Find the perimeter of a triangle, one of the angles of which is equal to $\alpha$, and the radii of the inscribed and circumscribed circles are $r$ and $R$.

#
Let the angle $A$ of triangle $A B C$ be equal to $\alpha, O$ - the center of the inscribed circle, $K$ - the point of tangency of the inscribed circle with side $A B, M$ - with side $A C$. Then

$$
B C=2 R \sin \alpha, A K=O K \operatorname{ctg} \frac{\alpha}{2}=r \operatorname{ctg} \frac{\alpha}{2}
$$

Since $A M=A K$ and $B K+C M=B C$, then

$$
A B+A C+B C=2 A K+2 B C=2 r \operatorname{ctg} \frac{\alpha}{2}+4 R \sin \alpha
$$

## Answer

$2\left(r \operatorname{ctg} \frac{\alpha}{2}+2 R \sin \alpha\right)$