BMO 2022 shortlist

Firstly observe that $OAXB$ is cyclic, with diameter $OX$, and $Y$ also lies on this circle since $OY\perp XC$. Hence: $$\angle AZC=\angle XAB=\angle ABX=\angle AYX$$ and so $CYAZ$ is cyclic.



Let $M$ be the intersection of $YZ$ and $AC$ and let $CY$ intersect $\omega$ again at $W$. Using the new cyclic relation we get $\angle CYZ=\angle CAZ$ and then using that $ZA$ is tangent to $\omega$ we get $\angle CAZ=\angle CWA$, so $\angle CYM=\angle CWA$. Therefore the triangles $CWA$ and $CYM$ are similar. But $CW$ is a chord of $\omega$, and $Y$ is the foot of the perpendicular from $O$, hence $Y$ is the midpoint of $CW$. It follows from the similarity relation that $M$ is the midpoint of $AC$, as required.

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Alternative solution.

Let $M$ be the midpoint of $AC$. We have $\angle CAZ=\angle CBA$ and $\angle ZCA=\angle BAC$ so the triangles $CAZ$ and $ABC$ are similar. The line $CYX$ is the $C$-symmedian of triangle $ABC$, and $ZM$ is the corresponding median in triangle $CAZ$, hence by isogonality $\angle AZM=\angle ACY$. So $$\angle ZMA=180^{\circ }-\angle AZM-\angle MAZ=180^{\circ }-\angle ACY-\angle CBA(1)$$ Now observe $\angle OMC=\angle OYC=90^{\circ }$, so $CMYO$ is cyclic. Thus: $$\angle CYM=\angle COM=\frac{1}{2}\angle COA=\angle CBA.$$ This shows that $$\angle YMC=180^{\circ }-\angle MCY-\angle CYM=180^{\circ }-\angle ACY-\angle CBA$$ Combining this with (1) we get that $\angle YMC=\angle ZMA$ and as $A,C,M$ are collinear, it follows that $Z,M,Y$ are collinear as required.

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Alternative solution.

As in Solution 2 we have that $CX$ is the A-symmedian of triangle $ABC$ and that triangle $ABC$ is similar to triangle $CAZ$. Let $f$ be the spiral similarity which maps $AC$ onto $AB$ and let $g$ be the reflection on the perpendicular bisector of $AB$. Note that $f$ is a rotation about $A$ by an angle of $\angle CAB$ (clockwise in our figure) followed by a homothety centered at $A$ by a factor of $AB/AC$. By the similarity of triangles $ABC$ and $CAZ$ we have that $g(f(Z))=C$, so actually $f(Z)$ is the other point of intersection, say $C^{\prime }$, of $CZ$ with $\omega$. As in Solution 1 we have that $CYAZ$ is cyclic. Therefore, letting $W$ be the other point of intersection of $CY$ with $\omega$, we have $\angle WAB=\angle WCB=\angle CAY$. We also have $\angle ACY=\angle ABW$. It follows that $f(Y)=W$. Let $W^{\prime }=g(W)$. Then $W^{\prime }\in \omega$ and since $CW$ is the A-symmedian, then $C{W}^{\prime }$ passes through the midpoint $N$ of $AB$. Now $C{W}^{\prime }$ and $C^{\prime }W$ intersect on the perpendicular bisector of $AB$ and therefore they intersect on $N$. It follows that $N=AB\cap C^{\prime }W=Af(C)\cap f(Z)f(Y)$ is the image of $M=AC\cap ZY$ under $f$. Since $N$ is the midpoint of $AB$, then $M$ is the midpoint of $AC$.