Putnam 2000 A1. Let A be a positive real number. WhatPare the possible values of x0 , x1 , x2 ,. . . are positive numbers for which ∞ j=0 xj = A? P∞ j=0 x2j , given that Solution. Since 0 < xj < A, ∞ X x2j j=0 < ∞ X Axj = A2 . j=0 P 2 2 2 Thus, the possible values of ∞ j=0 xj belong to (0, A ). We show that each point in (0, A ) is a possible value using the following argument due to Ken Rogers. Let xi = ay i for y ∈ (0, 1) and a constant a > 0. Then ∞ X j=0 We have ∞ X j=0 xj = A ⇔ x2j = ∞ X j=0 ∞ X a = ay j = A ⇔ a = A (1 − y) . 1−y j=0 ¡ ¢j a y2 = 2 a2 A2 (1 − y)2 1−y 2 = = A. 2 2 1−y 1−y 1+y Then, as y varies from 1 to 0, we have that 1−y 1+y varies from 0 to 1. A2. Prove that there exist infinitely many integers n such that n, n+1, and n+2 are each the sum of two squares of integers. [Example: 0 = 02 + 02 , 12 = 02 + 12 , and 2 = 12 + 12 .] Solution. Note that for any integer a (a + 1)2 − 1 = a2 + 2a, (a + 1)2 + 02 , (a + 1)2 + 12 will be a triple of such numbers if 2a is a square. Thus choose a = 2m2 , m = 0, 1, · · · . A3. The octagon P1 P2 P3 P4 P5 P6 P7 P8 is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon P1 P3 P5 P7 is a square of area 5 and the polygon P2 P4 P6 P8 is a rectangle of area 4, find the maximum possible area of the octagon. √ √ Solution. A square of area 5, has edge length 5, and diagonal length 10 which is the diameter of the circle. If x and y (say 0 < x < y) are the dimensions of the rectangle P2 P4 P6 P8 , then x2 + y 2 = 10 and xy = 4, √ √ ¡ √ ¢ and the relevant solution is x = 2, y = 2 2. We may assume that P1 = 12 10, 0 , and ³ √ ´ √ 1 1 P2 = 2 10 cos θ, 2 10 sin θ 1 √ ¡ ¢ for some θ ∈ 0, π2 , and P2 P4 = 2 (otherwise rotate all by 90◦ ). Then ³ √ ´ √ P2 = 12 10 cos (θ + α) , 12 10 sin (θ + α) ¡ ¢ where α is the angle subtended by the side P2 P4 . Since y = 2x, α = 2 arctan 12 . Thus, ¡ ¡ ¢¢ ¡ ¡ ¢¢ ¡ ¡ ¢¢ cos α = cos 2 arctan 12 = cos2 arctan 12 − sin2 arctan 12 ³ ´2 ³ ´2 3 √2 = − √15 = and 5 5 4 sin α = . 5 The area of an isosceles triangle with equal sides r at an angle β is 12 r2 sin β. Thus, the area of the octagon is ³ √ ´2 ¡ ¢ ¡ ¡ ¢ ¢ A (θ) := 2 · 12 · 12 10 sin θ + sin π2 − θ + sin α + θ − π2 + sin (π − (α + θ)) = 52 (sin θ + cos θ − cos (α + θ) + sin (α + θ)) = 52 (sin θ + cos θ − (cos α cos θ − sin α sin θ) + (sin α cos θ + cos α sin θ)) = 52 ((1 + sin α + cos α) sin θ + (1 + sin α − cos α) cos θ) ¡¡ ¢ ¡ ¢ ¢ = 52 1 + 45 + 35 sin θ + 1 + 45 − 35 cos θ = 6 sin θ + 3 cos θ. There is a further restriction on θ, namely θ ≥ π2 − α, or else P4 will come before P3 . For θ = π2 − α, sin θ = cos α = 3/5 and cos θ = 4/5, and ¡ ¢ A π2 − α = 6 · 35 + 3 · 45 = 6, while ¡ ¢ A π2 = 6 · 1 + 3 · 0 = 6. Now A0 (θ) = d dθ (6 sin θ + 3 cos θ) = 6 cos θ − 3 sin θ = 0 ⇒ θ = arctan 2, and √ A (arctan 2) = 6 sin (arctan 2) + 3 cos (arctan 2) = 6 √25 + 3 √15 = 3 5 > 6. √ As θ = arctan 2, the rectangle is “vertical” and the maximum area is 3 5. A4. Show that the improper integral Z lim B→∞ B sin(x) sin(x2 ) dx 0 converges. Solution. We use sin a sin b = 12 (cos (a − b) − cos (a + b)) 2 to get ¡ ¡ ¢ ¡ ¢¢ cos x − x2 − cos x + x2 ¡ ¡ 2 ¢ ¡ ¢¢ cos x − x − cos x2 + x ³ ³¡ ´ ³¡ ´´ ¢2 ¢2 = 12 cos x − 12 − 14 − cos x + 12 − 14 ³ ³¡ ´ ³¡ ´´ ¢2 ¢2 = 12 cos x − 12 − 14 − cos x + 12 − 14 ³¡ ³¡ ¢´ ¢´ 1 1 2 1 1 2 cos 4 cos x − 2 + sin 4 sin x − 2 ³¡ ´ ³¡ = 12 ³ ¢ ¢2 ´´ 2 − cos 14 cos x + 12 + sin 14 sin x + 12 sin(x) sin(x2 ) = = Note that Z 1 2 1 2 B cos 0 Z B sin 0 ³¡ ³¡ x± x± ¢ 1 2 2 ¢ 1 2 2 ´ ´ dx = dx = Z Z B± ± 1 2 B± ± 1 2 1 2 1 2 ¡ ¢ cos u2 du, and ¡ ¢ sin u2 du. Thus, it suffices to show the existence of improper integrals of the form Z B Z B 2 lim sin(x ) dx and lim cos(x2 ) dx. B→∞ B→∞ 0 Consider the parametric curve r (u) = (x (u) , y (u)) = µZ 0 u 2 cos(t ) dt, 0 Z 0 u ¶ sin(t ) dt . 