China-TST-2025A

For any point $X$ on line $A_{i+2}{B}_i$, we have: $$\frac{d_{X-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}}=\frac{d_{X-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}}.$$ For any point $X$ on line $A_i{B}_i$, we have: $$\frac{d_{X-A_{i-1}{A}_i}}{A_{i-1}{A}_i}+\frac{d_{X-A_i{A}_{i+1}}}{A_i{A}_{i+1}}=0.$$ Therefore, for point $B_i$: $$\frac{d_{B_i-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}}=\frac{d_{B_i-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}},(6)$$ and $$\frac{d_{B_i-A_{i-1}{A}_i}}{A_{i-1}{A}_i}+\frac{d_{B_i-A_i{A}_{i+1}}}{A_i{A}_{i+1}}=0.(7)$$ Combining these gives: $$\frac{d_{B_i-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}}=\frac{d_{B_i-A_{i-1}{A}_i}}{A_{i-1}{A}_i}+\frac{d_{B_i-A_i{A}_{i+1}}}{A_i{A}_{i+1}}+\frac{d_{B_i-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}}.$$ Similarly for $B_{i+1}$: $$\frac{d_{B_{i+1}-A_{i-1}{A}_i}}{A_{i-1}{A}_i}=\frac{d_{B_{i+1}-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}},$$ and $$\frac{d_{B_{i+1}-A_i{A}_{i+1}}}{A_i{A}_{i+1}}+\frac{d_{B_{i+1}-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}}=0.$$ Thus: $$\frac{d_{B_{i+1}-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}}=\frac{d_{B_{i+1}-A_{i-1}{A}_i}}{A_{i-1}{A}_i}+\frac{d_{B_{i+1}-A_i{A}_{i+1}}}{A_i{A}_{i+1}}+\frac{d_{B_{i+1}-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}}.$$ Since these relations are linear in the coordinates, point $C_i$ must also satisfy: $$\frac{d_{C_i-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}}=\frac{d_{C_i-A_{i-1}{A}_i}}{A_{i-1}{A}_i}+\frac{d_{C_i-A_i{A}_{i+1}}}{A_i{A}_{i+1}}+\frac{d_{C_i-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}}.(8)$$

Noting that $C_i$ lies on $A_i{A}_{i+1}$ (so $d_{C_i-A_i{A}_{i+1}}=0$), we can rewrite this as: $$\frac{d_{C_i-A_{i+2}{A}_{i+3}}}{A_{i+2}{A}_{i+3}}+\frac{d_{C_i-A_i{A}_{i+1}}}{A_i{A}_{i+1}}=\frac{d_{C_i-A_{i-1}{A}_i}}{A_{i-1}{A}_i}+\frac{d_{C_i-A_{i+1}{A}_{i+2}}}{A_{i+1}{A}_{i+2}}.(9)$$ This equation is completely symmetric for $C_1,C_2,C_3,C_4$. If it held identically for all points, then substituting $A_1$ would yield: $$\frac{A_1{A}_2\cdot \sin A_2}{A_2{A}_3}=\frac{A_1{A}_4\cdot \sin A_4}{A_3{A}_4},$$ and substituting $A_3$ would give: $$\frac{A_2{A}_3\cdot \sin A_2}{A_1{A}_2}=\frac{A_3{A}_4\cdot \sin A_4}{A_1{A}_4}.$$ This would imply $\sin A_2=\sin A_4$, and similarly $\sin A_1=\sin A_3$, meaning $A_1{A}_2{A}_3{A}_4$ would either be a parallelogram or cyclic - contradicting the given conditions. Therefore, the equation is not identically satisfied. Since it is linear in coordinates, the set of points satisfying it must form a straight line. Hence $C_1,C_2,C_3,C_4$ are collinear. $\square$

Second Proof: Define the cross ratio of four lines $PA,PB,PC,PD$ through point $P$ as: $$P(A,B;C,D):=\frac{\sin \angle APC\cdot \sin \angle BPD}{\sin \angle APD\cdot \sin \angle BPC}.$$ First, we prove a lemma. Lemma: Let $A,B,C,D$ be four points in the plane with no three collinear, and let $M$ be another point. Then a point $P$ lies on the conic through $A,B,C,D,M$ if and only if $P(A,B;C,D)=M(A,B;C,D)$. Proof of Lemma: Let $a,b,c,d,m,p$ be the complex coordinates of points $A,B,C,D,M,P$ respectively. Then: $$\begin{array}{rl}\frac{PA\cdot PB}{PC\cdot PD}& =\frac{\sin \angle APB\cdot \sin \angle BPD}{\sin \angle ABP\cdot \sin \angle BPD}\\ & =\frac{\mathrm{Im}\left(\frac{c-p}{a-p}/\left|\frac{c-p}{a-p}\right|\right)\cdot \mathrm{Im}\left(\frac{d-p}{b-p}/\left|\frac{d-p}{b-p}\right|\right)}{\mathrm{Im}\left(\frac{c-p}{b-p}/\left|\frac{c-p}{b-p}\right|\right)\cdot \mathrm{Im}\left(\frac{d-p}{a-p}/\left|\frac{d-p}{a-p}\right|\right)}\\ & =\frac{[(c-p)(\bar{a}-\bar{p})-(\bar{c}-\bar{p})(a-p)][(d-p)(\bar{b}-\bar{p})-(\bar{d}-\bar{p})(b-p)]}{[(c-p)(\bar{b}-\bar{p})-(\bar{c}-\bar{p})(b-p)][(d-p)(\bar{a}-\bar{p})-(\bar{d}-\bar{p})(a-p)]}\end{array}$$ Since each bracket is linear in $p$ and $\bar{p}$, the condition "cross ratio equals a constant" defines a quadratic equation in $p$ and $\bar{p}$, which corresponds to a conic section. (Note: This cannot be identically constant, as that would imply $A,B,C$ are collinear, a contradiction.) $\square$

