Asian Pacific Mathematics Olympiad

Let $N$ be the radical center of the circumcircles of triangles $ABC$, $BMP$ and $CMP$. The pairwise radical axes of these circles are $BD$, $CE$ and $PM$, and hence they concur at $N$. Now, note that in directed angles:
$$\angle MCE = \angle MPE = \angle MPY = \angle MBY.$$

It follows that $BY$ is parallel to $CE$, and analogously that $CX$ is parallel to $BD$. Then, if $L$ is the intersection of $BY$ and $CX$, it follows that $BNC L$ is a parallelogram. Since $BM = MC$ we deduce that $L$ is the reflection of $N$ with respect to $M$, and therefore $L \in AM$. Using power of a point from $L$ to the circumcircles of triangles $BPM$ and $CPM$, we have
$$LY \cdot LB = LP \cdot LM = LX \cdot LC.$$
Hence, $BYXC$ is cyclic. Using the cyclic quadrilateral we find in directed angles:
$$\angle LXY = \angle LBC = \angle BCN = \angle NDE.$$
Since $CX \parallel BN$, it follows that $XY \parallel DE$.
Let $Q$ and $R$ be two points in $\Gamma$ such that $CQ$, $BR$, and $AM$ are all parallel. Then in directed angles:
$$\angle QDB = \angle QCB = \angle AMB = \angle PMB = \angle PDB.$$
Then $D$, $P$, $Q$ are collinear. Analogously $E$, $P$, $R$ are collinear. From here we get $\angle PRQ = \angle PDE = \angle PXY$, since $XY$ and $DE$ are parallel. Therefore $QRYX$ is cyclic. Let $S$ be the radical center of the circumcircle of triangle $ABC$ and the circles $BCYX$ and $QRYX$. This point lies in the lines $BC$, $QR$ and $XY$ because these are the radical axes of the circles. Let $T$ be the second intersection of $AS$ with $\Gamma$. By power of a point from $S$ to the circumcircle of $ABC$ and the circle $BCXY$ we have
$$SX \cdot SY = SB \cdot SC = ST \cdot SA.$$
Therefore $T$ is in the circumcircle of triangle $AXY$. Since $Q$ and $R$ are fixed regardless of the choice of $P$, then $S$ is also fixed, since it is the intersection of $QR$ and $BC$. This implies $T$ is also fixed, and therefore, the circumcircle of triangle $AXY$ goes through $T \neq A$ for any choice of $P$.

Let the lines $DP$ and $EP$ meet the circumcircle of $ABC$ again at $Q$ and $R$, respectively. Then $\angle DQC = \angle DBC = \angle DPM$, so $QC \parallel PM$. Similarly, $RB \parallel PM$.

Now, $\angle QCB = \angle PMB = \angle PXC = \angle(QX, CX)$, which is half of the arc $QC$ in the circumcircle $\omega_{C}$ of $QXC$. So $\omega_{C}$ is tangent to $BS$; analogously, $\omega_{B}$, the circumcircle of $RYB$, is also tangent to $BC$. Since $BR \parallel CQ$, the inscribed trapezoid $BRQC$ is isosceles, and by symmetry $QR$ is also tangent to both circles, and the common perpendicular bisector of $BR$ and $CQ$ passes through the centers of $\omega_{B}$ and $\omega_{C}$. Since $MB = MC$ and $PM \parallel BR \parallel CQ$, the line $PM$ is the radical axis of $\omega_{B}$ and $\omega_{C}$.
However, $PM$ is also the radical axis of the circumcircles $\gamma_{B}$ of $PMB$ and $\gamma_{C}$ of $PMC$. Let $CX$ and $PM$ meet at $Z$. Let $p(K, \omega)$ denote the power of a point $K$ with respect to a circumference $\omega$. We have
$$p\left(Z, \gamma_{B}\right) = p\left(Z, \gamma_{C}\right) = ZX \cdot ZC = p\left(Z, \omega_{B}\right) = p\left(Z, \omega_{C}\right).$$
Point $Z$ is thus the radical center of $\gamma_{B}$, $\gamma_{C}$, $\omega_{B}$, $\omega_{C}$. Thus, the radical axes $BY$, $CX$, $PM$ meet at $Z$. From here,
$$\begin{aligned} & ZY \cdot ZB = ZC \cdot ZX \Rightarrow BCXY \text{ cyclic} \\ & PY \cdot PR = PX \cdot PQ \Rightarrow QRXT \text{ cyclic.} \end{aligned}$$
We may now finish as in Solution 1. $\square$