Using the proof of Theorem 32.11, show that the regular 9-gon is not constructible. 

Data from Theorem 32.11

Trisecting the angle is impossible; that is, there exists an angle that cannot be trisected with a straightedge and a compass. 

Proof Figure 32.12 indicates that the angle θ can be constructed if and only if a segment of length | cos θ| can be constructed. Now 60° is a constructible angle, and we shall show that it cannot be trisected. Note that

Let θ = 20°, so that cos 30 = ½, and let α = cos 20°. From the identity 4 cos³θ - 3 cosθ = cos3θ, we see that

Thus α is a zero of 8x³ - 6x - 1. This polynomial is irreducible in Q[x], since, by Theorem 23.11, it is enough to show that it does not factor in .Z[x]. But a factorization in Z[x] would entail a linear factor of the form (8x ± 1), (4x ± 1), (2x ± 1), or (x ± 1). We can quickly check that none of the numbers ±_1/8, ±¼, ±½, and ±1 is a zero of 8x³ - 6x - 1. Thus [Q(α) : Q] = 3, so by Corollary 32.8, α is not constructible. Hence 60° cannot be trisected.

Transcribed Image Text:

cos 30 = cos(20 + 0) = cos 20 cos 0 - sin 20 sin = (2 cos² 0 - 1) cos 0 - 2 sin cos sin 1) cos 0 - 2 cos 0 (1 - cos² 0) = (2 cos² 0 − - = 4 cos³ 0 - 3 cos 0.