Iranian Mathematical Olympiad

We prove a stronger statement; there is an ordering $P_1, P_2, \dots, P_n$ such that for every $1 \le i < n$, the angle $\angle P_{i-1}P_iP_{i+1}$ is acute.

Assume that $P_nP_{n-1}$ is the longest segment between any two of these points. Now we start with $P_{n-1}$ and at each step we choose a point $P$ such that $P_i, P$ has the longest distance between the remaining points as $P_{i-1}$.

We claim that this order has the desired property. Assume the contrary, the angle $\angle P_{i-1}P_iP_{i+1}$ is greater than $\frac{\pi}{2}$, then on the triangle $P_{i-1}P_iP_{i+1}$, $P_{i-1}P_{i+1}$ is the longest edge. In particular, it is longer than $P_iP_{i+1}$, this contradicts the way we choose $P_i$. So all these angles are acute as desired. ■