Tetrahedron splitting/subdivision Given a regular Tetrahedron A (i.e. each edge of A has same length), is it possible to split A into several smaller regular tetrahedra of equal size? I.e. smaller tetrahedra should completely fill volume of A, and they should not overlap.
This can be done in 2D with a triangle and square, and it can be done in 3D with cube (i.e. you can split cube into several smaller cubes of equal size). But I see no way to do same thing in 3D with tetrahedron. 
If this can be done, how (how smaller tetrahedra should be positioned)?
If this cannot be done, is there a proof that this is impossible?
P.S. I'm not a mathematician, and this is not a homework, but I'd like to know how/if this can be done.
 A: The answer is: No.  There is a somehwat rambling discussion here.  Let $B$ be a smaller tetrahedron that is jammed into the apex of $A$.  It fills the solid angle there completely.
Let $e$ be a base edge of $B$. Then one cannot fill the neighborhood of $e$ by gluing in
further regular tetrahedra along it.  One way to see this is that the dihedral angle of the tetrahedron is $\delta = \cos^{-1}(1/3) \approx 70.5^\circ$, and the dihedral angle along $e$ to be filled
is $\pi - \delta \approx 109.5^\circ$, which cannot be formed from copies of $\delta$.
A: It is however possible to split a regular tetrahedron into four smaller regular tetrahedra and one regular octahedron, and there are other possibilities based on the tetrahedral-octahedral honeycomb.
A: Answer 1: Look what happens on a face of the big tetrahedron where some edges of small ones come together: you have to make angle 180° out of some dihedral angles of the tetrahedra (which is about 70°) --- that is impossible.
Answer 2: There is the so-called Dehn invariant. If a polyhedron $X$ is split into a number of polyhedra $Y_1,Y_2,\dots,Y_n$ then the Dehn invariant of $X$ is equal to the sum of the Dehn invariants of $Y_i$.
For the regular tetrahedron, the Dehn inveriant is nonzero and proportional to the length of a side.
Suppose you could split a regular tetrahedron with side $a$ into a number of tetrahedra with sides $a_1, a_2,\dots, a_n$. Then from the volume you have
$$a_1^3+a_2^3+\dots+a_n^3=a^3$$
and from the Dehn invariant you have
$$a_1+a_2+\dots+a_n=a.$$
It follows that there is no nontrivial splitting.
A: This question has already been pretty handily answered by Anton, but I figured I'd add another (somewhat ridiculous) proof to his list:
Answer 3:  Suppose that we had an infinite subdivision of regular tetrahedra.  Then we would also have a crystallographic lattice with tetrahedral symmetry, since this structure would have to tile across 3D space.  However, the classification of crystallographic lattices rules out this possibility!
http://en.wikipedia.org/wiki/Crystallographic_group
