, 2024

*Claim 1.* $A$, $X$, $P$, $E$ are concyclic.

*Proof.* $X$, $I$, $M$, $P$ are concyclic because $\angle XIP = \angle XMP = 90^\circ$. It is well known that $S$ lies on the incircle and since $DI = SI$ and $DM = EM$, we get $IM \parallel AE$. So, $\angle EAP = \angle PIM = \angle PXE$, which means that $A$, $X$, $P$, $E$ are concyclic.

Let $XP$ intersect the circumcircle a second time at $Z$.

Let $M$ be the midpoint of $BC$, let the incircle touch $BC$ at $D$ and let $E$ be the reflection of $D$ over $M$. Let $DI$ and $AE$ intersect at $S$.

*Claim 2.* $IZ \perp XP$

Proof. $\triangle XIB \sim \triangle XCI$ because $\angle XIB = 90^\circ - \angle BIP = \frac{1}{2} \angle C = \angle XCI$. So, $XI^2 = XB \cdot XC = XZ \cdot XP$. In $\triangle XIP$, by the Euclidean theorem, $IZ \perp XP$.

Let $J$ be the reflection of $I$ over $O$. Let $T$ be the midpoint of $PZ$.

*Claim 3.* $X$, $Y$, $J$, $P$, $E$ are concyclic.

Proof. $OT \perp PZ$ and by claim 2, $IZ \perp PZ$. Therefore, by Thales' theorem, $JP \perp PZ$. So, $\angle JPX = \angle JEX = \angle JYX = 90^\circ$, therefore $X$, $Y$, $J$, $P$, $E$ are concyclic.

By claims 1 and 3, it is concluded that $A$, $P$, $X$, $Y$ are concyclic, as desired.

Let $XY$ and $AP$ intersect at $K$. Let $R$ be the circumradius of $ABC$.

*Claim 1.* $XI^2 = XB \cdot XC$

*Proof.* $\triangle XIB \sim \triangle XCI$ because $\angle XIB = 90^\circ - \angle BIP = \frac{1}{2} \angle C = \angle XCI$.

*Claim 2.* $KI^2 = KA \cdot KP$

*Proof.* By Claim 1 and power of a point, $XI^2 = XB \cdot XC = XO^2 - R^2$. Also, $XO^2 - XI^2 = YO^2 - YI^2 = KO^2 - KI^2$. Hence, $KI^2 = KO^2 - R^2$. By power of point $K$, $KO^2 - R^2 = KA \cdot KP$.

By the Euclidean Theorem, $KI^2 = KX \cdot KY$ which is by Claim 2 also equal to $KA \cdot KP$. Therefore, $A$, $P$, $X$, $Y$ are concyclic, as desired.