Solution: Let us denote the number of points in A, which is obviously even, by 2s. Connect all integer points X with the points from AX by arrows so that the arrow points to the smaller number. Suppose no point X with the required property exists. Then there are at least s+1 arrows pointing out of any point X. For every natural
number n denote the square with vertices (n,n),(−n,n),(−n,−n) and (n,−n) by Kn.
Since A is a finite set, there exists a natural number d such that A⊂Kd. For every n denote the number of arrows within the square Kn by Sn. Since there are at least s+1 arrows pointing out of every integer point of Kn (and these arrows are within the square Kn+d ), it follows that (2n+1)2(s+1)≤Sn+d. On the other hand, since AX is a central-symmetric set, there are at most s−1 arrows pointing to every integer point of Kn+d. Therefore Sn+d≤(2n+2d+1)2(s−1). Thus (2n+1)2(s+1)≤(2n+2d+1)2(s−1), so s+1≤(1+ 2n+12d)2(s−1 ) for any n. When n→∞ one obtains s+1≤s−1, a contradiction. Therefore a point X with the required property does exist.
## WINTER MATHEMATICAL COMPETITION
## Grade 8 - First Group.