12.424 The base of a pyramid inscribed in a cone is a quadrilateral, one side of which is equal to $a$, and each of the other three sides is equal to $b$. The vertex of the pyramid lies at the midpoint of one of the generatrices. Find the volume of the pyramid if the angle between the generatrix and the height of the cone is $\alpha$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-302.jpg?height=333&width=378&top_left_y=117&top_left_x=117)

Let the base of the given pyramid $FABCD$ be a quadrilateral $ABCD$ inscribed in a circle of radius $R$, which is the base of the given cone, $AB = BC = CD = b, AD = a, P$ - the point of intersection of the diagonals $AC$ and $BD$. Let $\angle ADP = \beta$.

Then $\angle PDC = \angle PBC = \beta$ (Theorem 6) $\Rightarrow$ $\Rightarrow AD \parallel BC, \angle B = 180^\circ - 2\beta \Rightarrow$

$\Rightarrow S_{ABCD} = S_{ACD} + S_{ABC} = \frac{1}{2} ab \sin 2\beta + \frac{1}{2} b^2 \sin 2\beta$.

On the other hand,

$S_{ABCD} = \frac{AC \cdot BD}{2} \cdot \sin \angle APD = \frac{(AC)^2}{2} \sin 2\beta \quad \left(AC = BD, \angle APD = 180^\circ - 2\beta\right) \Rightarrow$

$\Rightarrow AC = \sqrt{b^2 + ab}$. Additionally, by the Law of Sines for $\triangle ACD, R = \frac{AC}{2 \sin 2\beta}$.

From the problem statement, it follows that the height of the pyramid is half the height of the cone, i.e., $\frac{1}{2} R \cot \alpha \Rightarrow$

$$
\Rightarrow V_{FABCD} = \frac{1}{6} R \cot \alpha \cdot S_{ABCD} = \frac{1}{6} \cdot \frac{AC}{2 \sin 2\beta} \cdot \cot \alpha \cdot \frac{(AC)^2}{2} \cdot \sin 2\beta
$$

Answer: $\frac{\cot \alpha}{24} \left(b^2 + ab\right)^{\frac{3}{2}}$.