Homology and Orthology with Triangles for Central Points of Variable Flanks

Journal for Geometry and Graphics Volume 7 (2003), No. 1, 1–21.

Zvonko Cerinˇ Kopernikova 7, HR 10010 Zagreb, Croatia email: [email protected]

Abstract. Here we continue from the paper [1] our study of the following geometric conﬁguration. Let BR1R2C, CR3R4A, AR5R6B be rectangles build on sides of a triangle ABC such that oriented distances BR , CR , AR are | 1| | 3| | 5| λ BC , λ CA , λ AB for some real number λ. We explore the homology and | | | | | | orthology relation of the triangle on central points of triangles AR4R5, BR6R1, CR2R3 (like centroids, circumcenters, and orthocenters) and several natural triangles associated to ABC (as its orthic, anticomplementary, and complementary triangle). In some cases we can identify which curves trace their homology and orthology centers and which curves envelope their homology axis. Key Words: triangle, extriangle, ﬂanks, central points, Kiepert parabola, Kiepert hyperbola, homologic, orthologic, envelope, anticomplementary, complementary, Brocard, orthic, tangential, Euler, Torricelli, Napoleon MSC 2000: 51N20

1. Introduction

This paper is the continuation of the author’s preprint [1] where an improvement of three recent papers [3], [6], and [7] by L. Hoehn, F. van Lamoen, and C.R. Pranesachar and B.J. Venkatachala was presented. These articles considered independently the classical geometric conﬁguration with squares BS1S2C, CS3S4A, and AS5S6B erected on sides of a triangle ABC and studied relationships among central points (see [4]) of the base triangle τ = ABC and of three interesting triangles τA = AS4S5, τB = BS6S1, τC = CS2S3 (called ﬂanks in [6] and extriangles in [3]). In order to describe their main results, recall that triangles ABC and XY Z are homologic provided lines AX, BY , and CZ are concurrent. The point P in which they concur is their homology center and the line ` containing intersections of pairs of lines (BC, Y Z), (CA, ZX), and (AB, XY ) is their homology axis. In this situation we P use the notation ABC ./ XY Z where ` or both ` and P can be omitted. Let Xi = Xi(τ), ` j A B C Xi = Xi(τj) (for j = A, B, C), and σi = Xi Xi Xi , where Xi (for i = 1, . . .) is any of the triangle central point functions from Kimberling’s lists [4] or [5].

ISSN 1433-8157/$ 2.50 c 2003 Heldermann Verlag ° 2 Z. Cerin:ˇ Homology and Orthology with Triangles Instead of homologic, homology center, and homology axis many authors use the terms perspective, perspector, and perspectrix. Also, it is customary to use letters I, G, O, H, F, K, and L instead of X1, X2, X3, X4, X5, X6, and X20 to denote the incenter, the centroid, the circumcenter, the orthocenter, the center of the nine-point circle, the symmedian (or Grebe- Lemoine) point, and the de Longchamps point (the reﬂection of H about O, respectively.

Xj In [3] Hoehn proved τ ./ σ3 and τ ./ σi for (i, j) = (1, 1), (2, 4), (4, 2). In [7] C.R. Pranesachar and B.J. Venkatachala add some new results because they show that Xj X τ ./ σi for (i, j) = (1, 1), (2, 4), (4, 2), (3, 6), (6, 3). Moreover, they observe that if τ ./ XAXBXC and Y, YA, YB, and YC are isogonal conjugates of points X, XA, XB, and XC with Y respect to triangles τ, τA, τB, and τC, respectively, then τ ./ YAYBYC. Finally, they also answer in negative the question by Prakash Mulabagal of Pune if τ ./ XY Z, where X, Y , and Z are points where incircles of triangles τA, τB, and τC touch the sides opposite to A, B, and C, respectively.

Xj Xj In [6] van Lamoen says that Xi befriends Xj when τ ./ σi and shows first that τ ./ σi Xn implies τ ./ σm where Xm and Xn are isogonal conjugates of Xi and Xj. Also, he proves that Xj Xi Xj τ ./ σi is equivalent to τ ./ σj and that τ ./ σi for (i, j) = (1, 1), (2, 4), (3, 6), (4, 2), (6, 3). K( π −φ) Then he notes that τ .2/ K(φ), where K(φ) denotes the homology center of τ and the Kiepert triangle formed by apexes of similar isosceles triangles with the base angle φ erected Xi on the sides of ABC. This result implies that τ ./ σi for i = 485, 486 (Vecten points — for X π j φ = 4 ) and τ ./ σi for (i, j) = (13, 17), (14, 18) (isogonic or Fermat points X13 and X14 § π π — for φ = and Napoleon points X and X — for φ = ). Finally, van Lamoen § 3 17 18 § 6 observed that the Kiepert hyperbola (the locus of K(φ)) befriends itself; so does its isogonal transform, the Brocard axis OK. The idea of our generalization in [1] was in replacing squares with rectangles whose ratio of nonparallel sides is constant (see Fig. 1). More precisely, let BR1R2C, CR3R4A, AR5R6B be rectangles build on sides of a triangle ABC such that the oriented distances BR1 , CR3 , λ | λ | | | AR5 are λ BC , λ CA , λ AB for some real number λ. Let τA = AR4R5, τB = BR6R1, | |λ | | | | j| | λ and τC = CR2R3 and let Xi (λ) (for j = A, B, C) and σi have obvious meaning. The most important central points have their traditional notations so that we shall often use these because they might be easier to follow. For example, H A(λ) is the orthocenter of the flank λ λ A B C τA and σG is the triangle G (λ)G (λ)G (λ) on the centroids of flanks.

