四十一角形

正四十一角形においては、中心角と外角は8.78…°で、内角は171.219…°となる。一辺の長さが a の正四十一角形の面積 S は

${\displaystyle S={\frac {41}{4}}a^{2}\cot {\frac {\pi }{41}}\simeq 133.50783a^{2}}$

関係式

${\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{41}}+2\cos {\frac {18\pi }{41}}+2\cos {\frac {6\pi }{41}}+2\cos {\frac {28\pi }{41}}=x_{1}\\&2\cos {\frac {4\pi }{41}}+2\cos {\frac {36\pi }{41}}+2\cos {\frac {12\pi }{41}}+2\cos {\frac {26\pi }{41}}=x_{2}\\&2\cos {\frac {8\pi }{41}}+2\cos {\frac {10\pi }{41}}+2\cos {\frac {24\pi }{41}}+2\cos {\frac {30\pi }{41}}=x_{3}\\&2\cos {\frac {16\pi }{41}}+2\cos {\frac {20\pi }{41}}+2\cos {\frac {22\pi }{41}}+2\cos {\frac {34\pi }{41}}=x_{4}\\&2\cos {\frac {32\pi }{41}}+2\cos {\frac {40\pi }{41}}+2\cos {\frac {14\pi }{41}}+2\cos {\frac {38\pi }{41}}=x_{5}\\\end{aligned}}}$

組を作ると

${\displaystyle {\begin{aligned}&\left(2\cos {\frac {2\pi }{41}}+2\cos {\frac {18\pi }{41}}\right)+\left(2\cos {\frac {6\pi }{41}}+2\cos {\frac {28\pi }{41}}\right)=x_{1}\\&\left(2\cos {\frac {4\pi }{41}}+2\cos {\frac {36\pi }{41}}\right)+\left(2\cos {\frac {12\pi }{41}}+2\cos {\frac {26\pi }{41}}\right)=x_{2}\\&\left(2\cos {\frac {8\pi }{41}}+2\cos {\frac {10\pi }{41}}\right)+\left(2\cos {\frac {24\pi }{41}}+2\cos {\frac {30\pi }{41}}\right)=x_{3}\\&\left(2\cos {\frac {16\pi }{41}}+2\cos {\frac {20\pi }{41}}\right)+\left(2\cos {\frac {22\pi }{41}}+2\cos {\frac {34\pi }{41}}\right)=x_{4}\\&\left(2\cos {\frac {32\pi }{41}}+2\cos {\frac {40\pi }{41}}\right)+\left(2\cos {\frac {14\pi }{41}}+2\cos {\frac {38\pi }{41}}\right)=x_{5}\\\end{aligned}}}$

和積の公式より

${\displaystyle {\begin{aligned}&\left(2\cos {\frac {8\pi }{41}}\cdot 2\cos {\frac {10\pi }{41}}\right)+\left(2\cos {\frac {24\pi }{41}}\cdot 2\cos {\frac {30\pi }{41}}\right)=x_{1}\\&\left(2\cos {\frac {16\pi }{41}}\cdot 2\cos {\frac {20\pi }{41}}\right)+\left(2\cos {\frac {22\pi }{41}}\cdot 2\cos {\frac {34\pi }{41}}\right)=x_{2}\\&\left(2\cos {\frac {32\pi }{41}}\cdot 2\cos {\frac {40\pi }{41}}\right)+\left(2\cos {\frac {14\pi }{41}}\cdot 2\cos {\frac {38\pi }{41}}\right)=x_{3}\\&\left(2\cos {\frac {2\pi }{41}}\cdot 2\cos {\frac {18\pi }{41}}\right)+\left(2\cos {\frac {6\pi }{41}}\cdot 2\cos {\frac {28\pi }{41}}\right)=x_{4}\\&\left(2\cos {\frac {4\pi }{41}}\cdot 2\cos {\frac {36\pi }{41}}\right)+\left(2\cos {\frac {12\pi }{41}}\cdot 2\cos {\frac {26\pi }{41}}\right)=x_{5}\\\end{aligned}}}$

組の積を考えると

${\displaystyle {\begin{aligned}&\left(2\cos {\frac {2\pi }{41}}+2\cos {\frac {18\pi }{41}}\right)\cdot \left(2\cos {\frac {6\pi }{41}}+2\cos {\frac {28\pi }{41}}\right)=x_{2}+x_{3}\\&\left(2\cos {\frac {4\pi }{41}}+2\cos {\frac {36\pi }{41}}\right)\cdot \left(2\cos {\frac {12\pi }{41}}+2\cos {\frac {26\pi }{41}}\right)=x_{3}+x_{4}\\&\left(2\cos {\frac {8\pi }{41}}+2\cos {\frac {10\pi }{41}}\right)\cdot \left(2\cos {\frac {24\pi }{41}}+2\cos {\frac {30\pi }{41}}\right)=x_{4}+x_{5}\\&\left(2\cos {\frac {16\pi }{41}}+2\cos {\frac {20\pi }{41}}\right)\cdot \left(2\cos {\frac {22\pi }{41}}+2\cos {\frac {34\pi }{41}}\right)=x_{5}+x_{1}\\&\left(2\cos {\frac {32\pi }{41}}+2\cos {\frac {40\pi }{41}}\right)\cdot \left(2\cos {\frac {14\pi }{41}}+2\cos {\frac {38\pi }{41}}\right)=x_{1}+x_{2}\\\end{aligned}}}$

