1982 IMO Problems/Problem 6
Problem
Let be a square with sides length . Let be a path within which does not meet itself and which is composed of line segments with . Suppose that for every point on the boundary of there is a point of at a distance from no greater than . Prove that there are two points and of such that the distance between and is not greater than and the length of the part of which lies between and is not smaller than .
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.
See Also
1982 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
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