12.216. The perpendicular dropped from the center of the base of a cone to its generatrix rotates around the axis of the cone. Find the angle between its generatrix and the axis, if the surface of revolution divides the volume of the cone in half.
Solution.

Let $\triangle A B C$ be the axial section of the given cone (Fig. 12.83), $B O$ be its height, and $O M$ be the perpendicular dropped from the center of the base of the cone to the generatrix $A B$. As a result of rotating $O M$ around the line $B O$, a cone is formed, the axial section of which is $\triangle M O L, M L \perp B O$. The plane of the base of the resulting cone cuts off a cone from the given cone, the axial section of which is $\triangle M B L$. The point $K$ of intersection of $M L$ and $B O$ is the center of the common base of the cone obtained as a result of rotating the segment $O M$ and the cone cut off from the given one. The sum of the volumes $V_{1}$ and $V_{2}$ of these cones is $1 / 2$ of the volume $V$ of the given cone.

Let $\angle A B O=\alpha, 0^{\circ}<\alpha<90^{\circ}, A O=R$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0965.jpg?height=458&width=717&top_left_y=76&top_left_x=320)

Fig. 12.83

$\angle K L O=\angle A O L=\angle A B O=\alpha$.

In $\triangle A L O\left(\angle A L O=90^{\circ}\right): O L=O A \cdot \cos \angle A O L=R \cos \alpha$.

In $\triangle L K O\left(\angle L K O=90^{\circ}\right): L K=O L \cdot \cos \angle K L O=R \cos ^{2} \alpha$.

$V_{1}+V_{2}=\frac{1}{3} \pi L K^{2} \cdot O K+\frac{1}{3} \pi L K^{2} \cdot B K=\frac{1}{3} \pi L K^{2}(O K+B K)=\frac{1}{3} \pi L K^{2} \cdot B O$. $V=\frac{1}{3} \pi \cdot O A^{2} \cdot B O$.

Since $V_{1}+V_{2}=\frac{1}{2} V$, then $\frac{1}{3} \pi L K^{2} \cdot B O=\frac{1}{2} \cdot \frac{1}{3} \pi \cdot O A^{2} \cdot B O ; L M^{2}=\frac{1}{2} O A^{2}$;

$R^{2} \cos ^{4} \alpha=\frac{1}{2} R^{2} ; \cos ^{4} \alpha=\frac{1}{2} ; \cos \alpha=\frac{1}{\sqrt[4]{2}} ; \alpha=\arccos \left(\frac{1}{\sqrt[4]{2}}\right)$

Answer: $\arccos \left(\frac{1}{\sqrt[4]{2}}\right)$