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Re: lucas circles and soddy line

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  • peter_mows
    ... circle ... circle ... are ... Brocard ... The radical circle of the Lucas circles is centered on X(1151)= a^2 (2 SA + S) :: with radius R / (Cot[w] + 2)
    Message 1 of 2 , Jan 31, 2006
      > Building Lucas circles in triangle (ABC), we have their radical
      circle
      > being the inscrite circle in the triangle of centers (LaLbLc)
      > The center of this circle is the center of the inversion in this
      circle
      > which let globaly invariant the Lucas circles
      > The lucas circles being outwardly tangent to circle (ABC), they
      are
      > inwardly tangent to the inverse of circle (ABC)
      > The centers of these circles belong to the line (UK), axes of
      Brocard
      > of triangle (ABC), which is Soddy line of the triangle of centers
      >
      > The lines parallel to (LaLb) in C, to (LaLc) in B,
      > to (LbLc) en A are boarding a triangle in perspective with the
      > triangle (ABC)
      > Is the perspector well-know?
      >

      The radical circle of the Lucas circles is centered on
      X(1151)= a^2 (2 SA + S) :: with radius R / (Cot[w] + 2)

      If La,Lb,Lc are the centers of the Lucas circles,
      La = {a^2 (2 S + SA), b^2 SB, c^2 SC} & etc ..

      then let (if I am understanding what you mean)

      a' =
      line parallel to Lc La through B /\ line parallel to La Lb through C

      b' =
      line parallel to La Lb through C /\ line parallel to Lb Lc through A

      c' =
      line parallel to Lb Lc through A /\ line parallel to Lc La through B

      Then, I do not reckon on a'b'c' being perspective to ABC. However,
      it does appear to be perspective to LaLbLc at

      a^2 (a^2 (c^2 (b^2 + 2 S) + b^2 (c^2 + 2 S)) SA +
      b^2 c^2 (a^2 + 2 S) (2 S + SA)) ::with SEARCH of 3.6947387980418975

      Probably, for the 'external' version of the Lucas circles, replace S
      with -S, SEARCH now -0.4034639023136933. Neither are in ETC.

      Best regards,
      Peter.
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