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Re: [EMHL] NPC. locus.

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  • Antreas Hatzipolakis
    A CIRCLE: Let ABC be a triangle, A B C the cevian triangle of I and N1, N2, N3 the NPC centers of IB C , IC A , IA B , resp. The points I, N1,N2,N3 are
    Message 1 of 9 , Apr 17 3:03 AM
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      A CIRCLE:

      Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
      the NPC centers of IB'C', IC'A', IA'B', resp.

      The points I, N1,N2,N3 are concyclic.

      Center of the circle?

      LOCUS:

      Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
      N2, N3
      the NPC centers of PB'C', PC'A', PA'B', resp.

      Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

      Antreas


      On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@...> wrote:

      > **
      >
      >
      > Let ABC be a triangle and P a point.
      >
      > Which is the locus of P such that the NPC center of PBC
      > lies on the line AP ?
      >
      > Ceva triangle variation:
      >
      > Let ABC be a triangle, P a point and A'B'C'
      > the cevian triangle of P.
      > Which is the locus of P such that the NPC center
      > of PB'C' lies on the line APA' ?
      > (The I is on the locus)
      >
      > APH
      >
      >
      >
      >
      >

      http://anopolis72000.blogspot.com/


      [Non-text portions of this message have been removed]
    • César Lozada
      Yes, they are concyclic. The center X of the circle has trilinears: 2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : : ETC search: 2.387773069046934..,
      Message 2 of 9 , Apr 17 6:04 AM
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        Yes, they are concyclic.



        The center X of the circle has trilinears:

        2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : :



        ETC search: 2.387773069046934.., 2.38593313995937.., 0.886815506990847..



        X=Midpoint of X(I),X(J) for these (I,J):

        (1,500)




        X lies on line X(I),X(J) for these (I,J):

        (1,30), (3,81), (5,581), (21,323), (30,79), (58,5428), (79,500),
        (140,3216),

        (186,2906), (237,1896), (285,1082), (358,1882), (386,549), (500,554),
        (511,1385),

        (550,991), (554,1081), (1036,4022), (1081,1464), (1154,2646), (1464,1717),


        (1675,4280), (1717,1836), (1836,3058), (2071,4309), (2072,3746),
        (2771,3743),

        (2943,4341), (3058,3649), (3108,3784), (3649,3655), (3655,3656),
        (3656,3782),

        (3704,3990), (3782,4654), (4654,4854), (4854,5160), (5160,5434),
        (5434,5441)



        Regards

        César Lozada



        _____

        De: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] En nombre
        de Antreas Hatzipolakis
        Enviado el: Miércoles, 17 de Abril de 2013 05:34 a.m.
        Para: Hyacinthos
        Asunto: Re: [EMHL] NPC. locus.





        A CIRCLE:

        Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
        the NPC centers of IB'C', IC'A', IA'B', resp.

        The points I, N1,N2,N3 are concyclic.

        Center of the circle?

        LOCUS:

        Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
        N2, N3
        the NPC centers of PB'C', PC'A', PA'B', resp.

        Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

        Antreas

        On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@...
        <mailto:anopolis72%40gmail.com> > wrote:

        > **
        >
        >
        > Let ABC be a triangle and P a point.
        >
        > Which is the locus of P such that the NPC center of PBC
        > lies on the line AP ?
        >
        > Ceva triangle variation:
        >
        > Let ABC be a triangle, P a point and A'B'C'
        > the cevian triangle of P.
        > Which is the locus of P such that the NPC center
        > of PB'C' lies on the line APA' ?
        > (The I is on the locus)
        >
        > APH
        >
        >
        >
        >
        >

        http://anopolis72000.blogspot.com/

        [Non-text portions of this message have been removed]





        [Non-text portions of this message have been removed]
      • César Lozada
        Correction: X lies on line X(I),X(J) for these (I,J): (1,30), (3,81), (5,581), (21,323), (58,5428), (140,3216), (186,2906), (386,549), (511,1385), (550,991),
        Message 3 of 9 , Apr 17 7:45 AM
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          Correction:



          X lies on line X(I),X(J) for these (I,J):

          (1,30), (3,81), (5,581), (21,323), (58,5428), (140,3216), (186,2906),

          (386,549), (511,1385), (550,991), (1154,2646), (2771,3743)



          César Lozada

          _____

          De: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] En nombre
          de César Lozada
          Enviado el: Miércoles, 17 de Abril de 2013 08:35 a.m.
          Para: Hyacinthos@yahoogroups.com
          Asunto: RE: [EMHL] NPC. locus.





