- Mar 27, 2014On Thu, Mar 27, 2014 at 9:54 PM, <amontes@...> wrote:Dear AntreasYou are absolutely right.I forgot the symmedianAngel M._Dear AngelAnd I had written centroid X(6) instead of Symmedian point X(6) .....We can generalize it in various ways.Another way is this:Let A'B'C' be the antipedal triangle of a point P .

Which is the locus of P such that O, P* [=isog. conj. of P], G' [=G of A'B'C']are collinear?Symmedian point is on the locus.

Another (and more simple, I think):Which is the locus of P such that G' (=G of A'B'C') is lyingon the Euler line of ABC?

APH