HYACINTHOS 26167

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of H.

Denote:

N1 = the NPC center of AB'C'

 

Na = the reflection of N1 in HA'. Similarly Nb, Nc.

1. ABC, NaNbNc are orthologic.

 

2. The centroid of  NaNbNc lies on the Euler line.

 [César Lozada]:

 

> 1. ABC, NaNbNc are orthologic.

Z(A->Na) = 1/((b^2+c^2)*a^6-(3*b^4-4*b^2* c^2+3*c^4)*a^4+(b^2+c^2)*(3*b^ 4-7*b^2*c^2+3*c^4)*a^2-(b^4+3* b^2*c^2+c^4)*(b^2-c^2)^2) : : (barycentrics)

= isogonal conjugate of {3,74}/\{30,52} (see Q below)

= [ 34.319511049911280, 32.74595740884502, -34.869388054944310 ]

 

Z(Na->A) = X(13474)

 

Q = {3,74}/\{30,52}

= a*((b^2+c^2)*a^6-(3*b^4-4*b^2* c^2+3*c^4)*a^4+(b^2+c^2)*(3*b^ 4-7*b^2*c^2+3*c^4)*a^2-(b^4+3* b^2*c^2+c^4)*(b^2-c^2)^2) : : (trilinears)

= (2*cos(2*A)+3)*cos(B-C)-6*cos( A) : : (trilinears)

= 11*X(3)-9*X(7998) = 9*X(3)-7*X(7999) = 7*X(3)-5*X(11444) = 5*X(3)-3*X(11459) = 3*X(3)-2*X(11591) = 9*X(3)-8*X(11592) = 2*X(4)-3*X(5946) = 3*X(4)-4*X(10095) = 5*X(5)-6*X(5892) = 11*X(5)-12*X(6688) = 3*X(5)-4*X(9729) = 7*X(5)-8*X(11695) = 11*X(5892)-10*X(6688) = 9*X(5892)-10*X(9729) = 21*X(5892)-20*X(11695) = 9*X(5946)-8*X(10095) = 9*X(6688)-11*X(9729)

= On lines:

{3,74}, {4,3521}, {5,2883}, {20,1154}, {26,10605}, {30,52}, {49,2071}, {51,3853}, {54,13445}, {64,7526}, {125,13406}, {140,12162}, {143,382}, {182,9968}, {184,11250}, {373,12811}, {376,10627}, {381,10574}, {389,3627}, {546,1514}, {548,5562}, {549,5907}, {550,6101}, {568,3146}, {974,10113}, {1181,12084}, {1204,1658}, {1216,8703}, {1350,9925}, {1493,13352}, {1498,6644}, {1657,5889}, {1986,11565}, {2772,5694}, {2888,12317}, {2935,11702}, {2937,8718}, {3060,5073}, {3091,13363}, {3357,10274}, {3529,6243}, {3530,5891}, {3534,11412}, {3567,3830}, {3581,12088}, {3843,11455}, {3845,5462}, {3851,11439}, {3858,5943}, {4846,6145}, {5055,11017}, {5059,13421}, {5076,9781}, {5446,13382}, {5449,10264}, {6642,12315}, {7502,7689}, {7527,13353}, {7530,9786}, {7728,11561}, {9818,13093}, {10226,13367}, {10625,12103}, {10733,13358}, {11003,12300}, {11465,12046}, {12085,12161}

= midpoint of X(i) and X(j) for these {i,j}: {3,6241}, {185,10575}, {382,12279}, {1657,5889}, {3529,6243}, {10620,12270}

= reflection of X(i) in X(j) for these (i,j): (382,143), (3627,389), (5446,13382), (5562,548), (5876,3), (6101,550), (6102,185), (7728,11561), (10113,974), (10263,6102), (10625,12103), (10733,13358), (11381,546), (11455,13364), (12111,11591), (12162,140), (13474,5462)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (3,156,1511), (3,7999,11592), (3,10620,11440), (3,11456,156), (3,12111,11591), (381,10574,12006), (382,5890,143), (5462,13474,3845), (5890,12279,382), (5944,12041,3), (9730,11381,546), (10574,12290,381), (11591,11592,7999), (11591,12111,5876)

= [ 17.944606722537000, 18.80690556734412, -17.661627090501710 ]

 

> 2. The centroid of  NaNbNc lies on the Euler line.

Gn = (3*cos(2*A)-2)*cos(B-C)-cos(A) *cos(2*(B-C))-cos(3*A) : : (trilinears)

= 2*X(5)+X(11819) = 5*X(5)-2*X(12362) = 2*X(140)+X(7553) = X(140)-4*X(13163) = 2*X(143)+X(12134) = 2*X(546)+X(3575) = 2*X(5462)+X(13419) = X(6102)-4*X(11745) = X(6146)-4*X(10095) = 4*X(10110)-X(12370)

= Shinagawa coefficients: (E-4*F, 9*E+12*F)

= On lines: {2,3}, {143,12134}, {206,5476}, {539,11808}, {542,9969}, {1503,5946}, {3564,9971}, {5462,13419}, {6102,11745}, {6146,10095}, {10110,12370}, {12824,13451}

= midpoint of X(i) and X(j) for these {i,j}: {2,7540}, {381,7576}, {7553,7667}

= reflection of X(i) in X(j) for these (i,j): (549,10127), (7667,140)

= 1^st Droz-Farny circle-inverse-of-X(5189)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (4,7506,13371), (5,26,7568), (5,6756,11819), (25,381,10201), (25,11818,5), (26,7528,5), (381,7426,547), (381,10201,5), (546,10020,5576), (546,10096,13413), (2070,5133,140), (3518,5576,10020), (3542,7564,5), (6997,7514,5), (10201,11818,381)

= [ -0.692982667055059, -1.56067448417941, 5.040969586518613 ]

 

César Lozada