IMO 1989 SL 8

Origin: FRA

Problem

Let R be a rectangle that is the union of a finite number of rectangles Ri, 1 \leqi \leqn, satisfying the following conditions: (i) The sides of every rectangle Ri are parallel to the sides of R. (ii) The interiors of any two different Ri are disjoint. (iii) Every Ri has at least one side of integral length. Prove that R has at least one side of integral length.

Solution

Let A, B, C, D denote the vertices of R. We consider the set S of all points E of the plane that are vertices of at least one rectangle, and its subset S′ consisting of those points in S that have both coordinates integral in the orthonormal coordinate system with point A as the origin and lines AB, AD as axes. First, to each E \inS we can assign an integer nE as the number of rectangles Ri with one vertex at E. It is easy to check that nE = 1 if E is one of the vertices A, B, C, D; in all other cases nE is either 2 or 4. Furthermore, for each rectangle Ri we define f(Ri) as the number of vertices of Ri that belong to S′. Since every Ri has at least one side of integer length, f(Ri) can take only values 0, 2, or 4. Therefore we have

n  i=1 f(Ri) \equiv0 (mod 2). On the other hand, n i=1 f(Ri) is equal to  E\inS′ nE, implying that  E\inS′ nE \equiv0 (mod 2). However, since nA = 1, at least one other nE, where E \inS′, must be odd, and that can happen only for E being B, C, or D. We conclude that at least one of the sides of R has integral length. Second solution. Consider the coordinate system introduced above. If D is a rectangle whose sides are parallel to the axes of the system, it is easy to prove that A D sin 2\pix sin 2\piy dx dy = 0 if and only if at least one side of D has integral length. This holds for all Ri’s, so that adding up these equalities we get

R sin 2\pix sin 2\piy dx dy = 0. Thus, R also has a side of integral length.