- Dear friends,

First of all I wish you a very happy and fruitful 2005.

To start the new year let me offer a little theorem:

If A'B'C' is a circumcevian triangle, and A"B"C" are the reflections of

A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C), (A"BC")

and (AB"C") are concurrent in one point.

Kind regards,

Floor. - Let ABC be a triangle, P a point and A'B'C' the circumcevian triangle of P.

Denote:

A", B", C" = the reflections of A', B', C' in BC, CA, AB, resp

ABC, A"B"C" are circumcyclologic.

iethe circumcircles of ABC, AB"C", BC"A", CA"B" are concurrent[at the isogonal conjugate of the direction of the perpendicular to HP line.FvL, BG, Hyacinthos #10974 ]the circumcircles of A"B"C", A"BC, B"CA, C"AB are concurrent.

Which is this cyclologic center in terms of P?

APH