olymmath

US-TST 2002

USA IMO Team Selection Test 2002.

Other problems

Problem 1

Let ABCABC be a triangle, and AA , BB , CC its angles. Prove that sin⁡3A2+sin⁡3B2+sin⁡3C2≤cos⁡A−B2+cos⁡B−C2+cos⁡C−A2.\sin\frac{3A}{2}+\sin\frac{3B}{2}+\sin\frac{3C}{2}\leq \cos\frac{A-B}{2}+\cos\frac{B-C}{2}+\cos\frac{C-A}{2}.

Problem 2

Let p>5p>5 be a prime number. For any integer xx , define fp(x)=∑k=1p−11(px+k)2{f_p}(x) = \sum_{k=1}^{p-1} \frac{1}{(px+k)^2} Prove that for any pair of positive integers xx , yy , the numerator of fp(x)−fp(y)f_p(x) - f_p(y) , when written as a fraction in lowest terms, is divisible by p3p^3 .

Problem 3

Let nn be an integer greater than 2, and P1,P2,⋯ ,PnP_1, P_2, \cdots , P_n distinct points in the plane. Let S\mathcal S denote the union of all segments P1P2,P2P3,…,Pn−1PnP_1P_2, P_2P_3, \dots , P_{n-1}P_{n} . Determine if it is always possible to find points AA and BB in S\mathcal S such that P1Pn∥ABP_1P_n \parallel AB (segment ABAB can lie on line P1PnP_1P_n ) and P1Pn=kABP_1P_n = kAB , where (1) k=2.5k = 2.5 ; (2) k=3k = 3 .

Problem 4

Let nn be a positive integer and let SS be a set of 2n+12^n+1 elements. Let ff be a function from the set of two-element subsets of SS to {0,…,2n−1−1}\{0, \dots, 2^{n-1}-1\} . Assume that for any elements x,y,zx, y, z of SS , one of f({x,y}),f({y,z}),f({z,x})f(\{x,y\}), f(\{y,z\}), f(\{z, x\}) is equal to the sum of the other two. Show that there exist a,b,ca, b, c in SS such that f({a,b}),f({b,c}),f({c,a})f(\{a,b\}), f(\{b,c\}), f(\{c,a\}) are all equal to 0.

Problem 5

Consider the family of nonisosceles triangles ABCABC satisfying the property AC2+BC2=2AB2AC^2 + BC^2 = 2 AB^2 . Points MM and DD lie on side ABAB such that AM=BMAM = BM and ∠ACD=∠BCD\angle ACD = \angle BCD . Point EE is in the plane such that DD is the incenter of triangle CEMCEM . Prove that exactly one of the ratios CEEM,EMMC,MCCE\frac{CE}{EM}, \frac{EM}{MC}, \frac{MC}{CE} is constant.

Problem 6

Find in explicit form all ordered pairs of positive integers (m,n)(m, n) such that mn−1mn-1 divides m2+n2m^2 + n^2 .