Subdivide/ Decompose a simplex into "congruent" simplices So my question is about subdividing/decomposing a simplex into simpleces that are shrunken versions of the original simplex. For example for both the regular simplex and the simplex with a right angle in 2D (so basically equilateral triangle and isosceles triangle with a right angle) this works fine. However, for 3D this seems not to work for the simplex with the right angle and I am not sure for the regular simplex due to the octahedron in the middle. Can someone either confirm that this does not work in 3D or give a valid solution? Also, does it work in general for either of the two simplices or does it not work? Thanks a lot!

This is not possible due to the Dehn invariant. If $T$ is a tetrahedron then a congruent tetrahedron $T'$ of half its side lengths satisfies $D(T)=2D(T')$ where $D$ denotes Dehn invariant. But for eight copies of $T'$ to cover $T$ we need $D(T)=8D(T')$ and so $6D(T')=0$ (which implies $6D(T)=0$). But in the group of Dehn invariants, $D(T)$ is a non-torsion element, so this is impossible.