Affine Transformations

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Solutions to Chapter Problems 435

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)a2bc (sin α + sin β cos γ + sin γ cos β) = (1/2)a2bc (sin α + sin(β + γ )) = (1/2)a2bc (sin α + sin(360◦ − α)) = (1/2)a2bc (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

−1 12.1 Suppose h1 = g ◦ f 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 transformations mapping {0,i,j} to {p, q, r}. But since an affine transformation is uniquely determined by where it maps 0, i, and j, h1 ◦ f = h2 ◦ f . Composing each of these com- −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). ←→ ←→ 12.3LetABCD be a trapezoid with bases AD and BC.LetP 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 AP 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 Theo- rem 12.7), ABCD is mapped to an isosceles trapezoid, ABCD, with AB = CD.Itis ←−→ easy to see that←→P Q bisects the bases of A B C D ; we leave the details to the reader. We conclude that PQbisects the bases of ABCD also. This problem also appeared as Problem 3.3.14 and as Problem 8.12.

P P'

B C B' C'

Q Q'

A D A' D'

FIGURE 136. P1: KpB MABK012-BOOK MABK012/Bayer Trim Size: 7.5in× 10.5in July 15, 2010 16:24

436 Solutions to Chapter Problems

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.Letf 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;letA designate the point on C that corresponds to f (A) and likewise for B, C, and D.

B' C M'

A' O' D'

N' C'

FIGURE 137. By Theorem 12.7(7), the area of the region bounded by the chord AB and the boundary of C will be equal to the area of the region bounded by the chord CD and the boundary of C. This implies that Area(ABO) = Area(CDO) and that AB = CD.LetM and N be the midpoints of AB and CD, respectively. By Theorem 4.6 of Chapter 4, OM ⊥ AB and ON ⊥ CD. Since Area(ABO) = Area(CDO) and AB = CD, we conclude that OM = ON . Hence, AB and CD are both tangent to the circle C(O,OM). The pre-image of C(O,OM) is an ellipse having center at O, which is tangent to AB and CD.

12.5LetA1, ..., An be the vertices of an n-gon, and A1, ..., An be their images under an affine transformation f .LetG be the centroid of {A1,...,An}. Then by Example 65, −−→ GAi = 0. = Let G f (G). −→ −→ → = + = + Suppose−−→ f−−→is defined by x x A(x) (b). Then f (OAi ) A(OAi ) (b), and 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 = ax2,

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Solutions to Chapter Problems 437

1/a 0 x x/a x Now apply a third afﬁne transformation, f (x) = = = . 01/a y y/a y The equation y = ax2 can now be written (y/a) = a(x/a)2,ory = (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.

12.7LetA2, B2, and C2 be the points of intersection of BB1 and CC1, AA1 and CC1, and AA1 and BB1, 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, F1ED1 = D1FE1 = E1DF1 by SAS, so D1E1F1 is equilateral.

B E

A1 D1

C2 F2 F E C1 B 1 2 2 G G' D A2 2 C D F E1 A B1

FIGURE 138.

Let D2, E2, and F2 be the points of intersections of the segments EE1 and FF1, FF1 ◦ and DD1, and DD1 and EE1, respectively. Rotating DEF clockwise by 120 around the centroid, G ,ofDEF (as we did in Example 76) we see that D1 → E1 → F1 → D1, where “→” means “is mapped to.” This implies that DD1 → EE1 → FF1 → DD1, and therefore D2 → E2 → F2 → D2. This proves that D2E2F2 is equilateral. ◦ Furthermore, under the 120 rotation about O , the equilateral triangle D1E1F1 is mapped to itself, as is the equilateral triangle D2E2F2. This will only happen if G is the centroid of both D1E1F1 and D2E2F2. 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, A1B1C1, and A1B2C2. 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 MNPQto a square MN P Q having side length a = 1. (See Figure 139.) We now wish to prove that 1/MR + 1/MS = 1/MT . Let m∠QMT = α. Then, from the right triangles MN R and MQS, respectively, we see that

1 1 = cos ∠N MR = cos (90◦ − α) = sin α and = cos α. MR MS P1: KpB MABK012-BOOK MABK012/Bayer Trim Size: 7.5in× 10.5in July 15, 2010 16:24

438 Solutions to Chapter Problems

S' S

l f N' R' P' N R P 45 T' T α M Q M' a=1 Q'

FIGURE 139.

