An equilateral triangle can be divided into smaller equilateral triangles by drawing $n$ lines parallel to each side with equal spacing, as this image shows:
(source: hermetic.com)
Moreover, the vertices can be labeled with $1,2,3$ so that each smaller triangle contains exactly one of $1,2,3$. To do this, we start with $1$ on the first row, $2,3$ on the second row, $3,1,2$ on the third row, and so on.
Is there an analogous division and labeling for higher dimensions? For instance, if we start with a regular tetrahedron?
Best Answer
As already pointed out in a comment by Moishe Cohen and especially the comment by Alex Ravsky, the answer is negative. Let attention be restricted to the regular tetrahedron: dividing the length of each edge by two would result in $2^3 = $ eight smaller instances of that regular tetrahedron. Suppose that we try to assemble the original shape with these smaller tetrahedrons, placing four of the $1/8$ ones at the corners of the original to begin with. Then there will be a void in the middle which has the shape of a regular octahedron. It is impossible to fill that void with the remaining four smaller tetrahedrons. I have no rigorous proof for this, but that doesn't serve as a disclaimer. Just use your imagination and you shall see:
The simplified 2-D case is shown in the picture on the left. The picture on the right is the top view of a 3-D wire frame model of the subdivided regular tetrahedron. The four smaller ones at the corners of the original shape ABCT are: AcbP , BacQ , CbaR at the bottom and PQRT at the first floor.