Conics Associated with Triangles, Or How Poncelet Meets Morley

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P2 W1 C2 U1

C1 V1

P3 P5 A C

(a) (b)

Figure 1.1: (a) The closed in 5 steps Poncelet chain P1 ...P5 for the nested ellipses C1 and C2; (b) Morley’s trisector theorem.

2 When six points are conconic?

It is well known (see [2], or [5]) that any five distinct points (or lines) in general position on the plane uniquely determine a non-degenerate conic passing through (respectively, touching) these points (respectively, lines). However, if three or more points are collinear (respectively, three or more lines intersect in the same point), then the conic can degenerate and can even fail to be unique. Everything said above also holds on the projective plane, that is the Euclidean plane equipped with the so-called line at infinity consisting of points at infinity in which intersect all families of parallel lines. For the detailed exposition of projective geometry we send the reader to [8]. Since five points determine a conic, the right question to ask is when a sixth point lie on a single conic. We now state one of the possible answers to this question. Recall that a ray from a vertex of a triangle through a point on the opposite side (or its continuation) called a cevian.

Theorem 1. Let AA1, BB1, CC1 be three cevians of △ABC mutually meeting at the points X1 = BB1 ∩ CC1, Y1 = AA1 ∩ CC1, and Z1 = AA1 ∩ BB1. Suppose also that AA2, BB2, CC2 is an another triple of cevians meeting at the similarly constructed points X2, Y2, and Z2. If we further put U1 := BB1 ∩CC2, V1 := CC1 ∩ AA2, and W1 := AA1 ∩ BB2 (see Fig. 2.1 (a)), then the following conditions are equivalent:

1. A1, A2, B1, B2, C1, C2 lie on a single conic C0;

2. X1, X2, Y1, Y2, Z1, Z2 lie on a single conic C1;

3. AA1, AA2, BB1, BB2, CC1, and CC2 touch the same conic C2;

4. AU1, BV1, and CW1 intersect in a single point. Remark. Some implications of Theorem 1 seem to be folkloric statements re- discovered independently by several authors. Although the author of the present

2 note was aware about the statement of the theorem in 2007, the equivalence of 1, 3, and 4 ﬁrst appeared in [9]. The equivalence of 1, 2, and 3 was proved in [19] using barycentric coordinates. Recently in [3] D. Barali´calso proved the equivalence of 1, 2, and 3 by using a smart application of Carnot’s theorem. Our approach will be more straightforward then in the previous works.

B ¯b a¯1

a¯2 ¯ ¯ U1(b1,c2) C2(0,c2) c W¯ , 2 O¯ 1 1 q C2 A1 C 0 X2 C¯ (0,1) Z 1 b C 1 V¯ 1,1 1 1 p W1 C1 C U Y1 2 1 ¯ ¯ ¯ c¯ A B2(1,0) B1(b1,0) O Y2 ¯ A2 Q(q,0)

V1 X Z2 1 A C B2 B1 P¯(0,p) (a) (b)

Figure 2.1: (a) The conﬁguration of Theorem 1; (b) The conﬁguration of The- orem 1 after a suitable projective transformation.

Proof. Let us start with showing the equivalence of 1 and 4. Since the construction of the theorem is invariant under projective transformations, we can apply a particular one to help us. More precisely, let us make the transformation that maps the line BC to the line at infinity. This is a standard trick in projective geometry that simplifies the whole picture. If we denote the resulting image of an arbitrary point X as X¯, then B¯, C¯, A¯1, A¯2 are the points at infinity (see Fig. 2.1 (b)). Put ¯b, c¯, a¯1, and a¯2 for the images of the lines, respectively, AB, AC, AA1, and AA2. Furthermore, we can assume that the lines ¯b and c¯ are perpendicular, and A¯C¯1 = A¯B¯2 =1. Introduce the coordinate system with the origin at A¯ and axises ¯b and c¯, and set the coordinates for the points as B¯1(b1, 0), C¯2(0,c2), P¯(0,p), Q¯(q, 0) with P := AB ∩ A2B1 and Q := AC ∩ A1C2 (possibly being points at infinity). Since the points A1 and A2 were sent to infinity, we get a¯1 k Q¯C¯2 and a¯2 k B¯1P¯. Using these relations, it is straightforward to show that the points U¯1, V¯1, and W¯ 1 have the following coordinates: U¯1 = (b1,c2), V¯1 = (b1/p, 1), and W¯ 1 = (1,c2/q). If O = BV1 ∩ CW1, then the condition 4 is equivalent to O ∈ AU1, or O¯ ∈ A¯U¯1. It is easy to see that O¯ = (b1/p,c2/q). Therefore O¯ ∈ A¯U¯1 if and only if p = q. This equality is equivalent to the fact that Q¯P¯ k C¯1B¯2. From here we can get the equivalence of 1 and 4 by incorporating Pascal’s theorem stating that on the projective plane the six vertexes of a hexagon lie on the same conic if and only if the three points of intersection of the opposite sides of the hexagon are collinear (see [2], or [8]). Using this fundamental fact, Q¯P¯ k C¯1B¯2 is, in turn, equivalent to the fact that

