Track / Stage 4 / 177 of 340 #917 of 2444

Problem 917

AMC 12 late, AIME early

Geometry Difficulty 4.7 Multiple choice CEMC Fermat · Canada · 2016

In the diagram, $PQRS$ represents a rectangular piece of paper. The paper is folded along a line $VW$ so that $\angle VWQ = 125^\circ$ . When the folded paper is flattened, points $R$ and $Q$ have moved to points $R'$ and $Q'$ , respectively, and $R'V$ crosses $PW$ at $Y$ .

The measure of $\angle PYV$ is

Pick one

Official solution

Since points $Y$ , $W$ and $Q$ form a straight line segment, then $\angle YWV = 180^\circ - \angle VWQ$ and so $\angle YWV = 180^\circ - 125^\circ = 55^\circ$ .

Since $Q'$ is the final position of $Q$ after folding, then $\angle Q'WV$ is the final position of $\angle QWV$ after folding, and so $\angle Q'WV = \angle QWV$ .

Thus, $\angle Q'WV = \angle QWV = 125^\circ$ and so $\angle Q'WY = \angle Q'WV - \angle YWV = 125^\circ - 55^\circ = 70^\circ$ .

[[IMAGE0]]

Since $Q'W$ and $R'Y$ are parallel sides of the piece of paper, then $\angle R'YW + \angle Q'WY = 180^\circ$ , and so $\angle R'YW = 180^\circ - \angle Q'WY = 180^\circ - 70^\circ = 110^\circ$ .

Finally, $\angle PYV$ is opposite $\angle R'YW$ so $\angle PYV = \angle R'YW = 110^\circ$ .

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