INTERNATIONAL JOURNAL OF GEOMETRY Vol. 8 (2019), No. 2, 84 - 98 SOME ARCHIMEDEAN INCIRCLES IN AN ARBELOS NGUYEN NGOC GIANG and LE VIET AN Abstract. In this paper, we show that some Archimedean incircles con- structed from the arbelos. As we known, \An arbelos is a plane region bounded by three semicircles with three apexes such that each corner of the semicircles is shared with one of the others (connected), all on the same side of a straight line (the baseline) that contains their diameters. One of the properties of the arbe- los noticed and proved by Archimedes in his Book of Lemmas is that the two small circles inscribed into two pieces of the arbelos cut off by the line perpendicular to the baseline through the common point of the two small semicircles are congruent. The circles have been known as Archimedes' twin circles. Consider an arbelos formed by semi-circles (O1), (O2), and (O) of radii a; b; and a + b. The semi-circles (O1) and (O) meet at A,(O2) and (O) at B,(O1) and (O2) at C. Let CD be the divided line of the smaller ab semi-circles. The radius of Archimedes' twin circles equals t := a+b . Circles of the same radius are said to be Archimedean.", [1]. In this article, we go to some new Archimedean incircles constructed from the arbelos. 0 0 Theorem 1. Let O1;O2 be points symmetric to the points O1;O2 with re- 0 spect to the points A; B, respectively. Tangent line to (O) at D meets (BO1) 0 at A1 and B1, and meets (AO2) at A2 and B2. Let C1 be the point of inter- 0 section of the tangent lines to (BO1) at A1 and B1, and C1 be the point of 0 intersection of tangent lines to (AO2) at A2 and B2. Then the incircles of triangles A1B1C1 and A2B2C2 are Archimedean. ||||||||||||{ Keywords and phrases: arbelos, Archimedean incircles, Inversion. (2010)Mathematics Subject Classification: Received: 22.04.2019. In revised form: 26.09.2019. Accepted: 11.09.2019 Some Archimedean incircles in an arbelos 85 Figure 1. 0 0 Proof. Let M1 be the center of (BO 1) then M1 is the midpoint of BO 1 ; 0 BD meets (BO 1) at I1 (I1 6= B), M1C1 meets A1B1 at H1 (see figure 1). We have M1C1?A1B1 at the midpoint H1 of A1B1 and the homothety BA 0 BO 1 0 HB transforms (BA) into (BO 1). Hence M1I1 k DO?A1B1 . It follows M1I1 ≡ M1C1 or I1 2 M1C1. Since M1C1 is the perpendicular bisector of A1B1 and I1 2 M1C1, triangle I1A1B1 is isosceless at I1 . It follows I1\A1B1 = I1\B1A1. Note that C1A1 is 0 tangent to (BO 1) so IA\1C = I1\B1A1 = I1\A1B1. Similarly, I1\B1C1 = I1\B1A1. It follows that I1 is the incenter of triangle A1B1C1. Since I1H1?A1B1, I1H1 is the inradius of triangle A1B1C1. We easily see that two triangles I1H1D and BCD are similar (a-a). It follows I1H1 = BC . Hence I1D BD BC I D M O BM − BO I H = · I D = BC · 1 = BC · 1 = BC · 1 1 1 BD 1 BD BO BO 2a+2b+a − (a + b) ab = 2b · 2 = = t (a + b) a + b It follows that the incircle I1(H1) of triangle A1B1C1 is Archimedean. Similarly, the incircle of triangle A2B2C2 is also Archimedean. 0 0 Theorem 2. Let O 1;O 2 be points symmetric to the points O1;O2 with respect to A; B, respectively. Let C1;C2 be points symmetric to the point C with respect to A; B, respectively. The common tangent lines to two semi- 0 0 circles (AO 2) and (BO 1) meet the circle (BC1) at A1 and B1 , and meet the circle (AC2) at A2 and B2. Let T1 be the point of intersection of tangent 86 Nguyen Ngoc Giang and Le Viet An lines to (BC1) at A1 and B1, and T2 be the point of intersection of tangent lines to (AC2) at A2 and B2. Then the incircle of triangles A1B1T1 and A2B2T2 are Archimedean. Figure 2. 