解と係数の関係より

${\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{41}}+2\cos {\frac {18\pi }{41}}={\frac {x_{1}+{\sqrt {x_{1}^{2}-4(x_{2}+x_{3})}}}{2}}\\&2\cos {\frac {16\pi }{41}}+2\cos {\frac {20\pi }{41}}=2\cos {\frac {2\pi }{41}}\cdot 2\cos {\frac {18\pi }{41}}={\frac {x_{4}+{\sqrt {x_{4}^{2}-4(x_{5}+x_{1})}}}{2}}\\\end{aligned}}}$

解と係数の関係より

${\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{41}}={\frac {{\frac {x_{1}+{\sqrt {x_{1}^{2}-4(x_{2}+x_{3})}}}{2}}+{\sqrt {\left({\frac {x_{1}+{\sqrt {x_{1}^{2}-4(x_{2}+x_{3})}}}{2}}\right)^{2}-4\left({\frac {x_{4}+{\sqrt {x_{4}^{2}-4(x_{5}+x_{1})}}}{2}}\right)}}}{2}}\\&\cos {\frac {2\pi }{41}}={\frac {1}{8}}\left({x_{1}+{\sqrt {x_{1}^{2}-4(x_{2}+x_{3})}}}+{\sqrt {\left({x_{1}+{\sqrt {x_{1}^{2}-4(x_{2}+x_{3})}}}\right)^{2}-8\left({x_{4}+{\sqrt {x_{4}^{2}-4(x_{5}+x_{1})}}}\right)}}\right)\\&\cos {\frac {2\pi }{41}}={\frac {1}{8}}\left({x_{1}+{\sqrt {8-x_{2}-2x_{3}+2x_{4}}}}+{\sqrt {\left({x_{1}+{\sqrt {8-x_{2}-2x_{3}+2x_{4}}}}\right)^{2}-8\left({x_{4}+{\sqrt {8-x_{5}-2x_{1}+2x_{2}}}}\right)}}\right)\\&\cos {\frac {2\pi }{41}}={\frac {1}{8}}\left({x_{1}+{\sqrt {8-x_{2}-2x_{3}+2x_{4}}}}+{\sqrt {16+2x_{2}+4x_{4}+{2x_{1}\cdot {\sqrt {8-x_{2}-2x_{3}+2x_{4}}}}-8\left({x_{4}+{\sqrt {8-x_{5}-2x_{1}+2x_{2}}}}\right)}}\right)\\&\cos {\frac {2\pi }{41}}={\frac {1}{8}}\left({x_{1}+{\sqrt {8-x_{2}-2x_{3}+2x_{4}}}}+{\sqrt {16+2x_{2}-4x_{4}+{2{\sqrt {46-6x_{1}+14x_{2}+5x_{3}+12x_{4}}}}-8{\sqrt {9-x_{1}+3x_{2}+x_{3}+x_{4}}}}}\right)\\\end{aligned}}}$

ここで、${\displaystyle x_{1},x_{2},x_{3},x_{4},x_{5}}$は以下の五次方程式の解である。

${\displaystyle x^{5}+x^{4}-16x^{3}+5x^{2}+21x-9=0}$

${\displaystyle z^{5}=1}$の複素数解を ${\displaystyle \sigma ,\sigma ^{2},\sigma ^{3},\sigma ^{4}}$ として${\displaystyle \lambda _{k}=x_{1}+\sigma ^{k}x_{2}+\sigma ^{2k}x_{3}+\sigma ^{3k}x_{4}+\sigma ^{4k}x_{5}}$と定義すると

${\displaystyle {\begin{aligned}&x_{1}={\frac {-1+\lambda _{1}+\lambda _{2}+\lambda _{3}+\lambda _{4}}{5}}\,\\&x_{2}={\frac {-1+\lambda _{1}\sigma ^{4}+\lambda _{2}\sigma ^{3}+\lambda _{3}\sigma ^{2}+\lambda _{4}\sigma }{5}}\,\\&x_{3}={\frac {-1+\lambda _{1}\sigma ^{3}+\lambda _{2}\sigma +\lambda _{3}\sigma ^{4}+\lambda _{4}\sigma ^{2}}{5}}\,\\&x_{4}={\frac {-1+\lambda _{1}\sigma ^{2}+\lambda _{2}\sigma ^{4}+\lambda _{3}\sigma +\lambda _{4}\sigma ^{3}}{5}}\,\\&x_{5}={\frac {-1+\lambda _{1}\sigma +\lambda _{2}\sigma ^{2}+\lambda _{3}\sigma ^{3}+\lambda _{4}\sigma ^{4}}{5}}\,\\\end{aligned}}}$

ここで ${\displaystyle \lambda _{1},\lambda _{2},\lambda _{3},\lambda _{4}}$ は、${\displaystyle \lambda _{k}^{5}}$を計算することにより${\displaystyle \sigma }$の多項式となる。

${\displaystyle {\begin{aligned}&\lambda _{1}={\sqrt[{5}]{41(289+95\sigma +75\sigma ^{3}+5\sigma ^{4})}}\,\\&\lambda _{2}={\sqrt[{5}]{41(289+95\sigma ^{2}+75\sigma +5\sigma ^{3})}}\,\\&\lambda _{3}={\sqrt[{5}]{41(289+95\sigma ^{3}+75\sigma ^{4}+5\sigma ^{2})}}\,\\&\lambda _{4}={\sqrt[{5}]{41(289+95\sigma ^{4}+75\sigma ^{2}+5\sigma )}}\,\\\end{aligned}}}$

• 円分多項式

• z^p=1 の解法（p:素数） | てっぃちMarshの数学（Mathematics)教室