Loading ...
Sorry, an error occurred while loading the content.
 

Re: Perspective?

Expand Messages
  • Angel
    Dear Randy, A B C is also perspective to N1N2N3 at non-ETC 3.427184657834269 (coordinates?) Its coordinates are very complicated; the first coordinate of its
    Message 1 of 23 , Feb 12, 2013
      Dear Randy,

      A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269 (coordinates?)

      Its coordinates are very complicated; the first coordinate of its center is:

      a^20(b^2+c^2)-
      2a^18(3b^4+7b^2c^2+3c^4)+
      a^16(b^2+c^2)(13b^4+37b^2c^2+13c^4)-
      2a^14(4b^8+39b^6c^2+68b^4c^4+39b^2c^6+4c^8)-
      a^12(b^2+c^2)(14b^8-71b^6c^2-106b^4c^4-71b^2c^6+14c^8)+
      2a^10(14b^12-3b^10c^2-64b^8c^4-74b^6c^6- 64b^4c^8-3b^2c^10+14c^12)-
      a^8(b^2+c^2)(14b^12+27b^10c^2-110b^8c^4+ 66b^6c^6- 110b^4c^8+27b^2c^10+14c^12)-
      2a^6(b^2-c^2)^2(4b^12-27b^10c^2- 8b^8c^4-2b^6c^6-8b^4c^8-27b^2c^10+4c^12)+
      a^4(b^2-c^2)^4(b^2+c^2)(13b^8-23b^6c^2+4b^4c^4- 23b^2c^6+13c^8) -
      2a^2(b^2-c^2)^6(b^2+c^2)^2(3b^4-2b^2c^2+3c^4)+
      (b^2-c^2)^8(b^2+c^2)^3


      Best regards.
      Angel M.

      --- In Hyacinthos@yahoogroups.com, "rhutson2" wrote:
      >
      > Dear Antreas and Angel,
      >
      > A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269 (coordinates?)
      >
      > N1N2N3 is inversely similar to ABC, with similitude center X(3).
      > N1N2N3 is perspective to ABC at X(3519).
      > X(3) of N1N2N3 = X(3)
      > X(186) of N1N2N3 = X(1511)
      > X(2070) of N1N2N3 = X(110)
      > The Euler line of N1N2N3 is line X(3)X(74).
      > Line IO of N1N2N3 is line X(3)X(191).
      > The Brocard axis of N1N2N3 is line X(3)X(67).
      >
      > Randy
      >
      > --- In Hyacinthos@yahoogroups.com, "Angel" wrote:
      > >
      > > Dear Antreas,
      > >
      > > The triangles ABC, A'B'C' are perspective.
      > >
      > > Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)
      > >
      > > Best regards.
      > > Angel M.
      > >
      > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
      > > >
      > > > 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
      > > > the reflections of the NPC (N) in the perp. bisectors
      > > > OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
      > > > axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
      > > > The triangles ABC, A'B'C' are perspective.
      > > >
      > > > Perspector?
      > > >
      > > > APH
      > > >
      > >
      >
    • Antreas
      Dear Randy and Angel And in this case I think that A B C , N1N2N3 are perspective: Denote: HaHbHc = the orthic triangle. (N1), (N2), (N3) = the reflections of
      Message 2 of 23 , Feb 12, 2013
        Dear Randy and Angel

        And in this case I think that A'B'C', N1N2N3 are perspective:

        Denote:
        HaHbHc = the orthic triangle.
        (N1), (N2), (N3) = the reflections of (N) in HaO, HbO, HcO, resp.

        A'B'C' = the triangle bounded by the radical axes of ((O),(N1)),
        ((O),(N2)), ((O),(N3)),

        GENERALIZATION:

        Denote:

        PaPbPc = the pedal triangle of a point P on the Euler line

        (N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO, resp.

        A'B'C' = the triangle bounded by the radical axes of
        ((O),(N1)), ((O),(N2)),((O),(N3)).

        A'B'C', N1N2N3 are perspective.

        True???

