XVI OBM

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 $V_2{V}_k$ that doesn't contain $V_1$ and let $V_{\ell }$ a point in the region between the arc $V_2{V}_k$ and the chord $V_2{V}_k$ such that $\angle V_2{V}_{\ell }{V}_k$ is minimal. Notice that the circumcircle of $V_2{V}_{\ell }{V}_k$ 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.