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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.

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See also here:… – Benjamin Dickman May 22 '13 at 0:07
up vote 5 down vote accepted

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 $\delta$.

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Did not see Anton's answer while I was writing mine. Mine is a version of his Answer 1. – Joseph O'Rourke Jun 18 '10 at 12:14
Accepting this one because of the link to discussion. – SigTerm Jun 18 '10 at 12:43

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.

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Anton, why does the first answer not work for cubes? Secondly, I am just curious whether subdivision of "1/3" of cube (halfs of three faces with one joint vertex) into equal congruent parts is possible. – Wadim Zudilin Jun 18 '10 at 12:03
The Dehn invariant of cube is $0$. – Anton Petrunin Jun 18 '10 at 12:04
And for the first answer, the dihedral angle of a cube is 90 degrees, which evenly divides 180 degrees. – Gerald Edgar Jun 18 '10 at 13:32
The volume should involve cubes of $a_i$s. – Victor Protsak Jun 19 '10 at 2:30

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.

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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!

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Classification of crystallographic lattices relies on some angles dividing $\pi$, so this is a convoluted way of phrasing Anton's Answer 1. – Victor Protsak Jun 19 '10 at 2:28

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