# Golden Ratio in Mixtilinear Circles

Tran Quang Hung has found the Golden Ratio among mixtilinear circles in equilateral triangle. I am surprised no more. However, the novel manifestation of the universality of the Golden Ratio has warranted some investigation. It appears that even a single mixtilinear circle in equilateral triangle is conducive to proliferation of that universal constant. Here's what I found: ($F$ is the center of the circle. The significance of other points is clear from the diagram.)

$\displaystyle\frac{GP}{JP}=\frac{GJ}{IJ}=\frac{KO}{JO}=\frac{GK}{KQ}=\frac{KO}{KL}=\frac{GQ}{CG}=\frac{KQ}{BQ}=\phi.$

As a matter of fact, rather often, elegant results in mathematics are established by proofs that lack appeal, let alone elegance, and provide no enlightenment into the nature of the statement. For me, this was the case with all of the above ratios. I confess to having proved the third one and getting a notion of how to proceed from there. I recoil from doing that.

$\displaystyle\frac{KO}{JO}=\phi.$

I'll work with $\Delta FJN:$ Assume the side of $\Delta ABC$ is $1.$ Then $AD=\displaystyle\frac{2}{3}$ and $\displaystyle FJ=DF=\frac{2\sqrt{3}}{9}.$ $DN$ is one third of the altitude from $C,$ i.e., $DN =\displaystyle\frac{\sqrt{3}}{6}.$ Thus, $JN^{2}=FJ^{2}-FN^{2},$ or more explicitly

$\displaystyle JN^{2}=\bigg(\frac{2\sqrt{3}}{9}\bigg)^{2}-\bigg(\frac{2\sqrt{3}}{9}-\frac{\sqrt{3}}{6}\bigg)^{2}=\frac{5}{36},$

implying $\displaystyle JN=\frac{\sqrt{5}}{6}.$ Further, $\displaystyle DO=KO=\frac{1}{3}$ and $\displaystyle NO=\frac{1}{6}.$ It follows that

\begin{align}\displaystyle \frac{KO}{JO}&=\frac{1/3}{\sqrt{5}/6-1/6}\\ &=\frac{2}{\sqrt{5}-1}\\ &=\frac{\sqrt{5}+1}{2}=\phi. \end{align}

(A solution by Leo Giugiuc to the original problem of two mixtilinear circles can be found in a separate file.) ### Golden Ratio 