Constructing the 11-gon by splitting an angle in five

Question

In "Angle Trisection, the Heptagon, and the Triskaidecagon", published in the American Mathematical Monthly in March 1988, Andrew Gleason discusses what regular polygons can be constructed with compass, straightedge and angle trisector. At the end of that article he notes that the angle p-sectors required for a regular n-gon are the odd primes p dividing $\varphi(n)$.

For the heptagon, which only requires an angle trisector, he gives the minimal polynomial of $2\cos(2\pi/7)$ $$x^3+x^2-2x-1$$ and transforms it into the Chebyshev polynomial expression $$7\sqrt{28}(4\cos^3\theta-3\cos\theta)=7$$ leading to the final identity $$\sqrt{28}\cos\left(\frac{\cos^{-1}(1/\sqrt{28})}{3}\right)=1+6\cos(2\pi/7).$$

I am interested in the hendecagon (11 sides), which requires an angle quinsector (that splits an angle into five equal parts).

Is there a similar transformation between the minimal polynomial for $2\cos(2\pi/11)$ $$x^5+x^4-4x^3-3x^2+3x+1$$ and the relevant Chebyshev polynomial $$\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$ and how do I find it? If I had such a transformation, I could construct an exact hendecagon with the quinsector.

I have tried Tschirnhaus transforms on the former polynomial, without success.

Answer 1

(Updated answer)

While the old answer used a cubic transformation, I remembered that the general quintic can reduced to a one-parameter form (the Brioschi quintic) using only quadratic Tschirnhausen transformations. So given a solvable quintic,

$$F(x)=0$$

to transform that to the solvable DeMoivre form,

$$y^5+5my^3+5m^2y+n = 0$$

the trick is to use a rational Tschirnhausen of form,

$$y = \frac{x^2+ax+b}{x+c}$$

and the four parameters ($a,b,c,m$) suffice to enable the transformation. For the minimal polynomial of $x = 2\cos(2\pi/11),$

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1=0\tag1$$

surprisingly the transformation just involves powers of the golden ratio $\phi = \frac{1+\sqrt5}2,$

$$y = \frac{x^2+(\phi^2/\sqrt5)x-(\phi^2/\sqrt5)}{x+\phi^2}\tag2$$

Eliminating $x$ between $(1)$ and $(2)$, after simplification, one gets the special DeMoivre quintic,

$$y^5+5\left(\frac{-7+\sqrt5}{\;10\phi^2}\right)y^3+5\left(\frac{-7+\sqrt5}{\;10\phi^2}\right)^2y+\left(\frac{-830+139\sqrt5}{\;11(\phi\sqrt5)^6}\right)=0\tag3$$

Like any DeMoivre, this has a solution in terms of arccos. So the roots $x$ and $y$,

$$x = 2\cos(2\pi/11)$$ $$y = \frac2{\phi}\sqrt{\frac{7-\sqrt5}{10}}\,\cos\big(z\big)$$ $$z =\frac15\arccos\left(\frac{-15+32\sqrt5}{110}\sqrt{\frac{10}{7-\sqrt5}}\right)$$

Since $x,y$ only have a quadratic relation,

$$y = \frac{x^2+(\phi^2/\sqrt5)x-(\phi^2/\sqrt5)}{x+\phi^2}$$

then one can solve for $x$ using the quadratic formula and express it in terms of $y$ without using cube roots as in the old answer.

$$\\$$

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(Old answer)

The question essentially asks about transforming solvable equations from one form to another.

I. Cubic

Using just a linear transformation, the general cubic $P(x)=0$ can be transformed to the form,

$$y^3+3ay+b = 0\tag1$$

with solution,

$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{3}+\tfrac{1}{3}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^3}}\big)\right)\tag2$$

for $k=0,1,2$. Undoing the transformation establishes a relation between the roots $x,y$.

II. Quintic

Similarly, an appropriate Tschirnhausen transformation can transform a solvable quintic $P(x)=0$ to the Demoivre form (essentially the Chebyshev polynomial mentioned by the OP),

$$y^5+5ay^3+5a^2y+b = 0\tag3$$

with analogous solution,

$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{5}+\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big)\right)\tag4$$

for all five roots $y_k$. A cubic Tschirnhausen gives us three degrees of freedom to transform a solvable quintic to Demoivre form.

