単位円の半分に収まる(?)正方形の面積は有理数(一辺の長さは無理数(sin(arctan(2))=2/√5))

点$C$の位置

$$\begin{array}{rcl}& & C(\overline{OF},\overline{CF})\\ & =& C(\cos (\theta ),\sin (\theta ))\end{array}$$

角$\theta$との関係

$$\begin{array}{rcl}\frac{\overline{CF}}{\overline{OF}}& =& 2\cdots \mathrm{正}\mathrm{方}\mathrm{形}\mathrm{と}\mathrm{な}\mathrm{る}\mathrm{た}\mathrm{め}\\ & =& \frac{\sin (\theta )}{\cos (\theta )}\\ & =& \tan (\theta )\\ {\tan }^{-1}(2)& =& \theta \end{array}$$

線分$OF$の長さ

$$\begin{array}{rcl}\overline{OF}& =& \cos ({\tan }^{-1}\left(2\right))\\ & =& \frac{1}{\sqrt{\tan {({\tan }^{-1}(2))}^2+1}}\\ & & \cdots \cos (\theta )=\frac{1}{\sqrt{{\tan }^2\left(\theta \right)+1}}\\ & =& \frac{1}{\sqrt{2^2+1}}\\ & =& \frac{1}{\sqrt{5}}\end{array}$$

線分$CF$の長さ

$$\begin{array}{r}\overline{CF}\\ & =& \sin (\theta )\\ & =& 2\cos (\theta )\\ & =& 2\frac{1}{\sqrt{5}}\\ & =& \frac{2}{\sqrt{5}}\end{array}$$

面積$A_1$

$$\begin{array}{rcl}{A}_1& =& {\overline{CF}}^2\\ & =& {\left(\frac{2}{\sqrt{5}}\right)}^2\\ & =& \frac{2^2}{{\left(\sqrt{5}\right)}^2}\\ & =& \frac{4}{5}(=0.8)\end{array}$$