Since triangles AS4S5 and AR4R5 are homothetic and the vertex A is the center of this homothety (and similarly for pairs BS6S1, BR6R1 and CS2S3, CR2R3) we conclude that A A B B C C A, Xi , Xi (λ) , B, Xi , Xi (λ) , and C, Xi , Xi (λ) are sets of collinear points so that { } { } { } λ most statements from [3], [7], and [6] concerning triangles σi are also true for triangles σi . But, since instead of a single square on each side we have a family of rectangles it is possible to get additional information. The results in [1] explored cases when the base triangle τ is λ either homologic or orthologic with the triangles σi . The purpose of this paper is to investigate the relations of both homology and orthology λ for triangles σi with some important triangles associated to τ like its anticomplementary triangle τa, the ﬁrst Brocard triangle τb, the Euler triangle τE, the complementary triangle τg, the orthic triangle τh, the tangential triangle τt, the Torricelli triangles τu and τv, and the Napoleon triangles τx and τy. Z. Cerin:ˇ Homology and Orthology with Triangles 3

R3 C-ﬂank

C R2 PSfrag replacements

A-ﬂank A B B-ﬂank

R5 R6

Figure 1: The triangle ABC with three rectangles and three ﬂanks

2. The anticomplementary triangle τa

Let τa denote the anticomplementary triangle AaBaCa of ABC whose vertices are intersections of parallels to sidelines through opposite vertices.

λ Theorem 1 For every λ R the triangles τa and σG are homologic and their homology ∈ centers trace the Kiepert hyperbola of τa (see Fig. 2). Proof: In our proofs we shall use trilinear coordinates. Recall that the actual trilinear coordinates of a point P with respect to the triangle ABC are signed distances f, g, h of P from the lines BC, CA, and AB. We shall regard P as lying on the positive side of BC if P lies on the same side of BC as A. Similarly, we shall regard P as lying on the positive side of CA if it lies on the same side of CA as B, and similarly with regard to the side AB. Ordered triples x : y : z of real numbers proportional to (f, g, h) (that is such that x = mf, y = mg, and z = mh, for some real number m different from zero) are called trilinear coordinates of P . The advantage of their use is that a high degree of symmetry is present so that it usually suffices to describe part of the information and the rest is self evident. For example, when we write X1(1) or I(1) or simply say I is 1 this indicates that the incenter has trilinear coordinates 1 : 1 : 1. We gave only the first coordinate while the other two are cyclic permutations of the 1 1 1 1 1 first. Similarly, X2 a or G a say that the centroid has trilinears a : b : c , where a, b, c are the lengths of sides¡of¢ABC.¡ ¢ The expressions in terms of sides a, b, c can be shortened using the following notation. d = b c, d = c a, d = a b, z = b + c, z = c + a, z = a + b, a − b − c − a b c t = a + b + c, t = b + c a, t = c + a b, t = a + b c, a − b − c − m = abc, ma = bc, mb = ca, mc = ab, T = √ttatbtc. 4 Z. Cerin:ˇ Homology and Orthology with Triangles

R2 GC(λ) R3 PSfrag replacements Ba C

Aa R1

R4 I G

B A GB(λ) GA(λ)

Ca R5 R6

λ Figure 2: The homology centers of triangles τa and σG trace the Kiepert hyperbola of τa (Theorem 1)

For an integer n, let t = an + bn + cn and d = bn cn and similarly for other cases. Instead n na − of t2, t2a, t2b, and t2c we write k, ka, kb, and kc. Let ω denote the Brocard angle of ABC. In order to achieve even greater economy in our presentation, we shall describe coordinates or equations of only one object from triples of related objects and use cyclic permutations ϕ and ψ to obtain the rest. For example, the ﬁrst vertex Aa of the anticomplementary triangle A B C of ABC has trilinears 1 : 1 : 1 . Then the trilinears of B and C need a a a − a b c a a not be described because they are easily ﬁgured out and memorized by relations Ba = ϕ(Aa) and Ca = ψ(Aa). One must remember always that transformations ϕ and ψ are not only permutations of letters but also of positions, i.e., ϕ(a, b, c, 1, 2, 3 b, c, a, 2, 3, 1) and → ψ(a, b, c, 1, 2, 3 c, a, b, 3, 1, 2). Therefore, the trilinears of B and C are 1 : 1 : 1 and → a a a − b c 1 : 1 : 1 . a b − c The trilinears of the points R1 and R2 are equal to

2λm : c(T + λk ) : λbk and 2λm : λck : b(T + λk ) − c b − c b

(while R3 = ϕ(R1), R4 = ϕ(R2), R5 = ψ(R1), and R6 = ψ(R2)). It follows that the centroid 3T + 2a2λ k λ k λ XA(λ) or GA(λ) of the triangle AR R is : c : b . 2 4 5 a b c A − Hence, the line AaG (λ) is

3T + 6d z T λ + 3(k + 2b2) T λ + 3(k + 2c2) a a x + a y + a z = 0. a b c − Z. Cerin:ˇ Homology and Orthology with Triangles 5

2 2 2 (T + 2kbkc)λ 6T kaλ 9T It follows that the homology center of τ and σλ is − − . This point a G a traces the conic with the equation

a2(b2 c2)x2 + b2(c2 a2)y2 + c2(a2 b2)z2 = 0 or (in shorter notation) a2d z x2 = 0. − − − a a P Since the vertices of τa, the common centroid G of τ and τa, and the orthocenter of the 2 T +2kbkc anticomplementary triangle X20 a (known also as the de Longchamps point L of ABC) are on this curve, we conclude³ that it´is the Kiepert hyperbola of the anticomplementary triangle.

λ Theorem 2 The homology axis of τa and σG envelope the Kiepert parabola of τa.

