HYACINTHOS 25087

[Tran Quang Hung]:

Let ABC be a triangle with circumcenter O.

 

A'B'C' is pedal triangle of O

 

A''B''C'' is cevian triangle of O.

 

K is radical center of circles (AA'A''),(BB'B''),(CC'C'').

 

The isogonal conjugate of K wrt triangle A'B'C' lies on Euler of ABC.

 

Which is this point ?

 

[Angel Montesdeoca]:

Dear Tran Quang Hung

The isogonal conjugate  W of  the radical center of circles (AA'A''),(BB'B''),(CC'C'')  wrt triangle A'B'C' is the reflection of X(235) in X(5).

W =  ( (b^2-c^2)^4 (b^2+c^2)
             - 2 (b^2-c^2)^2 (b^4+c^4) a^2
            - 4 b^2 c^2 (b^2+c^2) a^4
           + 2 (b^2+c^2)^2a^6
           + (-b^2-c^2) a^8 : ... : ...),

with (6-9-13)-search number (3.14847023153707,  2.27064456467082,  0.615539676425936).

W lies on lines:
{2, 3}, {11, 1062}, {12, 1060}, {68, 394}, {96, 2986}, {113, 2883}, {122, 131}, {125, 5562}, {141, 1209}, {155, 1899}, {184, 9820}, {185, 1568}, {195, 1353}, {216, 1506}, {230, 10316}, {339, 3933}, {343, 1216}, {524, 8538}, {577, 7746}, {590, 10897}, {615, 10898}, {895, 3519}, {1038, 7951}, {1040, 7741}, {1147, 6146}, {1181, 5654}, {1352, 8549}, {1503, 10539}, {1941, 6761}, {3284, 7755}, {3925, 8251}, {5972, 10282}, {6509, 10600}, {7723, 10264}.

Best regards,
Angel Montesdeoca