Let's consider three planes that are parallel to one of the faces of a unit volume tetrahedron and bisect the tetrahedron's volume. What is the volume of the tetrahedron that is bounded by these three planes and the fourth face of the tetrahedron?
Solution. It is known that if the faces of two tetrahedra are parallel, then the tetrahedra are similar to each other (Figure 1). The ratio of the volumes of similar tetrahedra is equal to the cube of the ratio of their corresponding sides.

![](https://cdn.mathpix.com/cropped/2024_05_02_459c254b9a5ac0c66c39g-1.jpg?height=314&width=569&top_left_y=266&top_left_x=759)

Figure 1

Consider the face of the unit volume tetrahedron $\mathcal{T}$ for which we did not draw a parallel plane. If this face is $ABC$, then the planes parallel to the faces of $\mathcal{T}$, which bisect its volume, intersect $ABC$ at the segments $DE$, $FG$, and $HI$. These segments (due to the parallelism of the planes) are parallel to the corresponding sides of the triangle $ABC$, and their lengths satisfy

$$
AB : DE = BC : FG = CA : HI = \sqrt[3]{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_459c254b9a5ac0c66c39g-1.jpg?height=447&width=588&top_left_y=1004&top_left_x=755)

Let the vertices of the triangle determined by the intersection points of the segments $DE$, $FG$, and $HI$ be denoted by $J$, $K$, and $L$ as shown in Figure 2. The triangle $JKL$ is one of the faces of the new tetrahedron in the problem. The quadrilaterals $AFED$ and $IBED$ are parallelograms, since, for example, $AF = \frac{1}{\sqrt[3]{2}} AB = ED$. Thus,

$$
AI = FB = AB - DE = AB \left(1 - \frac{1}{\sqrt[3]{2}}\right)
$$

The quadrilaterals $AIJD$ and $FBEK$ are also parallelograms, so

$$
\begin{aligned}
JK & = DE - (DJ + KE) = DE - (AI + FB) = \\
& = AB \left(\frac{1}{\sqrt[3]{2}} - 2 \left(1 - \frac{1}{\sqrt[3]{2}}\right)\right) = \frac{3 - 2 \sqrt[3]{2}}{\sqrt[3]{2}} \cdot AB
\end{aligned}
$$

Since the edge $JK$ of the new tetrahedron corresponds to the edge $AB$ of $\mathcal{T}$, the ratio of the volumes of the two similar tetrahedra is

$$
\left(\frac{3 - 2 \sqrt[3]{2}}{\sqrt[3]{2}}\right)^3 : 1
$$

Therefore, the volume of the new tetrahedron in the problem is

$$
\left(\frac{3 - 2 \sqrt[3]{2}}{\sqrt[3]{2}}\right)^3 = \frac{11}{2} - 27 \sqrt[3]{2} + 18 \sqrt[3]{4} \approx 0.0554
$$

volume units.

()

Based on the work of Gábor Mészáros (Budapest, ELTE Apáczai Csere J. Gyak. Gimn., 9th grade)