April 3rd, 2007, 09:17 PM  #1 
Guest Joined: Posts: n/a Thanks:  Polyhedra
There are 5 regular polyhedra  tetrahedron, hexahedron (cube), octahedron, dodecahedron and icosahedron. There is also a curious way that these 3dimensional objects fit together. Starting with a dodecahedron, fit around an icosahedron, so that the 20 vertices of the dodecahedron meet the centre of the 20 faces of the icosahedron. Then build an octahedron around the icosahedron so that the 12 vertices of the icosahedron meet with the 12 edges of the octahedron (according to the golden section). Next build a tetrahedron around the octahedron so that the 6 vertices of the octahedron meet with the mid points of the 6 edges of the tetrahedron. Finally put a cube around the tetrahedron so that the corners meet. Assuming that all 30 edges of the dodecahedron are 2 in length, and that all of the polyhedra are regular. What is the length of the cube's edge? See, here is what I thought for an answer: square((square8*1.5*2)²/2)=a . . . . . . square((square8*2*2)²/2)=a which would make the answer 8. But this is just not correct. Here is a picture, too: So what I have to know is the formula, in its simplest form. Any help is appreciated! 
April 4th, 2007, 04:54 AM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
Unfortunately, I'm not sure I'll be much good at solving your problem, but that sure is a neat way to arrange the Platonic solids. I'd never seen them put together like that before.

April 5th, 2007, 02:54 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2217 
Let's work inwards from the cube. To get the tetrahedron's edge from the cube's edge, multiply by sqrt(2). Then divide that by two to get the octahedron's edge. Then divide that by two to get the icosahedron's edge. The final step requires a little trigonometry. Have a go at it.

April 5th, 2007, 04:24 PM  #4 
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Ok I follow you here. But this one starts from the inside? Hmm...any ideas anyone? My problem is finding the relationship between the dodecahedron (yellow solid) and icosahedron (red wire frame)...

April 11th, 2007, 01:53 AM  #5 
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I've been trying this, no luck. Is there anyone that can help me? 
April 11th, 2007, 04:02 AM  #6  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
 
November 21st, 2010, 01:52 AM  #7 
Member Joined: Jun 2010 Posts: 80 Thanks: 0  Re: Polyhedra
Is it possible to find those polyhedra with the formula : face, edges, vertices F  E + V = 2, if the faces are all triangles, we have E=3F/2, and V=3F/3 6F  9F + 6F = 12 3F = 12, hence F=4, if vertices are shared by 4 faces : V=3F/4, hence 4F  6F + 3F = 8 = F if we take 2 several regular polygon type as faces, is it sufficient to apply this formula, since then a polyhedron with 4 heptagons and 8 square could exist, which but i cannot visualize if it's possible. 
November 21st, 2010, 04:59 AM  #8 
Member Joined: Jun 2010 Posts: 80 Thanks: 0  Re: Polyhedra
I tried the same for mixed vertex sharing : the number of faces per vertex are mixed (like complementary fullerenes) : suppose base as a triangle and n1 and n2 are the possible number of faces per vertex, with 3F/k vertices n2connected, and 3F(11/k) n1connected F(n2(6n1) + 6/k*(n1n2)) = 4n1n2 for n1=5, n2=3, : F=12, k=6 : 10 vertices 5connected, 2 vertices 3connected 

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