0 0 Proof. Let (M1); (N1) be the circles (BO 1); (BC1); BD; AD meet (BO 1), 0 (AO 2) at E; F , respectively (see figure 2). We easily see that the homothety BO H BM1 transforms (O) into (BO0 ) . It follows BD = BO = a+b = 2(a+b) B 1 BE BM1 3a+2b 3a+2b 2 so DE = BE−BD = a , we have BD = 2pb(a + b). Hence DE = p BD BD 2(a+bp) b(a+b)a a(a+b)b . Similarly DF = . Hence DE = p a . On the other hand, a+b p a+b DF b 2 a(a+b) AD = p = p a . It follows DE = DA or DA:DF = DB:DE . This BD 2 b(a+b) b DF DB thing proves that the inscribed quadrilateral ABF E has two perpendicular diagonals. Note that O is the midpoint of AB, applying the Brahmagupta BO BM1 theorem, we have OD?EF . Under the homthety HB , we have OD k 0 M1E . It follows M1E?EF so EF is the tangent line to (BO 1). Similarly, 0 EF is the tangent line to (AO 2). Hence EF is the common tangent of 0 0 (BO 1) and (AO 2). From that A1B1 ≡ EF . Let I1 be the point of intersection of BD and (BC1)(I1 6= B); H1 = N1T1 \A1B1. We easily prove that I1(H1) is the incircle of triangle A1B1T1. 3a+2b a Furthermore, I1 2 N1T1 and M1N1 = (2a + b) − 2 = 2 . From that, we easily see that two right triangles I1EH1 and BDC are similar so a I1H1 BC I1E M1N1 2 ab = ) I1H1 = · BC = · BC = · 2b = = t: I1E BD BD BO a + b a + b It follows that I1(H1) is Archimedean. Similarly, the incircle of triangle A2B2T2 is also Archimedean. Theorem 3. Let D0 be a point symmetric to the point D with respect to 0 AB. The tangent line to (O) at D meets the circles (OiDD ) with i = 1; 2 Some Archimedean incircles in an arbelos 87 0 at Ci (Ci 6= D): The tangent lines to (OiDD ) at D and Ci meet at Ti. Then the incircles of triangles T1C1D and T2C2D are Archimedean. Figure 3. 0 Proof. Let P1 be the incenter of circle (O1DD ); H1 = P1T1 \ DC1 and let I1 be the incenter of triangle T1C1D. Similarly to the above theorem, we obtain that I1(H1) is the incircle of triangle T1C1D and I1 is the point of 0 intersection of P1T1 and the small arc DC of (pODD ) (see figure 3). We easily see that O1C = a; O1O = b; CD = 2 ab and OD = a + b. Applying the Pythagorean theorem, we have p 2 2 p 2 O1D = O1C + CD = a + 4ab: Let K1 be the midpoint of DO1 then we easily see that two triangles O1CD and O1K1P1 are similar. It follows O C O K 1 = 1 1 : O1D O1P1 p p 2 2 Hence O P = O K · O1D = a +4ab · a +4ab = a+4b . 1 1 1 1 O1C 2 a 2 a+4b a+2b Hence OP1 = O1P1 − OO1 = 2 − b = 2 . Let L1 be the projection from P1 onto OD then two triangles OCD and OL1P1 are similar. It follows OC OL1 OC a − b a + 2b = ) OL1 = · OP1 = · : OD OP1 OD a + b 2 88 Nguyen Ngoc Giang and Le Viet An Hence a − b a + 2b DL = OD − OL = a + b − · 1 1 a + b 2 2(a + b)2 − (a2 − 2b2 + ab) a2 + 3ab + 4b2 = = : 2(a + b) 2(a + b) Since P1H1DL1 is rectangular, a2 + 3ab + 4b2 P H = DL = : 1 1 1 2(a + b) It follows I1H1 = P1I1 − P1H1 = P1O1 − P1H1 a + 4b a2 + 3ab + 4b2 ab = − = = t; 2 2(a + b) a + b this thing means that I1(H1) is Archimedean. Similarly, the incircle of triangle T2C2D is also Archimedean. Theorem 4. The common tangents of two semi-circles (AO2) and (BO1) meets the circle A(C) at A1 and B1, and meets the circle B(C) at A2 and B2. The tangent lines at A1 and B1 of circle A(C) meet at C1, and the tangents at A2 and B2 of B(C) meets at T2. Then the incircles of triangles T1A1B1 and T2A2B2 are Archimedean. Figure 4. Proof. Let I1 be the incircle of triangle T1A1B1; O3;O4 be the centers of circles (AO2); (BO1), respectively; and suppose that the line A1B1 is tangent to (AO2); (BO1) at T3;T4, respectively; and AT1 \ A1B1 = H1 (see figure 4). We easily have that AT1?A1B1 and I1(H1) is the incircle of triangle T1A1B1; I1 belongs to A(C) and two lines O3T3 and O4T4 are parallel. Some Archimedean incircles in an arbelos 89 a+b 2a+b Using the simple calculations, we have O3O4 = 2 , O3T3 = AO3 = 2 , 2b+a 3a+2b O4T4 = 2 and AO2 = AO3 + O3O2 = 2 . We have O3T3:AO4 = AH1:O3O4 + O4T4:AO3; 2a + b 3a + 2b a + b 2b + a 2a + b · = AN · + · ; 2 2 3 2 2 2 2a2 + ab AN = : 3 a + b 2a2+ab ab From that I1H1 = AI1 − AH1 = 2a − a+b = a+b = t. This thing means that the circle I1(H1) is Archimedean. Similarly, the incircle of triangle T2A2B2 is also Archimedean. _ Theorem 5. Let M be the midpoint of the arc AB of a semi-circle (O). 0 The circle (MO ) meets (MO1O2) at M and T .