[ Auxiliary circle ] $[$ Sine theorem $\quad]$

On the side $A B$ of triangle $A B C$, an equilateral triangle is constructed outward. Find the distance between its center and vertex $C$, if $A B=c$ and $\angle C=120^{\circ}$.
## Solution

Point $C$ lies on the circumcircle of the constructed equilateral triangle ($60^{\circ}+120^{\circ}=$ $\left.180^{\circ}\right)$, so the desired distance is equal to the radius of this circle, i.e.,

$$
R=\frac{c}{2 \sin 60^{\circ}}=\frac{c}{\sqrt{3}}=\frac{c \sqrt{3}}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-39.jpg?height=328&width=374&top_left_y=442&top_left_x=842)

## Answer

$\frac{c \sqrt{3}}{3}$.