The Arbelos with Overhang

KoG•18–2014 H. Okumura: The Arbelos with Overhang

Original scientiﬁc paper HIROSHI OKUMURA Accepted 20. 10. 2014. The Arbelos with Overhang

The Arbelos with Overhang Arbelos s privjeskom ABSTRACT SAZETAKˇ We consider a generalized arbelos consisting of three semi- Promatra se poopćeni arbelos koji se sastoji od tri circles with collinear centers, in which only two of the three polukruˇznice s kolinearnim srediˇstima, pri ˇcemu se dvije semicircles touch. Many Archimedean circles of the ordi- on njih dodiruju. Mnoge Arhimedove kruˇznice obićnog nary arbelos are generalized to our generalized arbelos. arbelosa su poopćene za poopćeni arbelos. Key words: arbelos, arbelos with overhang, Archimedean circles Kljuˇcne rijeˇci: arbelos, arbelos s privjeskom, Arhimedove MSC2010: 51M04, 51N20 kruˇznice

1 Introduction stated. Let A′ (resp. B′) be a point on the half line with endpoint O passing through A (resp. B), and let A′O = 2a′ The arbelos is a plane figure consisting of three mutually (resp. B O = 2b ) (see Figure 1). Let γ = (AB| ), and| let | ′ | ′ touching semicircles with collinear centers. It has three δ′α be the circle touching the semicircle (A′O) externally points of tangency. In [5], [7] and [9], we have considered γ internally and the perpendicular to AB passing through a generalized arbelos called a collinear arbelos consisting O from the side opposite to the point B. The circle δβ′ is of three circles with collinear centers, in which one of the defined similarly. circles touches the remaining two circles, but the two circles do not touch in general. Thereby the collinear arbelos γ has two points of tangency. δ β δα In this paper, we consider the remaining case. We consider a configuration consisting of three semicircles with collinear centers, in which only two semicircles touch, i.e., it has only one point of tangency. Many Archimedean circles of the ordinary arbelos are generalized to our generalized arbelos, but also several new Archimedean circles of B B OA A the ordinary arbelos are induced by this. Z79+.; Figure 1 2 An arbelos with overhang Proposition 1 The two circles δ′α and δβ′ are congruent if Let O be a point on the segment AB with AO = 2a and and only if a a = b b. ′ − ′ − BO = 2b. We use a rectangular coordinate| system| with | | origin O such that the coordinates of the points A and B are Proof. Let r be the radius of δ′α. The center of the circle (2a,0) and ( 2b,0), respectively. For two points P and Q, with a diameter A O or AB, the center of δα and the foot of ′ ′ (PQ) and P(−Q) denote the circle with diameter PQ and the perpendicular from this point to AB form a right triangle. circle with center P passing through Q, respectively. How- Hence by the Pythagorean theorem, we get ever if their centers lie on the line AB, we consider them 2 2 2 2 as semicircles lying in the region y 0 unless otherwise (r + a′) (r a′) = ((a + b) r) (r (a b)) . ≥ − − − − − −

19 KoG•18–2014 H. Okumura: The Arbelos with Overhang

Notice that if the foot of perpendicular coincides with the Proof. Let r and (0,c) be the radius and the coordinates of center of the circle with a diameter A′O or AB, then one the center of the touching circle. Then we get of the triangles degenerates to a segment. But the equation (a b)2 + c2 = (a + b r)2. (1) still holds. Solving the equation we get r = ab/(a′ + b). − − δ Similarly, β′ has radius ab/(a+b′). Therefore the two cir- Also by similar triangles, we get cles are congruent if and only if a′ + b = a + b′. r ab Let α = (AO), β = (BO), and let a = a + h, b = b + h = cos∠VOA = . (2) ′ ′ c (a + h)(b + h) with min(a,b) < h. We relabel a′, b′, A′, B′, δ′α and δβ′ as s − α β ah, bh, Ah, Bh, δ and δ , respectively and let αh = (AhO) Eliminating c from (1) and (2), and solving the resulting h h h and βh = (BhO). The conﬁguration consisting of the three equation for r with h > 0, we get r = 2rA. semicircles α , β and γ is denoted by α β γ . We call h h ( h, h, ) Let α f = (A f O) and β f = (B f O). Archimedean circles of α β γ an arbelos with overhang h, and α β γ is ( h, h, ) ( h, h, ) the ordinary arbelos (α f ,β,(A f B)) have radius said to have overhang h. The ordinary arbelos (α,β,γ) has overhang 0. The perpendicular to AB passing through O is (ab/bh)b ab h = = rA. called the axis, which overlaps with the y-axis. ab/bh + b a + bh Similarly Archimedean circles of the ordinary arbelos γ β (α,β f ,(AB f )) have the same radius. Hence we get: δh α W δh V Proposition 3 The ordinary arbeloi (α f ,β,(A f B)) and βh αh (α,β f ,(AB f )) share Archimedean circles with (αh,βh,γ). β α The circle touching the axis at the point O from the side opposite to the point B and also touching the tangent of β

