Σάββατο 19 Οκτωβρίου 2019

HYACINTHOS 4619

  • 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 Yiu

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