CONFIGURATIONS ON CENTERS OF BANKOFF CIRCLES ZVONKO CERINˇ Abstract. We study configurations built from centers of Bankoff circles of arbelos erected on sides of a given triangle or on sides of various related triangles. 1. Introduction For points X and Y in the plane and a positive real number λ, let Z be the point on the segment XY such that XZ : ZY = λ and let ζ = ζ(X; Y; λ) be the figure formed by three mj utuallyj j tangenj t semicircles σ, σ1, and σ2 on the same side of segments XY , XZ, and ZY respectively. Let S, S1, S2 be centers of σ, σ1, σ2. Let W denote the intersection of σ with the perpendicular to XY at the point Z. The figure ζ is called the arbelos or the shoemaker's knife (see Fig. 1). σ σ1 σ2 PSfrag replacements X S1 S Z S2 Y XZ Figure 1. The arbelos ζ = ζ(X; Y; λ), where λ = jZY j . j j It has been the subject of studies since Greek times when Archimedes proved the existence of the circles !1 = !1(ζ) and !2 = !2(ζ) of equal radius such that !1 touches σ, σ1, and ZW while !2 touches σ, σ2, and ZW (see Fig. 2). 1991 Mathematics Subject Classification. Primary 51N20, 51M04, Secondary 14A25, 14Q05. Key words and phrases. arbelos, Bankoff circle, triangle, central point, Brocard triangle, homologic. 1 2 ZVONKO CERINˇ σ W !1 σ1 !2 W1 PSfrag replacements W2 σ2 X S1 S Z S2 Y Figure 2. The Archimedean circles !1 and !2 together. In [1] Bankoff discovered that the circumcircle !3 = !3(ζ) of the triangle MNZ has the same radius as the Archimedean twin circles !1 and ! , where M = σ σ and N = σ σ and σ is the circle that 2 3 \ 1 3 \ 2 3 touches σ from inside and σ1 and σ2 from outside (see Fig. 3). σ σ3 σ2 PSfrag replacements S3 σ1 M N W3 !3 X S1 Z S S2 Y Figure 3. The Bankoff circle !3 and the circle σ3. The purpose of this paper is to study triangles on centers of the Bankoff circles of arbelos either on sides of an arbitrary triangle ABC or on sides of some of its associated triangles. More precisely, our first goal is to prove the following theorem (see Fig. 4). All other results in this paper are similar. Let τ denote the base triangle ABC. Then τb is a short notation for its first Brocard triangle AbBbCb. Its vertices are the orthogonal projections of the symmedian point K onto the perpendicular bisectors of sides (see [6]). Let ζa = ζa(τ) = ζ(B; C; λ), ζb = ζb(τ) = ζ(C; A; λ) CONFIGURATIONS ON CENTERS OF BANKOFF CIRCLES 3 a b c and ζc = ζc(τ) = ζ(A; B; λ). Let W , W , W denote centers of !3(ζa), !3(ζb), !3(ζc), respectively. Recall that triangles τ and θ = XY Z are homologic and we write τ θ provided lines AX, BY , and CZ are concurrent. The point Q in whichi h they concur is their homology center and the line ` containing intersections of pairs of lines (BC; Y Z), (CA; ZX), and (AB; XY ) is their homology axis. In this situation we also use the notation τ Q; ` θ where ` and/or Q can be omitted. For Q and ` sometimes we useh τ; iθ and τ; θ . In stead of homologic, homology center, and homologyh i axis hhmanyiiauthors use perspective, perspector, and perspectrix. Theorem 1. For every λ 0 the triangle W aW bW c on the centers of Bankoff circles of arbelos ≥on sides of a triangle τ is homologic to its Brocard triangle τb. The centre of this homology lies on the Brocard circle of τ (i.e., on the circumcircle of τb). The triangles τ, τb, and W aW bW c have the same centroid. C W a Bb b W Ab Cb A B W c PSfrag replacements Figure 4. The triangle W aW bW c and the first Brocard triangle AbBbCb are homologic. 2. Bankoff circle !3 x+y a+b Let X(x; a) and Y (y; b). Then S( 2 ; 2 ) is the midpoint of the XZ x+λ y a+λ b j j segment XY . Since ZY = λ, the point Z is λ+1 ; λ+1 . More- j j (λ+2)x+λ y (λ+2) a+λ b over, semicircles σ1 and σ2 have centres at 2(λ+1) ; 2(λ+1) and x+(2 λ+1) y a+(2 λ+1) b 2(λ+1) ; 2(λ+1) (the midpoints of segments XZ and ZY ). 