          Yes, they are concyclic.

          The center X of the circle has trilinears:

          2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : :

          ETC search: 2.387773069046934.., 2.38593313995937.., 0.886815506990847..

          X=Midpoint of X(I),X(J) for these (I,J):

          (1,500)

          X lies on line X(I),X(J) for these (I,J):

          (1,30), (3,81), (5,581), (21,323), (30,79), (58,5428), (79,500),
          (140,3216),

          (186,2906), (237,1896), (285,1082), (358,1882), (386,549), (500,554),
          (511,1385),

          (550,991), (554,1081), (1036,4022), (1081,1464), (1154,2646), (1464,1717),

          (1675,4280), (1717,1836), (1836,3058), (2071,4309), (2072,3746),
          (2771,3743),

          (2943,4341), (3058,3649), (3108,3784), (3649,3655), (3655,3656),
          (3656,3782),

          (3704,3990), (3782,4654), (4654,4854), (4854,5160), (5160,5434),
          (5434,5441)

          Regards

          César Lozada

          _____

          De: Hyacinthos@yahoogroups.com <mailto:Hyacinthos%40yahoogroups.com>
          [mailto:Hyacinthos@yahoogroups.com <mailto:Hyacinthos%40yahoogroups.com> ]
          En nombre
          de Antreas Hatzipolakis
          Enviado el: Miércoles, 17 de Abril de 2013 05:34 a.m.
          Para: Hyacinthos
          Asunto: Re: [EMHL] NPC. locus.

          A CIRCLE:

          Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
          the NPC centers of IB'C', IC'A', IA'B', resp.

          The points I, N1,N2,N3 are concyclic.

          Center of the circle?

          LOCUS:

          Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
          N2, N3
          the NPC centers of PB'C', PC'A', PA'B', resp.

          Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

          Antreas

          On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@...
          <mailto:anopolis72%40gmail.com>
          <mailto:anopolis72%40gmail.com> > wrote:

          > **
          >
          >
          > Let ABC be a triangle and P a point.
          >
          > Which is the locus of P such that the NPC center of PBC
          > lies on the line AP ?
          >
          > Ceva triangle variation:
          >
          > Let ABC be a triangle, P a point and A'B'C'
          > the cevian triangle of P.
          > Which is the locus of P such that the NPC center
          > of PB'C' lies on the line APA' ?
          > (The I is on the locus)
          >
          > APH
          >
          >
          >
          >
          >

          http://anopolis72000.blogspot.com/

          [Non-text portions of this message have been removed]

          [Non-text portions of this message have been removed]





          [Non-text portions of this message have been removed]
        • Antreas
          ... The most interesting line is the first one (X1,X30): If I0 is the center of the circle, then the line II0 is parallel to Euler line of ABC. There are three
          Message 4 of 9 , Apr 17 11:16 AM
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            [APH]:
            > A CIRCLE:
            >
            > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
            > the NPC centers of IB'C', IC'A', IA'B', resp.
            >
            > The points I, N1,N2,N3 are concyclic.
            >
            > Center of the circle?
            >
            > LOCUS:
            >
            > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of
            > P and N1,
            > N2, N3
            > the NPC centers of PB'C', PC'A', PA'B', resp.
            >
            > Which is the locus of P such that the points P, N1,N2,N3
            > are concyclic ?

            [César Lozada]:
            > Yes, they are concyclic.
            >
            > The center X of the circle has trilinears:
            >
            > 2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : :
            >
            > ETC search: 2.387773069046934.., 2.38593313995937.., 0.886815506990847..
            >
            > X=Midpoint of X(I),X(J) for these (I,J):
            >
            > (1,500)
            >

            > X lies on line X(I),X(J) for these (I,J):
            >
            > (1,30), (3,81), (5,581), (21,323), (58,5428), (140,3216), (186,2906),
            >
            > (386,549), (511,1385), (550,991), (1154,2646), (2771,3743)


            The most interesting line is the first one (X1,X30):
            If I0 is the center of the circle, then the line II0
            is parallel to Euler line of ABC.

            There are three more centers of circles (but not triangle centrers):
            I1, I2, I3 = the extraversions of I0.

            ie:

            Let Ia, Ib, Ic be the excenters of ABC.

            Denote:

            A1B1C1, A2B2C2, A3B3C3 = the cevian triangles of Ia, Ib, Ic, resp.

            I1 = the center of the circle (Ia, N11, N12, N13), where
            N11 = the NPC center of IaB1C1,
            N12 = the NPC center of IaC1A1,
            N13 = the NPC center of IaA1B1.

