olymmath

US-TST 2005

USA IMO Team Selection Test 2005.

Other problems

Problem 1

Let nn be an integer greater than 11 . For a positive integer mm , let Sm={1,2,…,mn}S_{m}= \{ 1,2,\ldots, mn\} . Suppose that there exists a 2n2n -element set TT such that (a) each element of TT is an mm -element subset of SmS_{m} ; (b) each pair of elements of TT shares at most one common element; and (c) each element of SmS_{m} is contained in exactly two elements of TT .

Determine the maximum possible value of mm in terms of nn .

Problem 2

Let A1A2A3A_{1}A_{2}A_{3} be an acute triangle, and let OO and HH be its circumcenter and orthocenter, respectively. For 1≤i≤31\leq i \leq 3 , points PiP_{i} and QiQ_{i} lie on lines OAiOA_{i} and Ai+1Ai+2A_{i+1}A_{i+2} (where Ai+3=AiA_{i+3}=A_{i} ), respectively, such that OPiHQiOP_{i}HQ_{i} is a parallelogram. Prove that OQ1OP1+OQ2OP2+OQ3OP3≥3.\frac{OQ_{1}}{OP_{1}}+\frac{OQ_{2}}{OP_{2}}+\frac{OQ_{3}}{OP_{3}}\geq 3.

Problem 3

We choose random a unitary polynomial of degree nn and coefficients in the set 1,2,...,n!1,2,...,n! . Prove that the probability for this polynomial to be special is between 0.710.71 and 0.750.75 , where a polynomial gg is called special if for every k>1k>1 in the sequence f(1),f(2),f(3),...f(1), f(2), f(3),... there are infinitely many numbers relatively prime with kk .

Problem 4

Consider the polynomials f(x)=∑k=1nakxkandg(x)=∑k=1nak2k−1xk,f(x) =\sum_{k=1}^{n}a_{k}x^{k} \text{and} g(x) =\sum_{k=1}^{n}\frac{a_{k}}{2^{k}-1}x^{k}, where a1,a2,…,ana_{1},a_{2},\ldots,a_{n} are real numbers and nn is a positive integer. Show that if 1 and 2n+12^{n+1} are zeros of gg then ff has a positive zero less than 2n2^{n} .

Problem 5

Find all finite sets SS of points in the plane with the following property: for any three distinct points A,B,A,B, and CC in S,S, there is a fourth point DD in SS such that A,B,C,A,B,C, and DD are the vertices of a parallelogram (in some order).

Problem 6

Let ABCABC be an acute scalene triangle with OO as its circumcenter. Point PP lies inside triangle ABCABC with ∠PAB=∠PBC\angle PAB = \angle PBC and ∠PAC=∠PCB\angle PAC = \angle PCB . Point QQ lies on line BCBC with QA=QPQA = QP . Prove that ∠AQP=2∠OQB\angle AQP = 2\angle OQB .