HongKong -23 IMO Selection Tests, 2022

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 $\triangle ABC$ 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 $\triangle ABC$ is $\frac{1}{2} \cdot BC \cdot AD = BD \cdot AD = 10\sqrt{2}$.