Here, I describe something I recently learned: a simple proof of the impossibility of trisecting an angle in Euclidean geometry.
This proof is given in chapter 19 of Edwin E. Moise, Elementary Geometry From an Advanced Standpoint, Second Edition (Addison-Wesley, 1974).
We begin with constructable numbers: Given a unit length (a line segment of length \(1\)), if we can use compass and straightedge to construct a line segment with the length, the number is said to be constructable. For example, \(\sqrt{2}\) is constructable, because we can construct a square of sides \(1\), and that square has diagonal of length \(\sqrt{2}\).
In chapter 19, Moise shows that if \(a\) and \(b\) are constructable numbers, then so are \(a+b\), \(ab\), \(1/a\), and \(\sqrt{a}\). These constructions are not obvious, but easy to set up and understand (section 19.2).
This leads to the definition of the surd field. A surd is any number that can be constructed from rational numbers using finitely many operations of addition, subtraction, multiplication, division, and taking square roots. The surd field is indeed a field. The idea is that a number is constructable with compass and straightedge exactly if it is in the surd field.
To show this, it must be shown that any number constructed using compass and straightedge is in the surd field. The key idea is to show that the intersection of two lines, a line and a circle, and two circles are points with coordinates in the surd field, provided the coefficients of these equations are in the surd field. The verification is straightforward just using algebra.
The next step is to define a quadratic exention of a field, which Moise does in section 19.8: Suppose \(F\) is a field, and suppose \(k\in F\) is such that \(\sqrt{k}\) is not in \(F\). A simple example: \(F=\mathbb Q\) and \(k=2\), where \(\sqrt{k}=\sqrt{2}\) is not in \(\mathbb Q\). Then the quadratic extension field \(F(k)\) is the field of all numbers of the form \(x+y\sqrt{k}\) where \(x,y\in F\). It is easy to show that this is a field (the field operations are closed for numbers of this kind).
Now the operation of conjugation is defined on \(F(k)\) in analogy with conjugation of complex numbers: If \(z=x+y\sqrt{k}\) then \(\overline{z} = x - y\sqrt{k}\).
It is easy to verify for \(z_1,z_2\in F(k)\) that \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\) and \(\overline{z_1z_2}=\overline{z_1}\overline{z_2}\). It follows that \(\overline{z^n}=\overline{z}^n\). It also follows that \(\overline{f(z)} = f(\overline{z})\) whenever \(f\) is a polynomial with all coefficients in \(F\). From this we get the key result that \(z_0\) solves \(f(z)=0\) exactly when \(\overline{z_0}\) solves that equation, provided \(f\) is a polynomial with all coefficients in \(F\).
In section 19.9 Moise defines a surd field of order \(n\) to be a surd field \(F_n\) formed by \(n\) quadratic field extensions beginning with \(F_0 = \mathbb Q\). So field \(F_{i+1}=F_i(k_{i+1})\) for some \(k_{i+1}\in F_i\) where \(\sqrt{k_{i+1}}\) is not in \(F_i\).
Now we come to cubic equtions with rational coefficients. This is the heart of the matter. We consider an equation
\[ x^3 + ax^2 + bx + c = 0\]
where \(a,b,c\in\mathbb Q\). Suppose we factor this:
\[ (x-z_1)(x-z_2)(x-z_3) = 0 \]
so
\[\begin{aligned}
a &= - z_1 - z_2 - z_3 \\
b &= z_1z_2 + z_1z_3 + z_2z_3 \\
c &= z_1z_2z_3
\end{aligned}\]
Now suppose \(z_1\in F(k)\). Then \(\overline{z_1}\) is also a solution. So \(z_2\) or \(z_3\) must be \(z_1\); say \(z_2=\overline{z_1}\). Then we must have \(z_3\in F\), since \(z_3 = -a - z_1 - z_2\) and \(z_1+z_2\in F\).
So if a cubic equation with rational cofficients has a solution \(z_1\) in \(F(k)\), it must have a solution in \(F\).
It immediately follows that if the equation has a solution in a surd field of order \(n\), it must have a solution in the rationals.
Example (not in Moise): \(x^3 - 9x + 21x - 5 = 0\) has \(z_1=2+\sqrt{3}\) as a solution. Then \(z_2=2-\sqrt{3}\) also solves the equation. The assertion is that there is a real solution. This is
\[z_3 = - a - z_1 - z_2 = 9 - (2 +\sqrt{3}) - (2 - \sqrt{3}) = 5\]
which indeed is a solution.
We now apply this to angle trisection. We will show that \(\theta = 60^\circ\) cannot be trisected.
Otherwise, we could construct an angle \(\theta=20^\circ\). If this were constructable, we would be able to draw a ray in the first quadrant of the coordinate plane, with a point on it with coordinates \((x,y)\) in the surd field, and \(r = \sqrt{x^2+y^2}\) is also in the surd field. It follows that \(u = x/r = \cos20^\circ\) is in the surd field. It must also be true that \(z = 2u = 2\cos 20^\circ\) is in the surd field.
Now if \(\theta=20^\circ\), the identity
\[ \cos3\theta = 4\cos^3\theta - 3\cos\theta\]
becomes
\[ 1/2 = \cos 60^\circ = 4\cos^320^\circ - 3\cos20^\circ \]
so \(u\) satisfies
\[ 4u^3 - 3u = 1/2\]
or
\[ 8u^3 - 6u - 1 = 0\]
Because the lead coefficient is not \(1\), we cannot use the algebra described above. But \(z = 2u = 2\cos20^\circ\) satisfies
\[ 8u^3 - 6u - 1 = (2u)^3 - 3(2u) - 1 = z^3 - 3z - 1 = 0\]
and the algebra above does apply to this equation: If it has a solution in the surd field, it has a rational solution. But it has no rational solutions. So \(z = 2\cos20^\circ\) is not in the surd field, and therefore \(u=\cos20^\circ\) is not in the surd field.
The exact same reasoning applies to the equation \(x^3=2\), which shows that the ancient problem of duplication of the cube is also impossible to solve using compass and straightedge.