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Chapter 6 the Arbelos

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Chapter 6 the Arbelos Chapter 6 The arbelos 6.1 Archimedes’ theorems on the arbelos Theorem 6.1 (Archimedes 1). The two circles touching CP on different sides and AC CB each touching two of the semicircles have equal diameters AB· . P W1 W2 A O1 O C O2 B A O1 O C O2 B Theorem 6.2 (Archimedes 2). The diameter of the circle tangent to all three semi- circles is AC CB AB · · . AC2 + AC CB + CB2 · We shall consider Theorem ?? in ?? later, and for now examine Archimedes’ wonderful proofs of the more remarkable§ Theorems 6.1 and 6.2. By synthetic reasoning, he computed the radii of these circles. 1Book of Lemmas, Proposition 5. 2Book of Lemmas, Proposition 6. 602 The arbelos 6.1.1 Archimedes’ proof of the twin circles theorem In the beginning of the Book of Lemmas, Archimedes has established a basic proposition 3 on parallel diameters of two tangent circles. If two circles are tangent to each other (internally or externally) at a point P , and if AB, XY are two parallel diameters of the circles, then the lines AX and BY intersect at P . D F I W1 E H W2 G A O C B Figure 6.1: Consider the circle tangent to CP at E, and to the semicircle on AC at G, to that on AB at F . If EH is the diameter through E, then AH and BE intersect at F . Also, AE and CH intersect at G. Let I be the intersection of AE with the outer semicircle. Extend AF and BI to intersect at D. Now in triangle ABD, AI and BF are two altitudes intersecting at E. It follows that E is the orthocenter of the triangle, and ED, being perpendicular to AB, passes through the point D. Therefore, HE : AC = HD : AD (HE//AC) = CB : AB (CG//BI). AC CB This gives HE = AB· . The symmetry of this expression indicates that the same reasoning would show that this is also the diameter of the circle touching CP and the semicircles (AB) and (CB). Hence, the two circles are congruent. We call these Archimedes’ twin circles. 3Proposition 1. 6.1 Archimedes’ theorems on the arbelos 603 6.1.2 Archimedes’ calculation of the radius of the incircle of the arbelos Since AE and CI are altitudes of triangle ACG, their intersection L is the ortho- center of the triangle. Therefore, GL is perpendicular to the base AB. Similarly, HM is also perpendicular to the base AB. D G I H E F K L M A N C P B Figure 6.2: It follows that AN : NP = AL : LH (LN//HP ) = AC : CB (IC//DB) = GM : MB (AG//CM) = NP : P B (GN//MP ). Archimedes noted that this is enough to calculate the diameter of the incircle. Although he illustrated this with AC : CB =3:2, his calculation clearly can be extended to the general case. Since AN + NP + P B = AB and AN : NP : P B = AC2 : AC CB : CB2, · it follows that AC CB NP = · AB. AC2 + AC CB + CB2 · · This is the diameter of the incircle. 604 The arbelos 6.1.3 Constructions of Archimedes’ twin circles 4 Proposition 6.1. Let G be the point of tangency of the Archimedean circle (W1) with the semicircle (AC). (a) The tangent to (AC) at G passes through B. (b) If (W1) touches CP at E, then CW = WE, where W is the intersection of BG and CP . (c) BG = BP . Construction 6.1. Let G be the intersection of the semicircle (AC) with the circle B(P ). Join B, G to intersect CP at W . Construct the circle W (C) to intersect CP again at E. Construct the perpendicular to CP at E, to intersect the line O1G at W1. The circle W1(G) is tangent to the semicircles (AC) and (AB), and also to the line CP . P W1 E G W A O1 K O C B Figure 6.3: Remarks. (1) G can also be constructed as the intersection of the semicircles (O1B) and (AC). (2) The center W1 can also be obtained as the intersection of the bisector of angle GW P and the line O1G. 4Bankoff’s statement 13. 