10
17
18
19
20
21
22
23
24
3
4
5
2
2
4
2
3
25
26
27
29
30
35
36
38
2
2
1
2
1
2
2
1
43
47
62
75
76
78
81
100
1
1
1
1
1
1
1
1
184 1
229 1
312 1
4. There are 1195 lines that goes through at least five named points.
March 30, 2017 14:09
published under the GNU Free Documentation License
Bibliography
Al-Khwarizmi M.i.M. The Compendious Book on Calculation by Completion and Balancing (i.e.,
Algebra) (London) (825), xvi, English=208, Arabic=124 pp. Pages 58/59 are missing, http:
//www.wilbourhall.org/pdfs/The_Algebra_of_Mohammed_Ben_Musa2.pdf.
. 2 citations located at sections 1 and 1.
Alberich-Carramiñana M. Geometry of the Plane Cremona Maps. No. 1769 in Lecture Notes in
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