Mixed Intersection of Cevians and Perspective Triangles

MIXED INTERSECTION OF CEVIANS AND PERSPECTIVE TRIANGLES

Boris ODEHNAL Dresden University of Technology, Germany

ABSTRACT: Any point P in the plane of a triangle ∆ = (A,B,C) can be considered as the intersection of certain Cevians, whether P is a triangle center or not. Taking two centers, say Y and Z, we find six Cevians with a total of 11 points of intersection, among them ∆’s vertices and Y and Z. The remaining six points form two triangles ∆1 and ∆2 which are both perspective to the base triangle ∆ and to each other. The three centers of perspectivity are triangle centers of ∆ and collinear, independent of the choice of Y and Z. Furthermore any pair out of ∆, ∆1, and ∆2 has the same perspectrix which yields a closed chain of Desargues’ (103,103) configurations. Then special affine appearances of Desargues’ configurations can be obtained by a suitable choice of Y and Z. Any choice of a fixed center Y leads to exactly one center Z as the fourth point of intersection of two conic sections circumscribed to ∆ such that the perspectors P1, P2, and consequently P12 are points at infinity. For fixed center Y we obtain a quadratic transformation in ∆’s plane which transforms central lines to central conic sections.

Keywords: triangle, Cevian, perspective triangles, Desargues conﬁguration, triangle center, crosspoint

1. INTRODUCTION C Many triangle centers appear as the intersection of certain Cevians. For example the incenter X1 X4 is the intersection of the interior angle bisectors h of a triangle ∆ with vertices A, B, C. The cen- B m troid X2 can be found as the common point of hA X2 B the medians, whereas the circumcenter X3 is the mA C C meet of the bisectors of ∆’s edges. The ortho- m h center X4 comes along as the intersection of the A B altitudes of ∆. Here and in the following, centers Figure 1: Is it a mistake? of ∆ are labelled according to C. Kimberling’s list, cf. [4, 6]. What happens if one makes the erroneous centers, say Y and Z, and intersect the respective construction by mixing Cevians of different cen- Cevians we obtain, besides the two centers Y and ters when intersecting them? Figure 1 shows Z, the three vertices of ∆. Further we ﬁnd six such a mistake: The mixed intersection of me- points which can be arranged in many ways in ∆ dians and altitudes in produces a triangle. two perspective triangles ∆1 and ∆2. Usually three arbitrarily chosen Cevians will In Section 2 we shall show that these triangles not meet in a common point, except maybe in are perspective to each other with perspector P12. cases where the side lengths of ∆ fulﬁl certain They are also perspective to ∆ with perspectors relations. However, if we start with two triangle P1 and P2, cf. Figure 6. All the three perspec-

