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

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

角\(\theta\)との関係

$$\begin{eqnarray} \frac{\overline{CF}}{\overline{OF}}&=&2\;\cdots\;正方形となるため \\&=&\frac{\sin{(\theta)}}{\cos{(\theta)}} \\&=&\tan{(\theta)} \\\tan^{-1}{(2)}&=&\theta \end{eqnarray}$$

線分\(OF\)の長さ

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

線分\(CF\)の長さ

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

面積\(A_1\)

$$\begin{eqnarray} A_1&=&\overline{CF}^2 \\&=&\left(\frac{2}{\sqrt{5}} \right)^2 \\&=&\frac{2^2}{\left(\sqrt{5}\right)^2} \\&=&\frac{4}{5}\;(=0.8) \end{eqnarray}$$