A right circular cone has a regular triangular pyramid inscribed in it, the apothem of which is equal to $k$, and the lateral face forms an angle of $\alpha$ with the base plane. A plane is drawn through one of the lateral edges of the pyramid, intersecting the conical surface. Find the area of the section of the cone by this plane, given that this area has the largest of all possible values.
Let $S O$ be the height of a regular pyramid $S A B C$ with base $A B C$. Then $S O$ is the height of the cone, and the lateral edges of the pyramid are the generators of the cone. If $M$ is the midpoint of edge $A B$, then $S M=k$ is the apothem of the pyramid, and $\angle O M S=\alpha$ is the linear angle of the dihedral angle between the lateral face and the base plane. Considering that $O A=2 O M$, from the right triangles $O M S$ and $A O S$ we find that

$$
S O=S M \sin \alpha=k \sin \alpha, M O=S M \cos \alpha=k \cos \alpha
$$

$$
\begin{gathered}
\operatorname{tg} \angle O A S=\frac{S O}{O A}=\frac{k \sin \alpha}{2 k \cos \alpha}=\frac{1}{2} \operatorname{tg} \alpha, \angle A S O=\frac{\pi}{2}-\angle O A S \\
S A=\sqrt{S O^{2}+A O^{2}}=\sqrt{k^{2} \sin ^{2} \alpha+4 k^{2} \cos ^{2} \alpha}=k \sqrt{1+3 \cos ^{2} \alpha}
\end{gathered}
$$

The section of the cone by a plane passing through its generator is an isosceles triangle, the lateral sides of which are equal to $k \sqrt{1+3 \cos ^{2} \alpha}$. If the angle at its vertex is $\phi$, and the area of the triangle is $S(\phi)$, then $S(\phi)=\frac{1}{2} k 2(1+3 \cos 2 \alpha) \sin \phi$. Thus, to find the maximum value of this area, we need to find the maximum value of $\sin \phi$. If the angle at the vertex of the axial section is less than $\frac{\pi}{2}$, then due to the increasing function $\sin \phi$ on the interval $\left(0 ; \frac{\pi}{2}\right)$, the maximum value of $\sin \phi$ is achieved at the maximum value of $\phi$, i.e., if $\angle O A S<\frac{\pi}{4}$, or $\operatorname{arctg}\left(\frac{1}{2} \operatorname{tg} \alpha\right)<\frac{\pi}{4}$. This means that $\operatorname{tg} \alpha<2$. In this case

$$
S_{\max }=S\left(\frac{\pi}{2}\right)=\frac{1}{2} S A 2 \sin \frac{\pi}{2}=\frac{1}{2} k 2(1+3 \cos 2 \alpha) \cdot 1=\frac{1}{2} k 2(1+3 \cos 2 \alpha)
$$

If the angle at the vertex of the axial section is not less than $\frac{\pi}{2}$, which occurs when $\operatorname{tg} \alpha \geqslant 2$, then the maximum area is that of the axial section:

$$
S_{\max }=O A \cdot S O=2 O M \cdot O S=k \cos \alpha \cdot k \sin \alpha=k 2 \sin 2 \alpha
$$

## Answer

$k 2 \sin 2 \alpha$, if $\operatorname{arctg} 2 \leqslant \alpha<\frac{\pi}{2} ; \frac{1}{2} k 2(1+3 \cos 2 \alpha)$, if $0<\alpha<\operatorname{arctg} 2$.