12.427 A plane is drawn through the vertex of the base of a regular quadrilateral pyramid, intersecting the opposite lateral edge at a right angle. The area of the section is half the area of the base of the pyramid. Find the angle between the lateral edge and the base plane.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-304.jpg?height=397&width=361&top_left_y=129&top_left_x=119)

Let $FO$ be the height of the given regular pyramid. In $\triangle ACF$, draw the height $AK$, then through the intersection point $T$ of it with the height $FO$, draw the segment $MN \parallel BD$. Since $FC \perp BD$, then $FC \perp MN$. Moreover, $FC \perp AK$ (by construction), $FC \perp AMKN$ (Theorem 11), i.e., $AMKN$ is the desired section. $FA \perp MN \Rightarrow MN \perp AK$ (Theorem 12) $\Rightarrow$

$\Rightarrow S_{AMNK} = \frac{1}{2} MN \cdot AK$. Let $a = AO, \angle FCO = x$.

Then $AK = 2a \cdot \sin x, FO = a \cdot \tan x; TO = a \cdot \tan \angle TAO = a \cdot \cot x \Rightarrow$ $\Rightarrow FT = FO - TO \doteq a(\tan x - \cot x). \triangle FMN \sim \triangle FBD \Rightarrow$

$\Rightarrow MN = BD \cdot \frac{FT}{FO} = \frac{2a(\tan x - \cot x)}{\tan x}; S_{AMNK} = -\frac{2a^2 \cos 2x}{\sin x}$.

But $S_{ABCD} = 2S_{AMNK} = 2a^2 \Rightarrow -2 \cos 2x = \sin x \Rightarrow 4 \sin^2 x - \sin x - 2 = 0 \Rightarrow$ $\Rightarrow \sin x = \frac{1 + \sqrt{33}}{8}$.

Answer: $\arcsin \frac{1 + \sqrt{33}}{8}$.