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

Locus

Expand Messages
  • Antreas P. Hatzipolakis
    Let ABC be a triangle, and PaPbPc the pedal triangle of P. The line IPa intersects the circumcircle of ABC at A1,A2 The line IPb intersects the circumcircle of
    Message 1 of 235 , Jan 1, 1970
    • 0 Attachment
      Let ABC be a triangle, and PaPbPc the pedal triangle of P.

      The line IPa intersects the circumcircle of ABC at A1,A2
      The line IPb intersects the circumcircle of ABC at B1,B2
      The line IPc intersects the circumcircle of ABC at C1,C2
      (where A1,B1,C1 lie on the positive side of the corresponding
      triangle sides, and A2,B2,C2 on the negative side).

      Which is the locus of P such that the triangles
      ABC, A1B1C1; ABC, A2B2C2 are perspective?

      In general: Instead of the incenter I, a fixed point Q = (f:g:h).

      Antreas

      PS: Sorry if this isn't "The first time that Antreas asked this
      question,"
    • Antreas Hatzipolakis
      Let ABC be a triangle and P a point Which is the locus of P such that PA + PB + PC = 0 (signed segments) ? APH PS see special case
      Message 235 of 235 , Apr 10
      • 0 Attachment
        Let ABC be a triangle and P a point

        Which is the locus of  P such that PA + PB + PC = 0
        (signed segments) ?

        APH

        PS see special case

        https://www.facebook.com/photo.php?fbid=811391552270276&set=p.811391552270276&type=1&theater
      Your message has been successfully submitted and would be delivered to recipients shortly.