AustriaMO2011

Let $R$ be the common point of $t$ and the tangent $s$ of $k$ in $S$. Since $S$ is the mid-point of the arc $UV$, $s$ is parallel to $g$ (and not to $t$). Since $MR$ is perpendicular to $TS$ and bisects $\angle SRT$, $QP$ bisects $\angle UPT$. Let $W$ be the common point of $PQ$ and $TS$. Because of the given symmetry, we have $PQ\perp TS$, and triangle $PWT$ is therefore right-angled.



We therefore have $$\begin{array}{rl}\angle WTP+\angle WPT& =90^{\circ }\\ \iff 2\cdot \angle WTP+2\cdot \angle WPT& =180^{\circ }\\ \iff \angle QTP+\angle UPT& =180^{\circ },\end{array}$$ and $UV$ is therefore parallel to $QT$. $QTUV$ is therefore a trapezoid, as claimed. qed