IMO 1982 SL 6

Problem

A6 (VIE 1)IMO6 Let S be a square with sides of length 100 and let L be a path within S that does not meet itself and that is composed of linear segments A0A1, A1A2, . . . , An−1An with A0 ̸= An. Suppose that for every point P of the boundary of S there is a point of L at a distance from P not greater than 1

Solution

Denote by d(U, V ) the distance between points or sets of points U and V . For P, Q \inL we shall denote by LPQ the part of L between points P and Q and by lPQ the length of this part. Let us denote by Si (i = 1, 2, 3, 4) the vertices of S and by Ti points of L such that SiTi \leq1/2 in such a way that lA0T1 is the least of the lA0Ti’s, S2 and S4 are neighbors of S1, and lA0T2 < lA0T4. Now we shall consider the points of the segment S1S4. Let D and E be the sets of points defined as follows: D = {X \in[S1S4] | d(X, LA0T2) \leq1/2} and E = {X \in[S1S4] | d(X, LT2An) \leq1/2}. Clearly D and E are closed, nonempty (indeed, S1 \inD and S4 \inE) subsets of [S1S4]. Since their union is a connected set S1S4, it follows that they must have a nonempty intersection. Let P \inD \capE. Then there exist points X \inLA0T2 and

Y \inLT2An such that d(P, X) \leq1/2, d(P, Y ) \leq1/2, and consequently d(X, Y ) \leq1. On the other hand, T2 lies between X and Y on L, and thus LXY = LXT2 +LT2Y \geqXT2 +T2Y \geq(PS2 −XP −S2T2)+ (PS2 −Y P − S2T2) \geq99 + 99 = 198.