3.85. 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 $k$.
3.85. By the condition, $S A B C D$ is a regular pyramid, $S O \perp(A B C)$, $S_{\triangle S A C}: \mathrm{S}_{\text {bok }}=k, S F$ and $S E$ are the apothems of the pyramid (Fig. 3.85). Let $A B=$

$=a, \angle E S F=\alpha ;$ we have $S F=\frac{a}{2 \sin \frac{\alpha}{2}}$

(from $\triangle S O F), \quad S O=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-554.jpg?height=418&width=510&top_left_y=291&top_left_x=645)

Fig. 3.85 Therefore, $S_{\triangle S A C}=\frac{1}{2} a \sqrt{2} \cdot \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}=\frac{a^{2} \sqrt{2} \operatorname{ctg} \frac{\alpha}{2}}{4}, S_{60 x}=\frac{1}{2} \cdot 4 a \cdot \frac{a}{2 \sin \frac{\alpha}{2}}=\frac{a^{2}}{\sin \frac{\alpha}{2}}$.

Then we get the equation

$$
k=\frac{a^{2} \sqrt{2} \operatorname{ctg} \frac{\alpha}{2} \sin \frac{\alpha}{2}}{4 a^{2}}, \text { or } k=\frac{\sqrt{2} \cos \frac{\alpha}{2}}{4}, \text { i.e. } \cos \frac{\alpha}{2}=2 k \sqrt{2}
$$

Thus, $\cos \alpha=2 \cos ^{2} \frac{\alpha}{2}-1=16 k^{2}-1$. In this case $16 k^{2}-1<1$; $k^{2}<\frac{1}{8}, 0<k<\frac{\sqrt{2}}{4}$.