AustriaMO

Geometry Difficulty 6.0 AIME, harder Prove it Austria

Let $k$ be a circle with mid-point $M$ . $T$ is a point on $k$ and $t$ the tangent of $k$ in $T$ . $P$ is a point on $t$ with $P \neq T$ and $g$ a line containing $P$ with $g \neq t$ . $g$ has the points $U$ and $V$ in common with $k$ ( $U \neq V$ ), and $S$ is the mid-point of the arc $UV$ not containing $T$ . $Q$ is the point symmetric to $P$ with respect to $TS$ . Prove that $QTUV$ is a trapezoid.

Solution

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

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.