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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 $\displaystyle A,B,C,D,E$. Let's call the vertex in the center $\displaystyle F$. Triangle ABE is similar to triangle CDF. You will want to show that BF=FE=1 and FC=FD=(x-1). Then $\displaystyle \dfrac{1}{x-1} = \dfrac{EA}{FD} = \dfrac{BE}{CD} = \dfrac{x}{1}$.
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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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