Equioptic Points of a Triangle

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 labeled with I1, I2, and I3. We assume that I1 is S1 opposite to A, i.e., it is the center of the excircle 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, respectively, 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 under 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 equioptic circles. 2. In [1, pp. 64 ff.] we ﬁnd Γ1 X4 = [cosec A : cosec B : cosec C] and X9 = [b+c a, c+a b, a+b c] and obviously these coordinate− vectors− annihilate− Eq. (4). By I3 I2 inserting the trilinears of the other points mentioned in the theorem we proof the incidence. E Γ2 The trilinears of points Xi with i 360 can be found in [1] whereas the trilinears≤ for i > 360 e31 can be found in [2]. e23

In Fig. 6 the equioptic circles as well as the I1 equioptic points of an acute triangle are de- e12 Γ picted. Fig. 4 shows some of the centers men- 3 tioned in Th. 1 located on r.

X1542 X1766 X40 C X1855 e31 Figure 5: The only equioptic point of an equilateral X2551 X573 triangle. X966 X1277 e12 X2550 X71 X3501 The case of a single equioptic point is il- X3496 X169 lustrated in Fig. 8 at hand of an isosceles tri- X9 X2345 E1 X1826 angle. It is easily shown that in this case X1869 X10 e23 ∠ √ X1842 X1839,X1890 one has to choose ACB = 2 arcsin( 3 A B − 1) 94.11719432◦ in order to have a unique ≈ X4 equioptic point. Thus the triangle is obtuse. X19 The unique (real) equioptic angle now equals 3 √3 X2333 X2354 X1753 X2270 X242 X1861 2 arcsin − 78.68794716◦. 2 ≈ A generic triangle ∆ has a unique equioptic Figure 4: Some of the triangle centers on the radical point if and only if axis, cf. Th. 1. 6a2b2c2 + 4 a3b(b2 ac) = − cyclicX The circles in a pencil of circles can share = (a6 + 2aa5 a4(a2 +b 4bc)). two real points, one real point with multiplic- − cyclicX ity two, or no real points. Thus a triangle has b b either two equioptic points, or a single equiop- Here and in the following we use the abbre- tic point. viations a = b + c, b = c + a, and c = a + b. We skip the proof since it is merely based on b Remark 2. simple andb straight foreward computations.b In case of an equilateral triangle ∆ there is only one equioptic point E that coincides with 3 EQUIOPTIC CIRCLES OF THE IN- the center of ∆. The three equioptic circle be- CIRCLE AND AN EXCIRCLE come straight lines: eij is the k-th interior an- We recall that the equioptic circles of a pair gle bisector and the -th altitude of . Fig. 5 k ∆ (Γ, Γi) is the Thales circle of the line segment illustrates this case. From E any excircle Γj bounded by the internal and external center of 1 can be seen under arccos 97.180756◦. − 8 ≈ similitude of the incircle Γ and the i-th excircle 4 S31

equioptic point e31

T12 T31

T23

I1 C

B equioptic point I2 e23 A e12

Figure 6: The equioptic circles and equioptic points of a triangle and the radical axis through some triangle centers.

Γ1 C

e31

Γ2 G

e23 Γ B A

Γ3 e1 e12

Figure 7: The six equioptic circles of the incircle and the excircle, the three concurrent radical axes, and the center G from Th. 3.

