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Thread: Golden ratio pentagon

  1. #1
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    Golden ratio pentagon

    So I need to show that the diagonal x satisfies the following ratio x/1=1/(x-1). I know this turns out to be the golden ratio, But I cant seem to find the similar triangles in order to find this ratio in this pentagon. Any help is appreciated, thanks

    Golden ratio pentagon-screen-shot-2013-11-06-1.42.27-pm.png
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  2. #2
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    Re: Golden ratio pentagon

    Let's label the vertices clockwise starting with the topmost vertex. Let's call them A,B,C,D,E. Let's call the vertex in the center F. Triangle ABE is similar to triangle CDF. You will want to show that BF=FE=1 and FC=FD=(x-1). Then \dfrac{1}{x-1} = \dfrac{EA}{FD} = \dfrac{BE}{CD} = \dfrac{x}{1}.
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  3. #3
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    Re: Golden ratio pentagon

    I suggest reading this article I wrote (it's also in the MHF Magazine, Issue 2 I think) from page 10.
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