三十九角形

正三十九角形においては、中心角と外角は9.23…°で、内角は170.769…°となる。一辺の長さが a の正三十九角形の面積 S は

${\displaystyle S={\frac {39}{4}}a^{2}\cot {\frac {\pi }{39}}\simeq 120.77542a^{2}}$

${\displaystyle \cos(2\pi /39)}$を平方根と立方根で表すと

${\displaystyle {\begin{aligned}\cos {\frac {2\pi }{39}}=&\cos \left({\frac {2\pi }{3}}-{\frac {8\pi }{13}}\right)\\=&\cos {\frac {2\pi }{3}}\cos {\frac {8\pi }{13}}+\sin {\frac {2\pi }{3}}\sin {\frac {8\pi }{13}}\\=&-{\frac {1}{2}}\cos {\frac {8\pi }{13}}+{\frac {\sqrt {3}}{2}}\sin {\frac {8\pi }{13}}\\=&-{\frac {1}{2}}\cos {\frac {8\pi }{13}}+{\frac {\sqrt {3}}{2}}{\sqrt {\frac {1+\cos {\frac {16\pi }{13}}}{2}}}\\=&-{\frac {1}{24}}\cdot \left(12\cos {\frac {8\pi }{13}}\right)+{\frac {\sqrt {3}}{24}}{\sqrt {72+72\cos {\frac {10\pi }{13}}}}\\=&-{\frac {1}{24}}\left({\sqrt {13}}-1+\omega {\sqrt[{3}]{104-20{\sqrt {13}}+12i{\sqrt {39}}}}+\omega ^{2}{\sqrt[{3}]{104-20{\sqrt {13}}-12i{\sqrt {39}}}}\right)\\&+{\frac {\sqrt {3}}{24}}{\sqrt {72+6\cdot (12\cos {\frac {10\pi }{13}})}}\\=&-{\frac {1}{24}}\left({\sqrt {13}}-1+\omega {\sqrt[{3}]{104-20{\sqrt {13}}+12i{\sqrt {39}}}}+\omega ^{2}{\sqrt[{3}]{104-20{\sqrt {13}}-12i{\sqrt {39}}}}\right)\\&+{\frac {\sqrt {3}}{24}}{\sqrt {72+6\left(-{\sqrt {13}}-1+\omega ^{2}{\sqrt[{3}]{104+20{\sqrt {13}}+12i{\sqrt {39}}}}+\omega {\sqrt[{3}]{104+20{\sqrt {13}}-12i{\sqrt {39}}}}\right)}}\end{aligned}}}$

関係式

${\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{39}}+2\cos {\frac {32\pi }{39}}+2\cos {\frac {34\pi }{39}}={\frac {1}{4}}\left(1-{\sqrt {13}}-{\sqrt {6\left(13-3{\sqrt {13}}\right)}}\right)=x_{1}\\2\cos {\frac {4\pi }{39}}+2\cos {\frac {14\pi }{39}}+2\cos {\frac {10\pi }{39}}={\frac {1}{4}}\left(1+{\sqrt {13}}+{\sqrt {6\left(13+3{\sqrt {13}}\right)}}\right)=x_{2}\\2\cos {\frac {8\pi }{39}}+2\cos {\frac {28\pi }{39}}+2\cos {\frac {20\pi }{39}}={\frac {1}{4}}\left(1-{\sqrt {13}}+{\sqrt {6\left(13-3{\sqrt {13}}\right)}}\right)=x_{3}\\2\cos {\frac {16\pi }{39}}+2\cos {\frac {22\pi }{39}}+2\cos {\frac {38\pi }{39}}={\frac {1}{4}}\left(1+{\sqrt {13}}-{\sqrt {6\left(13+3{\sqrt {13}}\right)}}\right)=x_{4}\\\end{aligned}}}$

さらに、以下のような関係式が得られる。

両辺の立方根を取ると

${\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{39}}+\omega \cdot 2\cos {\frac {32\pi }{39}}+\omega ^{2}\cdot 2\cos {\frac {34\pi }{39}}=&{\sqrt[{3}]{\tfrac {104+34{\sqrt {13}}+3{\sqrt {6\left(13-3{\sqrt {13}}\right)}}+15{\sqrt {6\left(13+3{\sqrt {13}}\right)}}+3{\sqrt {3}}\left(-2{\sqrt {13}}+{\sqrt {6\left(13-3{\sqrt {13}}\right)}}+{\sqrt {6\left(13+3{\sqrt {13}}\right)}}\right)i}{8}}}\\2\cos {\frac {2\pi }{39}}+\omega ^{2}\cdot 2\cos {\frac {32\pi }{39}}+\omega \cdot 2\cos {\frac {34\pi }{39}}=&{\sqrt[{3}]{\tfrac {104+34{\sqrt {13}}+3{\sqrt {6\left(13-3{\sqrt {13}}\right)}}+15{\sqrt {6\left(13+3{\sqrt {13}}\right)}}-3{\sqrt {3}}\left(-2{\sqrt {13}}+{\sqrt {6\left(13-3{\sqrt {13}}\right)}}+{\sqrt {6\left(13+3{\sqrt {13}}\right)}}\right)i}{8}}}\\\end{aligned}}}$

よって

• 十三角形