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Center of the circular hull of the excircles

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  • Paul Yiu
    Dear Clark and friends, Happy New Year! Consider the circle tangent internally to each of the excircles of triangle ABC. This circle can be constructed as the
    Message 1 of 2 , Jan 1, 2002
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      Dear Clark and friends,

      Happy New Year!
      Consider the circle tangent internally to each of the excircles of
      triangle ABC. This circle can be constructed as the inversive image of the
      nine-point circle in the radical circle of the excircles. It has long been
      known that the radius of this circle is (r^2+s^2)/(4r). See, for example,
      Problem 1864 and solution, Crux Math. 20 (1994) 174--175. What is interesting
      is that the center is the point X(970) in ETC. As such, this center lies on
      the Brocard axis. Is there a simple explanation of this fact?

      Best regards
      Sincerely
      Paul
    • jpehrmfr
      Happy New Year to all Hyacinthists. ... excircles of ... of the ... long been ... example, ... interesting ... lies on ... Your circle intersects BC at (in
      Message 2 of 2 , Jan 2, 2002
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        Happy New Year to all Hyacinthists.

        Paul wrote :
        > Dear Clark and friends,
        >
        > Happy New Year!
        > Consider the circle tangent internally to each of the
        excircles of
        > triangle ABC. This circle can be constructed as the inversive image
        of the
        > nine-point circle in the radical circle of the excircles. It has
        long been
        > known that the radius of this circle is (r^2+s^2)/(4r). See, for
        example,
        > Problem 1864 and solution, Crux Math. 20 (1994) 174--175. What is
        interesting
        > is that the center is the point X(970) in ETC. As such, this center
        lies on
        > the Brocard axis. Is there a simple explanation of this fact?

        Your circle intersects BC at (in barycentrics)
        (0,s-cc,cc) and (0,bb,s-bb) where s = -2abc/(a+b+c).
        Thus your circle is a Tucker circle and he is centered on OK.
        In fact, as the radical center of the excircles is the Spieker point
        S, the center of your circle is the common point of OK and SN
        (N=NPcenter), which is, as you've noticed X(970).
        Friendly. Jean-Pierre
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