Straightedge-and-Compass Constructions

Three geometry problems survived from antiquity without an answer for two thousand years: build a cube of twice a given volume, cut an arbitrary angle in three, and build a square equal in area to a given circle, each using only a straightedge and a compass. Each was eventually shown impossible, and the proof is not geometric. It translates ruler-and-compass steps into field operations, notices that every step lands in an extension of degree at most $2$, and reads the impossibility off the tower law.

Constructions as field operations

Fix a segment of length $1$. Using it to scale the axes, place every construction in the Cartesian plane ${\mathbb{R}}^2$. A point is constructible if it can be reached from the starting data by a finite sequence of the four legal moves, and a real number is constructible if it is the coordinate of a constructible point.

Elementary geometry realizes each arithmetic operation as a construction. Given lengths $a$ and $b$, laying off segments produces $a+b$ and $a-b$ directly; similar triangles built from parallel lines produce the product $ab$ and the quotient $a/b$; and a semicircle on a diameter of length $1+a$ produces $\sqrt{a}$ as the perpendicular half-chord.

So constructible numbers are closed under the four field operations and under square roots. Two consequences fix the algebraic setting.

• The constructible numbers form a subfield of $\mathbb{R}$ containing $\mathbb{Q}$, since starting from $1$ the field operations reach every rational.
• A construction cannot escape a quadratic step. Lines through constructed points have equations with coefficients in the field $F$ generated so far, and intersecting two lines stays in $F$. A circle contributes an equation $(x-h{)}^2+(y-k{)}^2=r^2$ with $h,k,r\in F$; intersecting it with a line gives a quadratic for one coordinate, and intersecting two circles reduces, by subtracting the equations, to a line meeting a circle. Every new coordinate therefore lies in an extension of $F$ of degree $1$ or $2$.

The degree obstruction

Iterating the second point along a construction produces a tower in which each step is quadratic.

In particular every constructible number is algebraic over $\mathbb{Q}$, since it lies in a finite extension. The obstruction is one-directional: a $2$-power degree is necessary for constructibility, not sufficient. Some numbers of degree $4$ over $\mathbb{Q}$ are not constructible, because the obstruction lies in the splitting field rather than in the degree of the number itself; the sharp criterion is that $\alpha$ is constructible exactly when the splitting field of its minimal polynomial has $2$-power degree, which the Galois theory of later lessons makes precise. What matters for the three classical problems is the crude contrapositive: if $[\mathbb{Q}(\alpha ):\mathbb{Q}]$ is not a power of $2$, then $\alpha$ is not constructible.

A constructive tower

The positive direction is equally concrete: a number built from rationals by a finite chain of square roots is constructible, and that chain is the tower itself. The regular pentagon is the clean case. Its central quantity is $\alpha =2\cos \frac{2\pi }{5}$, a root of $x^2+x-1=0$, so

The single quadratic step $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{5})$ carries $\alpha$, and the construction mirrors it: build $\sqrt{5}$ as a semicircle altitude, do one rational division and subtraction, and the pentagon's vertex coordinate is in hand. A number needing two nested roots, such as $\sqrt{2+\sqrt{2}}$, sits atop a two-step tower $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{2+\sqrt{2}})$ of degrees $2$ and $2$, so it is constructible with degree $4$ over $\mathbb{Q}$. Each nested radical is one more quadratic floor.

The three impossibilities

Each collapses to a degree that is not a power of $2$.

Squaring the circle. A square equal in area to the unit circle has side $\sqrt{\pi }$, whose constructibility would make $\pi$ constructible and in particular algebraic. But $\pi$ is transcendental, so $[\mathbb{Q}(\pi ):\mathbb{Q}]$ is infinite, let alone a power of $2$.²

The proof of one impossibility also produces a positive fact worth recording: the trigonometric functions of an integer-degree angle are constructible exactly when the angle is a multiple of $3^{\circ }$. The equilateral triangle gives $60^{\circ }$, the regular pentagon gives $72^{\circ }$, and the difference and half-angle formulas assemble $3^{\circ }$; but $1^{\circ }$ and $2^{\circ }$ are out of reach, since either would let the addition formulas produce the impossible $20^{\circ }$.

The marked ruler

The negative results are specific to the straightedge, an unmarked edge. A ruler carries marks, and the extra move of sliding a marked ruler until a fixed distance falls between two curves reaches beyond quadratic steps. An Archimedean construction with a marked ruler trisects any angle, and a related one doubles the cube. Both classical impossibilities disappear once the tool is allowed to solve certain cubics, which shows the theorem is a statement about the algebra the two idealized instruments generate, degree-$2$ steps only, rather than about geometry in general.

Constructible regular polygons

The regular $n$-gon is constructible exactly when $\cos \frac{2\pi }{n}$ is a constructible number, so the question is again one of degree, now for a root of unity. The pentagon works because $\alpha =2\cos \frac{2\pi }{5}$ satisfies $x^2+x-1=0$, degree $2$. The regular $7$-gon and $9$-gon fail: $2\cos \frac{2\pi }{7}$ satisfies an irreducible cubic, and constructing the $9$-gon would trisect $120^{\circ }$.

The full criterion belongs to the theory of cyclotomic fields: the degree of $\mathbb{Q}(\cos \frac{2\pi }{n})$ over $\mathbb{Q}$ is $\frac{1}{2}\phi (n)$, and constructibility of the $n$-gon amounts to that degree being a power of $2$, which happens precisely when $n$ is a power of $2$ times distinct Fermat primes. The degree $\frac{1}{2}\phi (n)$ comes from the cyclotomic polynomials; the $17$-gon entry, Gauss's discovery, follows from $\phi (17)=16=2^4$.