Chinese IMO National Team Selection Test, 2023

All subscripts are to be understood modulo $2n$. Denote $\angle PA_1A_2 = \angle PA_2A_3 = \cdots = \alpha$. For $1 \le i \le 2n$, extend $A_iP$ to intersect the circumference at point $B_i$. Note that
$$\angle A_iB_iB_{i-1} = \angle A_iA_{i-1}P = \angle A_{i+1}A_iB_i = \alpha.$$
Hence, $A_iA_{i+1}B_{i-1}B_i$ is an isosceles trapezoid. (In fact, $B_1B_2 \cdots B_{2n}$ is the polygon obtained by rotating $A_2A_3 \cdots A_{2n}A_1$ in the clockwise direction by $2\alpha$ on the circumference.) In particular, $A_iA_{i+1} = B_{i-1}B_i$. Also, by $\triangle A_{i-1}PA_i \sim \triangle B_iPB_{i-1}$, we get
$$(*) \qquad \frac{A_{i-1}P}{B_iP} = \frac{A_iP}{B_{i-1}P} = \frac{A_{i-1}A_i}{B_{i-1}B_i} = \frac{A_{i-1}A_i}{A_iA_{i+1}}.$$
Taking the product of the first and last equalities in (*) over all odd $i$, we get
$$\prod_{i=1}^{n} \frac{A_{2i-2}A_{2i-1}}{A_{2i-1}A_{2i}} = \prod_{i=1}^{n} \frac{A_{2i-2}P}{B_{2i-1}P}.$$

Taking the product of the second and last equalities in (*) over all even $i$, we get
$$\prod_{i=1}^{n} \frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}} = \prod_{i=1}^{n} \frac{A_{2i}P}{B_{2i-1}P}.$$
Noting that the right-hand sides of the two equations above are equal, we have
$$\prod_{i=1}^{n} \frac{A_{2i-2}A_{2i-1}}{A_{2i-1}A_{2i}} = \prod_{i=1}^{n} \frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}} \Rightarrow \prod_{i=1}^{n} (A_{2i}A_{2i+1})^2 = \prod_{i=1}^{n} (A_{2i-1}A_{2i})^2.$$
This completes the proof of the problem. $\Box$

Draw line $A_1A_4$ intersecting line $PA_3$ at point $H$. Given that points $A_1, A_2, A_3, A_4$ are concyclic, we have
$$\begin{aligned} \angle A_2A_1A_4 &= 180^\circ - \angle A_2A_3A_4 = 180^\circ - \angle A_2A_3P - \angle PA_3A_4 \\ &= 180^\circ - \angle A_2A_3P - \angle A_3A_2P = \angle A_2PA_3. \end{aligned}$$
Therefore, points $A_1, A_2, H, P$ are concyclic. In particular, $\angle A_2HA_3 = \angle A_2A_1P = \angle HA_3A_4$, we have $A_2H \parallel A_3A_4$. From this and the fact that $A_1, A_2, H, P$ are concyclic, we get
$$\angle A_3A_4H = \angle A_2HA_1 = \angle A_1PA_2.$$
Together with $\angle A_2A_1P = \angle A_4A_3H$, we find that $\triangle A_1A_2P \sim \triangle A_3HA_4$. From this we obtain
$$\frac{A_1A_2}{A_1P} = \frac{A_3H}{A_3A_4} \Rightarrow A_1A_2 \cdot A_3A_4 = A_1P \cdot A_3H.$$
Moreover, from $\angle A_3HA_2 = \angle A_2A_1P = \angle A_3A_2P$, we have $\triangle A_3HA_2 \sim \triangle A_3A_2P$. We get
$$\frac{A_2A_3}{A_3H} = \frac{PA_3}{A_2A_3} \Rightarrow (A_2A_3)^2 = PA_3 \cdot A_3H.$$

