[Tran Quang Hung]:
Let ABC be a triangle.
OI line meets BC, CA, AB at A', B', C', resp..
Let Oa, Ob, Oc be the circumcenters of the triangles AB'C', BC'A', CA'B' respectively.
Then the Feuerbach point of the triangle OaObOc lies on OI line of ABC.
Which is this point?
[César Lozada]:
Q = X(1)X(3) ∩ X(227)X(18340)
= a*((b+c)*a^2-2*b*c*a-(b^2-c^2)*(b-c))*(a^5-(b+c)*a^4-(2*b-c)*(b-2*c)*a^3+2*(b^2-c^2)*(b-c)*a^2+(b^2-b*c+c^2)*(b-c)^2*a-(b^4-c^4)*(b-c)) : : (barys)
= lies on these lines: {1, 3}, {227, 18340}, {676, 2804}, {1465, 1538}, {6767, 8071}, {12515, 15501}
= {X(3035), X(10271)}-harmonic conjugate of X(15252)
= [ 1.5599987577643150, 1.5827083410468640, 1.8249438960607130 ]
César Lozada
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