1982 IMO Problems/Problem 6


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 $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. 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 $\frac {1} {2}$. 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 $198$.


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.

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1982 IMO (Problems) • Resources
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