12.222. The base of the pyramid is a rhombus, one of the angles of which is $\alpha$. The lateral faces are equally inclined to the plane of the base. A plane is drawn through the midpoints of two adjacent sides of the base and the vertex of the pyramid, forming an angle $\beta$ with the plane of the base. The area of the resulting section is $S$. Find the side of the rhombus.
## Solution.

Let the rhombus $ABCD$ be the base of the pyramid $LABCD$ (Fig. 12.89), $\angle BAD=\alpha$, $LO$ - the height of the pyramid. Since the lateral faces are equally inclined to the base plane, $O$ is the center of the circle inscribed in the rhombus, the point of intersection of its diagonals. Let $E$ be the midpoint of $AB$, $F$ - the midpoint of $AD$, $S_{ELF}=S$, $K$ - the point of intersection of $EF$ and $AC$, $AB=a$.

$EF \| BD, EF=\frac{1}{2} BD=BO=a \sin \frac{\alpha}{2}, EF \perp KO, K-$ the midpoint of $EF. KO$ is the projection of $KL$ on the base plane. Then $KL \perp EF, \angle LKO$ is the angle between the plane $ELF$ of the section and the base plane of the pyramid, $\angle LKO=\beta. KO=\frac{1}{2} AO=\frac{1}{2} a \cos \frac{\alpha}{2}$.

In $\triangle KOL\left(\angle KOL=90^{\circ}\right): KL=\frac{KO}{\cos \angle LKO}=\frac{a \cos \frac{\alpha}{2}}{2 \cos \beta}$. $S_{\triangle ELF}=\frac{1}{2} EF \cdot LK \Rightarrow \frac{1}{2} a \sin \frac{\alpha}{2} \cdot \frac{a \cos \frac{\alpha}{2}}{2 \cos \beta}=S ; \frac{a^{2} \sin \alpha}{8 \cos \beta}=S ; a=2 \sqrt{\frac{2 S \cos \beta}{\sin \alpha}}$. Answer: $2 \sqrt{\frac{2 S \cos \beta}{\sin \alpha}}$.