Global Journal of Advanced Research on Classical and Modern ISSN: 2284-5569, Vol.9, (2020), Issue 2, pp.127-132

ON CONCURRENCE OF NINE EULER LINES ON THE MORLEY’S CONFIGURATION

CHERNG-TIAO PERNG AND TRAN QUANG HUNG

Abstract. We establish the concurrence of nine Euler lines on the configura- tion of Morley’s trisector theorem with proof using computer algebra.

Euler proved in 1765 that the , orthocenter, and circumcenter of a given are collinear [1]. There are a few directions that Euler has been studied. For one, it is known that passes through other interesting triangle centers, such as the nine center, the , the Schiffler point, the , and the Gosssard perspector [7]. For the direction that is most relevant to our work, it concerns system of with concurrent Euler lines. It is known that for a triangle ABC and its two Fermat points F1 and F2, the Euler lines of the 10 triangles with vertices chosen from A, B, C, F1, and F2 are concurrent at the centroid of the triangle ABC [2]. Furthermore, the Euler lines of the four triangles formed by an are concurrent at the nine-point center common to all of the triangles [3]. On the other hand, Morley’s trisector theorem was discovered by Anglo-American mathematician Frank Morley in 1899. Since its discovery, there are many proofs, some of which are very technical. Recent proofs include the algebraic proof by Alain Connes (1998, 2004, [4], [5]), John Conway’s elementary geometry proof [6] and a very recent one by the second author [8]. Our following result states that nine Euler lines in an extended Morley’s configuration are concurrent (see Figure 1).

Theorem 1. Let ABC be a triangle with Morley triangle XYZ. Let Ka,Kb and Kc be circumcenters of triangles AY Z, BZX, and CXY respectively. Let La, Lb, and Lc be circumcenters of triangles XBC, YCA, and ZAB respectively. Then the Euler lines of the eight triangles

KaKbKc, LaLbLc, KaLbLc, KbLcLa, 4 4 4 4 KcLaLb, LaKbKc, LbKcKa, LcKaKb 4 4 4 4 are concurrent at circumcenter of triangle ABC. Thus with Euler line of ABC, we have nine concurrent Euler lines.

Date: June 3, 2020. 2010 Mathematics Subject Classification. 51M04, 51-03. Key words and phrases. Morley’s trisector theorem, Morley’s triangle, Euler line, . 127 128 CHERNG-TIAO PERNG AND TRAN QUANG HUNG

ELbKcKa

EKaLbLc

Lb

A

ELcKaKb Ka Lc

Y Z

O ABC X E Kc

Kb EKaKbKc ELaLbLc B C

EKcLaLb EKbLcLa

La

ELaKbKc

Figure 1. Extended Morley’s configuration

Proof. Without loss of generality, denote B and C by

B = (0, 0),C = (1, 0), ∠B = 3β, ∠C = 3γ. Then line AB and line AC have slope tan(3β) and tan(3γ), respectively. For convenience, denote − tan(β) = s, tan(γ) = t. It follows that 3 tan(β) tan3(β) 3s s3 tan(3β) = − = − =: m 1 3 tan2(β) 1 3s2 ab − − and 3 tan(γ) tan3(γ) 3t t3 tan(3γ) = − = − =: mac. − − 1 3 tan2(γ) −1 3t2 − − ON CONCURRENCE OF NINE EULER LINES ON THE MORLEY’S CONFIGURATION 129

Lb

A

Lc Ka

Y Z

O X

Kc Kb

B Z0 Y 0 C

La

Figure 2. Proof of extended Morley’s configuration

Solving the equations for AB and AC as follows, one gets the coordinates for A = (a, b): y = mabx, y = mac(x 1),A = (a, b). − To solve for the Morley triangle XYZ, we need to determine the equations for the lines BZ, BX, CY, CX, AZ and AY . The first four are straightforward. For the last two, we apply some geometry: Let AZ intersect BC at Z0 and AY intersect BC at Y 0 (see Figure 2). Then π π  AZ0C = ABC + BAZ0 = 3β + (β + γ) = 2β + γ ∠ ∠ ∠ 3 − 3 − and π π  AY 0B = BCA + CAY 0 = 3γ + (β + γ) = 2γ + β . ∠ ∠ ∠ 3 − 3 − 0 Denoting the slope of BZ as mbz, the slope of AZ = AZ as maz, etc., we find that 2 tan(β) 2s m = tan(2β) = = bz 1 tan2(β) 1 s2 − − mbx = tan(β) = s, mcx = tan(γ) = t − − 2 tan(γ) 2t mcy = tan(2γ) = = − −1 tan2(γ) −1 t2 − − 130 CHERNG-TIAO PERNG AND TRAN QUANG HUNG

π  π  tan(2β) + tan( 3 γ) maz = tan 2β + γ = − , 3 − 1 tan(2β) tan( π γ) − 3 − where tan(2β) = 2s/(1 s2) and − π  tan(π/3) tan(γ) √3 t tan γ = − = − . 3 − 1 + tan(π/3) tan(γ) 1 + √3t Similarly, π  π  tan(2γ) + tan( 3 β) may = tan 2γ + β = − , − 3 − −1 tan(2γ) tan( π β) − 3 − where tan(2γ) = 2t/(1 t2) and − π  tan(π/3) tan(β) √3 s tan β = − = − . 3 − 1 + tan(π/3) tan(β) 1 + √3s Now we can write down all the linear equations in order to solve for X,Y and Z:

Lbz : y = mbzx

Lbx : y = mbxx

Lcy : y = mcy(x 1) − Lcx : y = mcx(x 1) − Laz : y b = maz(x a) − − Lay : y b = may(x a), − − where Lbz is the equation for line BZ, etc.

