四十一角形

四十一角形（よんじゅういちかくけい、よんじゅういちかっけい、tetracontahenagon）は、多角形の一つで、41本の辺と41個の頂点を持つ図形である。内角の和は7020°、対角線の本数は779本である。 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 S = 41 4 a 2 cot ⁡ π 41 ≃ 133.50783 a 2 {\displaystyle S={\frac {41}{4}}a^{2}\cot {\frac {\pi }{41}}\simeq 133.50783a^{2}} 関係式 2 cos ⁡ 2 π 41 + 2 cos ⁡ 18 π 41 + 2 cos ⁡ 6 π 41 + 2 cos ⁡ 28 π 41 = x 1 2 cos ⁡ 4 π 41 + 2 cos ⁡ 36 π 41 + 2 cos ⁡ 12 π 41 + 2 cos ⁡ 26 π 41 = x 2 2 cos ⁡ 8 π 41 + 2 cos ⁡ 10 π 41 + 2 cos ⁡ 24 π 41 + 2 cos ⁡ 30 π 41 = x 3 2 cos ⁡ 16 π 41 + 2 cos ⁡ 20 π 41 + 2 cos ⁡ 22 π 41 + 2 cos ⁡ 34 π 41 = x 4 2 cos ⁡ 32 π 41 + 2 cos ⁡ 40 π 41 + 2 cos ⁡ 14 π 41 + 2 cos ⁡ 38 π 41 = x 5 {\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 ( 2 cos ⁡ 2 π 41 + 2 cos ⁡ 18 π 41 ) + ( 2 cos ⁡ 6 π 41 + 2 cos ⁡ 28 π 41 ) = x 1 ( 2 cos ⁡ 4 π 41 + 2 cos ⁡ 36 π 41 ) + ( 2 cos ⁡ 12 π 41 + 2 cos ⁡ 26 π 41 ) = x 2 ( 2 cos ⁡ 8 π 41 + 2 cos ⁡ 10 π 41 ) + ( 2 cos ⁡ 24 π 41 + 2 cos ⁡ 30 π 41 ) = x 3 ( 2 cos ⁡ 16 π 41 + 2 cos ⁡ 20 π 41 ) + ( 2 cos ⁡ 22 π 41 + 2 cos ⁡ 34 π 41 ) = x 4 ( 2 cos ⁡ 32 π 41 + 2 cos ⁡ 40 π 41 ) + ( 2 cos ⁡ 14 π 41 + 2 cos ⁡ 38 π 41 ) = x 5 {\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 ( 2 cos ⁡ 8 π 41 ⋅ 2 cos ⁡ 10 π 41 ) + ( 2 cos ⁡ 24 π 41 ⋅ 2 cos ⁡ 30 π 41 ) = x 1 ( 2 cos ⁡ 16 π 41 ⋅ 2 cos ⁡ 20 π 41 ) + ( 2 cos ⁡ 22 π 41 ⋅ 2 cos ⁡ 34 π 41 ) = x 2 ( 2 cos ⁡ 32 π 41 ⋅ 2 cos ⁡ 40 π 41 ) + ( 2 cos ⁡ 14 π 41 ⋅ 2 cos ⁡ 38 π 41 ) = x 3 ( 2 cos ⁡ 2 π 41 ⋅ 2 cos ⁡ 18 π 41 ) + ( 2 cos ⁡ 6 π 41 ⋅ 2 cos ⁡ 28 π 41 ) = x 4 ( 2 cos ⁡ 4 π 41 ⋅ 2 cos ⁡ 36 π 41 ) + ( 2 cos ⁡ 12 π 41 ⋅ 2 cos ⁡ 26 π 41 ) = x 5 {\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 ( 2 cos ⁡ 2 π 41 + 2 cos ⁡ 18 π 41 ) ⋅ ( 2 cos ⁡ 6 π 41 + 2 cos ⁡ 28 π 41 ) = x 2 + x 3 ( 2 cos ⁡ 4 π 41 + 2 cos ⁡ 36 π 41 ) ⋅ ( 2 cos ⁡ 12 π 41 + 2 cos ⁡ 26 π 41 ) = x 3 + x 4 ( 2 cos ⁡ 8 π 41 + 2 cos ⁡ 10 π 41 ) ⋅ ( 2 cos ⁡ 24 π 41 + 2 cos ⁡ 30 π 41 ) = x 4 + x 5 ( 2 cos ⁡ 16 π 41 + 2 cos ⁡ 20 π 41 ) ⋅ ( 2 cos ⁡ 22 π 41 + 2 cos ⁡ 34 π 41 ) = x 5 + x 1 ( 2 cos ⁡ 32 π 41 + 2 cos ⁡ 40 π 41 ) ⋅ ( 2 cos ⁡ 14 π 41 + 2 cos ⁡ 38 π 41 ) = x 1 + x 2 {\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 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 2 cos ⁡ 2 π 41 = x 1 + x 1 2 − 4 ( x 2 + x 3 ) 2 + ( x 1 + x 1 2 − 4 ( x 2 + x 3 ) 2 ) 2 − 4 ( x 4 + x 4 2 − 4 ( x 5 + x 1 ) 2 ) 2 cos ⁡ 2 π 41 = 1 8 ( x 1 + x 1 2 − 4 ( x 2 + x 3 ) + ( x 1 + x 1 2 − 4 ( x 2 + x 3 ) ) 2 − 8 ( x 4 + x 4 2 − 4 ( x 5 + x 1 ) ) ) cos ⁡ 2 π 41 = 1 8 ( x 1 + 8 − x 2 − 2 x 3 + 2 x 4 + ( x 1 + 8 − x 2 − 2 x 3 + 2 x 4 ) 2 − 8 ( x 4 + 8 − x 5 − 2 x 1 + 2 x 2 ) ) cos ⁡ 2 π 41 = 1 8 ( x 1 + 8 − x 2 − 2 x 3 + 2 x 4 + 16 + 2 x 2 + 4 x 4 + 2 x 1 ⋅ 8 − x 2 − 2 x 3 + 2 x 4 − 8 ( x 4 + 8 − x 5 − 2 x 1 + 2 x 2 ) ) cos ⁡ 2 π 41 = 1 8 ( x 1 + 8 − x 2 − 2 x 3 + 2 x 4 + 16 + 2 x 2 − 4 x 4 + 2 46 − 6 x 1 + 14 x 2 + 5 x 3 + 12 x 4 − 8 9 − x 1 + 3 x 2 + x 3 + x 4 ) {\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、 x 1 , x 2 , x 3 , x 4 , x 5 {\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。 x 5 + x 4 − 16 x 3 + 5 x 2 + 21 x − 9 = 0 {\displaystyle x^{5}+x^{4}-16x^{3}+5x^{2}+21x-9=0} z 5 = 1 {\displaystyle z^{5}=1} の複素数解を σ , σ 2 , σ 3 , σ 4 {\displaystyle \sigma ,\sigma ^{2},\sigma ^{3},\sigma ^{4}} として λ k = x 1 + σ k x 2 + σ 2 k x 3 + σ 3 k x 4 + σ 4 k x 5 {\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 x 1 = − 1 + λ 1 + λ 2 + λ 3 + λ 4 5 x 2 = − 1 + λ 1 σ 4 + λ 2 σ 3 + λ 3 σ 2 + λ 4 σ 5 x 3 = − 1 + λ 1 σ 3 + λ 2 σ + λ 3 σ 4 + λ 4 σ 2 5 x 4 = − 1 + λ 1 σ 2 + λ 2 σ 4 + λ 3 σ + λ 4 σ 3 5 x 5 = − 1 + λ 1 σ + λ 2 σ 2 + λ 3 σ 3 + λ 4 σ 4 5 {\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 λ 1 , λ 2 , λ 3 , λ 4 {\displaystyle \lambda _{1},\lambda _{2},\lambda _{3},\lambda _{4}} は、 λ k 5 {\displaystyle \lambda _{k}^{5}} を計算することにより σ {\displaystyle \sigma } の多項式となる。 λ 1 = 41 ( 289 + 95 σ + 75 σ 3 + 5 σ 4 ) 5 λ 2 = 41 ( 289 + 95 σ 2 + 75 σ + 5 σ 3 ) 5 λ 3 = 41 ( 289 + 95 σ 3 + 75 σ 4 + 5 σ 2 ) 5 λ 4 = 41 ( 289 + 95 σ 4 + 75 σ 2 + 5 σ ) 5 {\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}}} 正四十一角形は定規とコンパスによる作図が不可能な図形である。 正四十一角形は折紙により作図が不可能な図形である。  円分多項式 ポータル 数学 z^p=1 の解法（p:素数） | てっぃちMarshの数学（Mathematics)教室 表示 編集