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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