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22079Re: [EGML] Re: A locus

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  • Antreas Hatzipolakis
    Mar 27, 2014
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      On Thu, Mar 27, 2014 at 9:54 PM, <amontes@...> wrote:

      Dear Antreas
      You are absolutely right.
      I forgot the symmedian

      Angel M.


      Dear Angel

      And 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 lying
      on the Euler line of ABC?