正四十一角形 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
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b5a058e421203684b0fbcc2c279144b30dfd4dd)
- 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/642c7785ff3a29cbe06badb240a653a88464a713)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c6f4061e35612d7b2f16c673dca9ffcc7908a26)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12947dc87bf75128b529167c77c38bc965653839)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8e680a1dbe0b7839e8d0d857d7cd8b5cafc45c3)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5af768876cdd7a56c2bb552f7898bcc7b18f1a03)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b82e910a3c022a097d0969d8ce3619f61440981)
ここで、
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/074db9b03740b6d801df0fbc13eef96e3727d1a6)
の複素数解を
として
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f04042659b95024d6ac933d1cc4a5cf747abdb41)
ここで
は、
を計算することにより
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe86b9856cd651129ee4f24b3216daaf77122f70)
正四十一角形の作図
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。
脚注
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関連項目
外部リンク
- z^p=1 の解法(p:素数) | てっぃちMarshの数学(Mathematics)教室
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非古典的 (2辺以下) | |
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辺の数: 3–10 | |
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辺の数: 11–20 | |
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辺の数: 21–30 | |
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辺の数: 31–40 | |
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辺の数: 41–50 | |
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辺の数: 51–70 (抜粋) | |
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辺の数: 71–100 (抜粋) | |
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辺の数: 101– (抜粋) | |
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無限 | |
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星型多角形 (辺の数: 5–12) | |
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多角形のクラス | |
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