12.418 Through the vertex of the base of a regular triangular pyramid, a plane is drawn perpendicular to the opposite lateral face and parallel to the opposite side of the base. This plane forms an angle $\alpha$ with the plane of the pyramid's base. Find the plane angle at the vertex of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-297.jpg?height=360&width=365&top_left_y=1323&top_left_x=129)

Let $FABC$ be the given regular pyramid, $O$ be the center of the base, and the plane $AMN \perp FBC$. By the condition, $AMN \| BC \Rightarrow MN \| BC$ (Theorem 14). The line $l$ of intersection of $AMN$ and $ABC$ (passing through point $A$) is also parallel to $BC$ (Theorem 14). Let $K$ be the midpoint of $BC$, and $L$ be the point of intersection of $MN$ and $FK$. Since $AL$ is the median of the isosceles $\triangle AMN$, then $AL \perp MN (FM = FN$, because $MN \| BC) \Rightarrow l \perp AL$.

But $BC \perp AK \Rightarrow l \perp AK$, i.e., $\angle LAK$ is the angle between the planes $AMN$ and $ABC \Rightarrow \angle LAK = \angle OFK = \alpha \Rightarrow$

$\Rightarrow FK = \frac{OK}{\sin \alpha} = \frac{BC}{2 \sqrt{3} \sin \alpha} \Rightarrow \sqrt{3} \sin \alpha = \frac{BC}{2 \cdot FK} = \tan \angle CFK \Rightarrow \angle CFB = 2 \arctan(\sqrt{3} \cdot \sin \alpha)$.

Answer: $2 \arctan(\sqrt{3} \cdot \sin \alpha)$.