# 四十二角形

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## 正四十二角形

${\displaystyle S={\frac {42}{4}}a^{2}\cot {\frac {\pi }{42}}\simeq 140.11276a^{2}}$

${\displaystyle \cos(2\pi /42)}$を平方根と立方根で表すことが可能である。

{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{42}}=&\cos {\frac {\pi }{21}}\\=&{\frac {1}{12}}{\sqrt {72+72\cos {\frac {2\pi }{21}}}}\\=&{\frac {1}{12}}{\sqrt {72+72\cdot {\frac {1+{\sqrt {21}}+{\sqrt[{3}]{154-30{\sqrt {21}}+\left(42{\sqrt {3}}-18{\sqrt {7}}\right)i}}+{\sqrt[{3}]{154-30{\sqrt {21}}+\left(18{\sqrt {7}}-42{\sqrt {3}}\right)i}}}{12}}}}\\=&{\frac {1}{12}}{\sqrt {72+6\left({1+{\sqrt {21}}+{\sqrt[{3}]{154-30{\sqrt {21}}+\left(42{\sqrt {3}}-18{\sqrt {7}}\right)i}}+{\sqrt[{3}]{154-30{\sqrt {21}}+\left(18{\sqrt {7}}-42{\sqrt {3}}\right)i}}}\right)}}\end{aligned}}}
{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{42}}=&\cos {\frac {2\pi }{3\cdot 14}}\\=&{\frac {1}{2}}\cdot \left({\sqrt[{3}]{\cos {\frac {2\pi }{14}}+i\cdot \sin {\frac {2\pi }{14}}}}+{\sqrt[{3}]{\cos {\frac {2\pi }{14}}-i\cdot \sin {\frac {2\pi }{14}}}}\right)\\=&{\frac {1}{2}}\cdot {\sqrt[{3}]{{\tfrac {\sqrt {3\left(20+2{\sqrt[{3}]{28-84i{\sqrt {3}}}}+2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}+i\cdot {\tfrac {\sqrt {3\left(28-2{\sqrt[{3}]{28-84i{\sqrt {3}}}}-2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}}}+{\frac {1}{2}}\cdot {\sqrt[{3}]{{\tfrac {\sqrt {3\left(20+2{\sqrt[{3}]{28-84i{\sqrt {3}}}}+2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}-i\cdot {\tfrac {\sqrt {3\left(28-2{\sqrt[{3}]{28-84i{\sqrt {3}}}}-2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}}}\end{aligned}}}

{\displaystyle {\begin{aligned}&\alpha =2\cos {\frac {2\pi }{42}}+2\cos {\frac {10\pi }{42}}+2\cos {\frac {34\pi }{42}}={\frac {-1+{\sqrt {21}}}{2}}\\&\beta =2\cos {\frac {22\pi }{42}}+2\cos {\frac {26\pi }{42}}+2\cos {\frac {38\pi }{42}}={\frac {-1-{\sqrt {21}}}{2}}\\\end{aligned}}}

${\displaystyle \alpha ,\beta }$は以下の関係式より求められる。

{\displaystyle {\begin{aligned}&\alpha +\beta =-1\\&(\alpha -\beta )^{2}=21\\\end{aligned}}}

{\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{42}}\cdot 2\cos {\frac {10\pi }{42}}+2\cos {\frac {10\pi }{42}}\cdot 2\cos {\frac {34\pi }{42}}+2\cos {\frac {34\pi }{42}}\cdot 2\cos {\frac {2\pi }{42}}=-\alpha -1\\&2\cos {\frac {2\pi }{42}}\cdot 2\cos {\frac {10\pi }{42}}\cdot 2\cos {\frac {34\pi }{42}}=\beta -2\\\end{aligned}}}

${\displaystyle x^{3}-\alpha x^{2}+(-\alpha -1)x-(\beta -2)=0}$

${\displaystyle x=y+\alpha /3,\quad \beta =-1-\alpha ,\quad \alpha ^{2}=5-\alpha ,\quad \alpha ^{3}=6\alpha -5}$

${\displaystyle y^{3}-{\frac {2\alpha +8}{3}}x+{\frac {15\alpha +46}{27}}=0}$

${\displaystyle x={\frac {\alpha }{3}}+{\frac {2{\sqrt {2\alpha +8}}}{3}}\cos \left({\frac {1}{3}}\arccos {\frac {-(15\alpha +46)}{2({2\alpha +8})^{\frac {3}{2}}}}\right)}$

{\displaystyle {\begin{aligned}&x={\frac {\alpha }{3}}+{\frac {\sqrt {2\alpha +8}}{3}}{\sqrt[{3}]{{\frac {-(15\alpha +46)}{2({2\alpha +8})^{\frac {3}{2}}}}+i{\frac {\sqrt {189(\alpha +3)}}{2({2\alpha +8})^{\frac {3}{2}}}}}}+{\frac {\sqrt {2\alpha +8}}{3}}{\sqrt[{3}]{{\frac {-(15\alpha +46)}{2({2\alpha +8})^{\frac {3}{2}}}}-i{\frac {\sqrt {189(\alpha +3)}}{2({2\alpha +8})^{\frac {3}{2}}}}}}\\&x={\frac {\alpha }{3}}+{\frac {1}{6}}{\sqrt[{3}]{{-4(15\alpha +46)}+i\cdot 4{\sqrt {189(\alpha +3)}}}}+{\frac {1}{6}}{\sqrt[{3}]{-4(15\alpha +46)-i\cdot 4{\sqrt {189(\alpha +3)}}}}\end{aligned}}}

αの値((-1+√21)/2)を代入して、整理すると

${\displaystyle \cos {\frac {2\pi }{42}}={\frac {-1+{\sqrt {21}}+{\sqrt[{3}]{-154-30{\sqrt {21}}+\left(42{\sqrt {3}}+18{\sqrt {7}}\right)i}}+{\sqrt[{3}]{-154-30{\sqrt {21}}-\left(42{\sqrt {3}}+18{\sqrt {7}}\right)i}}}{12}}}$

## 脚注

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