Year: 1966
Topic: Geometry

Problem:
Let a convex polygon $P$ be contained in a square of side 1. Find the upper bound of the sum of the squares of the sides of $P$.

Solution:
Let $P_1, P_2, \dots, P_n$ be the vertices of $P$. Let $P'_1, P'_2, \dots, P'_n$ be the projections of $P_1, P_2, \dots, P_n$ upon one of the sides of the squares, and let $P''_1, P''_2, \dots, P''_n$ be the projections of $P_1, P_2, \dots, P_n$ upon a side that is orthogonal to the previous one. Since $P$ is convex, the first side will be covered at most twice by the segments $P'_1P'_2, \dots, P'_{n-1}P'_n, P'_nP'_1$. We thus deduce the inequality \[ P'_1P'_2^2 + \dots + P'_nP'_1^2 \leq 2. \] Similarly \[ P''_1P''_2^2 + \dots + P''_nP''_1^2 \leq 2. \] Adding these two inequalities and using the Pythagorean theorem the assertion follows. Thus the upper bound is \boxed{4}.