14 th Annual Harvard-MIT Mathematics Tournament Saturday 12 February 2011

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1 14 th Annual Harvard-MIT Mathematics Tournament Saturday 1 February Let a, b, and c be positive real numbers. etermine the largest total number of real roots that the following three polynomials may have among them: ax + bx + c, bx + cx + a, and cx + ax + b. Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a 4bc, b 4ca, and c 4ab. Multiplying these inequalities gives (abc) 64(abc), a contradiction. Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example, the values (a, b, c) = (1, 5, 6) give, 3 as roots to x + 5x + 6 and 1, 1 5 as roots to 5x + 6x Let A be a triangle such that A = 7, and let the angle bisector of A intersect line at. If there exist points E and F on sides A and, respectively, such that lines A and EF are parallel and divide triangle A into three parts of equal area, determine the number of possible integral values for. Answer: 13 E 14 F A 7 Note that such E, F exist if and only if [A] =. (1) [A] ([ ] denotes area.) Since A is the angle bisector, and the ratio of areas of triangles with equal height is the ratio of their bases, A A = = [A] [A]. Hence (1) is equivalent to A = A = 14. Then can be any length d such that the triangle inequalities are satisfied: d + 7 > > d Hence 7 < d < 1 and there are 13 possible integral values for.

2 3. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner. At each step, he can either move one square to the right or simultaneously move one square to the left and one square up. In how many ways can he reach the center square of the topmost row? Answer: n 1 Note that Josh must pass through the center square of each row. There are ways to get from the center square of row k to the center square of row k + 1. So there are n 1 ways to get to the center square of row n. 4. Let H be a regular hexagon of side length x. all a hexagon in the same plane a distortion of H if and only if it can be obtained from H by translating each vertex of H by a distance strictly less than 1. etermine the smallest value of x for which every distortion of H is necessarily convex. Answer: 4 X A 1 X A 6 A A 5 Y A 3 Y A 4 Let H = A 1 A A 3 A 4 A 5 A 6 be the hexagon, and for all 1 i 6, let points A i be considered such that A i A i < 1. Let H = A 1A A 3A 4A 5A 6, and consider all indices modulo 6. For any point P in the plane, let (P ) denote the unit disk {Q P Q < 1} centered at P ; it follows that A i (A i). Let X and X be points on line A 1 A 6, and let Y and Y be points on line A 3 A 4 such that A 1 X = A 1 X = A 3 Y = A 3 Y = 1 and X and X lie on opposite sides of A 1 and Y and Y lie on opposite sides of A 3. If X and Y lie on segments A 1 A 6 and A 3 A 4, respectively, then segment A 1A 3 lies between the lines XY and X Y. Note that x is the distance from A to A 1 A 3.

3 X A 1 X A 6 A A 5 Y A 3 Y A 4 If x, then (A ) cannot intersect line XY, since the distance from XY to A 1 A 3 is 1 and the distance from XY to A is at least 1. Therefore, A 1A 3 separates A from the other 3 vertices of the hexagon. y analogous reasoning applied to the other vertices, we may conclude that H is convex. If x <, then (A ) intersects XY, so by choosing A 1 = X and A 3 = Y, we see that we may choose A on the opposite side of XY, in which case H will be concave. Hence the answer is 4, as desired. 5. Let a b = ab + a + b for all integers a and b. Evaluate 1 ( (3 (4... (99 100)...))). Answer: 101! 1 We will first show that is both commutative and associative. ommutativity: a b = ab + a + b = b a Associativity: a (b c) = a(bc + b + c) + a + bc + b + c = abc + ab + ac + bc + a + b + c and (a b) c = (ab + a + b)c + ab + a + b + c = abc + ab + ac + bc + a + b + c. So a (b c) = (a b) c. So we need only calculate ((... (1 ) 3) 4) ). We will prove by induction that ((... (1 ) 3) 4)... n) = (n + 1)! 1. ase case (n = ): (1 ) = = 5 = 3! 1 Inductive step: Suppose that Then, (((... (1 ) 3) 4)... n) = (n + 1)! 1. ((((... (1 ) 3) 4)... n) (n + 1)) = ((n + 1)! 1) (n + 1) = (n + 1)!(n + 1) (n + 1) + (n + 1)! 1 + (n + 1) = (n + )! 1 Hence, ((... (1 ) 3) 4)... n) = (n + 1)! 1 for all n. For n = 100, this results to 101! 1.

