197 14th International Conference on Geometry and Graphics August 5-9, 2010, Kyoto ISBN 978-4-9900967-1-7 ©2010 International Society for Geometry and Graphics EQUIOPTIC POINTS OF A TRIANGLE Boris ODEHNAL Vienna University of Technology, Austria ABSTRACT: The locus of points where two non-concentric circles c1 and c2 are seen under equal angles is the equioptic circle e. The equioptic circles of the excircles of a triangle ∆ have a common radical axis r. Therefore the excircles of a triangle share up to two real points, i.e., the equioptic points of ∆ from which the circles can be seen under equal angles. The line r carries a lot of known triangle centers. Further we find that any triplet of circles tangent to the sides of ∆ has up to two real equioptic points. The three radical axes of triplets of circles containing the incircle are concurrent in a new triangle center. Keywords: triangle, excircle, incircle, equioptic circle, equioptic points, center of similitude, radical axis. ................................................... ........................................ 1. INTRODUCTION I2 Let there be given a triangle ∆ with vertices C A, B, and C. The incenter shall be denoted S2 I1 by I, the incircle by Γ. The excenters are la- beled with I1, I2, and I3. We assume that I1 is S1 opposite to A, i.e., it is the center of the excir- cle Γ touching the line [B, C] from the outside I 1 A B of ∆, cf. Fig. 1. Sometimes it is convenient to S3 number vertices as well as sides of ∆: The side (lines) [B, C], [C, A], [A, B] shall be the first, second, third side (line) and A, B, C shall be the first, second, third vertex, respectively. Ac- cording to [1, 2] we denote the centers of ∆ with Xi. For example the incenter I is labeled with X1. I3 The set of points where two curves can be seen under equal angles is called equioptic Figure 1: Notations in and around the triangle ∆. curve, see [3]. It is shown that any pair (c1, c2) of non-concentric circles has a circle e for its equioptic curve [3]. The circle e is the Thales sector of the centers of c1 and c2, provided that circle of the segment bounded by the inter- c1 and c2 are not concentric. The four circles Γ, Γi (with i 1, 2, 3 ) nal and external centers of similitude of either ∈ { } given circle, i.e., the center of e is the midpoint tangent to the sides of a triangle ∆ can be ar- of the two centers of similitude, see Fig. 2. In ranged in six pairs and thus they define that case of two congruent circles e becomes the bi- much equioptic circles. Among them we find four triplets of equioptic circles which have a common radical axis instead of a radical cen- 2 EQUIOPTIC CIRCLES OF THE EX- ter, i.e., the three circles of such a triplet form CIRCLES a pencil of circles. These shall be the contents In order to construct the equioptic circles of of Sec. 2 and Sec. 3. a pair of excircles we determine the respective centers of similitude. First we observe that the internal centers of similitude of Γi and Γj is c1 3 e the k-th vertex of ∆, where (i, j, k) I := (1, 2, 3), (2, 3, 1), (3, 1, 2) . Second∈ we have to{ find the external centers} of similitude. For S2 S1 any pair the -th side of is an ex- c2 (Γi, Γj) k ∆ terior common tangent of both Γi and Γj, re- spectively, and thus [Ii,Ij] and the k-th side of ∆ meet in the external center of similitude Sij of Γi and Γj. Now we are able to show a first result: Figure 2: Equioptic circle of c1 and c2. Corollary 1. The external centers of similitude of the ex- We use homogeneous trilinear coordinates Sij circles and of a triangle are collinear. of points and lines, respectively. The ho- Γi Γj ∆ The line carrying these points is the polar of mogeneous triplet of real (complex) numbers X with respect to ∆ and the polar line with re- (x : x : x ) are said to be the homoge- 1 0 1 2 spect to the excentral triangle of ∆ at the same neous trilinear coordinates of a point X if x i time. are the oriented distances of X with respect to the sides [B, C], [C, A], and [A, B] up to a Proof. We construct the polar line of the in- common non vanishing factor, see. [1]. When center X1 with respect to ∆. For that pur- we deal with trilinear coordinates of points ex- pose we project I from C to the line [A, B]. pressed in terms of