Maths Olympiad Prep

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Problem 580

Geometry Difficulty 2.7 Find the answer CEMC Cayley

In a regular pentagon PQRSTPQRST, what is the measure of ∠PRS\angle PRS?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Join PP to RR. Since PQRSTPQRST is a regular pentagon, then ∠PQR=∠QRS=108∘\angle PQR = \angle QRS = 108^{\circ}. Since PQ=QRPQ = QR, then △PQR\triangle PQR is isosceles with ∠QPR=∠QRP\angle QPR = \angle QRP. Since ∠PQR=108∘\angle PQR = 108^{\circ}, then ∠PQR+∠QPR+∠QRP=180∘\angle PQR + \angle QPR + \angle QRP = 180^{\circ}, 108∘+2∠QRP=180∘108^{\circ} + 2\angle QRP = 180^{\circ}, 2∠QRP=72∘2\angle QRP = 72^{\circ}, ∠QRP=36∘\angle QRP = 36^{\circ}. Since ∠QRS=108∘\angle QRS = 108^{\circ}, then ∠PRS=∠QRS−∠QRP=108∘−36∘=72∘\angle PRS = \angle QRS - \angle QRP = 108^{\circ} - 36^{\circ} = 72^{\circ}.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.