HYACINTHOS 4621

• Dear Paul,Jean-Pierre and friends,
  *******
  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
  *******
  Let r* be the radius of this circle.Then we have the following inequality
  r* >= (R1 + 5r1)/2,where R1 and r1 are respectively circumradius and
  inradius of triangle with vertices in excircle-centers of tringle ABC.
  In proof we use the formulas 4r*r = s^2 + r^2, s = (r1xs1)/R1,
  2rxR1 = (s1)^2 - (2R1 + r1)^2, (x denotes multiplication),and two
  Gerretsen's inequalities s^2 <= 4(R^2) + 4Rr + 3(r^2), s^2 >= 16Rr - 5(r^2),
  applied on excircle-centers triangle,and following equality
  (8R1)r* = 4(r1^2) + {(4r1^2)(2R1 + r1)^2}/{s1^2 - (2R1 +r1)^2} + s1^2 - (2R1 + r1)^2.
  Weaker inequalities are,obviously, r* >= (7r1)/2 > = 7r as consequences of
  R >= 2r.Maybe,this is a new inequality.
  Kind regards,
  Sincerely,
  Milorad R.Stevanovic