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 Post subject: A small math problem.
PostPosted: Thu Nov 18, 2010 10:08 am 
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Suppose we have a dodecahedron that is marked on the faces like a Supernova puzzle:

http://twistypuzzles.com/cgi-bin/puzzle.cgi?pkey=651

In other words, we have a dodecahedron with twelve rings around it that meet at the midpoints of each edge.

If we slice away the twelve sides along those lines, we are left with a smaller dodecahedron. What is the volume of that smaller dodecahedron, as a percentage of the original?


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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 10:41 am 
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I'm not sure how this would be calculated but I made the two shapes in a CAD program and had it calculate the volumes of both solids.
The smaller one was approximately 16.725% volume of the larger one. A tiny bit more than one sixth.

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 11:41 am 
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I calculated the edge of the smaller dodecahedron as
(sin 54° + sin 18° - 1/2) = 0.618 of the bigger one, so the volume is 0.618^3 = 23.6%.

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 12:39 pm 
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Timur wrote:
I calculated the edge of the smaller dodecahedron as
(sin 54° + sin 18° - 1/2) = 0.618 of the bigger one, so the volume is 0.618^3 = 23.6%.


Image

Maybe I'm not awake yet, but don't you mean (sin36°/2sin72°)=0.30902...? (and then cube that for the ratio of the volumes, of course)

Odd that neither of our answers correspond with what CAD apparently says. What gives?


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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 12:59 pm 
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It should end up being the cube of the ratio of the face-to-face distance of the smaller (internal) vs larger (original) dodecahedra, on the basis that volume is proportional to the cube of linear scale, for a similar 3D form.

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 1:24 pm 
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Iranon wrote:
Maybe I'm not awake yet, but don't you mean (sin36°/2sin72°)=0.30902...? (and then cube that for the ratio of the volumes, of course)

The pentagon that you mean is not a face of the small dodecahedron. Faces of the small dodec are bigger and they lie not on the outer surface, but on the cutting layer.

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 1:37 pm 
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Did anyone just mention the word "golden ratio"?
(something like "I tot I taw a puttytat", but in a more mathematical version)

:lol: 8-)


Pantazis

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 1:39 pm 
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Timur wrote:
Iranon wrote:
Maybe I'm not awake yet, but don't you mean (sin36°/2sin72°)=0.30902...? (and then cube that for the ratio of the volumes, of course)

The pentagon that you mean is not a face of the small dodecahedron. Faces of the small dodec are bigger and they lie not on the outer surface, but on the cutting layer.


Oops! Of course it is, thank you. I was wondering what was wrong... Now I'm curious, how do you set up a simple calculation for that inner edge length?


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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 1:51 pm 
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Iranon wrote:
Now I'm curious, how do you set up a simple calculation for that inner edge length?

It was not simple at all, I had to write a whole page before I got the result. No wonder if I made a mistake somewhere :)

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 2:25 pm 
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Iranon wrote:
Maybe I'm not awake yet, but don't you mean (sin36°/2sin72°)=0.30902...?

I calculated it completely another way and came to the conclusion that the edge size to be found will be exactly double of the side of this small pentagon on a face that you originally meant, that makes 2*0.30902 = 0.618, and that just confirms my first answer :)

So the answer may be also written as (sin 36°/sin 72°)^3 or just (2*sin 18°)^3.

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 Post subject: Re: A small math problem.
PostPosted: Thu Nov 18, 2010 2:38 pm 
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My apologies guys. Don't ask how I messed up a dodecahedron in CAD but I did. I measured again and the small dodecahedron's volume is 0.236068 of the original. The edge length is .618034 of the original. You all were absolutely right. Sorry for the confusion :oops:

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