三十五角形

三十五角形（さんじゅうごかくけい、さんじゅうごかっけい、triacontapentagon）は、多角形の一つで、35本の辺と35個の頂点を持つ図形である。内角の和は5940°、対角線の本数は560本である。 正三十五角形においては、中心角と外角は10.285…°で、内角は169.714…°となる。一辺の長さが a の正三十五角形の面積 S は S = 35 4 a 2 cot ⁡ π 35 ≃ 97.22046 a 2 {\displaystyle S={\frac {35}{4}}a^{2}\cot {\frac {\pi }{35}}\simeq 97.22046a^{2}} cos ⁡ ( 2 π / 35 ) {\displaystyle \cos(2\pi /35)} を平方根と立方根で表すことが可能である。 cos ⁡ 2 π 35 = cos ⁡ ( π 5 − π 7 ) = cos ⁡ π 5 cos ⁡ π 7 + sin ⁡ π 5 sin ⁡ π 7 = 1 4 ( 5 + 1 ) ⋅ cos ⁡ π 7 + 1 4 ( 10 − 2 5 ) ⋅ sin ⁡ π 7 = 5 + 1 4 ⋅ 3 ( 20 + 2 28 − 84 i 3 3 + 2 28 + 84 i 3 3 ) 12 + 10 − 2 5 4 ⋅ 3 ( 28 − 2 28 − 84 i 3 3 − 2 28 + 84 i 3 3 ) 12 {\displaystyle {\begin{aligned}\cos {\frac {2\pi }{35}}=&\cos \left({\frac {\pi }{5}}-{\frac {\pi }{7}}\right)\\=&\cos {\frac {\pi }{5}}\cos {\frac {\pi }{7}}+\sin {\frac {\pi }{5}}\sin {\frac {\pi }{7}}\\=&{\frac {1}{4}}\left({\sqrt {5}}+1\right)\cdot \cos {\frac {\pi }{7}}+{\frac {1}{4}}\left({\sqrt {10-2{\sqrt {5}}}}\right)\cdot \sin {\frac {\pi }{7}}\\=&{\frac {{\sqrt {5}}+1}{4}}\cdot {\frac {\sqrt {3\left(20+2{\sqrt[{3}]{28-84i{\sqrt {3}}}}+2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}+{\frac {\sqrt {10-2{\sqrt {5}}}}{4}}\cdot {\frac {\sqrt {3\left(28-2{\sqrt[{3}]{28-84i{\sqrt {3}}}}-2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}\end{aligned}}} 関係式 2 cos ⁡ 2 π 35 + 2 cos ⁡ 32 π 35 + 2 cos ⁡ 22 π 35 = 1 4 ( 1 + 5 − 14 ( 5 − 5 ) ) = x 1 2 cos ⁡ 4 π 35 + 2 cos ⁡ 6 π 35 + 2 cos ⁡ 26 π 35 = 1 4 ( 1 − 5 + 14 ( 5 + 5 ) ) = x 2 2 cos ⁡ 8 π 35 + 2 cos ⁡ 12 π 35 + 2 cos ⁡ 18 π 35 = 1 4 ( 1 + 5 + 14 ( 5 − 5 ) ) = x 3 2 cos ⁡ 16 π 35 + 2 cos ⁡ 24 π 35 + 2 cos ⁡ 34 π 35 = 1 4 ( 1 − 5 − 14 ( 5 + 5 ) ) = x 4 {\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{35}}+2\cos {\frac {32\pi }{35}}+2\cos {\frac {22\pi }{35}}={\frac {1}{4}}\left(1+{\sqrt {5}}-{\sqrt {14\left(5-{\sqrt {5}}\right)}}\right)=x_{1}\\2\cos {\frac {4\pi }{35}}+2\cos {\frac {6\pi }{35}}+2\cos {\frac {26\pi }{35}}={\frac {1}{4}}\left(1-{\sqrt {5}}+{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)=x_{2}\\2\cos {\frac {8\pi }{35}}+2\cos {\frac {12\pi }{35}}+2\cos {\frac {18\pi }{35}}={\frac {1}{4}}\left(1+{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}\right)=x_{3}\\2\cos {\frac {16\pi }{35}}+2\cos {\frac {24\pi }{35}}+2\cos {\frac {34\pi }{35}}={\frac {1}{4}}\left(1-{\sqrt {5}}-{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)=x_{4}\\\end{aligned}}} さらに、以下のような関係式が得られる。 ( 2 cos ⁡ 2 π 35 + ω ⋅ 2 cos ⁡ 32 π 35 + ω 2 ⋅ 2 cos ⁡ 22 π 35 ) 3 = 3 x 1 + x 2 + 6 x 3 + 12 cos ⁡ 2 π 5 + 3 ω ( 2 x 1 + x 4 + 6 cos ⁡ 4 π 5 ) + 3 ω 2 ( 2 x 1 + x 2 + x 3 ) = − 2 + 20 5 + 3 14 ( 5 − 5 ) + 14 ( 5 + 5 ) + 3 ω ( − 3 − 5 5 − 2 14 ( 5 − 5 ) − 14 ( 5 + 5 ) ) + 3 ω 2 ( 4 + 2 5 − 14 ( 5 − 5 ) + 14 ( 5 + 5 ) ) 4 = − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) − 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 ( 2 cos ⁡ 2 π 35 + ω 2 ⋅ 2 cos ⁡ 32 π 35 + ω ⋅ 2 cos ⁡ 22 π 35 ) 3 = 3 x 1 + x 2 + 6 x 3 + 12 cos ⁡ 2 π 5 + 3 ω 2 ( 2 x 1 + x 4 + 6 cos ⁡ 4 π 5 ) + 3 ω ( 2 x 1 + x 2 + x 3 ) = − 2 + 20 5 + 3 14 ( 5 − 5 ) + 14 ( 5 + 5 ) + 3 ω 2 ( − 3 − 5 5 − 2 14 ( 5 − 5 ) − 14 ( 5 + 5 ) ) + 3 ω ( 4 + 2 5 − 14 ( 5 − 5 ) + 14 ( 5 + 5 ) ) 4 = − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) + 