mersenneforum.org

mersenneforum.org (https://www.mersenneforum.org/index.php)
-   Homework Help (https://www.mersenneforum.org/forumdisplay.php?f=78)
-   -   problem to do with golden ratio equation (https://www.mersenneforum.org/showthread.php?t=25112)

wildrabbitt 2020-01-14 12:38

problem to do with golden ratio equation
 
1 Attachment(s)
Can anyone explain what's wrong with my logic?


[URL]https://www.mersenneforum.org/attachment.php?attachmentid=21617&stc=1&d=1579005470[/URL]

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 [tex]\frac{x}{y}\;=\;\frac{x\;+\;y}{x}[/tex]; the right-hand side clearly is greater than 1. Taking y = 1 gives

[tex]\frac{x}{1}\;=\;\frac{x\;+\;1}{x}\text{, or }x^{2}\;-\;x\;-\;1\;=\;0\text{.}[/tex]

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.


All times are UTC. The time now is 05:01.

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.