Competition problem — Geometry — Maths Olympiad Prep

Library / /5 of 14

Geometry Difficulty 5.5 AIME, harder Prove it 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 $\angle AGO = 90^\circ$ .

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 $\angle AGO = \angle AMS = 90^\circ$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.