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The Square of the Arbelos Hiroshi Okumura

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The Square of the Arbelos Hiroshi Okumura Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.5, (2016), Issue 1, pp.46-55 THE SQUARE OF THE ARBELOS HIROSHI OKUMURA Abstract. A square with two points on each of the sides are naturally constructed from an ar- belos. It has relationships to the other squares associated with the arbelos and special type of generalized arbelos, and yields several dozens of Archimedean circles. 2010 Mathematical Subject Classification: 51M04, 51M15, 51N20 Keywords and phrases: arbelos, arbelos with overhang, square of an arbelos, Archimedean circles, metallic mean, golden arbelos, silver arbelos 1. INTRODUCTION Let us consider an arbelos (a, b, g) consisting of three semicircles a, b and g with diam- eters AO, BO and AB, respectively for a point O on the segment AB in the plane. Let a b a and b be the radii of and , respectively. Circles of radius rA = ab/(a + b) form a special class of circles and are called Archimedean circles of the arbelos. In this paper we consider a square with two points on each of the sides. We show that it has relation- ships to the squares associated with the arbelos considered in [7, 8] and to a special type of a generalized arbelos called the arbelos with overhangp [9] and yields several dozens of Archimedean circles. Special cases where b/a = ( n2 + 4 ± n)/2 for n = 1, 2 are investigated. 2. THE SQUARE In this section we construct a square from the arbelos. We use a rectangular coordinate system with origin O such that the points A and B have coordinates (2a, 0) and (−2b, 0), respectively, where we assume that all the semicircles a, b and g are constructed in the ≥ a b region y 0. Let P1 and S1 be the centers of and , respectively, and let C be the farthest point on the circle with a diameter P1S1 from the line AB in the region y > 0. ◦ Let s be the rotation about C through 90 . The center of the semicircle g is denoted by s s s S2 and let S3 = O (see Figure 1). Then we get P1 = S , and let P2 = S2 , P3 = S3 and s s 1 j j Qi = Pi , Ri = Qi for i = 1, 2, 3. Then PiQiRiSi is a square with center C, and S1S2 = a, j j j j j − j S2P1 = b and S2S3 = a b . a b g We call P1Q1R1S1 and C the square and the center of ( , , ). Notice that the lines P2R3, a b g P3R2 and Q1R1 are the tangents of , and parallel to AB with points of tangency P2, R2 and Q3, and the lines S1R1, S2Q3, S3Q2 and P1Q1 are the perpendiculars from R1, Q3, Q2 and Q1 to AB. The points P1, P2 and P3 (resp. P3 and P2) lie on P1Q1 in this order if a < b (resp. a > b). 46 The square of the arbelos R1 Q3 Q2 Q1 g R2 P3 b C P2 R3 a B S1 S2 O=S3 P1 A Figure 1. 3. SQUARES In this section we consider relationships between the squares PiQiRiSi (i = 1, 2, 3) and the squares considered in [7, 8]. Let the line P2R2 meet a and b at points P and R again, respectively (see Figure 2). Then the lines AP and BR meet in a point Q lying on g and OPQR is the inscribed square of (a, b, g) [8]. The points P, Q and R have coordinates (2jb, 2ja), (−2jd, 2je) and (−2ja, 2jb), (1) 0 respectively, where d = a − b, e = a + b and j = ab/(a2 + b2). Let O be the point of 0 0 intersection of OQ2 and g. Let Q˜ and R˜ be the points on the segments AO and BO , respectively such that P˜Q˜ R˜ S˜ is a square, where P˜ and S˜ are the feet of perpendiculars from Q˜ and R˜ to AB, respectively. Then Q˜pand R˜ have coordinates (2a f /g, 2e f /g) and (−2b f /g, 2e f /g), respectively, where f = ab and g = e + f [7]. R1 Q3 Q2 Q1 0 O Q R 2 R˜ P Q˜ 3 R R3 P2 P B S1 S˜ S2 O P˜ P1 A E 0 R3 0 P3 0 Q3 Figure 2. Theorem 3.1. The following statements hold. 0 0 0 (i) If P3,Q3 and R3 are the reflections of the points P3,Q3 and R3 in the line AB,respectively, 0 0 0 2 2 then the squares OPQR and OR3Q3P3 are homothetic with center O and ratio 2ab : (a + b ). (ii) If a 6= b and E is the point of intersection of the lines AB and P2R2, the squares OPQR and 2 2 S2P2Q2R2 are homothetic with center E and ratio 2ab : (a + b ). (iii) The squares P1Q1R1S1 and P˜Q˜ R˜ S˜ are homothetic with center O and ratio g : 2 f . 