B C Proof: The line BaCa has the equation by + cz = 0 while the line G (λ)G (λ) is

2 2 2 a T λ + 6(b + c )λ + 9T x + bλ(T λ + 3kc)y + cλ(T λ + 3kb)z = 0. ¡ ¢ 6d z T λ2 + 6(b2 + c2)λ + 9T T λ2 + 6(b2 + c2)λ + 9T It follows that their intersection is a a : : . a b c λ − Hence, the homology axis of τa and σG has the following equation a[T 2(81 λ4) 6T λ(k + a2)(λ2 + 9) + 18λ2(T 2 4a2(b2 + c2))] x = 0. − − − P It envelopes [a2(k a2)2 x2 + 2bc(T 2 + b2k + c2k + m2) y z] = 0. In order to see that a − c b a this is the KiepPert parabola of τa it suffices to check that lines BaCa, CaAa, AaBa, the line at infinity, and the Lemoine line of τa (the homology axis of τa and its tangential triangle) are its tangents (see [2]). Indeed, by + cz = 0, cz + ax = 0, ax + by = 0, ax = 0, and a3(b2 + c2) x = 0 are their equations. By solving in one variable any of themPand substitutingP into the left hand side of the equation of the above conic we get remaining variables in a complete square which means that these lines have a point of tangency with the conic and our proof is accomplished. Recall that triangles ABC and XY Z are orthologic provided the perpendiculars at vertices of ABC onto sides Y Z, ZX, and XY of XY Z are concurrent. The point of concurrence of these perpendiculars is denoted by [ABC, XY Z]. It is well-known that the relation of orthology for triangles is reflexive and symmetric. Hence, the perpendiculars at vertices of XY Z onto sides BC, CA, and AB of ABC are concurrent at the point [XY Z, ABC]. λ λ Since G befriends H it is clear that triangles τa and σG are orthologic and [σG, τa] = H λ (the orthocenter). Our next result shows that points [τa, σG] trace the Kiepert hyperbola of τa.

λ λ Theorem 3 The locus of the orthology centers [τa, σG] of τa and σG is the Kiepert hyperbola of AaBaCa (see Fig. 3).

B C Proof: The perpendicular from Aa onto the line G (λ)G (λ) has the equation

6ad x + b(T λ + 3k )y c(T λ + 3k )z = 0. 2a c − b 2 2 2 T λ + 6T λka + 9(T 2kbkc) Therefore, the orthology center [τ , σλ ] is − . This point traces the a G a 2 2 conic with the equation a d2ax = 0 that was recognized as the Kiepert hyperbola of τa in the proof of Theorem 1.P 6 Z. Cerin:ˇ Homology and Orthology with Triangles

R2 GC PSfrag replacements R3 (λ)

Ba C Aa

R4 I λ G [τa, σG]

B A GB(λ) GA(λ)

R R5 6 Ca

λ λ Figure 3: The orthology center [τa, σG] of triangles τa and σG traces the Kiepert hyperbola of τa (Theorem 3)

Ba C Aa

PSfrag replacements R3

R2 GC(λ)

R5 R6 G line GK K

GA(λ) GB(λ) A B R4 R1

λ Figure 4: The homology centers of τa and σO trace the line GK (Theorem 4) Z. Cerin:ˇ Homology and Orthology with Triangles 7

λ Theorem 4 For every λ R the triangles τa and σO are homologic and their homology ∈ centers trace the line GK (see Fig. 4).

T + z λ λ λ Proof: The point 2a : : is the circumcenter OA(λ) of the ﬂank AR R . The line m c b 4 5 − λk + T A OA(λ) is aλd x + b(T + λb2)y c(T + λc2)z = 0. Hence, a is the homology center a 2a a λ − 1 of τa and σO. It traces the line ad2ax = 0 that goes through points G a (the centroid) and K(a) (the symmedian or GrebPe-Lemoine point). ¡ ¢

λ Theorem 5 For every λ R cot ω , the triangles τa and σO are orthologic. The orthol- λ ∈ \ {− } ogy center [ τa, σO ] is the de Longchamps point L or X20 of τ (or the orthocenter of τa) while λ the orthology centers [ σO, τa ] trace the line HK (see Fig. 5).

GC(λ)

PSfrag replacements R2

Ba C Aa

line HK

R1 H S2 S1 R4 K

A B

A G (λ) GB(λ)

R5 R6

λ Figure 5: The orthology center S1 = [τa, σO] is X20 while the orthology centers λ S2 = [σO, τa] trace the line HK (Theorem 5)

λ Proof: The triangle σO degenerates to a point if and only if λ = cot ω. Since the triangles τ λ − and σO are homothetic and their center of similitude is the symmedian point K, the triangles λ λ τa and σO have parallel corresponding sides. It follows that τa and σO are orthologic and that λ A [ τa, σO ] = X20. The perpendicular from O (λ) onto the line BaCa is

aλd k x + b(λd k T k )y + c(λd k + T k )z = 0. 2a a 2a a − b 2a a c 8 Z. Cerin:ˇ Homology and Orthology with Triangles

2 (kkbkc a T )λ + T kbkc Hence, [ σλ , τ ] has coordinates − . We infer that this orthology center O a a 2 traces the line HK because we get its equation ad2akax = 0 by eliminating the parameter λ. P Since H befriends G and the line AHA(λ) is the median AG that goes through the point λ Aa, it is clear that triangles τa and σH are homologic and that their homology center is G (the centroid). The axis of these homologies envelope a complicated quartic.

λ λ Theorem 6 The locus of the orthology centers [τa, σH ] of τa and σH is the Kiepert hyperbola λ λ of τa. The locus of the orthology centers [σH , τa] of σH and τa is the line HK (see Fig. 6).

