12.002. The angle at the base of an isosceles triangle is equal to $\alpha$. Find the ratio of the radii of the inscribed and circumscribed circles.
## Solution.

Given $A B=B C, \angle B A C=\angle B C A=\alpha$. Let $A C=x$, then from $\triangle A E O, \angle A E O=90^{\circ}$ we have $O E=r=\frac{x}{2} \operatorname{tg} \frac{\alpha}{2}$ (Fig. 12.5), where $r$ is the radius of the inscribed circle. From $\triangle A E B \quad \angle A E B=90^{\circ}$ we have that $A B=\frac{A E}{\cos \alpha}=\frac{x}{2 \cos \alpha}$. Consider $\triangle B K O_{1}, \angle B K O_{1}=90^{\circ}$, $B O_{1}=R=\frac{B K}{\cos K B O_{1}} ; B K=\frac{A B}{2}=\frac{x}{4 \cos \alpha}\left(O_{1}\right.$ - the center of the circumscribed circle). $\angle A B E=90^{\circ}-\alpha$, then $\quad B O_{1}=R=\frac{x}{4 \cos \alpha \cos \left(90^{\circ}-\alpha\right)}=$ $=\frac{x}{4 \cos \alpha \sin \alpha}$. Therefore, $\frac{r}{R}=\frac{x \operatorname{tg} \frac{\alpha}{2} \cdot 2 \cdot \sin 2 \alpha}{2 \cdot x}=\operatorname{tg} \frac{\alpha}{2} \cdot \sin 2 \alpha$. Answer: $\operatorname{tg} \frac{\alpha}{2} \cdot \sin 2 \alpha$.