We want to break down a cube in such a way that each piece becomes a triangular pyramid (tetrahedron). What is the minimum number of pieces that will result?
The dissection is obviously carried out with planar cuts. Each tetrahedron has four triangular faces, and any two faces share an edge. Therefore, in the sought dissection, each of the 6 faces of the cube is divided into triangles, so at least 12 triangles are formed on the surface, and the cutting planes intersect the cube in triangles, or they dissect the intersections of each other into triangles. - On the other hand, a tetrahedron can be part of at most 3 faces of the cube, because among any 4 faces chosen from the cube's faces, there are 2 that are parallel, so the remaining face(s) are created by a cutting plane. Thus, the parts of the cube's faces will belong to at least $(12: 3=) 4$ tetrahedra.

![](https://cdn.mathpix.com/cropped/2024_05_02_4b0cc536ad3f13852decg-1.jpg?height=515&width=553&top_left_y=434&top_left_x=775)

It is possible to provide a dissection where exactly 4 tetrahedra share the surface of the cube: we divide the 3 faces meeting at a selected vertex of the cube into two triangles each with the diagonals opposite to this vertex - on the 1st figure, the faces meeting at $F$ are divided by the diagonals $E B, B G$, and $G E$ - then we divide the third face adjacent to the selected vertex by the diagonals starting from the adjacent vertices, the $B D, G D$, and $E D$ diagonals. Thus, no bisecting diagonal starts from the vertices $A, C$, and $H$ in any of the faces they are adjacent to, and the diagonals connecting the endpoints of the 3 edges starting from these vertices in pairs lie in the same planes as $E B, B G$, and $G E$, so the 3 right-angled triangles meeting at the vertices $F, A, C$, and $H$ are cut from the cube's surface by the planes $E B G, E B D, B G D$, and $G E D$, respectively, forming a tetrahedron with each, and together they indeed use the 12 triangles formed from the 6 cube faces.

The 4 triangular faces of the tetrahedron cut from the cube form the complete surface of the $B D E G$ tetrahedron inside the cube, so the cube is divided into 5 tetrahedra.

![](https://cdn.mathpix.com/cropped/2024_05_02_4b0cc536ad3f13852decg-1.jpg?height=510&width=483&top_left_y=1442&top_left_x=821)

If we choose 1 triangle from each of the two adjacent, diagonally divided faces of the cube such that the bisecting diagonals do not start from the same endpoint of the common edge (2nd figure, $E B F$, and $B C F$), these two faces already mark 4 vertices of a tetrahedron, and the other two faces of this tetrahedron are inside the cube, so 3 tetrahedra are not enough to use the 10 right-angled triangular faces remaining on the cube's surface, so we do not get a dissection into fewer tetrahedra. (Proceeding similarly, we get 6 tetrahedra, with the $A B C D E F$ half of the cube yielding $E C B A$ and $E C A D$.)

Péter Racskó (Budapest, Madách I. g. IV. o. t.) János Havas (Budapest, Berzsenyi D. g. II. o. t.)

Remarks. 1. By considering the division of the cube's edges into the parts of the tetrahedra, we also obtain that the number of tetrahedra is at least 4.

2. Many solutions contain the following line of thought. If we cut a tetrahedron from the cube, the number of vertices of the remaining solid can be at most 1 less, leaving 7 or more vertices. We need to cut off 1 tetrahedron at least 4 times to leave a 4-vertex solid, i.e., a tetrahedron, so the number of parts is at least 5. - The small flaw in this is that it is assumed without justification that during the dissection, there is always a part-tetrahedron that can be separated from the others by a single planar cut along one of its faces, although in the 2nd figure, every tetrahedron can be removed with only two cuts. (However, if we first cut the cube in half with the $C D E F$ diagonal plane, neither of the resulting parts is a tetrahedron.)