Bh B Bf OAAf Ah from the point A is an Archimedean circle of the ordinary arbelos α β γ , which is denoted by W in [4]. Hence by Figure 2 ( , , ) 6 Proposition 3, we get the following proposition. By this α β δ δ h proposition we can construct the point A f (also B f ) even in Now the circles h and h have the same radius rA = ab/(a + b + h) by Proposition 1. The two circles are a the case h < 0 (see Figures 7 and 14). generalization of the twin circles of Archimedes of the Proposition 4 The point A coincides with the point of in- α β γ h f ordinary arbelos ( , , ). Circles of radius rA are said tersection of the line AB and the external common tangent to be Archimedean circles of (αh,βh,γ) or Archimedean of β and the Archimedean circle of (αh,βh,γ) touching the with respect to (αh,βh,γ). Also we say that (αh,βh,γ) has h axis at the point O from the side opposite to the point B. Archimedean circles of radius rA. The common radius of Archimedean circles of (α,β,γ) is denoted by rA, i.e., Since AB f : ABh = a : ah holds, we get the following | | | | rA = ab/(a + b). proposition, which also enable us to construct the points We deﬁne A f and B f as the points with coordinates A f and B f in the case h < 0. (2ab/bh,0) and ( 2ab/ah,0), respectively. Let γ have − Proposition 5 The pointB f divides the segment ABh in the points V and W in common with the semicircles αh and β ratio a : h internally or externally, according as h > 0 or h respectively in the case h 0 (see Figure 2). The points | | V and W have coordinates ≥ h < 0.

(2ab/bh, f (a,b)/bh) and ( 2ab/ah, f (a,b)/ah), − 3 Several twin circles respectively, where f a b 2 abh a b h . There- ( , )= ( + + ) In this section we show that several twin circles exist for fore the points A f and B f are the feet of perpendiculars α p (αh,βh,γ), if h > 0. Let us assume h > 0, and let ε be the from V and W to the line AB, respectively. By the coordi- 1 circle touching the semicircles α externally α internally nates of V and W, we get tan∠WOB = tan∠VOA. There- h ∠ ∠ and the segment A f V from the side opposite to the point A fore WOB = VOA holds. εα γ (see Figure 3). Let 2 be the circle touching the semicircles The circle touching internally and the segment AB at the α α γ εα externally h and internally. Also let 3 be the circle point O has radius 2rA [11]. The fact is generalized as fol- touching αh and γ externally and the axis from the side op- lows. β β β ε ε ε posite to the point B. The circles 1, 2 and 3 are deﬁned Proposition 2 If h > 0, the radius of the circle touching γ similarly. The following proposition has a straightforward h internally and the segments OV and OW is 2rA. proof that is omitted.

