Let S be a square with sides length 100. Let L be a path within S which does not meet itself and which is composed of line segments A0A1, A1A2, A2A3, ... , An-1An with A0 = An. Suppose that for every point P on the boundary of S there is a point of L at a distance from P no greater than 1/2. Prove that there are two points X and Y of L such that the distance between X and Y is not greater than 1 and the length of the part of L which lies between X and Y is not smaller than 198.
Solution
Let the square be A'B'C'D'. The idea is to find points of L close to a particular point of A'D' but either side of an excursion to B'.
We say L approaches a point P' on the boundary of the square if there is a point P on L with PP' ≤ 1/2. We say L approaches P' before Q' if there is a point P on L which is nearer to A0 (the starting point of L) than any point Q with QQ' ≤ 1/2.
Let A' be the first vertex of the square approached by L. L must subsequently approach both B' and D'. Suppose it approaches B' first. Let B be the first point on L with BB' ≤ 1/2. We can now divide L into two parts L1, the path from A0 to B, and L2, the path from B to An.
Take X' to be the point on A'D' closest to D' which is approached by L1. Let X be the corresponding point on L1. Now every point on X'D' must be approached by L2 (and X'D' is non-empty, because we know that D' is approached by L but not by L1). So by compactness X' itself must be approached by L2. Take Y to be the corresponding point on L2. XY ≤ XX' + X'Y ≤ 1/2 + 1/2 = 1. Also BB' ≤ 1/2, so XB ≥ X'B' - XX' - BB' ≥ X'B' - 1 ≥ A'B' - 1 = 99. Similarly YB ≥ 99, so the path XY ≥ 198.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998