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 ∈ M1C1. Since M1C1 is the perpendicular bisector of A1B1 and I1 ∈ 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 = √ BD BD 2(a+b√) b(a+b)a a(a+b)b . Similarly DF = . Hence DE = p a . On the other hand, a+b √ a+b DF b 2 a(a+b) AD = √ = 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 ∈ 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 (√ODD ) (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 √ √ 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 . Draw the tangents TE,TF to (O0). Then the incircle of triangle TEF is Archimedean.

Figure 5.

0 Proof. Draw the diameter MM of (MO1O2); Let H be the orthocenter of 0 triangle MO1O2; I be the point of intersection of segment TO and semi- 0 circle (O1O2); and H be a point symmetric to the point H with respect to O1O2 (see figure 5). Since MTM\ 0 = MTO\0, three points M 0,T,O are collinear lying on the line perpendicular to MT at T . 0 0 0 We also have M O1⊥MO1⊥HO2, M O1 k HO2. Similarly, M O2 k HO1 . 0 Hence, the quadrilateral HO1MO2 is a parallelogram and O is the midpoint of M 0H. 90 Nguyen Ngoc Giang and Le Viet An

Thus, four points M 0,O0,H,T are collinear. Applying the Intersecting-chords theorem, we have 0 0 0 0 0 0 0 2 0 2 0 2 O H · O T = O H · O M = O O1 · O O2 = (a + b) = O E = O F . Note that the triangles TEO0,TFO0 are right at E,F , respectively and the above relation, we have that EH,FH are the altitudes of triangles TEO0,TFO0, respectively. Hence, three points E,H,F are collinear lying on the line perpendicular to O0T at H. 0 ◦ According to the symmetric property, we have O\1H O2 = O\1HO2 = 180 − 0 O\1MO2 so the quadrilateral MO1H O2 is inscribed. Applying the Intersecting- chord theorem, we have: ab ab OH0 · OM = OO · OO = ab ⇒ OH = OH0 = = = t. 1 2 OM a + b 0 |a−b| On the other hand, OO = 2 , applying the Pythagorean theorem, we have: s p (a − b)2  ab 2 a2 + b2 O0H = O0O2 + OH2 = + = . 4 a + b 2(a + b)

a+b a2+b2 ab It follows HI = OI − OH = 2 − 2(a+b) = a+b = t. We easily see that I(H) is the incircle of triangle MEF . Thus, the incircle of triangle MEF is Archimedean.  There is a pair of Archimedean circles in [2] by T. O. Dao. However, from the construction of midline triangle, we can build some other Archimedean circles as the following theorem 6

Figure 6. Some Archimedean incircles in an arbelos 91

Theorem 6. The outside common tangent line of two semi-circles (O1) and (O2) meets the semi-circle (O) at M and N. The tangent lines at M and N of circle (O) meet at P . Let M1,N1,P1 be the midpoints of NP,PM,MN, respectively. Then the incircles of triangles MN1P1,NP1M1,PM1N1 and M1N1P1 are Archimedean. Proof. (see figure 6) According to [2], the inradius of triangle MNP is 2t. 1 Hence, from the simillarity transformation with the ratio 2 , the triangle MNP goes into each of triangles MN1P1,NP1M1,PM1N1 and M1N1P1 then these triangles will have the same inradius t, hence they are Archimedean. 

Remark 1. Clearly, the circle, which its center is the midpoint of M1N1 and it passes through the two incenters of triangles PM1N1 and M1N1P1, is also Archimedean. Next, we go to an another result similar to theorem 7 as follows:

Theorem 7. The tangent lines of two semi-circles (O1) and (O2) meet the circle A(C) at E1 and F1, meet B(C) at E2 and F2. The tangent lines at E1 and F1 of A(C) meet at G1. The tangent lines at E2 and F2 meet B(C) at G2. Let Mi,Ni and Pi , i = 1, 2 , be the midpoints of FiGi,GiEi and EiFi. Then the incircles of triangles EiPiNi,FiNiMi, GMiNi and MiNiPi are Archimedean.

Figure 7.

Proof. Segment AG1 meets A(C) at I1 , meets E1F1 at P1 (see figure 7). Then, we easily see that I1(P1) is the incircle of triangle E1F1G1 and further- more, according to the result at [3] then the circle (I1P1) is Archimedean. Hence I1P1 = 2t. From that, the incircles of triangles E1P1N1,F1N1M1, GM1N1 and M1N1P1 are Archimedean. 92 Nguyen Ngoc Giang and Le Viet An

Similarly, the incircles of triangles E2P2N2,F2N2M2, GM2N2 and M2N2P2 are Archimedean.  Remark 2: Similarly to the remark 1, we also have that the circles, which have the centers being the midpoints of EiFi and pass through the incircles of triangles GiMiNi and MiNiPi, are Archimedean. Furthermore, we can build some Archimedean circles similar to the construction of pair of circles of T. O. Dao [2], which are dash circles, are the circles passing through Ii and tangent with sides EiGi and FiGi as the figure 7. We have the following result from the theorem 6 and theorem 7: Theorem 8. : Let D0 be a point symmetric to the point D with respect to the point C. The common tangent lines to two circles (AO2) and (BO1) 0 0 meet the circle (O1DD ) at E1 and F1, meet the circle (O2DD ) at E2 and 0 F2. The tangent lines at E1 and F1 to (O1DD ) meet at G1. The tangent 0 lines at E2 and F2 to (O2DD ) meet at G2. Let Mi,Ni,Pi , i = 1, 2 , be the midpoints of FiGi,GiEi,EiFi, respectively. Then the incircles of triangles EiPiNi,FiNiMi, GMiNi and MiNiPi are Archimedean.