            Similarly I2, I3.

            Antreas
          • rhutson2
            A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle
            Message 5 of 9 , Apr 17 6:33 PM
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              A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.

              Randy

              --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
              >
              > A CIRCLE:
              >
              > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
              > the NPC centers of IB'C', IC'A', IA'B', resp.
              >
              > The points I, N1,N2,N3 are concyclic.
              >
              > Center of the circle?
              >
              > LOCUS:
              >
              > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
              > N2, N3
              > the NPC centers of PB'C', PC'A', PA'B', resp.
              >
              > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?
              >
              > Antreas
              >
              >
              > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@...> wrote:
              >
              > > **
              > >
              > >
              > > Let ABC be a triangle and P a point.
              > >
              > > Which is the locus of P such that the NPC center of PBC
              > > lies on the line AP ?
              > >
              > > Ceva triangle variation:
              > >
              > > Let ABC be a triangle, P a point and A'B'C'
              > > the cevian triangle of P.
              > > Which is the locus of P such that the NPC center
              > > of PB'C' lies on the line APA' ?
              > > (The I is on the locus)
              > >
              > > APH
              > >
              > >
              > >
              > >
              > >
              >
              > http://anopolis72000.blogspot.com/
              >
              >
              > [Non-text portions of this message have been removed]
              >
            • Francisco Javier
              This is the tricircular sextic Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z) where S=twice the area of ABC. (tricircular = the circular points
              Message 6 of 9 , Apr 17 11:30 PM
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                This is the tricircular sextic

                Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z)

                where S=twice the area of ABC.

                (tricircular = the circular points are triple points of the curve).

                Francisco Javier.

                --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
                >
                > A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.
                >
                > Randy
                >
                > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@> wrote:
                > >
                > > A CIRCLE:
                > >
                > > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
                > > the NPC centers of IB'C', IC'A', IA'B', resp.
                > >
                > > The points I, N1,N2,N3 are concyclic.
                > >
                > > Center of the circle?
                > >
                > > LOCUS:
                > >
                > > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
                > > N2, N3
                > > the NPC centers of PB'C', PC'A', PA'B', resp.
                > >
                > > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?
                > >
                > > Antreas
                > >
                > >
                > > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@> wrote:
                > >
                > > > **
                > > >
                > > >
                > > > Let ABC be a triangle and P a point.
                > > >
                > > > Which is the locus of P such that the NPC center of PBC
                > > > lies on the line AP ?
                > > >
                > > > Ceva triangle variation:
                > > >
                > > > Let ABC be a triangle, P a point and A'B'C'
                > > > the cevian triangle of P.
                > > > Which is the locus of P such that the NPC center
                > > > of PB'C' lies on the line APA' ?
                > > > (The I is on the locus)
                > > >
                > > > APH
                > > >
                > > >
                > > >
                > > >
                > > >
                > >
                > > http://anopolis72000.blogspot.com/
                > >
                > >
                > > [Non-text portions of this message have been removed]
                > >
                >
              • rhutson2
                Francisco Javier, While Q014 mentions PU(5), it does not contain them, nor X(13) and X(14), which are on this curve. Also, X(80) which is on Q014, is not on
                Message 7 of 9 , Apr 18 1:20 AM
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                  Francisco Javier,

                  While Q014 mentions PU(5), it does not contain them, nor X(13) and X(14), which are on this curve. Also, X(80) which is on Q014, is not on this curve (X(80) and the NPCs of BCX(80), CAX(80), BCX(80) are not concyclic).

                  By the way, the point Qi on this curve is the point which is the incenter of its anticevian triangle, and has ETC search value 1.999434154060428. Coordinates? Its isogonal conjugate Qi* is the point which is the nine-point center of its pedal triangle (ETC search 1.142779079509848). The line QiQi* passes through X(5).

