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Circumcenter Problem (was: Re: A concyclic problem)

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  • Antreas
    [APH] ... Variation (easier?): Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), P a point, and A ,B , and C the reflections of P in
    Message 1 of 21 , Apr 13, 2010
      [APH]
      > Let ABC be a triangle, HaHbHc the pedal triangle of H
      > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
      > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
      > resp.
      > Which is the locus of P such that the circumcenter of
      > A2B2C2 is lying on the Euler Line of ABC?
      >
      > It is the Euler Line + ?????

      Variation (easier?):

      Let ABC be a triangle, HaHbHc the pedal triangle of H
      (orthic triangle), P a point, and A',B', and C' the reflections
      of P in Ha, Hb, Hc, resp.
      Which is the locus of P such that the circumcenter of
      A'B'C' is lying on the Euler Line of ABC?

      Is it Euler Line +???

      APH
    • Francisco Javier
      Thear Antreas, ... This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc Equation: -a^4 c^2 x^3 y^2 + 2 a^2 b^2 c^2 x^3 y^2 - b^4 c^2 x^3 y^2 + a^2
      Message 2 of 21 , Apr 13, 2010
        Thear Antreas,

        > [APH]
        > > Let ABC be a triangle, HaHbHc the pedal triangle of H
        > > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
        > > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
        > > resp.
        > > Which is the locus of P such that the circumcenter of
        > > A2B2C2 is lying on the Euler Line of ABC?
        > >
        > > It is the Euler Line + ?????

        This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc

        Equation:

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



        >
        > Variation (easier?):
        >
        > Let ABC be a triangle, HaHbHc the pedal triangle of H
        > (orthic triangle), P a point, and A',B', and C' the reflections
        > of P in Ha, Hb, Hc, resp.
        > Which is the locus of P such that the circumcenter of
        > A'B'C' is lying on the Euler Line of ABC?
        >
        > Is it Euler Line +???
        >

        This is Euler line + nothing



        > APH
        >
      • Alexey Zaslavsky
        Dear Linh Nguyen Van and Nikos! If triangle ABC is regular we obtain next interesting result. Given regular triangle ABC and point P. A_1B_1C_1 is the
        Message 3 of 21 , Apr 13, 2010
          Dear Linh Nguyen Van and Nikos!

          If triangle ABC is regular we obtain next interesting result.
          Given regular triangle ABC and point P. A_1B_1C_1 is the circucmevian
          triangle of P, points A_2, B_2, C_2 are the reflections of A_1, B_1, C_1 in
          the midpoints of BC, CA, AB
          Then the circumcircle of A_2B_2C_2 pass through the center O of ABC and its
          center is the reflection of P in O.

          Proof. Points O and A_2 lie on the circle with center in p[oint A_0 opposite
          to A. Thus the bisector of segment OA_2 is also the bisectrix of angle
          OA_0A_2 which is parallel to AP.
          It is clear that this line passes through point Q symmetric to P wrt O.
          Similarly the bisectors of OB_2 and OC_2 pass through Q.

          Sincerely
          Alexey
        • Antreas
          Another problem with circumcenters: Let ABC be a triangle and X,Y,Z three collinear points and XaXbXc, YaYbYc the pedal triangles of X,Y resp. Let A1,B1,C1 be
          Message 4 of 21 , Apr 13, 2010
            Another problem with circumcenters:

            Let ABC be a triangle and X,Y,Z three collinear points and
            XaXbXc, YaYbYc the pedal triangles of X,Y resp.

            Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
            and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.

            The circumcircle of A2B2C2 passes through Z and has center
            on the line XYZ.

            True??

            Antreas
          • Francisco Javier
            Yes, it is true, If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = -2 - 1/t. Francisco Javier.
            Message 5 of 21 , Apr 13, 2010
              Yes, it is true,

              If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = -2 - 1/t.

              Francisco Javier.





              --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
              >
              >
              > Another problem with circumcenters:
              >
              > Let ABC be a triangle and X,Y,Z three collinear points and
              > XaXbXc, YaYbYc the pedal triangles of X,Y resp.
              >
              > Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
              > and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.
              >
              > The circumcircle of A2B2C2 passes through Z and has center
              > on the line XYZ.
              >
              > True??
              >
              > Antreas
              >
            • Antreas Hatzipolakis
              ... Is it listed in Bernard s list of 5ics? APH [Non-text portions of this message have been removed]
              Message 6 of 21 , Apr 14, 2010
                >
                >
                > > [APH]
                > > > Let ABC be a triangle, HaHbHc the pedal triangle of H
                > > > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
                > > > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                > > > resp.
                > > > Which is the locus of P such that the circumcenter of
                > > > A2B2C2 is lying on the Euler Line of ABC?
                > > >
                > > > It is the Euler Line + ?????
                >
                > [Francisco]


                > This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc
                >
                > Equation:
                >

                Is it listed in Bernard's list of 5ics?

                APH


                [Non-text portions of this message have been removed]
              • Antreas
                [APH] ... Another variation: Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), Q a fixed point on the Euler line, and P a point. Let
                Message 7 of 21 , Apr 14, 2010
                  [APH]
                  > > Let ABC be a triangle, HaHbHc the pedal triangle of H
                  > > (orthic triangle), P a point, and A',B', and C' the reflections
                  > > of P in Ha, Hb, Hc, resp.
                  > > Which is the locus of P such that the circumcenter of
                  > > A'B'C' is lying on the Euler Line of ABC?
                  > >
                  > > Is it Euler Line +???

                  [Francisco]:
                  > This is Euler line + nothing

                  Another variation:

                  Let ABC be a triangle, HaHbHc the pedal triangle of H
                  (orthic triangle), Q a fixed point on the Euler line, and
                  P a point.

                  Let A1,B1,C1 be the reflections of Q in Ha,Hb,Hc, resp.
                  and A2,B2,C2 the reflections of A1,B1,C1 in P, resp.

                  Which is the locus of P such that the circumcenter
                  of A2B2C2 is on the Euler line of ABC?

                  Is it the Euler Line ??

                  Perspectivity:

                  Are the triangles HaHbHc, A2B2C2 perspective?

                  In general:

                  Let X,Y,Z be three collinear points, XaXbXc the
                  pedal triangle of X, and A1,B1,C1 the reflections
                  of Y in Xa,Xb,Xc resp. and A2,B2,C2 the reflections
                  of A1,B1,C1 in Z, resp.

                  Are the triangles A2B2C2, XaXbXc perspective
                  (with perspector on the line XYZ)?

                  APH
                • lovemathforever
                  Dear friends of Hyacinthos: The synthetics solutions of this problem can be found at here:
                  Message 8 of 21 , May 2, 2010
                    Dear friends of Hyacinthos:
                    The synthetics solutions of this problem can be found at here:
                    http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&p=1863882&sid=ce1cf3b9df63a1477cc4699f020559dc#p1863882

                    Best regard,
                    Linh Nguyen Van
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