On each edge of a tetrahedron, we select one of the trisection points. What can be the ratio of the volume of the convex body determined by these points and the volume of the tetrahedron?
If we change one of the three edges starting from a vertex of a tetrahedron to $\lambda$ times its original length, while keeping the other two edges unchanged, then the volume of the resulting new tetrahedron will be $\lambda$ times the volume of the original tetrahedron. This is because the two tetrahedra will share a common face, and the height corresponding to this face will also change by a factor of $\lambda$ according to the theorem of parallel chords.

![](https://cdn.mathpix.com/cropped/2024_05_02_6f60ffc06e325ed91d3fg-1.jpg?height=548&width=703&top_left_y=336&top_left_x=711)

The volume of the solid determined by the trisection points can be obtained by subtracting the volumes of the four small tetrahedra from the volume of the original tetrahedron. These small tetrahedra are formed by one vertex of the large tetrahedron and the trisection points on the three edges starting from that vertex. Let the volume of the original tetrahedron be 27 units. If we mark a trisection point on an edge, we write 1 next to the closer endpoint and 2 next to the farther endpoint. Thus, each vertex of the tetrahedron will have three numbers, and the product of these numbers gives the volume of the small tetrahedron cut off at that vertex, as described in the first paragraph.

Since we have written 6 ones in total for the 4 vertices of the tetrahedron, we can assume that for the tetrahedron $ABCD$, we have written 1s on the edges $AB$ and $AC$ for vertex $A$. We can choose $B$ and $C$ such that on the edge $BC$, 1 is written next to $B$. Let the numbers written next to $D$ on the edges $DA$, $DB$, and $DC$ be $(3-a)$, $(3-b)$, and $(3-c)$, respectively. Then the volumes of the small tetrahedra cut off at $A$, $B$, $C$, and $D$ are $a$, $2b$, $4c$, and $(3-a)(3-b)(3-c)$, respectively. Since $a$, $b$, and $c$ can independently be 1 or 2, the following eight cases are possible:

| $a$ | $b$ | $c$ | Total volume of cut-off tetrahedra |
| :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 15 |
| 1 | 1 | 2 | 15 |
| 1 | 2 | 1 | 13 |
| 1 | 2 | 2 | 15 |
| 2 | 1 | 1 | 12 |
| 2 | 1 | 2 | 14 |
| 2 | 2 | 1 | 12 |
| 2 | 2 | 2 | 15 |

Therefore, the volume of the convex solid determined by the trisection points, i.e., the volume of the remaining solid after the cuts, can be $27 - 15 = 12$, $27 - 13 = 14$, $27 - 12 = 15$, or $27 - 14 = 13$. Thus, the value of the desired ratio is $\frac{12}{27}$, $\frac{13}{27}$, $\frac{14}{27}$, or $\frac{15}{27}$. Based on the work of Bence Béky (Fazekas M. Főv. Gyak. Gimn., 10th grade).