# 百二角形

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## 正百二角形

${\displaystyle S={\frac {102}{4}}a^{2}\cot {\frac {\pi }{102}}\simeq 827.6622a^{2}}$

${\displaystyle \cos(2\pi /102)}$を有理数と平方根で表すことが可能である。

{\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{102}}=&{\frac {{\frac {{\frac {{\frac {-1-{\sqrt {17}}}{2}}+{\sqrt {\frac {17+{\sqrt {17}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {17-{\sqrt {153}}}{2}}+{\sqrt {\frac {85-{\sqrt {6137}}}{2}}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {{\frac {51+{\sqrt {153}}}{2}}+{\sqrt {\frac {153+{\sqrt {1377}}}{2}}}}{2}}-{\sqrt {\frac {{\frac {153-{\sqrt {12393}}}{2}}-{\sqrt {\frac {6885-{\sqrt {40264857}}}{2}}}}{2}}}}{2}}}}{2}}\\\cos {\frac {2\pi }{102}}=&{\frac {{\frac {{\frac {{\frac {-1-{\sqrt {17}}}{2}}+{\sqrt {\frac {17+{\sqrt {17}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {17-3{\sqrt {17}}}{2}}+{\sqrt {\frac {85-19{\sqrt {17}}}{2}}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {{\frac {51+3{\sqrt {17}}}{2}}+{\sqrt {\frac {153+9{\sqrt {17}}}{2}}}}{2}}-{\sqrt {\frac {{\frac {153-27{\sqrt {17}}}{2}}-{\sqrt {\frac {6885-1539{\sqrt {17}}}{2}}}}{2}}}}{2}}}}{4}}\\\end{aligned}}}

## 脚注

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