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21654Re: Concurrent circles

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  • rhutson2
    Mar 1, 2013
      Dear Antreas,

      For the orthic version (P=X(4), case 1.), the circumcircles do not concur, but their radical center is non-ETC -10.266788994809312, which lies on the Euler line, and is in fact the reflection of X(403) in X(4).

      For the medial version (P=X(2), case 1.), the circumcircles do not concur. Their radical center is non-ETC 0.675001328037596, for which I can find no special properties.

      Randy

      --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
      >
      > Dear Randy
      >
      > Thanks.
      >
      > I guess that the point for the orthic version (ie ABC is orthic triangle of
      > reference triangle)
      > is not in ETC as well.
      >
      > Now, as for generalizations (P instead of I):
      >
      > There are two possible cases since for P = I the reflection of B' in AA' is
      > lying on AC etc.:
      >
      > For A'B'C' = Cevian triangle of P.
      >
      > Denote:
      >
      > 1. B'a, C'a = the reflections of B',C' in AA', resp. etc
      >
      > or
      >
      > 2. B'a = (perpendicular to AA' from B') /\ AC
      > etc
      >
      > Locus of P such that the circumcircles of the triangles A'aB'aC'a,
      > B'bC'bA'b, C'cA'cB'c are concurrent ??.
      >
      > Quite complicated, I guess.....!
      >
      > APH
      >
      >
      > On Fri, Mar 1, 2013 at 12:44 AM, rhutson2 <rhutson2@...> wrote:
      >
      > > **
      > >
      > >
      > > Antreas,
      > >
      > > The circumcircles are concurrent at non-ETC 0.812149174855220, which is,
      > > equivalently:
      > >
      > > X(1)-Ceva conjugate of X(36)
      > > Antigonal conjugate, wrt incentral triangle, of X(1)
      > > The point P for which P of the 'orthocentroidal triangle' = X(1).
      > >
      > > I define the 'orthocentroidal triangle' as:
      > > Let A* be the intersection, other than X(4), of the A-altitude and the
      > > orthocentroidal circle, and define B*, C* cyclically.
      > >
      > > A*B*C* is inversely similar to ABC, with similitude center X(6).
      > > X(i) of A*B*C* = reflection of X(i) (of ABC) in the centroid of its pedal
      > > triangle.
      > >
      > > As to your second question, the triangles ABC, IaIbIc are not perspective.
      > >
      > > Best regards,
      > > Randy
      > >
      > >
      > > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
      > > >
      > > > Let ABC be a triangle and A'B'C' the cevian triangle of I.
      > > >
      > > > Denote:
      > > >
      > > > B'a, C'a = the reflections of B',C' in AA', resp.
      > > >
      > > > A'a = (perpendicular to BB' from B'a) /\ (perpendicular to CC' from
      > > > C'a).
      > > >
      > > > C'b, A'b = the reflections of C',A' in BB', resp.
      > > >
      > > > B'b = (perpendicular to CC' from C'b) /\ (perpendicular to AA' from
      > > > A'b).
      > > >
      > > > A'c, B'c = the reflections of A',B' in CC', resp.
      > > >
      > > > C'c = (perpendicular to AA' from A'c) /\ (perpendicular to BB' from
      > > > B'c).
      > > >
      > > > Are the circumcircles of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c
      > > > concurrent?.
      > > >
      > > > Let Ia,Ib,Ic be the incenters of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c,
      > > > resp.
      > > >
      > > > Are the triangles ABC, IaIbIc perspective?
      > > >
      > > >
      > > > APH
      > >
      >
      >
      > [Non-text portions of this message have been removed]
      >
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