1. Given a triangle $A B C$ with $\angle A=90^{\circ}, \angle C=60^{\circ}$ and $|A C|=6$. Three circles with centers at $A, B$ and $C$ touch each other (externally) on the sides of this triangle.

Calculate the area of the plane region enclosed by these three circles (the gray area).

![](https://cdn.mathpix.com/cropped/2024_04_17_4d017e2c8f63b16aa043g-1.jpg?height=379&width=309&top_left_y=790&top_left_x=1519)
1. In triangle $A B C$ we immediately recognize a half equilateral triangle. It follows that $|B C|=$ $2|A C|=12$. The Pythagorean theorem then gives: $|A B|=\sqrt{|B C|^{2}-|A C|^{2}}=\sqrt{108}=6 \sqrt{3}$.

Let the points of tangency of the circles be $D, E$ and $F$ (see picture) and the radii of the three circles $r_{A}, r_{B}$ and $r_{C}$. The strategy is to determine the areas of the three circular sectors and subtract them from the area of triangle $A B C$.

We see that $2 r_{A}=\left(r_{A}+r_{C}\right)+\left(r_{A}+r_{B}\right)-\left(r_{B}+r_{C}\right)=|A C|+|A B|-|B C|=$ $6 \sqrt{3}-6$, so $r_{A}=3 \sqrt{3}-3$. It follows that $r_{B}=6 \sqrt{3}-r_{A}=3 \sqrt{3}+3$ and $r_{C}=6-r_{A}=9-3 \sqrt{3}$.

The area of a circle with radius $r$ is $\pi r^{2}$. The area of circular sector $A F E$ is thus $\frac{90}{360} \cdot \pi r_{A}^{2}$, or $\frac{1}{4} \pi(36-18 \sqrt{3})=9 \pi-\frac{9}{2} \sqrt{3} \pi$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2e1c76316c39be9d48a3g-1.jpg?height=439&width=371&top_left_y=677&top_left_x=1505)
For the areas of circular sectors $B D F$ and $C E D$, we find respectively $\frac{30}{360} \pi r_{B}^{2}=$ $3 \pi+\frac{3}{2} \sqrt{3} \pi$ and $\frac{60}{360} \pi r_{C}^{2}=18 \pi-9 \sqrt{3} \pi$.

Since the area of triangle $A B C$ is $\frac{1}{2} \cdot|A B| \cdot|A C|=18 \sqrt{3}$, we find for the gray area the area $18 \sqrt{3}-\left(9 \pi-\frac{9}{2} \sqrt{3} \pi\right)-\left(3 \pi+\frac{3}{2} \sqrt{3} \pi\right)-(18 \pi-9 \sqrt{3} \pi)=$ $18 \sqrt{3}-30 \pi+12 \sqrt{3} \pi$.