Golden Ratio in Three Regular Pentagons II

Problem

Golden   Ratio in Three Regular Pentagons II, problem

Solution

The problem and all variants can be solved by explicit coordinates of the involved points, using $\cos 36^{\circ}= \displaystyle \frac{\varphi}{2}.\,$ For example, when $ABCDE\,$ is the unit pentagon with $A=(1,0),\,$ the circle of the problem is centered at $\displaystyle \left(\frac{3-\sqrt{5}}{2}, 0\right)\,$ with squared radius $4-\sqrt{5}\,$ and $\displaystyle T = \left(3 - 3\frac{\varphi}{2}, \sqrt{3-\varphi}\left(\varphi-\frac{1}{2}\right)\right),\,$ $\displaystyle B=\left(\frac{\varphi-1}{2}, \frac{\varphi}{2}\cdot\sqrt{3-\varphi}\right),\,$ $F=(1, \varphi\cdot\sqrt{3-\varphi}).$

Variants

The configuration sports several occurences of the Golden Ratio. The applet below highlights seven of them. #5 is the original one. ##1-3 are rather traditional and simple, #4 is intermediate, ##5-7 are difficult. (I apologize for a change of notations.)

Acknowledgment

The problem has been kindly posted on the CutTheKnotMath facebook page by Tran Quang Hung.

 

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