# 三十四角形

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## 正三十四角形

${\displaystyle S={\frac {34}{4}}a^{2}\cot {\frac {\pi }{34}}\simeq 91.72961a^{2}}$

${\displaystyle \cos(2\pi /34)}$を有理数と平方根で表すことが可能である。

{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{34}}=\cos {\frac {\pi }{17}}=&{\frac {{\sqrt {34-{\sqrt {68}}}}-{\sqrt {17}}+1+2{\sqrt {{\sqrt {34-{\sqrt {68}}}}+{\sqrt {17}}-1}}{\sqrt {\sqrt {17+{\sqrt {272}}}}}}{16}}\\=&{\frac {1-{\sqrt {17}}+{\sqrt {34-{\sqrt {68}}}}+{\sqrt {68+{\sqrt {2448}}+{\sqrt {2720+{\sqrt {6284288}}}}}}}{16}}\\=&{\frac {1-{\sqrt {17}}+{\sqrt {2\cdot 17-{\sqrt {2^{2}\cdot 17}}}}+{\sqrt {2^{2}\cdot 17+{\sqrt {2^{4}\cdot 3^{2}\cdot 17}}+{\sqrt {2^{5}\cdot 5\cdot 17+{\sqrt {2^{10}\cdot 17\cdot 19^{2}}}}}}}}{16}}\\=&{\frac {1-{\sqrt {17}}+{\sqrt {2\cdot 17-2\cdot {\sqrt {17}}}}+{\sqrt {2^{2}\cdot 17+2^{2}\cdot 3\cdot {\sqrt {17}}+2^{2}{\sqrt {2\cdot 5\cdot 17+2\cdot 19\cdot {\sqrt {17}}}}}}}{16}}\\=&{\frac {1}{16}}\left({1-{\sqrt {17}}+{\sqrt {2}}\cdot {\sqrt {17-{\sqrt {17}}}}+2\cdot {\sqrt {17+3{\sqrt {17}}+{\sqrt {2}}\cdot {\sqrt {5\cdot 17+19{\sqrt {17}}}}}}}\right)\\\end{aligned}}}

## 脚注

 [脚注の使い方]