12.313. The height of the cone is $H$, and the angle between the generatrix and the base plane is $\alpha$. The total surface area of this cone is divided in half by a plane perpendicular to its height. Find the distance from this plane to the base of the cone.
Solution.

Let $R$ be the radius of the base of the given cone, $L$ its slant height, $r, l, h$ - respectively the radius of the base, slant height, and height of the truncated cone.

Then $R=H \operatorname{ctg} \alpha, L=\frac{H}{\sin \alpha}, r=h \operatorname{ctg} \alpha, l=\frac{h}{\sin \alpha}$, the total surface area of the cone $S=\pi R(R+L)=\pi \cdot H \operatorname{ctg} \alpha\left(H \operatorname{ctg} \alpha+\frac{H}{\sin \alpha}\right)=\frac{\pi H^{2} \cos \alpha(\cos \alpha+1)}{\sin ^{2} \alpha}$, the lateral surface area of the truncated cone $S_{1}=\pi r l=\frac{\pi h^{2} \operatorname{ctg} \alpha}{\sin \alpha}=\frac{\pi h^{2} \cos \alpha}{\sin ^{2} \alpha}$.

Since by the condition $S=2 S_{1}$, then $H^{2}(\cos \alpha+1)=2 h^{2}$;

$$
h^{2}=H^{2} \cdot \frac{1+\cos \alpha}{2}=H^{2} \cos ^{2} \frac{\alpha}{2} ; h=H \cos \frac{\alpha}{2} .
$$

The distance from the cutting plane to the base of the cone $x=H-h=$ $=H\left(1-\cos \frac{\alpha}{2}\right)=2 H \sin ^{2} \frac{\alpha}{4}$.

Answer: $2 H \sin ^{2} \frac{\alpha}{4}$.