Higher dimensional plane tessellation Let $n$ be some finite positive integer value.  Now consider the plane $z=n-x-y$, so the points $(n, 0, 0), (0, n, 0), (0, 0, n)$ bound the region of the plane with strictly non-negative coordinates.
This region of the plane is an equilateral triangle, and can thus be broken into smaller equilateral triangles that can tessellate the bounded section of the plane via a finite number of reflections.  
My question is as follows.  Now consider a four dimensional plane passing through the points $(n, 0, 0, 0), (0, n, 0, 0), (0, 0, n, 0), (0, 0, 0, n)$.  Does there exist a shape that can tessellate the region of this plane bounded by the four points?  If you continue moving to higher dimensions in a similar manner, maintaining the symmetry on each axis, can you guarantee there must be a shape that can tessellate the bounded region of the plane?
 A: If you try the same technique in four dimensions it won't work. The shape bounded by those four point is a tetrahedron, not a triangle. One easy way to see that you should get a three dimensional object here is that you have 4 variables $x,y,z,w$, and 1 constraint $x+y+z+w=n$, and $4-1=3$.
Let's try to tesselate the tetrahedron in the same way as the triangle and see what happens. We can start by drawing tick marks on each edge dividing them into $n$ pieces. Then we can cut the tetrahedron with planes through the tick marks and parallel to each face. This is the analogue of the way we cut the triangle into smaller triangles. Now looking at this chopped up tetrahedron it looks like the pieces are all tetrahedra, but they're not quite. Every other piece is a tetrahedron and every other piece is an octohedron. Here's a related stack exchange thread with a helpful picture: Tetrahedra require octahedra; 5-cells require...?
So, the obvious tessellation method yields a tessellation with two shapes. I doubt there is a way to tesselate it with one shape, but that sounds tedious to prove.
I recently was looking for a way to tesselate tetrahedra for a reason having to do with computer physics. For my purposes, it was desirable to have a tessellation made only of tetrahedra although they didn't have to all be congruent, and and would be helpful if they were a similar size. You can break the octahedron into four smaller tetrahedra by cutting through most of the edges. Remarkably, these four tetrahedra each have the same volume as the regular tetrahedra from the tessellation. The regular tetrahedra have edge lengths $(1,1,1,1,1,1)$ and the tetrahedra made from the octahedron have edge lengths $(1,1,1,1,1,\sqrt{2})$ (after scaling). You can find calculators online that will give you the volume of a tetrahedron in terms of it's edge lengths, and you should find that these tetrahedra have the same volume. This gives a way to tesselate a larger tetrahedra entirely into tetrahedra of the same volume, but this tessellation has less symmetry than the tessellation into regular octahedra and tetrahedra.
Anyways, that was an applied detour. Let's ask what happens when we apply the same technique in higher dimensions. For simplicity, let $n=2$. So we want to consider the region defined by $x_1+\cdots+x_d=2$, $x_i\geq 0$ for all $i$, where $d$ is the number of dimensions. This shape is a regular $(d-1)$-simplex, the $d-1$ dimensional shape which has $d$ vertices, one edge connecting each pair of vertices, one face connecting every three vertices, one volume connecting any four vertices, and in general one $(k-1)$-simplex connecting any $k$ vertices. Additionally since it is regular each edge is the same length. Now let's mark the halfway point of each edge and chop off the corners by cutting through these marks with hyperplanes parallel to each hyper-face. This shape should have $d\choose 2$ vertices because the $(d-1)$-simplex had $d \choose 2$ edges.
I'm not really sure what shape this is, but maybe you can figure it out using the tools I laid out. It's not a regular polytope in general in higher dimensions, because none of the higher dimensional regular polytopes have $d \choose 2$ vertices. Think about the shape of the hyper-faces left by the hyperplanes we used to cut out the shape. Figuring out this shape is definitely a doable task, and you can definitely figure it out if you put your mind to it. I might keep thinking about it and come back.
