Golden Sections and Archimedean Circles in an Arbelos

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 point√ R PQ PR 5+1 if PR = RQ . In this case, the divided ratio is the golden ratio ϕ := 2 , which satisﬁes ϕ2 = ϕ + 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 = ∠FBE = ∠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 respectively. 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 √ We have AO = AB = a + b, AD = AB.AC = 2pa(a + b), DC = √ √ 2 CA.CB = 2 ab, and DC2 CA.CB 4ab 2apb(a + b) DF = = √ = = . DB BA.BC 2pb(a + b) a + b It follows that DF √ DC MO = AO = ab = . DA 2 Let N be the orthogonal√ 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 = |AC − AO| = |2a − (a + b)| = |a − b| . It follows that NL2 = ML2 − MN 2 = (a + b)2 + ab − (a − b)2 = 5ab √ √ =⇒ 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 √ √ =⇒ CK = = √ √ = ( 5 − 1) ab. CL ( 5 + 1) ab Hence, √ KC KC 5 − 1 = = √ = ϕ, KD CD − KC 3 − 5 and KL LC − CK 2 = = √ = ϕ. 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. √ Proof. Since the right triangles √CPN 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’ theorem, 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. We easily see that CEDF is a rectangular. Hence M0 is the mid-point of EF . Furthermore ∠CEF = ∠CDB = ∠CAE. It follows that EF is Golden Sections and Archimedean Circles in an Arbelos 31

Figure 4. tangent with (O1). Similarly, EF is also tangent with (O2). Hence EF is the external common tangent line of two semi-circles (O1) and (O2). Hence EF and PQ are coincident. Since two semi-circles (O) and (O1) are inner tangent at A, the homothety a a+b a HA center A, ratio a+b transforms (O) into (O1). Under the homothety a a+b HA , points O, D go into points O1, E. Hence OD and O1E are parallel. Since EF is tangent with (O1), O1E is perpendicular to EF . It follows that OD is perpendicular to P1P2. This thing proves that OD is the perpendicular bisector of P1P2. It follows DP1 = DP2. We have ∠DP1P2 = ∠DP2P1 = ∠DAP1. This thing proves that DP1 is tangent with the circumcircle of triangle AEP1. Applying the Power-of-a- 2 2 point theorem, we have DP1 = DA.DE = DC . It follows DP1 = DC. Since DP1 = DP2, DC = DP1 = DP2, it follows that

D is the circumcenter of triangle CP1P2.(7) 0 0 Let C be the point symmetric to C across D. Then DC = DP1 = DC . 0 ◦ 0 It follows ∠CP1C = 90 . Hence three points C , P1 and Q1 are collinear.√ Applying the Power-of-a-point theorem with note that CC0 = 2CD = 4 ab 32 Nguyen Ngoc Giang and Le Viet An √ √ 1 0 0 and CM0 = 2 CD = ab, and C M0 = C C − CM0 = 3 ab, we have 0 0 0 2 2 0 2 2 0 0 C P1.C Q1 = C O − DO = C C − DC = 12ab = C M0.C C.

Applying the Intersecting-chords theorem, we have quadrilateral CM0P1Q1 being concyclic. ◦ Hence ∠CM0Q1 = ∠CP1Q1 = 90 . Hence Q1M0 is perpendicular to CD at the mid-point M0 of CD so triangle CDQ1 is isosceles at Q1. Thus, C1 belongs to Q1M0. Hence E1 and F1 are symmetric about point M0 and CF1 = DE1. On the other hand, since triangle CDQ1 is isosceles at Q1 and D is the circumcenter of triangle CP1P2, ∠CDP2 = 2∠CP1P2 = 2∠CQ1M0 = ∠CQ1D. Hence DP2 is tangent with the circumcircle of triangle CDQ1. Thus, DC1 is perpendicular to DP2. Applying the Pythagorean theorem, we have 2 2 2 2 2 2 2 CD = P2D = P2C1 − C1D = E1C1 − C1M0 + M0D 2 2 2 = E1C1 − C1M0 − M0D 2 2 = E1M0 − M0D = (E1M0 − M0D) . (E1M0 + M0D)

= E1D. (E1M0 + M0C)

= E1D.E1C

= E1D. (E1D + CD) . 2 2 CD It follows CD = E1D + E1D.CD. The above equality proves that = √ E1D 5+1 2 = ϕ, is golden ratio. Furthermore, since CF = DE , it follows DC = ϕ and DF1 = ϕ+1 = ϕ. 1 1 F1C DC ϕ This means that point D divides F1C in the golden ratio. Since the ﬁguration of theorem 6, we obtain a following result on the Archimedean circle.