R2 PSfrag replacements

C HB(λ) Aa Ba R1 S1 A R4 H (λ) S2 HK

A B line HK HC(λ)

R 5 R6

λ λ Figure 6: The orthology centers S1 = [τa, σH ] and S2 = [σH , τa] trace the Kiepert hyperbola of τa and the line HK (Theorem 6)

T 2λka λ λ A Proof: The point − : : c is the orthocenter H (λ) of the ﬂank AR4R5. The line aka b HB(λ)HC(λ) is

2 2 2 3kbkcλ 4a T λ + T λkb(3kcλ T ) λkc(3kbλ T ) − x + − y + − z = 0. bc ca ab A B C The perpendiculars from Aa and H (λ) onto lines H (λ)H (λ) and BaCa have equations

2 2 2daza(T 2kλ) T kc + (T 2kkc)λ T kb + (T 2kkb)λ − x + − y − z = 0 bc ca − ab and 2dazakaλ T kb + 2dazakaλ T kc 2dazakaλ x + y + − z = 0. bc ca ab Z. Cerin:ˇ Homology and Orthology with Triangles 9

λ The orthology center [ τa, σH ] is

bc [(8a2kT 2 16a2k2k + 8kk T 2 T 4)λ2 2T ((6a2 k)T 2+ − a a − − − +4k (k k 4a4))λ + T 2(2k k T 2)] a b c − b c − λ 2 2 while the orthology center [ σH , τa ] has coordinates T kbkc + 2(a T kkbkc) λ. In order to see − λ what curves trace these orthology centers we must eliminate the parameter λ. For [ τa, σH ] λ we get the equation for the Kiepert hyperbola of τa as in Theorem 1 and for [ σH , τa ] the equation for the line HK as in Theorem 5.

3. The ﬁrst Brocard triangle τb

Let τb = AbBbCb denote the ﬁrst Brocard triangle of ABC. Its vertices are projections of the symmedian point K onto perpendicular bisectors of sides.

λ Theorem 7 For every λ R the triangles τb and σG are homologic and their homology ∈ centers trace the Kiepert hyperbola of τb (see Fig. 7).

GC(λ) R3 PSfrag replacements

R2 R6 R5

K Cb

GA(λ) GB(λ) A B

λ Figure 7: The homology centers of τb and σG trace the Kiepert hyperbola of τb (Theorem 7)

A Proof: The line AbG (λ) is

kd z λ a2kλ + 3b2T a2kλ + 3c2T a a x + y z = 0 bc ca − ab 10 Z. Cerin:ˇ Homology and Orthology with Triangles

Kiepert hyperbola of τb

GC(λ)

PSfrag replacements C λ [τb, σG] Bb

G GB(λ) Cb Ab B A

GA(λ)

λ Figure 8: The points [τb, σG] trace the Kiepert hyperbola of τb (Theorem 8)

λ [σG, τb]

PSfrag replacements GC(λ)

C O

G Cb A b GB(λ) A B S GA(λ)

line X2X98

λ Figure 9: The orthology centers [σG, τb] trace the line X2X98 (Theorem 8) Z. Cerin:ˇ Homology and Orthology with Triangles 11 a2k k2λ2 + 3kT (b4 + c4)λ + 9m2T 2 since A is abc : c3 : b3. Hence, a a is the homology center of b a λ 4 2 2 2 τb and σG. This center will trace the curve daza(a x + ma(b + c )yz) = 0 while the 1 parameter λ changes. Since the vertices of τb,Pthe common centroid G a of τ and τb, and kk k the orthocenter 2a(T 2 a2k 2m2) b c of the ﬁrst Brocard triangle¡τ ¢are on this curve, − a − a − a b we conclude that it is the Kiepert hyperbola of τb. Notice that the circumcenter O(a ka) and 1 the 3rd Brocard point X76 a3 are also on this hyperbola. ¡ ¢ λ Theorem 8 For every λ R the triangles τb and σG are orthologic. The locus of the λ ∈ λ orthology centers [τb, σG] and [σG, τb] is the Kiepert hyperbola of τb and the line X2X98 joining the centroid with the Tarry point of ABC, respectively (see Figs. 8 and 9). B C Proof: The perpendicular from Ab onto the line G (λ)G (λ) has the equation ad (T λ + 3k) x + b(T z λ + 3a2k) y c(T z λ + 3a2k) z = 0. 2a 2c − 2c 2 2 2 2 2 T (2 k a )λ + 6T λ(a ka + z4a) + 9a kka Therefore, the orthology center [τ , σλ ] is − . This point b G a 4 2 traces the conic with the equation d2a(a x + maz2ayz) = 0 that was recognized as the Kiepert hyperbola of τb in the proofPof Theorem 7. A The perpendicular from G (λ) onto the line BbCb has the equation ad T x + b d T λ 3(c4 a2 k ) y + c(d T λ + 3(b4 a2 k ))z = 0. 2a 2a − − a 2a − a ¡ 2¢ 2 2 2 4 2 4 2 T (kkbkc + a (a ka 2T ) + 2m )λ + 3(b a ka)(c a ka) So, the orthology center [ σλ , τ ] is − − − . This G b a point traces the line with the equation d (k k + a2k 2m2)x = 0. The points X 1 2a b c a − a 2 a 1 P a ¡ ¢ and X98 2 are on it. Note that the points X110 (the focus of the Kiepert µa(z4a a z2a)¶ d2a − 2 2 2 ³ ´ 2 (a k T )(z4a a z2a) d ka parabola), X − − (the Kiepert antipode), and X 2a (the center 114 a 125 a of the Jerabek h³yperbola) also belong ´to this line. ³ ´ λ Since the vertices of τb are on perpendicular bisectors of sides of τ and triangles τ and σO λ λ are homothetic it follows that τb and σO are orthologic and [ τb, σO ] = O. λ λ Theorem 9 The locus of the orthology centers [σO, τb] of the triangles σO and τb is the line X6X98 joining the symmedian point X6 with the Tarry point X98 (see Fig. 10). A Proof: The perpendicular from the point O (λ) onto the line BbCb has the equation

λd aka λd aka T kb λd aka + T kc 2 x + 2 − y + 2 z = 0. bc ca ab 2 2 λ(kkbkc a T ) + T kbkc Hence, − is the orthology center [σλ , τ ]. It traces the line with the a O b equation ad (a2 k + 2 m2)(z a2 z )x = 0. 2a a a 4c − 2c One can easily check that Pthe points X6 and X98 are on this line.

