12.211. The base of the pyramid is a rhombus with an acute angle $\boldsymbol{\alpha}$. Find the volume of the pyramid if its lateral faces form the same dihedral angle $\beta$ with the base and the radius of the inscribed sphere is $r$.
Solution.

Let the rhombus $ABCD$ be the base of the pyramid $SABCD$ (Fig. 12.78), $SO$ - its height, $\angle BCD=\alpha, 0^{\circ}<\alpha<90^{\circ}$.

From the vertex $S$ of the pyramid, drop a perpendicular $SE$ to the side $CD$ of the base. Then $OE \perp CD$ and $\angle SEO-$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0958.jpg?height=676&width=671&top_left_y=712&top_left_x=589)

Fig. 12.78 is the linear angle of the dihedral angle formed by the lateral face $CSD$ with the base, $\angle SEO=\beta$. Since all faces of the given pyramid form the same angle with the base, the point $O$ is the center of the circle inscribed in the rhombus $ABCD$, the intersection point of its diagonals $AC$ and $BD$.

Then $\angle DOE=\angle OCE=\frac{1}{2} \angle BCD=\frac{\alpha}{2}$.

Let $O_1$ be the intersection point of the bisector of $\angle SEO$ and the height $SO$ of the pyramid. Drop a perpendicular $O_1F$ to the line $SE$ in the plane $SOE$. Since $DC$ is perpendicular to the plane $SOE$, the planes $SOE$ and $DSC$ are perpendicular, and thus $O_1F$ is perpendicular to the plane $DSC$. $O_1F=O_1O$ and since the face $DSC$ is chosen arbitrarily, $O_1$ is equidistant from all faces of the pyramid, that is, $O_1$ is the center of the sphere inscribed in the given pyramid, and $O_1O=O_1F=r$.

In $\triangle OOE\left(\angle OOE=90^{\circ}\right): OE=O_1O \operatorname{ctg} \angle OEO_1=r \operatorname{ctg} \frac{\beta}{2}$.

From $\triangle SOE\left(\angle SOE=90^{\circ}\right): SO=OE \operatorname{tg} \angle SEO=r \operatorname{ctg} \frac{\beta}{2} \operatorname{tg} \beta$.

In $\triangle OEC\left(\angle OEC=90^{\circ}\right): CO=\frac{OE}{\sin \angle OCE}=\frac{r \operatorname{ctg} \frac{\beta}{2}}{\sin \frac{\alpha}{2}}$.

From $\triangle OED\left(\angle OED=90^{\circ}\right): DO=\frac{OE}{\cos \angle DOE}=\frac{r \operatorname{ctg} \frac{\beta}{2}}{\cos \frac{\alpha}{2}}$.
The area of the base of the pyramid

$$
\begin{aligned}
& S_{\text {base}}=\frac{1}{2} AC \cdot BD=2 CO \cdot DO=\frac{2 r^2 \operatorname{ctg}^2 \frac{\beta}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{4 r^2 \operatorname{ctg}^2 \frac{\beta}{2}}{\sin \alpha} \\
& \text { Volume of the pyramid } V=\frac{1}{3} S_{\text {base}} \cdot SO=\frac{4 r^3 \operatorname{ctg}^3 \frac{\beta}{2} \operatorname{tg} \beta}{3 \sin \alpha}=\frac{4 r^3 \operatorname{tg} \beta}{3 \sin \alpha \operatorname{tg}^3 \frac{\beta}{2}} \\
& \text { Answer: } \frac{4 r^3 \operatorname{tg} \beta}{3 \sin \alpha \operatorname{tg}^3 \frac{\beta}{2}}
\end{aligned}
$$