A right circular cone is inscribed in a three-sided pyramid. The base of the pyramid is an isosceles triangle, with the angle at the base being $\alpha\left(=58^{\circ}\right)$, and the volume of the pyramid is $V\left(=671.6 \mathrm{~m}^{3}\right)$. What is the volume of the cone?
![](https://cdn.mathpix.com/cropped/2024_05_02_e81362a6e85a2ce5ba01g-1.jpg?height=482&width=1092&top_left_y=152&top_left_x=511)

The apex of the cone and the apex of the pyramid are common, their height \( m \) is the same, so the ratio of their volumes \( V_{k}: V \) is equal to the ratio of the areas of their bases. The base circle of the cone is the incircle of the base triangle of the pyramid, so its radius is

\[
\varrho = \frac{a}{2} \operatorname{tg} \frac{\alpha}{2}
\]

where \( a \) is the base segment of the base. The height segment of the base triangle is \( m_{a} = (a / 2) \operatorname{tg} \alpha \), so the mentioned ratio, and finally the sought volume is

\[
\frac{V_{k}}{V} = \frac{\pi \varrho^{2}}{a m_{a} / 2} = \pi \operatorname{tg}^{2} \frac{\alpha}{2} \operatorname{cotg} \alpha, \quad V_{k} = V \pi \operatorname{tg}^{2} \frac{\alpha}{2} \operatorname{cotg} \alpha
\]

With the numerical values, \( V_{k} = 405.1 \, \text{m}^3 \).

Note. We did not use the fact that it was a right cone, i.e., a cone of revolution.

Gábor Perémy (Budapest, Szilágyi E. g. III. o. t.)