HYACINTHOS 28906

 

[ Angel Montesdeoca]:

 

 

Let ABC be a triangle of circumcenter O and a point D variable on the circumcircle. Let Ha, Hb, Hc be the orthocenters of the triangles DBC, DCA, DAB, respectively.

 The circle of diameter OHa cuts again the lines BHa and CHa in Hab and Hac, respectively.
 
 The perpendicular bisector of HabHac passes through a fixed point Fa, on the  perpendicular bisector of BC, when D varies over the circumcircle.
 
 Let p_a be the polar of Fa with respect to the circumcircle. The polar p_b and p_c are defined cyclically.
 
  The triangle A'B'C' formed by the lines p_a, p_b and p_c is perspective with ABC, with perspector

W = X(3)X(125)∩X(25)X(6344)

= a^2SA (SA^2 + 5S^2)/(4SA^2-b^2c^2) : :    

=  lies on these lines: {3,125}, {25,6344}, {94,7517}, {476,5899}, {1141,11815}, {1593,23956}, {1989,8573}, {11060,21309}, {11141,21310}, {11142,21311} 

 
 References:
 
 https://artofproblemsolving.com/community/c6t48f6h1750087
 http://amontes.webs.ull.es/otrashtm/HGT2019.htm#HG080319
 
 
 Angel Montesdeoca