In triangle $P Q R$, angle $Q R P$ is $60^{\circ}$. Find the distance between the points of tangency with side $Q R$ of the inscribed circle of radius 2 and the circle of radius 3 that is tangent to the extensions of sides $P Q$ and $P R$.

#
The center of a circle inscribed in an angle lies on the bisector of this angle.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of circles with radii 2 and 3, respectively, and $M$ and $N$ be their points of tangency with side RQ. Then

$$
\begin{gathered}
R M=O_{1} M \operatorname{ctg} \angle M R O_{1}=2 \operatorname{ctg} 30^{\circ}=2 \sqrt{3} \\
R N=O_{2} N \operatorname{ctg} \angle N R O_{2}=3 \operatorname{ctg} 60^{\circ}=\sqrt{3}
\end{gathered}
$$

Therefore,

$$
M N=R M-R N=2 \sqrt{3}-\sqrt{3}=\sqrt{3} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-23.jpg?height=452&width=992&top_left_y=1858&top_left_x=539)

Answer

$\sqrt{3}$.