12.330. The base of the pyramid is a right-angled triangle, with the radius of the inscribed circle being $r$, and one of the acute angles being $\alpha$. All lateral edges of the pyramid form the same angle $\beta$ with the base plane. Find the volume of the pyramid.
## Solution.

Let $EO$ be the height of the pyramid $EABC$, $\angle ACB = \frac{\pi}{2}$, $\angle BAC = \alpha$, $O_1$ be the center of the circle inscribed in $\triangle ABC$, $L$ and $K$ be the points of tangency of this circle with sides $AC$ and $BC$ respectively (Fig. 12.191).

Then $O_1L \perp AC$, $O_1K \perp BC$, $O_1L = O_1K = CL = CK = r$, $\angle O_1AL = \frac{\alpha}{2}$, $\angle O_1BK = \frac{\pi}{4} - \frac{\alpha}{2}$. In $\triangle ALO_1$: $AL = r \operatorname{ctg} \frac{\alpha}{2}$. In $\triangle BKO_1$: $BK = r \operatorname{ctg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$.

$AC = AL + LC = r\left(\operatorname{ctg} \frac{\alpha}{2} + 1\right) = r\left(\operatorname{ctg} \frac{\alpha}{2} + \operatorname{ctg} \frac{\pi}{4}\right) = r \cdot \frac{\sin \left(\frac{\pi}{4} + \frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2} \sin \frac{\pi}{4}} = \frac{r \sqrt{2} \cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2}}$.

$BC = BK + KC = r\left(\operatorname{ctg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + 1\right) = r\left(\operatorname{ctg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \operatorname{ctg} \frac{\pi}{4}\right) = r \cdot \frac{\sin \left(\frac{\pi}{2} - \frac{\alpha}{2}\right)}{\sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \frac{\pi}{4}} = \frac{r \sqrt{2} \cos \frac{\alpha}{2}}{\sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}$.

$S_{\triangle ABC} = \frac{1}{2} AC \cdot BC = \frac{1}{2} \cdot \frac{r \sqrt{2} \cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2}} \cdot \frac{r \sqrt{2} \cos \frac{\alpha}{2}}{\sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)} = r^2 \operatorname{ctg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \operatorname{ctg} \frac{\alpha}{2}$. In $\triangle ABC \left(\angle ACB = \frac{\pi}{2}\right)$:

$$
AB = \frac{BC}{\sin \angle BAC} = \frac{r \sqrt{2} \cos \frac{\alpha}{2}}{\sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \alpha} = \frac{r \sqrt{2}}{2 \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1088.jpg?height=698&width=632&top_left_y=61&top_left_x=96)

Fig. 12.191

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1088.jpg?height=656&width=491&top_left_y=100&top_left_x=784)

Fig. 12.192

Since all lateral edges of the pyramid form the same angle $\beta$ with the plane of the base, the point $O$ is the center of the circle circumscribed around $\triangle ABC$, and since $\triangle ABC$ is a right triangle, $O$ is the midpoint of $AB$, $OA = \frac{1}{2} AB = \frac{r \sqrt{2}}{4 \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}$.

In $\triangle AOE \left(\angle AOE = \frac{\pi}{2}\right): EO = AO \operatorname{tg} \angle EAO = \frac{r \sqrt{2} \operatorname{tg} \beta}{4 \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}$.

The volume of the pyramid $V = \frac{1}{3} S_{\triangle ABC} \cdot EO = \frac{r^3 \sqrt{2} \operatorname{tg} \beta \operatorname{ctg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \operatorname{ctg} \frac{\alpha}{2}}{12 \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}$. Answer: $\frac{r^3 \sqrt{2} \operatorname{tg} \beta \operatorname{ctg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \operatorname{ctg} \frac{\alpha}{2}}{12 \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}$.