λ Theorem 10 For every number λ R and j = 4, 5, 20 the triangles τb and σj are orthologic. ∈ λ λ The locus of the orthology centers [ τb, σj ] of the triangles σj and τb is the Kiepert hyperbola λ of τb. The orthology centers [ σj , τb ] trace the line X4X98 for j = 4, the line through X98 parallel to the line X3X66 for j = 5, and a line through X98 for j = 20. Proof: We leave proofs of the statements of this theorem to the reader as an exercise because they are similar to the above proofs. 12 Z. Cerin:ˇ Homology and Orthology with Triangles

λ [σO, τb]

GC(λ)

PSfrag replacements C T

Bb O K Cb

Ab A B

GA(λ) GB(λ)

line X6X98

λ Figure 10: The orthology centers [σO, τb] trace the line X6X98 (Theorem 9)

4. The Euler triangle

The Euler triangle τE = AEBECE has the midpoints AE, BE, CE of segments AH, BH, CH joining vertices with the orthocenter H as vertices. Since the lines AGA(λ), BGB(λ), CGC(λ) are altitude lines of ABC it is obvious that λ for every λ R the triangles τE and σG are homologic and their homology center is the ∈ orthocenter H of τ.

λ Theorem 11 For every λ R the triangles τE and σO are homologic and their homology centers trace the line HK (see∈ Fig. 11).

T kc kb Proof: Since a : : are trilinears of AE and from the proof of Theorem 4 we know a ka − 2b 2c T + (b2 + c2)λ λ λ that : : are trilinears of OA(λ), we infer that A OA(λ) is m c b E − 2 λad2akax + b [λM+ + T kakb] y + c[λM− + T kakc]z = 0,

4 2 2 A B where M§ = z2aa 2z2aa + d2a(d2az2a 4m2a). Note that the lines AEO (λ), BEO (λ), − § 4 2 2 2 kbkcT + (3z2aa 2d a z2ad )λ and C OC (λ) are concurrent at the point − 2a − 2a . This point traces E a(2kλ + T ) 2 the line HK whose equation is ad2akax = 0. P λ Theorem 12 For every λ R the triangles τE and σH are homologic and their homology ∈ centers trace the Kiepert hyperbola of τE (see Fig. 12). Z. Cerin:ˇ Homology and Orthology with Triangles 13

C line HK

λ homology center of σO and τE

OC(λ) PSfrag replacements

C OB(λ) E OA(λ)

A B

AE BE

λ Figure 11: The homology centers of σO and τE trace the line HK (Theorem 11)

Kiepert hyperbola of τE

OC(λ) C

PSfrag replacements

A B OA(λ) CE λ homology center of σH and τE BE OB(λ) AE

λ Figure 12: The homology centers of σH and τE trace the Kiepert hyperbola of τE (Theorem 12) 14 Z. Cerin:ˇ Homology and Orthology with Triangles T 2 1 1 Proof: Since : : are trilinears of HA(λ) and from the proof of the previous akaλ − a b c T kc kb A theorem we know that a : : are trilinears of AE, we infer that AEH (λ) is aka − 2b 2c 2λad k x M (b)y M (c)z = 0, where M (b) = b 2λ(d a2 d T ) T k . 2a a − + − − § 2a − 4a § ∓ b £ ¤ A B C The lines AEH (λ), BEH (λ), and CEH (λ) are concurrent at the point

4 2 2 2 4 2 2 2 4T (a + z aa 2d )λ 2(3z aa 2d a z ad )λ + T kbkc 2 − 2a − 2 − 2a − 2 2a a that traces the conic whose equation is

d [a2k2x2 + bc(a4 + 2z a2 3d2 )yz] = 0. 2a a 2a − 2a P 4 2 2 a + z2aa 2d Since it goes through the vertices of τ , its centroid − 2a , and the common ortho- E a center H of τ and τE, we conclude that it is the Kiepert hyperbola of τE. The proof of the following theorem is left to the reader.

λ Theorem 13 For every number λ R and j = 2, 3, 4, 5, 20 the triangles τE and σj are ∈ λ λ orthologic. The orthology centers [τE, σ3 ] and [σ2 , τE] are the orthocenter H. For i = λ 2, 4, 5, 20 the locus of orthology centers [τE, σi ] is the Kiepert hyperbola of τE. The locus of λ the orthology centers [σi , τE] is the line HK for i = 3, 4, 5, 20.

5. The complementary triangle

The complementary triangle τg = AgBgCg has the midpoints Ag, Bg, Cg of sides BC, CA, AB as vertices. It is also the Cevian triangle of the centroid G. Since the lines AHA(λ), BHB(λ), CHC(λ) are median lines of ABC it is obvious that for λ every λ R the triangles τg and σH are homologic and their homology center is the centroid ∈ G of τ.

λ Theorem 14 For every λ R the triangles τg and σG are homologic and their homology ∈ centers trace the Kiepert hyperbola of τg.

3T + 2a2λ k λ k λ Proof: Since from the proof of Theorem 1 we know that : c : b are trilinears a b c A − A of G (λ) while the trilinears of Ag are 0 : c : b, we infer that AgG (λ) is

2λad x + M(b)y M(c)z = 0, where M(b) = b(2λa2 + 3T ). 2a − 4k a2λ2 + 6T z λ + 9T 2 The lines A GA(λ), B GB(λ), and C GC (λ) concur at the point a 2a that g g g a 2 2 traces the conic whose equation is d2a[a x + bcyz] = 0. Since it goes through the vertices of τg, the common centroid G of τ Pand τg, and the orthocenter O of τg, it follows that this is the Kiepert hyperbola of τg.

