GOLDEN SECTIONS AND ARCHIMEDEAN CIRCLES IN AN ARBELOS

INTERNATIONAL JOURNAL OF GEOMETRYVol. 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 Archimedeancircles in this configuration.

Consider an arbelos formed by semi-circles (O1), (O2), and (O) of radiia, 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

if PQPR = PR

RQ . In this case, the divided ratio is the golden ratio ϕ :=√5+12 ,

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 CGand CH meet the semi-circle (O) at I and J respectively, K and L are theincenter and C-excenter of triangle CIJ , respectively, then K divides boththe segments CD and LC in the golden ratios (see Figure 1).

Proof. Since DC is the altitude of right triangle ABD and the quadrilateralACHE is concyclic,

∠BDC = ∠BAD = ∠CAE = ∠BHC.

It follows that quadrilateral BCHD is concyclic.Similarly, the quadrilateral ACGD is also concyclic.Since AD touches the circles (O1) and (O2), DA.DE = DC2 = DB.DF

by the Power-of-a-point theorem and by the Intersecting-chords theorem,the quadrilateral ABFE is concyclic. Chasing angles, we have

∠GCD = ∠GAD = ∠FAE = ∠FBE = ∠DBH = ∠DCH.

This means that CD is the internal angle bisector of ∠ICJ , and K belongsto 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 ′ is the reflection of J across AB. By symmetry, J ′ belongs to thecircle (O) and AJ = AJ ′.It follows that ∠JIA = ∠AIJ ′ = ∠AIC. Hence, AI is the internal anglebisector 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, andMO is perpendicular to AB.

Since KC touches (O1),

∠MAO = ∠HAC = ∠HCK = ∠GCD = ∠GAD = ∠FAD.

It follows that right triangles MAO and FAD are similar,

MO

AO=FD

AD=⇒MO = AO

DF

DA.

Golden Sections and Archimedean Circles in an Arbelos 27

We have AO = AB2 = a + b, AD =

√AB.AC = 2

√a(a+ b), DC =√

CA.CB = 2√ab, and

DF =DC2

DB=

CA.CB√BA.BC

=4ab

2√b(a+ b)

=2a√b(a+ b)

a+ b.

It follows that

MO = AODF

DA=√ab =

DC

2.

Let N be the orthogonal projection of M onto CD. Then N is themidpoint 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 −MN2 = (a+ b)2 + ab− (a− b)2 = 5ab

=⇒ LC = LN + CN = (√

5 + 1)√ab.

Let K ′ be the reflection of K across AB. By symmetry, CK = CK ′ and

∠AK ′B = ∠AKB = 180◦ − ∠ALB.

It follows that quadrilateral ALBK ′ is concyclic. By the Intersecting-chordstheorem,

CK.CL = CK ′.CL = CA.CB

=⇒ CK =CA.CB

CL=

2a.2b

(√

5 + 1)√ab

= (√

5− 1)√ab.

Hence,

KC

KD=

KC

CD −KC=

√5− 1

3−√

5= ϕ,

and

KL

KC=LC − CK

CK=

2√5− 1

= ϕ.

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 segmentsLK and CL in the golden ratios.

The famous Archimedean twin circles associated in the arbelos have equalradii t := ab

a+b (see [2] and [3]). Circles with radius t are called Archimedeanand they are congruent to the Archimedean twin circles.

Theorem 3. If the perpendicular bisector of CD meets CG and CH at Pand Q respectively, then the circle with diameter PQ is Archimedean (seeFigure 2).


Figure 2.

Proof. Since the right triangles CPN and AFD are similar, CN =√ab,

AD = 2√a(a+ b) and DF =

2a√

b(a+b)

a+b , we deduce that

PN

CN=FD

AD=⇒ PN = CN

FD

AD=

ab

a+ b= t.

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 meetsCG and CH at R and S, respectively, then the circles with diameters DRand DS are Archimedean.

For two points P and Q in the plane, the circle with center P passingthrough Q is denoted by P (Q).

Theorem 5. The circle D(O) meets the perpendicular bisector of AB againat U . The semi-circle (O1) and the segment AU meet at V , (O2) and BUat 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

CB=

AV

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 circlesA(V ) and B(W ).

Let the line BU and (O) meet again at B1, and let the lines CV and CWmeet two common external tangent lines of two circles A(V ) and B(W ) atX,Y, Z, and T as show in the Figure 3.


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, thepoints 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, Y A bisects angleXTY . Chasing angles, we have ∠AB1X = ∠AYX = ∠AY T = ∠B1AY . Itfollows 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 themidpoint of OU , we get

CD = OH1 =OU

2.

Let us denote by α := ∠ACV , then α = ∠BCW = ∠BAB1 = ∠BCZ.Since the right triangles BAB1 and BUO are similar,

AB1

BB1=UO

BO=

2CD

BO=

4CD

AB=⇒ AB1 = 4DC

BB1

BA= 4DC.sinα.


