[ Sphere inscribed in a pyramid. ] [ Angles between lines and planes ]

Find the volume of a regular triangular pyramid with the radius $r$ of the inscribed sphere and the angle $\alpha$ between the lateral edge and the base plane.
Let $Q$ be the center of a sphere of radius $r$, inscribed in a regular triangular pyramid $ABCD$ with vertex $D$, and let $K$ be the midpoint of $BC$ (Fig.1). The point $Q$ lies on the line $DM$, where $M$ is the center of the base $ABC$. Denote $AB = BC = AC = a, \angle DKM = \beta$. Then

$$
BM = \frac{a}{\sqrt{3}}, \quad KM = \frac{a}{2 \sqrt{3}}, \quad DM = BM \tan \angle MBD = \frac{a \tan \beta^2}{\sqrt{3}}
$$

Consider the section of the pyramid and the sphere by a plane passing through the points $D, K$, and $M$ (Fig.2). We obtain a circle of radius $r$ with center $Q$ on the line $DM$, touching the side $KM$ of the angle $DKM$ at point $M$. From the right triangle $KMQ$, we find that

$$
r = QM = MK \tan \angle MKQ = \frac{a}{2 \sqrt{3}} \tan \frac{\beta}{2} = \frac{a \tan \beta^2}{2 \sqrt{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-36.jpg?height=97&width=1306&top_left_y=2015&top_left_x=-1)

$$
\tan \beta = \tan \angle DKM = \frac{DM}{KM} = \frac{\frac{a \tan \alpha}{\sqrt{3}}}{\frac{a}{2 \sqrt{3}}} = 2 \tan \alpha
$$

Substituting $\tan \beta = 2 \tan \alpha$ into the formula

$$
\tan \beta = \frac{2 \tan \frac{\alpha^2}{2}}{1 - \tan^2 \frac{\alpha}{2}}
$$

we get the equation

$$
\tan \alpha = \frac{\tan \frac{\alpha^2}{2}}{1 - \tan^2 \frac{\alpha}{2}}
$$

or

$$
\tan \alpha = \frac{\frac{2 r \sqrt{3}}{a}}{1 - \frac{12 r^2}{a^2}} = \frac{2 a r \sqrt{3}}{a^2 - 12 r^2} \Leftrightarrow a^2 \tan \alpha - 2 a r \sqrt{3} - 12 r^2 \tan \alpha = 0
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-37.jpg?height=97&width=728&top_left_y=0&top_left_x=665)

Therefore,

$$
\begin{aligned}
& V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DM = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot DM = \frac{\sqrt{3}}{12} \cdot a^2 \cdot DM = \frac{\sqrt{3}}{12} a^2 \cdot \frac{a \tan \alpha}{\sqrt{3}} = \\
& = \frac{1}{12} a^3 \tan \alpha = \frac{1}{12} \left( \frac{r \sqrt{3} \left(1 + \sqrt{1 + 4 \tan^2 \alpha} \right)}{\tan^2 \alpha} \right)^3 \tan \alpha =
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-37.jpg?height=117&width=872&top_left_y=542&top_left_x=593)

## Answer

$\frac{r^3 \sqrt{3} \left(1 + \sqrt{1 + 4 \tan^2 \alpha} \right)^3}{4 \tan^2 \alpha}$