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Re: Orthologic Triangles - Locus

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  • Angel
    Writing correct geometric property of the quintic Q068 in the message #21755: Let ABC be a triangle, A B C the pedal triangle of point P and O the
    Message 1 of 18 , Mar 15, 2013
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      Writing correct geometric property of the quintic Q068 in the message #21755:

      Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the
      circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A', B', C' w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.


      A particular situation:

      If P=X(3) the orthology centers of ABC y A"B"C" (on NPC) are Q=X(113) and R, nine-point-circle-antipode of X(3258).


      R= (2*a^8 - a^4*(b^4-4*b^2*c^2+c^4)- 2*a^6*(b^2+c^2) + (b^2-c^2)^4 )*
      (a^6*(b^2 + c^2) + a^4*(-3*b^4 + 2*b^2*c^2 - 3*c^4)+
      a^2*(3*b^6 - 2*b^4*c^2 - 2*b^2*c^4 + 3*c^6)-
      (b^2 - c^2)^2*(b^4 + 3*b^2*c^2 + c^4)) : .... : ....



      Angel M.

      --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
      >
      >
      >
      > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
      > >
      > > 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P
      > > and A",B",C" the second intersections of the circles (P,PA'),
      > > (P,PB'), (P,PC') with the pedal circle of P, resp.
      > >
      > > Which is the locus of P such that the triangles ABC, A"B"C"
      > > are orthologic?
      > >
      > > O is on the locus.
      >
      > Dear Antreas
      >
      > The locus of real point P such that the triangles ABC, A"B"C" are orthologic is Q068 (a circular equilateral quintic) in Higher Degree Related Curves.
      > http://bernard.gibert.pagesperso-orange.fr/relatedcurves.html
      >
      >
      > It is a result does not appear between the propiedades of Q068 in http://bernard.gibert.pagesperso-orange.fr/curves/q068.html
      >
      >
      > We can express as:
      >
      > Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A, B, C w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.
      >
      >
      > (http://groups.yahoo.com/group/Hyacinthos/files/Hyacinthos21754A.eps)
      >
      >
      > Angel M.
      >
    • Nikolaos Dergiades
      Dear Angel, you wrote ... I think you know that more precisely P lies on Linf + Circumcircle + Q068 + sextic Qx = -b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4
      Message 2 of 18 , Mar 16, 2013
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        Dear Angel,
        you wrote
        > Let ABC be a triangle, A'B'C' the pedal triangle of point P
        > and O' the
        > circumcenter of A'B'C'. Let A", B", C" be the symmetrical
        > points of A', B', C' w/r to the line PO'. The triangles ABC,
        > A"B"C" are  orthologic if and only if P lies on Q068.

        I think you know that more precisely
        P lies on Linf + Circumcircle + Q068 + sextic
        Qx = -b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4 x^3 y^3 + c^6 x^3 y^3 -
        a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z - b^4 c^2 x^4 y z -
        b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z - b^4 c^2 x^3 y^2 z +
        b^2 c^4 x^3 y^2 z - a^4 c^2 x^2 y^3 z + a^2 b^2 c^2 x^2 y^3 z +
        a^2 c^4 x^2 y^3 z - a^4 c^2 x y^4 z + a^2 b^2 c^2 x y^4 z -
        a^2 c^4 x y^4 z - b^4 c^2 x^4 z^2 + a^2 b^2 c^2 x^3 y z^2 +
        b^4 c^2 x^3 y z^2 - b^2 c^4 x^3 y z^2 + 6 a^2 b^2 c^2 x^2 y^2 z^2 +
        a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2 - a^2 c^4 x y^3 z^2 -
        a^4 c^2 y^4 z^2 - a^2 b^4 x^3 z^3 + b^6 x^3 z^3 - b^4 c^2 x^3 z^3 -
        a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +
        a^4 b^2 x y^2 z^3 - a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +
        a^6 y^3 z^3 - a^4 b^2 y^3 z^3 - a^4 c^2 y^3 z^3 - a^2 b^4 x^2 z^4 -
        a^4 b^2 x y z^4 - a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 -
        a^4 b^2 y^2 z^4 = 0

        But if P lies on Circumcircle then the points A', B', C'
        are collinear and
        if P lies on Qx then the points A", B", C" are collinear.

        Best regards
        Nikos Dergiades
      • Antreas Hatzipolakis
        Dear Nikos If P is lying on your Qx, what kind of circle is the pedal circle of P, on which should be lying three collinear points A ,B ,C ? APH On Sat, Mar
        Message 3 of 18 , Mar 16, 2013
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          Dear Nikos

          If P is lying on your Qx, what kind of circle is the pedal circle of
          P, on which
          should be lying three collinear points A",B",C"?