Paper #89 tors are triangle centers of ∆, provided Y and Z b := CA, and c := AB denote the side lengths are centers of ∆. Moreover, the perspectors P1, of ∆. We denote the line joining two points P P2, and P12 are collinear and consequently the and Q by [P,Q]. A point Z = (z0 : z1 : z2) is perspectrix is the same for any pair of triangles a center exactly if z0 = f (a,b,c) is a homoge- out of the three. So we obtain a closed chain neous function of the side lengths a, b, and c of three Desarguesian (103,103) configurations. of the base triangle ∆ with z1 = f (b,c,a) and This will be discussed in Section 4. An Exam- z2 = f (c,a,b). Later we use the symbol ζ in or- ple is displayed in Figure 6. Special affine ver- der to indicate that a homogeneous function f is sions of (103,103) configurations are described transformed via the cyclic shift: If f = f (a,b,c), in Section 5. In Section 3 we show that the per- then f ζ = f (b,c,a). Thus a triangle center X = spector P is the crosspoint of Y and Z. This ξ ξ ξ ξ ξ ζ 12 ( 0 : 1 : 2) is characterized by i+1 = i for gives a new access to the crosspoints of triangle i ∈{0,1,2} and i is counted modulo 3. Simi- centers. larly a line in ∆’s plane is called a central line if its homogeneous trilinear coordinates follow C the same rules as those of centers. Note that ζ applies to any cyclically ordered triplet. 2.PERSPECTIVE TRIANGLES AND A1 THEIR PERSPECTORS B1 Assume Y =(ξ0 : ξ1 : ξ2) and Z =(η0 : η1 : η2) ∆ C2 ∆ ∆1 are triangle centers of . Thenwelookatthefol- lowing points of intersection of Cevians through Y Y and Z: Z C1 ∆2 A1:=[C,Y] ∩ [B,Z],A2:=[B,Y] ∩ [C,Z], B A2 2 B1:=[A,Y] ∩ [C,Z],B2:=[C,Y] ∩ [A,Z], (1) A B C1:=[B,Y] ∩ [A,Z],C2:=[A,Y] ∩ [B,Z]. ∆ ∆ Figure 2: The triangles 1 and 2 whose vertices We define two triangles collecting the intersec- are the mixed intersections of Cevians of Y and tion points of wrong pairs of Cevians by letting Z. ∆1 =(A1,B1,C1) and ∆2 =(A2,B2,C2) as illustrated in Figure 2. If we fix Y then the mapping q : Z → P12 is No we can show: quadratic and sends centers to centers and central lines are mapped to central conics. The map- Theorem 2.1. The triangles ∆1 and ∆2 are per- ping q is birational, i.e., its inverse is also ra- spective to the base triangle ∆. The perspectors tional. It turns out that q is a composition of P1 and P2 are triangle centers of ∆. the isogonal conjugation with a collineation. We pay our attention to q in Section 6. Proof. With the above prerequisites we find the In the following we use homogeneous trilin- vertices of ∆1 and ∆2 as ear coordinates (p0 : p1 : p2) in order to repre- sent points in the plane of ∆. The vertices of A1 =(ξ0η0 : ξ1η0 : ξ0η2), ∆ are the base points and thus their coordinate ξ η ξ η ξ η vectors are A =(1:0:0), B =(0:1:0), and B1 =( 1 0 : 1 1 : 2 1), (2) C =(0:0:1), cf. [4, 6]. Further we let a := BC, C1 =(ξ0η2 : ξ2η1 : ξ2η2), 2 C can be found as intersection of [A,Ai] and [B,Bi] and we arrive at

P1 =(ξ0ξ1η0η2 : ξ1ξ2η1η0 : ξ2ξ0η2η1) P12 = X6 and (4) P2 P2 =(ξ0ξ2η0η1 : ξ1ξ0η1η2 : ξ2ξ1η2η0).

