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Old 2010-07-03, 19:44   #1
davieddy
 
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Default Platonic solids and the Golden ratio (r)

As an erstwhile 3D-engine programmer (among other things)
I read a beautiful article about simple starting points for the
vertices of the five regular solids.

Before my video(visio?)spatial ability goes completely bonkers
let me try to remember them:

(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1) tetrahedron

(+/-1,+/-1,+/-1) cube

(+/-1,0,0) etc octahedron

(+/-r,+/-1,0) (permute cyclicly) icosohedron

HELP!

David

Last fiddled with by Prime95 on 2010-07-03 at 21:40 Reason: Watch the language please
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Old 2010-07-04, 13:57   #2
ccorn
 
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Quote:
Originally Posted by davieddy View Post
(+/-r,+/-1,0) (permute cyclicly) icosohedron

HELP!
Use the duality.

P.S.: And post the result
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Old 2010-07-04, 17:14   #3
davieddy
 
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Quote:
Originally Posted by ccorn View Post
Use the duality.

P.S.: And post the result
I thought of that, but I'm not sure that taking
the centre of the 20 triangles gives the simplest
orientation for the vertices of the dodecahedron.

This is "puzzles" - not "homework help"

David

PS Apologies for my French in the first post.
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Old 2010-07-04, 17:48   #4
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Quote:
Originally Posted by davieddy View Post
I thought of that, but I'm not sure that taking
the centre of the 20 triangles gives the simplest
orientation for the vertices of the dodecahedron.
I get face-center coordinates such as [s,s,s] and [s,-s,s] with s = (1+r)/3. This can be scaled to [1,1,1] etc. Should be simple enough. Will follow up.
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Old 2010-07-04, 17:52   #5
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Quote:
Originally Posted by ccorn View Post
This can be scaled to [1,1,1] etc.
Which suggests that we have cubes in there. Correct.
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Old 2010-07-04, 18:14   #6
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Quote:
Originally Posted by ccorn View Post
I get face-center coordinates such as [s,s,s] and [s,-s,s] with s = (1+r)/3. This can be scaled to [1,1,1] etc. Should be simple enough. Will follow up.
Here is it:

(+/-r-1, +/-r, 0) (permute cyclicly), (+/-1,+/-1,+/-1) dodecahedron

(I have used r = (1+sqrt(5))/2, hence 1/r = r-1, but the above scheme can be used with r's conjugate as well.)

Last fiddled with by ccorn on 2010-07-04 at 18:55 Reason: Explain r
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Old 2010-07-04, 19:35   #7
davieddy
 
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Quote:
Originally Posted by ccorn View Post
Here is it:

(+/-r-1, +/-r, 0) (permute cyclicly), (+/-1,+/-1,+/-1) dodecahedron

(I have used r = (1+sqrt(5))/2, hence 1/r = r-1, but the above scheme can be used with r's conjugate as well.)
Sounds right to me.
The edges of one are perpendicular to those of the dual.
I've just remembered why I brought this up:
World cup football!
In Mexico 1970 they first used a truncated icosohedron,
(20 white hexagons and 12 black pentagons).
Better known these days as C60 or Buckminsterfullerine.
I can't see why he found it so difficult to think of
a structure with 60 vertices.
I made one out of cardboard at the time, also
the great(?) stellated(?) dodecahedron which
makes a beautiful Christmas decoration.

David

Last fiddled with by davieddy on 2010-07-04 at 19:47
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Old 2010-07-04, 21:41   #8
XYYXF
 
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http://en.wikipedia.org/wiki/Dodecah...an_coordinates
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Old 2010-07-04, 21:48   #9
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Well, now I can confirm that statement in Wikipedia
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Old 2010-07-04, 22:21   #10
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Quote:
Originally Posted by XYYXF View Post
I'd like to add a link to Wolfram's Mathworld. Particularly interesting for me is the stuff beginning with equation (28).
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