The Arbelos and Nine-Point Circles

Transcription

1 Forum Geometricorum Volume 7 (2007) FORU GEO ISSN The rbelos and Nine-Point Circles uang Tuan ui bstract. We construct some new rchimedean circles in an arbelos in connection with the nine-point circles of some appropriate triangles. We also construct two new pairs of rchimedes circles analogous to those of Frank Power, and one pair of rchimedean circles related to the tangents of the arbelos. 1. Introduction We consider an arbelos consisting of three semicircles (O 1 ), ( ), (O), with points of tangency,, P. Denote by r 1, r 2 the radii of (O 1 ), ( ) respectively. rchimedes has shown that the two circles, each tangent to (O), the common tangent P of (O 1 ), ( ), and one of (O 1 ), ( ), have congruent radius r = r 1r 2 r 1 +r 2. See [1, 2]. Let C be a point on the half line P such that PC = h. We consider the nine-point circle (N) of triangle C. This clearly passes through O, the midpoint of, and P, the altitude foot of C on. Let C intersect (O 1 ) again at, and C intersect ( ) again at. Let O e and H be the circumcenter and orthocenter of triangle C. Note that C and H are on opposite sides of the semicircular arc (O), and the triangles C and H have the same nine-point circle. We shall therefore assume C beyond the point on the half line P. See Figure 1. In this paper the labeling of knowing rchimedean circles follows [2]. 2. rchimedean circles with centers on the nine-point circle Let the perpendicular bisector of cut (N) at O and e, and the altitude CP cut (N) at P and h. See Figure It is easy to show that PO e h is a rectangle so e is the reflection of P in N. ecause O e is also the reflection of H in N, HPO e e is a parallelogram, and we have O e e = PH. (1) Furthermore, from the similarity of triangles HP and P C, wehave PH P = P PC. Hence, PH = 4r 1r 2 h. (2) Publication Date: June 18, Communicating Editor: Paul Yiu. The author thanks Paul Yiu for his help in the preparation of this paper.

2 116. T. ui C h e H N O e O 1 P O Figure Since C is beyond on the half line P, the intersection F of O 1 and is a point F below the arbelos. Denote by (I) the incircle of triangle FO 1. See Figure 2. The line IO 1 bisects both angles O 1 F and O 1. ecause O 1 = O 1, IO 1 is perpendicular to C, and therefore is parallel to H. Similarly, I is parallel to H. From these, two triangles H and IO 1 are homothetic with ratio O 1 =2. It is easy to show that O is the touch point of (I) with and that the inradius is IO = 1 PH. (3) 2 In fact, if F is the reflection of F in the midpoint of O 1 then O 1 F F is a parallelogram and the circle (PH) (with PH as diameter) is the incircle of F O 1. It is the reflection of (I) in midpoint of O Now we apply these results to the arbelos. From (2), 1 2 PH = 2r 1r 2 rchimedean radius r 1r 2 r 1 +r 2 if and only if CP = h =2(r 1 + r 2 )=. h = In this case, point C and the orthocenter H of C are easy constructed and the circle with diameter PH is the ankoff triplet circle (W 3 ). From this we can also construct also the incircle of the arbelos. In this case F = incenter of the arbelos. From (3) we can show that when CP =, the incircle of FO 1 is also rchimedean. See Figure 3. Let be the intersection of OO e and the semicircle (O), i.e., the highest point of (O). When CP = h =2(r 1 + r 2 )=, OO e = h H = h C = h PH 2 =(r 1 + r 2 ) r 1r 2 r 1 + r 2.