λ Theorem 15 For every λ R the triangles τg and σO are homologic and their homology ∈ centers trace the line GK (see Fig. 13). Z. Cerin:ˇ Homology and Orthology with Triangles 15

OC(λ)

PSfrag replacements K Bg Ag G λ homology center of σO and τg A Cg B

OA(λ)

OB(λ) line GK

λ Figure 13: The homology centers of σO and τg trace the line GK (Theorem 15)

T + z λ λ λ Proof: Now we recall from the proof of Theorem 4 that 2a : : are trilinears of m c b A A − O (λ). Hence, the line AgO (λ) is λad2ax + M(b)y M(c)z = 0, where M(b) is equal to − z λ + T b(λz + T ). The lines A OA(λ), B OB(λ), and C OC(λ) concur at the point 2a that 2a g g g a 1 traces the line whose equation is ad2ax = 0. One can easily check that the points G a and K(a) are on this line. P ¡ ¢

λ Theorem 16 For every number λ R and j = 2, 3, 4, 5, 20 the triangles τg and σj are ∈ λ λ orthologic. The orthology center [τg, σ3 ] is the circumcenter O and [σ2 , τg] is the orthocenter λ H. For i = 2, 4, 5, 20 the locus of orthology centers [τg, σi ] is the Kiepert hyperbola of τg. λ The locus of the orthology centers [σi , τg] is the line HK for i = 3, 4, 5, 20. 2 (a ka)λ + 2T d2bλ d2cλ Proof for the locus of [τ , σλ ]. Since F A(λ) is − : : , the perpendicular g F a b c B C B C − p Ag, F (λ)F (λ) from the point Ag onto the line F (λ)F (λ) has the equation ¡ ad (kλ 2T )x ¢+ M(b)y M(c)z = 0, where M(b) = b[(2a4 z a2 + d2 )λ 2a2T ]. 2a − − − 2a 2a − Then the perpendiculars B C C A A B p Ag, F (λ)F (λ) , p Bg, F (λ)F (λ) , and p Cg, F (λ)F (λ) concur at the¡ point ¢ ¡ ¢ ¡ ¢ 2 4 2 2 2 6 4 2 2 2 (z a z aa )(2a z aa + d )λ + T (2a z aa z ad )λ + 2a kaT 4 − 2 − 2 2a − 2 − 2 2a a

that traces the Kiepert hyperbola of τg (see the proof of Theorem 14). 16 Z. Cerin:ˇ Homology and Orthology with Triangles 6. The orthic triangle

The orthic triangle τh = AhBhCh has the feet Ah, Bh, Ch of altitudes of ABC as vertices. It is also the Cevian triangle of the orthocenter H. Since the lines AGA(λ), BGB(λ), CGC(λ) are altitude lines of ABC it is obvious that λ for every λ R the triangles τh and σG are homologic and their homology center is the ∈ orthocenter H of τ. λ Theorem 17 For every λ R the triangles τh and σO are homologic and their homology ∈ centers trace the equilateral hyperbola that goes through the vertices of τh and the central points H (the orthocenter), K (the symmedian point), X52 (the orthocenter of the orthic triangle), and X113 (Jerabek antipode) of the triangle ABC. The homology axis trace a parabola.

1 1 A T + z2aλ λ λ A Proof: Since Ah is 0 : : and O (λ) is : : , the line AhO (λ) has the b kb c kc m c b equation − λad k x M(b)y + M(c)z = 0, where M(b) = bk (λz + T ). 2a a − b 2a A B C It follows that the lines AhO (λ), BhO (λ), and ChO (λ) concur at the point 2 2 2 2 2a z2aλ + T (k + a )λ + T a ka 2 2 that traces an equilateral hyperbola with the equation d2a(a x + bckbkcyz) = 0. One can 1 easily check that the vertices of τh, the orthocenter H P , the symmedian point K(a), the a ka 2 2 2 2 ³ ´ orthocenter X52 (a(a ka T )(2ma T )) of the orthic triangle, and the Jerabek antipode 2 2 4 − 2 −2 2 T 3a ka)(z2aa 2(z4a m )a + z2ad X − − − a 2a all lie on it. 113 µ¡ a ¢¶

λ Theorem 18 For every real number λ and for j = O, K the triangles τh and σj are ortho- λ λ logic. The orthology center [τh, σO] is the orthocenter H and [σK , τg] is the circumcenter O. λ The locus of orthology centers [τh, σK] is the rectangular hyperbola AhBhChH. The locus of λ the orthology centers [σO, τh] is the line OK (see Fig. 14). λ A Proof for the locus of [τh, σK]. Since the point K (λ) is 2 2 2 2[T (3ka 2a ) + (z aa d )λ] − 2 − 2a : M(b, c) : M(c, b) a − 2 2 B C where the function M(b, c) is btc[(kb + 2mb) T ], the perpendicular p Ah, K (λ)K (λ) B C − from the point Ah onto the line K (λ)K (λ) has the equation ¡ ¢ 2λad k T x M(b, c)y + M(c, b)z = 0, 2a a − where the function M(b, c) is bk [T (k + 2a2)λ + (3k 4b2)(3k 4c2)]. Then the lines b − − B C C A A B p Ah, K (λ)K (λ) , p Bh, K (λ)K (λ) , and p Ch, K (λ)K (λ) concur at the¡ point ¢ ¡ ¢ ¡ ¢ 2 2 2 4 2 2 2 2 2 T (k + 2a )λ + 2T (3a + 7z aa + 8m )λ + (3k 4a )(3k 4b )(3k 4c ) 2 a − − − ka that traces the rectangular hyperbola with the equation (3k 4a2)d [a2k2x2 + m k k yz] = 0. − 2a a a b c P It is easy to check that the vertices of τh and the orthocenter H lie on it. Z. Cerin:ˇ Homology and Orthology with Triangles 17

OC(λ)

λ λ [σO, τh] and [σO, τt]

PSfrag replacements C

O Ah

Bt K Ch A B OB(λ) OA(λ) Ct Bh

line KO

λ λ Figure 14: Orthology centers [σO, τh] and [σO, τt] trace the line OK (Theorems 18 and 19)

7. The tangential triangle

Let pa, pb, and pc be perpendiculars at vertices A, B, and C to segments AO, BO, and CO joining the vertices with the circumcenter. The tangential triangle τt = AtBtCt has the intersections p p , p p , and p p as vertices. It is also the antipedal triangle of the b ∩ c c ∩ a a ∩ b circumcenter O. Since the lines AOA(λ), BOB(λ), COC (λ) are the symmedians of ABC it is obvious λ that for every λ R the triangles τt and σO are homologic and their homology center is the symmedian point∈K of τ.

