Golden Ratio in Pentagon: Once More

Problem

Let $ABCDE\;$ be a regular pentagon and $F\;$ an arbitrary point on the small arc $AB.$

Golden Ratio in Pentagon: Once More - Problem and Solution

Then

$\displaystyle\frac{EF+CF}{DF}=\frac{DF}{AF+BF}=\varphi,$

the Golden Ratio.

Proof

Elsewhere we proved that either ratio remains constant, regardless of the position of $F\;$ on the arc $EC!\;$ Thus the calculations can be reduced to a particular case where, e.g., $F=C.\;$ Then

$\displaystyle\frac{EF+CF}{DF}=\frac{CE}{CD}=2\cdot\sin 54^{\circ}=\varphi,$

see a derivation.

We could have taken $F=A.\;$ We would have

$\displaystyle\begin{align} \frac{EF+CF}{DF}&=\frac{AE+AC}{AD}\\ &=\frac{1+2\sin 54^{\circ}}{2\cdot\sin 54^{\circ}}\\ &=\frac{1+\varphi}{\varphi}\\ &=\varphi, \end{align}$

as expected.

Acknowledgment

The problem Dao Thanh Oai, with a comment to the latter by Tim Robinson who observed that

$\displaystyle\varphi = \frac{FB + FD}{FC} = \frac{FA + FD}{FE} = \frac{FC - FA}{FB}=\frac{FE - FB}{FA}.$

which led to a simple but general statement.

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