problem to do with golden ratio equation

wildrabbitt · 2020-01-14, 12:38

Can anyone explain what's wrong with my logic?

https://www.mersenneforum.org/attach...1&d=1579005470

axn · 2020-01-14, 13:09

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.

Dr Sardonicus · 2020-01-14, 14:45

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.

wildrabbitt · 2020-01-14, 14:54

Thanks very much to both of you.

2020-01-14, 12:38	#1
wildrabbitt Jul 2014 3×149 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

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2020-01-14, 13:09	#2
axn Jun 2003 5,051 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 4,643 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.