出典: フリー百科事典『ウィキペディア(Wikipedia)』
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

- 関係式

組を作ると

和積の公式より

組の積を考えると

解と係数の関係より

解と係数の関係より

ここで、
は以下の五次方程式の解である。

の複素数解を
として
と定義すると

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。
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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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