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The Apollonius Problem on the Excircles and Related Circles

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The Apollonius Problem on the Excircles and Related Circles Forum Geometricorum b Volume 6 (2006) xx–xx. bbb FORUM GEOM ISSN 1534-1178 The Apollonius Problem on the Excircles and Related Circles Nikolaos Dergiades, Juan Carlos Salazar, and Paul Yiu Abstract. We investigate two interesting special cases of the classical Apollo- nius problem, and then apply these to the tritangent of a triangle to find pair of perspective (or homothetic) triangles. Some new triangle centers are constructed. 1. Introduction This paper is a study of the classical Apollonius problem on the excircles and related circles of a triangle. Given a triad of circles, the Apollonius problem asks for the construction of the circles tangent to each circle in the triad. Allowing both internal and external tangency, there are in general eight solutions. After a review of the case of the triad of excircles, we replace one of the excircles by another (possibly degenerate) circle associated to the triangle. In each case we enumerate all possibilities, give simple constructions and calculate the radii of the circles. In this process, a number of new triangle centers with simple coordinates are constructed. We adopt standard notations for a triangle, and work with homogeneous barycen- tric coordinates. 2. The Apollonius problem on the excircles 2.1. Let Γ=((Ia), (Ib), (Ic)) be the triad of excircles of triangle ABC. The sidelines of the triangle provide 3 of the 8 solutions of the Apollonius problem, each of them being tangent to all three excircles. The points of tangency of the excircles with the sidelines are as follows. See Figure 2. BC CA AB (Ia) Aa =(0:s − b : s − c) Ba =(−(s − b):0:s) Ca =(−(s − c):s :0) (Ib) Ab =(0:−(s − a):s) Bb =(s − a :0:s − c) Cb =(s : −(s − c):0) (Ic) Ac =(0:s : −(s − a)) Bc =(s :0:−(s − c)) Cc =(s − a : s − b :0) Publication Date: Month day, 2006. Communicating Editor: 2 N. Dergiades et al 2.2. The nine-point circle (N), by the famous Feuerbach theorem, is tangent ex- ternally to each of the excircles, and provides a fourth solution. The points of tangency are 2 2 2 Fa =(−(b − c) s :(c + a) (s − c):(a + b) (s − b)), 2 2 2 Fb =((b + c) (s − c):−(c − a) s :(a + b) (s − a)), 2 2 2 Fc =((b + c) (s − b):(c + a) (s − a):−(a − b) a). These points of tangency form a triangle perspective with ABC at the outer Feuerbach point 1 (b + c)2 (c + a)2 (a + b)2 X = : : . 12 s − a s − b s − c Fb A Fc Fc Fb N S B C Fa Fa Figure 1 2.3. A simple way to obtain solutions to the Apollonius problem from known solutions is to make use of the radical circle of the triad. Given a non-coaxial triad Γ, the radical circle is the unique circle orthogonal to all three circles. Its center, the radical center of the triad, is the intersection of the three pairwise radical axes, which has equal powers with respect to the circles in the triad. Inversion in the radical circle leaves the circles in the triad invariant, but preserves tangency. 1The labeling of triangle centers follows Kimberling’s Encyclopedia of Triangle Centers [3]. The Apollonius problem on the excircles and related circles 3 Therefore the inversive image of one solution to the Apollonius problem of the triad is another solution. 2.4. The radical circle of the excircles is the Spieker radical circle,√ which has 1 2 2 center the Spieker center S =(b + c : c + a : a + b) and radius 2 r + s . See [2, Theorem 4]. 2 a2yz+b2zx+c2xy+(x+y+z)((s−b)(s−c)x+(s−c)(s−a)y+(s−a)(s−b)z)=0. 2.5. Inversion of the nine-point circle in the Spieker radical circle gives the Apol- lonius circle which is tangent to the excircles internally. The points of tangency are the second intersections of the lines NFa, NFb, NFc with the corresponding excircles. These are 2 2 2 2 2 2 2 2 Fa =(−a (a(b + c)+(b + c )) :4b (c + a) s(s − c): 4c (a + b) s(s − b)), 2 2 2 2 2 2 2 2 Fb =(4a (b + c) s(s − c): −b (b(c + a)+(c + a )) :4c (a + b) s(s − a)), 2 2 2 2 2 2 2 2 Fc =(4a (b + c) s(s − b): 4b (c + a) s(s − a): −c (c(a + b)+(a + b )) ). These form a triangle perspective with ABC at the Apollonius point a2(b + c)2 b2(c + a)2 c2(a + b)2 X = : : . 