12.385. The side of a regular triangle is equal to $a$. The triangle rotates around a line lying in the plane of the triangle outside it, passing through the vertex of the triangle and forming an angle $\alpha$ with the side. Find the volume of the solid of revolution and determine at what value of $\alpha$ this volume is the greatest.
Solution.

Points $B_{1}$ and $A_{1}$ are symmetric to points $B$ and $A$ with respect to the line $l$ passing through the vertex $C$ of the equilateral $\triangle ABC (AB = a)$ and serving as the axis of rotation in the problem's condition (Fig. 12.235), $O$ is the midpoint of $BB_{1}$, $O_{1}$ is the midpoint of $AA_{1}$, $\angle BCO = \alpha$. Then $\angle ACO_{1} = 120^{\circ} - \alpha$.

$V_{1}$ is the volume of the frustum of a cone with the axial section $AB B_{1} A_{1}$,

$V_{2}$ and $V_{3}$ are the volumes of the cones with the axial sections $BC B_{1}$ and $AC A_{1}$.

The volume of the solid of revolution $V = V_{1} - (V_{2} + V_{3})$.

If $OC = h_{1}$, $O_{1}C = h_{2}$, $OB = r$, $O_{1}A = R$, then

$$
V = V_{1} - (V_{2} + V_{3}) = \frac{1}{3} \pi (h_{1} + h_{2})(R^{2} + Rr + r^{2}) - \frac{1}{3} \pi r^{2} h_{1} -
$$

$$
\begin{aligned}
& - \frac{1}{3} \pi R^{2} h_{2} = \frac{\pi}{3} (R^{2} h_{1} + Rr h_{1} + r^{2} h_{1} + R^{2} h_{2} + Rr h_{2} + r^{2} h_{2} - r^{2} h_{1} - R^{2} h_{2}) = \\
& = \frac{\pi}{3} (R^{2} h_{1} + Rr h_{1} + r^{2} h_{2} + Rr h_{2}) = \frac{\pi}{3} (R + r)(h_{1} R + h_{2} r)
\end{aligned}
$$

In $\triangle BOC (\angle BOC = 90^{\circ})$: $r = a \sin \alpha$; $h_{1} = a \cos \alpha$.

In $\triangle AO_{1}C (\angle AO_{1}C = 90^{\circ})$: $R = a \sin (120^{\circ} - \alpha) = h_{2} = a \cos (120^{\circ} - \alpha)$.

Thus

$$
\begin{aligned}
& V = \frac{1}{3} \pi a^{3} (\sin \alpha + \sin (120^{\circ} - \alpha)) (\cos \alpha \sin (120^{\circ} - \alpha) + \cos (120^{\circ} - \alpha) \sin \alpha) = \\
& = \frac{1}{3} \pi a^{3} \cdot 2 \sin 60^{\circ} \cos (60^{\circ} - \alpha) \sin 120^{\circ} = \\
& = \frac{2}{3} \pi a^{3} \sin^{2} 60^{\circ} \cos (60^{\circ} - \alpha) = \frac{1}{2} \pi a^{3} \cos (60^{\circ} - \alpha)
\end{aligned}
$$

The volume reaches its maximum value when $\cos (60^{\circ} - \alpha) = 1$, $\alpha = 60^{\circ}$. Answer: $\frac{\pi a^{3}}{2} \cos (60^{\circ} - \alpha) \alpha = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1149.jpg?height=646&width=601&top_left_y=997&top_left_x=97)

Fig. 12.236