A tetrahedron inscribed in a parallelepiped cuts off four tetrahedra from the parallelepiped with its face planes. What fraction of the volume of the parallelepiped is the volume of the tetrahedron whose vertices are the centroids of the cut-off tetrahedra?
Let's use the notation from the figure. It is known that two tetrahedra can be inscribed in any parallelepiped, which are symmetric with respect to the center $O$ of the parallelepiped, and the volume of either is $\frac{1}{3}$ of the volume of the parallelepiped.

![](https://cdn.mathpix.com/cropped/2024_05_02_2ba3af16983606cf3665g-1.jpg?height=472&width=642&top_left_y=306&top_left_x=728)

We will see that, from the perspective of solving the problem, it does not matter which tetrahedron we choose. In the figure, we have drawn the tetrahedron $B E G D$. We will show that its volume is $\frac{1}{3}$ of the volume of the parallelepiped. Since, for example, the area of the face $A B D$ of the cut-off tetrahedron $A B D E$ is half the area of the face $A B C D$ of the parallelepiped, and the height corresponding to these faces is the same, the volume of one cut-off tetrahedron is $\frac{1}{6}$ of the volume of the parallelepiped. The sum of the volumes of the four cut-off tetrahedra is $4 \cdot \frac{1}{6}$, so the volume of the inscribed tetrahedron is indeed $\frac{1}{3}$ of the volume of the parallelepiped.

Let $S$ be the centroid of the tetrahedron $A B D E$. We will write the vector $\overrightarrow{G S}$. Considering that $\overrightarrow{G A}=\mathbf{a}+\mathbf{b}+\mathbf{c}, \overrightarrow{G B}=\mathbf{a}+\mathbf{b}, \overrightarrow{G D}=\mathbf{b}+\mathbf{c}$, and $\overrightarrow{G E}=\mathbf{a}+\mathbf{c}$, the centroid of the tetrahedron $A B D E$ can be obtained as follows:

$$
\overrightarrow{G S}=\frac{\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G D}+\overrightarrow{G E}}{4}=\frac{3}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})
$$

Therefore,

$$
\overrightarrow{O S}=\overrightarrow{G S}-\overrightarrow{G O}=\overrightarrow{G S}-\frac{\overrightarrow{G A}}{2}=\frac{3}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})-\frac{1}{2}(\mathbf{a}+\mathbf{b}+\mathbf{c})=\frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})
$$

From this, we can see that $\overrightarrow{O S}=-\frac{1}{2} \overrightarrow{O G}$, which means that $G$ is mapped to $S$ under a central similarity with center $O$ and ratio $-\frac{1}{2}$. Similarly, the centroids of the other three cut-off tetrahedra are the images of the other three vertices of the inscribed tetrahedron under the same central similarity. Therefore, the volume of the tetrahedron determined by the centroids is $\left(\frac{1}{2}\right)^{3}$ times the volume of the inscribed tetrahedron, which is $\left(\frac{1}{2}\right)^{3} \cdot \frac{1}{3}=\frac{1}{24}$ of the volume of the parallelepiped.

Marianna Csörnyei (Fazekas M. Fóv. Gyak. Gimn., II. o. t.)

Remarks 1. Some of our solutions found the central similarity with ratio $-\frac{1}{2}$ using elementary geometric tools (properties of parallelograms, the ratio in which medians divide each other). These otherwise elegant solutions are somewhat longer than the above vector solution.

2. In her second solution, Marianna Csörnyei used the fact that the parallelepiped can be transformed into a cube by an affinity. Since the affinity preserves ratios, it is sufficient to solve the problem for a cube. She then calculated the edge length of the regular tetrahedron determined by the centroids and subsequently its volume.