二十六角形

正二十六角形

二十六角形︵にじゅうろくかくけい、にじゅうろっかっけい、icosihexagon︶は、多角形の一つで、26本の辺と26個の頂点を持つ図形である。内角の和は4320°、対角線の本数は299本である。

正二十六角形

正二十六角形においては、中心角と外角は13.846…°で、内角は166.153…°となる。一辺の長さがaの正二十六角形の面積Sは

$S={\frac {26}{4}}a^{2}\cot {\frac {\pi }{26}}\simeq 53.53232a^{2}$$S={\frac {26}{4}}a^{2}\cot {\frac {\pi }{26}}\simeq 53.53232a^{2}$
$\cos(2\pi /26)$$\cos(2\pi /26)$を平方根と立方根で表すと

$\cos {\frac {2\pi }{26}}=\cos {\frac {\pi }{13}}={\frac {1}{12}}{\sqrt {72+72\cdot \cos {\frac {2\pi }{13}}}}={\frac {1}{12}}{\sqrt {72+72\cdot {\frac {1}{12}}\left({\sqrt[{3}]{104-20{\sqrt {13}}+12{\sqrt {-39}}}}+{\sqrt[{3}]{104-20{\sqrt {13}}-12{\sqrt {-39}}}}+{\sqrt {13}}-1\right)}}=0.970941...$$\cos {\frac {2\pi }{26}}=\cos {\frac {\pi }{13}}={\frac {1}{12}}{\sqrt {72+72\cdot \cos {\frac {2\pi }{13}}}}={\frac {1}{12}}{\sqrt {72+72\cdot {\frac {1}{12}}\left({\sqrt[{3}]{104-20{\sqrt {13}}+12{\sqrt {-39}}}}+{\sqrt[{3}]{104-20{\sqrt {13}}-12{\sqrt {-39}}}}+{\sqrt {13}}-1\right)}}=0.970941...$
関係式

${\begin{aligned}&\alpha =2\cos {\frac {2\pi }{26}}+2\cos {\frac {6\pi }{26}}+2\cos {\frac {18\pi }{26}}={\frac {1+{\sqrt {13}}}{2}}\\&\beta =2\cos {\frac {14\pi }{26}}+2\cos {\frac {10\pi }{26}}+2\cos {\frac {22\pi }{26}}={\frac {1-{\sqrt {13}}}{2}}\\\end{aligned}}$${\begin{aligned}&\alpha =2\cos {\frac {2\pi }{26}}+2\cos {\frac {6\pi }{26}}+2\cos {\frac {18\pi }{26}}={\frac {1+{\sqrt {13}}}{2}}\\&\beta =2\cos {\frac {14\pi }{26}}+2\cos {\frac {10\pi }{26}}+2\cos {\frac {22\pi }{26}}={\frac {1-{\sqrt {13}}}{2}}\\\end{aligned}}$
三次方程式の係数を求めると

${\begin{aligned}&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}+2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}+2\cos {\frac {18\pi }{26}}\cdot 2\cos {\frac {2\pi }{26}}=-1\\&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}=\beta -2\end{aligned}}$${\begin{aligned}&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}+2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}+2\cos {\frac {18\pi }{26}}\cdot 2\cos {\frac {2\pi }{26}}=-1\\&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}=\beta -2\end{aligned}}$
解と係数の関係より

$x^{3}-\alpha x^{2}-x-(\beta -2)=0$$x^{3}-\alpha x^{2}-x-(\beta -2)=0$
変数変換

$x=y+\alpha /3$$x=y+\alpha /3$
整理すると

$y^{3}-{\frac {13+{\sqrt {13}}}{6}}y+{\frac {26+5{\sqrt {13}}}{27}}=0$$y^{3}-{\frac {13+{\sqrt {13}}}{6}}y+{\frac {26+5{\sqrt {13}}}{27}}=0$
三角関数、逆三角関数を用いて解は

$x={\frac {1+{\sqrt {13}}}{6}}+{\frac {2}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}\cos \left({\frac {1}{3}}\arccos {\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}\right)$$x={\frac {1+{\sqrt {13}}}{6}}+{\frac {2}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}\cos \left({\frac {1}{3}}\arccos {\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}\right)$
平方根、立方根を用いて

$x={\frac {1+{\sqrt {13}}}{6}}+{\frac {1}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}+i{\frac {3{\sqrt {39}}}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}}}+{\frac {1}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}-i{\frac {3{\sqrt {39}}}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}}}$$x={\frac {1+{\sqrt {13}}}{6}}+{\frac {1}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}+i{\frac {3{\sqrt {39}}}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}}}+{\frac {1}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}-i{\frac {3{\sqrt {39}}}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}}}$
$x={\frac {1+{\sqrt {13}}}{6}}+{\frac {1}{3}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}+i{\frac {3{\sqrt {39}}}{2}}}}+{\frac {1}{3}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}-i{\frac {3{\sqrt {39}}}{2}}}}$$x={\frac {1+{\sqrt {13}}}{6}}+{\frac {1}{3}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}+i{\frac {3{\sqrt {39}}}{2}}}}+{\frac {1}{3}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}-i{\frac {3{\sqrt {39}}}{2}}}}$
$\cos(2\pi /26)$$\cos(2\pi /26)$を平方根と立方根で表すと

$\cos {\frac {2\pi }{26}}={\frac {1+{\sqrt {13}}}{12}}+{\frac {1}{6}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}+i{\frac {3{\sqrt {39}}}{2}}}}+{\frac {1}{6}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}-i{\frac {3{\sqrt {39}}}{2}}}}$$\cos {\frac {2\pi }{26}}={\frac {1+{\sqrt {13}}}{12}}+{\frac {1}{6}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}+i{\frac {3{\sqrt {39}}}{2}}}}+{\frac {1}{6}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}-i{\frac {3{\sqrt {39}}}{2}}}}$

正二十六角形の作図

正二十六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。

正二十六角形は折紙により作図可能である。

関連項目

• 十三角形