X4 P1 and P2 are centers for ξi and ηi (with i ∈ X ∆ ∆ 2 2 {0,1,2}) are center functions, i.e., they are homogeneous and cyclic symmetric in a, b, and c. ∆1 ζ For example (ξ0ξ1η0η2) = ξ1ξ2η1η0 and like- P1 wise for all the other coordinate functions of P1 and P2, respectively. A B Obviously P1 and P2 are triangle centers for Figure 3: The three perspectors P1, P2, and P12 any choice of centers Y and Z. It can be seen at constructed out of the centroid Y = X2 and the once that Y = Z results in P1 = P2. Furthermore ∆ orthocenter Z = X4. if Y = X1, i.e., the incenter of and Z is an arbi- η η C trary center not equal to X1, then P1 =( 0 2 : η1η0 : η2η0) and P2 = (η0η1 : η1η2 : η2η0). The mappings Z 7→ P and Z 7→ P are birational P 1 2 2 for they are compositions of the isogonal conju- P12 gation (x0 : x1 : x2) 7→ (x1x2 : x2x0 : x0x1) with collineations in ∆’s plane with X1 for a fixed ∆2 point and shifting ∆’s vertices in clockwise or X 28 counter clockwise direction, respectively. X 3 It is easy to show that the following holds: ∆ ∆ ∆1 Theorem 2.2. The triangles 1 and 2 are per- ∆ P 1 spective. The perspector P12 is a triangle center that is collinear with P1 and P2 from Theorem A B 2.1, except Y = Z, where P1 = P2 = P12 = Y = Z. Proof. The lines [A ,A ], [B ,B ], and [C ,C ] Figure 4: The three perspectors P , P , and P 1 2 1 2 1 2 1 2 12 are concurrent. In order to prove this, we com- constructed out of the circumcenter Y = X and 3 pute the trilinear coordinates of the three lines the center Z = X . 28 and show that they are linearly dependent. The point of concurrency can be computed as and the intersection of any pair of the above given ξ η ξ η ξ η A2 =( 0 0 : 0 1 : 2 0), lines which gives B2 =(ξ0η1 : ξ1η1 : ξ1η2), (3) P12 =(ξ0η0(ξ2η1 + ξ1η2) : C2 =(ξ2η0 : ξ1η2 : ξ2η2). ξ1η1(ξ0η2 + ξ2η0) : (5) Then we show that the lines [A1,A2], [B1,B2], ξ2η2(ξ1η0 + ξ0η1)), and [C1,C2] are concurrent by computing their trilinears and showing the linear dependency. which is obviously a triangle center of ∆ for the The perspectors Pi of ∆ and ∆i (with i ∈{1,2}) coordinate function i + 1 is the ζ -image of the 3 coordinate function i, for i ∈{0,1,2} and i is of P and Q, respectively. Then define ∆3 = counted modulo 3. (A′′′,B′′′,C′′′) by letting The three perspectors P1, P2, and P12 are A′′′ :=[A,A′′] ∩ [B′,C′], collinear, for their trilinear coordinate vectors ′′′ ′′ ′ ′ are linearly dependent. B :=[B,B ] ∩ [C ,A ], C′′′ :=[C,C′′] ∩ [A′,B′]. If Y = Z, then P1 = P2 = Z as outlined above. ξ η Let i = i (for i ∈{0,1,2}) then P12 = Z. Now it turns out that ∆1 and ∆3 are perspective and the perspector is called the crosspoint At this point we shall remark that P from 12 of P and Q. Somehow this definition seems to Equation (5) is the bicentric sum of P and P 1 2 be unsymmetric. However, if we define ∆ = given in Equation (4) as defined in [3, 5]. 