正八十四角形
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2021/09/10 21:57 UTC 版)
正八十四角形においては、中心角と外角は4.285…°で、内角は175.714…°となる。一辺の長さが a の正八十四角形の面積 S は S = 84 4 a 2 cot π 84 ≃ 561.23682 a 2 {\displaystyle S={\frac {84}{4}}a^{2}\cot {\frac {\pi }{84}}\simeq 561.23682a^{2}} 関係式 2 cos 2 π 84 + 2 cos 74 π 84 + 2 cos 50 π 84 = 3 − 7 2 = x 1 2 cos 38 π 84 + 2 cos 62 π 84 + 2 cos 58 π 84 = − 3 − 7 2 = x 2 2 cos 34 π 84 + 2 cos 82 π 84 + 2 cos 10 π 84 = − 3 + 7 2 = x 3 2 cos 26 π 84 + 2 cos 46 π 84 + 2 cos 22 π 84 = 3 + 7 2 = x 4 {\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{84}}+2\cos {\frac {74\pi }{84}}+2\cos {\frac {50\pi }{84}}={\frac {{\sqrt {3}}-{\sqrt {7}}}{2}}=x_{1}\\2\cos {\frac {38\pi }{84}}+2\cos {\frac {62\pi }{84}}+2\cos {\frac {58\pi }{84}}={\frac {-{\sqrt {3}}-{\sqrt {7}}}{2}}=x_{2}\\2\cos {\frac {34\pi }{84}}+2\cos {\frac {82\pi }{84}}+2\cos {\frac {10\pi }{84}}={\frac {-{\sqrt {3}}+{\sqrt {7}}}{2}}=x_{3}\\2\cos {\frac {26\pi }{84}}+2\cos {\frac {46\pi }{84}}+2\cos {\frac {22\pi }{84}}={\frac {{\sqrt {3}}+{\sqrt {7}}}{2}}=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 84 + ω ⋅ 2 cos 74 π 84 + ω 2 ⋅ 2 cos 50 π 84 ) 3 = 3 x 1 + 2 cos 2 π 28 + 2 cos 6 π 28 + 2 cos 18 π 28 + 6 ( x 4 ) + 3 ω ( 2 x 1 + 6 cos 10 π 12 + 2 cos 10 π 28 + 2 cos 26 π 28 + 2 cos 22 π 28 ) + 3 ω 2 ( 2 x 1 + x 4 + 2 cos 2 π 28 + 2 cos 6 π 28 + 2 cos 18 π 28 ) = 11 3 + 9 7 − 3 3 ( 7 3 + 5 7 ) i 4 ( 2 cos 2 π 84 + ω 2 ⋅ 2 cos 74 π 84 + ω ⋅ 2 cos 50 π 84 ) 3 = 3 x 1 + 2 cos 2 π 28 + 2 cos 6 π 28 + 2 cos 18 π 28 + 6 ( x 4 ) + 3 ω 2 ( 2 x 1 + 6 cos 10 π 12 + 2 cos 10 π 28 + 2 cos 26 π 28 + 2 cos 22 π 28 ) + 3 ω ( 2 x 1 + x 4 + 2 cos 2 π 28 + 2 cos 6 π 28 + 2 cos 18 π 28 ) = 11 3 + 9 7 + 3 3 ( 7 3 + 5 7 ) i 4 {\displaystyle {\begin{aligned}\left(2\cos {\frac {2\pi }{84}}+\omega \cdot 2\cos {\frac {74\pi }{84}}+\omega ^{2}\cdot 2\cos {\frac {50\pi }{84}}\right)^{3}=&3x_{1}+2\cos {\frac {2\pi }{28}}+2\cos {\frac {6\pi }{28}}+2\cos {\frac {18\pi }{28}}+6(x_{4})+3\omega \left(2x_{1}+6\cos {\frac {10\pi }{12}}+2\cos {\frac {10\pi }{28}}+2\cos {\frac {26\pi }{28}}+2\cos {\frac {22\pi }{28}}\right)+3\omega ^{2}\left(2x_{1}+x_{4}+2\cos {\frac {2\pi }{28}}+2\cos {\frac {6\pi }{28}}+2\cos {\frac {18\pi }{28}}\right)\\=&{\frac {11{\sqrt {3}}+9{\sqrt {7}}-3{\sqrt {3}}(7{\sqrt {3}}+5{\sqrt {7}})i}{4}}\\\left(2\cos {\frac {2\pi }{84}}+\omega ^{2}\cdot 2\cos {\frac {74\pi }{84}}+\omega \cdot 2\cos {\frac {50\pi }{84}}\right)^{3}=&3x_{1}+2\cos {\frac {2\pi }{28}}+2\cos {\frac {6\pi }{28}}+2\cos {\frac {18\pi }{28}}+6(x_{4})+3\omega ^{2}\left(2x_{1}+6\cos {\frac {10\pi }{12}}+2\cos {\frac {10\pi }{28}}+2\cos {\frac {26\pi }{28}}+2\cos {\frac {22\pi }{28}}\right)+3\omega \left(2x_{1}+x_{4}+2\cos {\frac {2\pi }{28}}+2\cos {\frac {6\pi }{28}}+2\cos {\frac {18\pi }{28}}\right)\\=&{\frac {11{\sqrt {3}}+9{\sqrt {7}}+3{\sqrt {3}}(7{\sqrt {3}}+5{\sqrt {7}})i}{4}}\\\end{aligned}}} 両辺の立方根を取ると 2 cos 2 π 84 + ω ⋅ 2 cos 74 π 84 + ω 2 ⋅ 2 cos 50 π 84 = 11 3 + 9 7 − 3 3 ( 7 3 + 5 7 ) i 4 3 2 cos 2 π 84 + ω 2 ⋅ 2 cos 74 π 84 + ω ⋅ 2 cos 50 π 84 = 11 3 + 9 7 + 3 3 ( 7 3 + 5 7 ) i 4 3 {\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{84}}+\omega \cdot 2\cos {\frac {74\pi }{84}}+\omega ^{2}\cdot 2\cos {\frac {50\pi }{84}}=&{\sqrt[{3}]{\frac {11{\sqrt {3}}+9{\sqrt {7}}-3{\sqrt {3}}(7{\sqrt {3}}+5{\sqrt {7}})i}{4}}}\\2\cos {\frac {2\pi }{84}}+\omega ^{2}\cdot 2\cos {\frac {74\pi }{84}}+\omega \cdot 2\cos {\frac {50\pi }{84}}=&{\sqrt[{3}]{\frac {11{\sqrt {3}}+9{\sqrt {7}}+3{\sqrt {3}}(7{\sqrt {3}}+5{\sqrt {7}})i}{4}}}\\\end{aligned}}} よって cos 2 π 84 = 1 6 ( 3 − 7 2 + 11 3 + 9 7 − 3 3 ( 7 3 + 5 7 ) i 4 3 + 11 3 + 9 7 + 3 3 ( 7 3 + 5 7 ) i 4 3 ) {\displaystyle {\begin{aligned}\cos {\frac {2\pi }{84}}=&{\frac {1}{6}}\left({\frac {{\sqrt {3}}-{\sqrt {7}}}{2}}+{\sqrt[{3}]{\frac {11{\sqrt {3}}+9{\sqrt {7}}-3{\sqrt {3}}(7{\sqrt {3}}+5{\sqrt {7}})i}{4}}}+{\sqrt[{3}]{\frac {11{\sqrt {3}}+9{\sqrt {7}}+3{\sqrt {3}}(7{\sqrt {3}}+5{\sqrt {7}})i}{4}}}\right)\\\end{aligned}}}
※この「正八十四角形」の解説は、「八十四角形」の解説の一部です。
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