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The Yamato Ratio Theorem of the Regular Pentagon
Simon Takahashi

$x^2+y^2=1$

$A\quad(0,1)$
$B\quad(\frac{2\sqrt{2}}{3},\frac{1}{3})$
$C\quad(\frac{1}{\sqrt{3}},-\frac{\sqrt{2}}{\sqrt{3}})$
$D\quad(-\frac{1}{\sqrt{3}},-\frac{\sqrt{2}}{\sqrt{3}})$
$E\quad(-\frac{2\sqrt{2}}{3},\frac{1}{3})$

$P\quad(0,-1)$

$x^2+y^2=1$
$x^2+(y-1)^2=(\frac{2}{\sqrt{3}})^2$
$y=\frac{1}{3}$

$x^2+y^2=1$
$x^2+(y+1)^2=(\frac{2}{\sqrt{3}})^2$
$y=-\frac{1}{3}$

$\vert\frac{1}{3}\vert+\vert-\frac{1}{3}\vert=\frac{2}{3}$

Assuming that $CP=\frac{2}{3},$

$AB=BC=CD=DE=EA=\frac{2}{\sqrt{3}}$

$\angle{}A=\angle{}B=\angle{}C=\angle{}D=\angle{}E=108^\circ$

$AP^2-CP^2=2^2-(\frac{2}{3})^2=(\frac{4\sqrt{2}}{3})^2$
$AC=\frac{4\sqrt{2}}{3}$

$AC=BD=CE=DA=EB=\frac{4\sqrt{2}}{3}$

$CP=\frac{2}{3}\Longleftrightarrow{}ABCDE$ is a regular pentagon.

$CP=\frac{2}{3}\land{}ABCDE$ is a regular pentagon.

$1^2-(\frac{\sqrt{2}}{\sqrt{3}})^2=(\frac{1}{\sqrt{3}})^2$

The assumption is correct.

$ABCD\equiv{}BCDE\equiv{}CDEA\equiv{}DEAB\equiv{}EABC$

$\frac{EB}{CD}=\frac{4\sqrt{2}}{3}\cdot{}\frac{\sqrt{3}}{2}=\frac{2\sqrt{2}}{\sqrt{3}}$
$\phi=\frac{2\sqrt{2}}{\sqrt{3}}$

$cf.$

$AP\neq{}1+\frac{\sqrt{2}}{\sqrt{3}}+\frac{1}{3}$

The Fibonacci golden number is not the golden ratio of regular pentagon.
$\because$
$\frac{1+\sqrt{5}}{2}$ is not a line segment.

2025.6.1