4 (A′′′′,B′′′′,C′′′′) with A′′′′ =[A,A′]∩[B′′,C′′], and Figures 3 and 4 show the perspectors P , P , 1 2 B′′′′ and C′′′′ cyclic, then ∆ is perspective to and P for different choices of Y and Z. 4 12 ∆ . Surprisingly, the crosspoint also serves as Note that Y and Z are also perspectors of the 2 the perspector for the latter two triangles. following four pairs of triangles: Y connects Now we have a simple geometric meaning of (∆,ζ (∆ )) and (∆,ζ −1(∆ )), whereas Z con- 1 2 the crosspoint of two centers. It can be found nects (∆,ζ (∆ )) and (∆,ζ −1(∆ )), with ζ ap- 2 1 as the perspector of the two triangles ∆ and ∆ plied to the vertices of the triangles. 1 2 defined in Equations (2) and (3) constructed via 3. THE RELATION TO CROSSPOINTS mixed intersection of the Cevians of two cer- Equation (5) allows us to define a mapping tain centers. Note that ∆1 and ∆2 differ from ⊕ : (Y,Z) 7→ Y ⊕ Z = P12 for any pair of points the triangles which are usually involved in the Y and Z in the plane of ∆. In Table 1 and Table construction of the crosspoint as defined in [6, 2 the ⊕-images of some pairs of triangle centers p. 202] and [7]. So we have found a simple ac- Xi and Xj are given. A ⋆ indicates a yet innom- cess to the crosspoints. inate center, the numbers in the Tables 1 and 2 are the Kimberling numbers of Xi ⊕Xj. We have C the following result: crosspoint of P and Q Theorem 3.1. The pointY ⊕Z thatis assignedto any pair (Y,Z) of points in ∆’s plane via Equa- B′′ tion (5) coincides with the crosspoint of Y and Z. A′′ Q B′ C′′′ Proof. We compare the coordinate representa- A′ tion of Y ⊕ Z given in Equation (5) with the ex- ′′′ A′′′ B pressions for crosspoints given in [6, p. 202] or P [7]. We shall emphasize that our access to the A C′ C′′ B crosspoint differs from the definition given in [6] or [7]. Figure 5: The usual way of finding a crosspoint Figure 5 shows how the crosspoint of two ar- of two point. bitrary points P and Q with respect to a triangle ∆ is usually found: Let P and Q be any Figure 3 illustrates the case where Y = X2 and ′ ′ ′ two points in ∆’s plane. Let ∆1 = (A ,B ,C ) Z = X4. The line carrying the three perspectors ′′ ′′ ′′ and ∆2 = (A ,B ,C ) be the Cevian triangles P1, P2, and P12 = X6 is also shown. To the best 4 of the authors knowledge this is the only pair Table 1: ⊕-compositions of some centers be- (Xi,Xj) where Xi ⊕ Xj = Xi+ j holds. In Figure sides X1 ⊕ X2 = X37. 4 we have chosen Y = X3 and Z = X28. ⊕ 3 4 5 6 7 8 9 10 The mapping ⊕ gives a tool for the construc- 1 73 65 2599 42 354 3057 55 2292 2 216 6 233 39 1 9 1212 1213 tion of triangle centers as perspectors of triangles 3 3 185 ⋆ 184 ⋆ ⋆ ⋆ ⋆ ∆1 and ∆2. Therefore we find a way to construct 4 4 3574 51 1836 1837 1864 1834 some triangle centers for which elementary con- 5 5 ⋆ ⋆ ⋆ ⋆ ⋆ 6 6 ⋆ ⋆ 2347 ⋆ structions are missing until now, especially those 7 7 497 ⋆ 4854 with large Kimberling number. For example 8 8 210 ⋆ 9 9 ⋆ X4854 = X7 ⊕ X10, i.e., the center X4854 can be 10 10 found as the perspector P12 with Y = X7 (Ger- ∆ gonne point of ) and Z = X10 (Spieker point of Table 2: More ⊕-compositions of centers. ∆), cf. Table 1. ⊕ 1 2 3 4 6 8 19 31 19 31 ⋆ ⋆ 1824 1400 ⋆ ⋆ 2179 20 ⋆ 1249 ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ 21 2646 ⋆ ⋆ 1858 ⋆ 960 ⋆ ⋆ 31 1964 ⋆ ⋆ ⋆ 213 ⋆ 2179 31 32 ⋆ ⋆ 682 ⋆ 3051 ⋆ ⋆ 1918

