30th Turkish Mathematical Olympiad

Let $S$ and $M$ be the second intersections of the circumcircle with the line $AH$ and the circle $(AHO)$, respectively. Since $OA=OM$, one has $\angle MHO=\angle OHS$, hence $M$ and $S$ are reflections of one another over the line $OH$. Therefore, the perpendicular bisector of $[MH]$, the perpendicular bisector of $[SH]$ which is the line $BC$, and the line $OH$ are concurrent; in other words, the perpendicular bisector of $[MH]$ passes through $T$. Therefore, the reflection of $H$ over the line $TX$ is just the point $M$.