Golden Ratio in Two Squares, Or, Perhaps in Three

Source

The following problem by Spt Ân (Vietnam) has been posted at the Peru Geometrico facebook group.

Golden Ratio in Two Squares, Or, Perhaps in Three, source

Problem

I prefer to reformulate the original problem by embedding it into an older configuration of three squares

Given three squares $ABCD,$ $DCEF,$ $FEGH$ and the circumcircle $\omega=(DCEF).$ Let $J=\omega\cap AG$ and $K=EJ\cap FH.$

Golden Ratio in Two Squares, Or, Perhaps in Three, illustration

Prove that $\displaystyle \frac{FK}{KH}=\frac{\sqrt{5}+1}{2}=\varphi,$ the Golden Ratio.

Proof

Let's denote $\angle CAG=\alpha.$ Then $\angle AGB=45^{\circ}-\alpha$ and $\displaystyle \tan(45^{\circ}-\alpha)=\frac{1}{3}.$ Setting $\tan\alpha=x,$ we get,

$\displaystyle \frac{1}{3}=\frac{1-x}{1+x},$

from which $\displaystyle x=\frac{1}{2}.$ By angle chasing, $\displaystyle \angle FEK=\angle FEJ=45^{\circ}-\frac{\alpha}{2}.$ If $\displaystyle y=\tan\frac{\alpha}{2},$ then $\displaystyle \frac{1}{2}=x=\frac{2y}{1-y^2}.$ Thus, choosing the positive value, $\displaystyle y=-2+\sqrt{5}.$ Further,

$\displaystyle \tan\angle FEK=\frac{1-y}{1+y}=\frac{3-\sqrt{5}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{2}.$

The meaning of this is that $\displaystyle \frac{FK}{EF}=\frac{FK}{FH}=\frac{\sqrt{5}-1}{2},$ so that $\displaystyle \frac{FK}{KH}=\frac{\sqrt{5}+1}{2}=\varphi,$ the Golden Ratio.

Extra

As the picture below (and above) shows, there are additional occurrences of the Golden Ratio.

Golden Ratio in Two Squares, Or, Perhaps in Three, illustration

 

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