Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
258 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.8: Construction 45.7, construction of a parallel line through apoint that is not on the line.2. Find the midpoint M of AC.3. Construct circle γ = C(M, AM).4. Construct a line l through B that is perpendicular to AC.5. γ intersects l at two points. Name one of these points D. (It doesnot matter which one).6. Let c = BD. Then c = √ a.Theorem 45.13 All constructible numbers are algebraic numbers, i.e., everyconstructible number is a root of the algebraic equationwhere a 0 , a 1 , ..., a n are integers.a n x n + a n−1 x n−1 + ... + a 0 = 0A number is not algebraic is said to be transcendental. Since π is transcendental1 , we have the following result.Theorem 45.14 π is not constructible.Corollary 45.15 The construction of squaring the circle with a straightedgeand a compass is impossible.Theorem 45.16 The general construction of trisecting an angle cannot beperformed with a straight-edge an compass alone.1 This result was proven by T. Lindemann in 1882.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
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.
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258 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.8: Construction 45.7, construction of a parallel line through apoint that is not on the line.2. Find the midpoint M of AC.3. Construct circle γ = C(M, AM).4. Construct a line l through B that is perpendicular to AC.5. γ intersects l at two points. Name one of these points D. (It doesnot matter which one).6. Let c = BD. Then c = √ a.Theorem 45.13 All constructible numbers are algebraic numbers, i.e., everyconstructible number is a root of the algebraic equationwhere a 0 , a 1 , ..., a n are integers.a n x n + a n−1 x n−1 + ... + a 0 = 0A number is not algebraic is said to be transcendental. Since π is transcendental1 , we have the following result.Theorem 45.14 π is not constructible.Corollary 45.15 The construction of squaring the circle with a straightedgeand a compass is impossible.Theorem 45.16 The general construction of trisecting an angle cannot beperformed with a straight-edge an compass alone.1 This result was proven by T. Lindemann in 1882.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
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