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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 42 , 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
      Let ABC be a triangle, P a point and A B C the pedal triangle of P. Denote: L = the Euler line of A B C A , B , C = orthogonal projections of A, B, C, on
      Message 42 of 42 , Jan 30

        Let ABC be a triangle, P a point and  A'B'C' the pedal triangle of P.

        Denote:

        L = the Euler line of A'B'C'

        A", B", C" = orthogonal projections of A, B, C, on L, resp.

        AaAbAc, BaBbBc, CaCbCc = the pedal triangles of A", B", C", wrt triangle A'B'C' resp.

        The circumcircles of AaAbAc, BaBbBc, CaCbCc are concurrent on the NPC circle of A'B'C'.
        The point of concurrence is the reflection point of L wrt the medial triangle of A'B'C'.

        Problem:

        Let ABC be a triangle and L a line.

        For which points P's the L is the Euler line of the pedal triangle of P ?

        APH
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