IMO 1963 - Problem 3

Problem: Prove that if all the angles of a convex $n-gon$ are equal and the lengths of consecutive edges $a_1,...,a_n$ satisfy $a_1\ge a_2\ge ...\ge a_n$, then $a_1=a_2=...=a_n$.
Solution: Let $\vec{OA_1},\vec{OA_2},...,\vec{OA_n}$ be the vectors corresponding respectively to the edges $a_1,a_2,...,a_n$ of the polygon. By the conditions of the problem, these vectors satisfy $\vec{OA_1}+...+\vec{OA_n}=\vec{0},\angle{A_1OA_2}=\angle{A_2OA_3}=...=\angle{A_nOA_1}=\frac{2\pi}{n}$ and $OA_1\ge OA_2\ge ...\ge OA_n$. Our task is to prove that $OA_1=...=OA_n$.
 Let $l$ be the line through $O$ perpendicular to $OA_n$ and $B_1,...,B_{n-1}$ the projections of $A_1,...,A_{n-1}$ onto $l$ respectively. By the assumptions, the sum of the $\vec{OB_i}'s$ is $\vec{0}$. On the other hand, since $OB_i\le OB_{n-i}$ for all $i\le \frac{n}{2}$, all the sums $\vec{OB_i}+\vec{OB_{n-i}}$ lie on the same side of the point $O$. Hence all these sums must be equal to $\vec{0}$. Consequently, $OA_i=OA_{n-i},$ from which the result immediately follows.