Planes through the vertices of a tetrahedron, parallel to the opposite faces, and the centroids of the tetrahedron's faces, form similar tetrahedra to the original with the same orientation. What is the relationship between the volume measures of these tetrahedra?

(Csillag Vilmos, university teaching assistant.)
Let $S A B C$ be the original tetrahedron.

![](https://cdn.mathpix.com/cropped/2024_05_02_e14c1e82abd0a756c58ag-1.jpg?height=369&width=477&top_left_y=192&top_left_x=802)

Let $A^{\prime}, B^{\prime}$, and $C^{\prime}$ be the midpoints of the base edges of the tetrahedron; $A_{1} B_{1} C_{1}$ and $S_{1}$ be the centroids of the faces of the tetrahedron. Under these conditions, we have:

$$
A^{\prime} A_{1}: A^{\prime} S = A^{\prime} S_{1}: A^{\prime} A = 1: 3
$$

The triangles $A^{\prime} A_{1} S_{1}$ and $A^{\prime} S A$ are therefore similar, and thus $A_{1} S_{1}$ is parallel to the edge $A S$, and similarly:

$$
A_{1} S_{1}: A S = 1: 3
$$

Similarly, it can be shown that $B_{1} S_{1} \| B S$ and $C_{1} S_{1} \| C S$.

Furthermore, since

$$
S A^{\prime}: S C^{\prime} = S A_{1}: S C_{1}
$$

$A_{1} C_{1}$ is parallel to $A^{\prime} C^{\prime}$, and since

$$
B A^{\prime}: B C^{\prime} = B C: B A
$$

$A C$ is also parallel to $A^{\prime} C^{\prime}$, which implies that $A_{1} C_{1}$ is parallel to $A C$. Similarly, it can be shown that $B_{1} C_{1}$ is parallel to $B C$ and $A_{1} B_{1}$ is parallel to $A B$; thus, the edges of the centroid tetrahedron are parallel to the corresponding edges of the original tetrahedron, which means they are similar, and the ratio of their volumes is equal to the cube of the ratio of the corresponding edges.

$$
\frac{S(A B C)}{S_{1}(A_{1} B_{1} C_{1})} = \left(\frac{A B}{A_{1} B_{1}}\right)^{3} = 3^{3}
$$

If we draw planes through the vertices of the original tetrahedron parallel to the opposite faces, the resulting tetrahedron $S_{2}(A_{2} B_{2} C_{2})$ has its centroid tetrahedron as the image of the original $S(A B C)$ tetrahedron, and thus

$$
\frac{S_{2}(A_{2} B_{2} C_{2})}{S(A B C)} = 3^{3}
$$

The product of (1) and (2) gives the desired ratio:

$$
\frac{S_{2}(A_{2} B_{2} C_{2})}{S_{1}(A_{1} B_{1} C_{1})} = 3^{6}
$$

Thus, the ratio of the volumes of the three tetrahedra is:

(Weisz Lipót, fóreálisk. VIII. o. t., Pécs.)

The problem was solved by: 1 1 Strasser G. Sándor. I. year philosophy student and Visnya Aladár I. year philosophy student; and also: Bálint Béla, Berger Hugó, Devecis Mihály, Freund Antal, Friedmann Bernát, Goldstein Zsigmond, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Kunsch Mátyás, Monoky Gyula, Piovarcsy Jenő, Riesz Frigyes, Schólcz Károly, Szabó István, Szigeth Gábor.[^0]


[^0]:    ${ }^{1}$ Note. Most solvers assume a regular tetrahedron, although the problem applies to any tetrahedron.