[math-fun] Non-standard "polyhedra" ...
Forgive me if this is obvious and well-known. Because of Euler we know that for polyhedra without holes we have V+F=E+2. As a result we can't have a polyhedron made entirely of hexagons, and need at least 12 pentagons or an equivalent set of shapes. Etc. But things change if we allow intersecting faces. Clearly we have a lot of scope for variations, but we can always retain limits to make things reasonable. But ... Can we make a "polyhedron" consisting only of hexagonal "faces"? Thanks Colin -- Every mathematical proof is a one liner if you start far enough on the left -- Anon
https://en.wikipedia.org/wiki/Szilassi_polyhedron On Wed, May 6, 2020 at 1:36 PM Colin Wright <math_fun@solipsys.co.uk> wrote:
Forgive me if this is obvious and well-known.
Because of Euler we know that for polyhedra without holes we have V+F=E+2. As a result we can't have a polyhedron made entirely of hexagons, and need at least 12 pentagons or an equivalent set of shapes. Etc.
But things change if we allow intersecting faces. Clearly we have a lot of scope for variations, but we can always retain limits to make things reasonable.
But ...
Can we make a "polyhedron" consisting only of hexagonal "faces"?
Thanks
Colin -- Every mathematical proof is a one liner if you start far enough on the left -- Anon
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