homogeneous polynomials This gives S3 := [A, B] [I,C]. Then we de- in ∆’s side lengths a = BC , b = CA , ∩ termine a fourth point C′ on [A, B] such that and c = AB we need tok havek a functionk kζ (A, B, S3,C′) is a harmonic quadrupel. The which producesk k a homogeneous function out of four lines [C, A], [C, B], [C, I], and [I1,I2] ob- a homogeneous function by cyclically replac- viously form a harmonic quadrupel and thus ing a b, b c, and c a. However, ζ → → → any line (which is not passing through C) is more powerful. It also applies to homoge- meets these four lines in four points of a har- neous polynomials depending on xi and does monnic quadrupel. So we have C′ = [I1,I2] xi xi (indices mod 3). It also changes ∩ → +1 [A, B] and obviouisly C′ = S12. Cycli- sin A to sin B and similar for other trigonomet- cally shifting labels of points gives S23, S31 ric functions. which are collinear with S12 and lie on the Another function which frequently appears polar of X1. On the other hand we have the will be denoted by σ. It acts on (homoge- harmonic quadruples (I1,I2, C, S12) (cyclic) neous coordinate) triplets in the following way which shows that [S12,S23] is the polar of X1 σ(x0 : x1 : x2) = (x2 : x0 : x1). In this paper with respect to the excentral triangle, see Fig. mappings will be written as superscripts, e.g., 3. σ ζ(X) = Xσζ = Xζσ if applied to points. ◦ Note that ζ σ = σ ζ, provided that xi are The centers Tij of the equioptic circles eij homogeneous◦ functions◦ in a, b, c. are the midpoints of the line segments bounded by Sij and the k-th vertex of ∆ with (i, j, k) ∈ 2 3 I . Their trilinears are ρijof the equioptic circles cij and find S12 = ( 1 : 1 : 0), ρ12 = dist(C, T12) = dist(S12,T12) − ab C S = Sσ ,S = Sσ . = sin , (2) 23 12 31 23 a b 2 − ρ23 = ζ(ρ12), ρ31 = ζ(ρ23). The centers Tij are thus By means of the distance formula from [1, p. T = (c : c : a b), 12 − − 31] or equivalently by means of the more com- σζ σζ (1) T23 = T12 ,T31 = T23 , plicated equation for a circle given by center and radius from [1, p. 223] we write down the and we can easily prove: equations of the equioptic circles Corollary 2. e : c x2 + c x2+ 12 − A 0 B 1 The centers Tij of the equioptic circles eij of +(1 cC)x2(x1 x0) − − (3) any pair (Γi, Γj) of excircles of ∆ are collinear. +(c c )x x = 0, B − A 0 1 e23 = ζ(e12), e31 = ζ(e23), where cA, cB, and cC are shorthand for cos A, S31 S23 S12 cos B, and cos C, respectively. Now it is easily I1 verified that the following holds: C I2 B Theorem 1. X1 1. The three equioptic circles of the excircles A of generic triangle ∆ have a common radical I3 axis r and thus they have up to two common Figure 3: Centers of similtude and harmonic real points, i.e., the equioptic points of the ex- quadruples. circles from which the excircles can be seen un- der equal angles. 2. The radical axis r contains X4 (ortho cen- Proof. The coordinate vectors of Tij given in ter), X9 (Mittenpunkt), X10 (Spieker center), (1) are linearly dependent. and further Xi with Remark 1. i 19, 40, 71, 169, 242, 281, 516, 573, ∈ { The line t connecting any Tij with any Tjk 966, 1276, 1277, 1512, 1542, 1544, (with (i, j, k) I3) has trilinear coordinates 1753, 1766, 1826, 1839, 1842, 1855, ∈ [λ0 : λ1 : λ2], where λ0 = a( a + b + c) and 1861, 1869, 1890, 2183, 2270, 2333, i − λi = ζ (λ0) with i 0, 1, 2 . Obviously t is 2345, 2354, 2550, 2551, 3496, 3501 . ∈ { } } the polar of X55 with respect to ∆. The center Proof. 1. Let P1 := µe12 + νe23, P2 := µe23 + X55 is the center of homothety of the tangential triangle, the intangent triangle, and the extan- νe31, and P3 := µ e31 + νe12 be the equations gent triangle, see [1]. Further it is the internal of the conic sections in the pencils panned by center of similitude of the incircle and the cir- any pair of equioptic circles eij. We compute cumcircle of the base triangle. the singular conic sections in the pencils and find that for all pencils the real singular conic We use the formula for the distance of two sections consist of the ideal line ω : ax0 + points given by their actual trilinear coordi- bx1 + cx2 and the line nates given in [1, p. 31] and compute the radii (b c) c x = 0, (4) − A 0 cyclicX 3 which is the radical axis of any pair of equiop- tic circles.