# On cutting tetrahedrons into mutually congruent pieces

Simple observations: A regular tetrahedron can be cut into 2 mutually congruent pieces (in 3 obvious ways which are all basically the same way, giving one and same pair of congruent pieces). The regular tet can be cut into 3 mutually congruent pieces (infinitely many sets of 3 resultant pieces) and into 6 pieces (1 obvious set of resultant pieces). It also allows congruent partition into 4 pieces (only one obvious set of pieces) and also into 12 pieces (infinitely many different sets of pieces based on first cutting into 4), 8 pieces (1 obvious set of pieces) and 24 pieces (1 obvious set).

• Is there any tetrahedron (regular or otherwise) that can be cut into n mutually congruent connected pieces (no other constraint on the pieces, not even convexity) for any other value of n, say, 16 or 20...?
• Do you know whether this is possible for triangles? Commented May 20 at 11:41
• @M.Winter Isosceles triangle admits a filling by $kn^2$ congruent pieces for $k = 1, 2, 3, 6$. For any number which is a sum of 2 squares there's a right-angled triangle which can be cut into that many congruent triangles. As far as I know, the question whether there's a dissection of an isosceles triangle into 7 congruent pieces is still open (they cannot be triangles, which was proved by Beeson quite recently). Commented May 20 at 12:11
• One simple note: the regular tetrahedron cannot be cut into any number of regular tetrahedra, and generically 'most' tetrahedra can't be cut into any number of tetrahedra congruent to the original, by non-zero Dehn invariant. (Of course, this does nothing to solve OP's question since this involves a strong additional constraint, but I figured it was worth noting) Commented May 20 at 16:39
• One probably crucial difference between triangles and tets could be: although tets do not have an obvious partition into arbitrarily large number of congruent pieces (all triangles allow such partition for n any perfect square), there is in 3D (unlike on plane), more 'room' for non-convex solids to curve and wind around one another so one might just have some non-trivial congruent partitions of the tet for n values like perhaps 16 or 20. Commented May 20 at 17:42
• I asked this question on Math StackExchange in the regular case three years ago here, along with some investigations into possible strategies for subdivision. Commented May 27 at 0:16

## 1 Answer

This MSE question exhibits two non-regular tetrahedra which can be decomposed into 8 smaller copies congruent to themselves; this yields $$8^n$$ for any $$n$$.

• This certainly is a partial answer - at least, I don't see a way to cut a regular tet into n pieces where n is any power of 8. Remark: It is not clear to me if allowing non-convex pieces would help at all. thanks for the pointer to your question. Commented May 29 at 4:51