          APH

          On Sat, Mar 16, 2013 at 10:15 AM, Nikolaos Dergiades
          <ndergiades@...> wrote:
          > Dear Angel,
          > you wrote
          >> Let ABC be a triangle, A'B'C' the pedal triangle of point P
          >> and O' the
          >> circumcenter of A'B'C'. Let A", B", C" be the symmetrical
          >> points of A', B', C' w/r to the line PO'. The triangles ABC,
          >> A"B"C" are orthologic if and only if P lies on Q068.
          >
          > I think you know that more precisely
          > P lies on Linf + Circumcircle + Q068 + sextic
          > Qx = -b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4 x^3 y^3 + c^6 x^3 y^3 -
          > a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z - b^4 c^2 x^4 y z -
          > b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z - b^4 c^2 x^3 y^2 z +
          > b^2 c^4 x^3 y^2 z - a^4 c^2 x^2 y^3 z + a^2 b^2 c^2 x^2 y^3 z +
          > a^2 c^4 x^2 y^3 z - a^4 c^2 x y^4 z + a^2 b^2 c^2 x y^4 z -
          > a^2 c^4 x y^4 z - b^4 c^2 x^4 z^2 + a^2 b^2 c^2 x^3 y z^2 +
          > b^4 c^2 x^3 y z^2 - b^2 c^4 x^3 y z^2 + 6 a^2 b^2 c^2 x^2 y^2 z^2 +
          > a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2 - a^2 c^4 x y^3 z^2 -
          > a^4 c^2 y^4 z^2 - a^2 b^4 x^3 z^3 + b^6 x^3 z^3 - b^4 c^2 x^3 z^3 -
          > a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +
          > a^4 b^2 x y^2 z^3 - a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +
          > a^6 y^3 z^3 - a^4 b^2 y^3 z^3 - a^4 c^2 y^3 z^3 - a^2 b^4 x^2 z^4 -
          > a^4 b^2 x y z^4 - a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 -
          > a^4 b^2 y^2 z^4 = 0
          >
          > But if P lies on Circumcircle then the points A', B', C'
          > are collinear and
          > if P lies on Qx then the points A", B", C" are collinear.
          >
          > Best regards
          > Nikos Dergiades
        • Nikolaos Dergiades
          Dear Antreas, this Qx comes from the calculations but I said nonsense because I thik it is obvious that this sextic is valid only for P = I. This point I lies
          Message 4 of 18 , Mar 16, 2013
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            Dear Antreas,
            this Qx comes from the calculations
            but I said nonsense because I thik it is
            obvious that this sextic is valid only for
            P = I.
            This point I lies on Q068 but it must be
            excluded because the line PO' is not defined.
            There is only one line passing through I
            such that the reflections A",B",C"
            give a triangle orthologic with ABC.

            Thank you Antreas.
            Regards
            Nikos


            --- Στις Σάβ., 16/03/13, ο/η Antreas Hatzipolakis <anopolis72@...> έγραψε:

            > Από: Antreas Hatzipolakis <anopolis72@...>
            > Θέμα: Re: [EMHL] Re: Orthologic Triangles - Locus
            > Προς: Hyacinthos@yahoogroups.com
            > Ημερομηνία: Σάββατο, 16 Μάρτιος 2013, 10:33
            > Dear Nikos
            >
            > If P is lying on your Qx, what kind of circle is the pedal
            > circle of
            > P,  on which
            > should be lying three collinear points A",B",C"?
            >
            > APH
            >
            > On Sat, Mar 16, 2013 at 10:15 AM, Nikolaos Dergiades
            > <ndergiades@...>
            > wrote:
            > > Dear Angel,
            > > you wrote
            > >> Let ABC be a triangle, A'B'C' the pedal triangle of
            > point P
            > >> and O' the
            > >> circumcenter of A'B'C'. Let A", B", C" be the
            > symmetrical
            > >> points of A', B', C' w/r to the line PO'. The
            > triangles ABC,
            > >> A"B"C" are  orthologic if and only if P lies
            > on Q068.
            > >
            > > I think you know that more precisely
            > > P lies on Linf + Circumcircle + Q068 + sextic
            > > Qx = -b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4 x^3
            > y^3 + c^6 x^3 y^3 -
            > >  a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z - b^4 c^2
            > x^4 y z -
            > >  b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z - b^4 c^2
            > x^3 y^2 z +
            > >  b^2 c^4 x^3 y^2 z - a^4 c^2 x^2 y^3 z + a^2 b^2
            > c^2 x^2 y^3 z +
            > >  a^2 c^4 x^2 y^3 z - a^4 c^2 x y^4 z + a^2 b^2 c^2
            > x y^4 z -
            > >  a^2 c^4 x y^4 z - b^4 c^2 x^4 z^2 + a^2 b^2 c^2
            > x^3 y z^2 +
            > >  b^4 c^2 x^3 y z^2 - b^2 c^4 x^3 y z^2 + 6 a^2 b^2
            > c^2 x^2 y^2 z^2 +
            > >  a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2 - a^2
            > c^4 x y^3 z^2 -
            > >  a^4 c^2 y^4 z^2 - a^2 b^4 x^3 z^3 + b^6 x^3 z^3 -
            > b^4 c^2 x^3 z^3 -
            > >  a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2
            > c^2 x^2 y z^3 +
            > >  a^4 b^2 x y^2 z^3 - a^2 b^4 x y^2 z^3 + a^2 b^2
            > c^2 x y^2 z^3 +
            > >  a^6 y^3 z^3 - a^4 b^2 y^3 z^3 - a^4 c^2 y^3 z^3 -
            > a^2 b^4 x^2 z^4 -
            > >  a^4 b^2 x y z^4 - a^2 b^4 x y z^4 + a^2 b^2 c^2 x
            > y z^4 -
            > >  a^4 b^2 y^2 z^4 = 0
            > >
            > > But if P lies on Circumcircle then the points A', B',
            > C'
            > > are collinear and
            > > if P lies on Qx then the points A", B", C" are
            > collinear.
            > >
            > > Best regards
            > > Nikos Dergiades
            >
            >
            > ------------------------------------
            >
            > Yahoo! Groups Links
            >
            >
            >     Hyacinthos-fullfeatured@yahoogroups.com
            >
            >
          • Nikolaos Dergiades
            Sorry there are two lines. ND
            Message 5 of 18 , Mar 16, 2013
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              Sorry there are two lines.

