Σάββατο 19 Οκτωβρίου 2019

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

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου