Solutions to hapter Problems 435 Then, using α + β + γ = 360, we obtain: ( ) x a = (/2) bc sin α a + ac sin β b + ab sin γ c a ( ) = (/2) bc sin α a 2 + (ac sin β)(ab cos γ ) + (ab sin γ )(ac cos β) = (/2)a 2 bc (sin α + sin β cos γ + sin γ cos β) = (/2)a 2 bc (sin α + sin(β + γ )) = (/2)a 2 bc (sin α + sin(360 α)) = (/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.7. ffine Transformations 2. Suppose h = g f and h 2 are both affine transformations sending p, q, and r to p, q, and r, respectively. We need to show h = h 2.Now,h f and h 2 f are both affine transformations mapping { 0, i, j} to { p, q, r }. ut since an affine transformation is uniquely determined by where it maps 0, i, and j, h f = h 2 f. omposing each of these compositions on the right with f, we obtain h = h 2, which proves the uniqueness of the composition h. 2.2 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). 2.3 LetD be a trapezoid with bases D and.letp be the intersection of and D, and let Q be the intersection of the diagonals, and D. y orollary 2.9, there is an affine transformation f mapping P D to an isosceles triangle P D. (See Figure 36.) ecause 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 2.7), D is mapped to an isosceles trapezoid, D, with = D.Itis easy to see that P Q bisects the bases of D ; we leave the details to the reader. We conclude that PQbisects the bases of D also. This problem also appeared as Problem 3.3.4 and as Problem 8.2. f P P' ' ' Q Q' D ' D' FIGURE 36.
436 Solutions to hapter Problems 2.4 ssume that and D are chords of an ellipse, E, having the property that the area determined by and the boundary of E is the same as the area determined by D and the boundary of E.Letf be an affine transformation mapping E to a circle, = (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 ;let designate the point on that corresponds to f () and likewise for,, and D. M' ' ' O' D' N' ' FIGURE 37. y Theorem 2.7(7), the area of the region bounded by the chord and the boundary of will be equal to the area of the region bounded by the chord D and the boundary of. This implies that rea( O ) = rea( D O ) and that = D.LetM and N be the midpoints of and D, respectively. y Theorem 4.6 of hapter 4, O M and O N D. Since rea( O ) = rea( D O ) and = D, we conclude that O M = O N. Hence, and D are both tangent to the circle (O,O M ). The pre-image of (O,O M ) is an ellipse having center at O, which is tangent to and D. 2.5 Let,..., n be the vertices of an n-gon, and,..., n be their images under an affine transformation f.letg be the centroid of {,..., n }. Then by Example 65, Gi = 0. Let G = f (G). Suppose f is defined by x x = ( x) + ( b). Then f ( O i ) = ( O i ) + ( b), and f ( OG) = OG + b,so G i = f ( G i ) = G i. This implies (by a generalization of Theorem 2.2(5)) that G i = 0, which is equivalent to G being the centroid of {,..., n }. 2.6 onsider a parabola, [ P, ] with vertex at (h, k). s 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 where a is a positive real number. y = ax 2,
Solutions to hapter Problems 437 [ ][ ] [ ] [ ] /a 0 x x/a x Now apply a third affine transformation, f ( x) = = = 0 /a y y/a y. The equation y = ax 2 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, P and P 2, are affine equivalent. The proof is similar to that of orollary 2.4. 2.7 Let 2, 2, and 2 be the points of intersection of and, and, and and, respectively. Let f be an affine transformation mapping to an equilateral triangle DEF, as illustrated in Figure 38. ecause affine transformations preserve ratios of parallel line segments, the points D = f ( ), E = f ( ), and F = f ( ) will divide the sides of DEF in the same fixed ratio as,, and divide the sides of. onsequently, F ED = D FE = E DF by SS, so D E F is equilateral. f E D F 2 G 2 2 D E 2 F 2 G' D 2 E F FIGURE 38. Let D 2, E 2, and F 2 be the points of intersections of the segments EE and FF, FF and DD, and DD and EE, respectively. Rotating DEF clockwise by 20 around the centroid, G,of DEF (as we did in Example 76) we see that D E F D, where means is mapped to. This implies that DD EE FF DD, and therefore D 2 E 2 F 2 D 2. This proves that D 2 E 2 F 2 is equilateral. Furthermore, under the 20 rotation about O, the equilateral triangle D E F is mapped to itself, as is the equilateral triangle D 2 E 2 F 2. This will only happen if G is the centroid of both D E F and D 2 E 2 F 2. Since any affine transformation maps the centroid of one triangle to the centroid of another, the pre-image of G, f (G ) = G, is the centroid of,, and 2 2. 2.8 y Theorem 2.7, 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 MNPQ to a square M N P Q having side length a =. (See Figure 39.) We now wish to prove that /M R + /M S = /M T. Let m Q M T = α. Then, from the right triangles M N R and M Q S, respectively, we see that M R = cos N M R = cos (90 α) = sin α and M S = cos α.