2 The components are integrals of Fresnel type. Note that ¡ ¢ r0 (u) = (x0 (u) , y 0 (u)) = cos(u2 ), sin(u2 ) and kr0 (u)k = 1. The curvature is given by κ (u) = x0 (u) y 00 (u) − y 0 (u) x00 (u) = cos(u2 ) cos(u2 )2u + sin(u2 ) sin(u2 )2u ¡ ¢ = 2u sin2 (u2 ) + cos2 (u2 ) = 2u. Since the curvature increases with arc length, the curve spirals inward to a limit point and the integrals then converge. Indeed, the center of curvature of r at r (u) is 1 c (u) = r (u) + (−y 0 (u) , x0 (u)) 2u µZ u ¶ Z u ¢ 1 ¡ 2 2 cos(t ) dt, sin(t ) dt + = − sin(u2 ), cos(u2 ) , and 2u 0 0 ¡ ¢ ¡ ¢ ¢ 1 ¡ c0 (u) = cos(u2 ), sin(u2 ) + − cos(u2 ), − sin(u2 ) + 2 − sin(u2 ), cos(u2 ) 2u ¢ 1 ¡ 2 2 = − sin(u ), cos(u ) . 2u2 3 Thus, Z 1 ∞ 0 kc (u)k du ≤ Z ∞ 1 1 1 du = <∞ 2u2 2 and the length of the curve c| [1, ∞) is finite, implying that L := limu→∞ c (u) exists. Now kr (u) − Lk ≤ kr (u) − c (u)k + kL − c (u)k = 1 + kL − c (u)k → 0 2u as u → ∞. Thus, limu→∞ r (u) = L, as desired. A5. Three distinct points with integer coordinates lie in the plane on a circle of radius r > 0. Show that two of these points are separated by a distance of at least r1/3 . Solution. Consider the triangle T with sides of length a, b, and c connecting these points. We first show the standard (?) fact that the area A of T is given by A= abc , 4r where r is the radius of the circumcircle of T . Let α be the angle of T opposite a. Then (from the cross product) we have A = 12 bc sin α. On the hand, the central angle opposite a is known to be 2α and so a = 2r sin α. Thus, A = 12 bc sin α = 12 bc 1 abc (2r sin α) = . 2r 4r If d = max (a, b, c), then d3 abc ≥ = A. 4r 4r But, using the cross-product again, we know that 2A2 is a determinant of a 2 × 2 integer √ 2 matrix, and hence a positive integer. Thus, 2A ≥ 1 or A ≥ 1/ 2. Hence, √ √ ¡ ¢1/3 d3 ≥ 4r/ 2 = 2 2r ⇒ d ≥ 23/2 r = (2r)1/3 > r1/3 . A6. Let f(x) be a polynomial with integer coefficients. Define a sequence a0 , a1 , ... of integers such that a0 = 0 and an+1 = f (an ) for all n > 0. Prove that if there exists a positive integer m for which am = 0 then either a1 = 0 or a2 = 0. Solution. Assume that a1 6= 0. We must then show that a2 = 0. Note that a1 = f (a0 ) = f (0) and so a1 is the nonzero constant term in f (x). We have f (am−1 ) = am = 0. Thus, am−1 is an integer zero of f(x). Since f (x) = (x − am−1 ) g(x) for some g(x) with integer coefficients, we have that am−1 divides the constant term of f (x), namely a1 . Since a1 is the constant term of f , we know that a1 divides all the iterates (f ◦ · · · ◦ f ) (a1 ). In particular, a1 divides am−1 , and we have already shown that am−1 divides a1 . Thus, am−1 = ±a1 . If am−1 = a1 , then a2 = f (a1 ) = f (am−1 ) = am = 0. 