12.423 The side of the base of a regular triangular pyramid is equal to $a$, and the plane angle at the vertex is $\alpha$. A sphere is inscribed in the pyramid. Find the area of the section of this sphere by a plane passing through the center of the base and perpendicular to its lateral edge.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-300.jpg?height=434&width=421&top_left_y=1264&top_left_x=461)

Let $F P$ be the height of the given regular pyramid $F A B C, O$ be the center of the inscribed sphere, and $K N M$ be the section of the pyramid by a plane passing through $P$ and perpendicular to the edge $F A$. Then $M N \perp F A \Rightarrow M N \perp A P$ (Theorem 12). $F P K \perp K N M$ (Theorem 15) $\Rightarrow$ the perpendicular $O L$, dropped from $O$ to the plane $K N M$, lies in the plane $F P K$. Moreover, $M N \perp K P$ (Theorem 12), since $M N \perp F P$. Therefore, $L$ is the center of the desired circle, which is the section of the inscribed sphere by the plane $K N M, L P$ is its radius. $O L \| F A \Rightarrow \frac{L P}{K P}=\frac{O P}{F P}$.

Let $D$ be the midpoint of $B C$. Then the radius $O E$ of the inscribed sphere lies in the plane $A F D$, since $A F D \perp F B C$. Hence, $O D$ is the bisector

$$
\triangle F D P \Rightarrow \frac{O P}{O F+O P}=\frac{-P D}{F D+P D} (\text{Theorem } 4) \Rightarrow L P=K P \cdot \frac{O P}{F P}=\frac{K P \cdot F D}{F D+P D}
$$

But $K P^{2}=A K \cdot K F \quad$ (Theorem 5),

$$
A K=A M \cdot \cos \angle F A C=\frac{2}{3} a \cdot \sin \frac{\alpha}{2}, K F=F A-A K, F A=F C=\frac{\alpha}{2 \sin \frac{\alpha}{2}}
$$

$$
\begin{gathered}
F D=\frac{a}{2} \cdot \operatorname{ctg} \frac{\alpha}{2} ; P D=\frac{a}{2 \sqrt{3}} \Rightarrow K P^{2}=\frac{a^{2}}{9}\left(3-4 \sin ^{2} \frac{\alpha}{2}\right) \Rightarrow \\
\Rightarrow L P^{2}=\frac{K P^{2} \cdot P D^{2}}{(P D+F D)^{2}}=\frac{a^{2}\left(3-4 \sin ^{2} \frac{\alpha}{2}\right)}{9\left(1+\sqrt{3} \operatorname{ctg} \frac{\alpha}{2}\right)^{2}}=\frac{a^{2}\left(3 \cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}\right)}{9\left(\sin \frac{\alpha}{2}+\sqrt{3} \cos \frac{\alpha}{2}\right)^{2}} \cdot \sin ^{2} \frac{\alpha}{2}= \\
=\frac{a^{2}\left(\sqrt{3} \cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)}{9\left(\sqrt{3} \cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)} \cdot \sin ^{2} \frac{\alpha}{2}=\frac{a^{2}}{9} \cdot \frac{\sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right)}{\sin \left(\frac{\pi}{3}+\frac{\alpha}{2}\right)} \cdot \sin ^{2} \frac{\alpha}{2}
\end{gathered}
$$

Answer: $\pi \cdot L \dot{P}^{2}=\frac{\pi a^{2}}{9} \cdot \sin ^{2} \frac{\alpha}{2} \cdot \frac{\sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right)}{\sin \left(\frac{\pi}{3}+\frac{\alpha}{2}\right)}$.