三十一角形
整理すると
v 3 − 6 − x 1 + x 2 − x 3 12 v − 24 − 6 x 1 + 12 x 2 − 6 x 3 − 3 x 4 − x 5 216 = 0 {\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
cos  2 π 31 = x 1 6 + 1 3 6 − x 1 + x 2 − x 3 ⋅ cos  ( 1 3 arccos  24 − 6 x 1 + 12 x 2 − 6 x 3 − 3 x 4 − x 5 2 ( 6 − x 1 + x 2 − x 3 ) 3 2 ) {\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
x 1 = 2 cos  2 π 31 + 2 cos  10 π 31 + 2 cos  12 π 31 = − 1 + λ 1 + λ 2 + λ 3 + λ 4 5 x 2 = 2 cos  4 π 31 + 2 cos  20 π 31 + 2 cos  24 π 31 = − 1 + λ 1 σ 4 + λ 2 σ 3 + λ 3 σ 2 + λ 4 σ 5 x 3 = 2 cos  8 π 31 + 2 cos  22 π 31 + 2 cos  14 π 31 = − 1 + λ 1 σ 3 + λ 2 σ + λ 3 σ 4 + λ 4 σ 2 5 x 4 = 2 cos  16 π 31 + 2 cos  18 π 31 + 2 cos  28 π 31 = − 1 + λ 1 σ 2 + λ 2 σ 4 + λ 3 σ + λ 4 σ 3 5 x 5 = 2 cos  6 π 31 + 2 cos  26 π 31 + 2 cos  30 π 31 = − 1 + λ 1 σ + λ 2 σ 2 + λ 3 σ 3 + λ 4 σ 4 5 {\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}}}