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. 



Question about tetrahedron decomposition
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](http://link.springer.com/article/10.1007/s00454-011-9334-z)." In particular, see **Theorem 1.1** (p. 600, pdf 2/11) for the citation above; alternatively, see page 2 of the arxiv version [here](http://arxiv.org/pdf/1006.1807.pdf).


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.


