USA Mathematics Talent Search

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Shannon Sparks
6 years ago
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1 ID#: We begin by noting that a convex regular polygon has interior angle measures (in degrees) that are integers if and only if the exterior angle measures are also integers. Since the sum of the exterior angles in any convex polygon is 360, an n-sided regular polygon has an exterior angle of ( 360 n ). Thus, for this to be an integer, n must divide 360. Since 360 = has (3 + 1)( + 1)(1 + 1) = 4 factors, any of these 4 will make 360 n an integer. However, a regular convex polygon cannot have 1 or sides, and these are also included in our count. Thus, our final answer becomes 4 = values of n. Page 1 of Problem 1

2 ID#: Squaring the given equation, we obtain a + b + c + ab + bc + ca = Since there are three radical terms on the right side (which are not integers), the three radicals on the left side must match up correspondingly. Also note that because a, b, c are positive integers, this implies a + b + c = 19. Without loss of generality, we may assume b a c ab bc ca to obtain the following system of equations: 1 ab = = bc = 1600 = ca = 3580 = 4 5 Multiplying the three gives = = Thus, we can solve for a, b, c: a = bc = = 1 a = 84. b = ca = = 30 b = 30. c = ab = = 105 c = 105. Checking, we note that a + b + c = = 19 still holds. Finally, we conclude that a = 84, b = 30, c = 105 satisfy the given equation. Page 1 of Problem

3 ID#: We claim that the polynomial f(a, b, c) = a 3 + b 3 + 4c 3 6 has the desired properties. Our proof begins with the following lemma: Lemma: The expressions s = p 3 + q 3 + r 3 3pqr and t = p + q + r have the same sign for real numbers p, q, r that are not all equal. Proof: We note the following identity: p 3 + q 3 + r 3 3pqr = (p + q + r)(p + q + r pq qr rp) = 1 (p + q + r)((p q) + (q r) + (r p) ), or equivalently, s = t ((p q) + (q r) + (r p) ). Note that (p q) +(q r) +(r p) 0, with equality if and only if p = q = r. By hypothesis, this cannot hold, so (p q) + (q r) + (r p) > 0. Thus, t = 0 if and only if s = 0. Moreover, when s, t 0, we may divide by t to get s t = 1 ((p q) + (q r) + (r p) ) > 0, and the result follows. Now, we set p = a, q = b 3, and r = c 3 4, so by our lemma, p 3 + q 3 + r 3 3pqr = a 3 + b 3 + 4c 3 6 has the same sign as p+q +r = a+b 3 +c 3 4, provided that p = q = r does not hold. If p = q = r does hold, then a = b 3 = c 3 4, which implies a = b = c = 0 because a, b, c are integers. Thus, f(a, b, c) = a 3 + b 3 + 4c 3 6 = a + b + c = 0, and this case is covered as well. This concludes our proof. Page 1 of Problem 3

4 ID#: For convenience, we subsitute a = 3x, b = 4y, c = 5z. Thus, we are trying to find all integers n such that there exists nonnegative reals a, b, c satisfying a + b + c = 3 and 9 a b c = n. When n is viewed as a function of a, b, c, it is continuous, so it suffices to find the minimum and maximum values of n. It will follow that n can assume any value inbetween. To find a lower bound, we apply Cauchy s inequality, ( 9 a b + 6 ) ( 9 5 c ) 6 (a + b + c), so 14n 3 n Note that equality is possible, if a = 3 b = 6 c, and since n is an integer, we must have n 38. For an upper bound, note the following, 9 a b c = 6 5 (a + b + c) 4 5 a b 1 5 (ab + bc + ca) 6 5 (a + b + c), and thus n = , with equality if and only if a = b = 0 and c = 3. Since n is an integer, we conclude that n 16. Thus we must have 38 n 16, and n can assume any integer value in this interval. Page 1 of Problem 4

5 ID#: For convenience, let r P QR, s P QR, and [P QR] denote the inradius, semiperimeter, and area, respectively, of an arbitrary triangle P QR. It is well-known that r P QR s P QR = [P QR]. Additionally, let a, b, c and m a, m b, m c be the corresponding sides and medians of triangle ABC. We proceed with the following lemma. Lemma: If two medians of triangle are equal, then the corresponding sides are also equal. Proof: Without loss of generality, we prove that m b = m c b = c in triangle ABC. We use the well-known relations m b = 1 4 (c + a b ) and m c = 1 4 (a + b c ). Thus, 0 = m c m b = 3 4 (b c ) b = c, as desired. We are given that r DGC = r DGB, and since the medians of a triangle divide the area into six equal parts, we have [DGC] = [DGB]. Thus, s DGC = s DGB, so DG + GC + CD = DG + GB + BD GC = GB. Because the medians of a triangle divide each other in a : 1 ratio, it follows that 3 m c = 3 m b, or m c = m b. By our lemma, we conclude that b = c. Next, we prove that a = b, c. Assume, for the sake of contradiction, that a b, c. Without loss of generality, we may set a = 1, and b = c = 1 + k, where k 0. The triangle inequality implies that b + c > a k > 1. Since r GAF = r GCD and [GAF ] = [GCD], we have GA + AF + F G = GC + CD + DG, or (m a m c ) = 3(a c). Writing this in terms of k and solving, we have b + c a a + b c = 3(a c) 4k + 8k + 3 k + k + 3 = 3k 4k + 8k + 3 = ( k + k + 3 3k) 4k + 8k + 3 = k + k + 3 6k k + k k 6k k + k + 3 = 6k 6k k + k + 3 = k 1 k + k + 3 = (k 1) 4k =, and thus the only nonzero real solution is k = 1. However, this is implies that b = c = 1, which is impossible, since this gives a degenerate triangle. Finally, we conclude that k = 0 and consequently a = b = c, so triangle ABC is equilateral, as desired. Page 1 of Problem 5

USA Mathematics Talent Search

USA Mathematics Talent Search 16 1 1 Let S denote the sum of three numbers on any side of the pentagon. We first note that if we add up S for every side of the pentagon, we add up the numbers on the vertices twice. However, we only