FORUM GEOM ISSN 1534-1178
Some Triangle Centers Associated with the Circles
Tangent to the Excircles
Boris Odehnal
Abstract. We study those tritangent circles of the excircles of a triangle which enclose exactly one excircle and touch the two others from the outside. It turns out that these three circles share exactly the Spieker point. Moreover we show that these circles give rise to some triangles which are in perspective with the base triangle. The respective perspectors turn out to be new polynomial triangle centers.
1. Introduction
LetT := ABC be a triangle in the Euclidean plane, and Γa,Γb,Γcits excircles, lying opposite toA, B, C respectively, with centers Ia,Ib,Ic and radiira,rb,rc. There are eight circles tangent to all three excircles: the side lines ofT (considered as circles with infinite radius), the Feuerbach circle (see [2, 4]), the so-called Apol-lonius circle (enclosing all the three excircles (see for example [3, 6, 9]), and three remaining circles which will in the following be denoted byKa,Kb,Kc. The circle Kais tangent toΓaand externally toΓbandΓc; similarly forKbandKc. The radii
of these circles are computed in [1]. These circles have the Spieker centerX10as a common point. In this note we study these circles in more details, and show that the triangle of contact pointsKa,aKb,bKc,cis perspective withT. Surprisingly, the triangleMaMbMc of the centers these circles is also perspective withT.
2. Main results
The problem of constructing the circles tangent to three given circles is well studied. Applying the ideas of J. D. Gergonne [5] to the three excircles we see that the construction of the circlesKaetc can be accomplished simply by a ruler. Let Ka,b be the contact point of circle Γa with Kb, and analogously define the remaining eight contact points. The contact pointsKa,a,Kb,a,Kc,aare the inter-sections of the excirclesΓa,Γb,Γcwith the lines joining their contact points with the sidelineBC to the radical center radical center of the three excircles, namely, the Spieker point
X10= (b + c : c + a : a + b)
Aa Ba Ca Ab Bb Ac Bc Cc X10 A B C Ia Ib Ic Ka,a Kb,a Kc,a
Figure 1. The circleKa
Lets := 12(a + b = c) be the semiperimeter. The contact points of the excircles with the sidelines are the points
Aa=(0 : s − b : s − c), Ba=(−(s − b) : 0 : s), Ca =(−(s − c) : s : 0);
Ab=(0 : −(s − a) : s), Bb=(s − a : 0 : s − c), Cb =(s : −(s − c) : 0);
Ac=(0 : c : −(s − a)), Bc =(s : 0 : −(s − b)), Cc=(s − a : s − b : 0).
A conic is be represented by an equation in the formxTMx = 0, where xT =
A B C Ia Ib Ic Ma Ka,a Kb,a Kc,a Mb Ka,b Kb,b Kc,b Mc Ka,c Kb,c Kc,c X10 Figure 2.
pointX, and M is a symmetric 3 × 3-matrix. For the excircles, these matrices are Ma= ⎛ ⎝ s 2 s(s − c) s(s − b) s(s − c) (s − c)2 −(s − b)(s − c) s(s − b) −(s − b)(s − c) (s − b)2 ⎞ ⎠ , Mb = ⎛ ⎝ (s − c) 2 s(s − c) −(s − a)(s − c) s(s − c) s2 s(s − a) −(s − a)(s − c) s(s − a) (s − a)2 ⎞ ⎠ , Mc = ⎛ ⎝ (s − b) 2 −(s − a)(s − b) s(s − b) −(s − a)(s − b) (s − a)2 s(s − a) s(s − b) s(s − a) s2 ⎞ ⎠ .
It is elementary to verify that the homogeneous barycentrics of the contact points are given by:
Kb,c = ((bs + ac) : −a s(s − c) : (a + b) (s − a)(s − c));
Kc,a = ((b + c)2(s − a)(s − b) : (cs + ab)2 : −b2s(s − a)),
Kc,b = ((cs + ab)2 : (c + a)2(s − a)(s − c) : −a2s(s − b)),
Kc,c = (b2s(s − a) : a2s(s − b) : −(a + b)2(s − a)(s − b)).
Theorem 1.
The triangleKa,aKb,cKc,cof contact points is perspective withT at a point with homogeneous barycentric coordinates
s − a a2 : s − b b2 : s − c c2 . (2)
Proof. The coordinates ofKa,a,Kb,b,Kc,ccan be rewritten as Ka,a = −(b+c)2b(s−b)(s−c)2c2s : s−bb2 : s−cc2 , Kb,b = s−a a2 : −(c+a) 2(s−a)(s−c) c2a2s : s−cc2 , Kc,c = s−a a2 : s−bb2 : −(a+b) 2(s−a)(s−b) a2b2s . (3)
From these, it is clear that the lines AKa,a, BKb,b, CKc,c meet in the point
given in (2).
Remark. The triangle centerPKis not listed in [7]. Theorem 2.
The linesAKa,a,BKa,b, andCKa,care concurrent.
Proof. The coordinates of the pointsKa,a,Ka,b,Ka,ccan be rewritten in the form Ka,a = −(b+c)2b(s−b)(s−c)2c2s : s−bb2 : s−cc2 ) , Ka,b = −s(s−b)(s−c)(as+bc)2 : (c+a) 2(s−b)(s−c)2 c2(as+b2)2 : s−cc2 , Ka,c = −s(s−b)(s−c)(as+bc)2 : s−bb2 : (a+b) 2(s−b)2(s−c) b2(as+bc)2 . (4)
From these, the linesAKa,a,BKa,b, andCKa,cintersect at the point
−s(s − b)(s − c) (as + bc)2 : s − b b2 : s − c c2 .
Theorem 3.
The triangleM1M2M3 is perspective withT at the point
1 −a5− a4(b + c) + a3(b − c)2+ a2(b + c)(b2+ c2) + 2abc(b2+ bc + c2) + 2(b + c)b2c2 : · · · : · · · . A B C Ia Ib Ic Ma Ka,a Kb,a Kc,a Mb Ka,b Kb,b Kc,b Mc Ka,c Kb,c Kc,c X10 PM Figure 3.
Proof. The center of the circleKais the point
Ma= −2a4(b + c) − a3(4b2+ 4bc + 3c2) + a2(b + c)(b2+ c2) − (b + c − a)(b2− c2)2 : −c5− c4(a + b) + c3(a − b)2+ c2(a + b)(a2+ b2) + 2abc(a2+ ab + b2) + 2a2b2(a + b) : −b5− b4(c + a) + b3(c − a)2+ b2(c + a)(c2+ a2) + 2abc(c2+ ca + a2) + 2c2a2(c + a).
[3] N. Dergiades and J. C. Salazaar, Some triangle centers associated with the tritangent circles,
Forum Geom., 9 (2009) 259–270.
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appl., 4 (1813–1814) 349–359.
[6] D. Grinberg and P. Yiu, The Apollonius circle as a Tucker circle, Forum Geom., 2 (2002) 175– 182.
[7] C. Kimberling, Encyclopedia of Triangle Centers, available at:
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.
[8] C. Kimberling, Triangle centers and central triangles, Congressus numerantium, 128 (1998) 1–295.
[9] M. R. Stevanovi´c, The Apollonius circle and related triangle centers, Forum Geom., 3 (2003) 187–195.
Boris Odehnal: Vienna University of Technology, Institute of Discrete Mathematics and Geome-try, Wiedner Hauptstraße 8-10, A-1040 Wien, Austria
