# 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.

### 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 $K=EJ\cap FH.$

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.