12.220. The lateral edge of a regular quadrilateral pyramid forms an angle $\boldsymbol{\alpha}$ with the plane of the base. A plane is drawn through the vertex of the base and the midpoint of the opposite lateral edge, parallel to one of the diagonals of the base. Find the angle between this plane and the plane of the base of the pyramid.
Solution.

Let SO be the height of the regular pyramid

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0968.jpg?height=666&width=666&top_left_y=1175&top_left_x=610)

Fig. 12.87
$S A B C D$ (Fig. 12.87), $L$ is the midpoint of $S C, \angle S C A=\alpha, M$ and $K$ are the points of intersection of the section passing through points $A$ and $L$ parallel to $B D$ with the edges $S B$ and $S D, M K$ is the line of intersection of the section and the plane $B S D$, hence $M K \|_{B D}$.

$B D \perp A C, B D \perp S O$. Thus, the line $B D$ is perpendicular to the plane $A S C$.

$A L$ is the line of intersection of the section and the plane $A S C$. Then $B D \perp A L$.

$A L \perp B D, A C \perp B D, B D$ is parallel to the line of intersection of the section and the base plane passing through point $A$. Therefore, $A L$ and $A C$ are perpendicular to this line, and $\angle L A C$ is the required angle of inclination of the cutting plane to the base plane.

Let $\angle L A C=x, E$ be the point of intersection of the medians $S O$ and $A L$ of triangle $A S C$. Then $E O=\frac{1}{3} S O$. From $\triangle A O E\left(\angle A O E=90^{\circ}\right): E O=O A \operatorname{tg} x$.

In $\triangle \operatorname{SOC}\left(\angle S O C=90^{\circ}\right): S O=O C \operatorname{tg} \angle S O C=O A \operatorname{tg} \alpha \Rightarrow$

$\Rightarrow O A \operatorname{tg} x=\frac{1}{3} O A \operatorname{tg} \alpha ; \operatorname{tg} x=\frac{1}{3} \operatorname{tg} \alpha ; x=\operatorname{arctg} \frac{\operatorname{tg} \alpha}{3}$.

Answer: $\operatorname{arctg} \frac{\operatorname{tg} \alpha}{3}$.