Figure 5. Golden Sections and Archimedean Circles in an Arbelos 33

Theorem 7. Line P1Q1 meets the circumcircle of triangle CDQ1 at K1. Let L1 be the projection from K1 onto P1P2. Then the circle with diameter K1L1 is Archimedean (see Figure 5).

Proof. Let H0 be the point of intersection of DO and P1P2. Since (7), we have OD perpendicular to P1P2 at H0, quadrilateral CM0P1Q1 is concyclic. ◦ Hence ∠DK1P1 = 180 − ∠DCQ1 = ∠M0P1Q1. It follows that DK1 and M0P1 are parallel. This means that K1L1 = DH0. On the other hand, the circle with diameter DH0 is Archimedean (it is the circle (W4) in [2], also the circle (A3) in [3]). Thus, the circle with diameter K1L1 is Archimedean.

Theorem 8. The external tangent line of two semi-circles (O1) and (O2) meets the semi-circle (O) at P1 and P2, and meets CD at M0. Line AM0 meets DO2 at A0. Let O0 be the point symmetric to O across D. Circle O0(A0) meets DP1 at X1 and Y1 such that P1, X1, D and Y1 lie on DP1 in that order. Then point X1 divides DP1 in the golden ratio, and point Y1 divides P1D in the golden the ratio.

Proof. (see Figure 6). Let H0 be the point of intersection of OD and P1P2. Then OD is perpendicular to P1P2 at H0 and DH0 = 2t.

Figure 6.

Since O2, M0 are the midpoints of CB,CD, respectively, O2M0 is parallel to BD, from BD is perpendicular to AD, O2M0 is perpendicular to AD. 34 Nguyen Ngoc Giang and Le Viet An

It follows that M0 is the orthocenter of triangle ADO2. Hence AA0 is perpendicular to√DO2 at A0. Since CD = 2 ab and CO2 = b,

p 2 2 p 2 (8) O2A0 + DA0 = DO2 = O2C + CD = b + 4ab. √ p Note that AO2 = 2a + b, AD = AC.AB = 2 a(a + b), we have 2 2 2 2 2 2 O2A0 − DA0 = O2A − DA = (2a + b) − 4a(a + b) = b 2 =⇒(O2A0 + DA0)(O2A0 − DA0) = b . Combining with (8), we obtain b2 b2 (9) O2A0 − DA0 = = √ . O2A0 + DA0 b2 + 4ab Since (8) and (9), it follows 2ab b2 + 2ab (10) DA0 = √ v`a O2A0 = √ . b2 + 4ab b2 + 4ab

Let R1 be the projection from O onto DO2. Applying the Thales’ theorem with the note (10), we have O2R1 = O2O . Since O O = a, O2A0 O2A 2 2 O2O b + 2ab a ab (11) O2R1 = O2A0 = √ · = √ . O2A b2 + 4ab 2a + b b2 + 4ab

Since (10) and (11), it follows DA0 = 2O2R1. Let S1 be the point symmetric to A0 across D, and let T1 be the point symmetric to D across O1. Then

R1S1 = R1D + DS1 = DA0 + (DO2 − O2R1) = DA0 − O2R1 + DO2

= O2R1 + DO2 = O2R1 + O2T1

= R1T1.