        APH


        On Wed, Feb 13, 2013 at 12:13 AM, Angel amontes1949@...> wrote:

        > Dear Randy,
        >
        >
        > A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269
        > (coordinates?)
        >
        > Its coordinates are very complicated; the first coordinate of its center
        > is:
        >
        > a^20(b^2+c^2)-
        > 2a^18(3b^4+7b^2c^2+3c^4)+
        > a^16(b^2+c^2)(13b^4+37b^2c^2+13c^4)-
        > 2a^14(4b^8+39b^6c^2+68b^4c^4+39b^2c^6+4c^8)-
        > a^12(b^2+c^2)(14b^8-71b^6c^2-106b^4c^4-71b^2c^6+14c^8)+
        > 2a^10(14b^12-3b^10c^2-64b^8c^4-74b^6c^6- 64b^4c^8-3b^2c^10+14c^12)-
        > a^8(b^2+c^2)(14b^12+27b^10c^2-110b^8c^4+ 66b^6c^6-
        > 110b^4c^8+27b^2c^10+14c^12)-
        > 2a^6(b^2-c^2)^2(4b^12-27b^10c^2- 8b^8c^4-2b^6c^6-8b^4c^8-27b^2c^10+4c^12)+
        > a^4(b^2-c^2)^4(b^2+c^2)(13b^8-23b^6c^2+4b^4c^4- 23b^2c^6+13c^8) -
        > 2a^2(b^2-c^2)^6(b^2+c^2)^2(3b^4-2b^2c^2+3c^4)+
        > (b^2-c^2)^8(b^2+c^2)^3
        >
        > Best regards.
        > Angel M.
        >
        >
        > --- In Hyacinthos@yahoogroups.com, "rhutson2" wrote:
        > >
        > > Dear Antreas and Angel,
        > >
        > > A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269
        > (coordinates?)
        > >
        > > N1N2N3 is inversely similar to ABC, with similitude center X(3).
        > > N1N2N3 is perspective to ABC at X(3519).
        > > X(3) of N1N2N3 = X(3)
        > > X(186) of N1N2N3 = X(1511)
        > > X(2070) of N1N2N3 = X(110)
        > > The Euler line of N1N2N3 is line X(3)X(74).
        > > Line IO of N1N2N3 is line X(3)X(191).
        > > The Brocard axis of N1N2N3 is line X(3)X(67).
        > >
        > > Randy
        > >
        > > --- In Hyacinthos@yahoogroups.com, "Angel" wrote:
        > > >
        > > > Dear Antreas,
        > > >
        > > > The triangles ABC, A'B'C' are perspective.
        > > >
        > > > Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)
        > > >
        > > > Best regards.
        > > > Angel M.
        > > >
        > > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
        > > > >
        > > > > 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
        > > > > the reflections of the NPC (N) in the perp. bisectors
        > > > > OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
        > > > > axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
        > > > > The triangles ABC, A'B'C' are perspective.
        > > > >
        > > > > Perspector?
        > > > >
        > > > > APH
      • Antreas Hatzipolakis
        As locus problem: Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. Denote: (N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO,
        Message 3 of 23 , Feb 12, 2013
          As locus problem:

          Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

          Denote:
          (N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO, resp.

          A'B'C' = the triangle bounded by the radical axes of
          ((O),(N1)), ((O),(N2)),((O),(N3)).

          Which is the locus of P such that A'B'C', N1N2N3 are perspective?

          Euler Line (?) + ??

          APH

          GENERALIZATION:

          >
          > Denote:
          >
          > PaPbPc = the pedal triangle of a point P on the Euler line
          >
          >
          > (N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO, resp.
          >
          > A'B'C' = the triangle bounded by the radical axes of
          > ((O),(N1)), ((O),(N2)),((O),(N3)).
          >
          > A'B'C', N1N2N3 are perspective.
          >
          > True???
          >
          >
          >


          [Non-text portions of this message have been removed]
        • Antreas
          GENERALIZATION Let ABC be a triangle, Q a point, QaQbQc the pedal triangle of Q, (X) the circumcircle of QaQbQc (=pedal circle of Q), P a point on the OQ line
          Message 4 of 23 , Feb 13, 2013
            GENERALIZATION

            Let ABC be a triangle, Q a point, QaQbQc the pedal triangle of Q,
            (X) the circumcircle of QaQbQc (=pedal circle of Q), P a point on
            the OQ line and PaPbPc the pedal triangle of P.

            Denote:

            (X1), (X2), (X3) the reflections of (X) in OPa,OPb,OPc, resp.

            A'B'C' = the triangle bounded by the radical axes of ((O),(X1)),((O),(X2)),((O),(X3))

            Conjecture:
            The triangles A'B'C', X1X2X3 are perspective.

            Figure:
            http://anthrakitis.blogspot.gr/2013/02/reflecting-pedal-circle.html

            Antreas

            [APH]
            >
            > Denote:
            >
            > PaPbPc = the pedal triangle of a point P on the Euler line
            >
            > (N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO, resp.
            >
            > A'B'C' = the triangle bounded by the radical axes of
            > ((O),(N1)), ((O),(N2)),((O),(N3)).
            >
            > A'B'C', N1N2N3 are perspective.
            >
            > True???
          • Antreas
            Let HaHbHc be the orthic triangle of ABC, (N1),(N2),(N3) the reflections of the NPC (N) in the sidelines HbHc, HcHa, HaHb of HaHbHc resp. and and A B C the
            Message 5 of 23 , Feb 14, 2013
              Let HaHbHc be the orthic triangle of ABC, (N1),(N2),(N3) the
              reflections of the NPC (N) in the sidelines HbHc, HcHa, HaHb of
              HaHbHc resp. and and A'B'C' the triangle bounded by the radical
              axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.

