IMO 1982 SL 2

Problem

A2 (YUG 1) Let K be a convex polygon in the plane and suppose that K is positioned in the coordinate system in such a way that area (K \capQi) = 1 4 area K (i = 1, 2, 3, 4, ), where the Qi denote the quadrants of the plane. Prove that if K contains no nonzero lattice point, then the area of K is less than 4.

Solution

Since K does not contain a lattice point other than O(0, 0), it is bounded by four lines u, v, w, x that pass through the points U(1, 0), V (0, 1), W(−1, 0), X(0, −1) respectively. Let PQRS be the quadrilateral formed by these lines, where U \inSP, V \inPQ, W \inQR, X \inRS. If one of the quadrants, say Q1, contains no vertices of PQRS, then K\capQ1 is contained in \triangleOUV and hence has area less than 1/2. Consequently the area of K is less than 2. Let us now suppose that P, Q, R, S lie in different quadrants. One of the angles of PQRS is at least 90◦: let it be \angleP. Then SUPV \leqPU \cdot PV/2 \leq (PU 2 + PV 2)/4 \leqUV 2/4 = 1/2, which implies that SK\capQ1 < SOUPV \leq

1. Hence the area of K is less than 4.