mersenneforum.org problem to do with golden ratio equation
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 2020-01-14, 12:38 #1 wildrabbitt   Jul 2014 1101111112 Posts problem to do with golden ratio equation Can anyone explain what's wrong with my logic? https://www.mersenneforum.org/attach...1&d=1579005470 Attached Thumbnails
 2020-01-14, 13:09 #2 axn     Jun 2003 2×13×181 Posts Golden ratio is an increasing ratio (i.e > 1). The first equation uses x as a decreasing ratio (i.e. x < 1). So you get 1/gr when you solve that.
 2020-01-14, 14:45 #3 Dr Sardonicus     Feb 2017 Nowhere 3·7·132 Posts The usual formulation for x and y being in golden proportion is $\frac{x}{y}\;=\;\frac{x\;+\;y}{x}$; the right-hand side clearly is greater than 1. Taking y = 1 gives $\frac{x}{1}\;=\;\frac{x\;+\;1}{x}\text{, or }x^{2}\;-\;x\;-\;1\;=\;0\text{.}$ An illustration is given by the 72-72-36 degree isosceles triangle. The bisector of one of the 72-degree angles divides the opposite side in golden ratio; calling x the length of the base and y the length of the smaller segment of the side opposite the angle bisector, gives the above proportion.
 2020-01-14, 14:54 #4 wildrabbitt   Jul 2014 3×149 Posts Thanks very much to both of you.

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