Returning to the problem, since $A_3{B}_1$ and $A_3{M}_1$ are symmetric about the angle bisector of $\angle A_2{A}_3{A}_4$, $A_3{B}_1$ is the symmedian of $△A_2{A}_3{A}_4$. Let $S$ be the intersection of $B_1{A}_1$ and $A_2{A}_4$, and $T$ the intersection of $B_1{A}_3$ and $A_2{A}_4$. Then: $$\begin{array}{rl}{B}_1(A_1,A_2;A_3,A_4)& =(S,A_2;T,A_4)=1-(S,T;A_2,A_4)\\ & =1-\frac{S{A}_2}{S{A}_4}\cdot \frac{T{A}_4}{T{A}_2}=1-\frac{A_1{A}_2^2\cdot A_3{A}_4^2}{A_1{A}_4^2\cdot A_3{A}_2^2}.\end{array}$$ Similarly, this holds for $B_2,B_3,B_4$. Thus all eight points $A_1,\dots ,A_4,B_1,\dots ,B_4$ lie on the same conic.

Considering $A_1,A_2,A_3,B_1,B_2,B_3$, by Pascal's Theorem, $C_1,C_2$ and $X:=A_3{B}_1\cap A_1{B}_3$ are collinear. Similarly, considering $A_1,A_4,A_3,B_1,B_4,B_3,C_3,C_4$ and $X$ are collinear. Considering $A_2,A_3,A_4,B_2,B_3,B_4,C_2,C_3$ and $Y:=A_4{B}_2\cap A_2{B}_4$ are collinear. Considering $A_4,A_1,A_2,B_4,B_1,B_2,C_1,C_4$ and $Y$ are collinear.

Let $W$ be the intersection of $A_1{B}_1$ and $A_1{B}_2$. Applying Menelaus' Theorem to triangle $△W{B}_1{B}_2$ with transversal $A_1{A}_2{C}_1$ gives: $$\frac{B_1{C}_1}{B_2{C}_1}=\frac{B_1{A}_1}{B_2{A}_2}\cdot \frac{A_{1}W}{A_{2}W}=\frac{B_1{A}_1}{B_2{A}_2}\cdot \frac{\sin \angle W{A}_2{A}_1}{\sin \angle W{A}_1{A}_2}.$$ By tangent properties, $\angle W{A}_2{A}_1=\angle A_2{A}_3{A}_1$, so: $$\frac{B_1{C}_1}{B_2{C}_1}=\frac{B_1{A}_1}{B_2{A}_2}\cdot \frac{\sin \angle A_2{A}_3{A}_1}{\sin \angle A_1{A}_4{A}_2}.$$ Taking the cyclic product (indices modulo 4): $$\begin{array}{rl}\prod _{i=1}^4\frac{B_i{C}_i}{B_{i+1}{C}_i}& =\prod _{i=1}^4\frac{B_i{A}_i}{B_{i+1}{A}_{i+1}}\cdot \prod _{i=1}^4\frac{\sin \angle A_{i+1}{A}_{i+2}{A}_i}{\sin \angle A_i{A}_{i-1}{A}_{i+1}}\\ & =\prod _{i=1}^4\frac{\sin \angle A_{i+1}{A}_{i+2}{A}_i}{\sin \angle A_{i+1}{A}_i{A}_{i+2}}=\prod _{i=1}^4\frac{A_i{A}_{i+1}}{A_{i+1}{A}_{i+2}}=1.\end{array}\text{\hspace{1em}(10)}$$

Thus: $$\frac{B_1{C}_1}{B_2{C}_1}\cdot \frac{B_2{C}_2}{B_3{C}_2}=\frac{B_1{C}_4}{B_4{C}_4}\cdot \frac{B_3{C}_4}{B_3{C}_3}.$$ Let this ratio be $\alpha$. On the extension of $B_3{B}_1$, take point $P$ such that $\frac{B_{3}P}{B_{1}P}=\frac{1}{\alpha }$. By the converse of Menelaus' Theorem applied to $△B_1{B}_2{B}_3$ and points $P,C_1,C_2$, these are collinear. Similarly, $P,C_3,C_4$ are collinear. Thus we have $P,C_1,C_2$ collinear and $P,C_3,C_4$ collinear, while $X,C_1,C_2$ and $X,C_3,C_4$ are also collinear. Since $P$ lies on $B_1{B}_3$ and $X=A_1{B}_3\cap A_3{B}_1$, if $P=X$ then $A_1=B_1$ or $A_3=B_3$, making $A_1{A}_3$ a symmedian for both $△A_1{A}_2{A}_4$ and $△A_2{A}_3{A}_4$, which would imply $A_1{A}_2{A}_3{A}_4$ is harmonic - contradicting the non-cyclic condition. Therefore $P\ne X$, and consequently $C_1,C_2,C_3,C_4$ all lie on the line $PX$. $\square$