6.2 Archimedes’ twin circles 605 6.2 Archimedes’ twin circles Theorem 6.3 (Archimedes). The two circles each tangent to CP , the largest semi- circle AB and one of the smaller semicircles have equal radii t, given by ab t = . a + b Q A O1 O P O2 B A O1 O P O2 B Proof. Consider the circle tangent to the semicircles O(a+b), O1(a), and the line P Q. Denote by t the radius of this circle. Calculating in two ways the height of the center of this circle above the line AB, we have (a + b t)2 (a b t)2 =(a + t)2 (a t)2. − − − − − − From this, ab t = . a + b The symmetry of this expression in a and b means that the circle tangent to O(a + b), O2(b), and P Q has the same radius t. 606 The arbelos 1 1 1 6.2.1 Harmonic mean and the equation a + b = t 2ab The harmonic mean of two quantities a and b is a+b . In a trapezoid of parallel sides a and b, the parallel through the intersection of the diagonals intercepts a segment whose length is the harmonic mean of a and b. See Figure 8A. We shall 1 1 1 write this harmonic mean as 2t, so that a + b = t . See Figure 8B. D a C b a t A b B (c) Figure 8A (d) Figure 8B Here is another construction of t, making use of the formula for the length of an angle bisector in a triangle. If BC = a, AC = b, then the angle bisector CZ has length 2ab C A t = cos = 2t cos . c a + b 2 2 The length t can therefore be constructed by completing the rhombus CXZY (by constructing the perpendicular bisector of CZ to intersect BC at X and AC at Y ). See Figure 9A. In particular, if the triangle contains a right angle, this trapezoid is a square. See Figure 9B. C X t Y b M t t t A Z B a (e) Figure 9A (f) Figure 9B 6.2 Archimedes’ twin circles 607 Construction 6.2. Construct the circle P (C3) to intersect the diameter AB at P1 and P2 so that P1 is on AP and P2 is on P B. The center C1 respectively C2 is the intersection of the circle O1(P2) respec- tively O2(P1) and the perpendicular to AB at P1 respectively P2. Q C1 Q1 C2 Q2 C3 A O1 P1 O P P2 O2 B Figure 6.9: 608 The arbelos 6.3 The incircle Theorem 6.4 (Archimedes). The incircle of the arbelos has radius r1r2(r1 + r2) ρ = 2 2 . r1 + r1r2 + r2 X Y θ A O1 O C O2 B Proof. Let ∠COO2 = θ. By the law of cosines, we have (r + ρ)2 =(r + r ρ)2 + r2 + 2r (r + r ρ) cos θ, 1 1 2 − 2 2 1 2 − (r + ρ)2 =(r + r ρ)2 + r2 2r (r + r ρ) cos θ. 2 1 2 − 1 − 1 1 2 − Eliminating θ, we have r (r + ρ)2 + r (r + ρ)2 =(r + r )(r + r ρ)2 + r r2 + r r2. 1 1 2 2 1 2 1 2 − 1 2 2 1 The coefficients of ρ2 on both sides are clearly the same. This is a linear equation in ρ: r3 + r3 + 2(r2 + r2)ρ =(r + r )3 + r r (r + r ) 2(r + r )2ρ, 1 2 1 2 1 2 1 2 1 2 − 1 2 from which 4(r2 + r r + r2)ρ =(r + r )3 + r r (r + r ) (r3 + r3) = 4r r (r + r ), 1 1 2 2 1 2 1 2 1 2 − 1 2 1 2 1 2 and ρ is as above. 6.3 The incircle 609 6.3.1 Construction of the incircle of the arbelos In [?], Bankoff published the following remarkable theorem which gives a con- struction of the incircle of the arbelos of the incircle, much simpler than the one we designed before from Archimedes’ proof. The simplicity of the construction is due to the existence of a circle congruent to Archimedes’ twin circles. Theorem 6.5 (Bankoff). The points of tangency of the incircle of the arbelos with the semicircles (AC) and (CB), together with C, are the points of tangency of the incircle (W3) of triangle O1O2O3 with the sides of the triangle. This circle (W3) is congruent to Archimedes’ twin circles (W1) and (W2). P O3 Q R W3 A O1 O C O2 B Proof. Since O1Q = O1C, O2C = O2R, and O3R = O3Q, the points C, Q, R are the points of tangency of the incircle of triangle O1O2O3 with its sides. The semi-perimeter of the triangle is 3 2 r1r2(r1 + r2) (r1 + r2) (r1 + r2) ρ s = r1 + r2 + ρ = r1 + r2 + 2 2 = 2 2 = . r1 + r2r2 + r2 r1 + r2r2 + r2 r1r2 The inradius of the triangle is the square root of 2 2 r1r2ρ r1r2 2 = 2 = t . s (r1 + r2) It follows that this inradius is t. The incircle of triangle O1O2O3 is congruent to Archimedes’ twin circles. 610 The arbelos Construction 6.3. Let M and N be the midpoints of the semicircles (AC) and (CB) respectively.
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