BMO 2022 shortlist

Claim 1. $PK=KQ$. Proof of Claim 1. Let $L$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Since $L$ and $K$ are midpoints of $AB$ and $AH$, then $LK\parallel BH$ and so $LK\perp AC$. Since also $LO\perp AB$, then $\angle KLO=\angle BAC=\alpha$. Also, $\angle OLP=90^{\circ }=\angle OKP$, so the quadrilateral $OKLP$ is cyclic and therefore $\angle KPO=\angle KLO=\alpha$. Similarly, $\angle KQO=\alpha$. Therefore, the triangle $OPQ$ is isosceles, and since $OK\perp PQ$ then $K$ is the midpoint of $PQ$. $\square$



Claim 2. The intersection of $YP$ and $AK$ lies on $\omega$. Proof of Claim 2. Let $D$ be the other point of intersection of $AK$ with $\omega$. Since $OK\perp PQ$, then $K$ is the midpoint of the chord $\ell$ of $\omega$ through $P,Q$. Since $AD$ and $YC$ intersect at $K$, by the Butterfly theorem the points $P^{\prime }=YD\cap \ell$ and $Q=AC\cap \ell$ are equidistant from $K$. Thus $P^{\prime }=P$ and $YP\cap AK=D\in \omega$. $\square$

Now let $A^{\prime }$ be the other point of intersection of $AO$ with $\omega$ and let $S$ be the other point of intersection of $A^{\prime }H$ with $\omega$.

Claim 3. $PQ$ is the perpendicular bisector of $HS$. Proof of Claim 3. We have $\angle HSA=\angle A^{\prime }SA=90^{\circ }$ so $S$ lies on the circle ${\omega }^{\prime }$ with diameter $AH$ centered at $K$. So $KS=KH$. Since $O$ and $K$ are the circumcenters of $\omega$ and ${\omega }^{\prime }$ respectively, and $AS$ is their common chord, then $OK\perp AS$. But we also have $HS\perp AS$ and $PQ\perp OK$, thus $HS\perp PQ$. Since $KS=KH$, then $K$ belongs on the perpendicular bisector of $HS$ and the result follows. $\square$

From Claim 3 we have $\angle KSP=\angle KHP$. From Claim 1 and the fact that $KP=KQ$ we have that $AQPH$ is a parallelogram and so (using Claim 2 as well) $$\angle KHP=\angle KAC=\angle DAC=\angle DYC=\angle PYK.$$



Since $\angle KSP=\angle KYP$ we get that $S$ belongs on the circumcircle of triangle $KPY$. Similarly it belongs to the circumcircle of triangle $KQX$ and therefore the result follows.