4 ZVONKO CERINˇ Our goal now is to find the center and the radius of the unique circle !3 in the arbelos ζ which touches the semicircles σ1 and σ2 from outside and the semicircle σ from inside. Let its centre be the point S3(p; q) and the radius a positive real number %. Let A = x y and B = a b. − − Since σ3 touches σ from inside, the distance S3S is equal to the 1 p 2 2 j j difference 2 A + B % of their radii. This condition leads to the relation − 2 2 x + y 2 a + b pA2 + B2 (1) p + q = % : − 2 − 2 2 − Since σ3 touches σ1 and σ2 from outside, the distances S3S1 and S3S2 λ pA2+B2 pA2+B2 j j j j are equal to the sums % + 2 (λ+1) and % + 2 (λ+1) of their radii. It follows that 2 2 2 (λ + 2) x + λ y (λ + 2) a + λ b λ pA2 + B2 (2) p + q = % + ; − 2 (λ + 1) − 2 (λ + 1) 2 (λ + 1) 2 2 2 x + (2 λ + 1) y a + (2 λ + 1) b pA2 + B2 (3) p + q = % + : − 2 (λ + 1) − 2 (λ + 1) 2 (λ + 1) These equations have two solutions in p, q, and %. We shall use the λ pA2+B2 solution % = 2 (λ2+λ+1) and (λ + 2) x + λ (2 λ + 1) y 2 λ (a b) (λ + 2) a + λ (2 λ + 1) b + 2 λ (x y) S3 − − ; − : 2 (λ2 + λ + 1) 2 (λ2 + λ + 1) It is now easy to find the coordinates of the points M and N because we know that they divide the segments S1S3 and S2S3 in the ratio of the radii of the circles σ1, σ3 and σ2, σ3. Hence, (λ + 2) x + λ (λ + 1) y λ (a b) (λ + 2) a + λ (λ + 1) b + λ (x y) M − − ; − ; λ2 + 2 λ + 2 λ2 + 2 λ + 2 (λ + 1) x + λ (2 λ + 1) y λ (a b) (λ + 1) a + λ (2 λ + 1) b + λ (x y) N − − ; − : 2 λ2 + 2 λ + 1 2 λ2 + 2 λ + 1 λ pA2+B2 The circumcircle !3 of the triangle MNZ has the radius 2 (λ+1)2 (of the Archimedean twin circles) and its center is at the point x + λ y λ (a b) a + λ b λ (x y) W − ; + − : 3 λ + 1 − 2 (λ + 1)2 λ + 1 2 (λ + 1)2 Remark 1. It is possible to prove the above by inversion (see [9, p. 224]). But, since in this paper we use analytic approach, we need functions that describe coordinates of the center of the Bankoff's cir- cle. Behind the scene we work in rectangular coordinates with compli- cated expressions and then transfer the results into trilinear coordinates where the expressions are often much simpler. CONFIGURATIONS ON CENTERS OF BANKOFF CIRCLES 5 3. Proof of Theorem 1 In our proofs we shall use trilinear coordinates. Recall that the actual trilinear coordinates of a point P with respect to the triangle ABC are signed distances f, g, and h of P from the lines BC, CA, and AB. We shall regard P as lying on the positive side of BC if P lies on the same side of BC as A. Similarly, we shall regard P as lying on the positive side of CA if it lies on the same side of CA as B, and similarly with regard to the side AB. Ordered triples x : y : z of real numbers proportional to (f; g; h) (that is such that x = µf, y = µg, and z = µh, for some real number µ different from zero) are called trilinear coordinates of P . The advantage of their use is that a high degree of symmetry is present so that it usually suffices to describe part of the information and the rest is self evident. For example, when we write X1(1) or I(1) or simply say I is 1 this indicates that the incenter has trilinear coordinates 1 : 1 : 1. We gave only the first coordinate while the other two are cyclic permutations of the first. Similarly, 1 1 1 1 1 X2( a ) or G( a ) say that the centroid has has trilinears a : b : c , where a, b, c are lengths of sides of ABC. We use Xn to denote the n-th central point of the triangle ABC (see [8]). The expressions in terms of sides a, b, and c can be shortened using the following notation. da = b c; db = c a; dc = a b; za = b + c; zb = c + a; zc = a + b; − − − t = a + b + c; ta = b + c a; tb = c + a b; tc = a + b c; − − − 1 m = abc; ma = bc; mb = ca; mc = ab; T = 4 pttatbtc; n n n n n For an integer n, let tn = a + b + c and dna = b c and similarly for other cases.