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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 39 , Feb 28, 2013
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      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
      Dear Randy Sorry for not being clear. Let Q be the point of concurrence of the circles wrt ABC. Now, let Q be the point of concurrence wrt orthic triangle.
      Message 39 of 39 , Mar 1, 2013
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        Dear Randy

        Sorry for not being clear.

        Let Q be the point of concurrence of the circles
        wrt ABC. Now, let Q' be the point of concurrence
        wrt orthic triangle.
        (ie we make the construction in the orthic, whose bisectors
        are the altitudes of ABC, so we take the cevian triangle of H =
        I(orthic) wrt orthic)
        The point Q' of orthic is the point Q of ABC,
        ie the coordinates of Q'(wrt orthic) = coordinates of Q(wrt ABC).
        Which are the coordinates of Q'(wrt ABC)?


        APH

        --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
        >
        > 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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