Figure 8.

0 Proof. Segment G2P2 meets (O2DD ) at I2. The tangent line to (O) at D 0 meets (O2DD ) again at C2 and meets G2P2 at J2 (see figure 8). Some Archimedean incircles in an arbelos 93

We easily see that I2(P2) is the incircle of triangle E2F2G2. From the resut of the article in [4], we have that OD is perpendicular to E2F2 and the distance from D to E2F2 is equal to t. Hence E2F2 is parallel to DC2 and J2P2 = t. By the theorem 2, we have I2J2 = t. If follows I2P2 = 2t. Hence, the inradius of triangle E2F2G2 is equal to 2t. From that, we follow the incircles of triangles E2P2N2,F2N2M2, GM2N2 and M2N2P2 being Archimedean. Similarly, the incircles of triangles E1P1N1,F1N1M1, GM1N1 and M1N1P1 are also Archimedean.  Remark 3: The dash circles in the figure are also Archimedean. We go to an another result based on [1] as follows :

Theorem 9. The circles A(C) and B(C) meet the semi-circle (O) at E1 and E2, respectively. Tangent line at E1 to (O) meets A(C) at F1, and tangent line at E2 meets B(C) at F2. Tangent lines at E1 and F1 to A(C) meet at G1, tangent lines at E2 and F2 to B(C) meet at G2. Points Mi,Ni,Pi, i = 1, 2, are the midpoints of FiGi,GiEi,EiFi, respectively. Then the incircles of triangles EiNiPi,FiPiMi,GiMiNi v`a MiNiPi are Archimedean.

Figure 9.

Proof. Let K1 be the point of intersection of two semi-circles (O1) and (O1O2); The tangent line at K1 to (O1O2) meets (O1) at L1; and tangent lines at K1 and L1 to (O1) meet at H1 (see figure 9). We easily see that, un- 2 der the homothety HC center H, with ratio 2, K1,L1,H1 go into the points E1,F1,G1. Hence, the incircle of triangle K1L1H1 goes into the incircle of triangle E1F1G1. According to the result at [1], the inradius of triangle K1L1H1 is equal to t, so the inradius of triangle E1F1G1 is equal to 2t. From that, it follows that the incircles of triangles E1N1P1,F1P1M1,G1M1N1 and 94 Nguyen Ngoc Giang and Le Viet An

M1N1P1 are Archimedean. Similarly, the incircles of triangles E2N2P2,F2P2M2,G2M2N2 and M2N2P2 are Archimedean.  Remark 4: We also have that the red circles in figure 9 are also Archimedean.

Theorem 10. Let C1,C2 be points symmetric to the point C with respect to A, B, respectively. The tangent line to (O) at D meets the circles (BC1) at E1,F1, and (AC2) at E2,F2. The tangent lines to (BC1) at E1 and F1 meet at G1. The tangent lines to (AC2) at E2 and F2 meet at G2. Let Mi,Ni,Pi , i = 1, 2 , be the midpoints of FiGi,GiEi,EiFi, respectively. Then the incircles of triangles EiPiNi,FiNiMi, GMiNi and MiNiPi are Archimedean.

Figure 10.

0 0 Proof. Let O1 be the center of circle (BC1); segment O1G1 meets the circle (BC1) at I1 and passes through the midpoint M1 of E1F1; drop AH1 per- pendicular to E1F1 at H1 (see figure 10). It is easy to have that I1(M1) is the incircle of triangle E1F1G1. We have that two triangles ADH1 and ADC are equal according to the case 0 a.s.a. It follows AH1 = AC = 2a. Note that O1M1, OD and AH1 are pairs 0 0 of parallel lines, OD = a + b, OO1 = a and AO1 = b. It follows that OO0 AO0 O0M = 1 · AH + 1 · OD 1 OA 1 AO a b = · 2a + · (a + b) a + b a + b 2a2 + ab + b2 = . a + b Some Archimedean incircles in an arbelos 95

0 From that, note that O1I1 = 2a + b so 0 0 I1M1 = O1I1 − O1M1 2a2 + ab + b2 = 2a + b − a + b 2ab = a + b = 2t

Hence, the inradius of triangle E1F1G1 is equal to 2t. From that, we fol- low the incircles of triangles E1P1N1,F1N1M1, GM1N1 and M1N1P1 being Archimedean. Similarly, the incircles of triangles E2P2N2,F2N2M2, GM2N2 and M2N2P2 are also Archimedean.  Remark 5: We also have that the red circles in figure 10 are also Archimedean.

