Singapore Mathematical Olympiad (SMO)

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