perspectrix p according to the perspectors. Since the per- C 12 spectors are collinear we have: X11 Theorem 4.1. The perspectrices p1,p2,and p12 fall in one line, i.e.,p1 = p2 = p12. The common perspectors perspectrix is a central line. Proof. The fact that all three perspectrices co-

X6 incide is caused by the collinearity of the three perspectors and the fact that the perspectrix is the same for two different pairs of triangles of the triplet (∆,∆ ,∆ ), for the set of perspective AB 1 2 collineations with common axis and collinear centers constitute a group, cf. Figure 7. It is also Figure 6: A closed chain of Desargues configu- possible to compute the perspectrix for any pair rations with Y being the Lemoine point X and Z 6 of triangles from our triplet. being the Feuerbach point X . 11 We only have to show that the perspectrix is a central line. The computation of its homo- 4.CHAINS OF DESARGUES' CONFIGU- geneous trilinear coordinates is straightforward RATIONS and we find its first coordinate From Theorem 2.1 we know that any pair of tri- ξ1ξ2η1η2 angles out of the triplet (∆,∆ ,∆ ) is perspective l0 = 1 2 η ξ − η ξ to a certain point. According to Desargues’ two 1 2 2 1 triangle theorem, any pair of triangles which is ζ ζ 2 and further p12 =(l0 : l0 : l0 ). Therefore it is a perspective to a point, is also perspective to a central line. line. Consequently, any pair of triangles out of the triplet (∆,∆1,∆2) is also perspective to a line. The Theorems 2.2 and 4.1 tell us that there The perspectrices shall be labelled p1, p2, and is a closed chain of Desargues configurations 5 Z31 numbers of centers are written. We have fixed Z 12 Y = X2 and thus c2 is the Steiner circumellipse Z23 of ∆, cf. [6]. The perspector P12 again moves to the ideal line and P12 = X524. X291 C X513 a ∆2 ∆3 X80

c1 ∆2 X1 X513

∆ A ∆1 X100 ∆1 Figure 7: Chain of Desargues configurations. Figure 8: Special affine appearance of Desar- with three mutually perspective triangles, three gues configuration: X1 ⊕ X100 = X513 which lies collinear perspectors, and one common perspec- on the line at infinity. trix. Figure 6 shows such a chain of Desargues configurations built from Y = X6 and Z = X11.

5.SPECIAL AFFINE VERSIONS OF DE- 2481 666 1121 903 X524 SARGUES CONFIGURATIONS ∆1 X524 671 2966 Equation (5) allows us to determine special ∆2 1494 2480 affine versions of closed Desargues’ chains. A X524 c2 perspectors P12 is an ideal point, if it is contained X2 in the ideal line with the equation ∆ 648 670 886 2479 L 190 ∞ : ax0 + bx1 + cx2 = 0. (6) 668 664 99 If we fix one center, say Y , then the locus of points Z such that Y ⊕ Z is an ideal point is a conic section kY . This isimmedeatelyseen, if we Figure 9: Special affine appearance of Desar- derive the incidence condition for (Y ⊕Z) ∈ L∞. gues configuration: X2 ⊕ X671 = X524 which is We insert Equation (5) into Equation (6) and find an ideal point.

kY : ∑ aξ0η0(ξ2η1 + ξ1η2) = 0, (7) Is it possible to find Desarguesian configura- cyclic tions with three perspectors on the ideal line? which is a homogeneous and quadratic equation The answer to this question is: yes. The condition for P and P to lie on the ideal line L∞ is in ηi for fixed Y =(ξ0 : ξ1 : ξ2). We observe that 1 2 given by kY is a circumconic of ∆, i.e., kY passes through ∆ ’s vertices. l1 : ∑ aξ0ξ1η0η2 = 0 Figure 8 shows c1 with some centers on it. cyclic The perspector of ∆1 and ∆2 is the ideal point and (8) X513. In Figure 9 we have simplified the nota- l2 : ∑ aξ0ξ2η0η1 = 0, tion in order to save space, only the Kimberling cyclic 6 where we have inserted the coordinates of P and 1 P12 P2 as given in Equation (4) into the Equation (6) P of L∞. For a fixed center Y =(ξ0 : ξ1 : ξ2) Equa- 12 ∆2 tions (8) are the equations of two conic sections. C ∆1 Both pass through the vertices A, B, C of the base P12 Z triangle and thus they have only one further common point. This yields: l2 Y l B 1 Theorem 5.1. To any fixed center Y there exists A ∆ exactly one center Z such that the perspectors P1,P2, andP12 lie on the ideal line. P2 P P Proof. We only have to show that l1 and l2 have 2 2 precisely one center in common. Obviously, three common points fall into the vertices of ∆. Figure 10: All three perspectors are ideal points It is elementary and straightforward to find the for a suitable choice of Z (precisely that of Equa- fourth intersection: tion (9)) depending on Y . ξ ξ 2 2 ξ ξ ξ 2 2 ξ ξ Z =( 0( 1 b − 0 2ca)( 2 c − 0 1ab) : of P12 from Equation (5) and we have 2 2 2 2 : (ξ1(ξ c − ξ1ξ0ab)(ξ a − ξ1ξ2bc) : 2 0 ξ η ξ η ξ η 2 2 2 2 q0 : 0 0( 2 1 + 1 2) = 0, : (ξ2(ξ a − ξ2ξ1bc)(ξ b − ξ2ξ0ca)), 0 1 ξ η ξ η ξ η q1 : 1 1( 0 2 + 2 0) = 0, (10) which is obviously a center of ∆. q2 : ξ2η2(ξ1η0 + ξ0η1) = 0.

