I can answer my own question: Let the vectors u, v, w, and x be (1,1,1), (-1,-1,1), (-1,1,-1), and (1,-1,-1). Then u+v = (0,0,2), u+w = (0,2,0), u+x = (2,0,0), v+w = (-2,0,0), v+x = (0,-2,0), w+x = (0,0,-2), u+v+w = (-1,1,1), u+v+x = (1,-1,1), u+w+x = (1,1,-1), and v+w+x = (-1,-1,-1), which exactly correspond to the coordinates given in https://en.wikipedia.org/wiki/Rhombic_dodecahedron#Cartesian_coordinates . Jim On Sat, May 12, 2018 at 10:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Coxeter's "Regular Polytopes" has a lot about this and similar zonohedral projections of orthotopes. WFL
On 5/13/18, James Propp <jamespropp@gmail.com> wrote:
But the rhombic dodecahedron has 14 vertices, and I thought mine had 15. Did I miscount?
... Oh wait, 0 is interior to the tetrahedron with vertices u, v, w, and x, so it can't be a vertex!
Can someone explain to me why Jeannine is right (now that I believe her)?
Jim
On Sat, May 12, 2018 at 7:22 PM, Jeannine Mosely <j9mosely@gmail.com> wrote:
I believe you are talking about the rhombic dodecahedron.
--Jeannine
On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
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