Iranian Mathematical Olympiad

Suppose that $P = P_1P_2 \dots P_n$, $P' = P'_1P'_2 \dots P'_n$ and that $AP_i$'s are a permutation of $A'P'_i$'s. Let $O$ and $O'$ be the centers of $P$ and $P'$, respectively. Without loss of generality it can be assumed that $A$ is inside or on the perimeter of triangle $OP_1P_2$ and $AP_1 \le AP_2$; $A'$ is inside or on the perimeter of $O'P'_1P'_2$ and $A'P'_1 \le A'P'_2$. Using the following lemma $AP_i$'s and $A'P'_i$'s can be sorted in ascending order.

Lemma. In a situation as described above, if $n$ is odd, then
$$\begin{aligned} & AP_1 \le AP_2 \le AP_n \le AP_3 \le AP_{n-1} \le AP_4 \le \dots \\ & \le AP_{\frac{n+3}{2}-1} \le AP_{\frac{n+3}{2}+1} \le AP_{\frac{n+3}{2}}, \end{aligned}$$
and if $n$ is even, then
$$AP_1 \le AP_2 \le AP_n \le AP_3 \le \dots \le AP_{\frac{n}{2}+2} \le AP_{\frac{n}{2}+1}.$$
Proof. The assertion can be easily proved by looking at the position of $A$ with respect to the perpendicular bisector of $P_2$ and $P_n$ (which is $OP_1$), then the perpendicular bisector of $P_n$ and $P_3$, then the perpendicular bisector of $P_3$ and $P_{n-1}$ and so on.

It can be assumed that similar inequalities hold for $A'P'_i$'s. So the problem statement is equivalent to,
$$AP_1 = A'P'_1 \le AP_2 = A'P'_2 \le AP_n = A'P'_n \le AP_3 = A'P'_3 \le \dots$$
Assume that $P$ is greater than $P'$. For each $1 \le i \le n$, ($P_{n+1} = P_1$, $P'_{n+1} = P'_1$)
$$\begin{aligned} & AP_i = A'P'_i,\ AP_{i+1} = A'P'_{i+1},\ P_iP_{i+1} > P'_iP'_{i+1} \\ & \qquad \Rightarrow \angle P_iAP_{i+1} > \angle P'_iA'P'_{i+1}. \end{aligned}$$
Hence
$$360^{\circ} = \sum_{i=1}^{n} \angle P_i AP_{i+1} > \sum_{i=1}^{n} \angle P'_i A'P'_{i+1} = 360^{\circ},$$
which is a contradiction. This shows that $P$ and $P'$ are equal.