Theorem 11. The tangent line that draws from O2 to the semi-circle (O1) meets the circle A(C) at E1 and F1; The tangent line that draws from O1 to the semi-circle (O2) meets the circle B(C) at E2 and F2. The tangent lines of A(C) meet E1 and F1 at G1; The tangent of circles B(C) at E2 and F2 meet at G2. Then the in-circles of triangles E1F1G1 and E2F2G2 are Archimedean.

Figure 11.

Proof. Let T1 be the tangent point of E1F1 and (O1); segment AG1 meets A(C) at I1 and meets E1F1 at H1 then I1(H1) is the incircle of triangle E1F1G1 (see figure 11). It is easy to see that O1T1 is parallel to AH1 so by the theorem Thales’ theorem, we have 2 a O1T1 O1O1 a + b 2a + b 2a + ab = = =⇒ AH1 = · a = . AH1 AH1 O1A 2a + b a + b a + b 96 Nguyen Ngoc Giang and Le Viet An

It follows 2a2 + ab ab I H = AI − AH = 2a − = = t. 1 1 1 1 a + b a + b

Hence the circle I1(H1) is Archimedean. Similarly, the incircle of triangle E2F2G2 is also Archimedean.  We continue to go to an interesting result that we are received by the inversion as follows: Theorem 12. Let (I) be the incircle of an abelos. The tangent lines at A to (I) meet CD at A1 and A2. The tangent lines at B to (I) meet CD at B1 and B2. Then the incircles of triangles AA1A2 and BB1B2 are Archimedean.

Figure 12.

Proof. Draw a circle A(C) that meets (O1) at E1 and F1 and consider the inversion I1 for this circle (see figure 12). From the transformation I1 then (O),CD, AA1, AA2 go into E1F1, (O1), AA1, AA2, respectively. Let (I1) be the image of (I) under I1. Since (I) is tangent to (O), (O1), AA1, AA2, (I1) is tangent to E1F1, CD, AA1, AA2. Note that, E1F1 is paralle to CD because both of them are perpendicular to AC. This thing proves that (I1)is the incircle of trapezoid formed by four lines E1F1, CD, AA1, AA2 so (I1) is the incircle of triangle AA1A2. Draw E1H1 that is perpendicular to CD at H1. Appying the result at [5], we have E1H1 = 2t. And hence the circle (I1) is equal to the circle (E1H1) so (I1) is Archimedean. Similarly, the incircle of triangle BB1B2 is also Archimedean.  Some Archimedean incircles in an arbelos 97

Remark 6: If the circle (T ) is tangent to the semi-circle (O) and tangent to the outside of semi-circle (O1) then the tangent lines at A to (T ) and the line CD form a triangle that has the incircle being Archimedean. We go to a triangle which its incircle is Archimedean as follows:

Theorem 13. The common tangent lines of (O1) and (O2) meet the per- pendicular bisectors of CA, CB at K1,K2, respectively. The tangent lines, which are different from K1K2 , at K1 to (O2) and at K2 to (O1) meet at K. Then the incircle of triangle KK1K2 is Archimedean.

Figure 13.

Proof. Suppose that K1K2 is tangent to (O1),(O2) at T1,T2, respectively; K1O2 meets K2O1 at P , and CD meets K1K2 at L; Draw PH that is perpendicular to K1K2 at H (see figure 13). We easily see that P is the incenter of triangle KK1K2. Hence, we will prove that PH = t. Since two triangles O1K1T1 and O2K2T2 are similar, O1K1 = O1T1 = a . O2K2 O2T2 b Hence, applying the Thales’ theorem, we have PK1 = O1K1 = a = CO1 . PO2 O2K2 b CO2 It follows that CP is parallel to O1K1 so CP is perpendicular to AB or P belongs to CD. We easily see that each pair of three triangles OLH, O1K1T1 and O2K2T2 are similar. It follows PH = PL and PH = PL . O1T1 P1K1 O2T2 P2K2 98 Nguyen Ngoc Giang and Le Viet An

From that, applying the Thales’ theorem, we have PH PH PL PL K L K L + = + = 2 + 1 = 1. O1T1 O2T2 O1K1 O2K2 K2K1 K1K2 It follows PH = O1T1.O2T2 = ab = t. O1T1+O2T2 a+b Thus, the incircle of triangle KK1K2 is Archimedean. 

References [1] V. A. Le and E. A. J. Garc´ıa(2019), A Pair of Archimedean Incircles, , Sangaku J. Math., 3 (2019), 1–2. [2] T. O. Dao, Two pairs of Archimedean circles in the arbelos, Forum Geom., 14 (2014), 201–202. [3] F. M. van Lamoen, Online catalogue of Archimedean circles, available at http://home.wxs.nl/˜lamoen/wiskunde/Arbelos/33Bro.htm. [4] E. A. J. Garc´ıa, Another Archimedean circle in an arbelos, Forum Geome., 15 (2015), 127–128. [5] F. M. van Lamoen, Online catalogue of Archimedean circles, available at http://home.wxs.nl/˜lamoen/wiskunde/Arbelos/10Schoch.htm.

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