Taiwan IMO Selection Camp, 2023

Solution 2. Below is another proof that the line $APA'$ is the radical axis of circles $ABD$ and $ACE$. This follows from showing that the second intersection point of $ABD$ and $ACE$ lies on line $AP$.

Define the point $N$ as the second intersection of circle $PDE$ with line $AP$. From $\angle DNA = \angle DNP = \angle DEP = \angle DBA$, we know that $N$ lies on circle $ABD$; similarly, one can show that $N$ lies on circle $ACE$. This completes the proof. □

Solution 1. Let $A'$ be the intersection of line $BX$ and line $CY$. By the power of a point, it suffices to prove that $A'B \cdot A'X = A'C \cdot A'Y$, or equivalently, that $A'$ lies on the radical axis of circles $ABDX$ and $ACEY$.

From $DA = DX$, we know that on circle $ABDX$, the point $D$ bisects one of the two arcs with endpoints $A, X$. Therefore, depending on the order of the points, line $BC$ is either the internal or external angle bisector of $\angle ABX$. In either case, line $BX$ is the reflection of line $BA$ over line $BC$. Similarly, line $CY$ is the reflection of line $CA$ over line $BC$. Hence $A'$ is the reflection of $A$ over line $BC$, from which it follows that $A, F, A'$ are collinear.

From $PD \parallel AC$ and $PE \parallel AB$, we know that $\frac{FD}{FC} = \frac{FP}{FA} = \frac{FE}{FB}$, which gives $FD \cdot FB = FE \cdot FC$.

Therefore, the point $F$ has equal power with respect to circles $ABDX$ and $ACEY$.

The point $A$ lies on both circles, so it also has equal power with respect to them. Therefore, the radical axis of circles $ABDX$ and $ACEY$ is the altitude $AF$, which passes through $A'$. This completes the proof. □