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[EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)

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  • Antreas
    Dear Randy I am not sure what you mean by confirming computationally. If you used a coordinate system or other general algebraic method it is a proof! Not nice
    Message 1 of 9 , Jan 4, 2013
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      Dear Randy

      I am not sure what you mean by confirming computationally.
      If you used a coordinate system or other general algebraic method
      it is a proof! Not nice as a synthetic one, but a proof!

      APH

      PS I had quoted not the right message!!
      The message that I should have quoted (with homothetic eq. triangles)
      is #21349 ANOTHER CONJECTURE ([EMHL] Re: X(5390))


      --- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
      >
      > Antreas,
      >
      > I was able to confirm this computationally.  Of course, not the same as a proof.
      >
      > Randy Hutson
      >
      >
      >
      >
      >
      > >________________________________
      > > From: Antreas
      > >To: Hyacinthos@yahoogroups.com
      > >Sent: Friday, January 4, 2013 12:56 PM
      > >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
      > >
      > >
      > > 
      > >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
      > >
      > >Conjecture:
      > >
      > >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
      > >(and of A"B'C', B"C'A', C"A'B')
      > >
      > >True?
      > >
      > >APH
      > >
      > >--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
      > >>
      > >> This topic is very rich.............!!!!
      > >>
      > >> Let A'B'C', A"B"C" be the 1st Morley and 1st Morley Adjunct triangles, resp.
      > >> [see definition http://mathworld.wolfram.com/FirstMorleyAdjunctTriangle.html ]
      > >>
      > >> Conjecture:
      > >>
      > >> The Euler Lines of A"B'C', B"C'A', C"A'B' are concurrent.
      > >>
      > >> If true, which point with respect A"B"C" is the point of concurrence?
      > >>
      > >> APH
    • Randy Hutson
      Dear Antreas, I constructed random homothetic equilateral triangles and the Euler lines of A B C , B C A , C A B (and of A B C , B C A , C A B ) are
      Message 2 of 9 , Jan 4, 2013
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        Dear Antreas,

        I constructed random homothetic equilateral triangles and the Euler lines of A'B"C", B'C"A", C'A"B" (and of A"B'C', B"C'A', C"A'B') are concurrent (the three intersections coincide to at least 15 decimal places).

        Randy





        >________________________________
        > From: Antreas <anopolis72@...>
        >To: Hyacinthos@yahoogroups.com
        >Sent: Friday, January 4, 2013 5:41 PM
        >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
        >
        >

        >Dear Randy
        >
        >I am not sure what you mean by confirming computationally.
        >If you used a coordinate system or other general algebraic method
        >it is a proof! Not nice as a synthetic one, but a proof!
        >
        >APH
        >
        >PS I had quoted not the right message!!
        >The message that I should have quoted (with homothetic eq. triangles)
        >is #21349 ANOTHER CONJECTURE ([EMHL] Re: X(5390))
        >
        >--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
        >>
        >> Antreas,
        >>
        >> I was able to confirm this computationally.  Of course, not the same as a proof.
        >>
        >> Randy Hutson
        >>
        >>
        >>
        >>
        >>
        >> >________________________________
        >> > From: Antreas
        >> >To: Hyacinthos@yahoogroups.com
        >> >Sent: Friday, January 4, 2013 12:56 PM
        >> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
        >> >
        >> >
        >> > 
        >> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
        >> >
        >> >Conjecture:
        >> >
        >> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
        >> >(and of A"B'C', B"C'A', C"A'B')
        >> >
        >> >True?
        >> >
        >> >APH
        >> >
        >> >--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
        >> >>
        >> >> This topic is very rich.............!!!!
        >> >>
        >> >> Let A'B'C', A"B"C" be the 1st Morley and 1st Morley Adjunct triangles, resp.
        >> >> [see definition http://mathworld.wolfram.com/FirstMorleyAdjunctTriangle.html ]
        >> >>
        >> >> Conjecture:
        >> >>
        >> >> The Euler Lines of A"B'C', B"C'A', C"A'B' are concurrent.
        >> >>
        >> >> If true, which point with respect A"B"C" is the point of concurrence?
        >> >>
        >> >> APH
        >
        >
        >
        >
        >

        [Non-text portions of this message have been removed]
      • Francisco Javier
        Dear Randy, in this case the concurrency can be completely proven by using computer algebra and suitable cartesian coordintates.
        Message 3 of 9 , Jan 5, 2013
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          Dear Randy, in this case the concurrency can be completely proven by using computer algebra and suitable cartesian coordintates.

