Question about tetrahedron decomposition 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. 
 A: 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.
