HYACINTHOS 24692

[Tran Quang Hung]:
  
    Let ABC be a triangle with incenter I.

    A'B'C' is cevian triangle of I.

    Na,Nb,Nc are NPC center of triangles IB'C',IC'A',IA'B'.

    N1,N2,N3 are NPC center of triangles INbNc,INcNa,INaNb.

    Then NPC center of triangle N1N2N3 lies on OI line of triangle ABC.

    Is this new point ?

 

[Angel Montesdeoca]:

   
    *** The  NPC center of triangle N1N2N3  is  N' = (2r^2+5rR+4s^2 ) I + r(2r+3R) O, of  barycentric coordinates:

(a (7a^5(b+c)
      +a^4(b^2+4b c+c^2)
      -a^3(14b^3+9b^2c+9b c^2+14c^3)
      -2a^2(b^4+5b^3c+7b^2c^2+5b c^3+c^4)
     +a(b-c)^2(7b^3+16b^2c+16b c^2+7c^3)
      +(b^2-c^2)^2(b^2+6b c+c^2)) : ... : ...),

with (6-9-13)-search numbers (1.93658689736328, 1.89366701233398, 1.43585490535472).

Angel Montesdeoca