BMO 2022 shortlist

Claim 1. $AX$ is the $A$-symmedian of $△ABC$. Proof of Claim 1. Let $Y\in \omega$ such that $AY$ is the $A$-symmedian of triangle $ABC$. We want to prove that $Y=X$. We have that $\angle BAY=\angle CAM$ and $\angle BYA=\angle BCA=\angle MCA$, therefore the triangles $ABY$ and $AMC$ are similar. It follows that $(AY)(AM)=(AB)(AC)$. Since $AE$ is the bisector of $\angle BAC$, then $\angle BAF=\angle CAE$. We also have $\angle ABF=\angle ABC=\angle AEC$, therefore the triangles $BAF$ and $EAC$ are similar. It follows that $(AE)(AF)=(AB)(AC)$. We get $(AY)(AM)=(AE)(AF)$ and since also $\angle YAF=\angle EAM$, then the triangles $YAF$ and $EAM$ are similar. So $\angle AFY=\angle AME$ and $\angle YFE=\angle EMV$. But as $E$ is the midpoint of the arc $YV$, it follows that $\angle EMV=\angle YME$. So $\angle YFE=\angle YME$ from which it follows that the quadrilateral $YFME$ is cyclic. But since $Y\in \omega$, we finally get that $Y=X$. $\square$ From Claim 1 we conclude that the triangles $XBC$ and $VCB$ are equal. Thus $MX=MV=M{X}^{\prime }$. So the triangle $X^{\prime }XV$ is a right-angled triangle and $X^{\prime }X$ is perpendicular to $XV$ and therefore also to $BC$. Thus $X^{\prime }$ is the reflection of $X$ on $BC$. Claim 2. The quadrilateral $AS{A}^{\prime }M$ is cyclic. Proof of Claim 2. We have $\angle AS{A}^{\prime }=\angle ASX=\angle ABX$. But from Claim 1 we also have $\angle ABX=\angle AMC$. So $\angle AS{A}^{\prime }=\angle AMC$ and the result follows. $\square$ Claim 3. The quadrilateral $XS{X}^{\prime }M$ is cyclic. Proof of Claim 3. From Claim 2 we have $\angle XSM=\angle A^{\prime }SM=\angle A^{\prime }AM$. Since $X{X}^{\prime }$ is parallel to $A{A}^{\prime }$ we have $\angle A^{\prime }AM=\angle X{X}^{\prime }M$. So $\angle XSM=\angle X{X}^{\prime }M$ and the result follows. $\square$

Now from Claim 3 we have $$\angle XS{X}^{\prime }=\angle XMV=\angle X{X}^{\prime }M+\angle X^{\prime }XM=2\angle X{X}^{\prime }M=2\angle A{A}^{\prime }M.$$ So to conclude the proof it is enough to also show that $\angle XSZ=2\angle A{A}^{\prime }M$. From Claim 1 we have $\angle ACX=\angle MAB$ and therefore $$\angle AZX=\angle ACX=\angle AMB=90^{\circ }-\angle A^{\prime }AM.$$ Since the triangle $XAZ$ is isosceles, we deduce that $$\angle XSZ=\angle XAZ=180^{\circ }-2\angle AZX=2\angle A^{\prime }AM$$ thus completing the proof.