Selection tests for the Gulf Mathematical Olympiad 2013

Let $P_1{P}_2\cdots P_n$ be a convex cyclic non-regular polygon with the measures of all its internal angles equal and let $O$ be its circumcenter. Because all the angles are equal, all the arcs $\widehat{P_i{P}_{i+2}}$, for $i=1,\dots ,n$, have the same length. Hence, $\angle P_{i}O{P}_{i+2}=\frac{4\pi }{n}$, for all $i=1,\dots ,n$, since the polygon is convex.



Let $\theta =\angle P_{1}O{P}_2$. We have $$\angle P_{2i-1}O{P}_{2i}=\angle P_{1}O{P}_2+\angle P_{2}O{P}_{2i}-\angle P_{1}O{P}_{2i-1}=\angle P_{1}O{P}_2=\theta .$$ If $n$ is an odd integer, $$\theta =\angle P_{n}O{P}_1=\frac{n+1}{2}\angle P_{n}O{P}_2=\frac{n+1}{2}\cdot \frac{4\pi }{n}=\frac{2\pi }{n}=\angle P_{1}O{P}_2\mathrm{mod}2\pi .$$ This means that the polygon is regular, which contradicts the hypothesis.

If $n$ is even, any value of $\theta$ with $0<\theta <\frac{4\pi }{n}$ and $\theta \ne \frac{2\pi }{n}$ defines a unique non-regular such polygon.

Therefore, the possible values of $n$ are all even positive integers $n\ge 4$.