Once we have all these equations, it is straightforward to solve for X,Y and Z, and then all the circumcenters, and Euler lines. Note that solving these and checking for concurrency is easy by a computer algebra system such as SAGE [9]. For com- pleteness, we record the results as follows.

t(4s3t s2t2 5s2 4st t2 + 3) st(s2t2 + s2 + 8st + t2 7)  Ka = − − − − , − (3s2t2 s2 8st t2 + 3)(s + t) −(3s2t2 s2 8st t2 + 3)(s + t) − − − − − −   p1(s, t) p2(s, t) Kb = , , q1(s, t) q2(s, t) where 4 4 4 3 3 4 4 2 3 3 2 4 4 3 2 2 3 p1(s, t) = t(3s t +2√3s t +8√3s t 8s t 16s t +10s t 6√3s t 16√3s t 28√3s t − − − − − 3s4+16s3t+40s2t2 16st3 t4+8√3s3+20√3s2t+32√3st2+2√3t3 18s2 48st 6√3t+9) − − − − − −

4 4 3 5 4 3 3 4 2 5 4 2 3 3 2 4 5 q1(s, t) = 2(3√3s t +3√3s t 6s t 3s t +3s t 4√3s t 18√3s t 15√3s t √3st − − − − − − +2s4t+26s3t2+18s2t3 7st4 t5+√3s4+11√3s3t+30√3s2t2+22√3st3+2√3t4 3s3 − − − 33s2t 30st2 3√3s2 9√3st 6√3t2 + 9s + 9t) − − − − −

4 4 4 3 4 2 3 3 2 4 4 3 2 2 3 4 4 p2(s, t) = t(3√3s t 6s t 4√3s t 16√3s t 6√3s t +2s t+32s t +4s t 8st +√3s − − − − − +16√3s3t + 32√3s2t2 + 16√3st3 √3t4 44s2t 16st2 + 10t3 10√3s2 16√3st − − − − − ON CONCURRENCE OF NINE EULER LINES ON THE MORLEY’S CONFIGURATION 131

12√3t2 + 24s + 18t 3√3) − − q2(s, t) = q1(s, t)

  r1(s, t) r2(s, t) Kc = , , s1(s, t) s2(s, t) where 5 4 5 3 4 4 5 2 4 3 3 4 5 4 2 3 3 r1(s, t) = (3s t +2√3s t 4√3s t +4s t 10s t +4s t +2√3s t+2√3s t +20√3s t − − − +2√3s2t4+s5 2s4t+4s3t2 36s2t3 7st4 2√3s4 12√3s3t 40√3s2t2 14√3st3 2√3t4 − − − − − − − − +12s2t + 48st2 + 6t3 + 6√3s2 + 18√3st + 6√3t2 9s 18t) − −

5 3 4 4 5 2 4 3 3 4 5 4 2 3 3 2 4 5 s1(s, t) = 2(3√3s t +3√3s t +3s t 3s t 6s t √3s t 15√3s t 18√3s t 4√3s t s − − − − − − − 7s4t+18s3t2+26s2t3+2st4+2√3s4+22√3s3t+30√3s2t2+11√3st3+√3t4 30s2t 33st2 3t3 − − − − 6√3s2 9√3st 3√3t2 + 9s + 9t) − − −

4 4 3 4 4 2 3 3 2 4 4 3 2 2 3 4 4 r2(s, t) = s(3√3s t 6s t 6√3s t 16√3s t 4√3s t 8s t+4s t +32s t +2st √3s − − − − − − +16√3s3t + 32√3s2t2 + 16√3st3 + √3t4 + 10s3 16s2t 44st2 12√3s2 − − − 16√3st 10√3t2 + 18s + 24t 3√3) − − − s2(s, t) = s1(s, t)

1 st 1  L = , − a 2 2(s + t)   l1(s, t) l2(s, t) Lb = , , m1(s, t) m2(s, t) where 7 4 7 3 6 4 7 2 6 3 5 4 7 6 2 5 3 l1(s, t) = (3s t 2√3s t 10√3s t 12s t +10s t +37s t 2√3s t+42√3s t +18√3s t − − − − − 18√3s4t4+s7+34s6t 156s5t2 154s4t3 7s3t4 4√3s6 78√3s5t+54√3s4t2+106√3s3t3 − − − − − − +18√3s2t4+15s5+270s4t+156s3t2 18s2t3 9st4 150√3s3t 162√3s2t2 42√3st3 6√3t4 − − − − − − 45s3 + 54s2t + 108st2 + 18t3 + 36√3s2 + 54√3st + 18√3t2 27s 54t) − − −