4 6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. If Nathaniel goes first, determine the probability that he ends up winning. 5 Answer: 11 For 1 k 6, let x k be the probability that the current player, say A, will win when the number on the tally at the beginning of his turn is k modulo 7. The probability that the total is l modulo 7 after his roll is 1 6 for each l k (mod 7); in particular, there is a 1 6 chance he wins immediately. The chance that A will win if he leaves l on the board after his turn is 1 x l. Hence for 1 k 6, x k = 1 (1 x l ) l 6, l k Letting s = 6 l=1 x l, this becomes x k = x k s or 5x k 6 = s Hence x 1 = = x 6, and 6x k = s for every k. Plugging this in gives 11x k 6 = 1, or x k = Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7. Hence Obediah s chance of winning is 6 11 and Nathaniel s chance of winning is Find all integers x such that x + x 6 is a positive integral power of a prime positive integer. Answer: 3,, 5 Let f(x) = x + x 6 = (x 3)(x + ). Suppose a positive integer a divides both x 3 and x +. Then a must also divide (x + ) (x 3) = 7. Hence, a can either be 1 or 7. As a result, x 3 = 7 n or 7 n for some positive integer n, or either x + or x 3 is ±1. We consider the following cases: (x 3) = 1. Then x =, which yields f(x) = 4, a prime power. (x 3) = 1. Then x = 1, which yields f(x) = 3, not a prime power. (x + ) = 1). Then x = 1, which yields f(x) = 5 not a prime power. (x + ) = 1. Then x = 3, which yields f(x) = 9, a prime power. (x 3) = 7. Then x = 5, which yields f(x) = 49, a prime power. (x 3) = 7. Then x =, which yields f(x) = 0, not a prime power. (x 3) = ±7 n (x 3) + 7, for n. Then, since x + =, we have that x + is divisible by 7 but not by 49. Hence x + = ±7, yielding x = 5, 9. The former has already been considered, while the latter yields f(x) = 147. So x can be either -3, or 5. (Note: In the official solutions packet we did not list the answer -3. This oversight was quickly noticed on the day of the test, and only the answer 3,, 5 was marked as correct. 8. Let AEF be a regular hexagon of area 1. Let M be the midpoint of E. Let X be the intersection of A and M, let Y be the intersection of F and AM, and let Z be the intersection of A and F. If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate [X] + [AY F ] + [AZ] [MXZY ]. Answer: 0

5 A Z Y X F O E M Let O be the center of the hexagon. The desired area is [AEF ] [AM] [F EM]. Note that [AM] = [AE]/ = [OE] = [A], where the last equation holds because sin 60 = sin 10. Thus, [AM] = [A] + [AM] = [A] + [A] = [A], but the area of A is half the area of the hexagon. Similarly, the area of [F EM] is half the area of the hexagon, so the answer is zero. 9. For all real numbers x, let f(x) = x 011. Evaluate (f(f(... (f(011))...))) 011, where f is applied 010 times. 011 Answer: x irect calculation shows that f(f(x)) = 011 and f(f(f(x))) = x. x Hence (f(f(... (f(x))...))) = x, where f is applied 010 times. So (f(f(... (f(011))...))) 011 = Let A be a square of side length 13. Let E and F be points on rays A and A, respectively, so that the area of square A equals the area of triangle AEF. If EF intersects at X and X = 6, determine F. Answer: 13 F x y Y 7 X 6 A E

