12.416 A sphere is inscribed in a cone such that their surfaces touch. The volume of the body enclosed between them is 8 times smaller than the volume of the sphere. Find the angle at the vertex of the axial section of the cone.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-296.jpg?height=338&width=350&top_left_y=257&top_left_x=132)

Thus, applying known formulas for volume calculations, we have:

$$
V_{C}=\frac{2}{3} \pi R^{2} \cdot M K, V_{T_{1}}=\frac{1}{3} \pi \cdot A K^{2} \cdot F K, \quad V_{T_{2}}=\frac{1}{3} \pi \cdot A K^{2} \cdot O K
$$

But $A K=R \cdot \cos \angle O A K=R \cdot \cos \alpha, M K=R-O K=R-R \sin \alpha$,

$$
F K+O K=F O=\frac{R}{\sin \alpha} \Rightarrow V_{T} \approx V_{T_{1}}+V_{T_{2}}-V_{C}=
$$

$$
=\frac{1}{3} \pi \cdot A K^{2}(F K+O K)-\frac{2}{3} \pi R^{2} \cdot M K=\frac{\pi R^{3}}{3}\left(\frac{\cos ^{2} \alpha}{\sin \alpha}-2(1-\sin \alpha)\right)=\frac{1}{6} \pi R^{3} \text { (the last }
$$

equality follows from the problem statement) $\Rightarrow$

$$
\Rightarrow 2 \sin ^{2} \alpha-5 \sin \alpha+2=0 \Rightarrow \sin \alpha=\frac{1}{2}, \alpha=\frac{\pi}{6}, \angle A F B=\frac{\pi}{3}
$$

Answer: $\frac{\pi}{3}$.