SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $M$, $N$, $P$ be the midpoints of $BC$, $CA$, $AB$ and $D$ be the projection of $A$ on $BC$. Denote $(E)$ as the Euler circle of $△ABC$; then $M$, $N$, $P$, $D$ lie on $(E)$ and $E$ is the midpoint of $OH$. Similarly, let $H_a$, $O_a$ be the midpoints of $AH$, $AO$ and $K$ be the projection of $A$ on $OH$; then $H_a$, $O_a$, $K$, $E$ lie on the Euler circle of $△AOH$.

Denote $J$ as the symmetric point of $K$ through $NP$. Foremost, we will prove that $J$ is the intersection of $(K{H}_a{O}_a)$ and $(E)$.

Indeed, it is easy to see that $O$, $E$ are the circumcenter and orthocenter of $△MNP$. From the midsegment, we have $$\overrightarrow{E{O}_a}=\frac{1}{2}\cdot \overrightarrow{HA}=\overrightarrow{MO},$$ so $O_a$ is symmetric with $E$ through $NP$. Then $E{O}_{a}JK$ is an isosceles trapezoid, which means that $J\in (E{O}_{a}K)$.

On the other hand, since $O$ is the orthocenter of $△MNP$, if we denote $O^{\prime }$ as the symmetric point of $O$ through $NP$, then $O^{\prime }\in (E)$. In addition, $M{O}^{\prime }\perp MD$, so $D{O}^{\prime }$ is the diameter of $(E)$. And by the symmetry, we get $$\angle DJ{O}^{\prime }=\angle AKO=90^{\circ },$$ which implies that $J\in (E)$.

From (1) and (2), we get $J$ is the intersection of $(K{H}_a{O}_a)$ and $(E)$. Note that $J\in (E)$ and $O^{\prime }J$ is symmetric to $OH$ through $NP$, which implies that $J$ is the anti-Steiner point of $OH$ with respect to $△MNP$. Similarly, $J$ lies on the Euler circles of $△BOH$, $△COH$. $\square$