INTERNATIONAL JOURNAL OF GEOMETRY Vol. 7 (2018), No. 2, 25 - 36 GOLDEN SECTIONS AND ARCHIMEDEAN CIRCLES IN AN ARBELOS NGUYEN NGOC GIANG and LE VIET AN Abstract. We construct some golden ratios in the arbelos and Archimedean circles in this configuration. Consider an arbelos formed by semi-circles (O1), (O2), and (O) of radii a; b; and a+b. The semi-circles (O1) and (O) meet at A,(O2) and (O) at B, (O1) and (O2) at C. Let CD be the divided line of the smaller semi-circles. A segment PQ is called to be divided in the golden ratio by a pointp R PQ PR 5+1 if PR = RQ . In this case, the divided ratio is the golden ratio ' := 2 , which satisfies '2 = ' + 1. Theorem 1. Let the segment AD and semi-circle (O1) meet again at E, BD and (O2) at F , AF and (O2) at G, BE and (O1) at H. If the rays CG and CH meet the semi-circle (O) at I and J respectively, K and L are the incenter and C-excenter of triangle CIJ, respectively, then K divides both the segments CD and LC in the golden ratios (see Figure 1). Proof. Since DC is the altitude of right triangle ABD and the quadrilateral ACHE is concyclic, \BDC = \BAD = \CAE = \BHC: It follows that quadrilateral BCHD is concyclic. Similarly, the quadrilateral ACGD is also concyclic. 2 Since AD touches the circles (O1) and (O2), DA:DE = DC = DB:DF by the Power-of-a-point theorem and by the Intersecting-chords theorem, the quadrilateral ABF E is concyclic. Chasing angles, we have \GCD = \GAD = \F AE = \F BE = \DBH = \DCH: This means that CD is the internal angle bisector of \ICJ, and K belongs to CD. ||||||||||||{ Keywords and phrases: Golden ratio, Archimedean circles, Arbelos. (2010)Mathematics Subject Classification: Received: 15.03.2018. In revised form: 05.09.2018. Accepted: 13.09.2018. 26 Nguyen Ngoc Giang and Le Viet An Figure 1. Let J 0 is the reflection of J across AB. By symmetry, J 0 belongs to the circle (O) and AJ = AJ 0. 0 It follows that \JIA = \AIJ = \AIC. Hence, AI is the internal angle bisector of \CIJ and K belongs to AI. Similarly, K also belongs to BJ. Since AJ and BJ are perpendicular, AJ is the external angle bisector of \CJI. We deduce that L belongs to AJ. Similarly, L also belongs to BI. Then the quadrilaterals ACIL and BCJL are concyclic. Let the lines AH and BG meet at M. Since CI and CJ are the anti- parallel of triangle ABL and they are perpendicular to BM and AM re- spectively. This means that M is the circumcenter of triangle ABL, and MO is perpendicular to AB. Since KC touches (O1), \MAO = \HAC = \HCK = \GCD = \GAD = \F AD: It follows that right triangles MAO and F AD are similar, MO FD DF = =) MO = AO : AO AD DA Golden Sections and Archimedean Circles in an Arbelos 27 p We have AO = AB = a + b, AD = AB:AC = 2pa(a + b), DC = p p 2 CA:CB = 2 ab, and DC2 CA:CB 4ab 2apb(a + b) DF = = p = = : DB BA:BC 2pb(a + b) a + b It follows that DF p DC MO = AO = ab = : DA 2 Let N be the orthogonalp projection of M onto CD. Then N is the midpoint of CD and CN = ab. By the Pythagorean theorem, ML2 = MA2 = AO2 + MO2 = (a + b)2 + ab; and since CNMO is the rectangle, MN = CO = jAC − AOj = j2a − (a + b)j = ja − bj : It follows that NL2 = ML2 − MN 2 = (a + b)2 + ab − (a − b)2 = 5ab p p =) LC = LN + CN = ( 5 + 1) ab: Let K0 be the reflection of K across AB. By symmetry, CK = CK0 and 0 ◦ \AK B = \AKB = 180 − \ALB: It follows that quadrilateral ALBK0 is concyclic. By the Intersecting-chords theorem, CK:CL = CK0:CL = CA:CB CA:CB 2a:2b p p =) CK = = p p = ( 5 − 1) ab: CL ( 5 + 1) ab Hence, p KC KC 5 − 1 = = p = '; KD CD − KC 3 − 5 and KL LC − CK 2 = = p = ': KC CK 5 − 1 These prove that K divides both CD and LC in the golden ratios. Remark 2. It is easy to see that CK = LD and D divides both segments LK and CL in the golden ratios. The famous Archimedean twin circles associated in the arbelos have equal ab radii t := a+b (see [2] and [3]). Circles with radius t are called Archimedean and they are congruent to the Archimedean twin circles. Theorem 3. If the perpendicular bisector of CD meets CG and CH at P and Q respectively, then the circle with diameter PQ is Archimedean (see Figure 2). 