Figure 10 illustrates the contents of Theorem Obviously these are three pairs of lines, each 5.1 with Y = X1. The resulting center containing a side line of ∆ and a further line ′ passing through the harmonic conjugates YA, 2 2 ′ ′ Z = ((b − ca)(c − ab) : YB, YC of the vertices of Y ’s Cevian triangle ∆ : (c2 − ab)(a2 − bc) : (9) (YA,YB,YC) with respect to ’s vertices, i.e., the pairs of lines are : (a2 − bc)(b2 − ca)), ′ q0 := [B,C] ∪ [A,YA], which is not yet named, yields triangles ∆1 and ′ q1 := [C,A] ∪ [B,YB], ∆ as defined in Equation (1), such that all the 2 ′ q2 := [A,B] ∪ [C,Y ]. three perspectors P1, P2, and P12 are ideal points. C 6. A QUADRATIC TRANSFORMATION ′ ′ ′ Note that the harmonic conjugates YA, YB, YC The coordinates of P12 given in Equation (5) are gather on the trilinear polar line of Y with respect quadratic in ξi and ηi. to the base triangle ∆. Figure 11 shows the base Assume Y =(ξ0 : ξ1 : ξ2) is an arbitrary point points and exceptional lines of q for an arbitrary in ∆’s plane, not necessarily a center. Then q is a point Y . quadratic mapping in ∆’s plane. The base points It is a well-known result from algebraic geo- of q are the vertices of ∆. The associated net metry that q is birational, i.e., its inverse is also of conic spanned by three pairs of lines whose rational, for it has three base points, which ap- equations are given by the coordinate functions pear as the intersections of the three degenerate 7 ′ YA with (λ : µ) 6=(0:0) is a parametrization of g. We substitute this parametrization into (5) and q0 obtain a parametrization which is obviously homogeneous and quadratic in (λ : µ). Thus it de- Y′ B scribes a conic section. We keep in mind that ηi C q1 are fixed values for a fixed center Y . By elimi- nating λ and µ we arrive at Equation (11). YB YA Y Figure 12 shows the q-image of the Euler-line e with Y = X1. The quadratic mapping q is also applied to some of the triangle centers on e. e eq C q2 ′ ABYC YC q Figure 11: Base points and exceptional lines of q e X2 q for an arbitrary point Y . X1 q X2 q X4 X5 X5 B conic sections qi defined by the three coordinate A q X functions (10) of q, cf. [2]. 20 Since Equation (5) can be rewritten in the X4 form η η η η ξ ξ 0 0 1 0 2 1 2 Figure 12: The quadratic mapping q and the im-     P12 = η0η1 0 η1η2 · ξ2ξ0 , age of the Euler-line e with Y = X1 together with     some centers on e and their q-images.  η0η2 η1η2 0   ξ0ξ1  we can conclude that q is a composition of the Any conic section through the base points of isogonal conjugation and a collinear transforma- q is mapped to a straight line. This is shown in ∆ tion that depends on Y . Figure 13, where Y = X2. ’s circumcircle u is The mapping q transforms central lines to cen- mapped to aline. In this case the line at infinity tral conics. We show that the image of a central is mapped to the Steiner inellipse. line is given by the following conic section: The determinant of the hessian of (11) equals 5 (2η0η1η2) g0g1g2. So the conic section is sin- Theorem 6.1. Assume that g =(g0 : g1 : g2) is a gular (splits off into two lines) if Y lies on one central line, i.e., its coordinate functions gi are side line of ∆. This is the case for example for a ∑ cyclic symmetric in a, b, c and cyclic X0g0 = 0 right angled triangle and Y = X3 or even Y = X4. η η η is its equation. Further let Y =( 0 : 1 : 2) be In general neither gi = 0, for g is a central line. a fixed triangle center for ∆. Then q(g) has the Exceptions occur for special triangles. equation In Figure 14 the action of q on ∆’s incricle i is illustrated. We have chosen four different cen- ∑ η2η2(g η − g η − g η )X 2 1 2 0 0 1 1 2 2 0 ters Y as pivot for q, namely X (incenter), X cyclic (11) 1 2 3 (centroid), X (circumcenter), and X (orthocen- −2g0η η1η2X1X2 = 0. 3 4 0 ter). The four quartic curves showing up as as Proof. Assume g =(g0 : g1 : g2) is a central line. the four different q-images of i are labelled ac- Then g(λ : µ) = λ(g2 :0: −g0)+µ(g1 : −g0 :0) cording to Y . Note that each of these quartics has 8 C u