λ Theorem 19 For every number λ R and j = O, K the triangles τt and σj are orthologic. λ ∈ λ The orthology centers [τt, σO] and [σK , τt] are the circumcenter O. The locus of orthology λ centers [τh, σK] is the rectangular hyperbola AtBtCtO. The locus of the orthology centers λ [σO, τt] is the line OK (see Fig. 14). T + z λ λ λ Proof for the locus of [σλ , τ ]. Since the point OA(λ) is 2a : : and A has trilinears O t m c b t a : b : c, the perpendicular p OA(λ), B C from the poin− t OA(λ) onto the line B C has − t t t t the equation ¡ ¢ 2λabcd x + c[2λb2d + k T ]y + b[2λc2d k T ]z = 0. 2a 2a c 2a − b Then the lines

A B C p O (λ), BtCt , p O (λ), CtAt , and p O (λ), AtBt ¡ ¢ ¡ ¢ ¡ ¢ concur at the point a(2λ(z a2 z ) k T ) that traces the line with the equation m d x = 2a − 2a − a a 2a 0. It is easy to check that the circumcenter O and the symmedian point K lie onPit. 18 Z. Cerin:ˇ Homology and Orthology with Triangles 8. The Torricelli triangles

Let Au, Bu, and Cu be vertices of equilateral triangles built on sides BC, CA, and ABb of ABC towards inside. When they are built towards outside then their vertices are denoted Av, Bv, and Cv. The negative Torricelli triangle τu is AuBuCu while AvBvCv is the positive Torricelli triangle τv of ABC.

λ Theorem 20 For every λ R the triangles τu and σG are homologic and their homology ∈ centers trace the Kiepert hyperbola of τu that goes through the vertices of τu and the central points G (the centroid), O (the circumcenter), and X14 (the negative isogonic point) of the triangle ABC.

Kiepert hyperbola of τv

OC(λ) PSfrag replacements C

OB(λ) G B λ Bv A homology center of σG and τv

OA(λ)

λ Figure 15: The homology centers of σG and τv trace the Kiepert hyperbola of τv (analogue of Theorem 20)

T √3 3kc T √3 3kb Proof: Since the point A has trilinears 1 : − : − and the point GA(λ) is u − 6ab 6ca 3T + 2λa2 λk λk : c : b , the line A GA(λ) has the equation a b c u − λad (3k + T √3)x M(b)y + M(c)z = 0, 2a a − where the function M(b) is equal to b[λ(3kad2a z2aT √3) + (3kb T √3)T ]. Hence, the lines A B C − − AuG (λ), BuG (λ), and CuG (λ) concur at the point

2 2 2 2 4 2 2 2kaa λ + 3[z aT (z aa d )√3]λ + 9[a + (z a T √3)a 2d ] 2 − 2 − 2a 2 − − 2a a Z. Cerin:ˇ Homology and Orthology with Triangles 19 that traces the conic d [a2(3k T √3)x2 m (3k +T √3)yz] = 0. It is easy to check that 2a a − − a a this is the Kiepert hypPerbola of τu because it goes through the vertices of τu, the common cen- 2 4 2 2 (ka 2a )T √3 + a + 2z2aa 3d troid G of τ and τ, and the orthocenter of τ with trilinears − − 2a . u u a In the same way one can prove that O (the circumcenter), and X14 (the negative isogonic point) of the triangle ABC are also on it.

λ Theorem 21 For every λ R the triangles τu and σ18 are homologic and their homology ∈ center is the negative isogonic point X14.

A Proof: The point X18(λ) is f : g(b, c) : g(c, b), where

g(b, c) = λm t (3k + 6m + T √3)(k + 2m T √3), b b c c c c − and 2 2 2 2 2 2 4T ma (ka 3a )T √3 + 3z aa 3d λ + 3(3ka + 2a )T 5T √3 f = − 2 − 2a − . £¡ ta(ka + T √3)¢ ¤

One can easily check that this point lies on the line AuX14 and thus complete the proof.

λ Theorem 22 The triangle τu is orthologic to σi for i = 2, 3, 4, 5, 14, 20. The orthology λ λ centers [τu, σO] and [σ14, τu] are the circumcenter O and the second Napoleon point X18, λ respectively. The locus of the orthology centers [τu, σj ] is the Kiepert hyperbola of AuBuCu for λ j = 2, 4, 5, 20. The orthology centers [τu, σ14] trace a hyperbola that goes through the vertices λ of τu and the circumcenter O. The locus of the orthology centers [σk , τu] for k = 2, 3, 4, 5, 20 are the lines GX18, KX18, HX18, X15X18, and a line through X18, respectively.

A Proof of the case i = 14. The point X14(λ) has trilinear coordinates f : g(b) : g(c), where g(b) = λm (3b2k + T (2k b2)√3) and b b a − 2 2 2 2 2 f = ma (3ka a )T √3 3d2a + 3z2aa λ + 3(3ka + 2a )T 3T √3 . h³ − − ´ − i A A We infer easily that the perpendicular p X14(λ), BuCu from the point X14(λ) onto the line BuCu has the equation b(kb T √3)y c¡(kc T √3)z =¢ 0 and it goes through the point X18 − 4 − − 2 2 a + (T √3 3z2a)a + 2d with the ﬁrst trilinear coordinate − 2a . This shows that the triangles σλ a 14 λ and τu are orthologic and that [σ14, τu] = X18. B C On the other hand, the perpendicular p Au, X14(λ)X14(λ) from the point Au onto the B C line X (λ)X (λ) has the equation fxx g+¡(b, c)y + g−(c, b)z =¢ 0, with 14 14 −

fx = ad2a(g1λ + g2), g = a4 + 10z a2 5z + 16m2 + (2a2 3k)T √3, 1 2a − 4a a − g = (3a4 + 6z a2 3z + 8m2)√3 3(k + 2a2)T, and 2 2a − 4a a − g (b, c) = b [9a6 + 4(d b2)a4 + (2m2 7d 4c4)a2 + 2(c2 d )d2 + § { 2a − a § 4a − ∓ 2a 2a +( a4 + (2c2 d )a2 2b2d )T √3]λ + g a2 . − ∓ 2a § 2a 2 } The lines