4 Thus, we are done in the case where an ≥ 0 for all n. Let an0 = min0≤n≤m {an } and let g(x) = f (x + an0 ) − an0 . Defining b0 = 0 and bn+1 = g(bn ), we have b1 = g(b0 ) = g(0) = f (0 + an0 ) − an0 = an0 +1 − an0 ≥ 0, b2 = g(b1 ) = g(an0 +1 − an0 ) = f(an0 +1 ) − an0 = an0 +2 − an0 ≥ 0, .. . bn = an0 +n − an0 ≥ 0 for all n. Since an0 +m − an0 = 0, we may then apply the case we have shown to deduce that if bm = 0 for some m > 0, then 0 = b1 = an0 +1 − an0 or 0 = b2 = an0 +2 − an0 = 0. If b1 = 0, then g(0) = 0 and bn = an0 +n − an0 = 0 for all n, in which case an0 +n = an0 and {an } is a constant sequence (necessarily zero). If 0 = b2 = an0 +2 − an0 = 0, then {bn } is periodic of period 2 in n and so is an , in which case a2 = a0 = 0. B1. Let aj , bj , cj , be integers for 1 ≤ j ≤ N. Assume, for each j, that at least one of aj , bj , cj is odd. Show that there exist integers r, s, t such that raj + sbj + tcj is odd for at least 4N/7 values of j, 1 ≤ j ≤ N. Solution. Note that for fixed j, the evenness or oddness of raj + sbj + tcj depends on the evenness or oddness of r, s, t and aj , bj , cj . Thus, it suffices to consider (r, s, t) ∈ S := {0, 1} × {0, 1} × {0, 1}, and we have a map F : {(j; aj , bj , cj ) : j ∈ {1, . . . , N }} → S For (r, s, t) ∈ S, let (r, s, t)1 := {(ρ, σ, τ ) ∈ S : (r, s, t) · (ρ, σ, τ ) = 1} . Clearly, (0, 0, 0)1 is empty, and it is easy to check that each of the remaining 7 sets (1, 0, 0)1 , (0, 1, 0)1 , (0, 0, 1)1 , (1, 1, 0)1 , (0, 1, 1)1 , (1, 0, 1)1 , (1, 1, 1)1 has exactly 4 elements and their union is S − {(0, 0, 0)}. Call these S1 , . . . , S7 . We need to show that for some i ∈ {1, . . . , 7} , #F −1 (Si ) ≥ 4N/7. Suppose that for all i, #F −1 (Si ) < 4N/7. Then, adding we have #F −1 (S1 ) + · · · + #F −1 (S7 ) < 4N. (∗) Now ∪7i=1 F −1 (Si ) = F −1 (S), since F −1 ((0, 0, 0)) = φ by assumption. Each F ((aj , bj , cj )) belongs to exactly 4 of the Si , namely those (r, s, t)1 such that (r, s, t) ∈ F ((aj , bj , cj ))1 . Thus, the left side of (∗) is 4N, and we have a contradiction. 5 B2. Prove that the expression µ ¶ gcd(m, n) n n m is an integer for all pairs of integers n ≥ m ≥ 1. [Here greatest common divisor of m and n.] ¡n¢ m = n! , m!(n−m)! and gcd(m, n) is the Solution. Note gcd(m, n) lcm(m, n) = mn. Thus, µ ¶ n n divides gcd(m, n) m µ ¶ n mn ⇔ n divides lcm(m, n) m µ ¶ n ⇔ lcm(m, n) divides m m µ ¶ µ ¶ n n ⇔ m divides m , and n divides m , m m ¡n¢ but certainly m divides m m while µ ¶ µ ¶ n n! (n − 1)! n−1 =n =n m =m m m! (n − m)! (m − 1)! (n − m)! m−1 ¡n¢ . and so n divides m m P B3. Let f (t) = N j=1 aj sin(2πjt), where each aj is real and aN 6= 0. Let Nk denote the k number of zeros (including multiplicities) of dtd k f(t). Prove that N0 ≤ N1 ≤ N2 ≤ . . . and lim Nk = 2N. k→∞ Solution. Here, they must have meant to restrict the domain to [0, 1), or equivalently, to the circle R/Z. By Rolle’s Theorem, between any two zeros of f (k) (t) there is at least one zero of f (k+1) (t). Since f (k) is defined on a circle, we then have Nk ≤ Nk+1 . Note that f (4k) 4k (t) = (2π) N X j 4k aj sin(2πjt) j=1 and eventually the maximum and minimum values of the term N 4k aN sin(2πN t) will dominate the value of the sum of the lower terms, since N−1 X j=1 j 4k |aj | ≤ (N − 1)4k 6 N−1 X j=1 |aj | ≤ N 4k |aN | , for k sufficiently large. Thus, f (4k) has at least 2N zeros for k sufficiently large. Also, for