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[EMHL] Re: Simson line of Feuerbach point

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  • Antreas
    Well... let s take three other points on the circumcircle.... The perpendicular bisectors of the sides of ABC intersect the circumcircle at A ,B ,C on the
    Message 1 of 12 , Mar 6, 2013
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      Well... let's take three other points on the circumcircle....

      The perpendicular bisectors of the sides of ABC intersect
      the circumcircle at A',B',C' on the positive sides of BC,CA,AB, resp.
      (ie both A,A' are on the same side of BC etc)

      The Simson lines of A',B',C' are concurrent at a point P.

      Note: If A",B",C" are the other intersections of the perp. bisectors
      withthe circumcircle, then the Simson Lines of A",B",C" bound a
      triangle which is perspective with the pedal triangle of O (medial
      triangle) at P.

      Generalization (Locus):

      Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.
      The line PPa intersects the circumcircle at A' on the positive side
      of BC (ie both A,A' are on the same side of BC). Similarly B',C'.

      Which is the locus of P such that the Simson lines of A',B',C' are
      concurrent?

      APH

      On Thu, Mar 7, 2013 at 7:56 AM, rhutson2 <rhutson2@...> wrote:
      >
      > They are not perspective. A*B*C* is also the circumcevian triangle of
      > X(56). If mixtilinear excircles are used instead, touching the circumcircle
      > at A**,B**,C**, then A**B**C** is the circumcevian triangle of X(55). The
      > triangle bounded by the Simson lines of A**,B**,C** is also not perspective
      > to A'B'C' or A"B"C".
      >
      > Randy
      >
      >
      > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
      > wrote:
      > >
      > > Let ABC be a triangle and A'B'C' the cevian triangle of I.
      > >
      > > The three mixtilinear incircles touch the circumcircle at A*,B*,C*
      > > respective to A,B,C.
      > >
      > > The Simson lines of A*,B*,C* bound a triangle A"B"C".
      > >
      > > Are the triangles A'B'C', A"B"C" perspective?
      > >
      > > APH
    • Antreas Hatzipolakis
      Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle) . Denote: IaaIabIac = the pedal triangle of Ia (wrt triangle ABC) IbbIbcIba =
      Message 2 of 12 , Oct 24, 2014
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        Let ABC be a triangle and IaIbIc the antipedal triangle of I
        (excentral triangle)
        .
        Denote:
        IaaIabIac = the pedal triangle of Ia (wrt triangle ABC)
        IbbIbcIba = the pedal triangle of Ib
        IccIcaIcb = the pedal triangle of Ic.

        Fa, Fb, Fc= the Feuerbach points relative to (Ia), (Ib), (Ic)
        (ie the points of contact of the NPC with the excircles (Ia), (Ib), (Ic),
        resp.)

        sFa, sFb, sFc =  the Simson lines of Fa, Fb, Fc wrt triangles
        IaaIabIac,IbbIbcIba, IccIcaIcb.

        A*B*C* = the triangle bounded by sFa, sFb, sFc

        I think the triangles IaIbIc, A*B*C* are parallelogic. and the parallelogic center
        (IaIbIc, A*B*C*) is the O of ABC = N of IaIbIc
        (ie the parallels to sFa, sFb, sFc through Ia, Ib,Ic, resp. concur at O)

        The other parallelogic center if exists ?
        (quite complicated, I guess :-)

        Note:
        sFa, sFb, sFc are not concurrent.
        See Hyacinthos,  messages ##21663, 21664


        Antreas
      • Antreas Hatzipolakis
        [APH] ... [Telv Cohl]: In general: Let P, Q be isogonal conjugate of ABC Let P be the orthopole of OP with respect to ABC Then the Simson line of P WRT the
        Message 3 of 12 , Oct 24, 2014
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          [APH]
          Let ABC be a triangle and IaIbIc the antipedal triangle of I
          (excentral triangle)
          .
          Denote:
          IaaIabIac = the pedal triangle of Ia (wrt triangle ABC)
          IbbIbcIba = the pedal triangle of Ib
          IccIcaIcb = the pedal triangle of Ic.

          Fa, Fb, Fc= the Feuerbach points relative to (Ia), (Ib), (Ic)
          (ie the points of contact of the NPC with the excircles (Ia), (Ib), (Ic),
          resp.)

          sFa, sFb, sFc =  the Simson lines of Fa, Fb, Fc wrt triangles
          IaaIabIac,IbbIbcIba, IccIcaIcb.

