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 poop´ceni 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 obi´cnog nary arbelos are generalized to our generalized arbelos. arbelosa su poop´cene za poop´ceni 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 cir- cles 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 con- sider 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 cir- cles of the ordinary arbelos are generalized to our general- ized 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.