正六十五角形
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2021/08/21 21:56 UTC 版)
正六十五角形においては、中心角と外角は5.538…°で、内角は174.461…°となる。一辺の長さが a の正六十五角形の面積 S は S = 65 4 a 2 cot π 65 ≃ 335.95298 a 2 {\displaystyle S={\frac {65}{4}}a^{2}\cot {\frac {\pi }{65}}\simeq 335.95298a^{2}} 関係式 x 1 = 2 cos 2 π 65 + 2 cos 32 π 65 + 2 cos 8 π 65 = 1 + 13 2 + 35 + 325 2 2 + 65 + 2925 2 − 715 + 494325 2 2 2 x 2 = 2 cos 4 π 65 + 2 cos 64 π 65 + 2 cos 16 π 65 = 1 − 13 2 + 35 − 325 2 2 + 65 − 2925 2 + 715 − 494325 2 2 2 x 3 = 2 cos 6 π 65 + 2 cos 34 π 65 + 2 cos 24 π 65 = 1 + 13 2 − 35 + 325 2 2 + 65 + 2925 2 + 715 + 494325 2 2 2 x 4 = 2 cos 12 π 65 + 2 cos 62 π 65 + 2 cos 48 π 65 = 1 − 13 2 − 35 − 325 2 2 − 65 − 2925 2 − 715 − 494325 2 2 2 x 5 = 2 cos 18 π 65 + 2 cos 28 π 65 + 2 cos 58 π 65 = 1 + 13 2 + 35 + 325 2 2 − 65 + 2925 2 − 715 + 494325 2 2 2 x 6 = 2 cos 36 π 65 + 2 cos 56 π 65 + 2 cos 14 π 65 = 1 − 13 2 + 35 − 325 2 2 − 65 − 2925 2 + 715 − 494325 2 2 2 x 7 = 2 cos 54 π 65 + 2 cos 46 π 65 + 2 cos 44 π 65 = 1 + 13 2 − 35 + 325 2 2 − 65 + 2925 2 + 715 + 494325 2 2 2 x 8 = 2 cos 22 π 65 + 2 cos 38 π 65 + 2 cos 42 π 65 = 1 − 13 2 − 35 − 325 2 2 + 65 − 2925 2 − 715 − 494325 2 2 2 {\displaystyle {\begin{aligned}&x_{1}=2\cos {\frac {2\pi }{65}}+2\cos {\frac {32\pi }{65}}+2\cos {\frac {8\pi }{65}}={\frac {{\frac {{\frac {1+{\sqrt {13}}}{2}}+{\sqrt {\frac {35+{\sqrt {325}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {65+{\sqrt {2925}}}{2}}-{\sqrt {\frac {715+{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{2}=2\cos {\frac {4\pi }{65}}+2\cos {\frac {64\pi }{65}}+2\cos {\frac {16\pi }{65}}={\frac {{\frac {{\frac {1-{\sqrt {13}}}{2}}+{\sqrt {\frac {35-{\sqrt {325}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {65-{\sqrt {2925}}}{2}}+{\sqrt {\frac {715-{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{3}=2\cos {\frac {6\pi }{65}}+2\cos {\frac {34\pi }{65}}+2\cos {\frac {24\pi }{65}}={\frac {{\frac {{\frac {1+{\sqrt {13}}}{2}}-{\sqrt {\frac {35+{\sqrt {325}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {65+{\sqrt {2925}}}{2}}+{\sqrt {\frac {715+{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{4}=2\cos {\frac {12\pi }{65}}+2\cos {\frac {62\pi }{65}}+2\cos {\frac {48\pi }{65}}={\frac {{\frac {{\frac {1-{\sqrt {13}}}{2}}-{\sqrt {\frac {35-{\sqrt {325}}}{2}}}}{2}}-{\sqrt {\frac {{\frac {65-{\sqrt {2925}}}{2}}-{\sqrt {\frac {715-{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{5}=2\cos {\frac {18\pi }{65}}+2\cos {\frac {28\pi }{65}}+2\cos {\frac {58\pi }{65}}={\frac {{\frac {{\frac {1+{\sqrt {13}}}{2}}+{\sqrt {\frac {35+{\sqrt {325}}}{2}}}}{2}}-{\sqrt {\frac {{\frac {65+{\sqrt {2925}}}{2}}-{\sqrt {\frac {715+{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{6}=2\cos {\frac {36\pi }{65}}+2\cos {\frac {56\pi }{65}}+2\cos {\frac {14\pi }{65}}={\frac {{\frac {{\frac {1-{\sqrt {13}}}{2}}+{\sqrt {\frac {35-{\sqrt {325}}}{2}}}}{2}}-{\sqrt {\frac {{\frac {65-{\sqrt {2925}}}{2}}+{\sqrt {\frac {715-{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{7}=2\cos {\frac {54\pi }{65}}+2\cos {\frac {46\pi }{65}}+2\cos {\frac {44\pi }{65}}={\frac {{\frac {{\frac {1+{\sqrt {13}}}{2}}-{\sqrt {\frac {35+{\sqrt {325}}}{2}}}}{2}}-{\sqrt {\frac {{\frac {65+{\sqrt {2925}}}{2}}+{\sqrt {\frac {715+{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\&x_{8}=2\cos {\frac {22\pi }{65}}+2\cos {\frac {38\pi }{65}}+2\cos {\frac {42\pi }{65}}={\frac {{\frac {{\frac {1-{\sqrt {13}}}{2}}-{\sqrt {\frac {35-{\sqrt {325}}}{2}}}}{2}}+{\sqrt {\frac {{\frac {65-{\sqrt {2925}}}{2}}-{\sqrt {\frac {715-{\sqrt {494325}}}{2}}}}{2}}}}{2}}\\\end{aligned}}} 三次方程式の係数を求めると 2 cos 2 π 65 ⋅ 2 cos 32 π 65 + 2 cos 32 π 65 ⋅ 2 cos 8 π 65 + 2 cos 8 π 65 ⋅ 2 cos 2 π 65 = x 3 + 2 cos 2 π 13 + 2 cos 6 π 13 + 2 cos 8 π 13 = x 3 + − 1 + 13 2 2 cos 2 π 65 ⋅ 2 cos 32 π 65 ⋅ 2 cos 8 π 65 = x 8 + 2 cos 2 π 5 = x 8 + − 1 + 5 2 {\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{65}}\cdot 2\cos {\frac {32\pi }{65}}+2\cos {\frac {32\pi }{65}}\cdot 2\cos {\frac {8\pi }{65}}+2\cos {\frac {8\pi }{65}}\cdot 2\cos {\frac {2\pi }{65}}\\&=x_{3}+2\cos {\frac {2\pi }{13}}+2\cos {\frac {6\pi }{13}}+2\cos {\frac {8\pi }{13}}\\&=x_{3}+{\frac {-1+{\sqrt {13}}}{2}}\\&2\cos {\frac {2\pi }{65}}\cdot 2\cos {\frac {32\pi }{65}}\cdot 2\cos {\frac {8\pi }{65}}=x_{8}+2\cos {\frac {2\pi }{5}}=x_{8}+{\frac {-1+{\sqrt {5}}}{2}}\\\end{aligned}}} 解と係数の関係より u 3 − x 1 u 2 + ( x 3 + − 1 + 13 2 ) u − ( x 8 + − 1 + 5 2 ) = 0 {\displaystyle u^{3}-x_{1}u^{2}+(x_{3}+{\frac {-1+{\sqrt {13}}}{2}})u-(x_{8}+{\frac {-1+{\sqrt {5}}}{2}})=0}
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