III. Transformations

For $p=7$:

$$x=2\cos\big(\tfrac{2\pi}{7}\big)\tag5$$

$$\color{blue}{y=3x+1} = 2\sqrt{7}\cos\left(\tfrac{1}{3}\,\cos^{-1}\big(\tfrac{1}{2\sqrt{7}}\big)\right)$$

then $x,y$ solves,

$$x^3+x^2-2x-1=0$$ $$y^3-21y-7=0$$

For $p=11$:

Let $\phi=\tfrac{1+\sqrt{5}}{2}$ be the golden ratio.

$$x=2\cos\big(\tfrac{2\pi}{11}\big)\tag6$$

$$\color{blue}{y =x^3-\phi\,x^2-\tfrac{7+\sqrt{5}}{2}x+\tfrac{5+4\sqrt{5}}{5}} = 2\,\phi\sqrt{\tfrac{11}{5}}\cos\left(-\tfrac{6\pi}{5}+\tfrac{1}{5}\,\cos^{-1}\big(\tfrac{-89-25\sqrt{5}}{44\sqrt{11}}\big)\right)$$

then $x,y$ solves,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1=0$$ $$y^5-5ay^3+5a^2y+b=0$$

where $a=\tfrac{11}{5}\phi^2,\;\;b=\tfrac{11(125+89\sqrt{5})}{250}\phi^5$.

Thus as you can see, the transformation (in blue) which relates the quintic roots $x,y$ is more complicated than the cubic version, but is nonetheless doable in radicals.

Answer 2

If the object is to construct a regular hendecagon, it can be done more simply without going through an angle quinsection. Benjamin and Snyder proved the existence of a construction using a marked ruler and compasses in 2014 (BENJAMIN, ELLIOT; SNYDER, C. Mathematical Proceedings of the Cambridge Philosophical Society156.3 (May 2014): 409-424.; http://dx.doi.org/10.1017/S0305004113000753 ).

The basic premises of the construction are as follows:

1. It's based on the properties of "conchoid-circle" constructions where we position the marked straightedge to paas through a fixed point $P$, with one mark on a line $l$ and the other on a circle $K$.

2. With this type of construction, we define a "signed distance" $z$. This is the distance between $P$ and the mark on $K$, with a negative sign if that mark lies vetween $P$ and the other mark which is on $l$, a positive sign otherwise.

3. Then $z$ satisfies a sextic equation whose coefficients satisfy certain relationships called the "verging theorem".

4. The quintic equation given here for the hendecagonal cosines is converted to a sextic equation that satisfies the verging theorem by (4.1) defining $z=ux$ for some scale factor $u$, and (4.2) introducing an additional root $\eta$ with the appropriate value relative to $u$. Then, all the geometric parameters needed to determine $l$ and $K$ may be expressed and constructed in terms of this scale factor $u$.

5. Now to the heart of the matter. It looks like we have to solve a seventh-degree equation for that parameter $u$. But, "a miracle occurs" (the authors' own words); the equation for $u$ is reducuble and all we are left with is a cubic factor equation (with integer coefficients), which can be solved by an auxiliary marked ruler construction. (The "miracle" appears to be related to the Gauss sum for hendecagonal angles, see here.)

6. So $z=ux$ has a construction with a marked ruler and compasses because it solves a sextic equation that satisfies the verging theorem, and $u$ has such a construction as well because it is obtained from a cubic equation in $Z[u]$; and so $x=2 cos(2\pi m/11)$ has one too.

7. Now for the parameters. For $u$, choose the one real root of $u^3+2u^2+2u+2=0$. For the construction of $z=ux$: $P=(0,0)$, $l$ is the line $x=-u-1$ where the length unit is the distance between the marks (conventional in this type of construction), $K$ is centeted at $(u(u-1)/2,-(u^2+3u+1)/2)$ and passes through $(-u-2,0)$. One orientation of the ruler is along the $x$ axis, that is the "extra" root of the sextic; the other roots on $K$ with proper distance signs, see (2), give the roots for $z$. Note that the authors do not give the formulas that way, I did some algebra of my own to get everything in terms of $u$.