22 1. Trisecting Angles solved simultaneously. If (x, y) is a point of intersection of the two circles, then x = h1 ± r2 1 − (y − k1)2 = h2 ± r2 2 − (y − k2)2. Equating these quantities yields ± r2 1 − (y − k1)2 = h2 − h1 ± r2 2 − (y − k2)2. Squaring both sides and performing some algebraic manipulation gives the equation Ay + B = ±H r2 2 − (y − k2)2, where A = −2k2 + 2k1 ∈ F , B = r2 1 − r2 2 − k2 1 + k2 2 − (h2 − h1)2 ∈ F , and H = h2 − h1 ∈ F . Squaring both sides again and doing some more algebra gives the equation Py2 + Qy + R = 0, where P = A2 + h2 ∈ F , Q = 2AB − 2H2k2 ∈ F , and R = B2 − H2r2 2 + H2k2 2 ∈ F . The quadratic formula shows that the solution y ∈ F ( Q2 − 4PR)). Thus x = h1 ± r2 1 − (y − k1)2 ∈ F ( Q2 − 4PR) r2 1 − (y − k1)2 . Proof of Theorem 1.6.6. To facilitate matters, say that x is a construct- ible number of order n if it is a coordinate of some n-th order constructible point (1.5.1). Let K(n) be the constructible numbers of order n. So K = n K(n). To prove the theorem, we need to show that every x in K(n) belongs to a quadratic extension of Q. The proof proceeds by induction. Recall the sets Pn, Ln, and Cn defined in (1.5.1). For n = 1, K(1) = {0, 1}. So there is nothing to prove. Assume that n ≥ 1 and every number in K(n) belongs to some quadratic extension of Q. Let a ∈ K(n + 1). There is a point b in K(n + 1) such that (a, b) ∈ Pn+1. Thus (a, b) is the point of intersection of two lines from Ln, a line from Ln and a circle from Cn, or two circles from Cn. Suppose that (a, b) is the point of intersection of two lines 1 and 2 in Ln. By definition there are points (s, t) and (u, v) in Pn that determine 1 . Since s, t, u, v ∈ K(n), there are quadratic extensions Fs,Ft,Fu,Fv of Q containing these numbers. By Proposition 1.6.5, there is a single quadratic extension of Q that contains all four of these numbers. Similarly there is a quadratic extension of Q that contains all four of the coordinates of the two points in Pn that determine

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.