Regional Competition

Since $\angle AT$ and $\angle BT$ are perpendicular to $\angle AO$ and $\angle BO$, the points $A$, $B$, $T$ and $O$ lie on a circle $k_1$ by Thales' theorem. By the central angle theorem we have $\angle AOB=2\gamma$. Since $BCS$ is an isosceles triangle, we find $\angle BCS=\angle CBS=\gamma$. Now $\angle ASB=2\gamma$, because an exterior angle of a triangle equals the sum of the other two interior angles. Thus $$\angle ASB=2\gamma =\angle AOB.$$ and by the inscribed angle theorem we find that the points $A$, $B$, $S$ and $O$ lie on a circle $k_2$. Since the circles $k_1$ and $k_2$ have the three points $A$, $B$ and $O$ in common, we have $k_1=k_2$ and the points $A$, $B$, $O$, $S$ and $T$ lie on a circle.

Finally we have $\angle TSB=\angle TOB=\gamma =\angle SBC$, from which it follows that $ST$ is parallel to $BC$.