I've been puzzled about a related question forever: Suppose we cover a sphere with (spherical) polygons that are all congruent. What's the largest number of polygons possible? (Ans: unbounded, using the bipyramid construction.) Changing the question, suppose we want the polygons to be as small as possible. We can still get area->0 with bipyramids, so we interpret "small" to mean diameter (longest side, for triangles). My best results are (a) 120 triangles, by splitting icosahedron faces into 6 small triangles, or (equivalently) splitting dodecahedron faces into 10 small triangles; and (b) slicing the sphere surface into orange wedges, and splitting the wedges into 4 triangles: A / \ / \ B-----C |\ | Assume we've chosen angle BAC = DFE = 2pi/N. | \ | A and F are poles, ABDF and ACEF are great semicircles, | \ | AB = BE = EF by suitable adjustment, ABE and BEF are | \ | isosceles triangles, AC=CE=BD=DF by bisecting angles | \| ABE and BEF. BC=DE, but they are not quite latitude D-----E lines. \ / \ / F But neither (a) nor (b) can achieve tiny diameter polygons. Since the number of faces is 4N, the "odd?" question remains. Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rwg@sdf.lonestar.org Sent: Tuesday, November 18, 2008 2:53 AM To: math-fun Cc: math-fun@mailman.xmission.com Subject: Re: [math-fun] polyhedra with all faces congruent
On Tuesday 18 November 2008, Mike Reid wrote:
[me, about making a polyhedron with many faces, all congruent:]
No. Take a skew prism -- lots of congruent triangular faces -- and put a cap on each end made out of triangles of the same dimensions.
[Mike:]
just a minor nit. this cannot always be done; it depends upon the antiprism. er, ... did i interpret this correctly "skew prism" = "(what i've always known as an) antiprism"?
Er, yes, "antiprism" was the word I was looking for. And I didn't mean to suggest that it can be done for an arbitrary antiprism -- of course it can't -- but only that there's a construction that works that way. Which there is :-).
anyway, it can be done if the isosceles triangles are sufficiently tall, and that suffices for the "no" answer above. this construction gives a polyhedron with 4n faces. however, the "belt" around the antiprism is not needed!
D'oh, of course. Silly me.
YOU feel silly? I implemented http://gosper.org/congru.htm < 4yrs ago! Of course, my original interest was "ball shaped" polyhedra, if we can make that precise. I think I can do 216. --rwg
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).
For a convex polyhedron, the number of sides per face must be less than 6 (think curvature)[1], so if it's even it must be 4. So, v-e+f=2; 4e=2f by counting (face,edge) pairs, so e=2f and v-f=2. Counting (face,vertex) pairs we have 4f = (avg vertex degree).v, so since v is bigger we have avg vertex degree < 4 but (for large face count) it's very close to 4. I suspect that this is impossible on the grounds that (handwaving again) many vertices must have at least 4 "average" angles at them, and therefore must be either flat (no!) or concave.
[1] Note: I'm handwaving here and haven't actually checked that what I say is true, but I'm pretty sure it's obvious :-).
Anyone fancy either filling in the details or refuting my handwaving?
-- g
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