12.318. Two cones with a common base are inscribed in a sphere of radius $R$; the vertices of the cones coincide with the opposite ends of the sphere's diameter. The spherical segment containing the smaller cone has an arc in the axial section equal to $\alpha^{\circ}$. Find the distance between the centers of the spheres inscribed in these cones.
## Solution.

The given set of bodies has an axis of symmetry; let's consider its axial section (Fig. 12.179): $O$ - the center of the given sphere, $\triangle A B C$ - the axial section of the smaller cone, $\triangle A D C$ - the axial section of the larger cone, $E$ - the center of their common base, $O_{1}$ and $O_{2}$ - the centers of the spheres inscribed in them, $\angle A O C=\alpha^{\circ}$.

From $\triangle A E O\left(\angle A E O=90^{\circ}\right): A E=O A \sin \angle A O E=R \sin \frac{\alpha}{2}$.

$\angle B C A$ is an inscribed angle and subtends an arc equal to $\frac{\alpha}{2}$. Thus, $\angle B C A=\frac{\alpha}{4}, \angle O_{1} C E=\frac{1}{2} \angle B C A=\frac{\alpha}{8}$.

From $\triangle O_{1} E C\left(\angle O_{1} E C=90^{\circ}\right): O_{1} E=E C \operatorname{tg} \angle O_{1} C E=R \sin \frac{\alpha}{2} \operatorname{tg} \frac{\alpha}{8}$.

$\angle B C D$ is an inscribed angle and subtends the diameter.

Therefore, $\angle B C D=90^{\circ}, \angle E C D=\angle B C D-\angle B C A=90^{\circ}-\frac{\alpha}{4}, \angle E C O_{2}=$

$= \frac{1}{2} \angle E C D=45^{\circ}-\frac{\alpha}{8}$. 

From $\triangle O_{2} E C\left(\angle O_{2} E C=90^{\circ}\right):$

$$
O_{2} E=E C \operatorname{tg} \angle E C O_{2}=R \sin \frac{\alpha}{2} \operatorname{tg}\left(45^{\circ}-\frac{\alpha}{8}\right)
$$

The required distance:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1074.jpg?height=682&width=327&top_left_y=79&top_left_x=206)

Fig. 12.180

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1074.jpg?height=680&width=329&top_left_y=76&top_left_x=784)

Fig. $12.180, \mathrm{a}$

$$
\begin{aligned}
& O_{1} O_{2}=O_{1} E+O_{2} E=R \sin \frac{\alpha}{2}\left(\operatorname{tg} \frac{\alpha}{8}+\operatorname{tg}\left(45^{\circ}-\frac{\alpha}{8}\right)\right)=\frac{R \sin \frac{\alpha}{2} \sin 45^{\circ}}{\cos \frac{\alpha}{8} \cos \left(45^{\circ}-\frac{\alpha}{8}\right)}= \\
& =\frac{R \sqrt{2} \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{8} \cos \left(45^{\circ}-\frac{\alpha}{8}\right)} \\
& \text { Answer: } \frac{R \sqrt{2} \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{8} \cos \left(45^{\circ}-\frac{\alpha}{8}\right)}
\end{aligned}
$$