James Buddenhagen wrote:
> "hermans" <sas### [at] telenet_invalidbe> wrote in message
> news:4241c745$1@news.povray.org...
> 
>>James Buddenhagen wrote:
>>
>>>"Le Forgeron" <jgr### [at] freefr> wrote in message
>>>news:42411a22$1@news.povray.org...
>>>
>>>
>>>>-----BEGIN PGP SIGNED MESSAGE-----
>>>>Hash: SHA1
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>>>>James Buddenhagen wrote:
>>>>
>>>>
>>>>>I'm interested in artistic images with a math flavor.
>>>>>Here is a simple one (at this point it has to be simple
>>>>>or I can't create it!) which had its inspiration in
>>>>>an arrangement Kottwitz found around 1995 for 20 points
>>>>>on a sphere whose convex hull consists entirely of
>>>>>triangles, and those triangles are as close to equilateral
>>>>>as possible.
>>>>
>>>>20 points on a sphere... sound like a dodecahedron to me...
>>>
>>>
>>>Except, the faces of the regular dodecahedron are not triangles...
>>>
>>>In the picture the 20 points are the centers of the red
>>>circular caps and the yellow polyhedron (which has 20 faces)
>>>is the dual of the convex hull.
>>>
>>>Jim
>>>
>>
>>How the 20 points are arranged? The yellow polyhedron looks rather
>>strange: hexagons and pentagons as faces and at a first sight rather
>>arbitrary ones. Can you give a link to the paper of Kottwitz?
>>
>>http://cage.ugent.be/~hs
> 
> 
> There was no paper.  Kottwitz and I looked at looked at the problem of
> point arragments on spheres giving all triangular faces for the convex
> hull, with the triangles "as close to equilateral as possible" for a
> while and then abandoned it.  I don't even remember how we defined "as
> close to equilateral as possible".  In any case, the 20 points of this
> example are given approximately by:
> 
> p1:=[  0.4579,   0.0000,   0.8890 ];
> p2:=[ -0.4579,   0.0000,   0.8890 ];
> p3:=[  0.0000,  -0.6915,   0.7224 ];
> p4:=[  0.0000,   0.6915,   0.7224 ];
> p5:=[ -0.6684,  -0.6414,   0.3766 ];
> p6:=[  0.6684,  -0.6414,   0.3766 ];
> p7:=[  0.6684,   0.6414,   0.3766 ];
> p8:=[ -0.6684,   0.6414,   0.3766 ];
> p9:=[  0.9897,   0.0000,   0.1435 ];
> p10:=[ -0.9897,   0.0000,   0.1435 ];
> p11:=[  0.0000,  -0.9897,  -0.1435 ];
> p12:=[  0.0000,   0.9897,  -0.1435 ];
> p13:=[ -0.6414,  -0.6684,  -0.3766 ];
> p14:=[  0.6414,  -0.6684,  -0.3766 ];
> p15:=[  0.6414,   0.6684,  -0.3766 ];
> p16:=[ -0.6414,   0.6684,  -0.3766 ];
> p17:=[  0.6915,   0.0000,  -0.7224 ];
> p18:=[ -0.6915,   0.0000,  -0.7224 ];
> p19:=[  0.0000,  -0.4579,  -0.8890 ];
> p20:=[  0.0000,   0.4579,  -0.8890 ];
> 
> I think it has S4 point symmetry, but better check.
> 
> Jim Buddenhagen ( jbuddenh at gmail dot com)
> 
Why don't you take the 20 points on the sphere as the vertices of a 
regular icosahedron? In that case all the triangles aren't "as close to 
equilateral as possible" but equilateral. The dual polyhedron is a 
regular dodecahedron (as can be seen on my site and on many other 
sites). For similar arrangments of more points in many cases one can use 
the theory of geodesic spheres (domes). Also in povray a lot of examples 
exist.
Herman Serras
http://cage.ugent.be/~hs

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