Olympiad Maths Prep

Track / Stage 6 / 67 of 400 #1067 of 2000

Problem 1067

National olympiad, first round
Combinatorics Difficulty 6.1 Prove it

Problem 11.3. Given an orthogonal coordinate system with origin OO in the plane. Distinct real numbers are written at the points with integer coordinates. Let AA be a nonempty finite set of integer points which is central-symmetric regarding OO and OAO \notin A. Prove that there exists an integer point XX such that if AXA_{X} is the image of AA under translation defined by OX\overrightarrow{O X}, then at least half of the numbers written at the points of AXA_{X} are greater than the number written at XX.

Avgustin Marinov

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution: Let us denote the number of points in AA, which is obviously even, by 2s2 s. Connect all integer points XX with the points from AXA_{X} by arrows so that the arrow points to the smaller number. Suppose no point XX with the required property exists. Then there are at least s+1s+1 arrows pointing out of any point XX. For every natural
number nn denote the square with vertices (n,n),(n,n),(n,n)(n, n),(-n, n),(-n,-n) and (n,n)(n,-n) by KnK_{n}.

Since AA is a finite set, there exists a natural number dd such that AKdA \subset K_{d}. For every nn denote the number of arrows within the square KnK_{n} by SnS_{n}. Since there are at least s+1s+1 arrows pointing out of every integer point of KnK_{n} (and these arrows are within the square Kn+dK_{n+d} ), it follows that (2n+1)2(s+1)Sn+d(2 n+1)^{2}(s+1) \leq S_{n+d}. On the other hand, since AXA_{X} is a central-symmetric set, there are at most s1s-1 arrows pointing to every integer point of Kn+dK_{n+d}. Therefore Sn+d(2n+2d+1)2(s1)S_{n+d} \leq(2 n+2 d+1)^{2}(s-1). Thus (2n+1)2(s+1)(2n+2d+1)2(s1)(2 n+1)^{2}(s+1) \leq(2 n+2 d+1)^{2}(s-1), so s+1(1+s+1 \leq(1+ 2d2n+1)2(s1\left.\frac{2 d}{2 n+1}\right)^{2}(s-1 ) for any nn. When nn \rightarrow \infty one obtains s+1s1s+1 \leq s-1, a contradiction. Therefore a point XX with the required property does exist.

## WINTER MATHEMATICAL COMPETITION

## Grade 8 - First Group.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.