Steiner inellipse = q(L∞) C

q(u) n 4 X 2 2 3 1

AB AB

Figure 13: The quadratic mapping q with Y = X2 Figure 15: Four different q-images of the nine- shows the Steiner inellipse as the image of the point circle n. ideal line. 7. CONCLUSION three ordinary cusps in points on the exceptional The study of mixed intersections of Cevians of lines. The cusps are the q-images of the contact triangle centers of a triangle ∆ led directly to ∆ points of i with ’s side lines. closed chains of Desarguesian (103,103) configurations. These chains consist of three such C configurations and it is still an open question, if there are such chains involving more then three Desargues figures. However, these chains where byproducts. The circumcircle u of ∆ is only one promi- i nent example of conic sections circumscribed to ∆. It is more or less an easy task to find the n q-images with arbitrary center as a pivot that map u or other circumscribed conic sections to 32 1 4 well-known central lines. We have skipped this lengthy and less fruitful enumeration and tried to give an example. A B The quadratic mapping q appears with pivots X1 and X2. Any center can serve as a pivot of Figure 14: Four different q-images of the incir- the transformation. Any conic section, whether cle i. it is circumscribed or inscribed to ∆, any central conic can be transformed via a certain q. All Figure 15 shows the image of the ninepoint the image curves are rational. Therefore none of circle n of ∆ under the same four quadratic trans- the well-known triangle cubics (see for example formations. Here we observe ordinary nodes on [1]) can be obtained as some q-image of a certain the image curves corresponding to the two dif- conic section, for all these are elliptic. If a line ferent transversal intersection of n with the ex- splits off from the q-image of a conic section, ceptional lines of q. then there remains at most a rational cubic curve. 9 REFERENCES under the postal address: TU Dresden, Institute [1] H.M. CUNDY, C.F. PARRY: Some cu- of Geometry, 01062 Dresden, Germany. bic curved associated with the triangle. J. Geom. 53/1-2 (1955), 41-66.

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ABOUT THE AUTHOR Boris Odehnal, PD PhD, is currently associate professor at the TU Dresden. His research in- terests are line and sphere geometry, higher di- mensional geometries, non-Euclidean geometries, projective geometries, algebraic curves and varieties, differential geometry of special curves and surfaces, and triangle geometry. He can be reached by e-mail: [email protected] dresden.de or [email protected] or 10