181 s − a s − b s − c r2+s2 The circle through them has center X970 and radius 4r . It can be constructed simply as the intersection of the lines NS and IX181. 2.6. Inversion of the line BC gives a circle tangent internally to (Ia) and exter- nally to (Ib) and (Ic). It is the circle through S and the intersections of BC with the Spieker radical circle. The points of tangency are 2 2 2 (Ia): (−(b + c) (s − b)(s − c): c s(s − b): b s(s − c)), 2 2 2 (Ib): ((b + c) (s − a)(s − c): −c s(s − a): (bs + ca) ), 2 2 2 (Ic): ((b + c) (s − a)(s − c): (cs + ab) : −b s(s − a)). The equation of the circle is 2 2 2 (a yz + b zx + c xy)+(x + y + z)fa(x, y, z)=0, where s(a(b + c)+(b2 + c2)) f (x, y, z):=− x +(s − a)(s − c)y +(s − a)(s − b)z. a 2(b + c) a r2+s2 This circle has radius b+c · 4r . 2 The Spieker center S appears as X10 in [3] Here is an easy way to identify this radical center. The radical axis of (Ib) and (Ic) is the perpendicular from the midpoint of BC to the line IbIc.It has equation (b − c)x +(b + c)(y − z)=0; similarly for the other two radical axes. The coordinates of the radical center can be found by solving these three equations simultaneously. 4 N. Dergiades et al A S B C Figure 2 2.7. Similarly, inversion of the lines CA and AB gives two more circles. The points of tangency of has perspector s − a s − b s − c : : . a2 b2 c2 The centers of the circles also form a triangle with perspector 1 a5 + a4(b + c) − a3(b − c)2 − a2(b + c)(b2 + c2) − 2abc(b2 + bc + c2) − 2b2c2(b + c) : ···: ···) . 3. Two excircles and a vertex 3.1. The Apollonius problem for two circles with disjoint closures and a point outside their closures has in general 4 solutions. Nik’s description. The Apollonius problem on the excircles and related circles 5 3.2. In the special case of the triad Γa := (A, (Ib), (Ic)), two of these circles are the sidelines AC and AB. The remaining two circles can be easily constructed as follows. The radical axis of (Ib) and (Ic) is the A-angle bisector of the medial triangle (which contains the Spieker center S). Let X be the intersection with the A-altude of triangle ABC. Then the circle with diameter AX is tangent to both (Ib) and (Ic). The reflection of this circle in the line IbIc is the fourth one. The line joining the centers of these two circles passes through the midpoint of IS, which is the triangle center X1125 =(2a + b + c : a +2b + c : a + b +2c). Cb Ib X Bc A Ic Cc Bb I S B C Ac Ab Figure 3. 3 The radical axis of (Ib) and (Ic) and the A-altitude intersect at X =(−(b + 2 2 2 c) : SC : SB). The circle with diameter AX has center = (a − 2(b + c) : SC : a A R SB) and radius = 4 tan 2 = 2 (1 − cos A). Its equation 1 a2yz+b2zx+c2xy− (x+y+z)((a2−b2+2bc+3c2)y+(a2+3b2+2bc−c2)z)=0. 4 Points of tangency with the excircles are 2 (Ib): ((b + c) : −s(s − c):(s − a)(s − c)) 2 (Ic): ((b + c) :(s − a)(s − b):−s(s − b)) 3See Footnote 2. 6 N. Dergiades et al The line joining these two points: (s − b)(s − c)x +4(b + c)(cy + bz)=0. Bound a triangle with perspector b + c : ···: ··· . (s − a)(as +2bc) 3.3. The reflection of the circle in the line IbIc 2 2 2 2 2 has center (2s (s − a) + a SA : b SB : c SC ). This means that the circle is also tangent to the circumcircle. s(s − a) a2yz + b2zx + c2xy − (x + y + z)(c2y + b2z)=0. bc 2 2 2 (Ib): ((b + c) s(s − a):−b (s − a)(s − c):c s(s − c)) 2 2 2 (Ic): ((b + c) s(s − a):b s(s − b):−c (s − a)(s − b)) x y z The line joining these is a(s−a) + b(s−b) + c(s−c) =0. a(b−c)(s−a) perspector: a3−a2(b+c)−a(b−c)2+(b−c)3 . 4. The triangle bounded by the polars of the vertices with respect to the excir- cles 4.1. The triangle UVW and its perspector. Consider the triangle bounded by the polars of the vertices of ABC with respect to the corresponding excircles. The polar of A with respect to the excircle (Ia) is the line BaCa; similarly for the other two polars.
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