              ND

              > Dear Antreas,
              > this Qx comes from the calculations
              > but I  said nonsense because I thik it is
              > obvious that this sextic is valid only for
              > P = I.
              > This point I lies on Q068 but it must be
              > excluded because the line PO' is not defined.
              > There is only one line passing through I
              > such that the reflections A",B",C"
              > give a triangle orthologic with ABC.
              >
              > Thank you Antreas.
              > Regards
              > Nikos
              >
              >
              > --- Στις Σάβ., 16/03/13, ο/η Antreas Hatzipolakis
              > <anopolis72@...>
              > έγραψε:
              >
              > > Από: Antreas Hatzipolakis <anopolis72@...>
              > > Θέμα: Re: [EMHL] Re: Orthologic Triangles - Locus
              > > Προς: Hyacinthos@yahoogroups.com
              > > Ημερομηνία: Σάββατο, 16 Μάρτιος
              > 2013, 10:33
              > > Dear Nikos
              > >
              > > If P is lying on your Qx, what kind of circle is the
              > pedal
              > > circle of
              > > P,  on which
              > > should be lying three collinear points A",B",C"?
              > >
              > > APH
              > >
              > > On Sat, Mar 16, 2013 at 10:15 AM, Nikolaos Dergiades
              > > <ndergiades@...>
              > > wrote:
              > > > Dear Angel,
              > > > you wrote
              > > >> Let ABC be a triangle, A'B'C' the pedal
              > triangle of
              > > point P
              > > >> and O' the
              > > >> circumcenter of A'B'C'. Let A", B", C" be the
              > > symmetrical
              > > >> points of A', B', C' w/r to the line PO'. The
              > > triangles ABC,
              > > >> A"B"C" are  orthologic if and only if P lies
              > > on Q068.
              > > >
              > > > I think you know that more precisely
              > > > P lies on Linf + Circumcircle + Q068 + sextic
              > > > Qx = -b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4
              > x^3
              > > y^3 + c^6 x^3 y^3 -
              > > >  a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z - b^4 c^2
              > > x^4 y z -
              > > >  b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z - b^4
              > c^2
              > > x^3 y^2 z +
              > > >  b^2 c^4 x^3 y^2 z - a^4 c^2 x^2 y^3 z + a^2 b^2
              > > c^2 x^2 y^3 z +
              > > >  a^2 c^4 x^2 y^3 z - a^4 c^2 x y^4 z + a^2 b^2
              > c^2
              > > x y^4 z -
              > > >  a^2 c^4 x y^4 z - b^4 c^2 x^4 z^2 + a^2 b^2 c^2
              > > x^3 y z^2 +
              > > >  b^4 c^2 x^3 y z^2 - b^2 c^4 x^3 y z^2 + 6 a^2
              > b^2
              > > c^2 x^2 y^2 z^2 +
              > > >  a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2 - a^2
              > > c^4 x y^3 z^2 -
              > > >  a^4 c^2 y^4 z^2 - a^2 b^4 x^3 z^3 + b^6 x^3 z^3
              > -
              > > b^4 c^2 x^3 z^3 -
              > > >  a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2
              > > c^2 x^2 y z^3 +
              > > >  a^4 b^2 x y^2 z^3 - a^2 b^4 x y^2 z^3 + a^2 b^2
              > > c^2 x y^2 z^3 +
              > > >  a^6 y^3 z^3 - a^4 b^2 y^3 z^3 - a^4 c^2 y^3 z^3
              > -
              > > a^2 b^4 x^2 z^4 -
              > > >  a^4 b^2 x y z^4 - a^2 b^4 x y z^4 + a^2 b^2 c^2
              > x
              > > y z^4 -
              > > >  a^4 b^2 y^2 z^4 = 0
              > > >
              > > > But if P lies on Circumcircle then the points A',
              > B',
              > > C'
              > > > are collinear and
              > > > if P lies on Qx then the points A", B", C" are
              > > collinear.
              > > >
              > > > Best regards
              > > > Nikos Dergiades
              > >
              > >
              > > ------------------------------------
              > >
              > > Yahoo! Groups Links
              > >
              > >
              > >     Hyacinthos-fullfeatured@yahoogroups.com
              > >
              > >
              >
              >
              > ------------------------------------
              >
              > Yahoo! Groups Links
              >
              >
              >     Hyacinthos-fullfeatured@yahoogroups.com
              >
              >
            • Angel
              ... Made the correction: Let A , B , C be the symmetrical points of A , B , C instead of: Let A , B , C be the symmetrical points of A, B, C This is the
              Message 6 of 18 , Mar 16, 2013
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                --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
                >
                >
                > Writing correct geometric property of the quintic Q068 in the message #21755:
                >
                > Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the
                > circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A', B', C' w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.
                >
                >
                > A particular situation:
                >
                > If P=X(3) the orthology centers of ABC y A"B"C" (on NPC) are Q=X(113) and R, nine-point-circle-antipode of X(3258).
                >
                >
                > R= (2*a^8 - a^4*(b^4-4*b^2*c^2+c^4)- 2*a^6*(b^2+c^2) + (b^2-c^2)^4 )*
                > (a^6*(b^2 + c^2) + a^4*(-3*b^4 + 2*b^2*c^2 - 3*c^4)+
                > a^2*(3*b^6 - 2*b^4*c^2 - 2*b^2*c^4 + 3*c^6)-
                > (b^2 - c^2)^2*(b^4 + 3*b^2*c^2 + c^4)) : .... : ....
                >
                >
                >
                > Angel M.
                >

                Made the correction:

                Let A", B", C" be the symmetrical points of A', B', C'

                instead of:

                Let A", B", C" be the symmetrical points of A, B, C


                This is the corrected EPS:

                http://groups.yahoo.com/group/Hyacinthos/files/Hyacinthos21754Acorrected.eps




                Angel M.
              • Angel
                Dear Antreas 2. Let ABC be a triangle, A B C the pedal triangle of point P, A*B*C* the circumcevian triangle of P wrt A B C and A ,B ,C the second
                Message 7 of 18 , Mar 16, 2013
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                  Dear Antreas

                  2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,
                  A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the
                  second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with
                  the pedal circle of P, resp.

                  The locus of P such that the triangles ABC, A"B"C" are orthologic is the septic Qnnn:

                  -a^2 c^4 x^4 y^3 + b^2 c^4 x^4 y^3 + c^6 x^4 y^3 -
                  a^2 c^4 x^3 y^4 + b^2 c^4 x^3 y^4 - c^6 x^3 y^4 +
                  2 b^2 c^4 x^4 y^2 z - 2 a^2 c^4 x^2 y^4 z -
                  2 b^4 c^2 x^4 y z^2 - 2 b^4 c^2 x^3 y^2 z^2 +
                  2 b^2 c^4 x^3 y^2 z^2 + 2 a^4 c^2 x^2 y^3 z^2 -
                  2 a^2 c^4 x^2 y^3 z^2 + 2 a^4 c^2 x y^4 z^2 +
                  a^2 b^4 x^4 z^3 - b^6 x^4 z^3 - b^4 c^2 x^4 z^3 - 2 a^4 b^2 x^2
                  y^2 z^3 + 2 a^2 b^4 x^2 y^2 z^3 + a^6 y^4 z^3 - a^4 b^2 y^4 z^3 +
                  a^4 c^2 y^4 z^3 + a^2 b^4 x^3 z^4 + b^6 x^3 z^4 -
                  b^4 c^2 x^3 z^4 + 2 a^2 b^4 x^2 y z^4 - 2 a^4 b^2 x y^2 z^4 -
                  a^6 y^3 z^4 - a^4 b^2 y^3 z^4 + a^4 c^2 y^3 z^4=0,

                  (+ line at infinity (A'B'C' undefined triangle) + circumcircle (A', B' and C' points aligned) + a sextic ).

                  Points of the septic Qnnn:

                  A, B, C (triple points)
                  X(1) double , X(4)
                  excenters (which are double)
                  feet of the altitudes.



                  The sextic (without real points ???)is:

                  -b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4 x^3 y^3 +
                  c^6 x^3 y^3 - a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z -
                  b^4 c^2 x^4 y z - b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z -
                  b^4 c^2 x^3 y^2 z + b^2 c^4 x^3 y^2 z - a^4 c^2 x^2 y^3 z +
                  a^2 b^2 c^2 x^2 y^3 z + a^2 c^4 x^2 y^3 z - a^4 c^2 x y^4 z +
                  a^2 b^2 c^2 x y^4 z - a^2 c^4 x y^4 z - b^4 c^2 x^4 z^2 +
                  a^2 b^2 c^2 x^3 y z^2 + b^4 c^2 x^3 y z^2 - b^2 c^4 x^3 y z^2 +
                  6 a^2 b^2 c^2 x^2 y^2 z^2 + a^4 c^2 x y^3 z^2 +
                  a^2 b^2 c^2 x y^3 z^2 - a^2 c^4 x y^3 z^2 - a^4 c^2 y^4 z^2 -
                  a^2 b^4 x^3 z^3 + b^6 x^3 z^3 - b^4 c^2 x^3 z^3 -
                  a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +
                  a^4 b^2 x y^2 z^3 - a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +
                  a^6 y^3 z^3 - a^4 b^2 y^3 z^3 - a^4 c^2 y^3 z^3 - a^2 b^4 x^2 z^4 - a^4 b^2 x y z^4 - a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 -
                  a^4 b^2 y^2 z^4=0.



                  Best regards
                  Angel Montesdeoca





                  --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
                  >
                  > 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P
                  > and A",B",C" the second intersections of the circles (P,PA'),
                  > (P,PB'), (P,PC') with the pedal circle of P, resp.
                  >
                  > Which is the locus of P such that the triangles ABC, A"B"C"
                  > are orthologic?
                  >
                  > O is on the locus.
                  >
                  > 2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,
                  > A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the
                  > second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with
                  > the pedal circle of P, resp.
                  >
                  > Which is the locus of P such that the triangles ABC, A"B"C"
                  > are orthologic?
                  >
                  > H is on the locus.
                  >
                  > Figures:
                  > http://anthrakitis.blogspot.gr/2013/03/orthologic-triangles-locus.html
                  >
                  > APH
                  >
                • Antreas Hatzipolakis
                  Let ABC be a triangle, P a point and A B C the pedal triangle of P. Denote: Bc = the orthogonal projection of B on PC B3 = the reflection of Bc in PB Cb =
                  Message 8 of 18 , Apr 29, 2014
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                    Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

                    Denote:

                    Bc = the orthogonal projection of B' on PC'
                    B3 = the reflection of Bc in PB'

                    Cb = the orthogonal projection of C' on PB'
                    C2 = the reflection of Cb in PC'

                    Similarly C1, A3 and A2, B1

                    Which is the locus of P such that ABC and triangle bounded by
                    (B3C2, C1A3, A2B1) are 1. orthologic 2. parallelogic?

                    For P = H
                    They are orthologic (the orth. center (ABC, (B3C2, C1A3, A2B1)) is (X(74))
                    and parallelogic (the parallelogic center (ABC, (B3C2, C1A3, A2B1)) is X(110)

                    APH
                  • Antreas Hatzipolakis
                    Starting point: Let ABC be a triangle. Denote: La, Lb, Lc = the reflections of the Euler line in BC,CA,AB, resp. Oa, Ob, Oc = the reflections of O in La, Lb,
                    Message 9 of 18 , Nov 4, 2014
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                      Starting point:

                      Let ABC be a triangle.

