Loading ...
Sorry, an error occurred while loading the content.

Re: [EMHL] mc points

Expand Messages
  • Bernard Gibert
    Dear Antreas, ... there are several such points on the McCay cubic. at most 4 real?, at least 2 real? I didn t find a construction for them but I suspect they
    Message 1 of 18 , May 31, 2004
    • 0 Attachment
      Dear Antreas,

      > [BG]:
      > >consider the circumcenters O1 and O2 of the pedal and antipedal
      > >triangles of a point P.
      > >for which P do O1 and O2 coincide ?
      >
      > [APH] Nice problem!!
      > If P is real, then it lies on the McCay cubic.
      > Now, no one of the simple centers on McCay (ie O,H,I) has
      > this property.
      > So, we have a new center on the McCay cubic???

      there are several such points on the McCay cubic. at most 4 real?, at
      least 2 real?

      I didn't find a construction for them but I suspect they are only
      conic-constructible.

      Jean-Pierre, help...

      Best regards

      Bernard

      [Non-text portions of this message have been removed]
    • Paul Yiu
      Dear Bernard and Antreas, [BG]: consider the circumcenters O1 and O2 of the pedal and antipedal triangles of a point P. For which P do O1 and O2 coincide ? ***
      Message 2 of 18 , Jun 1, 2004
      • 0 Attachment
        Dear Bernard and Antreas,

        [BG]: consider the circumcenters O1 and O2 of the pedal and antipedal
        triangles of a point P. For which P do O1 and O2 coincide ?

        *** I have a problem related to antipedal triangles.
        By calculations, the antipedal triangle of P and the circumcevian
        triangle of Q are perspective if and only if O, P, Q are collinear.
        Is it possible to give a synthetic proof?

        Best regards
        Sincerely
        Paul
      Your message has been successfully submitted and would be delivered to recipients shortly.