Brazilian Mathematical Olympiad

Let $V_1$ and $V_2$ be consecutive vertices of the polygon. Let $V_k$ be a vertex such that $\angle V_1 V_k V_2$ is minimal. Since all angles $\angle V_1 V_j V_2 \ge \angle V_1 V_k V_2$ for all $j \ne k$ of the polygon, then all vertices are contained in the circumcircle of the triangle $V_1 V_2 V_k$.

If $k=3$, we are done. Otherwise, consider the arc $V_2V_k$ that doesn't contain $V_1$ and let $V_\ell$ a point in the region between the arc $V_2V_k$ and the chord $V_2V_k$ such that $\angle V_2V_\ell V_k$ is minimal. Notice that the circumcircle of $V_2V_\ell V_k$ still contains the polygon and that the number of vertices between 1 and $k$ is less than the number of vertices between 2 and $k$. Repeat the procedure exchanging $V_2$ and $V_k$ by $V_\ell$ and $V_k$. We obtain again a circle with less vertices between the edges of the triangle. Continuing in this fashion we eventually obtain three consecutive vertices.