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Then, using α + β + γ = 360◦ , we obtain:
x · a = (1/2) bc sin α a + ac sin β b + ab sin γ c · a
= (1/2) bc sin α ||
a ||2 + (ac sin β) (ab cos γ ) + (ab sin γ ) (ac cos β)
= (1/2)a 2 bc (sin α + sin β cos γ + sin γ cos β)
= (1/2)a 2 bc (sin α + sin(β + γ )) = (1/2)a 2 bc (sin α + sin(360◦ − α))
= (1/2)a 2 bc (sin α − sin(α)) = 0.
Similarly we can show that x · b = 0. Since x · a = 0 and x · b = 0, but a and b are not
parallel, x = 0 by Theorem 11.7.
Affine Transformations
12.1 Suppose h1 = g ◦ f −1 and h2 are both affine transformations sending p,
q, and r to p , q ,
and r , respectively. We need to show h1 = h2 . Now, h1 ◦ f and h2 ◦ f are both affine trans i, j} to {p , q , r }. But since an affine transformation is uniquely
formations mapping {0,
i, and j, h1 ◦ f = h2 ◦ f . Composing each of these comdetermined by where it maps 0,
−1
positions on the right with f , we obtain h1 = h2 , which proves the uniqueness of the
composition h1 .
12.2 No. Every affine transformation will preserve the ratio of lengths of the parallel sides. So if
these ratios are different in two trapezoids, then they are not affine equivalent (i.e., there is
no affine transformation mapping one to the other).
←
→
←
→
12.3 Let ABCD be a trapezoid with bases AD and BC. Let P be the intersection of AB and CD,
and let Q be the intersection of the diagonals, AC and BD. By Corollary 12.9, there is an
affine transformation f mapping AP D to an isosceles triangle A P D . (See Figure 136.)
Because f maps line segments to line segments, preserves the property of parallelism
among line segments, and preserves ratios of lengths of collinear segments (all by Theorem 12.7), ABCD is mapped to an isosceles trapezoid, A B C D , with A B = C D . It is
←−→
easy to see that P Q bisects the bases of A B C D ; we leave the details to the reader. We
←
→
conclude that P Q bisects the bases of ABCD also.
This problem also appeared as Problem 3.3.14 and as Problem 8.12.
f
P'
P
B
B'
C
Q
A
C'
Q'
D
FIGURE
A'
136.
D'
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12.4 Assume that AB and CD are chords of an ellipse, E, having the property that the area
determined by AB and the boundary of E is the same as the area determined by CD and the
boundary of E. Let f be an affine transformation mapping E to a circle, C = C(O , r). (Recall
that O = f (O), where O is the center of E.) The chords of E are mapped by f to chords of
C; let A designate the point on C that corresponds to f (A) and likewise for B , C , and D .
B'
C
M'
O'
A'
D'
N'
C'
FIGURE 137.
By Theorem 12.7(7), the area of the region bounded by the chord A B and the boundary
of C will be equal to the area of the region bounded by the chord C D and the boundary of C.
This implies that Area(A B O ) = Area(C D O ) and that A B = C D . Let M and N be the midpoints of A B and C D , respectively. By Theorem 4.6 of Chapter 4, O M ⊥ A B and O N ⊥ C D . Since Area(A B O ) = Area(C D O ) and A B = C D , we conclude
that O M = O N . Hence, A B and C D are both tangent to the circle C(O , O M ). The
pre-image of C(O , O M ) is an ellipse having center at O, which is tangent to AB and CD.
12.5 Let A1 , . . . , An be the vertices of an n-gon, and A1 , . . . , An be their images under an affine
transformation f . Let G be the centroid of {A1 , . . . , An }. Then by Example 65,
−−→
GAi = 0.
Let G = f (G).
→
−→
Then f (−
and
x ) + (b).
OAi ) = A(OAi ) + (b),
Suppose f is defined by x → x = A(
−−→
−−→ f (OG) = AOG + b, so
−−→
−→
−→
G Ai = f (GAi ) = AGAi .
This implies (by a generalization of Theorem 12.2(5)) that
−−→
G Ai = 0,
which is equivalent to G being the centroid of {A1 , . . . , An }.
12.6 Consider a parabola,
P, with vertex at (h, k). As was the case with ellipse and hyperbola,
−h
the translation x +
will map P to a parabola whose vertex is the origin. Then apply
−k
a rotation so that the axis of the parabola aligns with the positive y-axis. Under these two
affine transformations, P is mapped to a parabola P , represented by an equation of the form
y = ax 2 ,
where a is a positive real number.
