12.454 The side of the base of a regular quadrilateral pyramid is equal to $a$. The angle between the height of the pyramid and a lateral edge is $\alpha$ $\left(\alpha \leq \operatorname{arctg} \frac{\sqrt{2}}{2}\right)$. Find the area of the section of the pyramid by a plane passing through the midpoint of the height and perpendicular to one of its lateral edges.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-321.jpg?height=303&width=320&top_left_y=616&top_left_x=191)

Let $P$ be the midpoint of the height $FO$ of the given regular pyramid FABCD, and let KNLM be the section of the pyramid by a plane passing through $P$ and perpendicular to the edge $FC$. (We will prove below that $L$ lies between vertices $A$ and $F$.)

\[
\begin{aligned}
& \triangle FKM = \Delta FKN \Rightarrow FM = FN \Rightarrow MN \parallel BD \Rightarrow \\
& BD \perp ACF \Rightarrow BD \perp KL \Rightarrow MN \perp KL
\end{aligned}
\]

Next, $KL = FK \cdot \operatorname{tg} 2\alpha = FP \cdot \cos \alpha \cdot \operatorname{tg} 2\alpha = \frac{FO}{2} \cdot \cos \alpha \cdot \operatorname{tg} 2\alpha = \frac{OC}{2} \cdot \operatorname{ctg} \alpha \cdot \cos \alpha \cdot \operatorname{tg} 2\alpha = \frac{a}{2\sqrt{2}} \cdot \operatorname{ctg} \alpha \cdot \cos \alpha \cdot \operatorname{tg} 2\alpha$.

Now, let's prove that $FL < FA = FC$.

$FL = \frac{FK}{\cos 2\alpha} = \frac{FP \cdot \cos \alpha}{\cos 2\alpha} = \frac{FO}{2} \cdot \frac{\cos \alpha}{\cos 2\alpha} = \frac{FC \cdot \cos^2 \alpha}{2 \cos 2\alpha} = \frac{FC}{2(1 - \operatorname{tg}^2 \alpha)} < FC$, since by the condition $\operatorname{tg}^2 \alpha < \frac{1}{2}$.

Thus, in the section, we have a quadrilateral $KNML$ with perpendicular diagonals $\Rightarrow S_{KNLM} = \frac{MN \cdot KL}{2} = \frac{a^2}{8} \cdot \cos \alpha \cdot \operatorname{ctg} \alpha \cdot \operatorname{tg} 2\alpha$.

Answer: $\frac{a^2}{8} \cdot \cos \alpha \cdot \operatorname{ctg} \alpha \cdot \operatorname{tg} 2\alpha$.