[Sphere circumscribed around a pyramid] $[\quad$ Dihedral angle $\quad]$

Find the volume of a regular triangular pyramid with the radius $R$ of the circumscribed sphere and the angle $\gamma$ between the lateral faces.
Let $O$ be the center of the sphere of radius $R$, circumscribed around a regular triangular pyramid $ABCD$ with vertex $D$ (Fig.1). The point $O$ lies on the line $DM$, where $M$ is the center of the base $ABC$. By the problem's condition, $OA=R$. Let $F$ be the foot of the perpendicular dropped from vertex $B$ of the base to the lateral edge $AD$, and $K$ be the midpoint of side $BC$ of the base. Since the line $AK$ is the orthogonal projection of the inclined line $AD$ onto the plane of the base and $AK \perp BC$, by the theorem of three perpendiculars, $AD \perp BC$. Therefore, the line $AD$ is perpendicular to two intersecting lines $BC$ and $BF$ in the plane $BFC$. Thus, $AD$ is perpendicular to the plane $BFC$, and $BFC$ is the linear angle of the dihedral angle between the lateral faces $ABD$ and $ACD$. By the problem's condition, $\angle BFC = \gamma$. Let $\angle DAM = \alpha$, $AB = BC = AC = a$. Then $AM = \frac{a \sqrt{3}}{3}$. From the right triangle $BKF$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-35.jpg?height=103&width=668&top_left_y=1518&top_left_x=695)

From the right triangle $AFK$, we find that

$$
\sin \angle DAM = \sin \alpha = \frac{KF'}{AK} = \frac{\frac{\alpha \operatorname{tg} \frac{\gamma}{2}}{2}}{\frac{\alpha \sqrt{2}}{2}} = \frac{\operatorname{ctg} \frac{\gamma}{2}}{\sqrt{3}}
$$

Next, we find:

$$
\begin{aligned}
& \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{\operatorname{ctg}^2 \frac{\gamma}{2}}{3}} = \frac{\sqrt{3 - \operatorname{ctg}^2 \frac{\gamma}{2}}}{\sqrt{3}} \\
& \operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\operatorname{ctg} \frac{\gamma}{2}}{\sqrt{3}}}{\frac{\sqrt{3 - \operatorname{ctg}^2 \frac{\gamma}{2}}}{\sqrt{3}}} = \frac{\operatorname{ctg} \frac{\gamma}{2}}{\sqrt{3 - \operatorname{ctg}^2 \frac{\gamma}{2}}}
\end{aligned}
$$

Consider the section of the pyramid and the sphere by a plane passing through points $A, D$, and $M$ (Fig.2). We obtain a circle of radius $R$ with center $O$ on the line $MD$. Extend the segment $DM$ beyond point $M$ to intersect the circle at point $P$. From the right triangles $AMD$ and $APD$, we find that

$$
\begin{gathered}
DM = AM \operatorname{tg} \angle DAM = \frac{a \sqrt{3} \operatorname{tg} \alpha}{3}, \\
AM^2 = DM \cdot MP, \text{ and } \frac{a^2}{3} = \frac{a \sqrt{3} \operatorname{tg} \alpha}{3} \left(2R - \frac{a \sqrt{3} \operatorname{tg} \alpha}{3}\right),
\end{gathered}
$$

from which

$$
a = \frac{2R \sqrt{3} \operatorname{tg} \alpha}{1 + \operatorname{tg}^2 \alpha} = 2R \sqrt{3} \operatorname{tg} \alpha \cos 2\alpha = 2R \sqrt{3} \sin \alpha \cos \alpha =
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-36.jpg?height=123&width=740&top_left_y=70&top_left_x=659)

Therefore,

$$
\begin{aligned}
& V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DM = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot \frac{a \sqrt{3} \operatorname{tg} \alpha}{3} = \frac{1}{12} a^3 \operatorname{tg} \alpha = \\
& = \frac{1}{12} \cdot \left(\frac{2R \sqrt{3} \operatorname{ctg} \frac{\gamma}{2} \sqrt{3 - \operatorname{ctg}^2 \frac{\gamma}{2}}}{\sqrt{3}}\right)^3 \cdot \frac{\operatorname{ctg} \frac{\gamma}{2}}{\sqrt{3 - \operatorname{ctg}^2 \frac{\gamma}{2}}} = \\
& = \frac{1}{2} \cdot \frac{8}{3 \sqrt{3}} \cdot R^3 \operatorname{ctg}^4 \frac{\gamma}{2} \left(3 - \operatorname{ctg}^2 \frac{\gamma}{2}\right) = \frac{2}{27} R^3 \sqrt{3} \operatorname{ctg}^4 \frac{\gamma}{2} \left(3 - \operatorname{ctg}^2 \frac{\gamma}{2}\right).
\end{aligned}
$$

## Answer

$\frac{2}{27} R^3 \sqrt{3} \operatorname{ctg}^4 \frac{\gamma}{2} \left(3 - \operatorname{ctg}^2 \frac{\gamma}{2}\right) = \frac{2R \sqrt{3} \cos^4 \frac{\gamma}{2} (1 - 2 \cos \gamma)}{27 \sin^6 \frac{\gamma}{2}}$.