10.399 The side of a regular triangle is equal to a. A circle with a radius of a/3 is described from its center. Determine the area of the part of the triangle that lies outside the circle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-242.jpg?height=316&width=319&top_left_y=1032&top_left_x=113)

Let $O$ be the center of the given equilateral $\triangle ABC$, $M$ and $N$ be the points of intersection of the circle with center $O$ and radius $a / 3$ with sides $AB$ and $AC$, and $BD$ be the height of $\triangle ABC$. Then $\cos \angle NOD = \frac{OD}{ON} = \frac{\sqrt{3}}{2}$

$$
\left(OD = \frac{a}{2 \sqrt{3}}\right) \Rightarrow \angle NOD = 30^{\circ}, ON \parallel AB
$$

Similarly, $OM \parallel AC \Rightarrow AMON$ is a rhombus and $S_{AMON} = \frac{a^2}{9} \sin 60^{\circ}$,

$$
S_{ONPM} = \frac{\pi}{6} \cdot ON^2 = \frac{\pi a^2}{54} \Rightarrow S_{AMPN} = \frac{a^2 \sqrt{3}}{18} - \frac{\pi a^2}{54} = \frac{a^2 (3 \sqrt{3} - \pi)}{54}
$$

The area of the part of $\triangle ABC$ lying outside the circle $O$ is

$$
3 \cdot S_{AMPN} = \frac{a^2 (3 \sqrt{3} - \pi)}{18}
$$

Answer: $\frac{a^2 (3 \sqrt{3} - \pi)}{18}$.