Taiwan IMO Selection Camp, 2023

Let us first prove an important lemma.

Lemma. If $X$ is the intersection point of $BC$ and $YZ$, then $X$ lies on the perpendicular bisector of $OH$.

Proof of the lemma. Let $H'$ be the reflection of $H$ across $BC$, and let $O'$ be the reflection of $O$ across $YZ$. It is well known that $AO = H'O$ and $AH = O'H$. Therefore $\angle HH'O = \angle OAH = \angle HO'H$, so $HH'OO'$ are concyclic. Hence $X$ is the center of the circle $HH'OO'$, as desired. □

Another proof of the lemma. Take a point $P$ such that $AHPO$ is a parallelogram. It is well known that $BC$ is the perpendicular bisector of $OP$, and $YZ$ is the perpendicular bisector of $HP$. Then $X$ is the circumcenter of triangle $OHP$, so $X$ lies on the perpendicular bisector of $OH$. □

Returning to the original problem. Since $ABC$ is a scalene triangle, $YZ$ must be different from $BC$. Hence $Z = BC \cap YZ$, which by the Lemma lies on the perpendicular bisector of $OH$. Again let $O'$ be the reflection of $O$ across $YZ$; we have $ZO' = ZO = ZH = O'H$, so $\triangle ZO'H$ is an equilateral triangle. Thus $\angle AOH = 180^\circ - \angle O'OH = 180^\circ - \frac{1}{2}\angle O'ZH = 150^\circ$. We also have
$$\angle AYH = 90^\circ - \angle AZY = 90^\circ - \angle HZO' = 30^\circ,$$
where the first equality comes from the properties of the orthocenter $H$ and circumcenter $O$. Therefore $AHOY$ are concyclic, as desired. □

Alternative proof. As in the proof above, it suffices to show $ZO = ZH$. Here we give another proof of this equality.

Since $AO \perp YZ$ and $AH \perp BC$, we have
$$\angle AOZ = \angle ZYA = 90^\circ - \angle YAO = 90^\circ - \angle HAZ = \angle AZC.$$
Hence the line $BC$ is tangent to the circumcircle of triangle $AOZ$ at the point $Z$. Consider the circumcenter $H''$ of $AOZ$. We have $H''Z \perp BC$ and $H''Z = HZ = AH$, since $H''$ is the reflection of $H$ across $AZ$. Since the distance from $O$ to $BC$ is $AH/2$, we get $H''O = ZO$. Combining this with the fact that $H''$ is the circumcenter of $AOZ$, we obtain $ZO = H''O = ZH'' = ZH$, as desired. □