Forum Geometricorum Volume 9 (2009) 119 123. FORUM GEOM ISSN 1534-1178 Note on the nticomplements of the Fermat Points osmin Pohoata bstract. We show that each of the anticomplements of the Fermat points is common to a triad of circles involving the triangle of reflection. We also generate two new triangle centers as common points to two other triads of circles. Finally, we present several circles passing through these new centers and the anticomplements of the Fermat points. 1. Introduction The Fermat points F ± are the common points of the lines joining the vertices of a triangle T to the apices of the equilateral triangles erected on the corresponding sides. They are also known as the isogonic centers (see [2, pp.107, 151]) and are among the basic triangle centers. In [4], they appear as the triangle centers X 13 and X 14. Not much, however, is known about their anticomplements, which are the points P ± which divide F ± G in the ratio F ± G : GP ± = 1 : 2. Given triangle T with vertices,,, (i) let,, be the reflections of the vertices,, in the respective opposite sides, and (ii) for ε = ±1, let ε, ε, ε be the apices of the equilateral triangles erected on the sides,, of triangle respectively, on opposite or the same sides of the vertices according as ε = 1 or 1 (see Figures 1 and 1). Theorem 1. For ε = ±1, the circumcircles of triangles ε ε, ε ε, ε ε are concurrent at the anticomplement P ε of the Fermat point F ε. 2. Proof of Theorem 1 For ε = ±1, let O a,ε be the center of the equilateral triangle ε ; similarly for O b,ε and O c,ε. (1) We first note that O a, ε is the center of the circle through, ε, and ε. Rotating triangle O a,ε through by an angle ε π 3, we obtain triangle O a, ε ε. Therefore, the triangles are congruent and O a, ε ε = O a,ε. Similarly, O a, ε ε = O a,ε. learly, O a,ε = O a, ε. It follows that O a, ε is the center of the circle through, ε and ε. Figures 1 and 1 illustrate the cases ε = +1 and ε = 1 respectively. (2) Let 1 1 1 be the anticomplementary triangle of. Since 1 and + have a common midpoint, + 1 is a parallelogram. The lines ε and 1 ε are parallel. Since 1 is the anticomplement of, it follows that the line 1 ε is the anticomplement of the line ε. Similarly, 1 ε and 1 ε are the anticomplements of the lines ε and ε. Since ε, ε, ε Publication Date: June 16, 2009. ommunicating Editor: Paul Yiu.
120. Pohoata + O a, + O a, Figure 1 Figure 1 concur at F ε, it follows that 1 ε, 1 ε, 1 ε concur at the anticomplement of F ε. This is the point P ε. (3) 1 1 1 is also the ε-fermat triangle of ε ε ε. (i) Triangles ε ε and ε 1 are congruent, since ε = ε, ε = = 1, and each of the angles ε ε and ε 1 is min ( + 2π 3, + + π 3). It follows that ε ε = ε 1. (ii) Triangles ε ε and 1 ε are also congruent for the same reason, and we have ε ε = 1 ε. It follows that triangle 1 ε ε is equilateral, and ε 1 ε = π 3. (4) ecause P ε is the second Fermat point of 1 1 1, we may assume ε P ε ε = π 3. Therefore, P ε lies on the circumcircle of 1 ε ε, which is the same as that of ε ε. On the other hand, since the quadrilateral + 1 is a parallelogram (the diagonals 1 and + have a common midpoint D, the midpoint of segment ), the anticomplement of the line coincides with 1 +. It now follows that the lines 1 +, 1 +, 1 + are concurrent at the anticomplement P of the second Fermat points, and furthermore, + P + = π 3. Since + + = 2π ( + + + + ) = 4π 3 = π + + 3 = + + + M = + M,
note on the anticomplements of the Fermat points 121 + 1 1 + F + P O a 1 + Figure 2 it follows that the triangles + + and + M are congruent. Likewise, + + = M +, and so the triangles + + and M + are congruent. Therefore, the triangle + M + is equilateral, and thus + M + = π 3. ombining this with + P + = 60, yields that the quadrilateral MP + + is cyclic, and since M + + is also cyclic, we conclude that the anticomplement P of the second Fermat point F lies on the circumcircle of triangle + +. Similarly, P lies on the circumcircles of triangles + +, and + +, respectively. This completes the proof of Theorem 1. 3. Two new triangle centers y using the same method as in [6], we generate two other concurrent triads of circles. Theorem 2. For ε = ±1, the circumcircles of the triangles ε, ε, ε are concurrent. Proof. onsider the inversion Ψ with respect to the anticomplement of the second Fermat point. ccording to Theorem 1, the images of the circumcircles of triangles + +, + +, + + are three lines which bound a triangle + + +,
122. Pohoata where +, +, + are the images of +, +, and +, respectively. Since the images,, of,, under Ψ lie on the sidelines + +, + +, + +, respectively, by Miquel s theorem, we conclude that the circumcircles of triangles +, +, + are concurrent. Thus, the circumcircles of triangles +, +, + are also concurrent (see Figure 3). Similarly, inverting with respect to the anticomplement of the first Fermat point, by Miquel s theorem, one can deduce that the circumcircles of triangles,, are concurrent. + + U + + Figure 3. Javier Francisco Garcia apitan has kindly communicated that their points of concurrency do not appear in [4]. We will further denote these points by U +, and U, respectively. We name these centers U +, U the inversive associates of the anticomplements P +, P of the Fermat points. 4. ircles around P ± and their inversive associates We denote by O, H the circumcenter, and orthocenter of triangle. Let J +, J be respectively the inner and outer isodynamic points of the triangle. Though the last two are known in literature as the common two points of the pollonius circles, L. Evans [1] gives a direct relation between them and the Napoleonic configuration, defining them as the perspectors of the triangle of reflections
note on the anticomplements of the Fermat points 123 with each of the Fermat triangles + + +, and. They appear as X 15, X 16 in [4]. Furthermore, let W +, W be the Wernau points of triangle. These points are known as the common points of the folloing triads of circles: + +, + +, + +, and respectively,, [3]. ccording to the above terminology, W +, W are the inversive associates of the Fermat points F +, and F. They appear as X 1337 and X 1338 in [4]. We conclude with a list of concyclic quadruples involving these triangle centers. The first one is an immediate consequence of the famous Lester circle theorem [5]. The other results have been verified with the aid of Mathematica. Theorem 3. The following quadruples of points are concyclic: (i) P +, P, O, H; (ii) P +, P, F +, J + ; (ii ) P +, P, F, J ; (iii) P +, U +, F +, O; (iii ) P, U, F, O; (iv) P +, U, F +, W + ; (iv ) P, U +, F, W ; (v) U +, J +, W +, W ; (v ) U, J, W +, W. References [1] L. Evans, Some configurations of triangle centers, Forum Geom., 3 (2003) 49 56. [2] W. Gallatly, The Modern Geometry of the Triangle, 2nd ed., London: Hodgson, 1913. [3] D. Grinberg, Hyacinthos messages 6874, 6881, 6882, pril, 2003. [4]. Kimberling, Encyclopedia of Triangle enters, available at http://faculty.evansville.edu/ck6/encyclopedia/et.html. [5] J. Lester, Triangles III: omplex Triangle Functions, equationes Math., 53 (1997) 4 35. [6]. Pohoata, On the Parry reflection point, Forum Geom., 8 (2008) 43 48. osmin Pohoata: 13 Pridvorului Street, ucharest, Romania 010014 E-mail address: pohoata cosmin2000@yahoo.com