Let ABC be a triangle, A1B1C1 the circumcevian triangle of H and A2B2C2 the cevian triangle of G (medial triangle).
Denote:
A3 = the other than A1 intersection of A1A2 and the circumcircle.
B3 = the other than B1 intersection of B1B2 and the circumcircle.
C3 = the other than C1 intersection of C1C2 and the circumcircle.
N1 = The NPC center of A3BC
N2 = The NPC center of B3CA
N3 = The NPC center of C3AB.
The four NPC centers N, N1,N2,N3 are concyclic (??).
Center of the circle?
APH, 14 February 2012
It also follows from the Picasso Circles problem. Its easy to see that, if A'B'C' is the antipodal triangle of ABC, then A'A3,B'B3,C'C3 concurr at the isogonal conjugate of H wrt A'B'C'.
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