Fiting together triangle anologs in the dimensions

[anderscolingustafson]

In 2d the analog to the triangle is the triangle it self. In order to put some triangles together to form a bigger triangle you just need four triangles with one inner triangle in the middle and the three outer triangles are on each of it's sides.

In 3d the analog to the triangle is the tetrahedron as all of it's faces are triangles. In order to put some tetrahedrons together to form a bigger tetrahedron you need four tetrahedrons and one octahedron. The octahedron is in the middle between the tetrahedrons and the tetrahedrons are on every other one of it's faces.

Now in 4d the analog to the triangle is the 5 cell as all of it's cells are tetrahedrons. I've been trying to visualize the 16 cell and I think that in order to put together some 5 cells to form a bigger 5 cell that you need five 5 cells and one 16 cell but I'm not sure. I think that the five cells of the 16 cell that would have a 5 cell attached would each have all of their corners touching each other but only their corners would touch but I'm not sure. Could someone conform rather or not what I am saying about this is true?

[wendy]

You take a cubic lattice in N dimensions, and stretch it on the long diagonal of a cube, until the cube-corner becomes a 60° angle.

Slice this perpendicular to the long diagonal of a cube. One gets for example, the plane x+y+z+... = K.

Cubes whose vertex at of the lowest sum of K-1, K-2, .... give the simplex, and its various rectates.

In four dimensions, one gets a tiling of equal numbers of pentachora (for K-1, K-4), and rectified pentachora (K=2, K=3). This is because the slices of the 5d cube, vertex-first, is a pentachoron, rectified pentachoron, rectified pentachoron, pentachoron. For K, K-5 here we get vertices where cubes touch.