# Pentagon

2015-01-02

A regular pentagon has the following dimensions:

AB=BC=CD=DE=AE=BG=EG=1

AC=AD=BD=BE=CE=Φ

BF=EF=Φ/2 = cos 36 = sin 54 = $\frac{1+\sqrt5}{4}$

CG=DG= Φ-1 = 1/Φ = $\frac{\sqrt5-1}{2}$

AF=FG= cos 54 = sin 36 = $\sqrt{\frac{5-\sqrt5}{8}}$

AG= 2cos 54 = 2sin 36= $\sqrt{\frac{5-\sqrt5}{2}}$

CH=DH=BK=CK=1/2

Height AH=EK=$\frac12\cos 18 = \frac12\sin 72= \frac12\sqrt{5+2\sqrt5}$

Circumradius AJ=$\sqrt{\frac{5+\sqrt5}{10}}$=BJ=CJ=DJ=EJ

Inradius JK=HJ=$\sqrt{\frac{5+2\sqrt5}{20}}$=$\frac{AH}{\sqrt5}$

FH= cos 18 =sin 72 = $\sqrt{\frac{5+\sqrt5}{8}}$

FJ=JK/Φ

GJ=AJ/Φ

GH=AF/Φ=$\frac12\sqrt{5-2\sqrt5}$

ABGE is a equilateral rhombus, with angles {108°, 54°, 108°, 54°} and diagonals of length Φ and $\sqrt{\frac{5-\sqrt5}{2}}$.

The small pentagon in the center has sides $1/\Phi^2 = \frac{3-\sqrt5}{2}$ times that of the larger.

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