12.333. The base of the right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| \mathrm{CC}_{1}\right)$ is an isosceles triangle $A B C(A B=A C)$, with a perimeter of $2 p$ and the angle at vertex $A$ equal to $\alpha$. A plane is drawn through side $B C$ and vertex $A_{1}$, forming an angle $\beta$ with the base plane. Find the volume of the prism.
## Solution.

Let $A D$ be the height of $\triangle A B C$ (Fig. 12.194). Then $\angle A_{1} D A$ is the angle of inclination of the section $B A_{1} C$ to the base plane, $\angle A_{1} D A=\beta$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1091.jpg?height=678&width=681&top_left_y=75&top_left_x=271)

Fig. 12.194

In $\triangle A D B\left(\angle A D B=\frac{\pi}{2}\right): B D=A B \sin \frac{\alpha}{2}$.

Then $B C=2 A B \sin \frac{\alpha}{2} ; 2 p=2 A B+2 A B \sin \frac{\alpha}{2} ; A B=A C=\frac{p}{1+\sin \frac{\alpha}{2}}$.

$S_{\triangle A B C}=\frac{1}{2} A B^{2} \sin \alpha=\frac{p^{2} \sin \alpha}{2\left(1+\sin \frac{\alpha}{2}\right)^{2}} \cdot$ In $\triangle A D B: A D=A B \cos \frac{\alpha}{2}=\frac{p \cos \frac{\alpha}{2}}{1+\sin \frac{\alpha}{2}}$.

In $\triangle A_{1} A D\left(\angle A_{1} A D=\frac{\pi}{2}\right): A A_{1}=A D \operatorname{tg} \beta=\frac{p \cos \frac{\alpha}{2} \operatorname{tg} \beta}{1+\sin \frac{\alpha}{2}}$.

Volume of the prism:

$$
V=S_{\triangle A B C} \cdot A A_{1}=\frac{p^{3} \sin \alpha \cos \frac{\alpha}{2} \operatorname{tg} \beta}{2\left(1+\sin \frac{\alpha}{2}\right)^{3}}=\frac{p^{3} \sin \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2} \operatorname{tg} \beta}{\left(1+\sin \frac{\alpha}{2}\right)^{3}}=
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1092.jpg?height=819&width=702&top_left_y=57&top_left_x=317)

Fig. 12.195

$$
\begin{aligned}
& =\frac{p^{3} \sin \frac{\alpha}{2} \cos ^{3} \frac{\alpha}{2} \operatorname{tg} \beta}{\left(1+\sin \frac{\alpha}{2}\right)^{3} \cos \frac{\alpha}{2}}=p^{3} \operatorname{tg} \beta \operatorname{tg} \frac{\alpha}{2} \cdot\left(\frac{\sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)}{1+\cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)}\right)^{3}= \\
& =p^{3} \operatorname{tg}^{3} \frac{\pi-\alpha}{4} \operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \beta \\
& \text { Answer: } p^{3} \operatorname{tg}^{3} \frac{\pi-\alpha}{4} \operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \beta
\end{aligned}
$$