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Polyhedra

 
 
Reply Mon 26 Mar, 2007 04:44 pm
There are 5 regular polyhedra - tetrahedron, hexahedron (cube), octahedron, dodecahedron and icosahedron.

There is also a curious way that these 3-dimensional 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?

http://img2.freeimagehosting.net/uploads/3b618f9fe9.gif


Edit [Moderator]: Moved from Riddles to Science & Mathematics.
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Type: Discussion • Score: 0 • Views: 1,579 • Replies: 8
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markr
 
  1  
Reply Mon 26 Mar, 2007 09:49 pm
I've solved this one before, but I don't recall the (puzzle?) site where I found it.
0 Replies
 
official
 
  1  
Reply Fri 30 Mar, 2007 12:54 am
I found the img on another site, but not the puzzle...:S
0 Replies
 
official
 
  1  
Reply Sat 31 Mar, 2007 01:06 pm
Here was my idea:

square((square8*1.5*2)²/2)=a

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square((square8*2*2)²/2)=a

This does not work, however
0 Replies
 
official
 
  1  
Reply Mon 2 Apr, 2007 09:18 pm
Anyone can help?
0 Replies
 
official
 
  1  
Reply Wed 11 Apr, 2007 03:53 am
I've been trying this, no luck. Is there anyone that can help me? Very Happy
0 Replies
 
official
 
  1  
Reply Fri 27 Apr, 2007 01:41 am
Still haven't been able to make sense of this one, although the tip to work from the outside in seems like it may help some.
0 Replies
 
official
 
  1  
Reply Sun 6 May, 2007 12:24 am
My idea:

start with side of cube = 1.
side of tetra = root 2
side of octa = root 2 / 2
side of icosa = side of octa * 1 / Phi exp 3/2
side of dodeca = 1 / Phi * side of icosa

But I am told the answer is somewhere 9+ and cannot write out an equation out of this information to make this work. Can anyone else help?
0 Replies
 
troptrop
 
  1  
Reply Tue 12 May, 2015 02:11 am
I guess your formules aren't correct:

The ones I found are different:
side of tetra=side of cube * sqrt(2)
side of dodeca = (1 / Phi) * side of icosa => because dodeca =2 > icosa =2 * phi

The other formules I couldn't find yet :s
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