HYACINTHOS 29450

 

[Tran Quang Hung]:

 

Let ABC be a triangle with circumcenter O.

 

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

 

A*, B*, C* are the isogonal conjugates of A, B, C wrt triangles OB'C', OC'A', OA'B' resp.

Then the circumcenter of the triangle A*B*C* lies on the Euler line of triangle ABC.

Which is this point?


[Petrer Moses]:

Hi Antreas,

= a^2*(a^14 - 4*a^12*b^2 + 6*a^10*b^4 - 5*a^8*b^6 + 5*a^6*b^8 - 6*a^4*b^10 + 4*a^2*b^12 - b^14 - 4*a^12*c^2 + 12*a^10*b^2*c^2 - 12*a^8*b^4*c^2 + 2*a^6*b^6*c^2 + 6*a^4*b^8*c^2 - 6*a^2*b^10*c^2 + 2*b^12*c^2 + 6*a^10*c^4 - 12*a^8*b^2*c^4 + 7*a^6*b^4*c^4 - 2*a^4*b^6*c^4 + a^2*b^8*c^4 - 5*a^8*c^6 + 2*a^6*b^2*c^6 - 2*a^4*b^4*c^6 + 2*a^2*b^6*c^6 - b^8*c^6 + 5*a^6*c^8 + 6*a^4*b^2*c^8 + a^2*b^4*c^8 - b^6*c^8 - 6*a^4*c^10 - 6*a^2*b^2*c^10 + 4*a^2*c^12 + 2*b^2*c^12 - c^14) : :

= lies on these lines: {2, 3}, {5961, 25739}, {12219, 13557}
= {X(186),X(2071)}-harmonic conjugate of X(17511)

Best regards,
Peter Moses.