12.126. Find the angle in the axial section of a cone if a sphere with its center at the vertex of the cone, touching its base, divides the volume of the cone in half.
## Solution.

The volume of the sector $V_{\text {sec }}=\frac{2}{3} \pi R_{w}^{2} \cdot h$ (Fig. 12.126). The volume of the cone $V_{\mathrm{k}}=\frac{1}{3} \pi R^{2} \cdot H$. According to the condition $V_{\mathrm{sec}}=\frac{V_{\mathrm{k}}}{2} ; \frac{2}{3} \pi R_{\mathrm{m}}^{2} h=\frac{1}{6} \pi R^{2} H$. The height of the cone $H=R_{w}$; the height of the segment $h=R_{w}-S A$. From $\triangle S A B$ we find:
$S A=R_{\mathrm{m}} \cdot \cos \frac{\alpha}{2} \cdot$ From this, $h=R_{\mathrm{m}}\left(1-\cos \frac{\alpha}{2}\right) \cdot$ From $\triangle S O C$ we find:

$R=O C=R_{w} \operatorname{tg} \frac{\alpha}{2}$. Substituting: $\frac{2}{3} \pi R_{m}^{3}\left(1-\cos \frac{\alpha}{2}\right)=\frac{1}{6} \pi R_{w}^{3} \operatorname{tg}^{2} \frac{\alpha}{2} \Rightarrow$ $\Rightarrow 4\left(1-\cos \frac{\alpha}{2}\right)=\operatorname{tg}^{2} \frac{\alpha}{2} \cdot 4\left(1-\cos \frac{\alpha}{2}\right)=\frac{1}{\cos ^{2} \alpha / 2}-1 ; 4\left(1-\cos \frac{\alpha}{2}\right)=\frac{1-\cos ^{2} \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}} ;$ $4 \cos ^{2} \frac{\alpha}{2}\left(1-\cos \frac{\alpha}{2}\right)=\left(1-\cos \frac{\alpha}{2}\right)\left(1+\cos \frac{\alpha}{2}\right) \Rightarrow 4 \cos ^{2} \frac{\alpha}{2}=1+\cos \frac{\alpha}{2}$.

$4 \cos ^{2} \frac{\alpha}{2}-\cos \frac{\alpha}{2}-1=0$. Let $\cos \frac{\alpha}{2}=t$. Solving the equation $4 t^{2}-t-1=0$, we get $t_{1}=\frac{1+\sqrt{17}}{8}$ and $t_{2}=\frac{1-\sqrt{17}}{2}$ (does not fit). Therefore, $\cos \frac{\alpha}{2}=\frac{1+\sqrt{17}}{8} \Rightarrow \frac{\alpha}{2}=\arccos \frac{1+\sqrt{17}}{8}$. Then $\alpha=2 \arccos \frac{1+\sqrt{17}}{8}$. Answer: $\quad 2 \arccos \frac{1+\sqrt{17}}{8}$.