三十一角形

正三十一角形においては、中心角と外角は11.612…°で、内角は168.387…°となる。一辺の長さが a の正三十一角形の面積 S は

${\displaystyle S={\frac {31}{4}}a^{2}\cot {\frac {\pi }{31}}\simeq 76.21197a^{2}}$

${\displaystyle \cos(2\pi /31)}$は五次方程式、三次方程式を解くことにより求められる[1]。

${\displaystyle z^{5}=1}$の複素数解を ${\displaystyle \sigma ,\sigma ^{2},\sigma ^{3},\sigma ^{4}}$ として

以下には、中間結果（五次方程式を1回解いた際の関係式）を示す。

${\displaystyle {\begin{aligned}&x_{1}=2\cos {\frac {2\pi }{31}}+2\cos {\frac {10\pi }{31}}+2\cos {\frac {12\pi }{31}}={\frac {-1+\lambda _{1}+\lambda _{2}+\lambda _{3}+\lambda _{4}}{5}}\,\\&x_{2}=2\cos {\frac {4\pi }{31}}+2\cos {\frac {20\pi }{31}}+2\cos {\frac {24\pi }{31}}={\frac {-1+\lambda _{1}\sigma ^{4}+\lambda _{2}\sigma ^{3}+\lambda _{3}\sigma ^{2}+\lambda _{4}\sigma }{5}}\,\\&x_{3}=2\cos {\frac {8\pi }{31}}+2\cos {\frac {22\pi }{31}}+2\cos {\frac {14\pi }{31}}={\frac {-1+\lambda _{1}\sigma ^{3}+\lambda _{2}\sigma +\lambda _{3}\sigma ^{4}+\lambda _{4}\sigma ^{2}}{5}}\,\\&x_{4}=2\cos {\frac {16\pi }{31}}+2\cos {\frac {18\pi }{31}}+2\cos {\frac {28\pi }{31}}={\frac {-1+\lambda _{1}\sigma ^{2}+\lambda _{2}\sigma ^{4}+\lambda _{3}\sigma +\lambda _{4}\sigma ^{3}}{5}}\,\\&x_{5}=2\cos {\frac {6\pi }{31}}+2\cos {\frac {26\pi }{31}}+2\cos {\frac {30\pi }{31}}={\frac {-1+\lambda _{1}\sigma +\lambda _{2}\sigma ^{2}+\lambda _{3}\sigma ^{3}+\lambda _{4}\sigma ^{4}}{5}}\,\\\end{aligned}}}$

ここで ${\displaystyle \lambda _{1},\lambda _{2},\lambda _{3},\lambda _{4}}$ は

${\displaystyle {\begin{aligned}&\lambda _{1}={\sqrt[{5}]{31(36\sigma +201\sigma ^{2}+66\sigma ^{3}+106\sigma ^{4})}}\,\\&\lambda _{2}={\sqrt[{5}]{31(36\sigma ^{2}+201\sigma ^{4}+66\sigma +106\sigma ^{3})}}\,\\&\lambda _{3}={\sqrt[{5}]{31(36\sigma ^{3}+201\sigma +66\sigma ^{4}+106\sigma ^{2})}}\,\\&\lambda _{4}={\sqrt[{5}]{31(36\sigma ^{4}+201\sigma ^{3}+66\sigma ^{2}+106\sigma )}}\,\\\end{aligned}}}$

${\displaystyle \cos(2\pi /31)}$は ${\displaystyle x_{1},x_{2},x_{3}}$ を用いた以下の三次方程式の解の一つである。

${\displaystyle u^{3}-{\frac {x_{1}}{2}}u^{2}+{\frac {x_{1}+x_{3}}{4}}u-{\frac {x_{2}+2}{8}}=0}$

変数変換

${\displaystyle v=u+{\frac {x_{1}}{6}}}$

整理すると

${\displaystyle v^{3}-{\frac {6-x_{1}+x_{2}-x_{3}}{12}}v-{\frac {24-6x_{1}+12x_{2}-6x_{3}-3x_{4}-x_{5}}{216}}=0}$

三角関数、逆三角関数を用いた解は

${\displaystyle \cos {\frac {2\pi }{31}}={\frac {x_{1}}{6}}+{\frac {1}{3}}{\sqrt {6-x_{1}+x_{2}-x_{3}}}\cdot \cos \left({\frac {1}{3}}\arccos {\frac {24-6x_{1}+12x_{2}-6x_{3}-3x_{4}-x_{5}}{2(6-x_{1}+x_{2}-x_{3})^{\frac {3}{2}}}}\right)}$

上記三次方程式を変形すると

${\displaystyle v^{3}-{\frac {6-x_{1}+x_{2}-x_{3}}{12}}v-{\frac {(6-x_{1}+x_{2}-x_{3})(129-10x_{1}+81x_{2}-48x_{3}-30x_{4})}{216\cdot 67}}=0}$

三角関数、逆三角関数を用いた解は

${\displaystyle \cos {\frac {2\pi }{31}}={\frac {x_{1}}{6}}+{\frac {1}{3}}{\sqrt {6-x_{1}+x_{2}-x_{3}}}\cdot \cos \left({\frac {1}{3}}\arccos {\frac {129-10x_{1}+81x_{2}-48x_{3}-30x_{4}}{134{\sqrt {6-x_{1}+x_{2}-x_{3}}}}}\right)}$

平方根、立方根で表すと

${\displaystyle {\begin{aligned}\cos {\frac {2\pi }{31}}={\frac {x_{1}}{6}}+{\frac {\sqrt {6-x_{1}+x_{2}-x_{3}}}{6}}{\sqrt[{3}]{{\frac {129-10x_{1}+81x_{2}-48x_{3}-30x_{4}}{134{\sqrt {6-x_{1}+x_{2}-x_{3}}}}}+i{\frac {\sqrt {27(1091-96x_{1}-348x_{2}-367x_{3}+114x_{4})}}{134{\sqrt {6-x_{1}+x_{2}-x_{3}}}}}}}\\+{\frac {\sqrt {6-x_{1}+x_{2}-x_{3}}}{6}}{\sqrt[{3}]{{\frac {129-10x_{1}+81x_{2}-48x_{3}-30x_{4}}{134{\sqrt {6-x_{1}+x_{2}-x_{3}}}}}-i{\frac {\sqrt {27(1091-96x_{1}-348x_{2}-367x_{3}+114x_{4})}}{134{\sqrt {6-x_{1}+x_{2}-x_{3}}}}}}}\end{aligned}}}$

• z^31=1 の解法 | てっぃちMarshの数学（Mathematics)教室