12.064. A square with side length $a$ is inscribed in the base of a cone. A plane passing through one of the sides of this square and the vertex of the cone intersects the surface of the cone to form an isosceles triangle, with the angle at the vertex being $\alpha$. Find the volume of the cone.
Solution.

Let $ABCD$ be the given square with side $a$ (Fig. 12.66) and let the cutting plane pass through the side $AD$. Then, by the condition, $\angle ASD = \alpha$. Let $L$ be the midpoint of

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-762.jpg?height=590&width=360&top_left_y=513&top_left_x=778)

Fig. 12.66 $AD$. This means that $AL = \frac{a}{2}$. Since $\triangle ASD$ is isosceles, $AS = \frac{AL}{\sin \frac{\alpha}{2}} = \frac{a}{2 \sin \frac{\alpha}{2}}$. The diagonal $AC$ of the square is the diameter of the circle: $2a^2 = d^2$. Hence, $d = a\sqrt{2}$, $AO = r = \frac{d}{2} = \frac{a\sqrt{2}}{2}$. Then the height $SO = \sqrt{AS^2 - AO^2} = \sqrt{\frac{a^2}{4 \sin^2 \frac{\alpha}{2}} - \frac{a^2}{2}} = \sqrt{\frac{a^2 (1 - 2 \sin^2 \frac{\alpha}{2})}{4 \sin^2 \frac{\alpha}{2}}} = \frac{a \sqrt{\cos \alpha}}{2 \sin^2 \frac{\alpha}{2}}$.

The area of the base of the cone $S_{\text{base}} = \pi r^2 = \pi \frac{a^2}{2}$. Then the volume

\[
\begin{aligned}
& V = \frac{1}{3} S_{\text{base}} \cdot SO = \frac{1}{3} \cdot \frac{\pi a^2}{2} \cdot \frac{a \sqrt{\cos \alpha}}{2 \sin^2 \frac{\alpha}{2}} = \frac{\pi a^3 \sqrt{\cos \alpha}}{12 \sin \frac{\alpha}{2}} \\
& \text{Answer: } \frac{\pi a^3 \sqrt{\cos \alpha}}{12 \sin \frac{\alpha}{2}}
\end{aligned}
\]