CEMC Pascal, 2016

Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

In the diagram, points $Q$ and $R$ lie on $PS$ and $\angle QWR = 38^\circ$ .

If $\angle TQP = \angle TQW = x^\circ$ , $\angle VRS = \angle VRW = y^\circ$ , and $U$ is the point of intersection of $TQ$ extended and $VR$ extended, then the measure of $\angle QUR$ is

Pick one

Solution

Since $\angle TQP$ and $\angle RQU$ are opposite angles, then $\angle RQU = \angle TQP = x^\circ$ .

Similarly, $\angle QRU = \angle VRS = y^\circ$ .

Since the angles in a triangle add to $180^\circ$ , then $\angle QUR = 180^\circ - \angle RQU - \angle QRU = 180^\circ - x^\circ - y^\circ$ Now $\angle WQP$ and $\angle WQR$ are supplementary, as they lie along a line.

Thus, $\angle WQR = 180^\circ - \angle WQP = 180^\circ - 2x^\circ$ .

Similarly, $\angle WRQ = 180^\circ - \angle WRS = 180^\circ - 2y^\circ$ .

Since the angles in $\triangle WQR$ add to $180^\circ$ , then $\begin{aligned} 38^\circ + (180^\circ - 2x^\circ) + (180^\circ - 2y^\circ) & = 180^\circ \\ 218^\circ & = 2x^\circ + 2y^\circ \\ x^\circ + y^\circ & = 109^\circ\end{aligned}$ Finally, $\angle QUR = 180^\circ - x^\circ - y^\circ = 180^\circ - (x^\circ + y^\circ) = 180^\circ - 109^\circ = 71^\circ$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.