Forum Geometricorum b Volume 9 (2009) 271–274. b b FORUM GEOM ISSN 1534-1178

Ten Concurrent Euler Lines

Nikolai Ivanov Beluhov

Dedicated to Svetlozar Doichev

Abstract. Let F1 and F2 denote the Fermat-Toricelli points of a given ABC. We prove that the Euler lines of the 10 with vertices chosen from A, B, C, F1, F2 (three at a time) are concurrent at the of triangle ABC.

Given a (positively oriented) triangle ABC, construct externally on its sides three equilateral triangles BCTa, CATb, and ABTc with centers Na, Nb and Nc respectively (see Figure 1). As is well known, triangle NaNbNc is equilateral. We call this the first Napoleon triangle of ABC.

Ta

F C 2 ′ Tc

Na

Tb Nb ′ Nb ′ ′ M Nc T F b 1 ′ Na

A B

Nc ′ Ta

Tc

Figure 1.

′ The same construction performed internally gives equilateral triangles BCTa, ′ ′ ′ ′ ′ CATb and ABTc with centers Na, Nb, and Nc respectively, leading to the second ′ ′ ′ Napoleon triangle NaNbNc. The centers of both Napoleon triangles coincide with the centroid M of triangle ABC.

Publication Date: October 19, 2009. Communicating Editor: Paul Yiu. 272 N. I. Beluhov

The lines ATa, BTb and CTc make equal pairwise , and meet together with the circumcircles of triangles BCTa, CATb, and ABTc at the first Fermat- Toricelli point F1. Denoting by ∠XY Z the oriented ∠(YX,YZ), we have ◦ ∠AF1B = ∠BF1C = ∠CF1A = 120 . Analogously, the second Fermat- ◦ Toricelli point satisfies ∠AF2B = ∠BF2C = ∠CF2A = 60 . Clearly, the sides of the Napoleon triangles are the bisectors of the segments joining their respective Fermat-Toricelli points with the vertices of triangle ABC (as these segments are the common chords of the circumcircles of the the equilateral triangles ABTc, BCTa, etc). We prove the following interesting theorem.

Theorem The Euler lines of the ten triangles with vertices from the set {A, B, C, F1, F2} are concurrent at the centroid M of triangle ABC. Proof. We divide the ten triangles in three types: (I): Triangle ABC by itself, for which the claim is trivial. (II): The six triangles each with two vertices from the set {A,B,C} and the re- maining vertex one of the points F1, F2. (III) The three triangles each with vertices F1, F2, and one from {A,B,C}. For type (II), it is enough to consider triangle ABF1. Let Mc be its centroid and MC be the of the segment AB. Notice also that Nc is the circumcenter of triangle ABF1 (see Figure 2). C

M F1

Mc A B MC

Nc

Tc Figure 2.

Now, the points C, F1 and Tc are collinear, and the points M, Mc and Nc divide the segments MC C, MC F1 and MC Tc in the same ratio 1 : 2. Therefore, they are collinear, and the Euler of triangle ABF contains M. Ten concurrent Euler lines 273

For type (III), it is enough to consider triangle CF1F2. Let Mc and Oc be its centroid and circumcenter. Let also MC and MF be the of AB and ′ ′ F1F2. Notice that Oc is the intersection of NaNb and NaNb as perpendicular ′ ′ bisectors of F1C and F2C. Let also P be the intersection of NbNc and NcNa, and ′ F be the reflection of F1 in MC (see Figure 3). C

O c Na

Nb

′ Nb Mc

P F M N ′ 2 ′ c Na MF F1

A MC B

F ′ Nc

Figure 3.

◦ ′ ′ The rotation of center M and angle 120 maps the lines NaNb and NaNb into ′ ′ ◦ NbNc and NcNa respectively. Therefore, it maps Oc to P , and ∠OcMP = 120 . ′ ◦ ′ Since ∠OcNaP = 120 , the four points Oc, M, Na, P are concyclic. The ◦ containing them also contains Nb since ∠P NbOc = 60 . Therefore, ∠OcMNb = ′ ∠OcNaNb. ′ ′ ′ The same rotation maps angle OcNaNb onto angle P NcNc, yielding ∠OcNaNb = ′ ′ ′ ′ ∠P NcNc. Since P Nc ⊥ BF2 and NcNc ⊥ BA, ∠P NcNc = ∠F2BA. ′ ◦ ◦ ′ Since ∠BF A = ∠AF1B = 120 = 180 −∠AF2B, the AF2BF ′ ′ is also cyclic and ∠F2BA = ∠F2F A. Thus, ∠F2F A = ∠OcMNb. ′ ′ Now, AF kF1B ⊥ NaNc and NbM ⊥ NaNc yield AF kNbM. This, together ′ ′ with ∠F2F A = ∠OcMNb, yields F F2kMOc. Notice now that the points Mc and M divide the segments CMF and CMC in ratio 1 : 2, therefore McMkMC MF . The same argument, applied to the segments ′ ′ F1F2 and F1F with ratio 1 : 1, yields MC MF kF F2. ′ In conclusion, we obtain McMkF F2kMOc. The collinearity of the points Mc, M and Oc follows.  274 N. I. Beluhov

References [1] N. Beluhov, Sets of Euler lines, Matematika +, 2/2006. [2] I. F. Sharygin, Geometriya. Planimetriya, Drofa, 2001.

Nikolai Ivanov Beluhov: ”Bulgarsko opalchenie” str. 5, Stara Zagora 6000, Bulgaria E-mail address: [email protected]