6 First Solution Let Y be the point of intersection of lines EF and. Note that [A] = [AEF ] implies that [EX] + [Y F ] = [Y X]. Since EX Y X Y F, there exists some constant r such that [EX] = r X, [Y F ] = r X, and [Y X] = r F. Hence X + F = X, so F = X X = = 13. Second Solution Let x = F and y = Y. Since XE XY F Y, we have E X = Y X = Y F = y x. Using X = 6, X = 7 and Y = 13 y we get E = 6y 13 y x and 7 = y x. Solving this last equation for y gives y = 13x x+7. Now [A] = [AEF ] gives 169 = 1 AE AF = 1 ( y ) (13 + x). x 169 = 6y + 13x + 78y x 13 = 6x x x + 78 x = x 13. Thus x = Let f(x) = x + 6x + c for all real numbers x, where c is some real number. For what values of c does f(f(x)) have exactly 3 distinct real roots? Answer: Suppose f has only one distinct root r 1. Then, if x 1 is a root of f(f(x)), it must be the case that f(x 1 ) = r 1. As a result, f(f(x)) would have at most two roots, thus not satisfying the problem condition. Hence f has two distinct roots. Let them be r 1 r. Since f(f(x)) has just three distinct roots, either f(x) = r 1 or f(x) = r has one distinct root. Assume without loss of generality that r 1 has one distinct root. Then f(x) = x + 6x + c = r 1 has one root, so that x + 6x + c r 1 is a square polynomial. Therefore, c r 1 = 9, so that r 1 = c 9. So c 9 is a root of f. So (c 9) + 6(c 9) + c = 0, yielding c 11c + 7 = 0, or (c 11. This results to c = 11± 13. ) = 13 If c = 11 13, f(x) = x + 6x = (x )(x ). We know f(x) = 7 13 has a double root, -3. Now > 7 13 so the second root is above the vertex of the parabola, and is hit twice. If c = , f(x) = x + 6x = (x )(x ). We know f(x) = has a double root, -3, and this is the value of f at the vertex of the parabola, so it is its minimum value. Since 5 13 < 7+ 13, f(x) = 5 13 has no solutions. So in this case, f has only one real root. So the answer is c = Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversight on the day of the test and awarded points only for the correct answer. 1. Let AEF be a convex equilateral hexagon such that lines, A, and EF are parallel. Let H be the orthocenter of triangle A. If the smallest interior angle of the hexagon is 4 degrees, determine the smallest angle of the triangle HA in degrees. Answer: 3

7 A A F H E A F H E Note that A and EF A are isosceles trapezoids, so A = A and F A = EA. In order for the hexagon to be convex, the angles at,, E, and F have to be obtuse, so A = = 4. Letting s be a side length of the hexagon, A = A cos A + + cos A = s(1 + cos A), so A is uniquely determined by A. Since the same equation holds for trapezoid EF A, it follows that A = F A = A = EA =. Then = 180 = 178. Since is isosceles, = 1 and A = 1. (One may also note that A = 1 by observing that equal lengths A and must intercept equal arcs on the circumcircle of isosceles trapezoid A). Let A,, and be the feet of the perpendiculars from A,, and to, A, and A, respectively. Angle chasing yields AH = AH + H = (90 A A ) + (90 ) = A + A = 1 + = 3 HA = 90 AH = 89 HA = 90 H = 88 Hence the smallest angle in HA is 3. It is faster, however, to draw the circumcircle of EF A, and to note that since H is the orthocenter of triangle A, is the orthocenter of triangle HA. Then since F is the reflection of across A, quadrilateral HAF is cyclic, so AH = AF + AF = 1 + = 3, as desired.

8 13. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 P (x) < 10 for all x {0, 1,, 3, 4, 5}? Answer: For each nonnegative integer i, let x i = x(x 1) (x i + 1). (efine x 0 = 1.) Lemma: Each polynomial with integer coefficients f can be uniquely written in the form f(x) = a n x n a 1 x 1 + a 0 x 0, a n 0. Proof: Induct on the degree. The base case (degree 0) is clear. If f has degree m with leading coefficient c, then by matching leading coefficients we must have m = n and a n = c. y the induction hypothesis, f(x) cx n can be uniquely written as a n 1 x n 1 (x) a 1 x 1 + a 0 x 0. There are 10 possible choices for a 0, namely any integer in [0, 10). Once a 0,..., a i 1 have been chosen so 0 P (0),..., P (i 1) < 10, for some 0 i 5, then we have P (i) = a i i! + a i 1 i i a 0 so by choosing a i we can make P (i) any number congruent to a i 1 i i 1 + +a 0 modulo i!. Thus there are 10 i! choices for a i. Note the choice of a i does not affect the value of P (0),..., P (i 1). Thus all polynomials we obtain in this way are valid. The answer is 5 i=0 10 i! = Note: Their is also a solution involving finite differences that is basically equivalent to this solution. One proves that for i = 0, 1,, 3, 4, 5 there are 5! i! ways to pick the ith finite difference at the point Let A be a cyclic quadrilateral, and suppose that = =. Let I be the incenter of triangle A. If AI = as well, find the minimum value of the length of diagonal. Answer: 3 A I

