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CEMC Cayley, 2021

Geometry Difficulty 2.4 Junior Find the answer Canada

In the diagram, pentagon $TPSRQ$ is constructed from equilateral $\triangle PTQ$ and square $PQRS$ .

The measure of $\angle STR$ is equal to

Pick one

Solution

Join $S$ to $T$ and $R$ to $T$ .

Since $PQRS$ is a square, $\angle SPQ = 90^\circ$ .

Since $\triangle PTQ$ is equilateral, $\angle TPQ = 60^\circ$ .

Therefore, $\angle SPT = \angle SPQ + \angle TPQ = 90^\circ + 60^\circ$ .

Since $PQRS$ is a square, $SP= PQ$ .

Since $\triangle PTQ$ is equilateral, $TP = PQ$ .

Since $SP = PQ$ and $TP = PQ$ , then $SP = TP$ which means that $\triangle SPT$ is isosceles.

Thus, $\angle PTS = \frac{1}{2}(180^\circ - \angle SPT) = \frac{1}{2}(180^\circ - 150^\circ) = 15^\circ$ .

Using a similar argument, we can show that $\angle QTR = 15^\circ$ .

This means that $\angle STR = \angle PTQ - \angle PTS - \angle QTR = 60^\circ - 15^\circ - 15^\circ = 30^\circ$ .

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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.