四十一角形
解と係数の関係より
2 cos  2 π 41 + 2 cos  18 π 41 = x 1 + x 1 2 − 4 ( x 2 + x 3 ) 2 2 cos  16 π 41 + 2 cos  20 π 41 = 2 cos  2 π 41 ⋅ 2 cos  18 π 41 = x 4 + x 4 2 − 4 ( x 5 + x 1 ) 2 {\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