Τρίτη 14 Φεβρουαρίου 2012

EROTOKRITOS CIRCLE


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
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Ετικέτες

1 σχόλιο:

  1. Chandan14 Φεβρουαρίου 2012 στις 9:59 μ.μ.

    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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