3.86. In a regular triangular pyramid with an angle $\alpha$ between a lateral edge and a side of the base, a section is made through the midpoint of the lateral edge parallel to a lateral face. Given the area $S$ of this section, find the volume of the pyramid. What are the possible values of $\alpha$?
3.86. By the condition, $PABC$ is a regular triangular pyramid, $\angle PBC=\alpha, PM=MC, (KMN) \|(APB)$, $S_{\triangle KMN}=S$ (Fig. 3.86). Since $PC=2MC$, we have $S_{\triangle APB}=4S_{\triangle KMN}=4S$. Further, we have $S_{\triangle APB}=\frac{1}{2} AP^2 \sin (180^\circ - 2\alpha)=$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-554.jpg?height=477&width=418&top_left_y=1402&top_left_x=737)

Fig. 3.86
$=\frac{1}{2} AP^2 \sin 2\alpha$, i.e., $AP^2=\frac{8S}{\sin 2\alpha}$. From $\triangle APF$, we find $AF=AP \cos \alpha$, hence $AB=2AP \cos \alpha$ and $OC=\frac{2AP \cos \alpha}{\sqrt{3}}$. Now, from $\triangle POC$, we get

$$
PO=\sqrt{PC^2-OC^2}=\sqrt{\frac{8S}{\sin 2\alpha}-\frac{32S \cos^2 \alpha}{3 \sin 2\alpha}}=\sqrt{\frac{8S(3-4 \cos^2 \alpha)}{3 \sin 2\alpha}}=
$$

$$
\begin{gathered}
=2 \sqrt{\frac{2S(1-2 \cos 2\alpha)}{3 \sin 2\alpha}}=4 \sqrt{\frac{S(\cos \frac{\pi}{3}-\cos 2\alpha)}{3 \sin 2\alpha}}= \\
=4 \sqrt{\frac{2S \sin \left(\alpha+\frac{\pi}{6}\right) \sin \left(\alpha-\frac{\pi}{6}\right)}{3 \sin 2\alpha}}
\end{gathered}
$$

Finally, we find

$$
V=\frac{1}{3} S_{\triangle ABC} \cdot PO=\frac{1}{3} \cdot \frac{AB^2 \sqrt{3}}{4} \cdot PO=\frac{1}{3} \cdot \frac{4 \cdot 8 \sqrt{3} S \cos^2 \alpha}{4 \sin 2\alpha} \times
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-555.jpg?height=178&width=1085&top_left_y=1088&top_left_x=56)
where $\frac{\pi}{6}<\alpha<\frac{\pi}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-555.jpg?height=400&width=417&top_left_y=1444&top_left_x=45)

Fig. 3.87