20 KoG•18–2014 H. Okumura: The Arbelos with Overhang

Proposition 6 If h > 0, the following statements hold. and label the images with an overline (see Figure 5). The α β (i) The circles ε and ε have the same radius x-coordinates of the points A, B, A f and B f are 2b, 2a, 1 1 − 2bh and 2ah, respectively, and the circle γ f = (A f B f ) has 1 1 − 1 1 h 1 − 1 1 − center (b a,0) and radius ah + bh. Let (xδ,yδ) and rδ be + + + = + . − h the coordinates of the center of the circle δ and its radius. a b ab h rA h The circle δ touches α and β, which are the perpendicu- εα εβ lars to AB passing through the points A and B, respectively. (ii) The circles 2 and 2 have the same radius Therefore we get xδ = b a and rδ = a+b. Since δ touches 1 1 − 1 1 1 − 1 1 − γ f externally and the x-coordinates of their centers are the + + = + . a b h r h same, we get A β εα ε yδ = ah + bh + rδ = 2(a + b + h). (iii) The circles 3 and 3 have the same radius ab/h. ε α The proposition also shows that the sum of the curvatures Since is the line perpendicular to the line and passes α δ of the circles εα and εα equals the curvature of the circle through the point of tangency of and , it is parallel to AB 2 3 δ εα. and passes through the center of . Hence the distance be- 1 ε h tween AB and the farthest point on equals 4ab/yδ = 2rA. Therefore ε is Archimedean with respect to (αh,βh,γ). The part (ii) is proved similarly. βh εβ 3 α γ ε3 W αh γ f β ε2 εβ V βh 1 αh β εα εα 2 β βf 1 αf α

α Bh B Bf OAAf Ah Figure 4 B Bf O Af A Figure 3

ε 4 Bankoff circles The circle orthogonal to the semicircles α, β and to the circle touching α and β externally and γ internally is an δ Archimedean circle of (α,β,γ) called the Bankoff triplet circle, which is denoted by W3 in [4]. The maximal circle touching the external common tangent of α and β and the arc of γ cut by the tangent internally is an Archimedean cir- β α γf cle of (α,β,γ) called the Bankoff quadruplet circle, which is denoted by W4 in [4]. In this section we generalize the two circles (see Figures 4 and 6). Let γ f = (A f B f ). β γf Theorem 1 The following two circles are Archimedean δ with respect to (αh,βh,γ), and coincide. (i) The circle orthogonal to the semicircles α, β and to the α ε circle touching α and β externally and γ f internally. B Bf Af (ii) The circle orthogonal to the semicircles α , β and to f f Bf B O A A Af the circle touching α and β externally and γ internally. f f Figure 5 Proof. Let δ be the circle touching α and β externally and γ f internally, and let ε be the circle denoted by (i). We in- We call the circle in Theorem 1 the Bankoff triplet circle vert the ﬁgure in the circle with center O and radius 2√ab, of (αh,βh,γ).

21 KoG•18–2014 H. Okumura: The Arbelos with Overhang

Theorem 2 If E is the external common tangent of the Proof. Since Iα has coordinates 0,2√ahb , the point di- semicircles α and βh, or αh and β, then the maximal circle viding the segment A f Iα in (i) has coordinates touching E and the arc of γ cut by E (the part of γ between the two points of intersection of γ and E) internally bh 2ab/bh a 2√ahb h h ah · , · = 2r ,2r . is Archimedean with respect to (α ,β ,γ). a + b a + b A A b h h h h r Proof. We prove the case E being the common tangent of This proves (i). The point of intersection of γ and Am(O) α and β . The other case is proved similarly. Let d be the h has coordinates 2rh ,2 (a rh )(b + rh ) . This proves distance between E and the centers of γ, and let T be the A − A A point of intersection of the lines E and AB. If T lies in the (ii). q region x > 0, let AT = t. By similar triangles, we get | | γ

a/(t + a)= d/(t + a + b)= bh/(t + 2a + bh). Bm (O) Am (O) Eliminating t and solving the resulting equations for d, we Iα β α get d = a + b 2rh . Therefore δ is an Archimedean cir- − A cle of (αh,βh,γ). The case T lying in the region x < 0 is proved similarly. If E and AB are parallel, then a = Bm Am h b + h = d and rA = ab/(2a)= b/2. Therefore we also get Bf B Bh O Ah A Af d = a + b 2rh . − A Figure 7

γ For a circle or a semicircle δ, its center is denoted by Oδ. The farthest point on δ from AB lying in the region y 0 ≥ is denoted by Tδ. If the segments TαTβ and TγOγ intersect βh at a point P, the circle (PTγ) is an Archimedean circle of β α αh (α,β,γ), which is denoted by W20 in [4]. The fact is generalized (see Figure 8).