              Are the triangles ABC, A'B'C' perspective ?

              aph

              [APH]
              > 2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the reflections
              > of the NPC (N) in the altitudes HHa,HHb,HHc,
              > resp. and and A'B'C' the triangle bounded by the radical axes of
              > ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
              >
              > Are the triangles HaHbHc, A'B'C' perspective ?
            • Angel
              Dear Antreas, The triangles ABC, A B C are perspective. Perspector: (Conway notations) (SA^2-3S^2)(S^2-SB^2)(S^2-SC^2))/SA: ... : ... Best regards. Angel M.
              Message 6 of 23 , Feb 14, 2013
                Dear Antreas,

                The triangles ABC, A'B'C' are perspective.

                Perspector: (Conway notations)

                (SA^2-3S^2)(S^2-SB^2)(S^2-SC^2))/SA: ... : ...

                Best regards.
                Angel M.


                --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
                >
                > Let HaHbHc be the orthic triangle of ABC, (N1),(N2),(N3) the
                > reflections of the NPC (N) in the sidelines HbHc, HcHa, HaHb of
                > HaHbHc resp. and and A'B'C' the triangle bounded by the radical
                > axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
                >
                > Are the triangles ABC, A'B'C' perspective ?
                >
                > aph
              • Antreas
                I am not sure If I have already posted this: Let ABC be a triangle, (N1),(N2),(N3) the reflections of (N)[=NPC] in OA,OB,OC, resp. and A B C the triangle
                Message 7 of 23 , Feb 20, 2013
                  I am not sure If I have already posted this:

                  Let ABC be a triangle, (N1),(N2),(N3) the reflections of
                  (N)[=NPC] in OA,OB,OC, resp. and A'B'C' the triangle bounded
                  by the radical axes of ((O),(N1)),((O),(N2)),((O),(N3)).

                  Are N1N2N3, A'B'C' perspective?

                  APH
                • Antreas Hatzipolakis
                  One more.... HaHbHc = the orthic triangle. (N1) = the circle (Ha, HaN) ie the circle with center Ha and radius HaN =R/2 Similarly (N2),(N3) A B C = the
                  Message 8 of 23 , Feb 20, 2013
                    One more....

                    HaHbHc = the orthic triangle.
                    (N1) = the circle (Ha, HaN) ie the circle with center Ha and radius HaN =R/2
                    Similarly (N2),(N3)

                    A'B'C' = the triangle bounded by the radical axes of
                    ((O),(N1)),((O),(N2)),((O),(N3)).

                    Are the triangles HaHbHc, N1N2N3 perspective?

                    aph


                    On Wed, Feb 20, 2013 at 1:29 PM, Antreas <anopolis72@...> wrote:

                    > **
                    >
                    >
                    > I am not sure If I have already posted this:
                    >
                    > Let ABC be a triangle, (N1),(N2),(N3) the reflections of
                    > (N)[=NPC] in OA,OB,OC, resp. and A'B'C' the triangle bounded
                    > by the radical axes of ((O),(N1)),((O),(N2)),((O),(N3)).
                    >
                    >
                    > Are N1N2N3, A'B'C' perspective?
                    >
                    > APH
                    >
                    > __
                    >


                    [Non-text portions of this message have been removed]
                  • Antreas Hatzipolakis
                    Let ABC be a triangle with excenters Ia,Ib,Ic. The NPC of AHIa interscts the excircle (Ia) at A other than the Feuerbach point Fa. The NPC of BHIb
                    Message 9 of 23 , May 6, 2014

                      Let ABC be a triangle with excenters Ia,Ib,Ic.

                      The NPC of  AHIa interscts the excircle (Ia) at A' other than the
                      Feuerbach point Fa.

                      The  NPC of  BHIb intersects the excircle (Ib) at B' other than the
                      Feuerbach point Fb,

                      The  NPC of  CHIc intersects the excircle (Ic) at C' other than the
                      Feuerbach point Fc.

                      Are the triangles ABC, A'B'C' perspective?

                      In any case, has the triangle A'B'C' any interesting properties?

                      APH

                    Your message has been successfully submitted and would be delivered to recipients shortly.