HYACINTHOS 24888

 

• [Antreas P. Hatzipolakis]:

   

  Let ABC be a triangle, A'B'C' the pedal triangle of I and IaIbIc the antipedal triangle of I (exentral triangle).
  
  

  Denote:
  
  

  Na, Nb, Nc = the NPC centers of IaBC, IbCA, IcAB, resp.
  
  

  N1, N2, N3 = the reflections of Na, Nb, Nc in IA', IB', IC', resp.
  
  

  ABC, N1N2N3 are orthologic.

  The orthologic center (N1N2N3, ABC) lies on the OI line.
  
  
  

  [Peter Moses]:
  
  

  
  

  Hi Antreas,

   

  >The orthologic center (N1N2N3, ABC) lies on the OI line.

  a (2 a^3-5 a^2 b-2 a b^2+5 b^3-5 a^2 c+10 a b c-5 b^2 c-2 a c^2-5 b c^2+5 c^3)::
  on lines {{1,3},{5,3626},{8,3545},{10, 547},{30,3244},...}.

   

  the other is

  a (3 a^3-3 a^2 b-3 a b^2+3 b^3-a^2 c+5 a b c-b^2 c-3 a c^2-3 b c^2+c^3) (3 a^3-a^2 b-3 a b^2+b^3-3 a^2 c+5 a b c-3 b^2 c-3 a c^2-b c^2+3 c^3)::
  on lines {{80,5433},{499,7319},...}.

  This is the isogonal conjugate of a point on lines {{1,3},{4,9897},{79,10944},... }.

   

  Best regards,

  Peter Moses.