Plane Geometry - Bruce E. Shapiro

SECTION 45. EUCLIDEAN CONSTRUCTIONS 259Figure 45.9: Construction 45.8, the division of a segment into n equal lengthsub-segments, illustrated for n = 3.Proof. Otherwise, we would be able to construct a segment of length π.Proof. (outline) Suppose we can trisect the angle 60 into thirds.cos 60 = 1/2, using the trigonometric identitySincecos 3θ = 4 cos 3 θ − 3 cos θif we set x = cos θ then8x 3 − 6x − 1 = 0Then one must show that (a) there are no rational solutions 2 ; and (b)if there is any solution of the form a + b √ c as a solution, we obtain acontradiction. 3Other ResultsTheorem 45.17 (Poncelet-Steiner Theorem) All constructions thatcan be performed with the straight-edge and compass can be performedwith a straight-edge alone, given a single circle and its center.2 Assume p/q is a solution; plug it in and you get that q must be a factor of 8. Byexhaustive checking one eliminates all possibilities.3 Since a + b √ c is a solution, so is a − b √ c as well as a third solution r; the sum of thesethree solutions must be zero. We add them up and solve for r and obtain the resultthat r = 2a is a solution, which contradicts the assumption that a + b √ c represents theleast-inclusive extension field of solutions.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.