Some triangle centers associated with the circles tangent to the excircles

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1 Some triangle centers associated with the circles tangent to the excircles Boris Odehnal February 19, 010 Abstract We study those tritangent circles of the excirlces 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 Let := {A, B, C} be a triangle in the Euclidean plane. Let Γ i be the excircles with centers I i and radii ρ i with i {1,, }. We assume that Γ 1 lies opposite to A. There are eight circles tangent to all three excircles: the side lines of (considered as circles with infinitely large radius), the Feuerbach circle (cf. [, 4]), the so-called Apollonius circle (enclosing all the three circles Γ i, see for example [, 6, 9]) and three remaining circles which will in the following be denoted by K i. Any circle K i encloses a circle Γ i and touches the remaining circles from the outside. Until now only their radii are known, see [1]. In the following we focus on the circles K i and show that these circles share the Spieker point X 10. Further the triangle of contact points {K 1,1, K,, K, } of Γ i and K i is perspective to. Surprisingly the triangle built by the centers M i of circles K i is also perspective to. Main results The problem of constructing the tritangent circles to three given circles is well studied. Applying the ideas of M. Gergonne [5] to the three excircles Γ i we see that the construction of the tritangent circles K j is linear. At first we find the contact points K i,j of Γ i with K j (see Fig. 1): We project the contact points I i,j of the i-th excircle with the j-th side 1 of from the radical center of the excircles to the i-th excircle. The radical center of Γ i is the Spieker point, cf. [8]. Knowing the contact points of K i,j it is easy to find the centers M j of K j. 1 The lines [B, C], [C, A], [A, B] are the first, second, and third side of. 1

2 I Γ Γ 1 C I 1 K, X 10 I, A I, B K 1, I 1, K Γ K, I M Figure 1: Linear construction of the tritangent circle enclosing Γ. By simple and straight foreward computation following the construction of the contact points K i,j of either circle K i with the excircle Γ j we can show: Theorem.1 The triangle {K 1,1, K,, K, } built by the contact points of K i with the i-th excircle Γ i is in perspective with, see Fig.. The perspector P K is has homogeneous barycentrics ( s a P K = a : s b b : s c ) c. (1) Proof: We denote the contact points of Γ i with the j-th side of by I i,j. Their homogeneous barycentrics are I 1,1 =( 0 : s b : s c), I 1, =( b s : 0 : s), I 1, =( c s : s : 0), I,1 =( 0 : a s : s), I, =( s a : 0 : s c), I, =( s : c s : 0), I,1 =( 0 : c : a s), I, =( s : 0 : b s), I, =( s a : s b : 0), () where a = BC, b = CA, c = AB are the side lengths of and s = (a + b + c)/ is the half of the perimeter. Note that K i,j = [I i,j, X 10 ] Γ j \ I i,j and the Spieker point is given by X 10 = (b + c : c + a : a + b),

3 K Γ P K K, K 1,1 K,1 C K 1, K, A K, K.1 B K 1, Γ 1 K 1 Γ K K, Figure : Two triangles in perspective with : the triangle of contact points K i,i, the center P K from Th..1, and the triangle mentioned in Th... cf. [7, 8]. In the following we use the abbreviations â := b + c, b = a + c, and ĉ = a + b. The equations of the incircles in terms of homogeneous barycentric coordinates can be written in the form x T B i x = 0, where x T = (x 0, x 1, x ) is the vector collecting the homogeneous barycentric coordinates of a point X and B i is a (regular) symmetric -matrix. The three coefficient matrices are B 1 = B = B = 4 s s(s c) s(s b) s(s c) (s c) (s b)(s c) s(s b) (s b)(s c) (s b) 4 4 (s c) s(s c) (s a)(s c) s(s c) s s(s a) (s a)(s c) s(s a) (s a) (s b) (s a)(s b) s(s b) (s a)(s b) (s a) s(s a) s(s b) s(s a) s It is elementary to verify that the homogeneous barycentrics of the contact 5, 5, 5. ()

