Are there tetrahedra which can be subdivided into three non-overlapping parts similar to the original? I believe this would require splitting one face into three parts. I know some types of tetrahedra for which this decomposition is impossible. In 2d, for right triangles you get a decomposition into two similar parts by dropping a perpendicular from the right angle to the hypotenuse, and I would be surprised if there were other 2d solutions.

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Yes I meant three parts similar to the original which fit inside the original without overlap and fill the original. But if you can fill the original with three nonoverlapping similar parts of any kind, that would be interesting too. The 2d case is Pythagoras' Theorem, and that got me thinking about the 3d problem, via a story about a proof of Pythagoras by a 12-year-old Einstein in a book called 'Fractals, Chaos, Power Laws' by Schroeder (page 3). – Dennis Farr May 21 '13 at 1:51
'given tetrahedron' for 'original' above, sorry. – Dennis Farr May 21 '13 at 1:54
I'm going to search over a sample of tetrahedra and try various methods of splitting them into three tetrahedra: either by dropping a line from one vertex to somewhere on the opposite face, or by slicing off a tetrahedron with a plane through one vertex and then subdividing the remaining irregular 4-sided pyramid into two tetrahedra with another plane slice. This shouldn't take a long time to run, and I'll keep score on which cases are closest to similar to each other and similar to the parent. If I find any good candidates I can refine the search or guess an answer. Have to code it up still. – Dennis Farr May 29 '13 at 15:45
As I think about writing the code, I am led to tetrahedral mesh refinement - a topic in finite element work. I'm beginning to think it will be easier to find code that is about what I need rather than writng it. – Dennis Farr May 29 '13 at 23:33

In the case where the three parts are each congruent to one another, the answer to your question is no: there is no such decomposition of a tetrahedron.

The terminology needed to find such an answer in the literature is "reptile" or "$k$-reptile simplices."

Citation for proof:

Safernová, Z.: Perfect tilings of simplices. Bc. degree thesis. Charles University, Prague (2008).

Unfortunately (for many) this thesis is written in Czech.

Fortunately, though, there is a more general paper on this topic, entitled "On the Nonexistence of $k$-reptile Tetrahedra." In particular, see Theorem 1.1 (p. 600, pdf 2/11) for the citation above; alternatively, see page 2 of the arxiv version here.

The citation for this latter paper is:

Matoušek, J., & Safernová, Z. (2011). On the Nonexistence of k-reptile Tetrahedra. Discrete & Computational Geometry, 46(3), 599-609.

If you relax the condition and require the simplices be similar to one another but not necessarily congruent, then the term "irreptile" is sometimes used (at least in the $2D$ case). Sadly, I do not know of any work on $k$-irreptile tetrahedra.

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Thanks for the arxiv link. Hill simplices are pretty interesting of themselves. – Dennis Farr May 22 '13 at 7:40
Ah, this is cool! – Per Alexandersson Dec 2 '14 at 9:42
@PerAlexandersson Agreed! And if you like questions about tetrahedra, then you might also like this one: mathoverflow.net/q/142983/22971 (Separately: I'm not sure why the simplicial-stuff tag was removed in the most recent edit -- the answer involves simplices...) – Benjamin Dickman Dec 2 '14 at 20:12
Simplicial-stuff usually means simplicial complexes, and (co)homology stuff. – Per Alexandersson Dec 2 '14 at 20:13