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Austrian Mathematical Olympiad National Competition (Final Round, part 1)

Geometry Difficulty 5.4 AIME, harder Prove it Austria

Let $ABCDE$ be a regular pentagon with center $M$ . A point $P \neq M$ is chosen on the line segment $MD$ . The circumcircle of $ABP$ intersects the line segment $AE$ in $A$ and $Q$ and the line through $P$ perpendicular to $CD$ in $P$ and $R$ .
*Prove that $AR$ and $QR$ are of the same length.*

Solution

Let $S$ denote the common point of $RP$ and $AE$ , see Figure 1. Since we are given a regular pentagon, the angles in triangle $ABE$ are well known as $\angle BAE = 108^\circ$ and $\angle ABE = \angle AEB = 36^\circ$ . Since $BE$ and $CD$ are parallel, $RP$ is perpendicular to $BE$ , and we therefore have $\angle ASP = 126^\circ$ and $\angle QSP = 54^\circ = \angle ASR$ . From this,
$\angle SPA = 54^\circ - \angle SAP = \angle PAB - 54^\circ = \angle PBA - 54^\circ = 126^\circ - \angle AQP = 126^\circ - \angle SQP = \angle SPQ$
follows, since $ABPQ$ is inscribed. We therefore see that $SP$ (or $RP$ ) bisects the angle $\angle APQ$ , which implies that $AR$ and $QR$ must be of equal length, as claimed.

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