By the symmetry, XT and Y Z are both perpendicular to AB. Let theline CD and circle (O) meet again at D3, AB and XT at M1, AB and Y Zat M2. By the Intersecting-chords theorem and symmetry,

AB21 = 16CD2.sin2α = 16CD.CD3.sinα.sinα

= 16CX.CZM1X

CX

M2Z

CZ= 16M1X.M2Z = 4XT.Y Z.

It follows that

XY 2 = 4XT.Y Z.(1)

Note that D1D2, XT and Y Z are pairwise parallel. By the Thales’ theo-rem,

D1X

D1Y=CX

CZ=XT

Y Z=⇒ D1X

XY=

XT

XT + Y Z,(2)

and similarly,

D1Y

XY=

Y Z

XT + Y Z.(3)

Comparing (1), (2) with (3), we obtain

D1X.D1Y =4XT 2.Y Z2

(XT + Y Z)2.(4)

Again, by the Thales’ theorem,

CD1

XT+CD1

Y Z=XD1

XY+Y D1

Y X=XD1 +D1Y

XY= 1 =⇒ CD1 =

XT.Y Z

XT + Y Z,

and similarly,

CD2 =XT.Y Z

XT + Y Z.

It follows that

D1D2 = CD1 + CD2 =2XT.Y Z

XT + Y Z.(5)

And by the Intersecting-chords theorem and symmetry,

D1X.D1Y = D1D.D1D3 = D1D.D2D.(6)

From (4), (5) and (6), we deduce that D1D22 = 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 onthe semi-circle (O) in that order. The line passing through P1 perpendicularto CP1 meets (O) at P1 and Q1. Let C1 be the circumcenter of triangleCDQ1. 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 F1Cin the golden ratios.

Proof. (see Figure 4). Segment DA meets the semi-circle (O1) at E, andsegment DB meets the semi-circle (O2) at F ; let M0 be the mid-point ofCD.

We easily see that CEDF is a rectangular. Hence M0 is the mid-pointof EF . Furthermore ∠CEF = ∠CDB = ∠CAE. It follows that EF is


Figure 4.

tangent with (O1). Similarly, EF is also tangent with (O2). Hence EF isthe external common tangent line of two semi-circles (O1) and (O2). HenceEF and PQ are coincident.

Since two semi-circles (O) and (O1) are inner tangent at A, the homothety

Ha

a+b

A center A, ratio aa+b transforms (O) into (O1). Under the homothety

Ha

a+b

A , 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 thatOD is perpendicular to P1P2. This thing proves that OD is the perpendic-ular bisector of P1P2. It follows DP1 = DP2.

We have ∠DP1P2 = ∠DP2P1 = ∠DAP1. This thing proves that DP1 istangent with the circumcircle of triangle AEP1. Applying the Power-of-a-point theorem, we have DP 2

1 = DA.DE = DC2. It follows DP1 = DC.Since DP1 = DP2, DC = DP1 = DP2, it follows that

D is the circumcenter of triangle CP1P2.(7)

Let C ′ be the point symmetric to C across D. Then DC = DP1 = DC ′.It follows ∠CP1C

′ = 90◦. Hence three points C ′, P1 and Q1 are collinear.Applying the Power-of-a-point theorem with note that CC ′ = 2CD = 4

√ab


and CM0 = 12CD =

√ab, and C ′M0 = C ′C − CM0 = 3

√ab, we have

C ′P1.C′Q1 = C ′O2 −DO2 = C ′C2 −DC2 = 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 CDat 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 andCF1 = DE1.

On the other hand, since triangle CDQ1 is isosceles at Q1 and D isthe 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

CD2 = P2D2 = P2C

21 − C1D

2 = E1C21 −

(C1M

20 +M0D

2)

=(E1C

21 − C1M

20

)−M0D

2

= E1M20 −M0D

2

= (E1M0 −M0D) . (E1M0 +M0D)

= E1D. (E1M0 +M0C)

= E1D.E1C

= E1D. (E1D + CD) .

It follows CD2 = E1D2 +E1D.CD. The above equality proves that CD

E1D=

√5+12 = ϕ, is golden ratio.

Furthermore, since CF1 = DE1, it follows DCF1C

= ϕ and DF1DC = ϕ+1

ϕ = ϕ.

This means that point D divides F1C in the golden ratio. �

Since the figuration of theorem 6, we obtain a following result on theArchimedean circle.

Figure 5.


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 diameterK1L1 is Archimedean (see Figure 5).

Proof. Let H0 be the point of intersection of DO and P1P2. Since (7), wehave OD perpendicular to P1P2 at H0, quadrilateral CM0P1Q1 is concyclic.