          A*B*C* = the triangle bounded by sFa, sFb, sFc

          I think the triangles IaIbIc, A*B*C* are parallelogic. and the parallelogic center
          (IaIbIc, A*B*C*) is the O of ABC = N of IaIbIc
          (ie the parallels to sFa, sFb, sFc through Ia, Ib,Ic, resp. concur at O)

          The other parallelogic center if exists ?
          (quite complicated, I guess :-)

          Note:
          sFa, sFb, sFc are not concurrent.
          See Hyacinthos,  messages ##21663, 21664



          [Telv Cohl]:

          In general:

          Let P, Q be isogonal conjugate of ABC
          Let P' be the orthopole of OP with respect to ABC
          Then the Simson line of P' WRT the pedal triangle of Q is parallel to OP
          (O is the circumcenter of triangle ABC )

        • Antreas Hatzipolakis
          ... 2 a^13+2 a^12 b-12 a^11 b^2-10 a^10 b^3+29 a^9 b^4+21 a^8 b^5-36 a^7 b^6-24 a^6 b^7+24 a^5 b^8+16 a^4 b^9-8 a^3 b^10-6 a^2 b^11+a b^12+b^13+2 a^12 c-2 a^11
          Message 4 of 12 , Oct 25, 2014
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            [APH]:


            Let ABC be a triangle and IaIbIc the antipedal triangle of I
            (excentral triangle)
            .
            Denote:
            IaaIabIac = the pedal triangle of Ia (wrt triangle ABC)
            IbbIbcIba = the pedal triangle of Ib
            IccIcaIcb = the pedal triangle of Ic.

            Fa, Fb, Fc= the Feuerbach points relative to (Ia), (Ib), (Ic)
            (ie the points of contact of the NPC with the excircles (Ia), (Ib), (Ic),
            resp.)

            sFa, sFb, sFc =  the Simson lines of Fa, Fb, Fc wrt triangles
            IaaIabIac,IbbIbcIba, IccIcaIcb.

            A*B*C* = the triangle bounded by sFa, sFb, sFc

            I think the triangles IaIbIc, A*B*C* are parallelogic. and the parallelogic center
            (IaIbIc, A*B*C*) is the O of ABC = N of IaIbIc
            (ie the parallels to sFa, sFb, sFc through Ia, Ib,Ic, resp. concur at O)

            The other parallelogic center if exists ?
            (quite complicated, I guess :-)

            Note:
            sFa, sFb, sFc are not concurrent.
            See Hyacinthos,  messages ##21663, 21664




            [PETER MOSES]: 

            >The other parallelogic center if exists ?
            2 a^13+2 a^12 b-12 a^11 b^2-10 a^10 b^3+29 a^9 b^4+21 a^8 b^5-36 a^7 b^6-24 a^6 b^7+24 a^5 b^8+16 a^4 b^9-8 a^3 b^10-6 a^2 b^11+a b^12+b^13+2 a^12 c-2 a^11 b c-18 a^10 b^2 c+6 a^9 b^3 c+47 a^8 b^4 c-5 a^7 b^5 c-54 a^6 b^6 c-a^5 b^7 c+30 a^4 b^8 c+3 a^3 b^9 c-8 a^2 b^10 c-a b^11 c+b^12 c-12 a^11 c^2-18 a^10 b c^2+24 a^9 b^2 c^2+54 a^8 b^3 c^2+6 a^7 b^4 c^2-48 a^6 b^5 c^2-42 a^5 b^6 c^2+30 a^3 b^8 c^2+18 a^2 b^9 c^2-6 a b^10 c^2-6 b^11 c^2-10 a^10 c^3+6 a^9 b c^3+54 a^8 b^2 c^3+32 a^7 b^3 c^3-34 a^6 b^4 c^3-47 a^5 b^5 c^3-28 a^4 b^6 c^3+6 a^3 b^7 c^3+24 a^2 b^8 c^3+3 a b^9 c^3-6 b^10 c^3+29 a^9 c^4+47 a^8 b c^4+6 a^7 b^2 c^4-34 a^6 b^3 c^4-28 a^5 b^4 c^4-18 a^4 b^5 c^4-22 a^3 b^6 c^4-10 a^2 b^7 c^4+15 a b^8 c^4+15 b^9 c^4+21 a^8 c^5-5 a^7 b c^5-48 a^6 b^2 c^5-47 a^5 b^3 c^5-18 a^4 b^4 c^5-18 a^3 b^5 c^5-18 a^2 b^6 c^5-2 a b^7 c^5+15 b^8 c^5-36 a^7 c^6-54 a^6 b c^6-42 a^5 b^2 c^6-28 a^4 b^3 c^6-22 a^3 b^4 c^6-18 a^2 b^5 c^6-20 a b^6 c^6-20 b^7 c^6-24 a^6 c^7-a^5 b c^7+6 a^3 b^3 c^7-10 a^2 b^4 c^7-2 a b^5 c^7-20 b^6 c^7+24 a^5 c^8+30 a^4 b c^8+30 a^3 b^2 c^8+24 a^2 b^3 c^8+15 a b^4 c^8+15 b^5 c^8+16 a^4 c^9+3 a^3 b c^9+18 a^2 b^2 c^9+3 a b^3 c^9+15 b^4 c^9-8 a^3 c^10-8 a^2 b c^10-6 a b^2 c^10-6 b^3 c^10-6 a^2 c^11-a b c^11-6 b^2 c^11+a c^12+b c^12+c^13::
             
            Best regards,
            Peter.

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