Golden Ratio in 5-gon and 6-gon

The following has been posted by Tran Quang Hung at the CutTheKnotMath facebook page, with a proof by Leo Giugiuc.

Golden Ratio 5-gon & 6-gon, problem

Indeed, due to symmetry, suffice it to prove that $\displaystyle\frac{DR}{ER}=\varphi.\;$

Choose, WLOG, $A=0,\;$ $F=2,\;$ $E=3+i\sqrt{3},\;$ and $D=2+2i\sqrt{3}.\;$ We have,$\displaystyle\frac{A-F}{P-F}=\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5},\;$ implying

$\displaystyle\begin{align} P&=2-2\cos\frac{3\pi}{5}+2i\sin\frac{3\pi}{5}\\ &=2+2\sin\frac{2\pi}{5}+2i\sin\frac{3\pi}{5}\\ &=4\cos^2\frac{\pi}{5}+2i\sin\frac{3\pi}{5}\\ &=\varphi^2+2i\sin\frac{3\pi}{5}\\ &=1+\varphi+2i\sin\frac{3\pi}{5}. \end{align}$

From here, the line $PR\;$ is $x=1+\varphi,\;$ so that $R=1+\varphi+ki\;$ (we are not interested in the value of k.) It follows that $\displaystyle\overrightarrow{DR}=\varphi-1+mi=\frac{1}{\varphi}+mi\;$ and $\displaystyle\overrightarrow{RE}=2-\varphi+ni=\frac{1}{\varphi^2}+ni\;$ and, finally, $\displaystyle\frac{DR}{ER}=\frac{\displaystyle\frac{1}{\varphi}}{\displaystyle\frac{1}{\varphi^2}}=\varphi.$

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