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By placing a tetrahedron on a face and making vertical cuts centered at the "top" vertex, it is easy to dissect the tetrahedron into $1, 2, 3,$ or $6$ congruent pieces.

By cutting the tetrahedron into four identical pyramids meeting at the center, one for each face, we can further subdivide these pyramids to yield a total of $4, 8, 12,$ or $24$ congruent pieces.

However, it's not obvious to me how to do anything more interesting than this in a way that yields dissections into other numbers of pieces - unlike the triangle, which can be subdivided into many smaller equilateral triangles, the tetrahedron does not decompose into congruent smaller regular tetrahedra, so there is no natural way to 'bootstrap' these constructions to higher numbers of pieces. In particular, I'm most interested in the question of whether a tetrahedron can be dissected into arbitrarily high numbers of pieces.

Some options that come to mind:

  • Some of the pieces resulting from one of the above dissections might have a further dissection I haven't thought of.

  • Polyforms on the tetrahedral-octahedral honeycomb could potentially work to tile a tetrahedron of large side length, just as some polyominoes tile a square and some polyiamonds tile a larger triangle. (However, since the ratio of tetrahedra to octahedra in a given finite tetrahedral portion of the honeycomb monotonically decreases with the side length of said tetrahedron, any given polyform will work with at most one scale.) See also this question on MSE.

  • Perhaps something like Dehn invariants could to used to attempt a proof of impossibility, somehow? I'd be skeptical, though.

RavenclawPrefect
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  • A good starting point: divide each edge of the tetrahedron by $n$ and use that to divide the tetrahedron into smaller tetrahedra and octahedra. What is the sequence defining the number of each? For which n do these numbers work nicely with each other? – Ed Pegg May 08 '21 at 19:15
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    @EdPegg: Yes, that's the method I discussed in my second bullet point. The smallest case which offers some possibility for an interesting number of dissections is with $n=5$, where the numbers indicate that a polyform with $4$ octahedra and $9$ tetrahedra might tile the large tetrahedron using five copies. However, I've written some code to iterate through all $86299$ possible polyforms with those parameters which fit into a tetrahedron, and none of them tile the large tetrahedron. – RavenclawPrefect May 09 '21 at 04:09
  • Excellent. Next step, some of those 86299 shapes can be divided into identical shapes of 1 octahedron and 2+1/4 tetrahedra. – Ed Pegg May 09 '21 at 14:20
  • @EdPegg: Late followup, but considering the behavior at the corners makes it clear that only one such shape has a chance of working, and it's fairly straightforward to check that it doesn't actually tile. It seems plausible to me that such approaches might work at larger scales, but I'd want to check things programmatically there. – RavenclawPrefect Feb 06 '24 at 11:39

2 Answers2

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At the question tetrahedron into similar tetrahedron are two examples of non-regular tetrahedra that can be divided into 8 copies of themselves. They each have some symmetry, so they are examples of non-regular tetrahedra that can be divided into 16 identical shapes. And they can be extended to more pieces.

Since you're using regular tetrahedra, you won't find a nice orthoscheme to allow that trick.

Not a proof:
The pieces of any shape would need to have portions of the arccos(1/3)~70.5288 degree dihedral angle. To divide up the existing angle symmetrically I believe you're trapped by symmetries of the octahedron and need a subgroup of $O$. To place that angle internally you'd need the hypothetical shape to handle that, which implies portions of an octahedron. If such a single piece existed, it would likely be nice enough to be a space-filler. If that could be proven, then a look-up could be done in On Space Groups and Dirichlet-Voronoi Stereohedra to see if the shape exists.

Ed Pegg
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  • I don't see why the dihedral angle would have to be divided symmetrically - couldn't we have, say, a dihedral angle of $45$ degrees and a dihedral angle of $25.5288$ degrees that fit together? – RavenclawPrefect May 08 '21 at 18:39
  • The dihedral angles of the full figure do not necessarily need to be divided symmetrically in a hypothetical solution. But if you do follow total symmetry, you're trapped by group theory. However, it's definitely possible to come up with wacky new shapes that work well. https://blog.wolfram.com/2019/03/07/shattering-the-plane-with-twelve-new-substitution-tilings-using-2-phi-psi-chi-rho/ – Ed Pegg May 08 '21 at 18:45
  • Late follow-up, but it definitely can't be a space-filling polyhedron: they have Dehn invariant zero, while the tetrahedron doesn't. – RavenclawPrefect Aug 16 '22 at 18:09
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The easiest approach is probably to consider the symmetries of the spherical tetrahedron as a kaleidoscope. Here, lines of mirror symmetry on the sphere are treated as kaleidoscope mirrors.

The tetrahedral kaleidoscope is illustrated, along with the others, in Rouse Ball and Coxeter; Mathematical Recreations and Essays (13th Edn, Dover, 1987, p.158).

Coxeter devotes a chapter of Regular Polytopes to a fuller discussion of such kaleidoscopes.

Note that in the case of the tetrahedral kaleidoscope, the polyhedron's duality of faces and vertices means that what appear to be 24 isosceles triangles on the sphere are scalene when drawn on the actual tetrahedron and occur in alternating left- and right-handed forms.

These represent the bases of irregular triangular pyramids (i.e. irregular tetrahedra) with their apex at the centre. They are the characteristic tetrahedra or orthoschemes of the regular polyhedron, and cannot in general be further subdivided into congruent parts.

The tetrahedral kaleidoscope yields just the congruent sets which the question enumerates, and no more.

Are there any asymmetric or low-symmetry solutions? For example there is at least one such dissection of the cube into three congruent pieces. It is difficult if not impossible to prove that no such "black swan" solutions exist, but (to the best of my knowledge) none is known for the regular tetrahedron.

Guy Inchbald
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  • Your answer seems to be assuming that the pieces are all related via a rotation or reflection about the center of the tetrahedron, and that the dissection's symmetry group is transitive on the pieces. I am not making any such assumptions in the question. – RavenclawPrefect Jul 18 '21 at 01:52
  • Fair enough. I have added a paragraph on that. – Guy Inchbald Jul 18 '21 at 08:36
  • Thank you. Could you provide a reference for the claim that no such dissections are possible? I haven’t found any discussion of this question in the literature so wasn’t able to get a sense of the scope of dissections investigated. – RavenclawPrefect Jul 18 '21 at 12:59
  • Nobody says that none are possible, it is just that none have been found. At least, I suppose I should say, not to my knowledge; I'll add that too. – Guy Inchbald Jul 18 '21 at 13:11
  • Yeah, bad phrasing on my part - should have said “known” rather than “possible”. For what it’s worth, I don’t share the intuition that none should exist - when I consider e.g. the large variety of cube dissections (even just the polycube ones), or the extent to which interesting triangle dissections are possible, I don’t feel like there should be barriers to such things here. If you could elaborate on what seems to distinguish this case from others (if only intuitively), I’d love to hear it. – RavenclawPrefect Jul 18 '21 at 13:21
  • Briefly; when you dissect a cube into cuboids, they all fit. But when you cut tetrahedra off a tetrahedron you tend to end up with an octahedron; when it comes to solid angles, you need both to fill the angle and so you lose congruence. Ludwig Schläfli studied orthoschemes, Williams gives an overview in his Geometrical foundations of natural structure, and Bucky Fuller investigated some dissections in his own inimitable jargon, though he disliked chiral polyhedra. But I never saw a hint that they could in general be further dissected (though the orthoscheme of the cube can be cut in half). – Guy Inchbald Jul 18 '21 at 13:41