                      Denote:

                      La, Lb, Lc = the reflections of the Euler line in BC,CA,AB, resp.

                      Oa, Ob, Oc = the reflections of O in La, Lb, Lc resp.

                      ABC, OaObOc are orthologic.
                      Which is the other than O orthologic center?
                      (ie the orthologic center (OaObOc, ABC))

                      Generalization:

                      Let ABC be a triangle and P a point.

                      Denote:

                      La, Lb, Lc = the reflections of the Euler line in BC,CA,AB, resp.

                      Pa, Pb, Pc = the reflections of P in La, Lb, Lc resp.

                      Which is the locus of P such that ABC, PaPbPc are orthologic ?

                      APH


                    • Antreas Hatzipolakis
                      Let ABC be a triangle and P a point. Denote: A B C = the pedal triangle of P O = the circumcenter of A B C ( = the center of the pedal circle of P) A B C =
                      Message 10 of 18 , Dec 16, 2014
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                         Let ABC be a triangle and P a point.

                        Denote:

                        A'B'C' = the pedal triangle of P

                        O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)

                        A"B"C" = the pedal triangle of O'.

                        Which is the locus of P such that A'B'C', A"B"C"
                        are orthologic?

                        On the locus:

                        1. P = H (O' = N)

                        Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'

                        Orthologic center (A'B'C', A"B"C") = ?

                        2. P = Ο (O' = N)

                        Orthologic center (A"B"C", A'B'C') = the O of A'B'C' =  the NPC center of ABC

                        Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle
                        Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?

                        This is interesting because we can take as reference triangle the orthic
                        tiangle in order to get a point on the Euler line of ABC.

                        That is:

                        Let ABC be a triangle and A'B'C' the antipedal triangle of I.
                        (ABC is the pedal triangle of H of A'B'C'').
                        Now:
                        Let A1B1C1 be the pedal triangle of N of A'B'C' [ = O of ABC]
                        Let A2B2C2 be the pedall triangle of O of A'B'C'

                        A1B1C1, A2B2C2 are orthologic.

                        The orthologic center (A1B1C1, A2B2C2) is N of A'B'C' [= O of ABC]
                        The orthologic center (A2B2C2, A1B1C1) lies on the Euler line
                        of ABC. Point ?

                        APH









                      • Antreas Hatzipolakis
                        ... Circumcircle, infinity, K003 . McCay cubic 1) Orthologic center (A B C , A B C ) = the NPC
                        Message 11 of 18 , Dec 16, 2014
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                          [Antreas]:

                           Let ABC be a triangle and P a point.

                          Denote:

                          A'B'C' = the pedal triangle of P

                          O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)

                          A"B"C" = the pedal triangle of O'.

                          Which is the locus of P such that A'B'C', A"B"C"
                          are orthologic?

                          On the locus:

                          1. P = H (O' = N)

                          Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'

                          Orthologic center (A'B'C', A"B"C") = ?

                          2. P = Ο (O' = N)

                          Orthologic center (A"B"C", A'B'C') = the O of A'B'C' =  the NPC center of ABC

                          Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle
                          Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?

                          This is interesting because we can take as reference triangle the orthic
                          tiangle in order to get a point on the Euler line of ABC.

                          That is:

                          Let ABC be a triangle and A'B'C' the antipedal triangle of I.
                          (ABC is the pedal triangle of H of A'B'C'').
                          Now:
                          Let A1B1C1 be the pedal triangle of N of A'B'C' [ = O of ABC]
                          Let A2B2C2 be the pedall triangle of O of A'B'C'

                          A1B1C1, A2B2C2 are orthologic.

                          The orthologic center (A1B1C1, A2B2C2) is N of A'B'C' [= O of ABC]
                          The orthologic center (A2B2C2, A1B1C1) lies on the Euler line
                          of ABC. Point ?

                          APH


                          [Peter Moses]:

                          >Which is the locus of P such that A'B'C', A"B"C" are orthologic?
                          Circumcircle, infinity, K003. McCay cubic
                           
                          1)
                          Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
                          Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?
                          2)
                          Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)
                           
                          Best regards
                          Peter


                        • Antreas Hatzipolakis
                          ... ************************** Thanks, Peter, What a surprise! A new point in such a simple construction !!!! I wouldn t bet a cent for it as a new point :-)
                          Message 12 of 18 , Dec 16, 2014
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                            [Antreas]:

                             Let ABC be a triangle and P a point.

