Proof of the Existence of a Ruler-and-Compass Construction for the Regular 17-gon

正十七边形的尺规作图存在之证明 苏剑林 · 2009-09-20 · Mathematics proof polygon compass and straightedge construction

Found online — there seem to be three different versions, all reproduced here.

For the ruler-and-compass construction method for the regular 17-gon, please see:

http://kexue.fm/article.asp?id=104

This article only proves its existence (i.e., derives $\cos ({2\pi}/{17})$). more

$$\cos \frac{2\pi}{17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}$$

Version 1 (from the Mathematics Research Forum):

Let the central angle of the regular 17-gon be $\theta$, then $17\theta=2\pi$, i.e. $16\theta=2\pi-\theta$

Hence $\sin 16\theta=-\sin \theta$, and

$$\begin{aligned}\sin 16\theta=2\sin 8\theta \cos 8\theta=2^2\sin 4\theta \cos 4\theta \cos 8\theta \\ =2^4 \sin \theta \cos \theta \cos 2\theta \cos 4\theta \cos 8\theta\end{aligned}$$

Since $\sin \theta\neq 0$, dividing both sides by it gives:

$$16\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta=-1$$

Also, from $2\cos \theta \cos 2\theta=\cos \theta+\cos 3\theta$ etc., we have

$$2(\cos \theta+\cos 2\theta+...+\cos 8\theta)=-1$$

Noting that $\cos 15\theta=\cos 2\theta,\cos 12\theta=\cos 5\theta$, let

$$\begin{aligned}x=\cos \theta+\cos 2\theta+\cos 4\theta+\cos 8\theta \\ y=\cos 3\theta+\cos 5\theta+\cos 6\theta+\cos 7\theta\end{aligned}$$

Then:

$$x+y=-1/2$$

Also $xy=(\cos \theta+\cos 2\theta+\cos 4\theta+\cos 8\theta)(\cos 3\theta+\cos 5\theta+\cos 6\theta+\cos 7\theta)$

$$=1/2(\cos 2\theta+\cos 4\theta+\cos 4\theta+\cos 6\theta+...+\cos \theta+\cos 15\theta)$$

By computation we know $xy=-1$

And also

$$x=(-1+\sqrt{17})/4,y=(-1-\sqrt{17})/4$$

Next, let: $x_1=\cos \theta+\cos 4\theta,x_2=\cos 2\theta+\cos 8\theta$

$$y_1=\cos 3\theta+\cos 5\theta,y_2=\cos 6\theta+\cos 7\theta$$

So we have $x_1+x_2=(-1+\sqrt{17})/4$

$$y_1+y_2=(-1-\sqrt{17})/4$$

Solving this gives: (solve it yourselves~~~~)

Finally, from $\cos \theta+\cos 4\theta=x_1,\cos \theta\cos 4\theta=(y_1)/2$

we can obtain $\cos \frac{2\pi}{17}$, which is a combination of addition, subtraction, multiplication, division, and square roots of numbers. Hence the regular 17-gon can be constructed with ruler and compass.

Version 2 (source forum unknown):

Version 3 (PDF file, also sourced online):

Introduction to the Approach and Method for Constructing the Regular 17-gon.zip