69th Belarusian Mathematical Olympiad

Denote the foot of the altitude from $A$ by $E$ and the midpoint of $OH$ by $L$. Since $\angle ATS=\angle AES=90^{\circ }$, the quadrilateral $ATLE$ is cyclic. Thus it is enough to prove that the quadrilateral $ALES$ is cyclic. Let $K$ be the reflection of $H$ about $E$. It is well known that $K$ lies on the circumcircle of $△ABC$. The segment $LE$ is the midline of the triangle $OHK$, so $LE=0.5OK=0.5OA=TA$. The segment $TL$ is the midline of the triangle $OAH$, so $TL\parallel AE$. Clearly $TL<AH<AE$, hence $ATLE$ is an isosceles trapezium, which is always cyclic.