Math Is Fun Forum

simron

Real Member

Registered: 2006-10-07

Posts: 237

pentagon conjecture

You may have noticed this before, but here goes:
1. Draw a regular pentagon.
2. Make a star in there by connecting every other vertex
3. Draw a star in that one by the same method
The pattern is the larger star's length is about 2.013 times the smaller star's length.
This number might be irrational (though I have no way to find out. )

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MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: pentagon conjecture

The pentagon and pentagram have the Golden Ratio in them. I have just finished this page: The Pentagram

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

simron

Real Member

Registered: 2006-10-07

Posts: 237

Re: pentagon conjecture

OK, how does the Golden Ratio corrospond to 2.013?

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Patrick

Real Member

Registered: 2006-02-24

Posts: 1,005

Re: pentagon conjecture

Not really:

It's great to see you're experimenting though, keep on doing that!

edit: Misread what you said.. I don't think that the ratio you're describing has anything to do with the golden ratio(might be wrong), Rod just said the pentagon has the Golden Raito in it.

Last edited by Patrick (2006-12-08 07:52:05)

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MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: pentagon conjecture

Here is a quick sketch I just did. I drew only one line of the "inner" pentagram so I could get a length. The ratio of the numbers is 1.618....

I don't get the 2.013. Could you explain further?

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

simron

Real Member

Registered: 2006-10-07

Posts: 237

Re: pentagon conjecture

Sorry... I totally edited this message because I wasn't clear in my first post.
EDIT: I can believe MathsIsFun, but that wasn't the ratio I meant. It's the side of the inner pentagon I meant. (Not the diagonal) Sorry for the confusion. Also, I probably calculated it wrong. Shame on me for not using a calculator.

Last edited by simron (2008-01-04 16:13:16)

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MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: pentagon conjecture

Good question, though. There will certainly be some interesting ratios. Try some other regular polygons, too.

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

simron

Real Member

Registered: 2006-10-07

Posts: 237

Re: pentagon conjecture

OK, looks like my division by hand isn't reliable. The ratio is 2.6180738... when I used the values MathsIsFun used in his drawing.
It's very close to the Golden Ratio +1. I wonder if the two are connected to each other somehow. This will be an interesting project to dust off after 2 years.

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John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: pentagon conjecture

@simron, I noted on calculator that 1.618034^2 is 2.618034

1.618034^2 = ([ sqroot{5}/2 ] + [ 1/2 ])^2

Expands to:  5/4 + sqroot{5}/2 + 1/4

2.618034 = 1.5 + sqroot{5}/2, which is clearly 1.0 larger than the
original 1.618034 = 0.5 + sqroot{5}/2

Last edited by John E. Franklin (2008-04-14 15:20:31)

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cushydom

Member

Registered: 2008-04-23

Posts: 10

Re: pentagon conjecture

The ratio of the two pentagons is equal to sqrt((1-cos(108))/(1-cos(36)))

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: pentagon conjecture

Or (Sin72°/Sin36°)*(Sin72°/Sin36°)

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simron

Real Member

Registered: 2006-10-07

Posts: 237

Re: pentagon conjecture

Ah. I wonder how to prove this.

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cushydom

Member

Registered: 2008-04-23

Posts: 10

Re: pentagon conjecture

You can prove that the ratio is sqrt((1-cos(108))/(1-cos(36))) by using the cosine rule to get expressions for the lentghs of th e two pentagons