Serbian Mathematical Olympiad

Let $M$ and $H$ denote the centroid and orthocenter of triangle $ABC$ respectively, and let $T$ be a point such that $3\overrightarrow{MT} = \overrightarrow{AA_1} = \overrightarrow{BB_1} = \overrightarrow{CC_1}$. Let $\omega$ be the sphere with diameter $HT$. Since the line $MT$ is perpendicular to the plane $ABC$, the point $M$ lies on $\omega$. We will show that the orthocenter $H_1$ of triangle $A_1 BC$ also lies on $\omega$ (the proof for the other two triangles is analogous).
Let $N$ be the midpoint of segment $BC$. Since $NA = 3NM$, the point $T$ lies on segment $A_1 N$, and consequently lies in the

plane $A_1BC$. Let $AA'$ be the altitude of triangle $ABC$. As $AA_1$ is perpendicular to the plane $ABC$, we have $\angle A_1AA' = 90^\circ$, and by the Three Perpendiculars Theorem, $A'A_1 \perp BC$, meaning $H_1$ lies on segment $A_1A'$.
Since the reflection of $H$ over $BC$ lies on the circumcircle of $ABC$, we have $A'H \cdot A'A = A'B \cdot A'C$. Applying the same reasoning to triangle $A_1BC$ gives $A'H_1 \cdot A'A_1 = A'B \cdot A'C = A'H \cdot A'A$. Therefore, quadrilateral $AHH_1A_1$ is cyclic, so $\angle HH_1A_1 = 180^\circ - \angle A_1AH = 90^\circ$.
Furthermore, since $HA' \perp BC$ and $H_1A' \perp BC$, applying the Three Perpendiculars Theorem again shows that line $HH_1$ is perpendicular to the plane $A_1BC$. Thus, $\angle HH_1T = 90^\circ$, proving that $H_1$ lies on sphere $\omega$, as required.