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1/a
0
x
x/a
x
.
Now apply a third affine transformation, f (
x) =
=
=
y
0
1/a
y
y/a
The equation y = ax 2 can now be written (y /a) = a(x /a)2 , or y = (x )2 . This proves the
theorem.
Note that this establishes as a corollary that any two parabolas, P1 and P2 , are affine
equivalent. The proof is similar to that of Corollary 12.14.
12.7 Let A2 , B2 , and C2 be the points of intersection of BB 1 and CC 1 , AA1 and CC 1 , and AA1
and BB 1 , respectively. Let f be an affine transformation mapping ABC to an equilateral
triangle DEF , as illustrated in Figure 138. Because affine transformations preserve ratios
of parallel line segments, the points D1 = f (A1 ), E1 = f (B1 ), and F1 = f (C1 ) will divide
the sides of DEF in the same fixed ratio as A1 , B1 , and C1 divide the sides of ABC.
Consequently, F1 ED1 ∼
= D1 F E1 ∼
= E1 DF1 by SAS, so D1 E1 F1 is equilateral.
f
B
E
A1
C1
A
D1
C2
B2
F1
F2
G'
D2
G
A2
B1
E2
D
C
FIGURE
E1
F
138.
Let D2 , E2 , and F2 be the points of intersections of the segments EE1 and F F1 , F F1
and DD1 , and DD1 and EE1 , respectively. Rotating DEF clockwise by 120◦ around
the centroid, G , of DEF (as we did in Example 76) we see that D1 → E1 → F1 → D1 ,
where “→” means “is mapped to.” This implies that DD 1 → EE 1 → F F 1 → DD 1 , and
therefore D2 → E2 → F2 → D2 . This proves that D2 E2 F2 is equilateral.
Furthermore, under the 120◦ rotation about O , the equilateral triangle D1 E1 F1 is mapped
to itself, as is the equilateral triangle D2 E2 F2 . This will only happen if G is the centroid
of both D1 E1 F1 and D2 E2 F2 . Since any affine transformation maps the centroid of one
triangle to the centroid of another, the pre-image of G , f −1 (G ) = G, is the centroid of
ABC, A1 B1 C1 , and A1 B2 C2 .
12.8 By Theorem 12.7, collinear segments change their lengths under an affine transformation by
the same ratio k > 0, so any affine transformation preserves the equality we seek to prove.
Let f be an affine transformation mapping the parallelogram MN P Q to a square M N P Q
having side length a = 1. (See Figure 139.) We now wish to prove that 1/M R + 1/M S =
1/M T .
Let m∠Q M T = α. Then, from the right triangles M N R and M Q S , respectively, we
see that
1
1
= cos ∠N M R = cos (90◦ − α) = sin α and
= cos α.
MR
M S
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S'
S
f
l
R
N
R'
N'
P
45
P'
T'
T
α
M'
Q
M
FIGURE
a=1
Q'
139.
In addition, note that m∠M N Q = 45◦ since N Q is the diagonal of a square. Applying
the Sine theorem to M T N ,
1
M T 1
sin (45◦ + α)
=
, so
=
.
◦
◦
sin (45 + α)
sin 45
MT
sin 45◦
Thus, verifying that 1/M R + 1/M S = 1/M T reduces to checking that sin (45◦ +
α)/ sin 45◦ = cos α + sin α. The validity of the latter equation follows from the sine sum
formula:
(sin 45◦ )(cos α) + (cos 45◦ )(sin α)
sin (45◦ + α)
=
= cos α + sin α.
◦
sin 45
sin 45◦
An alternate solution is provided in Problem 3.3.12.
12.9 Suppose E is an ellipse given by the equation x 2 /a 2 + y 2 /b2 = 1; then E has semi-axes
of lengths a and b, as shown in Figure 140(i). Construct a rectangle, R, with center at the
center of E and with sides parallel to the axes of E. Hence, the side lengths of R are 2a
and 2b.
R
E
-5
S
4
2
C
b
a
O
10
5
O
r=a
15
-2
-4
(ii)
(i)
-6
FIGURE
140.
