HMMT-Feb 2022 guts, Problem 25

Let $A B C$ be an acute scalene triangle with circumcenter $O$ and centroid $G$. Given that $A G O$ is a right triangle, $A O=9$, and $B C=15$, let $S$ be the sum of all possible values for the area of triangle $A G O$. Compute $S^{2}$.

Solution: Solution: Note that we know that $O, H$, and $G$ are collinear and that $H G=2 O G$. Thus, let $O G=x$ and $H G=2 x$. We also have $\sin A=\frac{B C}{2 R}=\frac{5}{6}$, so $\cos A=\frac{\sqrt{11}}{6}$. Then, if $A G \perp O G$, then we have $x^{2}+A G^{2}=O G^{2}+A G^{2}=A O^{2}=81$ and $H G^{2}+A G^{2}=4 x^{2}+A G^{2}=A H^{2}=(2 R \cos A)^{2}=99$. Solving gives us $x=\sqrt{6}$ and $A G=5 \sqrt{3}$. Thus, the area of $A G O$ in this case is $\frac{1}{2} \cdot \sqrt{6} \cdot 5 \sqrt{3}=\frac{5 \sqrt{3}}{2}$. If we have $A O \perp O G$, then we have $99=A H^{2}=A O^{2}+O H^{2}=81+9 x^{2}$. This gives us $x=\sqrt{2}$. In this case, we have the area of $A G O$ is $\frac{1}{2} \cdot \sqrt{2} \cdot 9=\frac{9 \sqrt{2}}{2}$. Adding up the two areas gives us $S=12 \sqrt{2}$. Squaring gives $S^{2}=288$. $\fbox{288}$.

Answer: 288