Track / Stage 5 / 231 of 400 #831 of 1964

Problem 831

AIME late

Geometry Difficulty 5.5 Prove it Singapore Mathematical Olympiad (SMO) · Singapore

In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$ , the extension of $AG$ meets $\omega$ at $M$ ; lines $AB$ and $CM$ intersect at $P$ ; and lines $AC$ and $BM$ intersect at $Q$ . Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A$ , $O$ , $S$ are collinear. Prove that $\angle AGO = 90^\circ$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $AD$ , $BE$ and $CF$ be the medians of the triangle $ABC$ . Since $A$ , $O$ , $S$ are collinear, $AS$ is a diameter of $\omega$ with $AO = OS$ . The triangles $AOB$ and $ASP$ are similar isosceles triangles. Thus $OB \parallel SP$ . Since $O$ is the midpoint of $AS$ , we have $B$ is the midpoint of $AP$ . Similarly, $C$ is the midpoint of $AQ$ . Since $F$ is the midpoint of $AB$ and $C$ is the midpoint of $AQ$ , we have $FC \parallel BQ$ . Thus $GC \parallel BM$ . Since $F$ is the midpoint of $AB$ , $G$ is the midpoint of $AM$ . Therefore, $GO \parallel MS$ . Then $\angle AGO = \angle AMS = 90^\circ$ .

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