[ Volume Ratios

In the parallelepiped $A B C D A 1 B 1 C 1 D 1$, all edges are equal to each other, all planar angles at vertex $A$ are acute and equal to each other. Plane $P$ passes through vertex $A$ and intersects the lateral edges $B B 1, C C 1$, and $D D 1$ at points $K, L$, and $M$ respectively. The areas of figures $A K B, A M D, D M L C$, and the area of the lower base $A B C D$ form a geometric progression in the given order. Find the ratio of the volume of the cut-off part $A B C D K L M$ to the volume of the entire parallelepiped.

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All faces of the parallelepiped $A B C D A 1 B 1 C 1 D 1$ are equal rhombuses. Let the side of each of them be $a$, and the height $h$. Denote $B K=x, D M=y, C L=z$. Since the numbers $S_{\triangle A K B}=\frac{1}{2} x h, S_{\triangle A M D}=\frac{1}{2} y h, S_{D M L C}=\frac{1}{2}(y+z) h$ and $S_{A B C D}=$ $a h$ form a geometric progression in the given order, $S 2_{\triangle A M D}=S_{\triangle A K B} \cdot S_{D M L C}$ and $S 2_{D M L C}=S_{\triangle A M D} \cdot$ $S_{A B C D}$, or

$$
\begin{gathered}
\frac{\mathbf{1}}{4} y 2 h 2=\frac{1}{2} x h \cdot \frac{1}{2}(y+z) h, \frac{1}{\mathbf{4}}(y+z) 2 h 2=12 y h \cdot a h, \\
y 2=x(y+z),(y+z) 2=2 a y .
\end{gathered}
$$

When two parallel planes are intersected by a third plane, parallel lines are formed, so the quadrilateral $A K L M$ is a parallelogram. Draw a line through point $K$ parallel to $B C$. Let this line intersect the edge $C C 1$ at point $E$. Triangles $K L E$ and $A M D$ are equal, so $E L=M D=y$ and $x=B K=$ $C E=C L-E L=z-y$. Thus, considering $a$ known, we get the system

$$
\left\{\begin{array}{l}
y^{2}=x(y+z) \\
(y+z)^{2}=2 a y \\
x=z-y
\end{array}\right.
$$

where $x, y, z$ take only positive values. Substituting $x=z-y$ into the first equation, we get that $z=y \sqrt{2}$. From the second equation, we find that $y=2 a(3-2 \sqrt{2})$. Then $z=2 a(3 \sqrt{2}-4), x=2 a(5 \sqrt{2}-7)$. Since the volume $V$ of the parallelepiped is the product of the base area and the height, and all faces of the given parallelepiped are equal, all heights of the parallelepiped are also equal. The plane of the diagonal section $A A 1 C 1 C$ divides the polyhedron $A B C D K L M$ into two quadrilateral pyramids: $A B C L K$ with base $B C L K$ and $A C L M D$ with base $C L M D$, and the heights of these pyramids are equal to the height of the parallelepiped. Denote it by $H$, then

$$
\begin{gathered}
V_{A B C L K}=\frac{1}{3} S_{B C L K} \cdot H=\frac{1}{3} \cdot \frac{1}{2}(B K+C L) h \cdot H=\frac{1}{\overline{6}}(x+z) h \cdot H= \\
=\frac{1}{\overline{6}}(2 a(5 \sqrt{2}-7)+2 a(3 \sqrt{2}-4)) h \cdot H=\frac{1}{3} a h H(8 \sqrt{2}-11)=\frac{1}{3}(8 \sqrt{2}-11) V \\
V_{A C L M D}=\frac{1}{3} S_{C L M D} \cdot H=\frac{1}{3} \cdot \frac{1}{2}(M D+C L) h \cdot H=\frac{1}{6}(y+z) h \cdot H= \\
=\frac{1}{\overline{6}}(2 a(3-2 \sqrt{2})+2 a(3 \sqrt{2}-4)) h \cdot H=\frac{1}{3} a h H(\sqrt{2}-1)=\frac{1}{3}(\sqrt{2}-1) V, \\
V_{A B C D K L M}=V_{A B C L K}+V_{A C L M D}=\frac{1}{3}(8 \sqrt{2}-11) V+\frac{1}{3}(\sqrt{2}-1) V=(3 \sqrt{2}-4) V .
\end{gathered}
$$

Therefore,

$$
\frac{V_{A B C D K L M}}{V_{A B C D A_{1} B_{1} C_{1} D_{1}}}=3 \sqrt{2}-4
$$

## Answer

$3 \sqrt{2}-4$