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The Yamato Ratio Theorem of the Regular Pentagon
Simon Takahashi

$x^2+y^2=1$

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

$P(0,-1)$

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

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

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

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

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

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

$A{P}^2-C{P}^2=2^2-(\frac{2}{3}{)}^2=(\frac{4\sqrt{2}}{3}{)}^2$
$AC=\frac{4\sqrt{2}}{3}$

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

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

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

The assumption is correct.

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

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

$cf.$

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

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