3 A1, A2, B1, B2, C1, and C2 lie on a single conic C0. The equivalence 1 ⇔ 4 is proved. The equivalence 3 ⇔ 4 is the corollary of an another fundamental fact, namely, Brianchon’s theorem which states that on the projective plane the side- lines of a hexagon are tangent to a single conic if and only if three lines joining the opposite vertexes meet at a single point (see [2], [8]). Finally, let us prove the equivalence 2 ⇔ 3. Suppose 2 holds. Consider the ′ conic C2 that simultaneously touches all the cevians except CC2. We know that ′ ′ C2 is uniquely deﬁned. Hence, we have two conics C1 and C2 and the triangle ′ X1Y1Z1 whose vertexes lie on C1 and whose sides are tangent to C2. Therefore, by Poncelet’s theorem, starting from any point on C1 the corresponding Poncelet chain always closes up in three steps. Let us pick the point X2. Since the Poncelet chain starting at X2 must close in three steps, we get that Y2Z2 is also ′ ′ tangent to C2. Thus C2 = C2, and the implication 2 ⇒ 3 is proved. The reverse implication can be proved by reversing the arguments above. We note that this shortcut with Poncelet’s theorem also appeared in [3] (see [15] for exploiting this idea to give a proof for Poncelet’s theorem). The last thing we should note here is that we omitted possible degenerate cases since the computation in such cases can be easily restored by the reader. Theorem 1 is proved.

3 What about Morley?

Now we will see how both celebrated facts mentioned in the introduction can meet each other. Again, let U1, V1, and W1 be intersection points of the adjacent angle trisectors of △ABC from the Morley configuration. It is known that the lines AU1, BV1, CW1 intersect in a single point called the second Morley center (with the first Morley center being just the center of the Morley triangle) (see [16]). Thus, using conditions 2 and 3 of Theorem 1 in the configuration of Morley’s theorem we can construct two conics C1 and C2 for which there exist two closed Poncelet chains X1Y1Z1 and X2Y2Z2. And Poncelet’s theorem tells us that there are infinitely many such triangles for each starting point on C1! Hence, in the Morley configuration one can find two quite “rare” objects – a pair of conics for which Poncelet chains are closing up! For an arbitrary given pair of conics such a conclusion is not always the case. There is still much discussion about simple, like Fuss’s formulas, and powerful, like general Cayley’s theorem, criteria ensuring existence of a closed Poncelet chain for a given pair of conics (see [12, Ch. 10], [10] for further details).

4 Closing remarks

In fact, Theorem 1 in Morley’s case gives us more, namely, that the intersections of the angular trisectors with the sides of the triangle belong to a single conic, which is a consequence of condition 1 of Theorem 1. Surprisingly, the same holds for any triple of isogonally conjugate cevians, that is cevians symmetric with respect to the corresponding angle bisectors. More precisely, one can get the following

4 Theorem 2. If AA1, AA2, BB1,BB2, and CC1, CC2 are three pairs of isogonally conjugate cevians of △ABC, then condition 1 of Theorem 1 holds. This result can be easily proved using condition 3 of Theorem 1 via straightforward calculation in coordinates; see also [9]. The fact dual to Theorem 2 is the following

Theorem 3. Suppose AA1, AA2, BB1,BB2, and CC1, CC2 are three pairs of isotomically conjugate cevians, that is cevians for which the pairs of points A1, A2, B1, B2, and C1, C2 are symmetric with respect to the midpoints of the corresponding sides; then condition 1 of Theorem 1 holds. On account of projective equivalence of the mentioned types of conjugacy we refer the reader to [1]. Theorem 3 may be proved by a direct computation using Theorem 1, or can be obtained as a corollary of Carnot’s theorem (see [2]). Finally, we can get another corollary of Theorem 1 “for free” (see [13] and [14] for an alternative approach).