α α γ γ Bh B OA Ah Theorem 4 The segments T h Tβ, T Tβh and T O intersect at a point P, which divides Tα Tβ and TαTβ in the ratios Figure 6 h h bh : aandb : ah internally, respectively. The circle (PTγ) is Archimedean with respect to (αh,βh,γ). 5 Miscellaneous Archimedean circles

Proof. The points dividing Tαh Tβ in the ratio bh : a inter- In this section we consider miscellaneous Archimedean nally and TαTβ in the ratio b : a internally have the same α β γ h h circles of ( h, h, ) obtained from points dividing given h coordinates (a b,a + b 2rA). segments in the ratio a : bh or ah : b internally, some of − − which seem to be new even for the ordinary arbelos. Let Tγ I be the point of intersection of the axis and the semicir- T cle γ. The minimal circle touching the axis and passing βh through the point of intersection of the semicircle α and P T T the segment AI is an Archimedean circle of (α,β,γ), which β αh T is denoted by W9 in [4]. Also the minimal circle touching α the axis and passing through the point of intersection of the semicircles γ and A(O) is an Archimedean circle of (α,β,γ), which is denoted by W13 in [4]. The two facts Bh B Oγ O A Ah are generalized. Let Am and Bm be the midpoints of the Figure 8 segments AAh and BBh, respectively (see Figure 7). In the theorem, the endpoints of the diameter of (PTγ) par- Theorem 3 (i) If Iα is the point of intersection of the axis allel to AB divide the segments TγTα and TγTβ in the ratios and the semicircle (AhB), then the point dividing the seg- a : bh and b : ah internally, respectively. ment A f Iα in the ratio a : bh internally lies on the semicircle α and its distance from the axis is 2rh . Theorem 5 Let T and T be the reﬂected images of the A α′h β′ (ii) The distance between the axis and the point of inter- points Tαh and Tβ in the line AB, respectively. The follow- γ h section of the semicircles Am(O) and is 2rA. ing statements hold.

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(i) Let C be the internal center of similitude of the semi- Let α(z) and β(z) be the semicircles constructed in the re- circles γ and αh. If D is the point of intersection of the gion y 0 touching the axis at the point O and having the ≥ lines CTβ′ and ATα, then D divides ATα in the ratio b : ah centers with coordinates (za,0) and ( zb,0), respectively internally and the circle touching α or AB at the point for a real number z1. Let C(m,n) be the− circle touching the A and passing through D is Archimedean with respect to semicircles γ internally and α(m) and β(n) at points differ- (αh,βh,γ). ent from O such that the points of tangency on α(m), γ and α β (ii) The segments A f Tβ, ATβ f and T O intersect at a point (n) lie counterclockwise in this order for real numbers m C H, which divides A f Tβ in the ratio a : bh internally ATβ f and n. The radius of (m,n) is expressed as follows [12, and TαO in the ratio ah : b internally, respectively. The cir- Theorem 1]: cle touching α or AB at the point O and passing through H ab(ma + nb) is Archimedean with respect to (αh,βh,γ). 2 2 . (3) (iii) If E is the point of intersection of the segments T Tγ ma + nb + mnab α′h and OTα, then E divides T Tγ and OTα in the ratio a : b α β α′h h Let h(z) and h(z) be the semicircles constructed in the internally and the circle touching α or the line OαTα at region y 0 touching the axis at the point O and having ≥ the point Tα and passing through E is Archimedean with the centers with coordinates (zah,0) and ( zbh,0), respec- α β γ − respect to ( h, h, ). tively for a real number z. Let Ch(m,n) be the circle touching the semicircles γ internally and αh(m) and βh(n) at Proof. The point C has coordinates (2aah/(2a + bh),0) points different from O such that the points of tangency β (see Figure 9). If we regard as a circle, then C coin- on α (m), γ and β (n) lie counterclockwise in this order β h h cides with the internal center of similitude of and the for real numbers m and n. Archimedean circle of (αh,βh,γ) touching α at the point A internally. Hence D has coordinates (2a rh ,rh ). This Theorem 6 The circle C (m,n) has radius − A A h proves (i). The points dividing A f Tβ in the ratio a : bh in- ab ma nb β α ( h + h) ternally, AT f and T O in the ratio ah : b internally have . (4) h h maah + nbbh + mnahbh the same coordinates (rA,rA). This proves (ii). The points dividing OTα and T Tγ in the ratio a : b internally have α α β α′h h Proof. Notice that h(m) = (mah/a) and h(n) = h h β the same coordinates (a rA,a rA). This proves (iii). (nbh/b). Replacing m and n by mah/a and nbh/b respec- − − tively in (3), we get (4). Tγ Theorem 7 The circle Ch(m,n) is Archimedean with re- γ spect to (αh,βh,γ) if and only if Tα αh 1 1 α + = 1. (5) β Tβ h β E m n

