HYACINTHOS 25133

[Tran Quang Hung]:

Let ABC be a triangle with circumcenter O.

 

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

 

I' is incenter of A'B'C'.

 

Then isogonal conjugate of I' wrt ABC lies on Euler line of ABC.

 

Which is this point ?

 

[Peter Moses]:

Hi Antreas,

 

These are always tricky!

 

Let U^2 = (a^12-3 a^10 b^2+a^8 b^4+6 a^6 b^6-9 a^4 b^8+5 a^2 b^10-b^12-3 a^10 c^2+7 a^8 b^2 c^2-6 a^6 b^4 c^2+6 a^4 b^6 c^2-7 a^2 b^8 c^2+3 b^10 c^2+a^8 c^4-6 a^6 b^2 c^4+6 a^4 b^4 c^4+2 a^2 b^6 c^4-3 b^8 c^4+6 a^6 c^6+6 a^4 b^2 c^6+2 a^2 b^4 c^6+2 b^6 c^6-9 a^4 c^8-7 a^2 b^2 c^8-3 b^4 c^8+5 a^2 c^10+3 b^2 c^10-c^12) with V and W cyclic.

 

Then I’ = a^2 (a^2-b^2-c^2) (V+W)::

 

and so gI’ = (a^2+b^2-c^2) (a^2-b^2+c^2) (U+V) (U+W):: which is indeed on the Euler line, 3 (64 a^2 SA^3 SB SC+(U+V) (U+W)) X[2]-2 (64 S^2 SA^2 SB SC+(U+V) (U+W)) X[3], Search = 5.2311813629880077184

 

Best regards,

Peter Moses.