Task B-3.4. A sphere is circumscribed around a right circular cone. A plane containing the center of this sphere and parallel to the base of the cone divides the cone into two parts of equal volumes. Determine the cosine of the angle at the vertex of the axial section of the cone.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4f4efb97ca3723b253dag-15.jpg?height=784&width=762&top_left_y=535&top_left_x=627)

$$
\begin{aligned}
& \frac{r^{2} \pi(R+x)}{3}=2 \cdot \frac{r_{1}^{2} \pi R}{3} \\
& r^{2} \pi(R+x)=2 \cdot r_{1}^{2} \pi R
\end{aligned}
$$

From the similarity of triangles, we have: \( r_{1}: r = R: (R+x) \), i.e.,

$$
r_{1}=r \cdot \frac{R}{R+x}
$$

Substituting this into the above equations, we get: \( r^{2}(R+x)=2 r^{2} \cdot \frac{R^{2}}{(R+x)^{2}} \cdot R \).

$$
(R+x)^{3}=2 \cdot R^{3} \quad \Rightarrow \quad R+x=R \sqrt[3]{2} \quad \Rightarrow \quad x=R(\sqrt[3]{2}-1)
$$

According to the theorem of inscribed and central angles, \(\alpha = \beta\).

Then, \(\cos \alpha = \cos \beta = \frac{x}{R} = \frac{R(\sqrt[3]{2}-1)}{R} = \sqrt[3]{2}-1\).