# Trisecting the Angle by Compass and Straightedge Construction is Impossible

## Theorem

There is no compass and straightedge construction for the trisection of the general angle.

## Proof

Let $OA$ and $OB$ intersect at $O$.

It will be shown that there is no general method using a compass and straightedge construction to construct $OC$ such that $\angle AOB = 3 \times \angle AOC$.

It is sufficient to demonstrate that this is impossible for one specific angle.

Hence we choose $\angle AOB = 60 \degrees$.

Let $A$ and $B$ be points on the unit circle whose center is at $\tuple {0, 0}$.

Let $A$ lie on the $x$-axis.

Thus:

$O$ is the point $\tuple {0, 0}$
$A$ is the point $\tuple {1, 0}$
$B$ is the point $\tuple {\cos 60 \degrees, \sin 60 \degrees}$

These all belong to $\Q \sqbrk {\sqrt 3}$.

trisection of $AOB$ is equivalent to constructing the point $\tuple {\cos 20 \degrees, \sin 20 \degrees}$.

$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$

so:

$8 \cos^3 20 \degrees - 6 \cos 20 \degrees = 2 \cos 60 \degrees = 1$

Thus $\cos 20 \degrees$ is a root of the polynomial:

$8 x^3 = 6 x - 1$

which by Irreducible Polynomial: $8 x^3 - 6 x - 1$ in Rationals is irreducible over $\Q$.

Thus $\cos 20 \degrees$ is algebraic over $\Q$ with degree $3$.

Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos 20 \degrees$ is not an element of any extension of $\Q$ of degree $2^m$.

The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2.

$\blacksquare$