12.317. The generatrix of a truncated cone circumscribed around a sphere is equal to $a$, the angle between the generatrix and the plane of the base is $\alpha$. Find the volume of the cone whose base is the circle of contact between the spherical surface and the lateral surface of the truncated cone, and whose vertex coincides with the center of the larger base of the truncated cone.
Solution.

Consider the axial section of the given set of bodies (Fig. 12.178): an isosceles trapezoid \(A A_{1} B_{1} B \left(A A_{1}=a, \angle A_{1} A B=\alpha\right)\), a circle inscribed in it with center \(E\), touching \(A A_{1}, B B_{1}, A B\) and \(A_{1} B_{1}\) at points \(C, D, O\) and \(O_{1}\) respectively, and \(\triangle C O D\) - the axial section of the cone of the desired volume.

The midpoint \(O_{2}\) of segment \(C D\) is the center of the base of the cone of the desired volume \(V\). Then \(V=\frac{1}{3} \pi \cdot O_{2} C^{2} \cdot \mathrm{OO}_{2}\).

Let the radius of the sphere be \(R: E O=E C=R\).

In quadrilateral \(A C E O \angle C A O=\alpha, \angle A C E=\angle A O E=90^{\circ} \Rightarrow \angle C E O=180^{\circ}-\alpha\) and \(\angle C E O_{2}=\alpha\).

In \(\triangle C O_{2} E\left(\angle C O_{2} E=90^{\circ}\right): C O_{2}=R \sin \alpha ; E O_{2}=R \cos \alpha\).

\[
\begin{aligned}
& O O_{2}=O E+E O_{2}=R+R \cos \alpha=2 R \cos ^{2} \frac{\alpha}{2} \Rightarrow \\
& \Rightarrow V=\frac{1}{3} \pi \cdot R^{2} \sin ^{2} \alpha \cdot 2 R \cos ^{2} \frac{\alpha}{2}=\frac{2}{3} \pi R^{3} \sin ^{2} \alpha \cos ^{2} \frac{\alpha}{2} \\
& A_{1} K-\text { the height of the trapezoid } A A_{1} B_{1} B . \text { Then } A_{l} K=2 R
\end{aligned}
\]

From \(\triangle A K A_{1}\left(\angle A K A_{1}=90^{\circ}\right): A_{1} K=A A_{1} \sin \angle A=a \sin \alpha \Rightarrow R=\frac{1}{2} a \sin \alpha\) and \(V=\frac{2}{3} \pi \cdot \frac{1}{8} a^{3} \sin ^{3} \alpha \cdot \sin ^{2} \alpha \cos ^{2} \frac{\alpha}{2}=\frac{\pi a^{3}}{12} \sin ^{5} \alpha \cos ^{2} \frac{\alpha}{2}\).

Answer: \(\frac{\pi a^{3}}{12} \sin ^{5} \alpha \cos ^{2} \frac{\alpha}{2}\).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1073.jpg?height=506&width=464&top_left_y=173&top_left_x=127)

Fig. 12.179