Question 332688
<font face="Garamond" size="+2">


There is not a geometric proof of which I am aware.  You have to rely on algebra and trigonometry.  The following courtesy of Wikipedia:


Let *[tex \LARGE \mathbb{Q}] denote the rational numbers.


A number constructible in one step from field K is a solution of a second order polynomial.  Also, *[tex \LARGE \frac{\pi}{3}] radians *[tex \LARGE =\ 60^\circ] is constructible.


But *[tex \LARGE \frac{\pi}{3}] radians cannot be trisected.


Note:  *[tex \LARGE \cos(\frac{\pi}{3})\ =\ \cos(60^\circ)\ =\ \frac{1}{2}]


If *[tex \LARGE 60^\circ] could be trisected, the minimal polynomial of *[tex \LARGE \cos(20^\circ)] over *[tex \LARGE \mathbb{Q}] would be a second order polynomial.


Note the trigonometric identity:  *[tex \LARGE \cos(3\theta)\ =\ 4\cos^3(\theta)\ -\ \3cos(\theta)]


Let *[tex \LARGE y\ =\ \cos(20^\circ)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(60^\circ)\ =\ \frac{1}{2}\ =\ 4y^3\ -\ 3y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ 4y^3\ -\ 3y\ -\ \frac{1}{2}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ 8y^3\ -\ 6y\ -\ 1\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ (2y)^3\ -\ 3(2y)\ -\ 1\ =\ 0]


Let *[tex \LARGE x\ =\ 2y\ \Rightarrow\ \ x^3\ -\ 3x\ -\ 1\ =\ 0]


Let *[tex \LARGE p(x)\ =\ x^3\ -\ 3x\ -\ 1\ =\ 0]


The minimal polynomial for *[tex \LARGE x] (hence the minimal polynomial for *[tex \LARGE \cos(20^\circ)]) is a factor of *[tex \LARGE p(x)].


If *[tex \LARGE p(x)] has a rational root by the Rational Root Theorem, that root must be either *[tex \LARGE \pm 1].  But:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(1)\ =\ 1\ -\ 3\ -\ 1\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(-1)\ =\ -1\ +\ 3\ -\ 1\ =\ 1]


Hence *[tex \LARGE p(x)] is irreducible over *[tex \LARGE \mathbb{Q}] and the minimal polynomial for *[tex \LARGE \cos(20^\circ)] is of degree 3.


Therefore *[tex \LARGE 60^\circ] cannot be trisected.


In fact there are some angles that can be trisected.  But an angle can be trisected if and only if *[tex \LARGE q(t)\ =\ 4t^3\ -\ 3t\ -\ \cos(\theta)] is reducible over the field extension *[tex \LARGE \mathbb{Q}\left(\cos(\theta)\right)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
</font>