12.244. A sphere is inscribed in a cone. The radius of the circle of contact between the sphere and the lateral surface of the cone is $r$. A line passing through the center of the sphere and an arbitrary point on the circumference of the base of the cone makes an angle $\alpha$ with the height of the cone. Find the volume of the cone.
## Solution.

Let $S$ be the vertex of the cone, $O$ the center of its base (the point of tangency of the inscribed sphere with the base plane), $C$ the center of the sphere, $D$ the center of the circle of tangency of the sphere's surface and the lateral surface of the cone, $AD$ the radius of this sphere, $AD = r$, $SB$ the generatrix of the cone, on which point $A$ lies, $\angle OCB = \alpha$ (Fig. 12.111), the radius of the sphere $CO = CA = R$, $A$ the point of tangency of the line $SB$ and the sphere's surface. Then $CA \perp SB$, $\triangle COB = \triangle CAB$ by the leg and hypotenuse.

Thus, $\angle ACB = \angle OCB = \alpha$, $\angle DCA = 180^\circ - 2\alpha$, $\angle OBS = 2 \angle OBC = 2(90^\circ - \alpha) = 180^\circ - 2\alpha$.

In $\triangle CDA$ ($\angle CDA = 90^\circ$): $R = \frac{r}{\sin(180^\circ - 2\alpha)} = \frac{r}{\sin 2\alpha}$,

In $\triangle COB$ ($\angle COB = 90^\circ$): $OB = CO \tan \angle OCB = R \tan \alpha$.

From $\triangle SOB$ ($\angle SOB = 90^\circ$):

$SO = OB \tan \angle OBS = R \tan \alpha \tan(180^\circ - 2\alpha) = -R \tan \alpha \tan 2\alpha$.

Volume of the cone:

\[
\begin{aligned}
& V = \frac{1}{3} \pi \cdot OB^2 \cdot SO = -\frac{1}{3} \pi R^3 \tan^3 \alpha \tan 2\alpha = -\frac{\pi}{3} \cdot \frac{r^3}{\sin^3 2\alpha} \cdot \tan^3 \alpha \tan 2\alpha = \\
& = -\frac{\pi r^3 \sin^3 \alpha \tan 2\alpha}{24 \sin^3 \alpha \cos^3 \alpha \cos^3 \alpha} = -\frac{\pi r^3 \tan 2\alpha}{24 \cos^6 \alpha} \\
& \text{Answer: } -\frac{\pi r^3 \tan 2\alpha}{24 \cos^6 \alpha}
\end{aligned}
\]