9 Let T be the point where the incircle intersects A, and let r be the inradius and R be the circumradius of A. Since = =, is on the midpoint of arc on the opposite side of as A, and hence on the angle bisector of A. Thus A, I, and are collinear. We have the following formulas: AI = IM sin IAM = = R sin A = R sin A The last two equations follow from the extended law of sines on A and A, respectively. Using AI = = gives sin A = r R. However, it is well-known that R r with equality for an equilateral triangle (one way to see this is the identity 1+ r R = cos A+cos +cos ). Hence sin A 1 4 and A 30. Then r sin A ( = R sin A cos A ) ( = cos A with equality when A is equilateral. ) 3 = 3 Remark: Similar but perhaps simpler computations can be made by noting that if A intersects at X, then A/X = A/X =, which follows from the exterior angle bisector theorem; if I A is the A-excenter of triangle A, then AI A /XI A = since it is well-known that is the circumcenter of cyclic quadrilateral II A. 15. Let f(x) = x r x + r 3 for all real numbers x, where r and r 3 are some real numbers. efine a sequence {g n } for all nonnegative integers n by g 0 = 0 and g n+1 = f(g n ). Assume that {g n } satisfies the following three conditions: (i) g i < g i+1 and g i+1 > g i+ for all 0 i 011; (ii) there exists a positive integer j such that g i+1 > g i for all i > j, and (iii) {g n } is unbounded. If A is the greatest number such that A r for any function f satisfying these properties, find A. Answer: onsider the function f(x) x. y the constraints of the problem, f(x) x must be negative for some x, namely, for x = g i+1, 0 i 011. Since f(x) x is positive for x of large absolute value, the graph of f(x) x crosses the x-axis twice and f(x) x has two real roots, say a < b. Factoring gives f(x) x = (x a)(x b), or f(x) = (x a)(x b) + x. Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c b be the number such that f(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, c exists and b+c = r b+a as the vertex of the parabola. y the same token, = r+1 is the vertex of f(x) x. Hence c = a 1. If f(x) > b then x < c or x > b. onsider the smallest j such that g j > b. Then by the above observation, g j 1 < c. (If g i b then f(g i ) g i b so by induction, g i+1 g i for all i j. Hence j > 1; in fact j 405.) Since g j 1 = f(g j ), the minimum value of f is less than c. The minimum value is the value of f evaluated at its vertex, b+a 1, so ( ) b + a 1 f < c ( ) ( ) b + a 1 b + a 1 a b + b + a 1 < a 1 1 (b a) + b a + 1 < (b a) < b a < (b a 1).

