15TH INTERNATIONAL CONFERENCE ON GEOMETRY AND GRAPHICS ©2012 ISGG 1–5 AUGUST, 2012, MONTREAL, CANADA 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, cross- point 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 illus- trated 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 cen- tral 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 ho- mogeneous 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.