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

Re: Help, please

Expand Messages
  • jpehrmfr
    Dear Jean-Louis and other Hyacinthists this follows from two facts : if P,P are inverse in the circumcircle of ABC, then
    Message 1 of 2 , Feb 25, 2013
      Dear Jean-Louis and other Hyacinthists
      this follows from two facts :
      if P,P' are inverse in the circumcircle of ABC, then
      <BPC + <BP'C = 2<BAC
      if P,P* are isogonal conjugates wrt ABC, then
      <BPC + <BP*C = <BAC
      Kind regards. Jean-Pierre

      --- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
      > Dear Jean-Pierre and Mathlinkers,
      > working on a generalization of a Nagel's result, I found in the archive of Hyacinthos the message # 19640
      > Dear Chris
      > [JP]
      > > > Consider a quadrilateral A_1,A_2,A_3,A_4
      > > > For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is
      > the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate
      > of A_k wrt T_k
      > > > Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the
      > center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
      > > > Is there a special name for this point M? Do you know some references?
      > [Chris]
      > > I do not know a special name for point M.
      > > However I noticed this. Maybe you know it already.
      > > Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
      > > Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with
      > A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
      > > T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same
      > orientation as A_1,A_2,A_3,A_4.
      > Thank you for your nice remark.
      > In fact, if O_1' is the circumcenter of O_2O_3O_4,...,
      > the same homothecy maps A_i to O_i' and B_i to O_i
      > The point M is characterized by the angular relations
      > <A_iMA_j = <A_iA_kA_j +<A_iA_lA_j (oriented angles modulo Pi) where (i,j,k,l) is
      > any permutation of (1,2,3,4)
      > Ma question is: how can we prouve the last angular relations? Is there a simple way to prouve it? I try to discover a synthetic proof without success.
      > Sincerely
      > Jean-Louis
      > [Non-text portions of this message have been removed]
    Your message has been successfully submitted and would be delivered to recipients shortly.