A cube with an edge length of 10 units has points $A, B$, and $C$ as the endpoints of the three edges starting from one vertex. The planes $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$, parallel to the plane $ABC$, divide the cube into three parts with volume ratios of $251: 248: 251$. What is the distance between the two planes?
Let the common neighbor of points $A, B, C$ be denoted by $O$. The volume of the tetrahedron $OABC$ is $\frac{1}{6} \cdot OA \cdot OB \cdot OC = \frac{1000}{6} < 251$.

$\overline{251+248+251} \cdot 1000$, so the plane $\mathcal{S}_{1}$ is farther from $O$ than the plane $ABC$. Let the plane $\mathcal{S}_{1}$ intersect the edges of the cube starting from $O$ at points $A_{1}, B_{1}, C_{1}$, respectively. The edges of the tetrahedron $OA_{1}B_{1}C_{1}$ starting from $O$ are of equal length. Let $AA_{1} = BB_{1} = CC_{1} = x$. The volume of the part cut off from the cube by the plane $\mathcal{S}_{1}$ - that is, the common volume of the tetrahedron $OA_{1}B_{1}C_{1}$ and the cube - is obtained by subtracting the volumes of three small tetrahedra cut off from the large tetrahedron by the faces of the cube at points $A$, $B$, and $C$ (Figure 1). The edges of these small tetrahedra meeting at $A$, $B$, and $C$ are pairwise perpendicular, and their lengths are $x-10$, so the volume of the part cut off from the cube by the plane $\mathcal{S}_{1}$ is:

$$
\frac{1}{6} x^{3} - 3 \cdot \frac{1}{6} (x-10)^{3}
$$

According to the condition, this is $\frac{251}{750}$ of the volume of the cube, so

$$
\frac{1}{6} x^{3} - 3 \cdot \frac{1}{6} (x-10)^{3} = \frac{251}{750} \cdot 1000
$$

Rearranging this equation, we get

$$
x^{3} - 45 x^{2} + 450 x - 496 = 0
$$

There is a general solution formula for cubic equations, similar to the quadratic formula. However, this equation can be solved without this knowledge. If the equation has an integer root, it must divide the constant term, 496. Since $496 = 2^{4} \cdot 31$, by checking the divisors, we find that $x = 31$ is a root of the equation. Then the left side of the equation can be factored as:

$$
x^{3} - 45 x^{2} + 450 x - 496 = (x-31)(x^{2} - 14 x + 16)
$$

Thus, the roots of the equation are: $x_{1} = 31$, $x_{2,3} = 7 \pm \sqrt{33}$.

Since the plane $\mathcal{S}_{1}$ is farther from $O$ than the plane $ABC$, but closer to $O$ than the opposite vertex of the cube, we have $10 < x < \frac{10 \sqrt{3}}{2}$. Only the solution $x = 7 + \sqrt{33}$ satisfies these conditions. By the central symmetry of the cube, the plane $\mathcal{S}_{2}$ also cuts off equal segments from the edges starting from the opposite vertex of the cube. The distance between the planes $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ is obtained by subtracting twice the height of the tetrahedron $OA_{1}B_{1}C_{1}$ from the length of the space diagonal of the cube.

The side length of the equilateral triangle $A_{1}B_{1}C_{1}$ is $x \cdot \sqrt{2}$, so the area of the triangle is $\frac{x^{2} \cdot 2 \cdot \sqrt{3}}{4}$. If we denote the height from $O$ by $m$, then the volume of the tetrahedron can be written in two ways:

$$
\frac{1}{3} \cdot \frac{x^{2} \cdot 2 \sqrt{3}}{4} \cdot m = \frac{1}{6} \cdot x^{3}
$$

Thus, $m = \frac{x}{\sqrt{3}} = \frac{7 \sqrt{3}}{3} + \sqrt{11}$.

Therefore, the distance between the planes $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ is

$$
10 \sqrt{3} - 2 m = \frac{16 \sqrt{3}}{3} - 2 \sqrt{11} \approx 2.6044 \text{ units. }
$$

Based on the work of Illés Horváth (Fazekas M. Főv. Gyak. Gimn., 11th grade).

![](https://cdn.mathpix.com/cropped/2024_05_02_8a4a9385ade6adc9c230g-1.jpg?height=669&width=748&top_left_y=2056&top_left_x=732)

![](https://cdn.mathpix.com/cropped/2024_05_02_8a4a9385ade6adc9c230g-2.jpg?height=512&width=661&top_left_y=129&top_left_x=713)