5 Γi. We observe that the i-th vertex of ∆ is Corollary 4. the external center of similitude of the above The two centers Ti and Tj of equioptic circles given pair of circles. The internal center is the ei, ej of Γ and the i-th and j-th excircle are meet of a common internal tangent, i.e, ∆’s i-th collinear with the center Tij of the equioptic side and the line [I,Ii] connecting the respec- circle eij of Γi and Γj. tive centers. Proof. We simply show dependencies of vectors given in Eqs. (1) and (5). e23 e31 Again we use the formulae given in [1, p. Γ2 E 223] in order to compute the equations of the Γ1 equioptic circles ei of the incircle Γ and the i- I2 I1 th excircle Γi. Note that these homogeneous equations can always be written in the form T ei : x Ai x = 0 with a regular and symmetric · · 3 3-matrix Ai. The matrix A reads AB × 1 e12 2 2a s(a s) 2abs(a s) 2acs(a s) 2abs(a − s)(⋆⋆)(− ⋆)− , 2acs(a − s)(⋆)(⋆ ⋆ ⋆) (6) Γ3 − A2 = ζ(A1),A3 = ζ(A2), I3 where (⋆) = 2b2c2 + 3b3c + 3bc3 3a2bc + 2b4 + 2c4 2a2b2 2a2c2, (⋆⋆) =−b(a2(4c − − − Figure 8: An iscosceles triangle with a unique b) + a2b 8b2c), and (⋆ ⋆ ⋆) = c(2bc2 + 7b2c − − equioptic point. c3 + a2c + 4b3). Here s = (a + b + c)/b2 is the halfperimeter of ∆. Consequently the internal centers of simili- Theorem 2. tude are the points Si (i 1, 2, 3 ). We have ∈ { } The equioptic circles ei, ej, and eij deﬁned by σ σ the incircle Γ and the excircles Γ , Γ have a S1 = (0 : 1 : 1),S2 = S1 ,S3 = S2 . i j 3 common radical axis rk (with (i, j, k) I ) ∈ As a consequence of Cor. 1 we have: and thus Γ, Γi, and Γj have up to two real equioptic points. Corollary 3. The two internal centers Si, Sj of similitude of Γ and Γi, Γj are collinear with the exter- Proof. With Eq. (3) and (6) we compute the nal center Sij of similitude of Γi and Γj for radical axis rk of Γ, Γi, and Γj (where (i, j) (1, 2), (2, 3), (3, 1) . (i, j, k) I3) as the singular conic sections in ∈ { } the pencil∈ of conics spanned by either two cir- Proof. The collinearity is easily checked by cles, cf. the proof of Th. 1. The radical axis r3 showing the linear dependency of the respec- is given by tive coordinate vectors. 5 2 4 2 2 3 r3 = [ ba (ba + 2bc)a + (ba b + c(2ba bc))a − b−2 b2 2 2 2b2 − The centers Ti of equioptic circles ei of Γ +a (a + 6c )a + c a (b + 4c): 5 b2 4 b2 b2 3 : ab + 2(b + 2ac)b (ab + c(2b ac))b and Γ are the midpoints of ∆’s i-th vertex and b2 b2 2 2− 2b2 −b i b (b + 6c )b c b (a + 4c)b : (7) − 5 4− 3 3 Si. Thus we have :(b a)c + 4(a b)ccb + (a b)(7cb 3ab)c − +4(a− b)cb(cb2 + ab−)c2 − σζ +7ab(a b)−cb2c + 4a2b2(a b)cb)]. T1 = (b + c : a : a),Ti+1 = Ti . (5) − − σζ σζ Now we observe the following: Finally we have r1 = r3 and r2 = r1 .

6 equioptic point Available at: http://faculty.evansville.edu/ Γ1 ck6/encyclopedia/ECT.html C [3] B. ODEHNAL: Equioptic curves of conic

e2 sections. J. Geom. Graphics 14/1 (2010), 29–43. T1 T12 Γ2 Γ T2 ABOUT THE AUTHOR ABe1 Boris ODEHNAL, PhD, is an assistent pro- fessor at the Vienna University of Technology. equioptic point E-mail address: [email protected] r3 e12 Postal address: Institute of Discrete Mathematics and Geome- Γ Γ Γ Figure 9: Equioptic circles and points of , 1, 2. try, Vienna University of Technology, Wiedner Hauptsraße 8–10, A-1040, Vienna, Austria.

A certain triplet of equioptic circles is shown in Fig. 9. Finally we have:

Theorem 3. The three radical axes rk (cf. Th. 2) are concurrent in a new triangle center.

Proof. The homogeneous coordinate vectors of the lines ri given in (7) are linearly dependent. This proves the concurrency. We compute the meet G = (g0 : g1 : g2) of any pair (ri, rj) of radical axes and ﬁnd

5 2 2 4 2 2 2 g0 = bcba (b c) + 2bcba (2ba 10ba bc + 5b c )a − − +ba3(ba4 8ba2bc + 4b2c2)a2 − 2(ba6 + 3bcba4 10b2c2ba2 + b3c3)a − 4 − 2 2 2 4 (8) ba(8ba 23bcba + 4b c )a − − 2(ba4 8bcba2 + 5b2c2)a5 − − +ba(7ba2 4bc)a6 + 2(2ba2 bc)a7. − −

Since g1 = ζ(g0) and g2 = ζ(g1) we ﬁnd that G is a center of ∆ which is not mentioned in [2] and shown in Fig. 7.

REFERENCES [1] C. KIMBERLING: Triangle Centers and Central Triangles. Congressus Numeran- tium, 129, Canada, 1998.

[2] C. KIMBERLING: Encylcopedia of Tri- angle Centers and Central Triangles.