[ Right Pyramid ] $[$ Area of the Cross-section $]$

A plane intersects a right square pyramid, passing through the vertex of the base and perpendicular to the opposite lateral edge. The area of the resulting cross-section is half the area of the base of the pyramid. Find the ratio of the height of the pyramid to the lateral edge.
Let $P A B C D$ be a regular quadrilateral pyramid with vertex $P, \alpha$ - the angle between the height $P O$ and the lateral edge. Then the desired ratio of the height of the pyramid to its lateral edge is $\cos \alpha$. Let the plane of section pass through point $A$ perpendicular to line $P C$ and intersect edges $P C, P B$ and $P D$ at points $M, K$ and $L$ respectively. Since line $D B$ and the section plane are perpendicular to line $P C$, line $B D$ is parallel to the section plane. A plane through line $B D$ intersects the section plane along line $K L$. Therefore, $K L \| B D$, and since $A M \perp B D$ (by the theorem of three perpendiculars), $K L \perp A M$. Thus, the area of quadrilateral $A K M L$ is half the product of its diagonals $A M$ and $K L$. Let segments $A M$ and $K L$ intersect at point $N$. Denote the side of the square $A B C D$ by $a$. Since $\angle M A C=\angle O P C$ $=\alpha, \mathrm{a} \angle A C M=90^{\circ}-\alpha$, then

$$
\begin{aligned}
& A M=A C \cos \alpha=a \sqrt{2} \cos \alpha \\
& O N=O A \operatorname{tg} \alpha=\frac{1}{2} a \sqrt{2} \operatorname{tg} \alpha, P O=O C \operatorname{ctg} \alpha=\frac{1}{2} a \sqrt{2} \operatorname{ctg} \alpha \\
& P N=P O-O N=\frac{1}{2} a \sqrt{2} \operatorname{ctg} \alpha-\frac{1}{2} a \sqrt{2} \operatorname{tg} \alpha=\frac{1}{2} a \sqrt{2}(\operatorname{ctg} \alpha-\operatorname{tg} \alpha)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_41a6f93ed2bb2ffd440eg-17.jpg?height=129&width=675&top_left_y=2388&top_left_x=680)

$$
\begin{aligned}
& L K=B D(1-\operatorname{tg} 2 \alpha)=a \sqrt{2}(1-\operatorname{tg} 2 \alpha), \\
& S_{A K M L}=\frac{1}{2} A M \cdot K L=\frac{1}{2} a \sqrt{2} \cos \alpha \cdot a \sqrt{2}(1-\operatorname{tg} 2 \alpha)=
\end{aligned}
$$

Then

$$
\frac{1}{2}=\frac{S_{A K M L L}}{S_{A B C D}}=\cos \alpha(1-\operatorname{tg} 2 \alpha)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_41a6f93ed2bb2ffd440eg-18.jpg?height=94&width=1035&top_left_y=250&top_left_x=0)

$$
\cos \alpha\left(2-\frac{1}{\cos ^{2} \boldsymbol{r} \boldsymbol{z}}\right)=\frac{1}{2}
$$

or

$$
2 \cos \alpha-\frac{1}{\cos \beta \alpha_{1}}=\frac{1}{2}, 4 \cos 2 \alpha-\cos \alpha-2=0
$$

from which we find that $\cos \alpha=\frac{1+\sqrt{33}}{8}$ (the second solution does not satisfy the condition of the problem). Therefore,

$$
\frac{P O}{P C}=\cos \alpha=\frac{1+\sqrt{33}}{8}
$$

## Answer

$\frac{1+\sqrt{33}}{8}$