Tβf H D Proof. The theorem follows from Theorem 6, because Bh Ah B Bf O C Oα Af A ab(mah + nbh) h rA maah + nbbh + mnahbh − h T (m + n mn)ahbhr β = − A . maah + nbbh + mnahbh Corollary 1 The following circles are Archimedean with T αh respect to (αh,βh,γ). Figure 9 (i) The circle touching the semicircles Ah(O) and Bh(O) externally and γ internally. The circle touching AB at O and passing through H in (ii) (ii) The circle touching γ internally and the two distinct is the Bankoff triplet circle of (αh,βh,γ). circles of radius ah + bh touching the axis at the point O externally. 6 Archimedean circles touching γ Proof. The part (i) follows from the fact Ah(O)= αh(2) In [12], we gave necessary and sufﬁcient conditions that a and Bh(O)= βh(2) (see Figure 10). The part (ii) follows circle touching the semicircle γ internally is Archimedean from the fact that m = (ah + bh)/ah and n = (ah + bh)/bh with respect to (α,β,γ). In this section we generalize this. satisfy (5). 1The notations are slightly changed from [12]

23 KoG•18–2014 H. Okumura: The Arbelos with Overhang

The circle described in (i) is a generalization of Schoch 7 Woo's Archimedean circles circle of the ordinary arbelos which is denoted by W15 in [4]. The circle described in (ii) is a generalization of the We assume a = b in this section. Peter Woo found that the 6 α β Archimedean circle of the ordinary arbelos in [10]. circle touching (z) and (z) externally with center on the line x = (b a)r /(b + a) is Archimedean with respect to Let γ = (A B ). If m > 0 (resp. m < 0), let Pα (m) be the A h h h h α β γ − external (resp. internal) center of similitude of the semicir- ( , , ) for a non-negative real number z [4]. The line is α γ called the Schoch line of (α,β,γ). The fact was generalized cles h(m) and h. Similarly the point Pβh (m) is deﬁned. for a real number z rA/(a + b) in [12]. We generalize this. ≥− Theorem 8 The points Pα (m) and Pβ (n) coincide if and h h A circle is said to touch α (z) appropriately if it touches only if (5) holds. h αh(z) externally in the case z > 0 and it touches the re- ﬂected image of α (z) in the line AB internally in the case Proof. The semicircles α (m) and γ have radii ma and h h h h z < 0. The same notion of appropriate tangency applies to ah + bh and centers with x-coordinates mah and ah bh, β h − h(z). Let sh = (bh ah)rA/(bh + ah). We call the line respectively. Hence Pαh (m) has x-coordinate − x = sh the Schoch line of (αh,βh,γ).

(ah + bh)mah mah(ah bh) 2mahbh − − = . Schoch line of (αh , βh , γ ) mah + (ah + bh) ah + bh mah β − − h

Similarly Pβh (n) has x-coordinate αh

(ah + bh)( nbh) nbh(ah bh) 2nahbh − − − = − . nbh + (ah + bh) ah + bh nbh − − While ( ) sh , l δ 2mahbh 2nahbh − ah + bh mah − ah + bh nbh βh(z) − − αh (z) 2(m + n mn)(ah + bh)ahbh O = − . (ah + bh mah)(ah + bh nbh) − − Therefore the proof is complete. Figure 11: z < 0 Corollary 2 The circle C (m,n) is Archimedean with re- h Theorem 9 Let δ be the circle touching αh(z) and βh(z) α β γ α spect to ( h, h, ) if and only if the points P h (m) and appropriately and having its center on the Schoch line of Pβ (n) coincide. h (αh,βh,γ) for a real number z = 0. The following statements hold. 6 (i) The circle δ is Archimedean with respect to (αh,βh,γ). 2 (ii) The circle δ exists if and only if ahbh/(ah + bh) z 0 or 0 z. − ≤ Bh (O) = βh(2) < < αh (2) = Ah(O) Proof. If r is the radius of δ and l is the y-coordinate of its γh γ center (see Figure 11), then we get 2 2 2 2 2 βh (zah +r) (sh zah) = (zbh +r) (sh +zbh) = l . (6) αh − − − Solving the equation for r, we get r = sh(bh + ah)/(bh h − ah)= rA. This proves (i). From (6) we also get B B OA A h h 2 4ahbhsh(sh + (bh ah)z) l = 2− Figure 10: m = n = 2 (ah bh) 2 2 − 2 4ahbh(ahbh + (ah + bh) z) If the external common tangents from γh to both αh(m) and = 4 . (ah + bh) βh(n) exist, then the circle Ch(m,n) is Archimedean with respect to (αh,βh,γ) if and only if the two tangents coin- Therefore l satisfying (6) is real if and only if ahbh/(ah + 2 − cide (see Figure 10). bh) z < 0or0 < z. This proves (ii). ≤

