Golden Ratio In Three Circles And Common Secant

The following construction of the Golden Ratio has appeared in the Mathematical Gazette, volume 101,number 551, July 2017, page 303. The construction is by John Molokach.

Golden Ratio In Three Circles And Common Secant #2

There are two unit circles $(A)\,$ and $(B),\,$ and circle $\omicron,\,$ tangent to both. The vertical segments $AC\,$ and $BF\,$ are tangent to $(A)\,$ and $(B),\,$ respectively. Both are of length $1.\,$ $CF\,$ crosses $\omicron\,$ in $D\,$ and $E,\,$ as shown.

John proves that $CE=\varphi,\,$ the Golden ratio. Indeed, let $CE=x.\,$ Then, by the Intersecting Secants theorem, $CA^2=CD\times CE,\,$ i.e., $1=(x-1)x\,$ so that $x\,$ is the positive root of the equation $x^2-x-1=0,\,$ which is exactly $\displaystyle \varphi=\frac{1+\sqrt{5}}{2}.\,$ In addition, $\displaystyle CD=\varphi-1=\frac{-1+\sqrt{5}}{2}=\frac{1}{\varphi}.\,$ It follows that, too, $\displaystyle \frac{DE}{CD}=\varphi.$

It must be noted that the construction, obtained independently, embeds into the one by John Arioni.

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