12.365. The radius of the base of a cone is $R$, and the angle between the generatrix and the base plane is $\alpha$. A sphere is inscribed in this cone. Through a point $P$ lying on the circle of contact between the sphere and the conical surface, a tangent line to this circle is drawn, and through this line a plane is drawn parallel to the generatrix of the cone passing through the point diametrically opposite to point $P$. Find the area of the section of the sphere by this plane.
Solution.

Consider the axial section of the given set of bodies passing through point $P$, perpendicular to the tangent line from the problem statement (Fig. 12.225). $P D$ is the diameter of the section of the sphere of the required area.

$$
\angle B A C=\angle B C A=\alpha .
$$

Since $P D \| B A$, then $\angle C P D=\angle C B A=180^{\circ}-2 \alpha$.

$$
\begin{aligned}
& O P \perp B C \text { and } \angle O P D=90^{\circ}-\angle C P D=90^{\circ}-\left(180^{\circ}-2 \alpha\right)=2 \alpha-90^{\circ} \\
& \angle O C O_{2}=\frac{1}{2} \angle B C A=\frac{\alpha}{2}, C O_{2}=R, \text { from } \triangle O O_{2} C\left(\angle O O_{2} C=90^{\circ}\right): \\
& O O_{2}=R \operatorname{tg} \frac{\alpha}{2} . O P=O O_{2}=R \operatorname{tg} \frac{\alpha}{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1126.jpg?height=579&width=504&top_left_y=66&top_left_x=93)

Fig. 12.225

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1126.jpg?height=577&width=634&top_left_y=67&top_left_x=628)

Fig. 12.226

$O_{1}$ is the midpoint. $P D$ is the center of the section of the required area.

In $\triangle O O_{1} P\left(\angle O O_{1} P=90^{\circ}\right):$

$O_{1} P=O P \cos \angle O P D=R \operatorname{tg} \frac{\alpha}{2} \cos \left(2 \alpha-90^{\circ}\right)=R \operatorname{tg} \frac{\alpha}{2} \sin 2 \alpha$.

The area of the section $S=\pi O_{1} P^{2}=\pi R^{2} \operatorname{tg}^{2} \frac{\alpha}{2} \sin ^{2} 2 \alpha$.

Answer: $\pi R^{2} \operatorname{tg}^{2} \frac{\alpha}{2} \sin ^{2} 2 \alpha$.