12.331. A sphere is inscribed in a cone, and a tangent plane is drawn to the sphere, parallel to the base plane of the cone. In what ratio does this plane divide the lateral surface of the cone, if the angle between the generatrix and the base plane is $\alpha$?
## Solution.

Consider the axial section of the given set of bodies (Fig. 12.192): $\triangle A E C$ - the axial section of the given cone with lateral surface $S_{1}$, $O$ - the center of the given sphere - the center of the circle inscribed in the trapezoid $A B C D$, which is the axial section of the frustum of the cone with lateral surface $S_{2}, \triangle B E D$ - the axial section of the cone with lateral surface $S_{3}$, cut off from the given cone by the tangent plane of the sphere, $L-$ the midpoint of $A C, K$ - the midpoint of $B D, \angle E A C=\alpha$.

The required ratio $\frac{S_{2}}{S_{3}}=\frac{S_{1}-S_{3}}{S_{3}}=\frac{S_{1}}{S_{3}}-1=\left(\frac{L A}{K B}\right)^{2}-1$. $\angle L A O=\frac{1}{2} \angle E A C=\frac{\alpha}{2}, \angle K B O=\frac{1}{2} \angle A B D=90^{\circ}-\frac{\alpha}{2}$.

If $O A=R$, then from $\triangle A L O\left(\angle A L O=90^{\circ}\right): O L=R \operatorname{tg} \frac{\alpha}{2}$.

$$
O K=O L=R \operatorname{tg} \frac{\alpha}{2}
$$

In $\triangle B K O\left(\angle B K O=90^{\circ}\right)$ :

$B K=O K \operatorname{ctg} \angle K B O=R \operatorname{tg}^{2} \frac{\alpha}{2} \Rightarrow \frac{S_{2}}{S_{3}}=\left(\frac{R}{R \operatorname{tg}^{2} \frac{\alpha}{2}}\right)^{2}-1=\frac{\cos ^{4} \frac{\alpha}{2}}{\sin ^{4} \frac{\alpha}{2}}-1=$ $=\frac{\cos ^{4} \frac{\alpha}{4}-\sin ^{4} \frac{\alpha}{4}}{\sin ^{4} \frac{\alpha}{2}}=\frac{\cos \alpha}{\sin ^{4} \frac{\alpha}{2}}$.

Answer: $\frac{\cos \alpha}{\sin ^{4} \frac{\alpha}{2}}$