Given an isosceles triangle $A B C$, where $A B=B C, \angle A B C=120^{\circ}$. The distance from the midpoint of side $A B$ to the base $A C$ is $a$. Find the area of the circle inscribed in triangle $A B C$.
Find the sides of triangle $ABC$ and the distance from vertex $B$ to the nearest point of tangency with the inscribed circle.

## Solution

Let $O$ be the center of the circle inscribed in triangle $ABC$, $M$ be the midpoint of $AB$, and $K$ be the projection of point $M$ onto the base $AC$. Then

$$
A M=2 M K=2 a, A B=2 A M=4 a \text {. }
$$

If $P$ is the point of tangency of the inscribed circle with side $AB$, and $Q$ is with side $AC$, then

$$
\begin{gathered}
A Q=A B \cos 30^{\circ}=2 a \sqrt{3}, A P=A Q=2 a \sqrt{3} \\
B P=A B-A P=4 a-2 a \sqrt{3}=2 a(2-\sqrt{3}) \\
O P=r=B P \operatorname{tg} 60^{\circ}=2 a \sqrt{3}(2-\sqrt{3})
\end{gathered}
$$

Then the area of the inscribed circle is

$$
\pi r^{2}=12 \pi a^{2}(2-\sqrt{3})^{2}=12 \pi a^{2}(7-4 \sqrt{3})
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-31.jpg?height=323&width=1214&top_left_y=-1&top_left_x=422)

## Answer

$12 \pi a^{2}(7-4 \sqrt{3})$