12.141. The height of an isosceles triangle is $h$ and forms an angle $\alpha\left(\alpha \leq \frac{\pi}{6}\right)$ with the lateral side. Find the distance between the centers of the inscribed and circumscribed circles of the triangle.
## Solution.

Let $B D$ be the height of $\triangle A B C, A B=B C, B D=h, \angle A B D=\alpha, O_{1}$ and $O_{2}$ be the centers of the circumcircle and incircle of triangle $A B C$, respectively.

Since $\angle A B C=2 \alpha \leq \frac{\pi}{3}$, the points $O_{1}$ and $O_{2}$ are located on the segment $B D$ such that $O_{1}$ lies between points $B$ and $O_{2}$ (Fig. 12.14).

In $\triangle A D B\left(\angle A D B=\frac{\pi}{2}\right)$:

$A D=B D \operatorname{tg} \angle A B D=h \mathrm{tg} \alpha$.

$\angle B A D=\frac{\pi}{2}-\alpha, \angle D A O_{2}=\frac{1}{2} \angle B A D=\frac{\pi}{4}-\frac{\alpha}{2}$.

In $\triangle A D O_{2}\left(\angle A D O_{2}=\frac{\pi}{2}\right): A O_{2}=\frac{A D}{\cos \angle D A O_{2}}=\frac{h \operatorname{tg} \alpha}{\cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)}$.

Since $\angle A O_{1} O_{2}$ is the exterior angle of the isosceles triangle $A O_{1} B\left(O_{1} A=O_{1} B\right)$, then $\angle A O_{1} O_{2}=2 \angle A B D=2 \alpha$.
$\angle O_{1} A O_{2}=\angle B A D-\left(\angle B A O_{1}+\angle D A O_{2}\right)=\frac{\pi}{2}-\alpha-\left(\alpha+\frac{\pi}{4}-\frac{\alpha}{2}\right)=\frac{\pi}{4}-\frac{3 \alpha}{2}$. In $\triangle O_{1} A O_{2}: \frac{O_{1} O_{2}}{\sin \angle O_{1} A O_{2}}=\frac{A O_{2}}{\sin \angle A O_{1} O_{2}} \Rightarrow O_{1} O_{2}=A O_{2} \cdot \frac{\sin \angle O_{1} A O_{2}}{\sin \angle O_{2} O_{1} A}=$ $=\frac{h \operatorname{tg} \alpha}{\cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)} \cdot \frac{\sin \left(\frac{\pi}{4}-\frac{3 \alpha}{2}\right)}{\sin 2 \alpha}=\frac{h \cos \left(\frac{\pi}{4}+\frac{3 \alpha}{2}\right)}{2 \cos ^{2} \alpha \cos \left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$. Answer: $\frac{h \cos \left(\frac{\pi}{4}+\frac{3 \alpha}{2}\right)}{2 \cos ^{2} \alpha \cos \left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$.