10 Then either b a 1 < or b a 1 >, but b > a, so the latter must hold and (b a) > 9. Now, the discriminant of f(x) x equals (b a) (the square of the difference of the two roots) and (r + 1) 4r 3 (from the coefficients), so (r + 1) > 9 + 4r 3. ut r 3 = g 1 > g 0 = 0 so r >. We claim that we can make r arbitrarily close to, so that the answer is. First define G i, i 0 as follows. Let N 01 be an integer. For ε > 0 let h(x) = x ε, g ε (x) = x + + ε and G N+1 = + ε, and define G i recursively by G i = g ε (G i+1 ), G i+1 = h(g i ). (These two equations are consistent.) Note the following. (i) G i < G i+1 and G i+1 > G i+ for 0 i N 1. First note G N = 4 + ε > 4 + ε + ε = ε. Let l be the negative solution to h(x) = x. Note that ε < G N < l < 0 since h(g N ) > 0 > G N. Now g ε (x) is defined as long as x ε, and it sends ( ε, l) into (l, 0) and (l, 0) into ( ε, l). It follows that the G i, 0 i N are well-defined; moreover, G i < l and G i+1 > l for 0 i N 1 by backwards induction on i, so the desired inequalities follow. (ii) G i is increasing for i N + 1. Indeed, if x + ε, then x x = x(x 1) > + ε so h(x) > x. Hence + ε = G N+1 < G N+ <. (iii) G i is unbounded. This follows since h(x) x = x(x ) ε is increasing for x > + ε, so G i increases faster and faster for i N + 1. Now define f(x) = h(x + G 0 ) G 0 = x + G 0 x + G 0 G 0 ε. Note G i+1 = h(g i ) while g i+1 = f(g i ) = h(g i + G 0 ) G 0, so by induction g i = G i G 0. Since {G i } i=0 satisfies (i), (ii), and (iii), so does g i. We claim that we can make G 0 arbitrarily close to 1 by choosing N large enough and ε small enough; this will make r = G 0 arbitrarily close to. hoosing N large corresponds to taking G 0 to be a larger iterate of + ε under g ε (x). y continuity of this function with respect to x and ε, it suffices to take ε = 0 and show that (letting g = g 0 ) ut note that for 0 θ π, g (n) () = g( g() ) 1 as n. }{{} n g( cos θ) = cos θ = sin ( ) ( θ π = cos θ ). Hence by induction, g (n) ( cos θ) = cos ( π π ( )) ( 1)n θ π. Hence g (n) () = g (n 1) ( cos 0) n converges to cos( π π 4 + ) = cos( π 3 ) = 1, as needed. 16. Let A be a quadrilateral inscribed in the unit circle such that A is 30 degrees. Let m denote the minimum value of P + P Q + Q, where P and Q may be any points lying along rays A and A, respectively. etermine the maximum value of m. Answer:

11 G A R E H Q P F For a fixed quadrilateral A as described, we first show that m, the minimum possible length of P + P Q + Q, equals the length of A. Reflect,, and P across line A to points E, F, and R, respectively, and then reflect and F across AE to points G and H, respectively. These two reflections combine to give a 60 rotation around A, so triangle AH is equilateral. It also follows that RH is a 60 rotation of P around A, so, in particular, these segments have the same length. ecause QR = QP by reflection, P + P Q + Q = Q + QR + RH. The latter is the length of a broken path QRH from to H, and by the shortest path is a straight line principle, this total length is at least as long as H = A. (More directly, this follows from the triangle inequality: (Q+QR)+RH R+RH H). Therefore, the lower bound m A indeed holds. To see that this is actually an equality, note that choosing Q as the intersection of segment H with ray A, and choosing P so that its reflection R is the intersection of H with ray AE, aligns path QRH with segment H, thus obtaining the desired minimum m = A. We may conclude that the largest possible value of m is the largest possible length of A, namely : the length of a diameter of the circle. 17. Let z = cos π π + i sin, and let P (x) = x x x for all complex numbers x. Evaluate P (z)p (z )P (z 3 )... P (z 010 ). Answer: ( ) Multiply P (x) by x 1 to get x + or, P (x)(x 1) = x x x , Multiplying by x 1 once again: P (x)(x 1) = x x x

12 (x 1)(P (x)(x 1) ) = x x x 010, = (x x x + 1) 011. Hence, P (x) = (x x x + 1) x 1 x 1 Note that x 010 +x x+1 has z, z,..., z 010 as roots, so they vanish at those points. Plugging those 010 powers of z into the last equation, and multiplying them together, we obtain 010 i=1 P (z i ) = ( 011) 1005 (x ) (x 1). Note that (x z)(x z )... (x z 010 ) = x x Using this, the product turns out to be ( ). 18. ollinear points A,, and are given in the artesian plane such that A = (a, 0) lies along the x-axis, lies along the line y = x, lies along the line y = x, and A/ =. If = (a, a), the circumcircle of triangle A intersects y = x again at E, and ray AE intersects y = x at F, evaluate AE/EF. Answer: 7