Dividing the first equation by the second one yields
$$\frac{A_1 A_2 \cdot A_3 A_4}{(A_2 A_3)^2} = \frac{P A_1}{P A_3}.$$
By doing the same procedure with adding 2 to each index and multiplying the resulting equations, we obtain
$$\prod_{i=1}^{n} \frac{A_{2i-1}A_{2i} \cdot A_{2i+1}A_{2i+2}}{(A_{2i}A_{2i+1})^2} = \prod_{i=1}^{n} \frac{P A_{2i-1}}{P A_{2i+1}} = 1.$$
Here all indices are understood modulo $2n$. Rearranging the above equation gives
$$\prod_{i=1}^{n} |A_{2i-1}A_{2i}|^2 = \prod_{i=1}^{n} |A_{2i}A_{2i+1}|^2.$$
This completes the proof of the problem. $\Box$

Let the circle in the problem be the unit circle in the complex plane, and let $A_1, \dots, A_{2n}$ correspond to complex numbers $z_1, \dots, z_{2n}$. Without loss of generality (we can always rotate if necessary), let $P$ correspond to a non-negative real number $a < 1$, and let $\angle PA_1A_2 = \pi - \alpha \in (0, \pi)$.
From the given conditions, we know that $\frac{z_2 - z_1}{(z_1 - a)e^{i\alpha}} \in \mathbb{R}$, thus
$$\frac{z_2 - z_1}{(z_1 - a)e^{i\alpha}} = \frac{\overline{z_2} - \overline{z_1}}{(\overline{z_1} - a)e^{-i\alpha}},$$
Considering that $z_1, z_2$ are unit complex numbers, and $z_1 \neq z_2$, we can solve to get $z_2 = \frac{e^{2i\alpha}(z_1 - a)}{az_1 - 1}$, and similarly $z_{j+1} = \frac{e^{2i\alpha}(z_j - a)}{az_j - 1}$, ($j = 1, 2, \dots, 2n$, indices are understood modulo $2n$). If $a = 0$, then $z_{j+1} = -e^{2i\alpha}z_j$, and in this case $A_1A_2\cdots A_{2n}$ forms a regular $2n$-gon, and the conclusion is trivially true. Below we assume $0 < a < 1$.
Now consider the Möbius transformation $f(z) = \frac{e^{2i\alpha}(z-a)}{az-1}$, which has two fixed points (roots of the quadratic equation $ax^2 - (1+e^{2i\alpha})x + ae^{2i\alpha} = 0$), denoted by $x_1$ and $x_2$. It can be easily verified that $x_1 \neq x_2$.
Let $\frac{z_j - x_1}{z_j - x_2} = w_j$, from $f(z_j) = z_{j+1}$, we have that the ratios of consecutive terms among $w_1, w_2, \dots, w_{2n}, w_1$ are constants. Thus,
$$z_j = \frac{x_1 - x_2 w_j}{1 - w_j} = x_2 \cdot \frac{w_j - \frac{x_1}{x_2}}{w_j - 1},$$
hence
$$\begin{aligned} z_{j+1} - z_j &= \frac{x_2}{(w_j - 1)(w_{j+1} - 1)} \left( \left( w_{j+1} - \frac{x_1}{x_2} \right) (w_j - 1) - \left( w_j - \frac{x_1}{x_2} \right) (w_{j+1} - 1) \right) \\ &= \frac{(x_1 - x_2)(w_{j+1} - w_j)}{(w_j - 1)(w_{j+1} - 1)}. \end{aligned}$$
Therefore,
$$\prod_{j=1}^{n} \frac{z_{2j+1} - z_{2j}}{z_{2j} - z_{2j-1}} = \prod_{j=1}^{n} \frac{w_{2j+1} - w_{2j}}{w_{2j} - w_{2j-1}} = \prod_{j=1}^{n} \frac{w_{2j}}{w_{2j-1}}$$

is a complex number of modulus 1 (in fact it can be shown that its value is $-1$), hence
the proposition holds. $\square$