28 Nguyen Ngoc Giang and Le Viet An Figure 2. p Proof. Since the right triangles pCPN and AF D are similar, CN = ab, p 2a b(a+b) AD = 2 a(a + b) and DF = a+b , we deduce that PN FD FD ab = =) PN = CN = = t: CN AD AD a + b Similarly, QN = t. It follows that the circle with diameter PQ is Archimedean. Remark 4. It is easy to see that if the line perpendicular to CD at D meets CG and CH at R and S, respectively, then the circles with diameters DR and DS are Archimedean. For two points P and Q in the plane, the circle with center P passing through Q is denoted by P (Q). Theorem 5. The circle D(O) meets the perpendicular bisector of AB again at U. The semi-circle (O1) and the segment AU meet at V , (O2) and BU at W . If the common external tangent lines of two circles A(V ) and B(W ) meet CD at D1 and D2 such that D, D1, D2 are collinear in that order, then D divides D2D1 in the golden ratio. Proof. Since the right triangles CAV and CBW are similar, CA AV = : CB BW This means that C is the internal homothetic center of two circles A(V ), B(W ), and CV , CW are the common internal tangent lines of two circles A(V ) and B(W ). Let the line BU and (O) meet again at B1, and let the lines CV and CW meet two common external tangent lines of two circles A(V ) and B(W ) at X; Y; Z, and T as show in the Figure 3. Golden Sections and Archimedean Circles in an Arbelos 29 Figure 3. Since XA and XB are the internal and external angle bisectors of \CXY , ◦ we get \AXB = 90 . It follows that X belongs to circle (O). Similarly, the points Y; Z; T also belong to circle (O). Note that CY and AB1 are both perpendicular to BB1, they are parallel, and since AX = AT by the symmetry under the axis AB, YA bisects angle XTY . Chasing angles, we have \AB1X = \AY X = \AY T = \B1AY . It follows that XB1 and AY are parallel. Since AXB1Y is the isosceles trapezoid with two bases AY and XB1, AB1 = XY . Let H1 be the orthogonal projection of D onto OU. Since H1 is the midpoint of OU, we get OU CD = OH = : 1 2 Let us denote by α := \ACV , then α = \BCW = \BAB1 = \BCZ. Since the right triangles BAB1 and BUO are similar, AB1 UO 2CD 4CD BB1 = = = =) AB1 = 4DC = 4DC:sinα: BB1 BO BO AB BA 30 Nguyen Ngoc Giang and Le Viet An By the symmetry, XT and YZ are both perpendicular to AB. Let the line CD and circle (O) meet again at D3, AB and XT at M1, AB and YZ at M2. By the Intersecting-chords theorem and symmetry, 2 2 2 AB1 = 16CD :sin α = 16CD:CD3:sinα:sinα M X M Z = 16CX:CZ 1 2 = 16M X:M Z = 4XT:Y Z: CX CZ 1 2 It follows that (1) XY 2 = 4XT:Y Z: Note that D1D2;XT and YZ are pairwise parallel. By the Thales' theo- rem, D X CX XT D X XT (2) 1 = = =) 1 = ; D1Y CZ YZ XY XT + YZ and similarly, D Y YZ (3) 1 = : XY XT + YZ Comparing (1), (2) with (3), we obtain 4XT 2:Y Z2 (4) D X:D Y = : 1 1 (XT + YZ)2 Again, by the Thales' theorem, CD CD XD YD XD + D Y XT:Y Z 1 + 1 = 1 + 1 = 1 1 = 1 =) CD = ; XT YZ XY YX XY 1 XT + YZ and similarly, XT:Y Z CD = : 2 XT + YZ It follows that 2XT:Y Z (5) D D = CD + CD = : 1 2 1 2 XT + YZ And by the Intersecting-chords theorem and symmetry, (6) D1X:D1Y = D1D:D1D3 = D1D:D2D: 2 From (4), (5) and (6), we deduce that D1D2 = DD1:DD2. This proves that D1 divides D2D in the golden ratio. Theorem 6. The external tangent line of two semi-circles (O1) and (O2) meets the semi-circle (O) at P1 and P2 such that A, P1, D, P2 and B lie on the semi-circle (O) in that order. The line passing through P1 perpendicular to CP1 meets (O) at P1 and Q1. Let C1 be the circumcenter of triangle CDQ1. Circle C1(P2) meets CD at E1 and F1 such that E1, D, C and F1 lie on CD in that order. Then D divides both the segments CE1 and F1C in the golden ratios. Proof. (see Figure 4). Segment DA meets the semi-circle (O1) at E, and segment DB meets the semi-circle (O2) at F ; let M0 be the mid-point of CD.