1
0
. Then f maps a
0 a/b
vector x, y to x , y = x, ay/b. Observe that the origin is mapped to the origin, and
the coordinate axes are mapped to themselves. Now (x )2 + (y )2 = x 2 + a 2 y 2 /b2 = a 2 , so
E is mapped to a circle with radius a, as shown in Figure 140(ii). Obviously, f (R) is a
Consider the affine transformation f (p)
= Ap,
where A =
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parallelogram that circumscribes the circle and has sides parallel to the coordinate axes.
Therefore f (R) is a square with side length 2a.
Since an affine transformation preserves the ratio of areas of two figures,
Area E
Area C
Area E
π a2
=
=⇒
=
.
Area R
Area S
4ab
4a 2
From here, it follows that the area of E is π ab.
12.10 Let ABCD be a parallelogram with inscribed ellipse E tangent to the sides AB, BC, CD, and
DA at the points P , Q, R, and S, respectively. Let f be an affine transformation mapping E
to a circle C = C(O , r). Under f , ABCD is mapped to a parallelogram A B C D , which is
tangent to the sides A B , B C , C D , and D A of C at points P , Q , R , and S , respectively.
(See Figure 141.)
f
B
Q
Q'
B'
C'
C
P'
2
P
-5
10
5
-2
A
S
O'
15
R
R'
-4
A'
D
FIGURE
S'
D'
141.
Since the ratio of lengths of parallel segments is preserved, it is sufficient to show that
C Q /Q B = C R /B P . As tangent segments to a circle from a point are congruent, B P =
B Q and C R = C Q . This implies the required equality of the ratios.
Remark. As A B + C D = B C + D A , the parallelogram A B C D is a rhombus, but
that fact is not used in the solution above.
12.11 Since affine transformations preserve the ratio of areas, we may assume that ABC
is a right triangle. Furthermore, we will choose a coordinatization such that C : (0, 0),
A1 : (1, 0), B : (α + 1, 0), B1 : (0, β), and A : (0, β + 1), with intersections as shown in
Figure 142.
We first find the coordinates, xC1 , yC1 , of C1 . By similar triangles,
γ
γ +1
yC1
β +1
=
and
=
.
xC1
α+1
1
γ +1
Therefore,
xC1 = γ
α+1
β +1
and yC1 =
.
γ +1
γ +1
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A: (0, β +1)
1
B1 : (0, β )
γ
C1
E
β
F
1
D
C: (0,0)
1
A1 : (1,0)
FIGURE
α
B: (α +1, 0)
142.
We then have the following equations of lines (details omitted):
←→
AA1 : x +
←→
BB1 :
←→
CC1 :
y
= 1,
β +1
y
x
+ = 1,
α+1 β
(β + 1)x
y=
.
γ (α + 1)
Solving the three systems of equations corresponding to the pairwise intersections of the
lines above (details omitted), we obtain :
β +1
γ (α + 1)
,
,
D:
γ + αγ + 1 γ + αγ + 1
α+1
αβ(β + 1)
E:
,
,
α + αβ + 1 α + αβ + 1
β(β + 1)
βγ (α + 1)
,
.
F :
β + βγ + 1 β + βγ + 1
Now, to find the area of DEF , we use the results of Problem 8.14, where the area of a
triangle is expressed in terms of the coordinates of its vertices. Again, we omit the calculations
(if you wish, use a computer!):
1
| − xE yD − xD yF − xF yE + xD yE + xF yD + xE yF |
2
(αβγ − 1)2 (α + 1)(β + 1)
.
=
2(α + αβ + 1)(α + αγ + 1)(β + βγ + 1)
Area(DEF ) =
Since the area of the (right) triangle ABC is (α + 1)(β + 1)/2, we have
Area(DEF )
(αβγ − 1)2
=
.
Area(ABC)
(α + αβ + 1)(α + αγ + 1)(β + βγ + 1)
Note that when α = β = γ = 2, this ratio simplifies to 1/7, which is the value found in
Example 76.
Also, note that our calculation of the area of DEF gives us yet another proof of Ceva’s
theorem (see Theorem 3.17, Problem 5.17, Problem 7.23, and Problem 8.16), since segments
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O
B
P
A
C
Q
FIGURE
143.
BB1 and CC1 will be concurrent if and only if the area of DEF is 0. Obviously, this is the
case if and only if αβγ = 1.
