2. Each vertex of the right-angled triangle $A B C$ with legs $\overline{A B}=4 \sqrt{3}$ and $\overline{A C}=4$ is the center of a circle such that the three circles touch each other and no circle lies inside another circle. Calculate the area of the part of triangle $A B C$ that lies outside the three circles.
Solution. Since $\overline{B C}=\sqrt{4^{2}+(4 \sqrt{3})^{2}}=8=2 \overline{A C}$, the angles $\alpha, \beta$ and $\gamma$ at the vertices $A, B$ and $C$ will be $90^{\circ}, 30^{\circ}$ and $60^{\circ}$ respectively. Using the notations from the diagram, we will set

$$
x=\overline{A B_{1}}=\overline{A C_{1}}, y=\overline{B A_{1}}=\overline{B C_{1}} \quad, x=\overline{A C_{1}}=\overline{C B_{1}}
$$

According to this, we get the system of equations:

$$
x+y=4 \sqrt{3} ; y+z=8 ; z+x=4
$$

Its solution is

$x=2 \sqrt{3}-2, y=2 \sqrt{3}+2 \quad z=6-2 \sqrt{3}$ Therefore, the sum of the areas of the sectors with vertices at $A, B$ and $C$ will be:

$$
\begin{aligned}
\frac{x^{2} \pi}{4}+\frac{y^{2} \pi}{12}+\frac{z^{2} \pi}{6} & =\frac{\pi}{12}\left(3 x^{2}+y^{2}+2 z^{2}\right) \\
& =\frac{8 \pi}{3}(5-2 \sqrt{3})
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_ddd0b825d3ca510ab0f2g-2.jpg?height=457&width=644&top_left_y=262&top_left_x=844)

The required area $P$ will be

$$
P=8 \sqrt{3}-\frac{8 \pi}{3}(5-2 \sqrt{3})
$$