Theorem 4. Let P1 and P2 be any two points inside △ABC. If for any i ∈ {1, 2} cevians AAi, BBi, CCi are passing through Pi, then condition 1 of Theorem 1 holds.

Proof. Since in this case Xi = Yi = Zi = Pi for any i ∈ {1, 2}, condition 2 of Theorem 1 holds trivially for a degenerate conic C1 being the (doubly covered) line passing through P1 and P2.

Acknowledgment

The author would like to thank Yuliya V. Eremenko for her guidance during the preparation for the Regional scientiﬁc contest in 2007, during which a part of this work was done, and for all her help ever since.

References

[1] A. V. Akopyan, Conjugation of lines with respect to a triangle, Journal of Classical Geometry, 1 (2012) 23–31. [2] A. V. Akopyan, A. A. Zaslavsky, Geometry of conics, Amer. Math. Soc., Providence, RI, 2007. [3] D. Barali´c, A Short Proof of the Bradley Theorem, Amer. Math. Monthly (to appear, also available at arXiv:1308.6144 [math.MG]) [4] W. Barth, T. Bauer, Poncelet theorems, Expos. Math., 14 (1996) 125–144. [5] M. Berger, Geometry, Springer-Verlag, Berlin, 1987. [6] A. Connes, A new proof of Morley’s theorem, Publications Math´ematiques de l’IHES,´ S88 (1998), 43-46 [7] J. H. Conway, The power of mathematics, in Power, (eds. A. Blackwell, D. MacKay), Cambridge University Press, 2006; 36–50.

5 [8] H. S. M. Coxeter, Projective geometry, 2nd ed., Springer-Verlag, New York, 2003. [9] P. Dolgirev, On lines in a triangle tangent to a conic, Matematicheskoye Obrazovaniye, 59-60: 3-4 (2011) 15–24 (in Russian) (also available at arXiv:1101.3283 [math.MG]). [10] V. Dragović, M. Radnović, Poncelet porisms and beyond: integrable billiards, hyperelliptic Jacobians and pencils of quadrics, Frontiers in Mathe- matics, Springer Birkhäuser, Basel, 2011. [11] J. J. Duistermaat, Discrete integrable systems: QRT maps and elliptic sur- faces, Springer, 2010. [12] L. Flatto, Poncelet’s theorem, American Mathematical Society, Providence, RI, 2009. [13] D.Gale, Triangles and Proofs, Math. Intell., 18:1 (1996) 31–34. [14] B. Gibert, Bicevian Conics and CPCC Cubics, avaliable at http://bernard.gibert.pagesperso-orange.fr/. [15] L. Halbeisen, N. Hungerb´’uhler, A Simple Proof of Poncelet’s Theorem, Amer. Math. Monthly (to appear) [16] C. Kimberling, Central points and central lines in the plane of a triangle, Math. Mag. 67 (1994), 163–187. [17] M. Levi, S. Tabachnikov, The Poncelet grid and billiards in ellipses, Amer. Math. Monthly, 114 (2007), 895–908. [18] C. O. Oakley, J. C. Baker, The Morley trisector theorem, Amer. Math. Monthly, 85 (1978) 737–745. [19] Z. Szilasi,Two applications of the theorem of Carnot, Annales Mathemati- cae et Informaticae, 40 (2012), 135–144.

Kostiantyn Drach: V.N. Karazin Kharkiv National University, Geometry Department, Svobody Sq. 4, 61022, Ukraine; and Sumy State University, Department of Mathematical Analysis and Optimiza- tion, Rimskogo–Korsakova str. 2, 40007, Ukraine E-mail: [email protected]