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Equilateral Triangles and Archimedean Circles Related to an Arbelos

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Equilateral Triangles and Archimedean Circles Related to an Arbelos Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.8, (2019), Issue 1, pp.1-11 EQUILATERAL TRIANGLES AND ARCHIMEDEAN CIRCLES RELATED TO AN ARBELOS NGUYEN NGOC GIANG AND LE VIET AN ABSTRACT . In this article, we construct the Archimedean circles from one or some equi- lateral triangles related to arbelos. And its converse, we also construct an another equi- lateral triangle from Yiu’s circle. 1. INTRODUCTION We consider an arbelos with greater semi-circle (O) of radius r and smaller semicircles (O1) and (O2) of radii r1 and r2 respectively. The semi-circles (O1) and (O) meet at A, (O2) and (O) at B, (O1) and (O2) at C. The line through C perpendicular to AB meets (O) at D, and (O′) is the Midway semi-circle with diameter O1O2, all on the same side r1r2 AB . Circles with radius t := r are called Archimedean. They are congruent to the Archimedean twin circles [1, 2]. We shall make use of the Yiu’s circle (that is the circle (A26 ) in [3], is also the circle (K3) in [4], or the circle touches γ internally, and touch α[1] and β[1] externally at Corollary 3 in [5]), with center Y (See figure 1). FIGURE 1. Yiu’s circle Key words and phrases. Arbelos, Archimedean circle, equilateral triangle, Yiu’s circle. AMS 2010 Mathematics Subject Classification:51M04. 1 N. G. Nguyen; V. A. Le Proposition 1.1. (Yiu [3] , [4] ). The circle (Y) tangent to both semi-circles (O) and (O′) and its center on CD is Archimedean. For three non-collinear points P, Q and R, we use (PQ ), P(Q), (PQR ) to denote the circle with diameter PQ , the circle with center P passing through Q, and the circumcircle of triangle PQR . 2. PRELIMINARIES 2.1. Archimedean circles are constructed from one or some equilateral triangles releted to an arbelos. In this subsection, we refer to some new Archimedean circles as well as pairs of new Archimedean circles in an arbelos. Theorem 2.1. Consider an equilateral triangle A 1B1C1 such that the vertex C 1 lies on the ray CD and sides C 1 A1 and C 1B1 are all tangent to the semi-circle (O) at A 1 and B 1, respectively. Line passing through C perpendicular to A 1B1 at E and meet the semi-circle (O) at F. Then the circles (CE ) and (EF ) are Archimedean. FIGURE 2 Proof. (See figure 2). We easily see that O1O = O2C = r2, O2O = O1C = r1, CO = r1 r2 , O1O2 = r1 + r2 = r. | − ∠| ∠ ∠ Since C1CO = C1 A1O = C1B1O = 90 ◦, it follows that C1O is the diameter of (CA 1B1). Line A1B1 meets C1C, C1O at G, H, respectively. Since C1 A1 and C1B1 are tangent lines of (O), OC 1 is perpendicular to A1B1 at H. 2 Equilateral triangles and Archimedean circles related to an arbelos Since OA 1 = OB 1 = r and triangle A1B1C1 is equilateral, C1O = 2r, A1B1 = B1C1 = √ 3 1 C1 A1 = 3r, C1 H = 2 r and OH = C1O C1 H = 2 r. Applying the Pythagoras’ theorem, we have− C C2 = C O2 OC 2 = 4r2 (r r )2 = 4(r + r )2 (r r )2 1 1 − − 1 − 2 1 2 − 1 − 2 2 2 = 3r1 + 3r2 + 10 r1r2. 2 2 It follows C1C = 3r1 + 3r2 + 10 r1r2. q Since two right triangles C1GH and C1OC are similar, 2 C1G C1 H C1 H 3r = = C1G = C1O = . C1O C1C ⇒ C1C 2 2 3r1 + 3r2 + 10 r1r2 q Hence, 2 2 2 3r CG = C1C C1G = 3r1 + 3r2 + 10 r1r2 − q − 2 2 3r1 + 3r2 + 10 r1r2 q 2 2 2 3(r1 + r2) 4r1r2 = 3r1 + 3r2 + 10 r1r2 = . q − 2 2 2 2 3r1 + 3r2 + 10 r1r2 3r1 + 3r2 + 10 r1r2 q q Applying the Thales’ theorem with note that CE is parallel to C1O, we have 4r1r2 2 2 CE CG CG √3r +3r +10 r1r2 3 = = CE = C H = 1 2 (r + r ) = 2t. 1 3(r +r )2 1 2 C1 H C1G ⇒ C1G 1 2 2 2 2 √3r1+3r2+10 r1r2 This thing means that (CE ) is Archimedean. On the other hand, we easily see that the circumcenter of circle (A1B1C1) lies on the circle (O), note that two circles (O) and (A1B1C1) have the same radius r and O be- longs to (A1B1C1), they are symmetric across the line A1B1. Hence, under this symmetry A1B1, C goes into the point F. It follows EF = CE = 2t. This means that (EF ) is also Archimedean. Theorem 2.2. Consider the equilateral triangle A 2B2C2 such that the vertex C 2 lies on the ray CD and sides C 2 A2 and C 2B2 are tangent to the Midway semi-circle (O′) at A 2 and B 2, respectively. Line A 2B2 meets the circle A (O1) at K 1 and L 1, and the circle B (O2) at K 2 and L 2. The tangent lines at K 1 and L 1 of A (O1) meet at T 1, and tangent lines at K 2 and L 2 of B (O2) meet at T 2. Then the incircles of triangles K 1L1T1 and K 2L2T2 are Archimedean. Proof. (See figure 3). Let K, L be the projections from C, O onto A2B2. Since CA−→ = 2CO−−→ and CB−→ = 2CO−−→ , the homothety 2 with center C, ratio 2 transforms 1 2 HC (O′) into (O). Under this transformation 2 , the points A , B , C , K go into the points A , B , C , E. HC 2 2 2 1 1 1 It follows that K is the mid-point of CE and A1B1 is parallel to A2B2. Note that OL is perpendicular to A2B2, OH is perpendicular to A1B1. It follows that OL and OH are coincident or L belongs to OH . 3 N. G. Nguyen; V. A. Le FIGURE 3 r Using the theorem 2.1, we have CE = 2t and OH = 2 . It follows HL = KE = CK = t. Hence r r + r r r r2 + r2 OL = OH HL = t = 1 2 1 2 = 1 2 . − 2 − 2 − r1 + r2 2(r1 + r2) Segment AT 1 meets the circle A(O1) and the chord K1L1 at I1 and H1, respectively. Then 1 ∠I K T = 90 ◦ ∠AK I = 90 ◦ (180 ◦ ∠I AK ) 1 1 1 − 1 1 − 2 − 1 1 1 1 1 = ∠I AK = ∠H AK = ∠H K T . 2 1 1 2 1 1 2 1 1 1 It follows that I1K1 is the interior bisector of angle T1K1L1. Since I1 lies on the bisector of angle K1T1L1, I1 is the incenter of triangle K1L1T1. Furthermore, the inradius of triangle K1L1T1 is I1 H1 = AI 1 AH 1 = r1 AH 1. Without loss of generality,− suppose− that r r then OC = r r . Note that three lines 1 ≥ 2 1 − 2 AH 1, OL and CK are pairwise parallel, it follows 2 2 r1 + r2 OA OC r r1 r2 = OL = CK + AH 1 = t + − AH 1 2(r1 + r2) AC CA 2r1 2r1 r2 r1 r2 = + − AH 1. 2 2r1 4 Equilateral triangles and Archimedean circles related to an arbelos It follow AH 1 r1 (r1 r2) = 0. − r1 − r1 + r2 If r1 = r2 then r1 = r2 = 2t and O C. We easily see that A1B1 is parallel to AB and hence,• AH = OL = OC = t. It follows≡ I H = r AH = t. Conversely, if r > r then 1 1 1 1 − 1 1 2 2 AH 1 r1 r1 = 0 = AH 1 = . r1 − r1 + r2 ⇒ r1 + r2 2 r1 And hence I1H1 = r1 AH 1 = r1 = t. − − r1+r2 This thing means that, we always have I1 H1 = t in all of cases. It follows that the incircle of triangle K1L1T1 is Archimedean. Similarly, the incircle of triangle K2L2T2 is also Archimedean. Theorem 2.3. Consider the equilateral triangle O 1O2O3 lying inside the semi-circle (O). Con- struct two lines O 3 M1 and O 3N1 that they are tangent to (O1) at M 1 and N 1, and two lines O3 M2 and O 3N2 are all tangent to (O2) at M 2 and N 2. Then the incircles of triangles O 3 M1N1 and O 3 M2N2 are Archimedean. FIGURE 4 Proof. (See figure 4). Segment O1O3 meets the circle (O1) and the chord M1N1 at J1 and P1, respectively. Then, we easily have that J1 and J1P1 are the incenter an inradius of triangle O3 M1N1, respectively. 5 N. G. Nguyen; V. A. Le Since the triangle M1O1O3 is right at M1, M1P1 is perpendicular to O1O3 and note that O1O3 = O1O2 = r1 + r2, we get 2 2 2 O1 M1 r1 O1 M1 = O1P1.O1O3 = O1P1 = = . ⇒ O1O3 r1 + r2 Therefore, 2 r1 J1P1 = OJ 1 OP 1 = r1 = t. − − r1 + r2 This thing proves that the incircle of triangle O3 M1N1 is Archimedean. Similarly, the incircle of triangle O3 M2N2 is also Archimedean. 2 We known the homothety C transforms (O′) into (O). From our exploitation, we obtain the following result. H Theorem 2.4. Consider the equilateral triangle ABC ′ lying on the same side for AB as arbelos. Lines C ′Y1 and C ′Z1 are tangent to A (C) at Y 1 and Z 1; lines C ′Y2 and C ′Z2 are tangent to B (C) at Y 2 and Z 2. Segment AC ′ meets the circle A (C) at G 1, and Y 1Z1 at D 1; segment BC ′ meets B(C) at G 2, and Y 2Z2 at D 2. Let E 1, F1 be the projections from G 1 onto C ′Y1, C′Z1 respectively, and E 2, F2 be the projections from G 2 onto C ′Y2, C′Z2 respectively.
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