7 3 6 4 7 2 6 3 5 4 7 6 2 5 3 4 4 m1(s, t) = 2(3√3s t +3√3s t +3s t 21s t 24s t √3s t 21√3s t 3√3s t +17√3s t − − − − − s7 s6t+117s5t2+125s4t3+8s3t4+4√3s6+33√3s5t 51√3s4t2 95√3s3t3 15√3s2t4 15s5 − − − − − − 183s4t 159s3t2+9s2t3+117√3s3t+153√3s2t2+39√3st3+3√3t4+45s3 27s2t 81st2 9t3 − − − − − 36√3s2 45√3st 9√3t2 + 27s + 27t) − − −

4 4 4 3 3 4 4 2 3 3 2 4 3 2 2 3 4 3 l2(s, t) = s(√3s t 8s t 8s t 6√3s t +8√3s t +6√3s t +48s t +24s t +√3s +8√3s t − − − 36√3s2t2 24√3st3 3√3t4 8s3 56s2t+6√3s2+40√3st+18√3t2 24t 3√3)(s2 3) − − − − − − − − m2(s, t) = m1(s, t) 132 CHERNG-TIAO PERNG AND TRAN QUANG HUNG

  u1(s, t) u2(s, t) Lc = , , v1(s, t) v2(s, t) where

4 6 4 5 3 6 4 4 3 5 2 6 4 3 3 4 6 u1(s, t) = t(3s t 4√3s t +4√3s t 11s t 32s t 6s t +16√3s t +12√3s t 4√3st − − − − − +9s4t2+96s3t3+78s2t4+32st5 t6 12√3s4t 84√3s3t2 48√3s2t3 12√3st4+4√3t5 9s4 − − − − − − 162s2t2 96st3 15t4+36√3s3+144√3s2t+84√3st2 54s2+45t2 36√3s 36√3t+27) − − − − − −

4 6 3 7 4 5 3 6 2 7 4 4 3 5 2 6 7 v1(s, t) = 2(3√3s t +3√3s t 24s t 21s t +3s t +17√3s t 3√3s t 21√3s t √3st − − − − − +8s4t3+125s3t4+117s2t5 st6 t7 15√3s4t2 95√3s3t3 51√3s2t4+33√3st5+4√3t6+9s3t2 − − − − − 159s2t3 183st4 15t5+3√3s4+39√3s3t+153√3s2t2+117√3st3 9s3 81s2t 27st2+45t3 − − − − − − 9√3s2 45√3st 36√3t2 + 27s + 27t) − − −

4 6 4 5 3 6 4 4 3 5 2 6 4 3 3 4 2 5 4 2 u2(s, t) = t(√3s t 8s t 8s t +3√3s t +8√3s t 6√3s t +24s t +48s t +48s t 21√3s t − − − − 48√3s3t3 18√3s2t4+8√3st5+√3t6 72s3t2 144s2t3 56st4 8t5+9√3s4+72√3s3t+126√3s2t2 − − − − − − +16√3st3 + 3√3t4 + 144st2 + 24t3 54√3s2 120√3st 21√3t2 + 72s + 9√3) − − − v2(s, t) = v1(s, t).

Finally the circumcenter of ABC is 1 (s2t2 3s2 8st 3t2 + 1)(st 1) O = , − − − − , 2 2(3s2t2 s2 8st t2 + 3)(s + t) − − − and all the Euler lines concur at this point.  References [1] Euler, Leonhard, Solutio facilis problematum quorundam geometricorum difficillimorum (Easy solution of some difficult geometric problems), Novi Commentarii academiae scientiarum Petropolitanae, 11 (1767), pp. 103–123. [2] N. I. Beluhov, Ten concurrent Euler lines, Forum Geom., 9 (2009), pp. 271–274. [3] A.-C. Nathan, College Geometry, Dover Publications, 2007. [4] A. Connes, A new proof of Morley’s theorem, Publications Math´ematiquesde l’I.H.E.S.´ , 88 (1998), pp. 43–46. [5] A. Connes, Symmetries, EMS Newsletter, 54 (2004), pp. 11–18. [6] J. Conway, On Morley’s Trisector Theorem, Math. Intell., 36 (2014) No. 3, pp. 4–7. [7] C. Kimberling, Encyclopedia of Triangle Centers, available at http://faculty.evansville. edu/ck6/encyclopedia/ETC.html. [8] Q. H. Tran, A direct trigonometric proof of Morley’s theorem, International Journal of Geom- etry, 8 (2019), No. 2, pp. 46–48. [9] Sage Notebook v6.10, http://sagemath.org.

Department of Mathematics, Norfolk State University, Norfolk, Virginia, USA Email address: [email protected]

High School for Gifted Students, Vietnam National University, Hanoi, Vietnam Email address: [email protected]