12.184. In an isosceles triangle, the angle at the base is $\alpha$. The height dropped to the base is greater than the radius of the inscribed circle by $m$. Find the radius of the circumscribed circle.
## Solution.

In $\triangle A B C$ (Fig. 12.53) $A B=B C, \angle B A C=\alpha, B D$ is the altitude, $O$ is the center of the inscribed circle.

Then, by the condition, $B O=B D-O D=m . \angle B A O=\angle D A O=\frac{\alpha}{2}, \angle A O B=$ $=\angle O A D+\angle A D O=\frac{\alpha}{2}+90^{\circ}$ as the external angle of triangle $A O B$.

By the Law of Sines, from $\triangle A O B: \frac{A B}{\sin \angle A O B}=\frac{B O}{\sin \angle B A O} \Rightarrow A B=$ $=\frac{B O \sin \angle A O B}{\sin \angle B A O}=\frac{m \sin \left(\frac{\alpha}{2}+90^{\circ}\right)}{\sin \frac{\alpha}{2}}=m \operatorname{ctg} \frac{\alpha}{2}$.

Then the radius of the circle circumscribed around triangle $A B C$,

$$
R=\frac{A B}{2 \sin \angle A C B}=\frac{m \operatorname{ctg} \frac{\alpha}{2}}{2 \sin \alpha}=\frac{m \operatorname{ctg} \frac{\alpha}{2}}{4 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{m}{4 \sin ^{2} \frac{\alpha}{2}}
$$

Answer: $\frac{m}{4 \sin ^{2} \frac{\alpha}{2}}$.