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

15418Re: [EMHL] Cevian Trace Equal Area Points

Expand Messages
  • Quang Tuan Bui
    Aug 5, 2007
    • 0 Attachment
      Dear Francois and Nikos,

      We denote S(T1, T2) = equal area point of two inscribed triangles T1, T2. Of course S(T1, T2) = S(T2, T1).
      Suppose T1, T2, T3 are three inscribed triangles then three equal area points S(T1, T2), S(T2, T3), S(T3, T1) are always on one circumconic. Denote this circumconic as C(T1, T2, T3).

      Suppose now T1, T2, T3 are three Cevian triangles of P=X(2), U=X(4) and X respectively.
      There is one special locus:
      The locus of X such that circumconic C(T1, T2, T3) is passing also through X is one quintic:
      CyclicSum[ (b^2 - c^2)*y*z*(a^2*SA*(x^3 - y*z*(y + z - x)) + b^2*c^2*(x^3 - y*z*x)) ] = 0
      This circumquintic passes:
      - Vetices of antimedial triangle.
      - X(2), X(4), X(69), X(110), X(2574), X(2575)

      Best regards,
      Bui Quang Tuan


      Quang Tuan Bui <bqtuan1962@...> wrote: Given triangle ABC and two points P, U with barycentrics P = (p : q : r), U = (u : v : w). PaPbPc and UaUbUc are Cevian triangles of P, U respectively.
      We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U))
      here:
      c(P) = complement of P
      c(U) = complement of U
      t(cd(P, U)) = isotomic conjugate of cross difference of P, U
      * = barycentric product
      The results:
      a). Barycentrics of Q = (q + r)*(v + w)/(q*w - r*v) : :
      b). Area(QPaUa) = Area(QPbUb) = Area(QPcUc)


      ---------------------------------
      Luggage? GPS? Comic books?
      Check out fitting gifts for grads at Yahoo! Search.

      [Non-text portions of this message have been removed]
    • Show all 18 messages in this topic