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Soddy Gergonne Nagel Cubic Makes Circle Chain

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  • Quang Tuan Bui
    Dear All My Friends, Given triangle ABC, incenter I, point P with Cevian triangle A B C . Denote Ra(X) = reflection of X in IA. Similarly with Rb(X), Rc(X)
    Message 1 of 1 , Jul 30, 2010
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      Dear All My Friends,

      Given triangle ABC, incenter I, point P with Cevian triangle A'B'C'.
      Denote Ra(X) = reflection of X in IA. Similarly with Rb(X), Rc(X)
      Denote Sa(X) = symmetry of X in A'. Similarly with Sb(X), Sc(X)
      Denote X(Y) = circle centered at X passing Y
      We start with any point E on line CA
      Ca = Ra(E)
      Cb = Sc(Ca)
      Ab = Rb(Cb)
      Ac = Sa(Ab)
      Bc = Rc(Ac)
      Ba = Sb(Bc)
      Please note that all these points are on sidelines of ABC and we can construct six following chain of touching circles centered at vetices of ABC and Cevian triangle A'B'C':
      A(Ca), C'(Cb), B(Ab), A'(Ac), C(Bc), B'(Ba)

      The proccess can be finite if Ba = E. The result as following:

      The locus of P such that with any E on line CA, Ba = E is Soddy Gergonne Nagel cubic K200.

      Best regards,
      Bui Quang Tuan
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