, 2021

Solution:

Drop an altitude to point $D$ on $\overline{AB}$ from $C$ and let $x=AD$. Solving for $x$, we find
$$\begin{aligned} 12-x^{2}=18-(3+\sqrt{3}-x)^{2} &\Rightarrow 12=18-9-6\sqrt{3}-3+2(3+\sqrt{3})x-x^{2} \\ &\Rightarrow 6+6\sqrt{3}=(6+2\sqrt{3})x \\ &\Rightarrow x=\sqrt{3} \end{aligned}$$
So $AC=2AD$, from which we have $\angle CAD=60^{\circ}$. Also, $CD=AD\sqrt{3}=3$ and $BD=AB-AD=3+\sqrt{3}-\sqrt{3}=3$, so $\angle DBC=45^{\circ}$. Then, if $E$ is the intersection of the circle with $\overline{AC}$, $F$ is the intersection of the circle with $\overline{BC}$, and $O$ is the midpoint of $\overline{AB}$, $\angle AOE=60^{\circ}$ and $\angle BOF=90^{\circ}$. Then, letting $r=\frac{AB}{2}$, we get that the area of the part of $\triangle ABC$ that lies inside the semicircle is
$$\begin{aligned} \frac{1}{2}\pi r^{2}-\left(\frac{1}{4}+\frac{1}{6}\right)\pi r^{2}+\frac{1}{2}r^{2}\sin 60^{\circ}+\frac{1}{2}r^{2}\sin 90^{\circ} &=\frac{1}{12}\pi r^{2}+\frac{\sqrt{3}}{4}r^{2}+\frac{1}{2}r^{2} \\ &=\frac{1}{12}(\pi+3\sqrt{3}+6)r^{2} \end{aligned}$$
So the desired area is
$$$$
$$\begin{aligned} 3r-\frac{1}{12}(\pi+3\sqrt{3}+6)r^{2} &=\frac{9+3\sqrt{3}}{2}-\frac{1}{8}(\pi+3\sqrt{3}+6)(2+\sqrt{3}) \\ &=\frac{1}{2}(9+3\sqrt{3})-\frac{1}{8}(2+\sqrt{3})\pi-\frac{1}{8}(21+12\sqrt{3}) \\ &=\frac{15-(2+\sqrt{3})\pi}{8} . \end{aligned}$$