2.33. In a regular quadrilateral pyramid, the dihedral angle at the lateral edge is $120^{\circ}$. Find the lateral surface area of the pyramid if the area of its diagonal section is $S$.
### 2.33. The sought lateral surface is expressed as: $S_{\text {side}}=2 A B \cdot S M$ (Fig. 2.28). Let $A B=a, S O=h$; we need to find the relationship between $a$ and $h$. Draw $A K \perp S B$, $C K \perp S B$; since $\angle A K C=120^{\circ}$, then $\angle A K O=60^{\circ}$ and from $\triangle A K O$ we get $O K=A O \operatorname{ctg} 60^{\circ}=\frac{a}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}}=\frac{a}{\sqrt{6}}$. Next, in the right triangle $S O B$ we have $O K \perp S B$, therefore

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-472.jpg?height=530&width=492&top_left_y=1004&top_left_x=660)

Fig. 2.28 $S B: S O=S O: S K$, i.e., $h^{2}=S B \cdot S K$.

From $\triangle S K O$ it follows that $S K^{2}=S O^{2}-O K^{2}=h^{2}-\frac{a^{2}}{6}$, hence $S B^{2}=\frac{6 h^{4}}{6 h^{2}-a^{2}}$. On the other hand, from $\triangle S O B$ we find

$$
\begin{aligned}
& S B^{2}=h^{2}+\frac{a^{2}}{2} . \text { Thus, } \\
& \quad h^{2}+\frac{a^{2}}{2}=\frac{6 h^{4}}{6 h^{2}-a^{2}}, \text { or } 12 h^{4}-2 a^{2} h^{2}+6 a^{2} h^{2}-a^{4}=12 h^{4}
\end{aligned}
$$

from which $h=\frac{a}{2}$ and $S M=\frac{a}{\sqrt{2}}$. By the condition, $\frac{1}{2} A C \cdot S O=\frac{a \sqrt{2}}{2} \cdot h=S$, i.e., $S=\frac{a^{2} \sqrt{2}}{4}$. Considering that $S_{\text {side}}=2 A B \cdot S M=2 a \cdot \frac{a}{\sqrt{2}}=a^{2} \sqrt{2}$, we finally get $S_{\text {side}}=4 S$.

Answer: 4S.