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Problem 765

AMC 10/12, early questions

Geometry Difficulty 3.5 Multiple choice CEMC Fermat · Canada · 2026

In the diagram, point $P$ lies inside square $WXYZ$ so that $\triangle PXY$ is an equilateral triangle.The measure of $\angle WPZ$ is

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Official solution

In equilateral $\triangle PXY$ , the measure of $\angle PXY$ is $60\degree$ and $PX=XY$ . In square $WXYZ$ , the measure of $\angle WXY$ is $90\degree$ and $WX=XY$ . Therefore, $\$\angle WXP=\angle WXY-\angle PXY=90\degree-60\degree=30\degree$ $and$ WX=PX $. Thus$ $\triangle$ WXP $is isosceles, and so$ $\angle PWX=\angle WPX=\dfrac{180\degree-30\degree}{2}=75\degree$ $. The measure of$ $\angle$ PWZ $is$ $90\degree-\angle PWX=90\degree-75\degree=15\degree\$.$

In a similar way, we can show that $\angle PZW=15\degree$ , and so $\angle WPZ=180\degree-\angle PWZ-\angle PZW$ or $\angle WPZ=180\degree-15\degree-15\degree=150\degree$ .

Figure for this problem

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