# Golden Ratio in Square

Tran Quang Hung has posted on the CutTheKnotMath facebook page a simple construction of the golden ratio.

### Construction

In square $ABCD$ $M,N,P,Q$ are midpoints of the sides, as shown below; $F$ the center of the square. Then circle $(AP)$ on $AP$ as a diameter, cuts $NF$ (point $X$) in Golden Ratio. Chord $AX$ cuts $MF$ in Golden Ratio:

Then $\displaystyle\frac{FX}{NX}=\frac{MZ}{FZ}=\phi,$ the golden ratio.

### Proof

Assume, without loss of generality, that $AB=2.$ Then $AP=\sqrt{5}.$ Since $AP$ is a diameter of the circle, $E$ - the intersection of $AP$ and $NQ$ is its center, which shows that $EX=\frac{1}{2}\sqrt{5}$ and that $EF=\frac{1}{2}.$ It follows that $FX=\frac{1}{2}(\sqrt{5}-1)$ and $NX=1-FX=\frac{1}{2}(3-\sqrt{5}),$ with a consequence that $\displaystyle\frac{FX}{NX}=\frac{1+\sqrt{5}}{2}=\phi.$

Next, triangles $AQX$ and $ZFX$ are similar: $\displaystyle\frac{FZ}{AQ}=\frac{FX}{QX},$ implying

$\displaystyle FZ=\frac{FX}{FX+1}=\frac{\sqrt{5}-1}{\sqrt{5}+1}=NX.$

We thus also have $MZ=FX$ and $\displaystyle\frac{MZ}{FZ}=\frac{FX}{NX}=\phi.$

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

Copyright © 1996-2018 Alexander Bogomolny