12.284. In a regular triangular pyramid, a plane is drawn through a lateral edge and the height. The ratio of the area of the section to the total surface area of the pyramid is $k$. Find the dihedral angle at the base and the permissible values of $k$.

In a regular triangular pyramid, a plane is drawn through a lateral edge and the height. The ratio of the area of the section to the total surface area of the pyramid is $k$. Find the dihedral angle at the base and the permissible values of $k$.
## Solution.

Let $\triangle D E C$ be the section of the regular pyramid $D A B C$ (Fig. 12.147) by a plane passing through the lateral edge $D C$ and the height of the pyramid $D O$. Then $E$ is the midpoint of $A B, \angle D E O$ is the linear angle of the dihedral angle at the base.

If $\angle D E O=\alpha, A B=1$, then the area of the base $S_{\text {base }}=\frac{A B^{2} \sqrt{3}}{4}=$ $=\frac{\sqrt{3}}{4}$, the lateral surface area $S_{\sigma}=\frac{S_{\text {base }}}{\cos \alpha}$, the total surface area of the pyramid

$$
S_{1}=S_{\mathrm{base}}+S_{\sigma}=S_{\mathrm{base}}+\frac{S_{\mathrm{base}}}{\cos \alpha}=S_{\mathrm{base}} \cdot \frac{1+\cos \alpha}{\cos \alpha}=\frac{\sqrt{3}}{4} \cdot \frac{2 \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}=\frac{\sqrt{3} \cos ^{2} \frac{\alpha}{2}}{2 \cos \alpha} .
$$

$$
O E=\frac{A B \sqrt{3}}{6}=\frac{\sqrt{3}}{6}, C E=\frac{A B \sqrt{3}}{2}=\frac{\sqrt{3}}{2} \text {. }
$$

In $\triangle D O E\left(\angle D O E=90^{\circ}\right): D O=O E \operatorname{tg} \angle D E O=\frac{\sqrt{3}}{6} \operatorname{tg} \alpha$.

The area of the section $S_{2}=\frac{1}{2} C E \cdot D O=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{6} \operatorname{tg} \alpha=\frac{\operatorname{tg} \alpha}{8}$.

Since $\frac{S_{2}}{S_{1}}=k$, then $\frac{\operatorname{tg} \alpha}{8} \cdot \frac{2 \cos \alpha}{\sqrt{3} \cos ^{2} \frac{\alpha}{2}}=k ; \frac{\sin \alpha}{\cos ^{2} \frac{\alpha}{2}}=4 k \sqrt{3} ; \frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}}=$ $=4 k \sqrt{3} ; \operatorname{tg} \frac{\alpha}{2}=2 k \sqrt{3} ; \alpha=2 \operatorname{arctg}(2 k \sqrt{3}) ; 0<2 k \sqrt{3}<1$, i.e., $0<k<\frac{\sqrt{3}}{6}$. Answer: $2 \operatorname{arctg}(2 k \sqrt{3}) ; 0<k<\frac{\sqrt{3}}{6}$.