HYACINTHOS 25099

[Tran Quang Hung]:

Let ABC be a triangle with circumcenter O.

 

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

 

A'',B'',C'' are reflection of A,B,C through B'C',C'A',A'B'.

 

Then the circles (AA'A''),(BB'B''),(CC'C'') are coaxal.

 

Which are the intersection points ?

[Peter Moses];

Hi Antreas,

 

Intersection points are X(74) and

a^2 (a^2+b^2-c^2) (a^2-b^2-b c-c^2) (a^2-b^2+b c-c^2) (a^2-b^2+c^2) (a^12-3 a^10 b^2+3 a^8 b^4-2 a^6 b^6+3 a^4 b^8-3 a^2 b^10+b^12-3 a^10 c^2+7 a^8 b^2 c^2-2 a^6 b^4 c^2-6 a^4 b^6 c^2+5 a^2 b^8 c^2-b^10 c^2+3 a^8 c^4-2 a^6 b^2 c^4+6 a^4 b^4 c^4-2 a^2 b^6 c^4-5 b^8 c^4-2 a^6 c^6-6 a^4 b^2 c^6-2 a^2 b^4 c^6+10 b^6 c^6+3 a^4 c^8+5 a^2 b^2 c^8-5 b^4 c^8-3 a^2 c^10-b^2 c^10+c^12)::
on lines {{3,3462},{54,74},{112,216},{ 186,3258},{378,3438},{6104, 6116},{6105,6117}}.

On K073.

X(3)-ceva conjugate of X(186).

Inverse of X(5667) in the circumcircle.

 

The line joining the intersections is X{54,74,185,933,2914,3520, 7722,10628,11430}.

 

Best regards,

Peter Moses.