3.91. An obtuse isosceles triangle rotates around a line passing through the point of intersection of its altitudes parallel to the larger side. Find the volume of the solid of revolution if the obtuse angle is $\alpha$, and the side opposite to it is $a$.
3.91. By the condition, $\angle B A C=\alpha-$ is an obtuse angle, $A B=A C, B M \perp A C$, $C N \perp A B, A D \perp B C, O$ - the orthocenter (i.e., the point of intersection of the altitudes) of $\triangle A B C$, $O_{1} O_{2} \| B C, O \in O_{1} O_{2}, B C=a$ (Fig. 3.91). The volume of the solid of revolution of $\triangle A B C$ around the line $O_{1} O_{2}$ is found using the formula $V_{\text {solid of rev }}=V_{O_{1} O_{2} C B}-2 V_{O_{1} O A B}$, where $V_{O_{1} O_{2} C B}=V_{\text {cyl }}=\pi O_{1} B^{2} \cdot B C$. We have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-558.jpg?height=332&width=419&top_left_y=1480&top_left_x=721)

Fig. 3.91 $\angle O A M=\angle C A D=\frac{\alpha}{2}, \angle M O A=90^{\circ}-\frac{\alpha}{2}$,
and $\angle O B D=\frac{\alpha}{2}$. From $\triangle O D B$ and $\triangle A D B$ we find $O D=\frac{a}{2} \operatorname{tg} \frac{\alpha}{2}$, $A D=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2} ; \quad$ then $\quad O_{1} B=O D=\frac{a}{2} \operatorname{tg} \frac{\alpha}{2}, O A=O D-A D=$ $=\frac{a}{2}\left(\operatorname{tg} \frac{\alpha}{2}-\operatorname{ctg} \frac{\alpha}{2}\right)$. Finally, we get

$$
\begin{aligned}
& V_{\text {solid of rev }}= \pi \frac{a^{2}}{4} \operatorname{tg}^{2} \frac{\alpha}{2} \cdot a-\frac{2}{3} \pi \cdot \frac{a}{2}\left(\frac{a^{2}}{4} \operatorname{tg}^{2} \frac{\alpha}{2}+\frac{a^{2}}{4} \operatorname{tg} \frac{\alpha}{2}\left(\operatorname{tg} \frac{\alpha}{2}-\operatorname{ctg} \frac{\alpha}{2}\right)+\right. \\
&\left.+\frac{a^{2}}{4}\left(\operatorname{tg} \frac{\alpha}{2}-\operatorname{ctg} \frac{\alpha}{2}\right)^{2}\right)=\frac{\pi a^{3}}{4}\left(\operatorname{tg}^{2} \frac{\alpha}{2}-\frac{1}{3} \operatorname{tg}^{2} \frac{\alpha}{2}-\right. \\
&\left.-\frac{1}{3} \operatorname{tg}^{2} \frac{\alpha}{2}+\frac{1}{3}-\frac{1}{3} \operatorname{tg}^{2} \frac{\alpha}{2}+\frac{2}{3}-\frac{1}{3} \operatorname{ctg}^{2} \frac{\alpha}{2}\right)=\frac{\pi a^{3}}{12}\left(3-\operatorname{ctg}^{2} \frac{\alpha}{2}\right)
\end{aligned}
$$