HYACINTHOS 24908

 [Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

A', B', C' = the reflections of O in BC, CA, AB, resp.

Nab, Nac = the NPC centers of AOB', AOC', resp.
Nbc, Nba = the NPC centers of BOC', BOA', resp.
Nca, Ncb = the NPC centers of COA', COB', resp.

Oa, Ob, Oc = the circumcenters of ANabNac, BNbcNba, CNcaNcb

ABC, OaObOc are orthologic.
The orthologic center (OaObOc, ABC) is the N. The other one?

[Angrl Montesdeoca]:

The orthologic center (ABC,OaObOc) is

W =  ( (b^2-c^2)^6
  -4 (b^2-c^2)^4 (b^2+c^2) a^2
  +(5 b^8-2 b^6 c^2-7 b^4 c^4-2 b^2 c^6+5 c^8) a^4
  -4 b^2 c^2 (b^2+c^2) a^6
  +(-5 b^4-4 b^2 c^2-5 c^4) a^8
  +4 (b^2+c^2)a^10
  -a^12 :  ... :  ... ),
 
  with (6,9,13)-search numbers in ETC: (0.0855948998410226,  0.0857542228846969,  3.54179083536835).
 
  W lies on lines:
{324,9381}, {1370,10155}, {3153,9221}, {5189,7608}, {7394,7612}, {7533,7607}

Angrl Montesdeoca