                            Denote:

                            A'B'C' = the pedal triangle of P

                            O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)

                            A"B"C" = the pedal triangle of O'.

                            Which is the locus of P such that A'B'C', A"B"C"
                            are orthologic?

                            On the locus:

                            1. P = H (O' = N)

                            Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'

                            Orthologic center (A'B'C', A"B"C") = ?

                            2. P = Ο (O' = N)

                            Orthologic center (A"B"C", A'B'C') = the O of A'B'C' =  the NPC center of ABC

                            Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle
                            Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?




                            [Peter Moses]:

                            >Which is the locus of P such that A'B'C', A"B"C" are orthologic?
                            Circumcircle, infinity, K003. McCay cubic
                             
                            1)
                            Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
                            Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?
                            2)
                            Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)
                             
                            Best regards
                            Peter



                            **************************

                            Thanks, Peter,

                            What a surprise! A new point in such a simple construction !!!!
                            I wouldn't bet a cent for it as a new point :-)

                            Description of the point:

                            Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of H,N, resp.

                            A'B'C', A"B"C" are orthologic with orthologic center (A'B'C', A"B"C")
                            on {4,93},{6,24},{25,195},.. lines of ABC.

                            If you compute other "things" of the point (coordinates etc) please let me know
                            (to inform Clark)

                            Season's Greetings

                            APH
                          • Antreas Hatzipolakis
                            ... *************************************** Another observation in the same configuration. But let s first rename the pedal triangles. Let ABC be a triangle
                            Message 13 of 18 , Dec 16, 2014
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                              [Antreas]:

                               Let ABC be a triangle and P a point.

                              Denote:

                              A'B'C' = the pedal triangle of P

                              O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)

                              A"B"C" = the pedal triangle of O'.

                              Which is the locus of P such that A'B'C', A"B"C"
                              are orthologic?

                              On the locus:

                              1. P = H (O' = N)

                              Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'

                              Orthologic center (A'B'C', A"B"C") = ?

                              2. P = Ο (O' = N)

                              Orthologic center (A"B"C", A'B'C') = the O of A'B'C' =  the NPC center of ABC

                              Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle
                              Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?




                              [Peter Moses]:

                              >Which is the locus of P such that A'B'C', A"B"C" are orthologic?
                              Circumcircle, infinity, K003. McCay cubic
                               
                              1)
                              Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
                              Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?
                              2)
                              Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)
                               
                              Best regards
                              Peter



                              ***************************************

                              Another observation in the same configuration.
                              But let's first rename the pedal triangles.

                              Let ABC be a triangle and AhBhCh, AoBoCo, AnBnCn the pedal triangles
                              of H,O,N resp

                              The midpoint of the line segment joining the orthologic centers (AhBhCh, AnBnCn) and (AoBoCo, AnBnCn) is the orthocenter of AnBnCn.

                              Which is this point (the orthocenter of the pedal triangle of N)?

                              And is it a general property of all (pairs of isogonal conj.) points on the locus (the McCay cubic)?

                              ie that, that midpoint is the orthocenter of the pedal triangle of the center of the common pedal circle of the isog. conj. points?
                              (Or at least is it lying on the Euler line of the triangle)?


                              Hmmm....... neither I understand at first glance what I wrote ("the orthocenter
                              of the pedal triangle of the center of the common pedal circle of the isog. conj.
                              points") !  :-). So let me rewrit it

                              Let ABC be a triangle and P,P* two isogonal conjugate points on the McCay
                              cubic and let Q be the center of the common pedal circle of P,P* (= midpoint
                              of PP*)

                              Denote:
                              T1, T2, T3 = the pedal triangles of P,  P*, Q, resp.

                              Let Mp be the midpoint of the line segment joining the orthologcic centers (T1, T3) and (T2, T3).

                              Is Mp the orthocenter of T3 for all P's (on the cubic)?
                              Or in general, is it lying on the Euler line of T3 (orthocenter or not) ?

                              Which is its locus as P moves on the McCay cubic?

                              APH





                            • Antreas Hatzipolakis
                              From: Randy Hutson Dear Antreas and Peter, This point appears in César E. Lozada s Perspective-Orthologic-Parallelogic.pdf as the orthologic center of orthic
                              Message 14 of 18 , Dec 16, 2014
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                                From: Randy Hutson
                                 

                                Dear Antreas and Peter,

                                This point appears in César E. Lozada's Perspective-Orthologic-Parallelogic.pdf as the orthologic center of orthic and reflection triangles, also reflection of X(54) in X(973), with trilinears a*((S^2-SA^2)*(SW-R^2)+2*S^2*SA)/SA : : and search value 12.481825032289.

