2023 Chinese IMO National Team Selection Test

All subscripts are to be understood modulo $2n$. Denote $\angle P{A}_1{A}_2=\angle P{A}_2{A}_3=\cdots =\alpha$. For $1\le i\le 2n$, extend $A_{i}P$ to intersect the circumference at point $B_i$. Note that $$\angle A_i{B}_i{B}_{i-1}=\angle A_i{A}_{i-1}P=\angle A_{i+1}{A}_i{B}_i=\alpha .$$ Hence, $A_i{A}_{i+1}{B}_{i-1}{B}_i$ is an isosceles trapezoid. (In fact, $B_1{B}_2\cdots B_{2n}$ is the polygon obtained by rotating $A_2{A}_3\cdots A_{2n}{A}_1$ in the clockwise direction by $2\alpha$ on the circumference.) In particular, $A_i{A}_{i+1}=B_{i-1}{B}_i$. Also, by $△A_{i-1}P{A}_i\sim △B_{i}P{B}_{i-1}$, we get $$(\ast )\frac{A_{i-1}P}{B_{i}P}=\frac{A_{i}P}{B_{i-1}P}=\frac{A_{i-1}{A}_i}{B_{i-1}{B}_i}=\frac{A_{i-1}{A}_i}{A_i{A}_{i+1}}.$$ 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. $\square$

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Alternative solution.

Draw line $A_1{A}_4$ intersecting line $P{A}_3$ at point $H$. Given that points $A_1,A_2,A_3,A_4$ are concyclic, we have $$\begin{array}{rl}\angle A_2{A}_1{A}_4& =180^{\circ }-\angle A_2{A}_3{A}_4=180^{\circ }-\angle A_2{A}_{3}P-\angle P{A}_3{A}_4\\ & =180^{\circ }-\angle A_2{A}_{3}P-\angle A_3{A}_{2}P=\angle A_{2}P{A}_3.\end{array}$$ Therefore, points $A_1,A_2,H,P$ are concyclic. In particular, $\angle A_{2}H{A}_3=\angle A_2{A}_{1}P=\angle H{A}_3{A}_4$, we have $A_{2}H\parallel A_3{A}_4$. From this and the fact that $A_1,A_2,H,P$ are concyclic, we get $$\angle A_3{A}_{4}H=\angle A_{2}H{A}_1=\angle A_{1}P{A}_2.$$ Together with $\angle A_2{A}_{1}P=\angle A_4{A}_{3}H$, we find that $△A_1{A}_{2}P\sim △A_{3}H{A}_4$. From this we obtain $$\frac{A_1{A}_2}{A_{1}P}=\frac{A_{3}H}{A_3{A}_4}\Rightarrow A_1{A}_2\cdot A_3{A}_4=A_{1}P\cdot A_{3}H.$$ Moreover, from $\angle A_{3}H{A}_2=\angle A_2{A}_{1}P=\angle A_3{A}_{2}P$, we have $△A_{3}H{A}_2\sim △A_3{A}_{2}P$. We get $$\frac{A_2{A}_3}{A_{3}H}=\frac{P{A}_3}{A_2{A}_3}\Rightarrow (A_2{A}_3{)}^2=P{A}_3\cdot A_{3}H.$$

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. $\square$

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Alternative solution.

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 $\angle P{A}_1{A}_2=\pi -\alpha \in (0,\pi )$. 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 $z_1\ne z_2$, we can solve to get $z_2=\frac{e^{2i\alpha }(z_1-a)}{a{z}_1-1}$, and similarly $z_{j+1}=\frac{e^{2i\alpha }(z_j-a)}{a{z}_j-1}$, ($j=1,2,\dots ,2n$, indices are understood modulo $2n$). If $a=0$, then $z_{j+1}=-e^{2i\alpha }{z}_j$, and in this case $A_1{A}_2\cdots A_{2n}$ 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 $a{x}^2-(1+e^{2i\alpha })x+a{e}^{2i\alpha }=0$), denoted by $x_1$ and $x_2$. It can be easily verified that $x_1\ne x_2$. 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 $$\begin{array}{rl}{z}_{j+1}-z_j& =\frac{x_2}{(w_j-1)(w_{j+1}-1)}\left(\left(w_{j+1}-\frac{x_1}{x_2}\right)(w_j-1)-\left(w_j-\frac{x_1}{x_2}\right)(w_{j+1}-1)\right)\\ & =\frac{(x_1-x_2)(w_{j+1}-w_j)}{(w_j-1)(w_{j+1}-1)}.\end{array}$$ 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. $\square$