z = e(2πt)i , we have f (4k) N X (z j − z −j ) (t) = (2π) j aj 2i j=1 ¡ ¢ N N+j N−j X − z z = (2π)4k j 4k aj 2iz N j=1 4k = 4k p (z) zN for a polynomial p (z) of degree 2N . Thus, f (4k) also has at most 2N zeros. B4. Let f(x) be a continuous function such that f (2x2 − 1) = 2xf(x) for all x. Show that f (x) = 0 for −1 ≤ x ≤ 1. Solution. Note that cos (2u) = cos2 (u) − sin2 (u) = 2 cos2 u − 1. Thus, f(cos (2u)) = 2 cos (u) f (cos (u)) and f (cos v) = 2 cos (v/2) f (cos (v/2)) = sin (v) f(cos (v/2)) sin (v/2) sin (v) sin (v/2) f(cos (v/4)) sin (v/2) sin (v/4) ¡ k¢ sin (v) = ··· = f (cos v/2 ) sin (v/2k ) = Also − cos (2u) = sin2 (u) − cos2 (u) = 2 sin2 u − 1, so that f (− cos (2u)) = 2 sin (u) f (sin (u)) Note that f is odd since 2xf(x) = f (2x2 − 1) is even. Thus, 2 sin (u) f(sin (u)) = f (− cos (2u)) = −f(cos (2u)) = −2 cos (u) f (cos (u)) or sin (u) f(cos (u)) = − f (sin (u)) if cos (u) 6= 0. cos (u) Hence, as k → ∞ and f (0) = 0 due to the oddness of f, we have f (cos v) = ¡ k¢ ¡ k¢ sin (v) sin (v) ) = − f (cos v/2 f(sin v/2 ) → 0. sin (v/2k ) cos (v/2k ) B5. Let S0 be a finite set of positive integers. We define finite sets S1 , S2 , ... of positive integers as follows: Integer a is in Sn+1 , if and only if exactly one of a − 1 or a is in Sn . 7 Show that there exist infinitely many integers N for which SN = S0 ∪ {N + a : a ∈ S0 }. Solution. Let pn (x) be a polynomial with coefficients in Z2 , such that the coefficient ak (n) of xk in pn (x) is 1 when k ∈ Sk and 0 otherwise. Note that the coefficient ak (n + 1) of pn+1 (x) is given by ½ 0 if ak (n) + ak−1 (n) = 1 ak (n + 1) = 1 if ak (n) + ak−1 (n) = 0. This is also the coefficient of xk for the polynomial pn (x) + xpn (x) = (1 + x) pn (x) . Thus, pn+1 (x) = (x + 1) pn (x) and pn (x) = (1 + x)n p0 (x) . We must show that there are infinitely many N, such that ¡ ¢ pN (x) = p0 (x) + xN p0 (x) = 1 + xN p0 (x) . In other words, that there are infinitely many n, such that (1 + x)N = 1 + xN This, is true for N = 2 and note that if it is true for N , then it is true for 2N, since ³ ´2 ¡ ¢2 (1 + x)2N = (1 + x)N = 1 + xN = 1 + 2xN + x2N = 1 + x2N . Thus, SN = S0 ∪ {N + a : a ∈ S0 } for N equal to a power of 2. B6. Let B be a set of more than 2n+1 /n distinct points with coordinates of the form (±1, ±1, ..., ±1) in n-dimensional space, with n ≥ 3. Show that there are three distinct points in B which are the vertices of an equilateral triangle. Solution. Let C = {(x1 , x2 , ..., xn ) : xi = ±1} For a fixed point p ∈ C, let Sp be the set of points in C of minimal distance (namely 2) from p. A point q ∈ Sp if q differs from p in one coordinate. Note that two points q1 , q2 ∈ Sp agree in all but two√coordinates, namely the distinct coordinates in which they differ from p. Thus, kq1 − q2 k = 8, and hence all points in Sp are equidistant from each other. It suffices to show that there is some p ∈ C, with # (B ∩ Sp ) ≥ 3. Consider the set A = {(q, p) : q ∈ B ∩ Sp } Note that for each q ∈ B, there are n points p1 (q) , . . . , pn (q) with q ∈ Spi (q) . Thus, #A = n · # (B) . Also, for each p, the number of q ∈ B with (q, p) ∈ A is # (Sp ∩ B). Thus, X # (B ∩ Sp ) = #A = n · # (B) > n · 2n+1 /n = 2n+1 = 2 · 2n p∈C Since there are 2n terms on the left side, one of them must be bigger than 2. 8