It follows that OR1 is the perpendicular bisector of S1T1. Hence OS1 = OT1. Applying the property of rotation about center, we have A0O0 = OS1 = OT1, and BT1 is parallel and equal to DC, so BT1 is perpendicular to OB. Applying the Pythagorean theorem, we have (12) p 2 2 p 2 p 2 2 O0A0 = OT1 = T1B + BO = 4ab + (a + b) = a + b + 6ab. √ Since (7) we have DP1 = DC = 2 ab. Hence √ DH0 2t 2ab ab (13) cos ∠H0DP1 = = = √ = . DP1 DP1 (a + b)2 ab a + b

Applying the Cosine’s law to triangle O0DX1, we have 2 2 2 O0X1 = O0D + X1D − 2O0D.X1D. cos ∠O0DX1 2 2 = O0D + X1D + 2O0D.X1D. cos ∠H0DP1 2 2 = (a + b) + X1D + 2(a + b).X1D. cos ∠H0DP1. Golden Sections and Archimedean Circles in an Arbelos 35

Note that O0X1 = O0A0. Combining with (12) and (13), we deduce that √ ab a2 + b2 + 6ab = (a + b)2 + X D2 + 2(a + b).X D 1 1 a + b √ 2 2 = (a + b) + X1D + 2 ab.X1D. √ 2 It follows X1D + 2 abX1D − 4ab = 0. From this, we get √ √ (14) X1D = (−1 + 5) ab. Since (14), it follows √ √ √ √ √ (15) X1P1 = DP1 − X1D = 2 ab − (−1 + 5) ab = (3 − 5) ab. Since (14) and (15), it follows √ √ √ X D (−1 + 5) ab 1 + 5 1 = √ √ = = ϕ. X1P1 (3 − 5) ab 2

This means that X1 divides DP1 in the golden ratio. On the other hand, applying the Power-of-a-point theorem with the note that O0D = OD = a + b and (12), (14), we have √ √ 2 2 4ab DX1.DY1 = O0A0 − DO0 = 4ab =⇒ Y1D = = (1 + 5) ab. DX1 It follows √ √ Y D (1 + 5) ab Y P ϕ + 1 1 = √ = ϕ and 1 1 = = ϕ. DP1 2 ab Y1D ϕ Hence point Y1 divides segment P1D in the golden ratio. From the conﬁguration of theorem 8, we obtain a pair of Archimedean circles as follows

Theorem 9. The circle O0(A0) meets the semi-circle (O) at U1 and U2. Then the circles that are tangent with P1P2 and their centers are U1 and U2, respectively are Archimedean (see ﬁgure 7).

Proof. Let V1 be the projection from U1 onto OO0. We have

(16) O0V1 + OV1 = OO0 = 2OD = 2(a + b). √ 2 2 Note that U1O0 = O0A0 = a + b + 6ab (since (12)), DH0 = 2t and U1O = a + b. We have 2 2 2 2 2 2 2 V1O0 − V1O = U1O0 − U1O = a + b + 6ab − (a + b) = 4ab.

It follows OO0(O0V1 − OV1) = 4ab. From this, we get 4ab 4ab 2ab (17) O0V1 − OV1 = = = = 2t. OO0 2(a + b) a + b Since (16) and (17), it follows

O0V1 = a + b + t and OV1 = a + b − t. Hence

V1H0 = |OV1 − H0O| = |a + b − t − (a + b)| = t. 36 Nguyen Ngoc Giang and Le Viet An

Figure 7.

This prove that the distance from U1 to P1P2 is equal to t. Similarly, the distance from U2 to P1P2 is also equal to t. It follows that the circles are tangent with P1P2 and their centers are U1 and U2, respectively such that their radii are equal to t, so they are Archimedean. The theorem is proved.

References [1] T. O. Dao, Q. D. Ngo, and P. Yiu, Golden sections in an isosceles triangle and its circumcircle, Global Journal of Advanced Research on Classical and Modern Geometries, 5 (2016), 93—97. [2] C. W. Dodge, T. Schoch, P. Y. Woo and P. Yiu, Those ubiquitous Archimedean circles, Math. Mag., 72 (1999), 202-–213. [3] F. M. van Lamoen, Online catalogue of Archimedean circles, available at http://home.kpn.nl/lamoen/wiskunde/Arbelos/Catalogue.htm.

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