MathDB

2

Part of 2007 All-Russian Olympiad

Problems(4)

10 x 10 table

Source: All-Russian 2007

5/4/2007
The numbers 1,2,…,1001,2,\ldots,100 are written in the cells of a 10×1010\times 10 table, each number is written once. In one move, Nazar may interchange numbers in any two cells. Prove that he may get a table where the sum of the numbers in every two adjacent (by side) cells is composite after at most 3535 such moves. N. Agakhanov
geometryrectanglecombinatorics unsolvedcombinatorics
$100$ fractions are written on a board

Source: All-Russian 2007

5/4/2007
100100 fractions are written on a board, their numerators are numbers from 11 to 100100 (each once) and denominators are also numbers from 11 to 100100 (also each once). It appears that the sum of these fractions equals to a/2a/2 for some odd aa. Prove that it is possible to interchange numerators of two fractions so that sum becomes a fraction with odd denominator. N. Agakhanov, I. Bogdanov
number theory unsolvednumber theory
inequality on polynomials

Source: All-Russian 2007

5/4/2007
Given polynomial P(x)=a0xn+a1xn−1+⋯+an−1x+anP(x) = a_{0}x^{n}+a_{1}x^{n-1}+\dots+a_{n-1}x+a_{n}. Put m=min⁡{a0,a0+a1,…,a0+a1+⋯+an}m=\min \{ a_{0}, a_{0}+a_{1}, \dots, a_{0}+a_{1}+\dots+a_{n}\}. Prove that P(x)≥mxnP(x) \ge mx^{n} for x≥1x \ge 1. A. Khrabrov
inequalitiesalgebrapolynomialinductionalgebra unsolved
incircle of triangle

Source: All-Russian 2007

5/4/2007
The incircle of triangle ABCABC touches its sides BCBC, ACAC, ABAB at the points A1A_{1}, B1B_{1}, C1C_{1} respectively. A segment AA1AA_{1} intersects the incircle at the point Q≠A1Q\ne A_{1}. A line ℓ\ell through AA is parallel to BCBC. Lines A1C1A_{1}C_{1} and A1B1A_{1}B_{1} intersect ℓ\ell at the points PP and RR respectively. Prove that ∠PQR=∠B1QC1\angle PQR=\angle B_{1}QC_{1}. A. Polyansky
geometrycircumcirclegeometry proposed