          --- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
          >
          > Dear Antreas,
          >
          > I constructed random homothetic equilateral triangles and the Euler lines of A'B"C", B'C"A", C'A"B" (and of A"B'C', B"C'A', C"A'B') are concurrent (the three intersections coincide to at least 15 decimal places).
          >
          > Randy
          >
        • Antreas
          Dear Randy Note that the conjecture is not true if one of the six triangles A B C ,...., A B C is degenerated (if A ,B ,C etc are collinear). Figures here
          Message 4 of 9 , Jan 6, 2013
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            Dear Randy

            Note that the conjecture is not true if one of the six triangles A"B'C',...., A'B"C" is degenerated (if A",B',C' etc are collinear).

            Figures here
            http://tech.groups.yahoo.com/group/Hyacinthos/message/21357

            APH

            --- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
            >
            > Antreas,
            >
            > I was able to confirm this computationally.  Of course, not the same as a proof.
            >
            > Randy Hutson
            >
            >
            >
            >
            >
            > >________________________________
            > > From: Antreas
            > >To: Hyacinthos@yahoogroups.com
            > >Sent: Friday, January 4, 2013 12:56 PM
            > >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
            > >
            > >
            > > 
            > >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
            > >
            > >Conjecture:
            > >
            > >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
            > >(and of A"B'C', B"C'A', C"A'B')
            > >
            > >True?
            > >
            > >APH
          • Randy Hutson
            Dear Antreas, The conjecture still holds if we extend the definition of Euler line to include degenerate triangles (e.g. A ,B ,C collinear) as: the line
            Message 5 of 9 , Jan 6, 2013
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              Dear Antreas,

              The conjecture still holds if we extend the definition of Euler line to include degenerate triangles (e.g. A",B',C' collinear) as: the line perpendicular to the line of vertices and passing through the centroid of the vertices.  This line meets the line at infinity at a point which is the 'circumcenter', 'orthocenter', and 'nine-point center' of the degenerate triangle.  In this case, the three Euler lines still concur.

              Best regards,
              Randy





              >________________________________
              > From: Antreas <anopolis72@...>
              >To: Hyacinthos@yahoogroups.com
              >Sent: Sunday, January 6, 2013 5:05 AM
              >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
              >
              >

              >Dear Randy
              >
              >Note that the conjecture is not true if one of the six triangles A"B'C',...., A'B"C" is degenerated (if A",B',C' etc are collinear).
              >
              >Figures here
              >http://tech.groups.yahoo.com/group/Hyacinthos/message/21357
              >
              >APH
              >
              >--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
              >>
              >> Antreas,
              >>
              >> I was able to confirm this computationally.  Of course, not the same as a proof.
              >>
              >> Randy Hutson
              >>
              >>
              >>
              >>
              >>
              >> >________________________________
              >> > From: Antreas
              >> >To: Hyacinthos@yahoogroups.com
              >> >Sent: Friday, January 4, 2013 12:56 PM
              >> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
              >> >
              >> >
              >> > 
              >> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
              >> >
              >> >Conjecture:
              >> >
              >> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
              >> >(and of A"B'C', B"C'A', C"A'B')
              >> >
              >> >True?
              >> >
              >> >APH
              >
              >
              >
              >
              >