B C C A A B p Au, X14(λ)X14(λ) , p Bu, X14(λ)X14(λ) , and p Cu, X14(λ)X14(λ) ¡ ¢ ¡ ¢ ¡ ¢ 20 Z. Cerin:ˇ Homology and Orthology with Triangles concur at the point h λ2 + 2h λ√3 3a2k h , where 2 1 − a 0 h2 = h20√3 + 3T h21, h = 18a10 65z a8 + (68z + 49m2)a6 z (33z 40m2)a4+ 20 − 2a 4a a − 2a 4a − a +(14z 9m2z 22m )a2 2z d2 (z 5m2), 8a − a 4a − 4a − 2a 2a 4a − a h = 12a8 7z a6 (z + 7m2)a4 + z (2z + 3m2)a2 2d2 (3z + m2), 21 − 2a − 4a a 2a 4a a − 2a 4a a h1 = h10√3 + 3T h21, h = 12a10 32z a8 + (12z + m2)a6 + 3z (3z m2)a4+ 10 − 2a 4a a 2a 4a − a +(2z + 17m2)d2 a2 3z d2 (d2 m2), 4a a 2a − 2a 2a 2a − a 2 2 2 h0 = h00√3 + 21T (ma + mb + mc ), and at last h = 6a6 9z a4 (9z + 29m2)a2 + 3z (2d + c2)(d c2). 00 − 2a − 4a a 2a 2a 2a − In order to ﬁnd the curve which traces this point we must eliminate the parameter λ. We obtain an equilateral hyperbola that goes through the vertices of τu and the circumcenter O.

Of course, there are versions of the above three theorems for the positive Torricelli triangle τv of ABC (see Fig. 15). Instead of numbers 18 and 14 now the numbers 17 and 13 play important role.

9. The Napoleon triangles

Let Ax, Bx, and Cx be centers of equilateral triangles built on sides BC, CA, and AB of ABC towards inside. When they are built towards outside then their vertices are denoted Ay, By, and Cy. The negative Napoleon triangle τx is AxBxCx while AyByCy is the positive Napoleon triangle τy of ABC.

λ Theorem 23 For every λ R the triangles τx and σG are homologic and their homology ∈ centers trace the hyperbola that goes through the vertices of τx and the central points G (the centroid), O (the circumcenter), and X18 (the second Napoleon point) of the triangle ABC.

kc T √3 kb T √3 A Proof: Since Ax has coordinates 1 : − : − , the line AxGλ has the equation − 2mc 2mb 2 2 A 2λad2ax + b(2λa √3 kb + 3T )y c(2λa √3 kc + 3T )z = 0. It follows that the lines AxGλ , B C− − − BxGλ , and CxGλ concur at the point

2 2 2 2 2 4 2 2 2a kaλ + (3z aT √3(z aa d ))λ 3√3 a T 3a + 9z aa 6d 2 − 2 − 2a − − 2 − 2a . a This point traces the equilateral hyperbola with the equation d [a2(k T √3)x2 m (k + T √3)yz] = 0. 2a a − − a a P It goes through the vertices of τx, the centroid G, the circumcenter O, and the second Napoleon point X18.

λ Theorem 24 For every λ R the triangles τx and σ14 are homologic and their homology ∈ center is the second Napoleon point X18. A Proof: The point X14(λ) whose coordinates have been described in the proof of Theorem 22 is easily seen to lie on the line AxX18. Z. Cerin:ˇ Homology and Orthology with Triangles 21

λ Theorem 25 The triangle τx is orthologic to σi for i = 2, 3, 4, 5, 18, 20. The orthology λ λ centers [τx, σO] and [σ18, τx] are the circumcenter O and the second isogonic point X14. The λ locus of orthology centers [τx, σj ] is the hyperbola that goes through the vertices of τx and points G, O, and X18 of the triangle ABC for j = 2, 4, 5, 20. The locus of the orthology λ centers [σk , τx] for k = 2, 3, 4, 5, 20 are the lines GX14, KX14, HX14, X14X16, and a line λ through X14, respectively. The orthology centers [σ18, τx] trace a hyperbola that goes through the vertices of τx and O. Proof: The proofs of the claims in this theorem are left to the reader as an exercise (see the proof of Theorem 22). Of course, there are versions of the above three theorems for the positive Napoleon triangle τy of ABC. Instead of numbers 18 and 14 now the numbers 17 and 13 play important role.

References

[1] Z. Cerinˇ : Loci related to variable ﬂanks. Forum Geometricorum 2, 105–112 (2002). [2] R.H. Eddy, R. Fritsch: The conics of Ludwig Kiepert: A comprehensive lesson in the geometry of the triangle. Mathematics Magazine 67, 188–205 (1994). [3] L. Hoehn: Extriangles and excevians. Mathematics Magazine 74, no. 5, 384–388 (2001). [4] C. Kimberling: Central points and central lines in the plane of a triangle. Mathematics Magazine 67, 163–187 (1994). [5] C. Kimberling: Encyclopedia of Triangle Centers. 2000, http://cedar.evansville.edu /˜ck6/encyclopedia/. [6] F. van Lamoen: Friendship Among Triangle Centers. Forum Geometricorum 1, 1–6 (2001). [7] C. R. Pranesachar, B. J. Venkatachala: On a Curious Duality in Triangles. Samasy¯a 7 (2), 13–19 (2001).

Received September 23, 2002; ﬁnal form June 5, 2003