3.40. The ratio of one side of the base of a triangular pyramid to each of the other five edges of the pyramid is $k$. Find the dihedral angle between two equal lateral faces of the pyramid and the permissible values of $k$.


3.40. Given that $S A B C$ is a triangular pyramid, $\frac{A C}{A B}=\frac{A C}{B C}=$ $=\frac{A C}{S A}=\frac{A C}{S B}=\frac{A C}{S C}=k$ (Fig. 3.42). Since

$A B=B C=S A=S B=S C=\frac{A C}{k}$,

then $\triangle A S B=\triangle B S C$ (by three sides). Draw $A D \perp S B$ and connect points $D$ and $C$. Then $\triangle A S D=$ $=\triangle S D C$ (since $S A=S C$, side $S D$ is common and $\angle A S D=$ $=\angle D S C)$, from which it follows that $\angle S D C=\angle S D A=90^{\circ}$. Therefore, $\angle A D C$ is the linear angle of the dihedral angle $A S D C$, i.e., the required angle. Let $\angle A D C=x$. From $\triangle A D C$ by the cosine theorem, we find $A C^{2}=$ $=2 A D^{2}-2 A D^{2} \cos x$, where

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-524.jpg?height=558&width=510&top_left_y=413&top_left_x=645)

Fig. 3.42 $A D=\frac{\sqrt{3} S A}{2}=\frac{\sqrt{3} A C}{2 k}$, since $\triangle A S B$ is equilateral. Thus,

$$
\begin{gathered}
A C^{2}=2 \cdot \frac{3 A C^{2}}{4 k^{2}}(1-\cos x) ; 1-\cos x=\frac{2 k^{2}}{3} ; 2 \sin ^{2} \frac{x}{2}=\frac{2 k^{2}}{3} \\
\sin \frac{x}{2}=\sqrt{\frac{k^{2}}{3}}=\frac{k \sqrt{3}}{3}
\end{gathered}
$$

Therefore, $x=2 \arcsin \frac{k \sqrt{3}}{3}$, where $0<\frac{k \sqrt{3}}{3}<1$, i.e., $0<k<\sqrt{3}$.