IMO 2003 SL C5

Origin: ROM | Category: Combinatorics

Problem

Every point with integer coordinates in the plane is the center of a disk with radius 1/1000. (a) Prove that there exists an equilateral triangle whose vertices lie in different disks. (b) Prove that every equilateral triangle with vertices in different disks has side length greater than 96.

Solution

(a) By the pigeonhole principle there are two different integers x1, x2, x1 > x2, such that |{x1 \sqrt 3} −{x2 \sqrt 3}| < 0.001. Set a = x1 −x2. Consider the equilateral triangle with vertices (0, 0), (2a, 0), (a, a \sqrt 3). The points (0, 0) and (2a, 0) are lattice points, and we claim that the point (a, a \sqrt 3) is at distance less than 0.001 from a lattice point. Indeed, since 0.001 > |{x1 \sqrt 3}−{x2 \sqrt 3}| = |a \sqrt 3−([x1 \sqrt 3]−[x2 \sqrt 3])|, we see that the distance between the points (a, a \sqrt 3) and (a, [x1 \sqrt 3] − [x2 \sqrt 3]) is less than 0.001, and the point (a, [x1 \sqrt 3] −[x2 \sqrt 3]) is with integer coefficients. (b) Suppose that P ′Q′R′ is an equilateral triangle with side length l \leq96 such that each of its vertices P ′, Q′, R′ lies in a disk of radius 0.001 centered at a lattice point. Denote by P, Q, R the centers of these disks. Then we have l −0.002 \leqPQ, QR, RP \leql + 0.002. Since PQR is not an equilateral triangle, two of its sides are different, say

PQ ̸= QR. On the other hand, PQ2, QR2 are integers, so we have 1 \leq|PQ2 −QR2| = (PQ + QR)|PQ −QR| \leq0.004(PQ + QR) \leq (2l + 0.004) \cdot 0.004 \leq2 \cdot 96.002 \cdot 0.004 < 1, which is a contradiction.