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On Concurrence of Nine Euler Lines on the Morley's Global Journal of Advanced Research on Classical and Modern Geometries 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 centroid, orthocenter, and circumcenter of a given triangle are collinear [1]. There are a few directions that Euler line has been studied. For one, it is known that Euler line passes through other interesting triangle centers, such as the nine point center, the de Longchamps point, the Schiffler point, the Exeter point, and the Gosssard perspector [7]. For the direction that is most relevant to our work, it concerns system of triangles 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 orthocentric system 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, concurrent lines. 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) p3 t tan γ = − = − : 3 − 1 + tan(π=3) tan(γ) 1 + p3t 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) p3 s tan β = − = − : 3 − 1 + tan(π=3) tan(β) 1 + p3s 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 +2p3s t +8p3s t 8s t 16s t +10s t 6p3s t 16p3s t 28p3s t − − − − − 3s4+16s3t+40s2t2 16st3 t4+8p3s3+20p3s2t+32p3st2+2p3t3 18s2 48st 6p3t+9) − − − − − − 4 4 3 5 4 3 3 4 2 5 4 2 3 3 2 4 5 q1(s; t) = 2(3p3s t +3p3s t 6s t 3s t +3s t 4p3s t 18p3s t 15p3s t p3st − − − − − − +2s4t+26s3t2+18s2t3 7st4 t5+p3s4+11p3s3t+30p3s2t2+22p3st3+2p3t4 3s3 − − − 33s2t 30st2 3p3s2 9p3st 6p3t2 + 9s + 9t) − − − − − 4 4 4 3 4 2 3 3 2 4 4 3 2 2 3 4 4 p2(s; t) = t(3p3s t 6s t 4p3s t 16p3s t 6p3s t +2s t+32s t +4s t 8st +p3s − − − − − +16p3s3t + 32p3s2t2 + 16p3st3 p3t4 44s2t 16st2 + 10t3 10p3s2 16p3st − − − − − ON CONCURRENCE OF NINE EULER LINES ON THE MORLEY'S CONFIGURATION 131 12p3t2 + 24s + 18t 3p3) − − 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 +2p3s t 4p3s t +4s t 10s t +4s t +2p3s t+2p3s t +20p3s t − − − +2p3s2t4+s5 2s4t+4s3t2 36s2t3 7st4 2p3s4 12p3s3t 40p3s2t2 14p3st3 2p3t4 − − − − − − − − +12s2t + 48st2 + 6t3 + 6p3s2 + 18p3st + 6p3t2 9s 18t) − − 5 3 4 4 5 2 4 3 3 4 5 4 2 3 3 2 4 5 s1(s; t) = 2(3p3s t +3p3s t +3s t 3s t 6s t p3s t 15p3s t 18p3s t 4p3s t s − − − − − − − 7s4t+18s3t2+26s2t3+2st4+2p3s4+22p3s3t+30p3s2t2+11p3st3+p3t4 30s2t 33st2 3t3 − − − − 6p3s2 9p3st 3p3t2 + 9s + 9t) − − − 4 4 3 4 4 2 3 3 2 4 4 3 2 2 3 4 4 r2(s; t) = s(3p3s t 6s t 6p3s t 16p3s t 4p3s t 8s t+4s t +32s t +2st p3s − − − − − − +16p3s3t + 32p3s2t2 + 16p3st3 + p3t4 + 10s3 16s2t 44st2 12p3s2 − − − 16p3st 10p3t2 + 18s + 24t 3p3) − − − 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 2p3s t 10p3s t 12s t +10s t +37s t 2p3s t+42p3s t +18p3s t − − − − − 18p3s4t4+s7+34s6t 156s5t2 154s4t3 7s3t4 4p3s6 78p3s5t+54p3s4t2+106p3s3t3 − − − − − − +18p3s2t4+15s5+270s4t+156s3t2 18s2t3 9st4 150p3s3t 162p3s2t2 42p3st3 6p3t4 − − − − − − 45s3 + 54s2t + 108st2 + 18t3 + 36p3s2 + 54p3st + 18p3t2 27s 54t) − − − 7 3 6 4 7 2 6 3 5 4 7 6 2 5 3 4 4 m1(s; t) = 2(3p3s t +3p3s t +3s t 21s t 24s t p3s t 21p3s t 3p3s t +17p3s t − − − − − s7 s6t+117s5t2+125s4t3+8s3t4+4p3s6+33p3s5t 51p3s4t2 95p3s3t3 15p3s2t4 15s5 − − − − − − 183s4t 159s3t2+9s2t3+117p3s3t+153p3s2t2+39p3st3+3p3t4+45s3 27s2t 81st2 9t3 − − − − − 36p3s2 45p3st 9p3t2 + 27s + 27t) − − − 4 4 4 3 3 4 4 2 3 3 2 4 3 2 2 3 4 3 l2(s; t) = s(p3s t 8s t 8s t 6p3s t +8p3s t +6p3s t +48s t +24s t +p3s +8p3s t − − − 36p3s2t2 24p3st3 3p3t4 8s3 56s2t+6p3s2+40p3st+18p3t2 24t 3p3)(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 4p3s t +4p3s t 11s t 32s t 6s t +16p3s t +12p3s t 4p3st − − − − − +9s4t2+96s3t3+78s2t4+32st5 t6 12p3s4t 84p3s3t2 48p3s2t3 12p3st4+4p3t5 9s4 − − − − − − 162s2t2 96st3 15t4+36p3s3+144p3s2t+84p3st2 54s2+45t2 36p3s 36p3t+27) − − − − − − 4 6 3 7 4 5 3 6 2 7 4 4 3 5 2 6 7 v1(s; t) = 2(3p3s t +3p3s t 24s t 21s t +3s t +17p3s t 3p3s t 21p3s t p3st − − − − − +8s4t3+125s3t4+117s2t5 st6 t7 15p3s4t2 95p3s3t3 51p3s2t4+33p3st5+4p3t6+9s3t2 − − − − − 159s2t3 183st4 15t5+3p3s4+39p3s3t+153p3s2t2+117p3st3 9s3 81s2t 27st2+45t3 − − − − − − 9p3s2 45p3st 36p3t2 + 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(p3s t 8s t 8s t +3p3s t +8p3s t 6p3s t +24s t +48s t +48s t 21p3s t − − − − 48p3s3t3 18p3s2t4+8p3st5+p3t6 72s3t2 144s2t3 56st4 8t5+9p3s4+72p3s3t+126p3s2t2 − − − − − − +16p3st3 + 3p3t4 + 144st2 + 24t3 54p3s2 120p3st 21p3t2 + 72s + 9p3) − − − v2(s; t) = v1(s; t).
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