HYACINTHOS 29556

[Tran Quang Hung]:

 

Let ABC be a triangle.

OI line meets BC, CA, AB at A', B', C'. resp.

Let Oa, Ob, Oc be circumcenters of triangles AB'C', BC'A', CA'B' respectively.

[...]

Also X(56) of triangle OaObOc lies on OI line of ABC.

Which is this point ?

 

[César Lozada]:

 

Q = MIDPOINT OF X(56) AND X(23981)

= a^3*(a+b-c)*(a-b+c)*(a^5-(b+c)*a^4-2*(b-c)^2*a^3+2*(b^3+c^3)*a^2+(b^4+c^4-(4*b^2-b*c+4*c^2)*b*c)*a-(b+c)*(b^4+c^4-(2*b^2-b*c+2*c^2)*b*c))*((b+c)*a^2-2*b*c*a-(b^2-c^2)*(b-c)) : : (barys)

= lies on these lines: {1, 3}, {1403, 6769}, {6767, 8069}

= midpoint of X(56) and X(23981)

= [ 0.0183414481164834, 0.3097215998853760, 3.4177765520868950 ]

 

César Lozada