Golden Ratio in Two Squares

Tran Quang Hung has posted on the CutTheKnotMath facebook page a simple construction of the Golden Ratio in Two Squares.

Construction

Square $MNPC$ has point $M$ on the diagonal $BD$ of square $ABCD,$ $Q$ the intersection of $MP$ and $CD,$ extended.

golden ratio by Tran Quang Hung, construction #4

If $M$ is such that $2DQ=CD$ then $AN/BN=\phi.$

Proof

The proof is by Leo Giugiuc and uses complex numbers.

We choose $A=-1+i,$ $B=1+i,$ $C=1-i,$ $D=-1-i$ so that $Q=-2-i$ and $M=k+ki,$ $k\in (-1 ,1).$

$CN\perp QM$ and $QM$ has the slope $\displaystyle\frac{k+1}{k+2},$ implying $CN$ is described by the equation

$\displaystyle y+1=-\frac{k+2}{k+1}(x-1).$

Since $N\in AB$ then $N=x+i;$ so solving for $x,$ $x=-\displaystyle\frac{k}{k+2}$ and then for $N=-\displaystyle\frac{k}{k+2}+i.$ $NMCP$ is a positively oriented square, therefore, $\displaystyle\frac{N-M}{C-M}=i$ such that

$-k(k+3)+(1-k)(k+2)i=(k+2)[(k+1)+(1-k)i]$

which simplifies to $k^2+3k+1=0.$ Since $k\in (-1 ,1),$ $k=-1/\phi^{2},$ $N=(1/\phi^{2} )/\phi +i=1/\phi^{3} +i=\sqrt{5}-2+i$ so that, finally, $NA/NB=\phi.$

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]