In addition, note that m∠MN Q = 45◦ since N Q is the diagonal of a square. Applying the Sine theorem to MT N , 1 MT 1 sin (45◦ + α) = , so = . sin (45◦ + α) sin 45◦ MT sin 45◦ Thus, verifying that 1/MR + 1/MS = 1/MT 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◦ + α) (sin 45◦)(cos α) + (cos 45◦)(sin 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 x2/a2 + y2/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 S 4

E C 2 b

-5 O a 5 10 O 15r=a

-2

-4

(ii) -6(i)

FIGURE 140. 10 Consider the afﬁne transformation f (p) = Ap, where A = . 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 = x2 + a2y2/b2 = a2,so E is mapped to a circle with radius a, as shown in Figure 140(ii). Obviously, f (R)isa P1: KpB MABK012-BOOK MABK012/Bayer Trim Size: 7.5in× 10.5in July 15, 2010 16:24

Solutions to Chapter Problems 439

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 πa2 = =⇒ = . Area R Area S 4ab 4a2

From here, it follows that the area of E is πab. 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 toacircleC = C(O,r). Under f , ABCD is mapped to a parallelogram ABCD, which is tangent to the sides AB, BC, CD, and DA of C at points P , Q, R, and S, respectively. (See Figure 141.)

B Q B' Q' C' C P' P 2

-5 5 10 O' 15 R -2 R' A -4 D' S D A' S'

FIGURE 141.

Since the ratio of lengths of parallel segments is preserved, it is sufﬁcient to show that CQ/QB = CR/BP . As tangent segments to a circle from a point are congruent, BP = BQ and CR = CQ. This implies the required equality of the ratios.

Remark. As AB + CD = BC + DA, the parallelogram ABCD 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 ,ofC1. By similar triangles,

γ γ + 1 y β + 1 = C1 = . + and + xC1 α 1 1 γ 1

Therefore,

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440 Solutions to Chapter Problems

A: (0, β +1) 1 γ ) B1 : (0, β

E C1 β F 1 D

: (0,0) α (α+1, 0) C 1 A1 : (1,0) B:

FIGURE 142.

We then have the following equations of lines (details omitted): ←→ y AA : x + = 1, 1 β + 1 ←→ x y BB : + = 1, 1 α + 1 β ←→ (β + 1)x CC : y = . 1 γ (α + 1) Solving the three systems of equations corresponding to the pairwise intersections of the lines above (details omitted), we obtain : γ (α + 1) β + 1 D : , , γ + αγ + 1 γ + αγ + 1 α + 1 αβ(β + 1) E : , , α + αβ + 1 α + αβ + 1 βγ(α + 1) β(β + 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 Area(DEF) = |−x y − x y − x y + x y + x y + x y | 2 E D D F F E D E F D E F (αβγ − 1)2(α + 1)(β + 1) = . 2(α + αβ + 1)(α + αγ + 1)(β + βγ + 1)

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 P1: KpB MABK012-BOOK MABK012/Bayer Trim Size: 7.5in× 10.5in July 15, 2010 16:24

Solutions to Chapter Problems 441

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 OPQ: 1 π 1 √ π √ π (OC)(QP ) − 3 r2 = (2r)( 3r) − r2 = r2 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 2 area of one of the ellipses to one of the circles – which is ab : r , 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 P1: KpB MABK012-BOOK MABK012/Bayer Trim Size: 7.5in× 10.5in July 15, 2010 16:24

442 Solutions to Chapter Problems

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 = A1A2A3 ...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 A1A2 = A2A3, so that the three consecutive vertices are A1, A2, and A3. The area of A1A2A3 ...An is the area of the (n − 1)-gon A1A3 ...An together with the area of A1A2A3. The area of A1A2A3 is (A1A3)(A2X)/2, where X is the foot of the altitude from A2. For a ﬁxed value of A1A3, the area of the triangle is maximized when A2X is maximized, which occurs when A2X bisects A1A3. Since this is not the case for A1A2A3, 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.

O C M

FIGURE 144.

Since OAC is a right triangle and AM is an altitude to its hypotenuse, by Theo- rem 3.19(2), AM2 = (OM)(MC). Therefore,

OA2 = AM2 + OM2 = (OM)(MC) + OM2 = (OM)(OC).

But OA = r, which is the radius of inversion. Hence, r2 = (OM)(OC). This assures us that I(M) = C. 13.2LetI = I(O,r), and let line OM intersect C along the diameter AB. Then line OM intersects C along the diameter AB. Using Theorem 13.1(3) we obtain

r2 r2 AM = AM, and BM = BM. OA · OM OB · OM