3.81. A perpendicular, equal to $p$, is dropped from the base of the height of a regular triangular pyramid onto a lateral edge. Find the volume of the pyramid if the dihedral angle between its lateral faces is $\alpha$.
3.81. By the condition, $S A B C$ is a regular triangular pyramid, $O$ is the center of the base, $O K \perp S B, O K=p, \angle A S B C=\alpha$ (Fig. 3.82). We construct the linear angle of the dihedral angle $A S B C$. We have $A C \perp S B$, since $S B$ is a slant to ( $A B C$ ), $O B$ is its projection onto this plane, $A C \perp O B$. Draw a plane $A P C \perp S B$ through $A C$; then $\angle A P C=\alpha$. Since $M P$ and $O K$ belong to $(S M B)$ and $M P \perp S B, O K \perp S B$, then $M P \| O K$ and, therefore, $\frac{M P}{O K}=\frac{M B}{O B}$, i.e., $M P=\frac{3 p}{2}$. From $\triangle M P C$ we find

$$
M C=M P \operatorname{tg} \frac{\alpha}{2}=\frac{3 p}{2} \operatorname{tg} \frac{\alpha}{2}, \quad A C=3 p \operatorname{tg} \frac{\alpha}{2}, \quad \text { hence }
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-552.jpg?height=523&width=521&top_left_y=108&top_left_x=51)

Fig. 3.82 $S_{\triangle A B C}=\frac{9 p^{2} \sqrt{3} \operatorname{tg}^{2} \frac{\alpha}{2}}{4} . \quad$ Further, $\Delta S O K \sim \triangle S O B$ (right triangles with a common angle $O S B$ ); therefore $\frac{O K}{O S}=\frac{O B}{S B}$, From this, considering that $\quad O B=p \sqrt{3} \operatorname{tg} \frac{\alpha}{2}$, we get

$$
S B=\frac{O B \cdot O S}{O K}=\frac{p \sqrt{3} \operatorname{tg} \frac{\alpha}{2} \cdot O S}{p}=O S \sqrt{3} \operatorname{tg} \frac{\alpha}{2}
$$

In $\triangle S O B$ we have

$$
S B^{2}=O S^{2}+O B^{2} ; 3 O S^{2} \operatorname{tg}^{2} \frac{\alpha}{2}-O S^{2}=3 p^{2} \operatorname{tg}^{2} \frac{\alpha}{2} ; O S=\frac{p \sqrt{3} \operatorname{tg} \frac{\alpha}{2}}{\sqrt{3 \operatorname{tg}^{2} \frac{\alpha}{2}-1}}
$$

Thus,

$$
V=\frac{1}{3} \frac{9 p^{2} \operatorname{tg}^{2} \frac{\alpha}{2} \cdot \sqrt{3} \cdot \sqrt{3} p \operatorname{tg} \frac{\alpha}{2}}{4 \sqrt{3 \operatorname{tg}^{2} \frac{\alpha}{2}-1}}=\frac{9 p^{3} \operatorname{tg}^{3} \frac{\alpha}{2}}{4 \sqrt{3 \operatorname{tg}^{2} \frac{\alpha}{2}-1}}
$$