24 KoG•18–2014 H. Okumura: The Arbelos with Overhang

Notice that the circle δ in Theorem 9 is not uniquely de- overhang tah a. The rest of (ii) follows from (i). If t = − σ σ α termined if a = b. We get an inﬁnite set of Archimedean a/ah, the points A and Ah coincide, i.e., (Ah O)= . While circles of (αh,βh,γ), whose centers lie on the line x = a/ah > a/(ah +b)= a/(a+bh) holds. Thereforewe get an α τ σ sh by the theorem. However the Archimedean circle of ordinary arbelos ( ,(BhO),(AB )), whose Archimedean α β γ 2 h α β ( h, h, ) with center (sh,2ahbh√ahbh/(ah + bh) ) is not circles have radius rA by (ii). While ( , f ,(AB f )) is a member of this set. In fact, there are inﬁnitely many also an ordinary arbelos having Archimedean circles of the circles passing through O with center on this line. But it same radius by Proposition 3. Therefore the two ordinary seems to be natural to consider this circle as a member of arbeloi coincide. The rest of (iii) is obvious. this set. 9 New type of Archimedean circles 8 Dilation Quang Tuan Bui has found a pair of new type of In [8], we have shown that if σ is a dilation with center O, Archimedean circles such that the endpoints of the diam- σ σ the circle touching the semicircles (A O) externally (AB ) eter parallel to the line AB lie on a given circle [1], which internally and the axis from the side opposite to the point B has been rediscovered by us [6]. One of the circles is ob- α β γ σ σ is an Archimedean circle of ( , , ), where A and B are tained as follows: If the line TαOα intersects the semicircle the images of A and B by σ, respectively. In this section γ at a point S and the lines SA and SO intersect the semi- we generalize this fact. circle α at points T and U respectively, the circle (TU) is Archimedean with respect to (α,β,γ). The fact is general- h ized (see Figures 13 and 14). Notice that h + rA > 0 and γ a rh > 0. − A

βh αh S γ (Ah B)

σ τ σ Tα Bh BB Bh OAAh Ah β h α Figure 12: t = 2/3 U T Theorem 10 Let σ and τ be the dilations with center O and A with the same ratio of magniﬁcation t, respectively. B B B O OAO A A Then the following statements are true. h f β h α h f h σ (i) The circle touching the semicircles (Ah O) externally Figure 13 (ABσ) internally and the axis from the side opposite to the point B is Archimedean with respect to (αh,βh,γ). σ τ σ Theorem 11 (i) Let S be the point of intersection of the (ii) If t > a/(a + bh), then ((A O),(B O),(AB )) is an ar- h h semicircle (AhB) and the the line Tαh Oαh . IfT is thepoint h h belos with overhang tah a, and has Archimedean circles dividing the segment SA f in the ratio (h + r ) : (a r ) h − A − A of radius rA. σ τ σ internally and U is the point of intersection of the line SO (iii) If t = a/ah, then ((Ah O),(BhO),(AB )) coincides with and the semicircle α, then T lies on α and the line TU is α β τ σ ( , f ,(AB f )), andthepointsBh andB also coincide with parallel to AB and the circle (TU) is Archimedean with the point B f . respect to (αh,βh,γ). (ii) Let S be the point of intersection of the semicircle (AB ) Proof. Let r be the radius of the touching circle in (i) (see h and the line TαOα. If the line SO intersects the semicircle Figure 12). Then we get α f at a pointU and the line parallel to AB passing through 2 2 2 2 U intersects α f at a point T again, then the circle (TU) is (tah + r) (tah r) = (bt + a r) (( bt + a) r) . − − − − − − Archimedean with respect to (αh,βh,γ). h Solving the equation, we get r = rA. This proves(i). If t > τ a/(a + bh), then AB = 2t(a + bh) > 2a = AO . Hence Proof. The point S in (i) has coordinates (ah,g(a,b)), τ | h| | | the point Bh lies on the half line with endpoint O passing where g(a,b) = ah(ah + 2b). Hence the points T σ τ h h through B. While Ah O AO = 2(tah a) and BhO and U have coordinates (a + rA,g(a,b)rA/b) and (a σ | | − | | − | |− h h p − B O = 2t(a + bh) 2(tb a)= 2(tah a). Hence the rA,g(a,b)rA/b), respectively. This proves (i). The point | | σ − τ − σ − h conﬁguration ((A O),(B O),(AB )) is an arbelos with U in (ii) has x-coordinate ab/bh r . This proves (ii). h h − A

