MathLabs

Doubling the cube

Solved, 1837GeometryAlgebra
Statement

Given the edge of a cube of side length aa, construct in finitely many steps using only an idealized compass and an unmarked straightedge the edge length x=23 ax = \sqrt[3]{2}\,a of a second cube whose volume x3=2a3x^3 = 2a^3 is twice that of the given cube.

Proved impossible by Pierre Laurent Wantzel in 1837. Setting a=1a = 1, doubling the cube requires constructing the real number x=23x = \sqrt[3]{2}, which is a root of the polynomial x3−2=0x^3 - 2 = 0. By Eisenstein's criterion for the prime p=2p = 2 (or rational root inspection), x3−2x^3 - 2 is irreducible over Q\mathbb{Q}, so the field extension [Q(23):Q][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] has degree 33. Wantzel proved that any length constructible with an unmarked straightedge and compass lies in an iterated quadratic extension of Q\mathbb{Q} of degree 2k2^k; since 33 does not divide 2k2^k, 23\sqrt[3]{2} is not constructible.

While impossible with an unmarked straightedge and compass alone, 23\sqrt[3]{2} can be constructed exactly when the allowed operations are extended to solve cubic equations. Classical Greek geometers gave solutions using three-dimensional intersecting cylinders, cones, and tori (Archytas of Tarentum, c. 400 BCE), conic sections (Menaechmus), the cissoid of Diocles, the conchoid of Nicomedes, or a marked straightedge (neusis). In modern origami mathematics, Peter Messer (1986) showed that folding a square sheet of paper divided into three equal strips constructs 23\sqrt[3]{2} directly as a ratio of segments along the sheet's edge.

References

  1. Pierre Laurent Wantzel (1837). Recherches sur les moyens de reconnaître si un problème de géométrie peut se résoudre avec la règle et le compas
  2. Thomas Little Heath (1921). A History of Greek Mathematics, Volume 1: From Thales to Euclid
  3. David S. Richeson (2019). Tales of Impossibility: The 2000-Year Quest to Solve the Mathematical Problems of Antiquity