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South African Mathematical Olympiad Third Round

Geometry Difficulty 5.6 AIME, harder Prove it South Africa

Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$ . Let $D$ be a point on segment $BC$ such that $\angle BAD = \angle DAC$ , and let $M$ and $N$ be points on segments $BD$ and $CD$ , respectively, such that $\angle MAD = \angle DAN$ . Let $S, P$ and $Q$ (all different from $A$ ) be the intersections of the rays $AD, AM$ and $AN$ with $\Gamma$ , respectively. Show that the intersection of $SM$ and $QD$ lies on $\Gamma$ .

Solution

Let $X$ be the intersection of $SM$ with the circumcircle. Note that $\angle MXA = \angle SXA = \angle SBA = \angle SBC + \angle CBA$ . Since $AS$ bisects the angle $\angle BAC$ , we have $\angle SBC = \angle SAC = \angle BAS$ . It follows that
$\angle MXA = \angle BAS + \angle CBA = \angle BAD + \angle DBA = 180^\{\circ\} - \angle ADB = 180^\{\circ\} - \angle ADM,$
so $AXMD$ is a cyclic quadrilateral. Consequently, $\angle SXD = \angle MXD = \angle MAD = \angle DAN = \angle SAQ = \angle SXQ$ , which means that $X, D$ and $Q$ lie on a common straight line. Hence $X$ is also the intersection of $SM$ and $QD$ , which completes the proof.

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