                                This point is also the orthic isogonal conjugate of X(1594), and X(79) of orthic IF ABC is acute.

                                Best regards,
                                Randy



                              • Antreas Hatzipolakis
                                Dear Antreas and Peter, This point appears in César E. Lozada s Perspective-Orthologic-Parallelogic.pdf as the orthologic center of orthic and reflection
                                Message 15 of 18 , Dec 16, 2014
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                                  Dear Antreas and Peter,

                                  This point appears in César E. Lozada's Perspective-Orthologic-Parallelogic.pdf as the orthologic center of orthic and reflection triangles, also reflection of X(54) in X(973), with trilinears a*((S^2-SA^2)*(SW-R^2)+2*S^2*SA)/SA : : and search value 12.481825032289.

                                  This point is also the orthic isogonal conjugate of X(1594), and X(79) of orthic IF ABC is acute.

                                  Best regards,
                                  Randy


                                  ********************

                                  Thanks, Randy!!

                                  Yes!!!! since the pedal trangle of N and the reflection triangles are homothetic

                                  and later also appeared the point in question in the paper
                                  Jesus Torres, The triangle of reflections,Forum Geometricorum, 14 (2014) 265--294  [in Theorem 4.3].
                                  (published in 7 Oct. 2014)

                                  APH





                                • Antreas Hatzipolakis
                                  ... [Peter Moses]: A few extra properties of the point: Barys: a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^4-2 a^2 b^2+b^4-2 a^2 c^2-b^2 c^2+c^4) (a^4 b^2-2 a^2
                                  Message 16 of 18 , Dec 16, 2014
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                                    [Antreas]:

                                     Let ABC be a triangle and P a point.

                                    Denote:

                                    A'B'C' = the pedal triangle of P

                                    O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)

                                    A"B"C" = the pedal triangle of O'.

                                    Which is the locus of P such that A'B'C', A"B"C"
                                    are orthologic?

                                    On the locus:

                                    1. P = H (O' = N)

                                    Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'

                                    Orthologic center (A'B'C', A"B"C") = ?

                                    2. P = Ο (O' = N)

                                    Orthologic center (A"B"C", A'B'C') = the O of A'B'C' =  the NPC center of ABC

                                    Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle
                                    Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?




                                    [Peter Moses]:

                                    >Which is the locus of P such that A'B'C', A"B"C" are orthologic?
                                    Circumcircle, infinity, K003. McCay cubic
                                     
                                    1)
                                    Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
                                    Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?
                                    2)
                                    Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)
                                     
                                    Best regards
                                    Peter



                                    **************************

                                    Thanks, Peter,

                                    What a surprise! A new point in such a simple construction !!!!
                                    I wouldn't bet a cent for it as a new point :-)

                                    Description of the point:

                                    Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of H,N, resp.

                                    A'B'C', A"B"C" are orthologic with orthologic center (A'B'C', A"B"C")
                                    on {4,93},{6,24},{25,195},.. lines of ABC.

                                    If you compute other "things" of the point (coordinates etc) please let me know
                                    (to inform Clark)

                                    Season's Greetings

                                    APH


                                    [Peter Moses]:

                                    A few extra properties of the point:
                                     
                                    Barys: a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^4-2 a^2 b^2+b^4-2 a^2 c^2-b^2 c^2+c^4) (a^4 b^2-2 a^2 b^4+b^6+a^4 c^2-2 a^2 b^2 c^2-b^4 c^2-2 a^2 c^4-b^2 c^4+c^6)::
                                     
                                    On lines {{4,93},{6,24},{25,195},{49,143},{52,539},{70,6145},{113,5446},{155,3060},{403,3574},{648,1179},{1112,2914},{1209,1216},{1843,5965},{1986,3575}}.
                                    Reflection of X(i) in X(j) for these {i,j}: {54,973},{1493,143},{2914,1112}.
                                    3 X[1209] - 2 X[1216].
                                    3 X[54] - 5 X[3567].
                                    6 X[973] - 5 X[3567].
                                    X(4)-ceva conjugate of X(1594).
                                    X(4)-crosspoint of X(3518).
                                    X(3)-crosssum of X(3519).
                                    X(2216)-isoconjugate of X(3519).
                                    Trilinear product X(1594) X(2964).
                                    Barycentric product X(1594) X(1994).
                                    X(79) Orthic triangle.
                                    {{1,21},{7,79},...} of Tangential triangle.
                                     
                                    Best regards,
                                    Peter.





                                     








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