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 {\displaystyle {\begin{aligned}\left(2\cos {\frac {2\pi }{35}}+\omega \cdot 2\cos {\frac {32\pi }{35}}+\omega ^{2}\cdot 2\cos {\frac {22\pi }{35}}\right)^{3}=&3x_{1}+x_{2}+6x_{3}+12\cos {\frac {2\pi }{5}}+3\omega (2x_{1}+x_{4}+6\cos {\frac {4\pi }{5}})+3\omega ^{2}(2x_{1}+x_{2}+x_{3})\\=&{\tfrac {{-2+20{\sqrt {5}}+3{\sqrt {14\left(5-{\sqrt {5}}\right)}}+{\sqrt {14\left(5+{\sqrt {5}}\right)}}}+3\omega \left(-3-5{\sqrt {5}}-2{\sqrt {14\left(5-{\sqrt {5}}\right)}}-{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)+3\omega ^{2}\left(4+2{\sqrt {5}}-{\sqrt {14\left(5-{\sqrt {5}}\right)}}+{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)}{4}}\\=&{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}-3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}\\\left(2\cos {\frac {2\pi }{35}}+\omega ^{2}\cdot 2\cos {\frac {32\pi }{35}}+\omega \cdot 2\cos {\frac {22\pi }{35}}\right)^{3}=&3x_{1}+x_{2}+6x_{3}+12\cos {\frac {2\pi }{5}}+3\omega ^{2}(2x_{1}+x_{4}+6\cos {\frac {4\pi }{5}})+3\omega (2x_{1}+x_{2}+x_{3})\\=&{\tfrac {{-2+20{\sqrt {5}}+3{\sqrt {14\left(5-{\sqrt {5}}\right)}}+{\sqrt {14\left(5+{\sqrt {5}}\right)}}}+3\omega ^{2}\left(-3-5{\sqrt {5}}-2{\sqrt {14\left(5-{\sqrt {5}}\right)}}-{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)+3\omega \left(4+2{\sqrt {5}}-{\sqrt {14\left(5-{\sqrt {5}}\right)}}+{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)}{4}}\\=&{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}+3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}\\\end{aligned}}} 両辺の立方根を取ると 2 cos ⁡ 2 π 35 + ω ⋅ 2 cos ⁡ 32 π 35 + ω 2 ⋅ 2 cos ⁡ 22 π 35 = − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) − 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 3 2 cos ⁡ 2 π 35 + ω 2 ⋅ 2 cos ⁡ 32 π 35 + ω ⋅ 2 cos ⁡ 22 π 35 = − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) + 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 3 {\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{35}}+\omega \cdot 2\cos {\frac {32\pi }{35}}+\omega ^{2}\cdot 2\cos {\frac {22\pi }{35}}=&{\sqrt[{3}]{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}-3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}}\\2\cos {\frac {2\pi }{35}}+\omega ^{2}\cdot 2\cos {\frac {32\pi }{35}}+\omega \cdot 2\cos {\frac {22\pi }{35}}=&{\sqrt[{3}]{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}+3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}}\\\end{aligned}}} よって cos ⁡ 2 π 35 = 1 6 ( 1 + 5 − 14 ( 5 − 5 ) 4 + − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) − 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 3 + − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) + 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 3 ) {\displaystyle {\begin{aligned}\cos {\frac {2\pi }{35}}=&{\frac {1}{6}}\left({\tfrac {1+{\sqrt {5}}-{\sqrt {14\left(5-{\sqrt {5}}\right)}}}{4}}+{\sqrt[{3}]{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}-3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}}+{\sqrt[{3}]{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}+3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}}\right)\\\end{aligned}}} 正三十五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。 正三十五角形は折紙により作図可能である。  五角形 七角形 十角形 十四角形 ポータル 数学 表示 編集