HYACINTHOS 24699

[Antreas P. Hatzipolakis]:

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

Denote:

Ab, Ac = the reflections of A' in OB,OC, resp.
Bc, Ba = the reflections of B' in OC,OA, resp.
Ca, Cb = the reflections of C' in OA,OB, resp.

Na, Nb, Nc = the NPC centers of AAbAc, BBcBa, CCaCb, resp.

Oa, Ob, Oc = the circumcenters of AAbAc, BBcBa, CCaCb, resp.

 

ABC, NaNbNc are orthologic.
The orthologic center (NaNbNc, ABC) is the N.

 

ABC, OaObOc are orthologic.
The orthologic center (OaObOc, ABC) is the O.

Conjecture:

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

ABC, PaPbPc are orthologic.

The orthologic center (PaPbPc, ABC) is P on the Euler line of ABC same with Pa,Pb,Pc.

Which is the locus of the other orthologic center (ABC, PaPbPc) as Pa,Pb,Pc move on the Euler lines of AAbAc, BBcBa, CCaCb, resp. being same?

[César Lozada]:

 

Conjecture proved.

When P moves on the Euler lines of AAbAc, BBcBa, CCaC, the orthologic center (ABC,PaPbPc)=Za moves on the circumconic = isogonal conjugate of the line { 184, 9292 }. This conic passes through ETC’s 264 and 1975, its center is not related to any ETC center and its perspector is known to lie on the line {297,525}    

César Lozada

[APH]:

 

Dear César,

Thanks !!!

I am wondering if the orthologic centers (ABC, PaPbPc) are in ETC for simple points Pa,Pb,Pc = Na,Nb,Nc or Oa, Ob, Oc.

Thanks again

 

[César Lozada]:

ETC pairs (P, Za(P)): (2, 264), (3,1975)

 

Za( H ) =  (SA^2+SB*SC)/( a^2*(S^2*(17*R^2-4*SW)+2*S^2* SA+(R^2-2*SW)*SA^2)) : : (barycentrics)

= [ 27.749850650381050, -5.91478177369960, -5.072109974937919 ]

 

Za(N) = (SA^2+SB*SC)/( a^2*(S^2*(33*R^2-8*SW)+4*S^2* SA+(-4*SW+R^2)*SA^2)) : : (barycentrics)

= [ -4.102272890073460, 6.36779703601517, 1.125546329315473 ]

 

No relations with ETC centers were found for Za(H) and Za(N).

 

César Lozada