438 Solutions to hapter Problems S S' N T R l P f N' 45 R' T' P' M Q α M' a= Q' FIGURE 39. In addition, note that m M N Q = 45 since N Q is the diagonal of a square. pplying the Sine theorem to M T N, sin (45 + α) = M T sin 45, so = sin (45 + α). M T sin 45 Thus, verifying that /M R + /M S = /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 + α) = (sin 45 )(cos α) + (cos 45 )(sin α) = cos α + sin α. sin 45 sin 45 n alternate solution is provided in Problem 3.3.2. 2.9 Suppose E is an ellipse given by the equation x 2 /a 2 + y 2 /b 2 = ; then E has semi-axes of lengths a and b, as shown in Figure 40(i). onstruct 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 4 S E 2 b O a -5 5 0 5 O r=a -2-4 -6 (i) (ii) FIGURE 40. [ ] 0 onsider the affine transformation f ( p) = p, where =. 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 /b 2 = a 2,so E is mapped to a circle with radius a, as shown in Figure 40(ii). Obviously, f (R) isa
Solutions to hapter 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, rea E rea R = rea rea S = rea E 4ab = πa2 4a 2. From here, it follows that the area of E is πab. 2.0 Let D be a parallelogram with inscribed ellipse E tangent to the sides,, D, and D at the points P, Q, R, and S, respectively. Let f be an affine transformation mapping E toacircle = (O,r). Under f, D is mapped to a parallelogram D, which is tangent to the sides,, D, and D of at points P, Q, R, and S, respectively. (See Figure 4.) f P 2 Q P' ' Q' ' -5 5 0 O' 5-2 R R' -4 S ' S' D' D FIGURE 4. Since the ratio of lengths of parallel segments is preserved, it is sufficient to show that Q /Q = R / P. s tangent segments to a circle from a point are congruent, P = Q and R = Q. This implies the required equality of the ratios. Remark. s + D = + D, the parallelogram D is a rhombus, but that fact is not used in the solution above. 2. Since affine transformations preserve the ratio of areas, we may assume that is a right triangle. Furthermore, we will choose a coordinatization such that :(0, 0), :(, 0), :(α +, 0), :(0,β), and :(0,β + ), with intersections as shown in Figure 42. We first find the coordinates, ( x,y ),of. y similar triangles, γ x = γ + α + and y = β + γ +. Therefore, x = γ α + γ + and y = β + γ +.
440 Solutions to hapter Problems : (0, β +) : (0, β ) γ β E D F : (0,0) : (,0) α : (α+, 0) FIGURE 42. We then have the following equations of lines (details omitted): : x + : : y = y β + =, x α + + y β =, (β + )x γ (α + ). Solving the three systems of equations corresponding to the pairwise intersections of the lines above (details omitted), we obtain : ( ) γ (α + ) D : γ + αγ +, β +, γ + αγ + ( ) α + αβ(β + ) E :,, α + αβ + α + αβ + ( ) βγ(α + ) F : β + βγ +, β(β + ). β + βγ + Now, to find the area of DEF, we use the results of Problem 8.4, where the area of a triangle is expressed in terms of the coordinates of its vertices. gain, we omit the calculations (if you wish, use a computer!): rea( DEF) = 2 x E y D x D y F x F y E + x D y E + x F y D + x E y F = (αβγ ) 2 (α + )(β + ) 2(α + αβ + )(α + αγ + )(β + βγ + ). Since the area of the (right) triangle is (α + )(β + )/2, we have rea( DEF) rea( ) = (αβγ ) 2 (α + αβ + )(α + αγ + )(β + βγ + ). Note that when α = β = γ = 2, this ratio simplifies to /7, which is the value found in Example 76. lso, note that our calculation of the area of DEF gives us yet another proof of eva s theorem (see Theorem 3.7, Problem 5.7, Problem 7.23, and Problem 8.6), since segments
Solutions to hapter Problems 44 O Q P FIGURE 43. and will be concurrent if and only if the area of DEF is 0. Obviously, this is the case if and only if αβγ =. 2.2 Let E be one of the three congruent and similarly oriented ellipses. Let f be an affine transformation that maps E to a circle,. y Theorem 2.7, each of the other three ellipses will be mapped to a circle that is congruent to and the three circles will be externally tangent in pairs, as the three ellipses were. ssume that the three circles have centers O, P, and Q, and radius r; let the points of tangency be,, and. (See Figure 43.) The area of the curvilinear triangle can be found by subtracting the area of the three congruent circular sectors from the area of the triangle OPQ: ( π 2 (O)(QP ) 3 6 r2) = 2 (2r)( 3r) π 2 r2 = r 2( π ) 3. 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 2.9. In order to attain this ratio, the area of the elliptical curvilinear triangle must be ab( 3 π/2). 2.3 onsider an affine transformation f that maps the given ellipse, E, to the unit circle, which we denote. Then f will map the inscribed triangle T to a triangle T inscribed in. Furthermore, if T is a triangle with maximum area in, by Theorem 2.7(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
442 Solutions to hapter 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 nalysis 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. onsider a convex n-gon, P = 2 3... n, 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 2 2 3, so that the three consecutive vertices are, 2, and 3. The area of 2 3... n is the area of the (n )-gon 3... n together with the area of 2 3. The area of 2 3 is ( 3 )( 2 X)/2, where X is the foot of the altitude from 2. For a fixed value of 3, the area of the triangle is maximized when 2 X is maximized, which occurs when 2 X bisects 3. Since this is not the case for 2 3, 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 3. onsider the circle (O,r) with chord having midpoint M, as shown in Figure 44. learly segment O bisects chord. Hence, we are assured that points O, M, and are collinear. O M FIGURE 44. Since O is a right triangle and M is an altitude to its hypotenuse, by Theorem 3.9(2), M 2 = (OM)(M). Therefore, O 2 = M 2 + OM 2 = (OM)(M) + OM 2 = (OM)(O). ut O = r, which is the radius of inversion. Hence, r 2 = (OM)(O). This assures us that I(M) =. 3.2 LetI = I(O,r), and let line OM intersect along the diameter. Then line OM intersects along the diameter. Using Theorem 3.(3) we obtain M = r 2 O OM M, and M r 2 = O OM M.