A circle of radius 12 has an angle of $120^{\circ}$ inscribed in it such that the center of the circle lies on the bisector of the angle. Find the area of the part of the circle that is outside the angle.

#
Triangle $A B C$ is isosceles.

## Solution

Let angle $B A C$, equal to $120^{\circ}$, be inscribed in a circle with center $O$, and the bisector of this angle passes through point $O$.

When reflecting over the diameter $A A_{1}$, the circle maps to itself, and the rays $A B$ and $A C$ map to each other. Therefore, points $B$ and $C$ also map to each other. Hence, triangle $A B C$ is isosceles. Let $S$ be the desired area, $S_{1}$ be the area of sector $VOC$ containing point $A$, $S_{2}$ be the area of rhombus $O B A C$, which consists of two equilateral triangles $O A B$ and $O A C$, and $R=12$ be the radius of the circle. Then,

$$
S=S_{1}-S_{2}=\frac{1}{3} \pi R^{2}-R^{2} \cdot \sin 120^{\circ}=\frac{1}{3} \pi \cdot 144-144 \cdot \frac{\sqrt{3}}{2}=48 \pi-72 \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-44.jpg?height=337&width=360&top_left_y=0&top_left_x=855)

## Answer

$48 \pi-72 \sqrt{3}$.