Golden Ratio in Equilateral Triangle on the Shoulders of George Odom

Source

The following has been posted by Kadir Altintas at the Peru Geometrico facebook group.

Golden Ratio in  Equilateral Triangle on the Shoulders of George Odom

Problem

$ABC$ is an equilateral triangle; $D,F$ are intagent points on $AB,AC$ respectively; $N$ is the midpoint of $DE;$ $MP$ through $N$ is parallel to $AB$ where $M$ is on $AC,$ $P$ is on the incircle of $\Delta ABC,$ so that $N$ is in the segment $MP.$

Golden Ratio in  Equilateral Triangle on the Shoulders of George Odom,illustration

Prove that $\displaystyle \frac{PN}{NM}=\varphi,$ the Golden Ratio.

Proof

Let $G=MP\cap DF,$ where $F$ is the midpoint of $BC.$ Since $N$ is the midpoint of $DE,$ the two equilateral triangles $MNE$ and $DGN$ are equal.

Golden Ratio in  Equilateral Triangle on the Shoulders of George Odom,solution

This was one of the first and most surprising discoveries in modern era made by George Odom in the early 1980s that $\displaystyle \frac{NG}{PG}=\varphi.$ It follows that

$\displaystyle\begin{align} \frac{PN}{NM}&=\frac{PG+GN}{GN}=\frac{PG}{GN}+1=\frac{1}{\varphi}+1\\ &=\frac{\varphi+1}{\varphi}=\frac{\varphi^2}{\varphi}=\varphi. \end{align}$

 

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