Hence ∠DK1P1 = 180◦−∠DCQ1 = ∠M0P1Q1. It follows that DK1 andM0P1 are parallel. This means that K1L1 = DH0.

On the other hand, the circle with diameter DH0 is Archimedean (it isthe 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. CircleO0(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 Y1divides 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 parallelto BD, from BD is perpendicular to AD, O2M0 is perpendicular to AD.


It follows that M0 is the orthocenter of triangle ADO2. Hence AA0 isperpendicular to DO2 at A0.

Since CD = 2√ab and CO2 = b,

O2A0 +DA0 = DO2 =√O2C2 + CD2 =

√b2 + 4ab.(8)

Note that AO2 = 2a+ b, AD =√AC.AB = 2

√a(a+ b), we have

O2A20 −DA2

0 = O2A2 −DA2 = (2a+ b)2 − 4a(a+ b) = b2

=⇒(O2A0 +DA0)(O2A0 −DA0) = b2.

Combining with (8), we obtain

O2A0 −DA0 =b2

O2A0 +DA0=

b2√b2 + 4ab

.(9)

Since (8) and (9), it follows

DA0 =2ab√b2 + 4ab

va O2A0 =b2 + 2ab√b2 + 4ab

.(10)

Let R1 be the projection from O onto DO2. Applying the Thales’ theoremwith the note (10), we have O2R1

O2A0= O2O

O2A. Since O2O = a,

O2R1 = O2A0O2O

O2A=

b2 + 2ab√b2 + 4ab

· a

2a+ b=

ab√b2 + 4ab

.(11)

Since (10) and (11), it follows DA0 = 2O2R1. Let S1 be the point sym-metric 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

O0A0 = OT1 =√T1B2 +BO2 =

√4ab+ (a+ b)2 =

√a2 + b2 + 6ab.

(12)

Since (7) we have DP1 = DC = 2√ab. Hence

cos∠H0DP1 =DH0

DP1=

2t

DP1=

2ab

(a+ b)2√ab

=

√ab

a+ b.(13)

Applying the Cosine’s law to triangle O0DX1, we have

O0X21 = O0D

2 +X1D2 − 2O0D.X1D. cos∠O0DX1

= O0D2 +X1D

2 + 2O0D.X1D. cos∠H0DP1

= (a+ b)2 +X1D2 + 2(a+ b).X1D. cos∠H0DP1.


Note that O0X1 = O0A0. Combining with (12) and (13), we deduce that

a2 + b2 + 6ab = (a+ b)2 +X1D2 + 2(a+ b).X1D

√ab

a+ b

= (a+ b)2 +X1D2 + 2

√ab.X1D.

It follows X1D2 + 2

√abX1D − 4ab = 0. From this, we get

X1D = (−1 +√

5)√ab.(14)

Since (14), it follows

X1P1 = DP1 −X1D = 2√ab− (−1 +

√5)√ab = (3−

√5)√ab.(15)


X1D

X1P1=

(−1 +√

5)√ab

(3−√

5)√ab

=1 +√

5

2= ϕ.

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

DX1.DY1 = O0A20 −DO2

0 = 4ab =⇒ Y1D =4ab

DX1= (1 +

√5)√ab.

It follows

Y1D

DP1=

(1 +√

5)√ab

2√ab

= ϕ andY1P1

Y1D=ϕ+ 1

ϕ= ϕ.

Hence point Y1 divides segment P1D in the golden ratio. �

From the configuration of theorem 8, we obtain a pair of Archimedeancircles 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 andU2, respectively are Archimedean (see figure 7).

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

O0V1 +OV1 = OO0 = 2OD = 2(a+ b).(16)

Note that U1O0 = O0A0 =√a2 + b2 + 6ab (since (12)), DH0 = 2t and

U1O = a+ b. We have

V1O20 − V1O2 = U1O

20 − U1O

2 = a2 + b2 + 6ab− (a+ b)2 = 4ab.

It follows OO0(O0V1 −OV1) = 4ab. From this, we get

O0V1 −OV1 =4ab

OO0=

4ab

2(a+ b)=

2ab

a+ b= 2t.(17)


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

V1H0 = |OV1 −H0O| = |a+ b− t− (a+ b)| = t.


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 andU2, 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 cir-cumcircle, 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 athttp://home.kpn.nl/lamoen/wiskunde/Arbelos/Catalogue.htm.

BANKING UNIVERSITY OF HO CHI MINH CITY36 TON THAT DAM, district 1, HO CHI MINH CITY, VIETNAM.E-mail address: [email protected]

PHU VANG, THUA THIEN HUE, VIETNAM.E-mail address: [email protected]

GOLDEN SECTIONS AND ARCHIMEDEAN CIRCLES IN AN ARBELOS

Documents

Transcript of GOLDEN SECTIONS AND ARCHIMEDEAN CIRCLES IN AN ARBELOS