8,9

In what ratio does a plane, perpendicular to the diagonal of a cube and dividing the diagonal in the ratio: a) 2:1; b) $3: 1$, divide the volume of the cube?
It is known that a plane passing through the ends of three edges of a parallelepiped, emanating from one vertex, divides the diagonal of the parallelepiped, emanating from this vertex, in the ratio 1:2. Moreover, if the parallelepiped is a cube, then this plane is perpendicular to the specified diagonal. Let the volume of the cube ABCDA1B1C1D1 be $V$. 1) Consider the plane passing through points A1, B, and C1. This plane is perpendicular to the diagonal DB1 and divides it in the ratio 1:2, i.e., if $P$ is the point of intersection of the diagonal with the specified plane, then B1P:PD = 1:2. Then the volume of the tetrahedron A1BC1B1 is ${ }^{\frac{1}{6} \mathrm{~V}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-09.jpg?height=89&width=1929&top_left_y=2334&top_left_x=13)
The diagonal is intersected by the point $Q$ in the ratio $B 1 Q: Q D=1: 3$, and is parallel to the plane $A 1 B C 1$. Therefore, the plane $\alpha$ cuts off a tetrahedron from the cube, similar to the tetrahedron A1BC1B1, with a similarity coefficient of $\frac{\frac{B}{1}^{B_{1} Q}}{B_{1}^{P}}=\frac{\frac{1}{\frac{1}{3}}}{\frac{1}{3}}=\frac{3}{4}$. Therefore, the volume of the tetrahedron cut off by the plane $\alpha$ is $\stackrel{\left.\frac{3}{4}\right)}{\mathbf{4}} \cdot \frac{1}{6} \mathrm{~V}=\frac{9}{128}$. Thus, the plane $\alpha$ divides the volume of the cube in the ratio $\frac{g}{128-9}=\frac{9}{119}$.

## Answer

$1: 5 ; 9: 119$.

## Problem