[ Volume Ratios ]
$\qquad$ The base of the right prism $K L M N K 1 L 1 M 1 N 1$ is a rhombus $K L M N$ with an angle of $60^{\circ}$ at vertex $K$. Points $E$ and $F$ are the midpoints of edges $L L 1$ and $L M$ of the prism. The edge $S A$ of a regular square pyramid $S A B C D$ ( $S$ - apex) lies on the line $L N$, vertices $D$ and $B$ lie on the lines $M M 1$ and $E F$ respectively. Find the ratio of the volumes of the prism and the pyramid, if $S A=2 A B$
The line $L N$ is perpendicular to two intersecting lines $K M$ and $L L 1$ in the plane $M M 1 K 1 K$, so the line $L N$ is perpendicular to this plane. Therefore, any line passing through point $P$ and perpendicular to $N L$ (or coinciding with it, the line $S A$), lies in the plane $M M 1 K 1 K$. It is known that the lateral edge of a regular quadrilateral pyramid is perpendicular to the diagonal of the base that it intersects. Moreover, if a line $l$ and a plane $\alpha$ are perpendicular to the same line, then the line $l$ either lies in the plane $\alpha$ or is parallel to it. The intersecting lines $S A$ and $B D$ are perpendicular, and the plane $M M 1 K 1 K$ is perpendicular to the line $S A$, so the line $B D$ either lies in the plane $M M 1 K 1 K$ or is parallel to it. The second case is excluded because, according to the problem, point $D$ lies on the line $M M 1$, i.e., it is a common point of the line $B D$ and the plane $M M 1 K 1 K$. Therefore, the line $B D$ lies in the plane $M M 1 K 1 K$. At the same time, point $B$ lies in the plane $M M 1 L 1 L$, since it lies on the line $E F$ of this plane. Therefore, point $B$ lies on the line $M M 1$ of intersection of the planes $M M 1 K 1 K$ and $M M 1 L 1 L$. Then $M$ is the midpoint of the diagonal of the base $A B C D$ of the pyramid. Therefore, $M P$ is the common perpendicular of the intersecting lines $S A$ and $B D$. Let $A B=a$. Then

$$
S A=2 a, A M=M D=\frac{1}{2} B D=\frac{a \sqrt{2}}{2}, S M=\sqrt{S A^{2}-A M^{2}}=\sqrt{4 a^{2}-\frac{\alpha^{2}}{2}}=\frac{a \sqrt{7}}{\sqrt{2}},
$$

$$
\begin{gathered}
M P=\frac{\frac{A M \cdot S M}{S A}}{S A}=\frac{\frac{a}{\sqrt{2}} \cdot \frac{\cdot \sqrt{7}}{\sqrt{2}}}{2 a}=\frac{a \sqrt{7}}{4}, \\
L P=M P \operatorname{tg} \angle L M P=M P \operatorname{tg} 30^{\circ}=\frac{a \sqrt{7}}{4} \cdot \frac{1}{\sqrt{3}}=\frac{a \sqrt{7}}{4 \sqrt{3}}, \\
S_{K L M N}=\frac{1}{2} K M \cdot L N=\frac{1}{2} \cdot 2 M P \cdot 2 L P=2 M P \cdot L P=2 \cdot \frac{a \sqrt{7}}{4} \cdot \frac{a \sqrt{7}}{4 \sqrt{3}}=\frac{7 a^{2}}{8 \sqrt{3}} .
\end{gathered}
$$

From the equality of triangles $B M F$ and $E L F$, it follows that $E L=M B=M D=\frac{a}{\sqrt{2}}$, so $L L 1=2 E L=a \sqrt{2}$. Let $V 1$ and $V 2$ be the volumes of the prism KLMNK1L1M1N1 and the pyramid $S A B C D$. Then

$$
\begin{aligned}
& V 1=S_{K L M N} \cdot L L 1=\frac{7 a^{2}}{8 \sqrt{3}} \cdot a \sqrt{2}=\frac{7 a^{3} \sqrt{2}}{3 \sqrt{3}} \\
& V 2=\frac{1}{3} S_{A B C D} \cdot S M=\frac{1}{3} a 2 \cdot \frac{a \sqrt{7}}{\sqrt{2}}=\frac{a^{3} \sqrt{7}}{3 \sqrt{2}}
\end{aligned}
$$

Therefore,

$$
\frac{v_{1}}{v_{2}}=\frac{\frac{7 a^{3} \sqrt{2}}{\frac{3}{3}}}{\frac{a^{3} \sqrt{7}}{3 \sqrt{2}}}=\frac{\sqrt{21}}{4}
$$

## Answer

$\frac{\sqrt{21}}{4}$