IMO 1971 SL 14

Origin: USS

Problem

A broken line A1A2 . . . An is drawn in a 50\times50 square, so that the distance from any point of the square to the broken line is less than

Solution

Denote by V the figure made by a circle of radius 1 whose center moves along the broken line. From the condition of the problem, V contains the whole 50 \times 50 square, and thus the area S(V ) of V is not less than 2500. Let L be the length of the broken line. We shall show that S(V ) \leq2L+\pi, from which it will follow that L \geq1250 −\pi/2 > 1248. For each segment li = AiAi+1 of the broken line, consider the figure Vi obtained by a circle of radius 1 whose center moves along it, and let Vi be obtained by cutting offthe circle of radius 1 with center at the starting point of li. The area of Vi is equal to 2AiAi+1. It is clear that the union of all the figures Vi together with a semicircle with center in A1 and a semicircle with center in An contains V completely. Therefore S(V ) \leq\pi + 2A1A2 + 2A2A3 + \cdot \cdot \cdot + 2An−1An = \pi + 2L. This completes the proof.