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Re: [EMHL] Re: Concurrent circles

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  • Antreas Hatzipolakis
    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
    Message 1 of 44 , Feb 28, 2013
      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]
    • Antreas Hatzipolakis
      [Tran Quang Hung]:. Let ABC be a triangle with orthocenter H and centroid G. A ,B ,C are on HA, HB, HC such that GA _|_ GA, GB _|_ GB, GC _|_ GC. Then the
      Message 44 of 44 , Jul 26
        [Tran Quang Hung]:.

        Let ABC be a triangle with orthocenter H and centroid G.

        A',B',C' are on HA, HB, HC such that GA' _|_ GA, GB' _|_ GB, GC' _|_ GC.

        Then the circles (A',A'A), (B',B'B), (C',C'C) are concurrent at a point.

        1. Which is this point ?

        2. I see that A',B',C' and G are concyclic. Is this new circle ?


        [César Lozada]:


        1.  X(671)

        2.  This circle, with center X(8176) , was named the ANTICOMPLEMENTARY-VAN LAMOEN CIRCLE in the preamble of X(8176).

        Regards,

        César Lozada
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