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 $\triangle ABC$; 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 $\triangle AOH$.

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

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

On the other hand, since $O$ is the orthocenter of $\triangle MNP$, if we denote $O'$ as the symmetric point of $O$ through $NP$, then $O' \in (E)$. In addition, $MO' \perp MD$, so $DO'$ is the diameter of $(E)$. And by the symmetry, we get
$$\angle DJO' = \angle AKO = 90^\circ,$$
which implies that $J \in (E)$.

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