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

Re: [EMHL] AltiEulerian Triangles

Expand Messages
  • Quang Tuan Bui
    Dear Antreas, Interesting question generates interesting result: The rectangular circumhyperbola passing through A, B, C, H and P* has center at concurrent
    Message 1 of 5 , Aug 1, 2006
      Dear Antreas,
      Interesting question generates interesting result:
      The rectangular circumhyperbola passing through A, B, C, H and P* has center at concurrent point Q.
      Here H is orthocenter of ABC, P* is isogonal conjugate of P wrt ABC and P* is also orthocenter of triangle PaPbPc.
      Best regards,
      Bui Quang Tuan

      "Antreas P. Hatzipolakis" <xpolakis@...> wrote: Dear Tuan

      [BQT]
      > Let ABC be a triangle, HaHbHc its Euler triangle, P = (x:y:z) with
      >respect >ABC, a point, and Pa, Pb, Pc, the corresponding points with
      >respect AHbHc, >BHcHa, CHaHb, resp.
      > Now I try to connect Pa, Pb, Pc with the vertices of Euler
      >triangle Ea, Eb, >Ec, the circumcenters of AHbHc, BHcHa, CHaHb.
      > The result is very nice: with any P point = (x:y:z), three lines
      >PaEa, PbEb, >PcEc are always concurrent at one point, say Q, on the
      >nine point circle. The >locus is whole plane.

      and I am wondering the point of concurrence
      whose r. c/hyperbola the center is.

      APH
      --


      ---------------------------------
      Yahoo! Messenger with Voice. Make PC-to-Phone Calls to the US (and 30+ countries) for 2ยข/min or less.

      [Non-text portions of this message have been removed]
    Your message has been successfully submitted and would be delivered to recipients shortly.