13 Q P O F E A Let points O, P, and Q be located at (0, 0), (a, a), and (0, a), respectively. Note that /A = 1/ implies [O]/[OA] = 1/, so since [OP ] = [OA], [O]/[OP ] = 1/. It follows that [O] = [OP ]. Hence O = P. We may conclude that triangles OQ and P A are congruent, so = (a/, a). It follows that A is right, so the circumcircle of triangle A is the midpoint of A, which is located at (3a/4, a/). Let (3a/4, a/) = H, and let E = (b, b). Then the power of the point O with respect to the circumcircle of A is O OE = ab, but it may also be computed as OH HA = 13a/16 5a/16 = a/. It follows that b = a/4, so E = (a/4, a/4). We may conclude that line AE is x + 3y = a, which intersects y = x at an x-coordinate of a/7. Therefore, AE/EF = (a a/4)/(a/4 a/7) = (3a/4)/(3a/8) = 7. Remark: The problem may be solved more quickly if one notes from the beginning that lines OA, O, OP, and OQ form a harmonic pencil because is the midpoint of AP and lines OQ and AP are parallel. 19. Let {a n } and {b n } be sequences defined recursively by a 0 = ; b 0 =, and a n+1 = a n 1 + a n + b n b n ; b n+1 = b n 1 + a n + b n + a n. Find the ternary (base 3) representation of a 4 and b 4. Answer: and Note first that 1 + a n + b n = 3 n. The proof is by induction; the base case follows trivially from what is given. For the inductive step, note

14 that 1 + a n+1 + b n+1 = 1 + a n(1 + a n + b n) + b n a n b n 1 + a n + b n + b n(1 + a n + b n) + a n + a n b n 1 + a n + b n = 1 + (a n + b n)(1 + a n + b n) + a n + b n = (1 + a n + b n). Invoking the inductive hypothesis, we see that 1 + a n+1 + b n+1 = (3n ) = 3 n+1, as desired. The quickest way to finish from here is to consider a sequence of complex numbers {z n } defined by z n = a n + b n i for all nonnegative integers n. It should be clear that z 0 = + i and z n+1 = z n (3 n + i). Therefore, z 4 = ( + i)(3 0 + i)(3 1 + i)(3 + i)(3 3 + i). This product is difficult to evaluate in the decimal number system, but in ternary the calculation is a cinch! To speed things up, we will use balanced ternary 1, in which the three digits allowed are 1, 0, and 1 rather than 0, 1, and. Let x + yi = (3 0 + i)(3 1 + i)(3 + i)(3 3 + i), and consider the balanced ternary representation of x and y. For all 0 j 15, let x j denote the digit in the 3 j place of x, let y j denote the digit in the 3 j place of y, and let b(j) denote the number of ones in the binary representation of j. It should be clear that x j = 1 if b(j) (mod 4), x j = 0 if b(j) 1 (mod ), and x j = 1 if b(j) 0 (mod 4). Similarly, y j = 1 if b(j) 1 (mod 4), y j = 0 if b(j) 0 (mod ), and y j = 1 if b(j) 3 (mod 4). onverting to ordinary ternary representation, we see that x = and y = It remains to note that a 4 = x y and b 4 = x + y and perform the requisite arithmetic to arrive at the answer above. 0. Let ω 1 and ω be two circles that intersect at points A and. Let line l be tangent to ω 1 at P and to ω at Q so that A is closer to P Q than. Let points R and S lie along rays P A and QA, respectively, so that P Q = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenter of triangle ASR, and let and be the midpoints of major arcs AP and AQ, respectively. If AP Q is 45 degrees and AQP is 30 degrees, determine O in degrees. Answer: 14.5 Q P A R S T T We use directed angles throughout the solution. Let T denote the point such that T = 1/ AP Q and T = 1/ AQP. We claim that T is the circumcenter of triangle SAR. Since P = A, QP = RA, and P Q = P A + AP Q = P A + AP = AR, we have P Q = AR. y spiral similarity, we have P A QR. Let T denote the reflection of T across. Since T T = AP Q = AP, we have T T AP RQ. Again, by spiral similarity centered at, we have T R T Q. ut T = T, so T R = T Q and T R = T Q. Similarly, T T AQ, and spiral similarity centered at shows that T A = T Q. Thus T A = T Q = T R. We similarly have T A = T P = T S, so T is indeed the circumcenter. Therefore, we have O = T = =

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