2022 China Team Selection Test for IMO

Proof. Denote the centers of the two circles by $O_1,O_2$, and the radii by $r_1,r_2$, respectively, where $r_1>r_2$. We first show that there are two points $P,Q$ on the ray $O_1{O}_2$ such that $O_{1}P\cdot O_{1}Q=r_1^2$ and $O_{2}P\cdot O_{2}Q=r_2^2$. One can choose a point $K$ on the ray $O_1{O}_2$ such that $O_{1}K=\frac{O_1{O}_2^2+r_1^2-r_2^2}{2O_1{O}_2}$. Since $O_1{O}_2\le r_1-r_2$, we have $O_{1}K\ge r_1$. We choose two points $P,Q$ on the line $O_1{O}_2$ such that $KP=KQ=\sqrt{O_1{K}^2-r_1^2}$ so that $O_{1}P\cdot O_{1}Q=r_1^2$. On the other hand, $$\begin{array}{rl}{O}_{2}P\cdot O_{2}Q-O_{1}P\cdot O_{1}Q& =O_2{K}^2-O_1{K}^2=(O_{1}K-O_1{O}_2{)}^2-O_1{K}^2\\ & =O_1{O}_2^2-2O_1{O}_2\cdot O_{1}K=r_2^2-r_1^2,\end{array}$$ which implies that $O_{2}P\cdot O_{2}Q=r_2^2$. For any given line $\ell$ as in the problem, if it is perpendicular to $O_1{O}_2$, we certainly have $\angle APC=\angle DPB$ by symmetry. If not, from $O_2{C}^2=O_{2}Q\cdot O_{2}P$ we know that $△O_{2}CQ\sim △O_{2}PC$. So $\frac{CQ}{CP}=\frac{r_2}{O_{2}P}$. Similarly, we obtain $\frac{DQ}{DP}=\frac{r_2}{O_{2}P}$, and therefore $\frac{CQ}{CP}=\frac{DQ}{DP}$, which means that the bisectors of $\angle CPD$ and $\angle CQD$ meet $\ell$ at the same point, say $M$. A similar argument shows that the bisectors of $\angle APB$ and $\angle AQB$ intersect $\ell$ at the same point again, say $M^{\prime }$.

Note that the points $P,Q,M$ are all on an Apollonian circle with distance ratio $\frac{CQ}{DQ}$ to $C$ and $D$. The center of the Apollonian circle must be on $\ell$. Thus this center must be the intersection of $\ell$ with the perpendicular bisector of $PQ$. Similarly, the points $P,Q,M^{\prime }$ are on the Apollonius circle with distance ratio $\frac{AQ}{BQ}$ to $A$ and $B$, whose center is the same point as above. Note that $K$ does not lie in the interior of ${\Gamma }_1$, so the center of this circle, denoted by $L$, must be outside ${\Gamma }_1$ so that $M$ coincides with $M^{\prime }$. Therefore, $$\angle APC=\angle APM-\angle CPM=\angle BPM-\angle DPM=\angle BPD.$$ This completes the proof.