25 KoG•18–2014 H. Okumura: The Arbelos with Overhang

α Proof. Let K be the point in (i) lying on h and Tα f Oα f . 2 2 2 2 γ Then KOα = a (ah ab/bh) . Therefore KOγ | f | h − − | | − 2 γ α 2 α 2 2 S (a+b) = O O f + KO f (a+b) = (ab/bh (a 2 2 | | | 2 | − 2 − − b)) + ah (ah ab/bh) (a + b) = 2ab. Therefore U T K generates− Archimedean− − circles of (α−,β,γ) with γ by Lemma 1. The part (ii) follows from the fact that the O α powers of the point O with respect to (AOβ), (AhOβ f ) and Bf B Bh Oβ O Ah A Af (BhOα f ) are the same. Figure 14 B(I ) Notice that (h + rh ) : (a rh )= b : a if h = 0. A − A K α Ah (O) I 10 Power type Archimedean circles

From now on we consider all the semicircles with centers γ on the line AB as circles. If two congruent circles of radius K r touching at a point D also touch a given circle δ at points different from D, we say that D generates circles of radius αh r with δ, and the two circles are said to be generated by D with δ. If the two generated circles are Archimedean with respect to (αh,βh,γ), wesaythat D generates Archimedean B O F A Ah circles of (αh,βh,γ) with δ. Frank Power has found that the point Tα generates Figure 16 Archimedean circles of (α,β,γ) with the circle γ [13]. Recall that I is the point of intersection of the axis and the Quang Tuan Bui has found that the circles (AOβ), (BOα) circle γ lying in the region y > 0. Quang Tuan Bui has also and the axis belong to the same intersecting pencil of cir- found that the points of intersection of the circles (AO) and cles and the points of intersection generate Archimedean B(I) generate Archimedean circles of (α,β,γ) with the cir- circles of (α,β,γ) with γ [3]. We generalize the two facts. cle γ [2]. Let J be the point of intersection of the circle The following lemma is needed [7]. B(I) and the line AB lying in the region x > 0. If a < b, Lemma 1 For a circle δ of radius r, a point D generates we can choose h so that 4ah < OJ holds. Then the circles 2 2 | | circles of radius DOδ r /(2r) with δ. Ah(O) and B(I) have no points in common. Let Kα be the || | − | perpendicular to the line AB from the center of the circle The parts (i) and (ii) of the next theorem are generaliza- δα. Quang Tuan Bui’s result is generalized as follows. tions of Power’s result and Bui’s result, respectively (see h Figure 15). Theorem 13 (i) The circles Ah(O), B(I) and the line Kα belong to the same pencil of circles. If the pencil is inter- γ secting, the points of intersection generate Archimedean α β γ γ α (Bh Oαf ) circles of ( h, h, ) with each of the circles and h. W (A O ) K V h βf (ii) The circles Am(O),Bm(I) and the line Kα belong to the same intersecting pencil of circles, and the points of intersection generate Archimedean circles of (α,β,γ) with the circle γ.

