Solution. We can assume that the length of the edges of the cube is 1. Then $A K=1 / 3, B L=1 / 4$ and $C M=1 / 5$. Let $N$ be the intersection point of the edge $D H$ and the plane $K L M$, $\mathcal{S}$ the plane passing through $M$, and $\mathcal{T}$ the horizontal plane passing through $K$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c7bcc98c05b0333993edg-1.jpg?height=583&width=553&top_left_y=305&top_left_x=775)

Then $\mathcal{S}$ cuts a right prism $\mathcal{H}_{1}$ with base $A B C D$ from the cube, whose height is $C M$, so its volume is $V_{1}=1 / 5$. The plane $\mathcal{T}$ cuts a right prism $\mathcal{H}_{2}$ with base $E F G H$ from the cube, whose height is $E K=A E-A K$, so its volume is $V_{2}=2 / 3$. The part of the cube between the planes $\mathcal{S}$ and $\mathcal{T}$ is also a right prism, denoted by $\mathcal{H}_{3}$. Its volume is

$$
V_{3}=1-\left(V_{1}+V_{2}\right)=1-\left(\frac{1}{5}+\frac{2}{3}\right)=\frac{2}{15}
$$

We will show that the plane $K L M N$ bisects the volume of $\mathcal{H}_{3}$. The lines $K L$ and $M N$ are coplanar, but since they lie on two parallel faces of the cube, one on $A B F E$ and the other on $D C G H$, they cannot intersect. Therefore, the two lines are parallel. Similarly, we can show that the lines $L M$ and $K N$ are also parallel. Thus, the quadrilateral $K L M N$ is a parallelogram because its opposite sides are parallel. Then, however, its diagonals bisect each other. Let the intersection point of the diagonals of $K L M N$ be $P$. Then $P$ is also the midpoint of the body diagonal $K M$ of the right prism $\mathcal{H}_{3}$, which is the center of symmetry of the prism, because the midpoints of the four body diagonals of the prism coincide. The central reflection about $P$ therefore maps $\mathcal{H}_{3}$ onto itself, swapping the point pairs $(K, M)$ and $(L, N)$. Thus, it swaps the parts of $\mathcal{H}_{3}$ below and above the plane $K L M N$. Therefore, the two parts are congruent, and thus each part has a volume of $\frac{V_{3}}{2}=\frac{1}{15}$.

The volume of the part of the cube below the plane $K L M N$ is therefore

$$
V_{1}+\frac{V_{3}}{2}=\frac{1}{5}+\frac{1}{15}=\frac{4}{15}
$$

and the volume of the part above the plane $K L M N$ is

$$
V_{2}+\frac{V_{3}}{2}=\frac{2}{3}+\frac{1}{15}=\frac{11}{15}
$$

Thus, the ratio of the volumes of the two parts is $4: 11$.
The cube $A B C D E F G H$ on the table has three vertical edges divided by points $K, L$, and $M$ in the ratios $1: 2, 1: 3$, and $1: 4$ respectively, as shown in the figure. The plane $K L M$ cuts the cube into two parts. What is the ratio of the volumes of the two parts?

![](https://cdn.mathpix.com/cropped/2024_05_02_b7118c837d7a0c04ca29g-1.jpg?height=357&width=379&top_left_y=266&top_left_x=862)