Solution. The bases of the three tetrahedra, with the notation of the figure, are the triangles $ABH$, $BDC$, and $ACF$, and their fourth vertex coincides with the vertex $O$ of the regular tetrahedron. The common part of the three tetrahedra is thus the pyramid with the base $DEFGHI$ and vertex $O$, whose height is equal to the height of the original tetrahedron, so instead of the ratio of their volumes, it is sufficient to examine the ratio of their base areas.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e4e9f1e766e96e0b5adg-1.jpg?height=512&width=553&top_left_y=351&top_left_x=775)

We break down the area of the hexagon into the areas of four triangles: $T_{DEFGHI} = T_{DFH \triangle} + T_{DFE \triangle} + T_{FHG \triangle} + T_{HDI \triangle}$, where the last three areas are equal due to the congruence of the triangles.

Since the sides of $\triangle DFH$ are the midlines of $\triangle A_1B_1C_1$, and the sides of $\triangle A_1B_1C_1$ are the midlines of $\triangle ABC$, we have

$$
T_{DFH \triangle} = \frac{1}{4} \cdot \frac{1}{4} \cdot T_{ABC \triangle} = \frac{1}{16} \cdot T_{ABC \triangle}
$$

The distance between the lines $AB$ and $DF$, which is also the height of $\triangle DFC_1$, is one-fourth of the height of $\triangle ABC$, and this is also the sum of the heights of $\triangle ABE$ and $\triangle DFE$: $m_{ABE \triangle} + m_{DFE \triangle} = \frac{1}{4} m_{ABC \triangle}$. The triangles $\triangle DFE$ and $\triangle ABE$ are similar, with the similarity ratio:

$$
\begin{gathered}
\frac{DF}{AB} = \frac{1}{4} = \frac{m_{DFE \Delta}}{m_{ABE \triangle}}, \quad \text{so} \\
m_{ABE \triangle} + m_{DFE \triangle} = 4 \cdot m_{DFE \triangle} + m_{DFE \triangle} = 5 m_{DFE \triangle} = \frac{1}{4} m_{ABC \triangle}
\end{gathered}
$$

from which

$$
\begin{aligned}
m_{DFE} & = \frac{1}{20} m_{ABC} \\
T_{DFE \triangle} & = DF \cdot m_{DFE \triangle} = \frac{1}{4} AB \cdot \frac{1}{20} m_{ABC \triangle} = \frac{1}{80} T_{ABC \triangle} .
\end{aligned}
$$

Summing up the areas:

$$
T_{DEFGHI} = T_{DFH \triangle} + 3 \cdot T_{DFE \triangle} = \frac{1}{16} \cdot T_{ABC \triangle} + \frac{3}{80} T_{ABC \triangle} = \frac{1}{10} \cdot T_{ABC \triangle}
$$

Thus, due to the equality of the heights, the volume of the common part of the three tetrahedra is also $\frac{1}{10}$ of the volume of the tetrahedron.
We draw the midlines on the base of a regular tetrahedron. We connect the midpoint of each midline with the vertices of the side face parallel to the midline. What fraction of the volume of the tetrahedron is the volume of the common part of the three tetrahedra thus obtained?