                  Randy

                  --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
                  >
                  > This is the tricircular sextic
                  >
                  > Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z)
                  >
                  > where S=twice the area of ABC.
                  >
                  > (tricircular = the circular points are triple points of the curve).
                  >
                  > Francisco Javier.
                  >
                  > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
                  > >
                  > > A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.
                  > >
                  > > Randy
                  > >
                  > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@> wrote:
                  > > >
                  > > > A CIRCLE:
                  > > >
                  > > > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
                  > > > the NPC centers of IB'C', IC'A', IA'B', resp.
                  > > >
                  > > > The points I, N1,N2,N3 are concyclic.
                  > > >
                  > > > Center of the circle?
                  > > >
                  > > > LOCUS:
                  > > >
                  > > > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
                  > > > N2, N3
                  > > > the NPC centers of PB'C', PC'A', PA'B', resp.
                  > > >
                  > > > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?
                  > > >
                  > > > Antreas
                  > > >
                  > > >
                  > > > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@> wrote:
                  > > >
                  > > > > **
                  > > > >
                  > > > >
                  > > > > Let ABC be a triangle and P a point.
                  > > > >
                  > > > > Which is the locus of P such that the NPC center of PBC
                  > > > > lies on the line AP ?
                  > > > >
                  > > > > Ceva triangle variation:
                  > > > >
                  > > > > Let ABC be a triangle, P a point and A'B'C'
                  > > > > the cevian triangle of P.
                  > > > > Which is the locus of P such that the NPC center
                  > > > > of PB'C' lies on the line APA' ?
                  > > > > (The I is on the locus)
                  > > > >
                  > > > > APH
                  > > > >
                  > > > >
                  > > > >
                  > > > >
                  > > > >
                  > > >
                  > > > http://anopolis72000.blogspot.com/
                  > > >
                  > > >
                  > > > [Non-text portions of this message have been removed]
                  > > >
                  > >
                  >
                • rhutson2
                  My apologies. I misread Francisco Javier s post below. As he explained privately, he meant the difference: LOCUS = SEXTIC(Q014) - SEXTIC(S 4 S^2 x y z (x + y
                  Message 8 of 9 , Apr 18 3:58 AM
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                    My apologies. I misread Francisco Javier's post below. As he explained privately, he meant the difference:

                    LOCUS = SEXTIC(Q014) - SEXTIC(S 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z) )

                    Question: does this locus contain any ETC centers or bicentric pairs besides X(13), X(14), and PU(5)? Are the circumcircle intercepts of line X(5)X(523) triangle centers or another bicentric pair?

                    Randy

                    --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
                    >
                    > Francisco Javier,
                    >
                    > While Q014 mentions PU(5), it does not contain them, nor X(13) and X(14), which are on this curve. Also, X(80) which is on Q014, is not on this curve (X(80) and the NPCs of BCX(80), CAX(80), BCX(80) are not concyclic).
                    >
                    > By the way, the point Qi on this curve is the point which is the incenter of its anticevian triangle, and has ETC search value 1.999434154060428. Coordinates? Its isogonal conjugate Qi* is the point which is the nine-point center of its pedal triangle (ETC search 1.142779079509848). The line QiQi* passes through X(5).
                    >
                    > Randy
                    >
                    > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
                    > >
                    > > This is the tricircular sextic
                    > >
                    > > Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z)
                    > >
                    > > where S=twice the area of ABC.
                    > >
                    > > (tricircular = the circular points are triple points of the curve).
                    > >
                    > > Francisco Javier.
                    > >
                    > > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
                    > > >
                    > > > A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.
                    > > >
                    > > > Randy
                    > > >
                    > > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@> wrote:
                    > > > >
                    > > > > A CIRCLE:
                    > > > >
                    > > > > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3
                    > > > > the NPC centers of IB'C', IC'A', IA'B', resp.
                    > > > >
                    > > > > The points I, N1,N2,N3 are concyclic.
                    > > > >
                    > > > > Center of the circle?
                    > > > >
                    > > > > LOCUS:
                    > > > >
                    > > > > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,
                    > > > > N2, N3
                    > > > > the NPC centers of PB'C', PC'A', PA'B', resp.
                    > > > >
                    > > > > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?
                    > > > >
                    > > > > Antreas
                    > > > >
                    > > > >
                    > > > > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@> wrote:
                    > > > >
                    > > > > > **
                    > > > > >
                    > > > > >
                    > > > > > Let ABC be a triangle and P a point.
                    > > > > >
                    > > > > > Which is the locus of P such that the NPC center of PBC
                    > > > > > lies on the line AP ?
                    > > > > >
                    > > > > > Ceva triangle variation:
                    > > > > >
                    > > > > > Let ABC be a triangle, P a point and A'B'C'
                    > > > > > the cevian triangle of P.
                    > > > > > Which is the locus of P such that the NPC center
                    > > > > > of PB'C' lies on the line APA' ?
                    > > > > > (The I is on the locus)
                    > > > > >
                    > > > > > APH
                    > > > > >
                    > > > > >
                    > > > > >
                    > > > > >
                    > > > > >
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