The Greeks left three construction problems that nobody managed to solve with ruler and compass, the tools of Euclid’s Elements:

  1. Doubling the cube: constructing the edge of a cube with twice the volume of a given cube.
  2. Trisecting the angle: dividing any angle into three equal parts.
  3. Squaring the circle: constructing a square with the same area as a given circle.

Legend has it that, during a plague, the oracle of Delos told the Athenians to double the altar of Apollo, which had the shape of a cube. They doubled the edge, the volume became eight times larger, and the plague went on. For two thousand years mathematicians tried. The answer only came in the nineteenth century, with the ideas of Gauss and Galois: it is impossible.

What ruler and compass do

Start with a segment of length 1. With ruler and compass one can add, subtract, multiply and divide lengths, using similar triangles. And one can take square roots, with the geometric mean in a semicircle. But that is all.

The reason lies in analytic geometry. Each new point of a construction is the intersection of two lines, of a line and a circle, or of two circles. Lines have equations of degree 1 and circles of degree 2. Solving these intersections never requires more than a quadratic equation. So each step of the construction adds at most one square root:

A length constructible with ruler and compass can be written from 1 with the four operations and square roots. In particular, if it is a root of an equation with no rational factors, the degree of that equation is a power of 2.

The cube of Delos

Doubling a cube of edge 1 means constructing the edge $x$ of a cube of volume 2:

$$x^3=2,\qquad x=\sqrt[3]2\approx1.2599.$$

The equation $x^3-2=0$ has no rational root and does not factor into smaller rational polynomials. Its degree is 3, which is not a power of 2. Therefore $\sqrt[3]2$ cannot be constructed. The altar of Delos cannot be doubled with ruler and compass.

The 60° angle

Some angles are easy to trisect: a right angle divides into three angles of $30°$, which are constructed with an equilateral triangle. But the problem asks for a method for any angle, and one counterexample is enough. Take $60°$. Trisecting it means constructing $20°$, that is, the length $x=\cos20°$. By the triple-angle formula for the cosine,

$$\cos3\theta=4\cos^3\theta-3\cos\theta,$$

and with $\theta=20°$ and $\cos60°=\frac12$:

$$4x^3-3x=\frac12\qquad\Longrightarrow\qquad8x^3-6x-1=0.$$

The only possible rational roots would be $\pm1,\pm\frac12,\pm\frac14,\pm\frac18$, and none of them works. So the equation does not factor, and its degree is 3. The $60°$ angle cannot be trisected with ruler and compass.

Wantzel and Galois

Both proofs were published in 1837 by Pierre Wantzel, a 23-year-old Frenchman, a student at the École Polytechnique, in Liouville’s journal. Wantzel used ideas of Gauss and Lagrange; Galois’s works would only be published nine years later. Today the argument is explained with Galois theory: the degree of an equation measures the size of the world of numbers its roots create, and ruler and compass can only climb in steps of two.

The third problem is harder still. Squaring a circle of radius 1 means constructing $\sqrt\pi$. In 1882 the German Ferdinand von Lindemann proved that $\pi$ is not a root of any equation with rational coefficients: it is a transcendental number. With that, the circle cannot be squared either.

Exercise: can the regular heptagon be constructed with ruler and compass?

No. The number $x=\cos\frac{2\pi}7\approx0.6235$ is a root of $8x^3+4x^2-4x-1=0$. The possible rational roots would be $\pm1,\pm\frac12,\pm\frac14,\pm\frac18$, and none works: for example, at $x=\frac12$ the polynomial is $1+1-2-1=-1$. So the equation does not factor, it has degree 3, and $\cos\frac{2\pi}7$ is not constructible.

Exercise: why can a 90° angle be trisected, if the triple-angle equation is also of degree 3?

With $\cos90°=0$ the equation becomes $4x^3-3x=0$, which factors: $x(4x^2-3)=0$. The root we want, $x=\cos30°=\frac{\sqrt3}2$, comes from an equation of degree 2, and square roots can be constructed. The problem is only impossible when the cubic does not factor, as in the case of $60°$.

Sources: Pierre Wantzel, “Recherches sur les moyens de reconnaître si un problème de géométrie peut se résoudre avec la règle et le compas”, Journal de mathématiques pures et appliquées 2 (1837) · Ian Stewart, Galois Theory (CRC Press, 4th ed., 2015) · Felix Klein, Famous Problems of Elementary Geometry (1897).

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