# Vertex Coordinates of Regular Hexadecachoron/16-cell/4-orthoplex?

What are the vertex coordinates of a regular Hexadecachoron?

I suspect it's just all 8 combinations of $$(0,0,0, \pm\frac{1}{\sqrt{2}})$$ but I'm not sure how to be sure for sure.

• Um. The Wikipedia Hexadecachoron article you linked to says "The eight vertices of the 16-cell are $(\pm 1, 0, 0, 0)$, $(0, \pm 1, 0, 0)$, $(0, 0, \pm 1, 0)$, $(0, 0, 0, \pm 1)$. All vertices are connected by edges except opposite pairs." It is centered at origin, so you can simply use any real constant $r$ instead of $1$. – Nominal Animal Jan 10 at 1:27
• @NominalAnimal yep. I failed at reading. – guest Jan 10 at 10:07
• No worries, it happens to all of us. Me fail English often. – Nominal Animal Jan 10 at 18:06

Just as the vertices of the 3D cross-polytope, the octahedron, are the 6 ones obtained as permutations of $$(0, 0, \pm\frac1{\sqrt2})$$, or the vertices of the 2D cross-polytope, the square, are the 4 ones obtained as permutations of $$(0, \pm\frac1{\sqrt2})$$.
In fact, the general cross-polytope is just defined to be the convex hull of all points on all Cartesian axes (both directions each), which are $$\frac1{\sqrt2}$$ off.
Just as the general hypersphere is defined by the set of points, which bow to $$||x||_2=\sqrt{\sum_i x_i^2}=const$$, so the surface of the general cross-polytope would follow to $$||x||_1=\sum_i |x_i|=const$$. Within your chosen scaling the latter constant happens to be dimension dependent, $$const(D)=\frac D{\sqrt2}$$. Thereby it is chosen such that the edge length of the thus constructed cross-polytope in any dimension would become unity. Btw. the surface of the general measure-polytope similarily would follow $$||x||_{\infty}=\max_i |x_i|=const$$.