HongKong 2022-23 IMO Selection Tests

Answer: $10\sqrt{2}$

We use $[XYZ]$ to denote the area of $XYZ$. Suppose $[GDB]=b$ and $[GDC]=c$. We now make use of the side length ratios to get the following:

Using $AG:GD=4:1$, we have $[AGB]=4b$ and $[AGC]=4c$. Using $BG:GE=3:2$, we have $[AGE]=\frac{2}{3}(4b)$ and $[CGE]=\frac{2}{3}(b+c)$. As $[AGC]=[AGE]+[CGE]$, we have $4c=\frac{2}{3}(4b)+\frac{2}{3}(b+c)$, which gives $b=c$. We have $AF:FB=[AGC]:[BGC]=4c:2c=2:1$, so $[FGB]=\frac{1}{3}[AGB]=\frac{4}{3}b$. * It follows that $FG:GC=[FGB]:[CGB]=\frac{4}{3}b:2b=2:3$.

Since $CF=5$, the last point above gives $GC=3=GB$. Hence $△ABC$ must be isosceles with $AB=AC$ and hence $AD\perp BC$ (as $b=c$). It then follows that $$BD=\sqrt{3^2-1^2}=2\sqrt{2}$$ and so the area of $△ABC$ is $\frac{1}{2}\cdot BC\cdot AD=BD\cdot AD=10\sqrt{2}$.