12.12 Let E be one of the three congruent and similarly oriented ellipses. Let f be an affine
transformation that maps E to a circle, C. By Theorem 12.7, each of the other three ellipses
will be mapped to a circle that is congruent to C and the three circles will be externally tangent
in pairs, as the three ellipses were. Assume that the three circles have centers O, P , and Q,
and radius r; let the points of tangency be A, B, and C. (See Figure 143.)
The area of the curvilinear triangle ABC can be found by subtracting the area of the three
congruent circular sectors from the area of the triangle OP Q:
√
π 1
√
π
π
1
.
(OC)(QP ) − 3 r 2 = (2r)( 3r) − r 2 = r 2 3 −
2
6
2
2
2
This area is constant, determined only by the radii of the three congruent, externally tangent,
circles. Since an affine transformation preserves ratios of areas, the area of the original
curvilinear triangle bounded by the three congruent ellipses must also be independent of the
position of the three ellipses. Furthermore, the ratio of the area of the elliptical curvilinear
triangle to the area of the circular curvilinear triangle must be the same as the ratio of the
area of one of the ellipses to one of the circles – which is ab : r 2 , by Problem
√ 12.9. In order
to attain this ratio, the area of the elliptical curvilinear triangle must be ab( 3 − π/2).
12.13 Consider an affine transformation f that maps the given ellipse, E, to the unit circle, which
we denote C. Then f will map the inscribed triangle T to a triangle T inscribed in C.
Furthermore, if T is a triangle with maximum area in C, by Theorem 12.7(7), T will be a
triangle with maximum area in the original ellipse. Now, since f will map the centroid of T
to the centroid of T , it remains for us to show that the centroid of a T with maximal area
coincides with the center of the unit circle. This is true since T must be equilateral. See the
solution of Problem 5.28, the related discussion in [49] mentioned in the footnote, and the
solution of Problem 7.33.
We now turn to the general case of an n-gon inscribed in an ellipse. What we present below
is not a rigorous argument, but a very natural one which is often suggested by those who are
unaware of the hidden problem with this kind of reasoning. We show that a non-regular n-gon
P inscribed in a circle does not have the maximum area. Moreover, we describe a way in
which P can be modified to create another inscribed n-gon with a greater area. This solution
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becomes rigorous if one can explain that an inscribed n-gon of the maximum area exists, but
we do not know how to do this in a relatively short manner and without applying some facts
from Real Analysis that many students may not know. See the solution of Problem 5.29, the
related discussion in [49] mentioned in the footnote, and the solution of Problem 7.33. The
solution we gave for Problem 5.29 and the discussion in [49] circumvent the necessity for
such a proof by a direct comparison of the areas of P with the area of the regular inscribed
n-gon. Here is our promised nonrigorous argument.
Consider a convex n-gon, P = A1 A2 A3 . . . An , inscribed in the unit circle, and assume
that the n-gon is not regular. Then there must be a set of three consecutive vertices such
that the pair of side lengths with endpoints at these vertices are not congruent; assume
that A1 A2 = A2 A3 , so that the three consecutive vertices are A1 , A2 , and A3 . The area of
A1 A2 A3 . . . An is the area of the (n − 1)-gon A1 A3 . . . An together with the area of A1 A2 A3 .
The area of A1 A2 A3 is (A1 A3 )(A2 X)/2, where X is the foot of the altitude from A2 . For a
fixed value of A1 A3 , the area of the triangle is maximized when A2 X is maximized, which
occurs when A2 X bisects A1 A3 . Since this is not the case for A1 A2 A3 , we conclude that
P does not have maximum area among convex n-gons that are inscribed in the unit circle.
That is, if P is not regular, P does not have maximum area.
Inversions
13.1 Consider the circle C(O, r) with chord AB having midpoint M, as shown in Figure 144.
Clearly segment OC bisects chord AB. Hence, we are assured that points O, M, and C are
collinear.
A
O
M
C
B
FIGURE
144.
Since OAC is a right triangle and AM is an altitude to its hypotenuse, by Theorem 3.19(2), AM 2 = (OM)(MC). Therefore,
OA2 = AM 2 + OM 2 = (OM)(MC) + OM 2 = (OM)(OC).
But OA = r, which is the radius of inversion. Hence, r 2 = (OM)(OC). This assures us
that I (M) = C.
13.2 Let I = I (O, r), and let line OM intersect C along the diameter AB. Then line OM intersects
C along the diameter A B . Using Theorem 13.1(3) we obtain
A M =
r2
r2
AM, and B M =
BM.
OA · OM
OB · OM
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