Another Golden Ratio in Regular Pentagon

Problem

Another Golden Ratio in Regular Pentagon

Solution 1

WLOG, $OP=1.\;$ Recollect that $\displaystyle\varphi=2\cos\frac{\pi}{5}.\;$

In $\Delta OEM,\;$ $\displaystyle\frac{1}{OE}=\cos\frac{\pi}{5},\;$ so that $OE=\displaystyle\frac{2}{\varphi},\;$ implying $EP=\displaystyle\frac{2-\varphi}{\varphi}.\;$ Hence, too, $MR=\displaystyle\frac{2-\varphi}{\varphi}.\;$ From which

$\displaystyle NR=\frac{1}{\varphi}-\frac{2-\varphi}{\varphi}=\frac{\varphi-1}{\varphi}=\frac{1}{\varphi^2}.\;$

Triangles $QRN\;$ and $PQO\;$ are similar, such that

It follows that, $\displaystyle \frac{QR}{PR}=\varphi^2-1=\varphi.$

Solution 2

As above, $DO=\displaystyle\frac{2}{\varphi},\;$ $\displaystyle NO=\frac{1}{\varphi},\;$ $\displaystyle NQ=1-\frac{1}{\varphi}=\frac{\varphi-1}{\varphi}=\frac{1}{\varphi^2},\;$ $\displaystyle\frac{PR}{QR}=\frac{NO}{NQ}=\frac{\displaystyle\frac{1}{\varphi}}{\displaystyle\frac{1}{\varphi^2}}=\varphi.$

Acknowledgment

The problem has been posted on the CutTheKnotMath facebook page by Tran Quang Hung and commented on by Leo Giugiuc with Solution 1.

The new construction is a byproduct of the beautifully austere Tran Quang Hung's observation: $\displaystyle\frac{OF}{FG}=\varphi,\;$ where $F\;$ is the midpoint of $BO:$

Austere beauty in Regular Pentagon

 

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