I'm stuck in how can I construct a regular 19-sided polygon with angle trisection allowed. My interest came from the article "Angle Trisection, the Heptagon, and the Triskaidecagon", published in the American Mathematical Monthly in March 1988 by Andrew Gleason.
Gleason shows how can we construct a regular heptagon (7 sides) and a triskaidecagon (13 sides) using one angle trisection. At the end of his text, he commented that the enneadecagon (19-gon) can be construct using two trisections, and left the details to the reader.
Well, I tried to relate the eighteen complex roots of the equation $z^{19}-1=0$ with the problem. Gauss, in his work "Disquisitones Arithmeticae", shows that these roots can be divided in groups of six distincts roots.
If $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then the notation $(6,1)$, $(6,2)$ and $(6,3)$ correspond to
$(6,1) = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18}$
$(6,2) = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17}$
$(6,4)=\xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15}$
And these values are the three roots of the equation $x^3+x^2-6x-7=0$
Especifically, the root $(6,2)$ is equal to
$(6,2)=2\left(\cos\left(\frac {4\pi}{19}\right)+\cos\left(\frac {6\pi}{19}\right)+\cos\left(\frac {10\pi}{19}\right)\right)=\frac{2\sqrt{19}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{7}{2\sqrt{19}}}\right)}-\frac{1}{3}$
Which makes me think that this number is constructible by one trisection.
Moreover, Gauss shows that $(2,1)=2\cos\left(\frac{2\pi}{19}\right)$ is a root of the equation
$x^3-(6,1)x^2+[(6,1)+(6,4)]x-2-(6,2)=0$ (★)
And gave the relations:
$(6,4)=4-(6,2)^2$
$(6,1)=-5-(6,2)+(6,2)^2$
And I made it until here. I don't know how to compute the second trisection, since if I could substitute the values of $(6,4)$ and $(6,1)$ in the equation (★) to solve $x$ in terms of $(6,2)$, I couldn't find a satisfatory answer, like
$2\cos\left(\frac{2\pi}{19}\right)=?$
Am I missing something that could help me to solve this second trisection? Thanks.
EDIT:
After some computations, I found this result (checked by WA, though):
$2\cos\left(\frac{2\pi}{19}\right)=(2,1)=\frac{2}{3}\cdot\sqrt{2k+7}\cdot\cos\left(\frac{1}{3}\arccos\frac{(3k^2+17k+18)}{2\sqrt{19}\sqrt{4k^2+18k+21}}\right)+\frac{(-5-k+k^2)}{3} $
Where $k=(6,2)=\frac{2\sqrt{19}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{7}{2\sqrt{19}}}\right)}-\frac{1}{3}$
But, man, now I have no clue how to construct this.