Angle trisection with compass and straightedge

Given an arbitrary angle $\theta$, construct in finitely many steps using only an idealized compass and an unmarked straightedge two rays that divide $\theta$ into three equal angles of measure $\theta/3$.

Proved impossible by Pierre Laurent Wantzel in 1837. Wantzel showed that a real number $\alpha$ is constructible by compass and straightedge from a unit segment if and only if $\alpha$ lies in a tower of quadratic field extensions of $\mathbb{Q}$, which implies that the degree $[\mathbb{Q}(\alpha):\mathbb{Q}]$ of its minimal polynomial must be a power of $2$. For $\theta = 60^\circ$, the triple-angle identity $\cos(3\alpha) = 4\cos^3(\alpha) - 3\cos(\alpha)$ with $\cos(60^\circ) = 1/2$ shows that $x = 2\cos(20^\circ)$ satisfies $x^3 - 3x - 1 = 0$. Since this cubic polynomial has no rational roots, it is irreducible over $\mathbb{Q}$, so $[\mathbb{Q}(\cos 20^\circ):\mathbb{Q}] = 3 \neq 2^k$, proving that $60^\circ$ cannot be trisected.