4 points K i,j are given by: K 1,1 =( ba (s b)(s c) : c s(s b) : b s(s c)), K 1, =( c s(s b) : b (s b)(s c) : (as + bc) ), K 1, =( b s(s c) : (as + bc) : bc (s b)(s c)), K,1 =( ba (s a)(s c) : c s(s a) : (bs + ac) ), K, =( c s(s a) : b (s a)(s c) : a s(s c)), K, =( (bs + ac) : a s(s c) : bc (s a)(s c)), K,1 =( ba (s a)(s b) : (cs + ab) : b s(s a)), K, =( (cs + ab) : b (s a)(s c) : a s(s b)), K, =( b s(s a) : a s(s b) : bc (s a)(s b)). (4) Now it is easily verified that the three vectors A K 1,1, B K,,, and C K, are linearly dependent und thus the respective lines meet in the point given in (1). Obviously the point P K is a triangle center which is not listed in [7]. Moreover with the computations from the previous proof we can show: Theorem. The lines [K 1,1, A], [K 1,, B], [K 1,, C] are concurrent. The lines [K,1, A], [K,, B], [K,, C] are concurrent. The lines [K,1, A], [K,, B], [K,, C] are concurrent. Proof: Use Eq. (4) compute the barycentric coordinates of the lines and show the linear dependency. Remark: The loci of concurrency of the three triplets of lines mentioned in Th.. are three points building a triangle which is itself in perspective with. The pespector is the point P K, cf. Fig.. Theorem. The triangle {M 1, M, M } built by the centers of the circle K 1, K, and K is in perspective with, see Fig.. The perspector is the point P M with homogeneous barycentrics (α : β : γ), where α = (a 5 + a 4 â a (b c) a â(b + c ) abc(â bc) âb c ) 1. (5) Proof: The center M of the tritangent circle K (enclosing Γ ) is found as the meet of two straight lines M = [I 1, K 1, ] [I, K, ]. Analogously we find the remaining centers. Their barycentric coordinates are thus M 1 = ( baa 4 (ba bc)a + ba(b + c )a + ba (b c) a ba (b c) : : (ba + b )a + ba(b + c )a + bac(b c)(b + c)a + ba c (b c) : : (ba + c )a + ba(b + c )a bab(b c)(b + c)a ba b (b c)), M = (( b b + a )b + b b(a + c )b + b bc(a c)(a + c)b + b b c (a c) : : b bb 4 ( b b ac)b + b b(a + c )b + b b (a c) b b b (a c) : : ( b b + c )b + b b(a + c )b b ba(a c)(a + c)b b b a (a c)), (6) M = ((bc + a )c + bc(a + b )c + bcb(a b)(a + b)c + bc b (a b) : : (bc + b )c + bc(a + b )c bca(a b)(a + b)c bc a (a b) : : bcc 4 (bc ab)c + bc(a + b )c + bc (a b) c bc (a b) ). Now the concurrency of the lines [M 1, A], [M, B], and [M, C] is easily shown. The point P M is a center and it is not listed in [7]. 4

5 K M M 1 Γ P M Γ 1 K 1 M Γ K Figure : The three tritangent circles, the triangle of the respective centers, and the perspector P M. Theorem.4 The three tritangent circles K i enclosing the i-th excircle of and touching the others from the exterior share exactly the Spieker point X 10 of. Proof: We switch to trilinear coordinates of points and use the formula for the disctance between two points given in terms of their respective actual trilinear coordinates (cf. [8, p. 1] and find the radii R i of the tritangent circles K i âr 1 = br = ĉr = s(ab + bc + ca)/4f R, (7) where R is the circumradius of and F is the area of. Eq. (7) is a relation between the radii which is equivalent to those given in [1]. The equations of the circles K i in terms of homogeneous barycentrics are determined uniquely (up to a non-vanishing scalar) according to [8, p. ] and can thus be written in the form x T C i x = 0 with the following (regular) symmetric 5

6 -matrices [ s(b + c + aba) C 1 = ba(c a ) ab ba(b a ) ac [ 4 b b(s b)(s c) C = b b(c b ) a b b b bs [ 4bc(s b)(s c) C = cbcs bc(b c ) a c ba(c a ) ab 4ba(s a)(s c) abas b b(c b ) a b s(a + c + b b b) b b(a b ) bc cbcs 4bc(s a)(s c) bc(a c ) b c ] ba(b a ) ac abas, 4ba(s a)(s b) b b ] bs b b(a b ) bc, 4 b b(s a)(s b) ] bc(b c ) a c bc(a c ) b c. s(a + b + cbc) (8) By substituting the barycentrics of X 10 into the equations of K i we prove that the Spieker point is a common point of the three circles K i. References [1] A. Aeppli: Das Taktionsproblem von Apollonius angewandt auf die vier Berührungskreise eines Dreiecks. Elem. Math. 1 (1958), 5 0. [] C.B. Boyer: A History of Mathematics. J. Wiley, New York, [] N. Dergiades & J.C. Salazaar: Some triangle centers associated with the tritangent circles. Forum Geometricorum 9 (009), [4] K.W. Feuerbach: Eigenschaften einiger merkwürdigen Punkte des geradlinigen Dreiecks und mehrerer durch sie bestimmten Figuren. PhD thesis, Nürnberg, 18. [5] M. Gergonne: Recherche du cercle qui en touche trois autres sur une sphère. Ann. math. pures appl. 4 ( ), [6] D. Grinberg & P. Yiu: The Apollonius circle as a Tucker circle. Forum Geometricorum (00), [7] C. Kimberling: Encyclopedia of Triangle Centers. Available at: [8] C. Kimberling: Triangle centers and central triangles. Congressus numerantium 18, Utilitas Mathematica Publishing Inc., Winnipeg, [9] M.R. Stevanović: The Apollonius Circle and Related Triangle Centers. Forum Geometricorum (00), Boris Odehnal Vienna University of Technology Institute of Discrete Mathematics and Geometry Wiedner Hauptstraße 8-10 A-1040 Wien, Austria boris@geometrie.tuwien.ac.at 6