4. Given an equilateral triangle $ABC$ with side $12 \, \text{cm}$. Two circles with the same radius equal to $\sqrt{3} \, \text{cm}$ touch the side $BC$ from the outside of the triangle, touch each other, and one of the circles touches the line $AB$. Find the area of the triangle whose vertices are the centers of the circles and the midpoint of side $AB$.
Solution. Let $S$ and $T$ be the centers of the circles, and let $E$ be the midpoint of side $AB$ (see diagram). Then, we have $\overline{ST}=2 \sqrt{3}$, $x=\overline{EB} \sin 60^{\circ}=3 \sqrt{3}$, $h=x+\sqrt{3}=4 \sqrt{3}$.

According to this, the area $P$ of triangle $EST$ will be:

$$
P=\frac{2 \sqrt{3} \cdot 4 \sqrt{3}}{2}=12
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_a9dc3c06af03b935fe82g-2.jpg?height=356&width=493&top_left_y=1401&top_left_x=989)

## II year