# 二百五十五角形

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## 正二百五十五角形

${\displaystyle S={\frac {255}{4}}a^{2}\cot {\frac {\pi }{255}}\simeq 5174.26329a^{2}}$

${\displaystyle \cos(2\pi /255)}$は有理数と平方根の組み合わせのみで表せる。

{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{255}}=&\cos \left({\frac {\pi }{15}}-{\frac {\pi }{17}}\right)\\&=\cos {\frac {\pi }{15}}\cos {\frac {\pi }{17}}+\sin {\frac {\pi }{15}}\sin {\frac {\pi }{17}}\\&={\frac {1}{8}}\left({-1+{\sqrt {5}}+{\sqrt {30+{\sqrt {180}}}}}\right)\cdot \cos {\frac {\pi }{17}}+{\frac {1}{8}}\left({+{\sqrt {3}}-{\sqrt {15}}+{\sqrt {10+{\sqrt {20}}}}}\right)\cdot \sin {\frac {\pi }{17}}\\&={\frac {1}{8}}\left({-1+{\sqrt {5}}+{\sqrt {30+{\sqrt {180}}}}}\right)\cdot {\frac {1}{16}}\left({+1-{\sqrt {17}}+{\sqrt {34-{\sqrt {68}}}}+{\sqrt {68+{\sqrt {2448}}+{\sqrt {2720+{\sqrt {6284288}}}}}}}\right)+{\frac {1}{8}}\left({+{\sqrt {3}}-{\sqrt {15}}+{\sqrt {10+{\sqrt {20}}}}}\right)\cdot {\frac {1}{8}}\left({\sqrt {34-{\sqrt {68}}-{\sqrt {136-{\sqrt {1088}}}}-{\sqrt {272+{\sqrt {39168}}-{\sqrt {43520+{\sqrt {1608777728}}}}}}}}\right)\\\end{aligned}}}
${\displaystyle 255=16^{2}-1=(16+1)(16-1)=(16+1)(4+1)(4-1)=(16+1)(4+1)(2+1)(2-1)=17\cdot 5\cdot 3}$

## 脚注

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