12.422 A regular triangular pyramid is intersected by a plane passing through its lateral edge and height. In the section, a triangle with an angle $\pi / 4$ at the vertex of the pyramid is formed. Find the angle between the lateral edge and the base plane of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-300.jpg?height=355&width=393&top_left_y=548&top_left_x=128)

Let $F P$ be the height of the given regular pyramid $F A B C, K$ be the point of intersection of the plane $F C P$ with the edge $A B$. $K$ is the midpoint of $A B$ ($C K$ passes through the center of $\triangle A B C$) $\Rightarrow \angle F K C$ is the desired angle.

Let $\alpha = \angle F K C$. Then $\angle F C P = 135^{\circ} - \alpha$ (by the problem statement), $F P = P K \cdot \operatorname{tg} \alpha = P C \cdot \operatorname{tg}(135^{\circ} - \alpha) \Rightarrow$ $\Rightarrow 2 \operatorname{tg}(135^{\circ} - \alpha) = \operatorname{tg} \alpha (P C = 2 P K) \Rightarrow \operatorname{tg} \alpha = \frac{3 + \sqrt{17}}{2}$. Answer: $\operatorname{arctg} \frac{3 + \sqrt{17}}{2}$.