a constructible number
Start with a segment you declare to have length 1. Which other lengths can you then produce with compass and straightedge alone? A number is called constructible if a segment of exactly that length can be built starting from the unit. This single idea is the bridge that turns the geometry game of constructions into a question of algebra, and algebra is what finally decides what is possible.
From length 1 you can construct sums and differences (lay segments end to end), products and quotients (using similar triangles), and — the crucial extra power — square roots (a clean trick uses the altitude of a right triangle inscribed in a semicircle, giving the geometric mean sqrt(a) of a length a and 1). So the constructible numbers form a field that also contains the square root of anything already in it: starting from the rationals, you may take square roots over and over. In algebra terms, a number is constructible exactly when it lies in some tower of fields each obtained from the last by adjoining a square root — equivalently, it is algebraic and the degree of its minimal field extension over the rationals is a power of 2.
This algebraic criterion is the master key to every impossibility result. The cube root of 2 has degree 3 over the rationals (not a power of 2), so doubling the cube is out. cos(20 degrees) satisfies an irreducible cubic, so trisecting a 60-degree angle is out. And pi is transcendental — it satisfies no polynomial at all with rational coefficients, let alone one of degree a power of 2 — so squaring the circle is out. The geometry is impossible because the numbers are not constructible.
sqrt(2) is constructible: it is the diagonal of a unit square, or the geometric mean of 1 and 2 via the altitude of a right triangle in a semicircle. But the cube root of 2 is not, because its degree over the rationals is 3, not a power of 2.
Constructible numbers = a field closed under square roots; degree over the rationals must be a power of 2.
Being a power-of-2 degree is necessary but care is needed: it is not automatically sufficient for every algebraic number, though for the classical impossibility proofs the degree argument is exactly what does the work. Pi's case is different again — it is not algebraic at all (it is transcendental).