12.207. 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 $\alpha$.
Solution.

Let SO be the height of the regular pyramid $SABC$ (Fig. 12.74), $OM \perp SA, OM=p, AL$ - the height of $\triangle ABC$.

In the plane $ABC$, through point $O$, draw $FK$ parallel to $BC$.

The line $AL$ is the projection of $SA$ onto the plane $ABC, AL \perp BC$. Thus, $SA \perp BC. SA \perp BC, FK \| BC$, hence $SA \perp FK$.

$SA \perp FK, SA \perp OM$. Therefore, the plane $MFK$ is perpendicular to the edge $SA, O$ is the center of the regular $\triangle ABC$ and $FK \| BC \Rightarrow O$ is the midpoint of $FK \Rightarrow MO$ is the height and median of $\Delta FMK \Rightarrow \Delta FMK$ is isosceles, $\angle FMO=\frac{1}{2} \cdot \angle FMK=\frac{\alpha}{2}$.

In $\triangle FOM\left(\angle FOM=90^{\circ}\right): FO=MO \operatorname{tg} \angle FMO=p \operatorname{tg} \frac{\alpha}{2}$.

$\triangle ALC \sim \triangle AOF$ and since $AL=\frac{3}{2} AO$, then $CL=\frac{3}{2} FO=\frac{3 p \operatorname{tg} \frac{\alpha}{2}}{2}$.

$$
BC=2 CL=3 p \operatorname{tg} \frac{\alpha}{2}, S_{\triangle ABC}=\frac{BC^{2} \sqrt{3}}{4}=\frac{9 p^{2} \sqrt{3} \operatorname{tg}^{2} \frac{\alpha}{2}}{4}
$$

Let $\angle S A O=\beta$.

In $\triangle FOM\left(\angle FOM=90^{\circ}\right): \operatorname{tg} \frac{\alpha}{2}=\frac{FO}{MO}$.

From $\triangle AMO\left(\angle AMO=90^{\circ}\right): \sin \beta=\frac{MO}{AO}$.

Then $\sin \beta \operatorname{tg} \frac{\alpha}{2}=\frac{MO}{AO} \cdot \frac{FO}{MO}=\frac{FO}{AO}=\operatorname{tg} \angle FAO=\operatorname{tg} 30^{\circ}=\frac{1}{\sqrt{3}}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0955.jpg?height=777&width=620&top_left_y=31&top_left_x=336)

Fig. 12.75

$\sin \beta=\frac{1}{\sqrt{3} \operatorname{tg} \frac{\alpha}{2}}, \cos \beta=\sqrt{1-\sin ^{2} \beta}=\frac{\sqrt{3 \operatorname{tg}^{2} \frac{\alpha}{2}-1}}{\sqrt{3} \operatorname{tg} \frac{\alpha}{2}}$.

$OM$ is the height of the right $\triangle AOS \Rightarrow \angle MOS=\angle SAO=\beta$.

In $\triangle SMO\left(\angle SMO=90^{\circ}\right): SO=\frac{OM}{\cos \angle MOC}=\frac{p \sqrt{3} \operatorname{tg} \frac{\alpha}{2}}{\sqrt{3 \operatorname{tg}^{2} \frac{\alpha}{2}}-1}$.

The volume of the pyramid

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

Answer:

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