Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
230 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLESFigure 42.1: Construction of inscribed regular hexagon H 1 = P 1 P 2 · · · P 6 .We will define L 1 to be the perimeter of H 1 :L 1 = P 6 P 1 +5∑P i P i+1We will then construct a sequence of inscribed regular polygons H 2 , H 3 , . . .as follows:i=1a. Each vertex of H i is a vertex of H i+1 ;b. For k = 1, .., i − 1, choose point Q k ∈ Γ to be the intersection of theangle bisector of ∠P k OP k+1 and Γ;c. Choose point Q i to be the intersection of the angle bisector of ∠P n OP 1 .and Γ;d. Define H i+1 = P 1 Q 1 P 2 Q 2 · · · P n Q nEach H i+1 has twice as many sides as the H i ; hence H k has 3 × 2 k sides.H 1 has 6 sides; H 2 has 12 sides; H 3 has 24 sides; etc.Define L n as the perimeter of the 3 × 2 n -gon defined in this manner.Theorem 42.2 The sequence of perimeters L 1 , L 2 , L 3 , ... is strictly increasing.Proof. (Exercise.)Theorem 42.3 The perimeter of a circumscribed square exceeds theperimeter of each of the 3 × 2 n -gons defined above.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLES 231Figure 42.2: Construction of inscribed regular 12-gon from regular hexagonby bisecting each central angle.Theorem 42.4 The sequence of perimeters L 1 , L 2 , ... converges to somenumber C.Proof. The sequence is bounded above and increasing. From a theorem ofcalculus, the sequence converges.Definition 42.5 Let Γ = C(O, r). The circumference of Γ is defined tobeC = limn→∞ L nTheorem 42.6 The value of the circumference only depends on the radiusand not the center point.Proof. Let Γ = C(O, r) and Γ ′ = C(O ′ , r).We can construct the perimeter of each sequence of inscribed 3 × 2 n -gonsH n and H n.′ Applying SAS to each sector, the corresponding triangles△P k OP k+1∼ = △P′kO ′ Pk+1′Since each triangle is congruent, the corresponding edges have equal length.Hence the L n = L ′ n for each n. Hence each sequence must converge to thesame limit.Theorem 42.7 Let r, r ′ > 0 be any positive numbers; and let C and C ′be the circumferences of concentric circles Γ = C(O, r) and Γ ′ = C(O, r ′ ).Then C/r = C ′ /r ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.
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230 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLESFigure 42.1: Construction of inscribed regular hexagon H 1 = P 1 P 2 · · · P 6 .We will define L 1 to be the perimeter of H 1 :L 1 = P 6 P 1 +5∑P i P i+1We will then construct a sequence of inscribed regular polygons H 2 , H 3 , . . .as follows:i=1a. Each vertex of H i is a vertex of H i+1 ;b. For k = 1, .., i − 1, choose point Q k ∈ Γ to be the intersection of theangle bisector of ∠P k OP k+1 and Γ;c. Choose point Q i to be the intersection of the angle bisector of ∠P n OP 1 .and Γ;d. Define H i+1 = P 1 Q 1 P 2 Q 2 · · · P n Q nEach H i+1 has twice as many sides as the H i ; hence H k has 3 × 2 k sides.H 1 has 6 sides; H 2 has 12 sides; H 3 has 24 sides; etc.Define L n as the perimeter of the 3 × 2 n -gon defined in this manner.Theorem 42.2 The sequence of perimeters L 1 , L 2 , L 3 , ... is strictly increasing.Proof. (Exercise.)Theorem 42.3 The perimeter of a circumscribed square exceeds theperimeter of each of the 3 × 2 n -gons defined above.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012