In the tetrahedron $A B C D$, we have drawn planes through a point on the edge $A B$ that are parallel to the faces $A C D$ and $B C D$ of the tetrahedron. The volumes of the two smaller tetrahedrons thus formed are 8 and $1 \mathrm{~cm}^{3}$. What is the volume of the tetrahedron $A B C D$?
We denote the vertices of the resulting small tetrahedra as shown in the figure: $M, R, S$, and $M, P, Q$ (where $M$ is a point on the edge $AB$ of the original tetrahedron $ABCD$). Since the original tetrahedron was cut by planes parallel to each of its faces, the resulting small tetrahedra are similar to the original and to each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_25bbedb9623cdee58094g-1.jpg?height=623&width=550&top_left_y=304&top_left_x=777)

We know that the ratio of the volumes of similar polyhedra is equal to the cube of the ratio of their corresponding edges. Therefore:

$$
\frac{1}{8}=\frac{V_{A M R S}}{V_{M B P Q}}=\left(\frac{A M}{M B}\right)^{3}, \quad \frac{A M}{M B}=\frac{1}{2}
$$

From this, we can determine the ratio of a corresponding edge of the smallest and the original tetrahedron:

$$
\frac{A B}{A M}=\frac{A M}{A M}+\frac{M B}{A M}=1+2=3
$$

Therefore, the ratio of the volumes of the two tetrahedra is:

$$
\frac{V_{A B C D}}{V_{A M R S}}=\left(\frac{A B}{A M}\right)^{3}=27
$$

The volume of the tetrahedron $A M R S$ is $1 \mathrm{~cm}^{3}$, so the volume of the tetrahedron $A B C D$ is $27 \mathrm{~cm}^{3}$.

Remark. If the volumes of the two small tetrahedra are $V_{1}$ and $V_{2}$, then the volume of the original tetrahedron is

$$
V_{1} \cdot\left(1+\sqrt[3]{\frac{V_{2}}{V_{1}}}\right)^{3}=V_{1}+3 \sqrt[3]{V_{1} \cdot V_{2}^{2}}+3 \sqrt[3]{V_{1}^{2} \cdot V_{2}}+V_{2}
$$

Based on the work of Jurek Zoltán (Debrecen, Fazekas M. Gymnasium, II. grade)