Take a perfect cube. While keeping the cube intact and all the pieces together, make as many planar slices as possible through exactly three vertices.
After doing so, separate the resulting pieces. What shapes result and how many are there of each and in total?
To what do these numbers correspond?
Note: Please do NOT punch the problem into a 3D graphics program and then rush to post the solution here so you can be first. This remains my most enjoyable solution to date because I attempted it without pen and paper, computer or polyhedron.
... to picture the cube with what look like X's on each face representing the cuts.
If any two of the three vertices were on the same edge of the cube, there would be more than three vertices on that plane. If any two vertices were opposite each other, then in order to include any other vertex, a fourth vertex would need to be included as well.
So only vertices connected by diagonals of faces can be included in the dividing planes. Therefore each planar slice will produce a regular triangle with three vertices, surrounding one of the other 8 vertices of the cube. The planar slices intersect one another, so the right-angled tetrahedra formed by each individual slice and its respective corner overlap.
The overlap between two of these tetrahedra will each include one edge of the cube. The two slices themselves meet in a line segment within the cube, and so each edge has its own small tetrahedron.
Since there are 8 slices, and they don't go through the center of the cube, they form a regular octahedron in the center.
Note that the 12 edge tetrahedra each have an internal edge as mentioned above. Each of these is an internal edge of the central octahedron.
But the octahedron must adjoin something at each of its faces--not just its edges. The things they adjoin are the portions of the vertex tetrahedra that are not shared between two vertex tetrahedra. There are 8 of them. Each has a triangular face adjoining the central octahedron. That means there are three faces extending out from there, that are all interna, that coincide with faces of the irregular edge-tetrahedra, and come together at the vertices of the cube. Thus each of these pieces is a tetrahedron as well. The base (at the central octahedron) is a regular (equilateral) triangle. The angles at the vertices of the cube are also angles of the large equilateral triangles formed on the faces of the cube, and so also are equilateral triangles. Thus these triangles are also equilateral and the tetrahedra in question are regular.
Thus in total, there's one central regular octahedron, 12 irregular (but right-angled) tetrahedra around the edges of the cube, and 8 regular tetrahedra, each with a base on a face of the octahedron and the remaining vertex coincident with a vertex of the cube.
I see this disagrees with the previous post in that I think the set of 8 tetrahedra face-to-face with the central octahedron are regular.
Posted by Charlie
on 2005-07-27 20:42:16