12.281. A plane is drawn through the diagonal of the base and the height of a regular quadrilateral pyramid. The ratio of the area of the section to the lateral surface area of the pyramid is $k$. Find the cosine of the angle between the apothems of opposite lateral faces and the permissible values of $k$.
## Solution.

Let $SO$ be the height, $SE$ and $SF$ be the apothems of the opposite lateral faces of the regular pyramid $SABCD$ (Fig. 12.144), $AD=a, \angle SFO=\alpha$.

Then $AC=a \sqrt{2}$, and from $\triangle \operatorname{SOF} \left(\angle SOF=90^{\circ}\right)$:

$$
SO = OF \operatorname{tg} \angle SFO = \frac{a}{2} \operatorname{tg} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1036.jpg?height=773&width=778&top_left_y=76&top_left_x=271)

Fig. 12.144

The area of the section $\triangle ASC$:

$$
S_{1} = \frac{1}{2} AC \cdot SO = \frac{a^{2} \sqrt{2} \operatorname{tg} \alpha}{4}
$$

The area of the base of the pyramid $S_{\text{base}} = a^{2}$, the area of the lateral surface $S_{2} = \frac{S_{\text{base}}}{\cos \alpha} = \frac{a^{2}}{\cos \alpha}$.

Since $\frac{S_{1}}{S_{2}} = k$, then $\frac{\sqrt{2} \operatorname{tg} \alpha}{4} \cos \alpha = k, \sin \alpha = \frac{4k}{\sqrt{2}} = 2 \sqrt{2} k$.

The required $\cos \angle E S F = \cos \left(180^{\circ} - 2 \alpha\right) = -\cos 2 \alpha = 2 \sin^{2} \alpha - 1 = 2(2 \sqrt{2} k)^{2} - 1 = 16 k^{2} - 1$, where $16 k^{2} - 1 < 1$, or $k^{2} < 0.125$; i.e., $0 < k < 0.25 \sqrt{2}$.

Answer: $16 k^{2} - 1 ; 0 < k < 0.25 \sqrt{2}$.