---
tag: W8J42M1G28
kind: statement
role: theorem
name: Impossibility of the classical construction problems
msc: [12F05, 51M15]
uses: [0FMN2MN7TM, 0SSSM1D475, 0CQ357M0QS, 0EWXPZCSZ0, 0X7CVDE3BG, 0JNCBFMPXR]
about: [0FMN2MN7TM, 0DAECMZVRT]
sources:
  - url: https://en.wikipedia.org/wiki/Straightedge_and_compass_construction
---

## Statement

Start from two points at distance $1$ in the plane, and use only straightedge and compass.

1. It is impossible to construct a segment of length $\sqrt[3]{2}$, that is, to double the volume of a cube with edge $1$.
2. It is impossible to trisect every angle; in particular the angle $\pi/3$ cannot be trisected.
3. It is impossible to construct a square with the same area as a circle of radius $1$, that is, a segment of length $\sqrt{\pi}$.

## Remarks

A length is constructible exactly when it is a [[constructible number|0FMN2MN7TM]], and [[the degree of a constructible number over $\Q$ is a power of $2$|0SSSM1D475]].
For clause 1, $\sqrt[3]{2}$ has minimal polynomial $x^3 - 2$, of degree $3$; see [[impossibility of doubling the cube|0CQ357M0QS]].
For clause 2, $2\cos(\pi/9)$ has minimal polynomial $x^3 - 3x - 1$, of degree $3$; see [[impossibility of trisecting the angle|0EWXPZCSZ0]].
For clause 3, constructible numbers are algebraic, while $\pi$, and hence $\sqrt{\pi}$, is transcendental by the [[Lindemann–Weierstrass theorem|0JNCBFMPXR]]; see [[impossibility of squaring the circle|0X7CVDE3BG]].
Some angles can be trisected, such as $\pi/2$, since $\pi/6$ is constructible; clause 2 says that no method works for all angles.

## Explanations

### L1 human

The ancient Greeks posed three famous drawing problems: double a cube, divide any angle into three equal parts, and turn a circle into a square of the same area.
They allowed only an unmarked ruler and a compass.
Two thousand years later, algebra showed that all three are impossible.
The reason is that these tools can only produce numbers reached by repeated square roots.

### L2 human

Each straightedge-and-compass step solves linear or quadratic equations, so each new point lies in an extension of degree $1$ or $2$ of the field of the previous points.
A constructible number therefore lies at the top of a tower of quadratic extensions, and by the tower law its degree is a power of $2$.
Clauses 1 and 2 then follow from degree $3$ minimal polynomials, while clause 3 needs the much deeper transcendence of $\pi$.
The same method, with [[cyclotomic extensions|VD6VC3ZMMY]], decides which regular polygons are constructible.