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Re: Concurrent circles

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  • 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 1 of 44 , Mar 1, 2013
      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]
      > >
      >
    • 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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