Team Selection Test for IMO 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. $△XIB\sim △XCI$ because $\angle XIB=90^{\circ }-\angle BIP=\frac{1}{2}\angle C=\angle XCI$. So, $X{I}^2=XB\cdot XC=XZ\cdot XP$. In $△XIP$, 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.

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Alternative solution.



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

Claim 1. $X{I}^2=XB\cdot XC$

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

Claim 2. $K{I}^2=KA\cdot KP$

Proof. By Claim 1 and power of a point, $X{I}^2=XB\cdot XC=X{O}^2-R^2$. Also, $X{O}^2-X{I}^2=Y{O}^2-Y{I}^2=K{O}^2-K{I}^2$. Hence, $K{I}^2=K{O}^2-R^2$. By power of point $K$, $K{O}^2-R^2=KA\cdot KP$.

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