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Dear John, Paul, Milorad and other Hyacinthists
[PY]
> > > [The] inversive image of the nine-point circle in the
OK.
> > >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
> > 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
[JHC]
> > (N=NPcenter), which is, as you've noticed X(970).
> 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
There is no quotation.
> such way (which maybe Jean-Pierre is quoting).
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
Σάββατο 19 Οκτωβρίου 2019
HYACINTHOS 4623
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