HYACINTHOS 26677

 [Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.

The perpendicular from Na to BC intersects AB, AC at Ab, Ac, resp.
The perpendicular from Nb to CA intersects BC, BA at Bc, Ba, resp.

The perpendicular from Nc to AB intersects CA, CB at Ca, Cb, resp.

(Oa), (Ob), (Oc) = the circumcircles of AAbAc, BBcBa, CCbCa, resp.

1. The radical center of (Oa), (Ob), (Oc) is the O.

2. ABC, OaObOc are orthologic.

 

Conjecture:

Let Pa, Pb, Pc be same points on the Euler lines of AAbAc, BBcBa, CCbCa, resp.

ABC, PaPbPc are orthologic.

[César Lozada]:



Conjecture: Proved.

 

If Pa=P is such that OaPa=t*OaHa, orthologic centers are (barycentrics):

 

Qa=Q(A->Pa) = 1/(3*S^2*(-S^2+12*R^2*SA-3*SA^2-2*SB*SC)*t+SB*SC*(7*SA^2+3*S^2)) : :

 

Qp=Q(Pa->A) = 3*((8*R^2+3*SA+SW)*S^2+3*(4*R^2+SW)*(SA-SW)*SA)*t+2*(-40*R^2-3*SA+7*SW)*S^2-2*(28*R^2-SW)*(SA-SW)*SA: :

 

ETC pairs (P,Qa): (1597,2349), (3839,1799)

ETC pairs (P,Qp): (2, 5)

 

Some others (all barycentrics):

 

Qa( X(2) ) = isogonal conjugate of X(13367)

= SB*SC *(SB+8*R^2-3*SW)*(SC+8*R^2-3*SW) : :

=  On lines: {5, 8884}, {225, 7541}, {264, 7507}, {381, 1093}, {393, 3091}, {403, 1179}, {427, 1105}, {648, 3574}, {847, 7547}, {1300, 1594}, {1826, 7563}, {3832, 6526}, {6531, 7745}, {7566, 14249}

= [ 0.218068206221583, 0.40554983126844, 3.259252349696269 ]

 

Qa( X(3) ) = X(54)X(550) ∩ X(74)X(140)

= SA*(3*S^2+7*SB^2)*(3*S^2+7*SC^2) : :

= 7*X(4)-12*X(13566)

=  On Jerabek hyperbola and these lines: {2, 13452}, {4, 12006}, {6, 1657}, {20, 13472}, {30, 1173}, {54, 550}, {64, 1656}, {65, 4857}, {74, 140}, {185, 3519}, {3426, 3851}, {3431, 3522}, {3523, 11270}, {3527, 5073}, {3858, 13603}, {5056, 11738}, {5663, 5900}, {6697, 13093}

= [ 12.067806784031910, 12.04145000110140, -10.265480957639010 ]

 

Qp( X(3) ) = X(4)X(54) ∩ X(140)X(6000)
= 3*(4*R^2+SA-2*SW)*S^2+(28*R^2-SW)*SB*SC : :

= 5*X(4)+3*X(9833) = 7*X(4)+9*X(11206) = X(64)-3*X(10193) = 5*X(140)-3*X(6696) = 9*X(154)-X(1657) = X(550)+3*X(2883) = X(550)-3*X(10282) = 3*X(1498)+5*X(1656) = 7*X(9833)-15*X(11206)

= On lines: {4, 54}, {18, 10675}, {64, 10193}, {140, 6000}, {154, 1657}, {399, 3519}, {550, 1511}, {1498, 1656}, {1503, 3850}, {2393, 12002}, {3357, 3523}, {3522, 5878}, {5056, 14216}, {5073, 14530}, {5270, 10535}, {5562, 6053}, {5655, 13564}, {5972, 10575}, {6225, 10299}, {6747, 6750}, {8960, 12970}, {10112, 11799}, {10116, 11563}

= midpoint of X(i) and X(j) for these {i,j}: {2883, 10282}, {5656, 10182}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (4, 1614, 10619), (4, 10619, 13403)

= [ -5.649243121525654, -7.45696570265887, 11.410522178298510 ]

 

Qa( X(4) ) = X(20)X(2889) ∩ X(1249)X(3533)

= (3*S^4-(SB+6*SC)*SA*S^2+(SB+SC)*SA^2*SB)*(3*S^4-(SC+6*SB)*SA*S^2+(SC+SB)*SA^2*SC) : :

= On lines: {20, 2889}, {1249, 3533}, {3851, 14249}

= [ 6.731301948367092, 3.92155285583444, -2.181011471378122 ]

 

Qp( X(4) ) = X(4)X(51) ∩ X(140)X(1503)

= 3*(4*R^2-SA)*S^2-(4*R^2+5*SW)*SB*SC : :

= 7*X(4)-3*X(5878) = 11*X(4)-3*X(6225) = 5*X(4)+3*X(12324) = X(4)+3*X(14216) = 3*X(64)+X(5073) = 10*X(140)-9*X(10182) = 4*X(140)-3*X(10282) = 11*X(5878)-7*X(6225) = 5*X(5878)+7*X(12324) = X(5878)+7*X(14216) = 5*X(6225)+11*X(12324) = 6*X(10182)-5*X(10282)

= On lines: {4, 51}, {26, 11645}, {64, 5073}, {140, 1503}, {427, 12242}, {550, 6247}, {1498, 3851}, {1656, 1853}, {1657, 3357}, {2781, 13421}, {2883, 3858}, {3519, 10625}, {3523, 9833}, {3533, 11206}, {3818, 11695}, {3854, 5656}, {6240, 13399}, {6288, 14855}, {10619, 11430}

= {X(11457), X(11550)}-Harmonic conjugate of X(389)

= [ 12.957101467956020, 13.95961139268678, -12.003882544393860 ]

 

César Lozada