Re: [math-fun] polyhedra with all faces congruent
are there examples of polyhedra with all faces congruent, that have an odd number of faces? (i did not see any parity restriction, although the faces must have an even number of sides.)
If you don't mind nonconvexity, it's pretty easy to make one out of squares (all meeting at right angles).
i don't mind non-convex polyhedra, though a convex one would be better. however, i don't see your "pretty easy" way to do it. unless you intend to count faces in two different ways: first, as (connected) subsets of planes, and second, each individual unit square counts as its own face. for example, a "notched cube" (figure obtained by removing a 1x1x1 corner from a 2x2x2 cube) has 9 planar faces. but then it's unreasonable to say that all faces are congruent, by re-defining the faces by subdividing the previous 9 faces into 24 unit squares, which are indeed congruent. i doubt this is what you had intended anyway ... unless i missed it, your argument that faces must have fewer than 6 sides doesn't require convexity, (as long as we can't have a vertex of degree 2) but probably requires that the genus is 0 . the higher genus case opens another can of worms. for quadrilateral faces, the sides must occur in two congruent pairs (or all four congruent) by the same parity reasoning. this means either parallelograms or kite faces, and i think parallelogram faces will not work (such polyhedra decompose into parallelepipeds). so ... seems unlikely, but can it be proved? mike
On Tuesday 18 November 2008, Michael Reid wrote:
are there examples of polyhedra with all faces congruent, that have an odd number of faces? (i did not see any parity restriction, although the faces must have an even number of sides.)
If you don't mind nonconvexity, it's pretty easy to make one out of squares (all meeting at right angles).
i don't mind non-convex polyhedra, though a convex one would be better. however, i don't see your "pretty easy" way to do it.
OK. I'll build a polyhedron as a union of cubes from the obvious dissection of R^3 into equal cubes. Number the cubes with triples of integers. Start with a "spine" formed by cubes (n,0,0) for n=0,1,...,2m. Now add (n,+-1,0) and (n,0,+-1) for n=1,3,...,2m-1. We get 4 faces from each "even" cube on the spine other than the ends, 5 for each of the extra cubes attached at odd x, and 5 for each end cube: 4(m-1)+5.4.m+10 = 24m+6 faces. (This even gives the right answer for m=0.)
unless i missed it, your argument that faces must have fewer than 6 sides doesn't require convexity,
It does, because it relies on not being able to fit too much "angle" around each vertex. My polyhedron above has "crinkly" vertices where six faces meet, with a total angle of 540 degrees.
for quadrilateral faces, the sides must occur in two congruent pairs (or all four congruent) by the same parity reasoning. this means either parallelograms or kite faces, and i think parallelogram faces will not work (such polyhedra decompose into parallelepipeds). so ... seems unlikely, but can it be proved?
I'm sure it can, but I haven't time right now... -- g
participants (2)
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Gareth McCaughan -
Michael Reid