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.



Tetrahedron splitting/subdivision
The answer is: No. There is a somehwat rambling discussion [here](http://answers.google.com/answers/threadview/id/497054.html). 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 $\delta$.


