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