3.73. The height of an isosceles triangle is equal to $h$ and forms an angle $\alpha\left(\alpha \leq \frac{n}{6}\right)$ with the lateral side. Find the distance between the centers of the inscribed and circumscribed circles of the triangle.
3.73. By the condition, $A B=B C$, $B B_{1} \perp A C, B B_{1}=h, \angle A B B_{1}=\alpha$ ( $\alpha \leq \frac{\pi}{6}$ ), $O$ - the center of the circumcircle of $\triangle A B C, O_{1}$ the center of the incircle of $\triangle A B C$ (Fig. 3.74). From $\triangle A B_{1} B$ we find $A B_{1}=h \operatorname{tg} \alpha$; then $A C=$ $=2 h \operatorname{tg} \alpha=2 R \sin 2 \alpha$, where $R$ is the radius of the circumcircle, from which $R=\frac{h \operatorname{tg} \alpha}{\sin 2 \alpha}=\frac{h}{2 \cos ^{2} \alpha}$. Further,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-546.jpg?height=525&width=525&top_left_y=107&top_left_x=628)

Fig. 3.74 from $\triangle A B_{1} O_{1}$ we get

$$
O_{1} B_{1}=r=A B_{1} \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)=h \operatorname{tg} \alpha \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) .
$$

Therefore,

$$
\begin{aligned}
& O O_{1}=h-O B-O_{1} B_{1}=h-\frac{h}{2 \cos ^{2} \alpha}-h \operatorname{tg} \alpha \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)= \\
& =h\left(1-\frac{1}{2 \cos ^{2} \alpha}-\frac{\sin \alpha\left(1-\cos \left(\frac{\pi}{2}-\alpha\right)\right)}{\cos \alpha \sin \left(\frac{\pi}{2}-\alpha\right)}\right)= \\
& =h\left(1-\frac{1}{2 \cos ^{2} \alpha}-\frac{\sin \alpha(1-\sin \alpha)}{\cos ^{2} \alpha}\right)= \\
& =h\left(\frac{2 \cos ^{2} \alpha-1-2 \sin \alpha+2 \sin ^{2} \alpha}{2 \cos ^{2} \alpha}\right)=h \frac{1-2 \sin \alpha}{2 \cos ^{2} \alpha}= \\
& =h \frac{\sin \frac{\pi}{6}-\sin \alpha}{\cos ^{2} \alpha}=\frac{2 h \sin \left(\frac{\pi}{12}-\frac{\alpha}{2}\right) \cos \left(\frac{\pi}{12}+\frac{\alpha}{2}\right)}{\cos ^{2} \alpha} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-547.jpg?height=292&width=436&top_left_y=125&top_left_x=101)

Fig. 3.75