imc

Let the centers of the smaller and larger circles be $O_1$ and $O_2$ , respectively. Let their tangent points to $△ABC$ be $D$ and $E$, respectively. We can then draw the following diagram: We see that $△AD{O}_1\sim △AE{O}_2\sim △AFC$. Using the first pair of similar triangles, we write the proportion: $\frac{A{O}_1}{A{O}_2}=\frac{D{O}_1}{E{O}_2}⟹\frac{A{O}_1}{A{O}_1+3}=\frac{1}{2}⟹A{O}_1=3$ By the Pythagorean Theorem, we have $AD=\sqrt{3^2-1^2}=\sqrt{8}$. Now using $△AD{O}_1\sim △AFC$, $\frac{AD}{AF}=\frac{D{O}_1}{FC}⟹\frac{2\sqrt{2}}{8}=\frac{1}{FC}⟹FC=2\sqrt{2}$ Hence, the area of the triangle is $$\frac{1}{2}\cdot AF\cdot BC=\frac{1}{2}\cdot AF\cdot (2\cdot CF)=AF\cdot CF=8\left(2\sqrt{2}\right)=\boxed{16\sqrt{2}}$$