With only an unmarked straightedge and a compass, ancient geometers could bisect angles, build equilateral triangles and draw regular pentagons. Yet four simple-sounding tasks resisted every attempt for millennia. The reason turned out to be algebraic: once constructions are translated into equations, it becomes clear why squaring the circle, doubling the cube, trisecting an angle and building a regular heptagon are impossible.
The Two Tools
A straightedge draws a straight line through two marked points. A compass draws a circle centered at one marked point and passing through another. Whenever a line or circle meets another line or circle, the intersection becomes a new point. Fix two points and declare their distance to be one unit; every later length must come from these steps.
Constructible Numbers and Algebra
Using Cartesian coordinates, a line has the form ax + by = c, and a circle centered at (h, k) with radius r satisfies (x − h)² + (y − k)² = r². For example, the point (3, 2) lies on y + 2x = 8 because 2 + 6 = 8, and (4, 3) lies on x² + y² = 25 because 16 + 9 = 25. Intersecting lines and circles means solving linear or quadratic equations, so only square roots ever appear.
A number built from integers using addition, subtraction, multiplication, division and square roots is called a constructible number, and these numbers describe the coordinates of constructible points, while their nonnegative values are constructible lengths [1]. A constructible number must lie in a field reached by a sequence of quadratic extensions of the rationals, so its degree over the rationals is a power of 2 [1]. This one fact drives most of the impossibility proofs below.
Squaring the Circle
Draw a circle of radius 1. Its area is π × 1² = π. Squaring the circle asks for a square of the same area, whose side would have to be √π.
A number is algebraic if it is a root of a nonzero polynomial with integer coefficients. For instance, 3x² − 30 = 0 has roots ±√10. Numbers that are not algebraic are called transcendental. Every constructible number is algebraic, and the constructible numbers are closed under multiplication, so a constructible √π would make π = √π × √π constructible and hence algebraic. Lambert's 1761 proof that π is irrational was not enough, because some irrational algebraic numbers are constructible; Lindemann's 1882 proof that π is transcendental settled the matter [2]. Squaring the circle is therefore impossible. The full transcendence proof is too long to reproduce here.
Doubling the Cube
A cube with edge 1 has volume 1. A cube of volume 2 would need an edge of length ∛2. The number ∛2 is a root of the irreducible polynomial x³ − 2, so it has degree 3 over the rationals [3]. Since 3 is not a power of 2, ∛2 cannot be constructible [4]. Square roots, which circles provide, never produce a cube root in general, so doubling the cube is impossible.
The same algebraic criterion rules out all three of the classical Greek problems [1]. For more on how the circle and angles relate through trigonometry, see our explanation of angle units such as turns, degrees and radians.
Trisecting the Angle
Trisecting means splitting an arbitrary angle into three equal parts. To show it cannot always be done, one counterexample suffices. A 60° angle is easy to construct, but trisecting it would produce a 20° angle, and a 20° angle cannot be constructed [4].
Building a 20° angle amounts to constructing the length cos 20°. The triple-angle identity for cosine says:
cos 3θ = 4cos³θ − 3cos θ
Setting θ = 20° and cos 60° = ½ gives ½ = 4cos³20° − 3cos 20°. Multiplying by 2 and letting x = cos 20° gives:
8x³ − 6x − 1 = 0
By the rational root theorem, any rational root would be ±1, ±½, ±¼ or ±⅛. Substituting each shows that none works, so the cubic is irreducible over the rationals and cos 20° has degree 3. Because 3 is not a power of 2, cos 20° is not constructible, and a 60° angle cannot be trisected. A corollary is that a regular 18-gon is also impossible, since it would give a 20° angle [3]. For a related reminder of why degree, roots and polynomial factoring matter, see completing the square and solving quadratics.
The Regular Heptagon
The triangle, square, pentagon and hexagon are all constructible, but the regular heptagon is not. The argument follows the same pattern. The relevant length is cos(2π/7), and 2cos(2π/7) is a zero of the irreducible cubic x³ + x² − 2x − 1, so that cubic is its minimal polynomial [5]. A constructible number needs a minimal polynomial of power-of-2 degree, and 3 is not, so the heptagon is impossible [6].
Which Regular Polygons Can Be Built?
The general answer is the Gauss–Wantzel theorem: a regular n-gon is constructible exactly when n is a power of 2 times any number of distinct Fermat primes [7]. Fermat primes have the form 22ⁿ + 1, and the only known ones are 3, 5, 17, 257 and 65537 [8]. For n = 3 this formula gives 2⁸ + 1 = 257. Gauss established sufficiency of the general criterion, and Wantzel proved its necessity in 1837 [9]. The heptagon fails because 7 is not a Fermat prime, and the nonagon fails because 9 = 3 × 3 repeats the Fermat prime 3 [10].
Pierpont primes are a different family, of the form 2ᵘ × 3ᵛ + 1. The number 7 is one (2 × 3 + 1), and it is not a Fermat prime, so the heptagon cannot be built with straightedge and compass, although it can be built with a marked ruler and compass [5]. Pierpont primes are not the criterion for ordinary constructibility; the Fermat primes are. The number 17 = 2⁴ × 3⁰ + 1 = 22² + 1 is both, and the regular 17-gon really can be constructed. To see how related shapes behave in higher settings, explore the complex geometry shapes article.
Sources
- Math 5111 (Algebra 1) – Lecture #7 ∼ October 1, 2020
- Ferdinand von Lindemann (1852 – 1939) – Biography – MacTutor History of Mathematics
- COMPASS AND STRAIGHTEDGE APPLICATIONS OF FIELD THEORY SPENCER CHAN
- 21.3: Geometric Constructions – Mathematics LibreTexts
- Heptagon
- Compass and Straightedge Construction for Regular Heptagon does not exist – ProofWiki
- Constructible polygon
- 5.6. Regular Polygons 1 5.6. Regular Polygons
- Theorem of Wantzel
- Dividing the circle
Leave a Reply