47 Hiroshi Okumura 2 2 2 2 0 0 Proof. If s = 2ab/e and t = (a + b )/e , we have O = sP3 + tR = sR3 + tP. This proves (i). The part (ii) follows from (i) and the fact E being the homothety center of the triangles − − − OPR and S2P2R2. The part (iii) follows from O = (g/(g 2 f ))Q˜ + ( 2 f /(g 2 f ))Q1 = − − − □ (g/(g 2 f ))R˜ + ( 2 f /(g 2 f ))R1. g The part (ii) shows that the point Q2 lies(p on the line )EQ, which is the tangent of with F 1 2 point of tangency Q [8]. Let (n) = 2 n + 4 + n for a real number n. It is called a metallic mean if n is a natural number [11]. The first part of the following proposition follows from (1), and the last part is obvious. Proposition 3.1. The slopes of the lines OP and OR equal a/b and −b/a, respectively. If n is a j j F ±1 non-negative real number, then d = nrA if and only if b/a = (n) . Since F(n) is a monotonically increasing function of n and F(0) = 1, there exists a non- negative real number n such that b/a = F(n) if and only if b ≥ a. Theorem 3.2. The following statements are equivalent for a non-negative real number n. (i) b/a = F(n). (ii) The point R3 divides the segment QR in the ratio (2 + n) : n externally. (iii) The point P3 divides the segment PQ in the ratio n : j2 − nj internally if n < 2 externally if 2 < n, and coincides with Q if n = 2. Proof. The part (i) is equivalent to to each of the equations R3 = ((2 + n)/2)R + (−n/2)Q and P3 = ((2 − n)/2)P + (n/2)Q being true. □ 4. AN ARBELOS WITH OVERHANG In [9] we considered a generalization of the arbelos, called an arbelos with overhang. In this section we show that the square of (a, b, g) is closely related to a special type of 0 arbelos with overhang. Let Ah and Bh be point on the line AB with x-coordinates a and − 0 0 − 0 − − a b b , respectively such that a a = b b = h > min(a, b). Let h and h be the semicircles with diameters AhO and BhO, respectively erected on the same side of AB g a b g as . The configuration of the three semicircles h, h and is called an arbelos with a b g # # overhang h and is denoted by ( h, h, ) [9]. If h > 0, let a,1 (resp. b,1) be the circle g a b a b # # touching and h (resp. h) internally and (resp. ) externally, also let a,2 (resp. b,2) g a b be the circle touching and h (resp. h) externally and the axis from the side opposite to B (resp. A). We use the next proposition. a b g Proposition 4.1 ([9]). If h > 0, the following statements are true for ( h, h, ). # # −1 (i) The circles a,1 and b,1 are congruent and have radius (1/rA + 1/h) . # # (ii) The circles a,2 and b,2 are congruent and have radius ab/h. a b g Let us consider an arbelos with overhang ( h, h, ) with h = rA. The radius of the circle touching a externally and g internally is proportional to the distance between the center of this circle and the radical axis of a and g [3, p. 108]. Therefore the circle # a g touching externally internally and P1Q1 from the side opposite to B has radius rA/2. # # While the circle a,1 has radius rA/2 by (i) of the proposition. Therefore the circles and # # # a,1 coincide, i.e., a,1 touches P1Q1 from the side opposite to B. Similarly the circle b,1 # # touches R1S1 from the side opposite to A (see Figure 3). The circles a,2 and b,2 have radius e by (ii) of the proposition, i.e., they are congruent to g. Since the centers of g and #a,2 have x-coordinates d and e, their point of tangency, which is the midpoint of the 48 The square of the arbelos segment joining their centers, has x-coordinate a, i.e., the point lies on P1Q1. Similarly g # the point of tangency of and b,2 lies on R1S1. The perpendicular to AB at the center a b # # of h (resp, h) also touches a,1 (resp, b,1). If we erect a square on the same side of AB g a b as with a side joining the centers of h and h, the distance between Q1R1 and the side of this square parallel to AB equals 2rA.
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