12.339. The acute angle of the rhombus at the base of a quadrilateral pyramid is $\alpha$. The ratio of the total surface area of the pyramid to the square of the side of the base is $k$. Find the sine of the angle between the apothem and the height of the pyramid, given that all its lateral faces are equally inclined to the base plane. What are the permissible values of $k$?
Solution.

If $EO$ is the height of the pyramid $EABCD$ (Fig. 12.200), and $EF$ is the height of its lateral face $CED$, then $\angle EFO$ is the angle of inclination of this lateral face to the plane of the base, and since all lateral faces are equally inclined to it, $O$ is the center of the circle inscribed in the base of the pyramid, which is the point of intersection of the diagonals of the rhombus, $S_{\sigma}=\frac{S_{\text{base}}}{\cos \beta}$, where $\beta=\angle EFO$.

Let $AB=a$. Then $S_{\text{base}}=a^{2} \sin \alpha$ and the total surface area of the pyramid $S=S_{\text{base}}+S_{6}=S_{\text{base}}+\frac{S_{\text{base}}}{\cos \beta}=S_{\text{base}} \cdot \frac{1+\cos \beta}{\cos \beta}=\frac{a^{2}(1+\cos \beta) \sin \alpha}{\cos \beta} \Rightarrow k=$ $=\frac{S}{a^{2}}=\frac{(1+\cos \beta) \sin \alpha}{\cos \beta} \Leftrightarrow k \cos \beta=\sin \alpha+\cos \beta \sin \alpha ; \cos \beta=\frac{\sin \alpha}{k-\sin \alpha}$.

In $\triangle EOF\left(\angle EOF=90^{\circ}\right)$, the required $\sin \angle OEF=\cos \angle EFO=\frac{\sin \alpha}{k-\sin \alpha}$.

Since $\frac{\sin \alpha}{k-\sin \alpha} < 2 \sin \alpha$.

Answer: $\frac{\sin \alpha}{k-\sin \alpha} ; k>2 \sin \alpha$.