We draw a plane through each edge of a regular tetrahedron and the midpoint of the opposite edge. Into how many parts does this 6 planes divide the tetrahedron?

How many additional pieces can be fitted into the place of a selected piece

a) with a suitable translation,

b) with a translation and a suitable reflection?
I. A plane intersecting the tetrahedron $ABCD$ passes through its centroid $S$. Let the midpoints of the edges $BC$, $CA$, and $AB$ of the base triangle $ABC$ be $E$, $F$, and $G$ respectively (Figure 1). The planes $DAE$, $DBF$, and $DCG$ intersect the base triangle along the medians $AE$, $BF$, and $CG$ respectively. The common point of these medians is the centroid $S_D$ of the triangle $ABC$. Therefore, the three planes intersect along the line $DS_D$, which is a median of the tetrahedron.

![](https://cdn.mathpix.com/cropped/2024_05_02_b828d7de40674d2da513g-1.jpg?height=667&width=702&top_left_y=325&top_left_x=690)

Figure 1

Thus, these three planes divide the space and the tetrahedron into 6 parts (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_b828d7de40674d2da513g-1.jpg?height=461&width=371&top_left_y=1130&top_left_x=861)

Figure 2

The same can be said for any other set of three edges meeting at a vertex of the tetrahedron and the corresponding set of three planes. Any two medians, such as $AS_A$ and $DS_D$, as lines in the plane $DAE$, intersect at the centroid of the tetrahedron.

The tetrahedron is divided into 4 smaller tetrahedra by the 6 edges and the point $S$, each of which has one face as a face of the original tetrahedron, and the three faces meeting at $S$ are parts of the intersecting planes. Thus, each of these smaller tetrahedra is further divided by the remaining 3 intersecting planes. For example, the tetrahedron $SABC$ is divided into 6 parts by the planes $DAE$, $DBF$, and $DCG$, just as we have already stated for the original tetrahedron. The same applies to each of the smaller tetrahedra, so the 6 intersecting planes divide the tetrahedron into 24 parts.

So far, we have not used the fact that the original tetrahedron is regular, so our findings are valid for any tetrahedron.

II. Based on the regularity of our tetrahedron, $DA = DB = DC$. Therefore, the projection of $D$ onto the plane $ABC$ is equidistant from $A$, $B$, and $C$, which is the circumcenter of the triangle $ABC$. In a regular triangle, the circumcenter is also the centroid, so the projection is identical to $S_D$. Thus, $DS_D$ is perpendicular to the plane $ABC$, and a $120^\circ$ rotation about the axis $DS_D$ maps the plane and the triangle $ABC$ onto itself. This rotation also maps the tetrahedra $SDAB$, $SDBC$, and $SDCA$ onto each other, so these tetrahedra are congruent. A rotation about another median of the tetrahedron shows that $SABC$ is also congruent to them.

Applying the above rotations to the 6 parts of the tetrahedron $SABC$, we find that the parts $SS_DAG$, $SS_DBE$, and $SS_DCF$, as well as the parts $SS_DAF$, $SS_DBG$, and $SS_DCE$, can be mapped onto each other. Similarly, the corresponding 3 parts of each of the 3 smaller tetrahedra can also be mapped onto each other. Thus, 12 of the 24 parts can be mapped onto each other by some translation.

The parts of the two groups of 12 can be mapped onto each other by reflection, for example, $SS_DAG$ can be mapped onto $SS_DAF$ by reflection in the plane $DAE$, which maps $B$ and $C$ (and thus $F$ and $G$) onto each other, and is a symmetry plane of the original tetrahedron. Thus, any of the 24 parts can be mapped onto any other, so the answer to part b) is: 23 parts.

However, we will show that, for example, the part $SS_DCF$ cannot be mapped onto $SS_DAF$ by a translation. Indeed, the shortest edge of both parts is $SS_D$, the next longest is $S_DF$, because, as can be easily seen, $SD = 3 \cdot SS_D$, so

$$
SS_D = \frac{DS_D}{4} < \frac{DF}{4} < \frac{DF}{3} = S_BF = S_DF
$$

and

$$
\angle SS_DF = \angle S_DF A = \angle SS_DA = \angle S_FA = 90^\circ \text{ and } \angle S_DF A = 30^\circ
$$

(Figure 3) implies

$$
S_DF < SF < SA \quad \text{ and } \quad S_DF < FA < S_DA
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_b828d7de40674d2da513g-2.jpg?height=269&width=588&top_left_y=424&top_left_x=752)

Figure 3

Thus, in the desired mapping, the right-angled triangular face $SS_DF$ can only map onto itself. However, the faces of the parts (viewed from the outside) have opposite orientations. This proves our statement.

Therefore, the answer to part a) is: 11 parts.

Remark. The equality $SD = 3 \cdot SS_D$ in part II follows from the fact that $EA = 3 \cdot ES_D$, $ED = 3 \cdot ES_A$, and $S_A S_D \parallel DA$, and from the trapezoid $AD S_A S_D$, $SD : SS_D = DA : S_D S_A = 3$.