3 The arbelos and nine-point circles 117 F C H N O e O O 1 P I Figure 2. F Therefore, ( O e =(r 1 + r 2 ) r 1 + r 2 r ) 1r 2 = r 1r 2. r 1 + r 2 r 1 + r 2 From (1), O e e = PH = 2r 1r 2 r 1 +r 2. This means that is the midpoint of O e e, or the two circles centered at O e and e and touching (O) at are also rchimedean circles. See Figure 3. We summarize the results as follows. Proposition 1. In the arbelos (O 1 ), ( ), (O), ifc is any point on the half line Pbeyond and H is orthocenter of C, then the circle (PH) is rchimedean if and only if CP = =2(r 1 + r 2 ). In this case, we have the following results. (1). The orthocenter H of C is the intersection point of ankoff triplet circle (W 3 ) with P(other than P ). (2). The incircle of triangle FO 1 is an rchimedean circle touching at O; it is reflection of (W 3 ) in the midpoint of O 1. (3). The circle centered at circumcenter O e of C and touching (O) at its highest point is an rchimedean circle. This circle is (W 20 ). (4). The circle centered on nine point circle of C and touching (O) at is an rchimedean circle; it is the reflection of (W 20 ) in. (5). The reflection F of F in midpoint of O 1 is the incenter of the arbelos. Remarks. (a) The rchimedean circles in (2) and (4) above are new.

4 118. T. ui C h e F H N O e a b O 1 P O I F Figure 3. (b) There are two more obvious rchimedean circles with centers on the ninepoint circle. These are ( a ) and ( b ), where a and b are the midpoints of H and H respectively. See Figure 3. (c) The midpoints a, b of H, H are on nine point circle of C and are two vertices of Eulerian triangle of C. Two circles centered at a, b and touch at O 1, respectively are congruent with (W 3 ) so they are also rchimedean circles (see [2]). 3. Two new pairs of rchimedean circles If T is a point such that OT 2 = r r2 2, then there is a pair of rchimedean circles mutually tangent at T, and each tangent internally to (O). Frank Power [5]. constructed two such pairs with T = 1, 2, the highest points of (O 1 ) and ( ) respectively. llowing tangency with other circles, Floor van Lamoen [4] called such a pair Powerian. We construct two new Powerian pairs.

5 The arbelos and nine-point circles 119 I 2 1 O 1 P O Figure The triangle 1 2 has 1 = 2 r 2, 2 = 2 r 1, and a right angle at. Its incenter is the point I on O such that I = ( )=(r 1 + r 2 ) r1 2 + r2 2. Therefore, OI 2 = r1 2 + r2 2, and we have a Powerian pair. See Figure Consider also the semicircles (T 1 ) and (T 2 ) with diameters and O 1. The intersection J of (T 1 ) and (T 2 ) satisfies OJ 2 = OP 2 + PJ 2 =(r 1 r 2 ) 2 +2r 1 r 2 = r1 2 + r2 2. Therefore, we have another Powerian pair. See Figure 5. J O 1 T 1 P O T 2 Figure Two rchimedean circles related to the tangents of the arbelos We give two more rchimedean circles related to the tangents of the arbelos. Let L be the tangent of (O) at, and 1, 2 the orthogonal projections of O 1, on L. The lines O 1 1 and 2 intersect the semicircles (O 1 ) and ( ) at R 1 and R 2 respectively. Note that R 1 R 2 is a common tangent of the semicircles (O 1 ) and ( ). The circles (N 1 ), (N 2 ) with diameters 1 R 1 and 2 R 2 are rchimedean. Indeed, if (W 6 ) and (W 7 ) are the two rchimedean circles through

6 120. T. ui 2 1 N 2 N 1 R 2 R 1 W 6 P W 7 O 1 O Figure 6. P with centers on (see [2]), then N 1, N 2, W 6, W 7 lie on the same circle with center the midpoint of P. See Figure 6. We leave the details to the reader. References [1] L. ankoff, re the rchimedean circles really twin?, ath. ag., 47 (1974) [2] C. W. Dodge, T. Schoch, P. Y. Woo and P. Yiu, Those ubiquitous rchimedean circles, ath. ag., 72 (1999) [3] F.. van Lamoen, Online catalogue of rchimedean Circles, available at lamoen/wiskunde/rbelos/catalogue.htm [4] F.. van Lamoen, Some more Powerian pairs in the arbelos, Forum Geom., 7 (2007) [5] F. Power, Some more rchimedean circles in the rbelos, Forum Geom., 5 (2005) uang Tuan ui: 45, 296/86 by-street, inh Khai Street, Hanoi, Vietnam address: bqtuan1962@yahoo.com