HYACINTHOS 4623

• Dear John, Paul, Milorad and other Hyacinthists
  
  [PY]
  

  > > > [The] inversive image of the nine-point circle in the
  > > >radical circle of the excircles [has] radius (r^2+s^2)/(4r).
  > > > [The] center[,] X(970)[,] lies on the Brocard axis.
  > > > Is there a simple explanation of this fact?
  >
  > [Jean-Pierre(?):]
  >
  > > [...] 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).
  
  

  [JHC]
  

  > I suspect there must also be some connection with "the Conway
  

  circle"
  

  > (center Io), since the radius of that is root(rr+ss). There's
  

  probably
  

  > a vast generalization that describes all the Tucker circles in some
  > such way (which maybe Jean-Pierre is quoting).
  
  

  There is no quotation.
  As Paul had noticed that the circle was centered on OK, I've just
  looked whether he could be a Tucker circle (it's enough to check that
  bb BL + cc CL' = 0 where L,L' are the common points of BC and the
  circle) and the answer was yes but I'm unable to understand why.
  More over, I think that I'm not the first one to notice that this
  circle is a Tucker circle but I've never heard of that.
  Happy 2002 to all Hyacinthists.
  Friendly. Jean-Pierre Ehrmann