Oα Proof. The circles Ah(O) and B(I) are expressed by the Bh B Bf Oβ Oβ OAOαf Af Ah f equations Figure 15 (x 2a )2 + y2 = 4a2 (7) α α α h h Theorem 12 (i) If the circle h and the line T f O f have a − point in common, the point generates Archimedean circles and of (α,β,γ) with γ. α α 2 2 2 (ii) The circles (AOβ), (BO ), (AhOβ f ), (BhO f ) and the (x + 2b) + y = 4b + 4ab, (8) axis belong to the same intersecting pencil of circles, and the points of intersection generate Archimedean circles of respectively (see Figure 16). Subtracting (8) from (7) and α β γ γ h ( , , ) with . rearranging, we get x = rA. Therefore Ah(O), B(I) and

26 KoG•18–2014 H. Okumura: The Arbelos with Overhang

Kα belong to the same pencil of circles. Let us assume [4] C. W. DODGE, T. SCHOCH, P. Y. WOO, P. YIU, that the pencil is intersecting and K is one of the points Those ubiquitous Archimedean circles, Math. Mag. of intersection. Let F be the foot of the perpendicular 72 (1999), 202–213. 2 2 h 2 from K to AB. Then KF = 4ah (rA 2ah) . Then 2 2 h| | 2 − −2 2 [5] H.OKUMURA, Lamoenian circles of the collinear ar- KOγ (a + b) = (rA (a b)) + KF (a + b) = | 2(a|+−b)rh . Therefore −K generates− Archimedean| | − circles belos, KoG 17 (2013), 9–13. − A of (αh,βh,γ) with γ by Lemma 1. The rest of (i) follows 2 2 h 2 2 2 h [6] H. OKUMURA, Archimedean twin circles in the ar- from KOα a = (r ah) + KF a = 2ahr . We | h | − h A − | | − h A belos, Math. Gaz. 97 (2013), 512. prove (ii). The circles Am(O) and Bm(I) are expressed by the equations [7] H. OKUMURA, Archimedean circles of the collinear arbelos and the skewed arbelos, J. Geom. Graph. 17 (x (2a + h))2 + y2 = (2a + h)2 (9) − (2013), 31–52. and [8] H.OKUMURA, Dilations and the arbelos, Normat 60 (x + 2b + h)2 + y2 = (2b + h)2 + 4ab, (10) (2012), 4–8.

h [9] H. OKUMURA, Ubiquitous Archimedean circles of respectively. Subtracting (10) from (9), we get x = rA. Therefore Am(O), Bm(I) and Kα belong to the same pen- the collinear arbelos, KoG 16 (2012), 17–20. cil. Substituting x = rh in (9), and using a + h > 0, we get A [10] H. OKUMURA, M.WATANABE, Remarks on Woo’s y2 = (2a + h)2 (rh (2a + h))2 = rh (2(2a + h) rh ) > − A − A − A Archimedean circles, Forum Geom. 7 (2007), 125– rh (2a rh ) > 0. Therefore A (O) and Kα intersect. The A A m 128. rest of− (ii) can be proved similarly as the proof of (i). Let K be one of the points of intersection in (ii), and let [11] H. OKUMURA, M.WATANABE, The twin circles of F be the foot of the perpendicular from K to AB. Then Archimedes in a skewed arbelos, Forum Geom. 4 2 h h 2 2 KF = rA(2(2a+h) rA). Therefore KOγ (a+b) = (2004), 229–251. | h | 2 −2 2 | | − (rA (a b)) + KF (a + b) = 2ab. This proves (ii).− − | | − − [12] H. OKUMURA, M. WATANABE, The Archimedean circles of Schoch and Woo, Forum Geom. 4 (2004), 27–34. References [13] F. POWER, Some more Archimedean circles in the [1] Q. T. BUI, A newly born pair of siblings to arbelos, Forum Geom. 5 (2005), 133–134. Archimedes’ twins, http://www.cut-the-knot.org/Curriculum/ Geometry/ArbelosBui.shtml. Hiroshi Okumura e-mail: [email protected] [2] Q. T. BUI, Two more Powerian pairs in the arbelos, Department of Mathematics, Yamato University Forum Geom. 8 (2008), 149–150. 2-5-1 Katayama Suita Osaka 564-0082, Japan [3] Q.T.BUI, The arbelos and nine points circles, Forum Geom. 7 (2007), 115–120.