# 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.

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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