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五十二角形 .
正五十二角形 五十二角形 (ごじゅうにかくけい、ごじゅうにかっけい、pentacontadigon)は、多角形 の一つで、52本の辺 と52個の頂点 を持つ図形である。内角の和 は9000°、対角線 の本数は1274本である。
正五十二角形においては、中心角と外角は6.923076…°で、内角は173.076923…°となる。一辺の長さが a の正五十二角形の面積 S は
S
=
13
a
2
cot
π
52
a
2
{\displaystyle S=13a^{2}\cot {\frac {\pi }{52))a^{2))
関係式
x
1
=
2
cos
2
π
52
+
2
cos
18
π
52
+
2
cos
46
π
52
=
13
−
3
13
2
x
2
=
2
cos
10
π
52
+
2
cos
14
π
52
+
2
cos
22
π
52
=
13
+
3
13
2
x
3
=
2
cos
50
π
52
+
2
cos
34
π
52
+
2
cos
6
π
52
=
−
13
−
3
13
2
x
4
=
2
cos
42
π
52
+
2
cos
38
π
52
+
2
cos
30
π
52
=
−
13
+
3
13
2
{\displaystyle {\begin{aligned}&x_{1}=2\cos {\frac {2\pi }{52))+2\cos {\frac {18\pi }{52))+2\cos {\frac {46\pi }{52))={\sqrt {\frac {13-3{\sqrt {13))}{2))}\\&x_{2}=2\cos {\frac {10\pi }{52))+2\cos {\frac {14\pi }{52))+2\cos {\frac {22\pi }{52))={\sqrt {\frac {13+3{\sqrt {13))}{2))}\\&x_{3}=2\cos {\frac {50\pi }{52))+2\cos {\frac {34\pi }{52))+2\cos {\frac {6\pi }{52))=-{\sqrt {\frac {13-3{\sqrt {13))}{2))}\\&x_{4}=2\cos {\frac {42\pi }{52))+2\cos {\frac {38\pi }{52))+2\cos {\frac {30\pi }{52))=-{\sqrt {\frac {13+3{\sqrt {13))}{2))}\\\end{aligned))}
三次方程式の係数を求めると
2
cos
2
π
52
⋅
2
cos
18
π
52
+
2
cos
18
π
52
⋅
2
cos
46
π
52
+
2
cos
46
π
52
⋅
2
cos
2
π
52
=
2
cos
10
π
26
+
2
cos
14
π
26
+
2
cos
22
π
26
+
2
cos
4
π
13
+
2
cos
10
π
13
+
2
cos
12
π
13
=
−
13
2
cos
2
π
52
⋅
2
cos
18
π
52
⋅
2
cos
46
π
52
=
x
4
{\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{52))\cdot 2\cos {\frac {18\pi }{52))+2\cos {\frac {18\pi }{52))\cdot 2\cos {\frac {46\pi }{52))+2\cos {\frac {46\pi }{52))\cdot 2\cos {\frac {2\pi }{52))\\&=2\cos {\frac {10\pi }{26))+2\cos {\frac {14\pi }{26))+2\cos {\frac {22\pi }{26))+2\cos {\frac {4\pi }{13))+2\cos {\frac {10\pi }{13))+2\cos {\frac {12\pi }{13))=-{\sqrt {13))\\&2\cos {\frac {2\pi }{52))\cdot 2\cos {\frac {18\pi }{52))\cdot 2\cos {\frac {46\pi }{52))=x_{4}\\\end{aligned))}
解と係数の関係より
u
3
−
x
1
u
2
−
13
u
−
x
4
=
0
{\displaystyle u^{3}-x_{1}u^{2}-{\sqrt {13))u-x_{4}=0}
変数変換
u
=
v
+
x
1
/
3
{\displaystyle u=v+x_{1}/3}
整理すると
v
3
−
13
+
3
13
6
v
−
(
13
+
6
13
)
x
1
+
27
x
4
27
=
0
{\displaystyle v^{3}-{\frac {13+3{\sqrt {13))}{6))v-{\frac {(13+6{\sqrt {13)))x_{1}+27x_{4)){27))=0}
三角関数、逆三角関数を用いた解は
u
1
=
x
1
3
+
2
3
13
+
3
13
2
cos
(
1
3
arccos
(
(
13
+
6
13
)
x
1
+
27
x
4
(
13
+
3
13
)
13
+
3
13
2
)
)
{\displaystyle u_{1}={\frac {x_{1)){3))+{\frac {2}{3)){\sqrt {\frac {13+3{\sqrt {13))}{2))}\cos \left({\frac {1}{3))\arccos \left({\frac {(13+6{\sqrt {13)))x_{1}+27x_{4)){(13+3{\sqrt {13))){\sqrt {\frac {13+3{\sqrt {13))}{2))))}\right)\right)}
u
1
=
1
3
13
−
3
13
2
+
2
3
13
+
3
13
2
cos
(
1
3
arccos
(
−
5
13
26
)
)
{\displaystyle u_{1}={\frac {1}{3)){\sqrt {\frac {13-3{\sqrt {13))}{2))}+{\frac {2}{3)){\sqrt {\frac {13+3{\sqrt {13))}{2))}\cos \left({\frac {1}{3))\arccos \left({\frac {-5{\sqrt {13))}{26))\right)\right)}
平方根、立方根で表すと
u
1
=
1
3
13
−
3
13
2
+
1
3
13
+
3
13
2
−
5
13
26
+
i
3
39
26
3
+
1
3
13
+
3
13
2
−
5
13
26
−
i
3
39
26
3
{\displaystyle u_{1}={\frac {1}{3)){\sqrt {\frac {13-3{\sqrt {13))}{2))}+{\frac {1}{3)){\sqrt {\frac {13+3{\sqrt {13))}{2))}{\sqrt[{3}]((\frac {-5{\sqrt {13))}{26))+i{\frac {3{\sqrt {39))}{26))))+{\frac {1}{3)){\sqrt {\frac {13+3{\sqrt {13))}{2))}{\sqrt[{3}]((\frac {-5{\sqrt {13))}{26))-i{\frac {3{\sqrt {39))}{26))))}
cos
(
2
π
/
52
)
{\displaystyle \cos(2\pi /52)}
を平方根と立方根で表すと
cos
2
π
52
=
1
6
13
−
3
13
2
+
1
6
13
+
3
13
2
−
5
13
26
+
i
3
39
26
3
+
1
6
13
+
3
13
2
−
5
13
26
−
i
3
39
26
3
{\displaystyle \cos {\frac {2\pi }{52))={\frac {1}{6)){\sqrt {\frac {13-3{\sqrt {13))}{2))}+{\frac {1}{6)){\sqrt {\frac {13+3{\sqrt {13))}{2))}{\sqrt[{3}]((\frac {-5{\sqrt {13))}{26))+i{\frac {3{\sqrt {39))}{26))))+{\frac {1}{6)){\sqrt {\frac {13+3{\sqrt {13))}{2))}{\sqrt[{3}]((\frac {-5{\sqrt {13))}{26))-i{\frac {3{\sqrt {39))}{26))))}
正五十二角形は定規 とコンパス による作図 が不可能な図形である。
正五十二角形は折紙 により作図可能である。
非古典的 (2辺以下) 辺の数: 3–10
辺の数: 11–20 辺の数: 21–30 辺の数: 31–40 辺の数: 41–50 辺の数: 51–70 (selected) 辺の数: 71–100 (selected) 辺の数: 101– (selected) 無限 星型多角形 (辺の数: 5–12)多角形のクラス
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