              [Non-text portions of this message have been removed]
            • Nikolaos Dergiades
              Dear Antreas, If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics and A B C is the homothetic of ABC in the homothety PA = t.PA then the Euler
              Message 6 of 9 , Jan 6, 2013
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                Dear Antreas,
                If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics
                and A'B'C' is the homothetic of ABC in the homothety PA' = t.PA
                then the Euler lines of A'BC, B'CA, C'AB are concurrent at
                Q = {2 p^3 q^2 + p^2 q^3 + 5 p^3 q r + 8 p^2 q^2 r + 2 p q^3 r +
                2 p^3 r^2 + 8 p^2 q r^2 + 5 p q^2 r^2 + p^2 r^3 + 2 p q r^3 +
                2 p^4 q t + 2 p^2 q^3 t + p q^4 t + 2 p^4 r t + 4 p^3 q r t +
                4 p^2 q^2 r t + 4 p q^3 r t + 4 p^2 q r^2 t + 4 p q^2 r^2 t +
                q^3 r^2 t + 2 p^2 r^3 t + 4 p q r^3 t + q^2 r^3 t + p r^4 t -
                p^4 q t^2 + 2 p^3 q^2 t^2 - p^4 r t^2 - 2 p^3 q r t^2 +
                p q^3 r t^2 + q^4 r t^2 + 2 p^3 r^2 t^2 + 3 p q^2 r^2 t^2 +
                q^3 r^2 t^2 + p q r^3 t^2 + q^2 r^3 t^2 + q r^4 t^2 - p^3 q^2 t^3 -
                p^3 r^2 t^3 + q^3 r^2 t^3 + q^2 r^3 t^3,
                p^3 q^2 + 2 p^2 q^3 + 2 p^3 q r + 8 p^2 q^2 r + 5 p q^3 r +
                5 p^2 q r^2 + 8 p q^2 r^2 + 2 q^3 r^2 + 2 p q r^3 + q^2 r^3 +
                p^4 q t + 2 p^3 q^2 t + 2 p q^4 t + 4 p^3 q r t + 4 p^2 q^2 r t +
                4 p q^3 r t + 2 q^4 r t + p^3 r^2 t + 4 p^2 q r^2 t +
                4 p q^2 r^2 t + p^2 r^3 t + 4 p q r^3 t + 2 q^2 r^3 t + q r^4 t +
                2 p^2 q^3 t^2 - p q^4 t^2 + p^4 r t^2 + p^3 q r t^2 -
                2 p q^3 r t^2 - q^4 r t^2 + p^3 r^2 t^2 + 3 p^2 q r^2 t^2 +
                2 q^3 r^2 t^2 + p^2 r^3 t^2 + p q r^3 t^2 + p r^4 t^2 -
                p^2 q^3 t^3 + p^3 r^2 t^3 - q^3 r^2 t^3 + p^2 r^3 t^3,
                2 p^3 q r + 5 p^2 q^2 r + 2 p q^3 r + p^3 r^2 + 8 p^2 q r^2 +
                8 p q^2 r^2 + q^3 r^2 + 2 p^2 r^3 + 5 p q r^3 + 2 q^2 r^3 +
                p^3 q^2 t + p^2 q^3 t + p^4 r t + 4 p^3 q r t + 4 p^2 q^2 r t +
                4 p q^3 r t + q^4 r t + 2 p^3 r^2 t + 4 p^2 q r^2 t +
                4 p q^2 r^2 t + 2 q^3 r^2 t + 4 p q r^3 t + 2 p r^4 t + 2 q r^4 t +
                p^4 q t^2 + p^3 q^2 t^2 + p^2 q^3 t^2 + p q^4 t^2 + p^3 q r t^2 +
                3 p^2 q^2 r t^2 + p q^3 r t^2 + 2 p^2 r^3 t^2 - 2 p q r^3 t^2 +
                2 q^2 r^3 t^2 - p r^4 t^2 - q r^4 t^2 + p^3 q^2 t^3 + p^2 q^3 t^3 -
                p^2 r^3 t^3 - q^2 r^3 t^3}

                The locus of Q is

                -p^2 (p - q) q^2 (p - r) (q - r) r^2 (p + q + r)^11 (p q + p r +
                q r) x^3 (p^6 q^2 x^3 - 5 p^4 q^4 x^3 + 4 p^2 q^6 x^3 -
                2 p^6 q r x^3 + 5 p^4 q^3 r x^3 + 2 p^3 q^4 r x^3 + 4 p q^6 r x^3 +
                p^6 r^2 x^3 - 2 p^3 q^3 r^2 x^3 + q^6 r^2 x^3 + 5 p^4 q r^3 x^3 -
                2 p^3 q^2 r^3 x^3 - 8 p^2 q^3 r^3 x^3 - 4 p q^4 r^3 x^3 -
                5 p^4 r^4 x^3 + 2 p^3 q r^4 x^3 - 4 p q^3 r^4 x^3 -
                2 q^4 r^4 x^3 + 4 p^2 r^6 x^3 + 4 p q r^6 x^3 + q^2 r^6 x^3 -
                3 p^6 q^2 x^2 y + 9 p^5 q^3 x^2 y + 12 p^4 q^4 x^2 y -
                18 p^3 q^5 x^2 y - 9 p^5 q^2 r x^2 y + 3 p^4 q^3 r x^2 y -
                9 p^2 q^5 r x^2 y + 6 p q^6 r x^2 y + 3 p^6 r^2 x^2 y -
                6 p^3 q^3 r^2 x^2 y + 3 q^6 r^2 x^2 y - 12 p^4 q r^3 x^2 y +
                24 p^3 q^2 r^3 x^2 y - 9 p^2 q^3 r^3 x^2 y + 6 p q^4 r^3 x^2 y -
                3 p^4 r^4 x^2 y + 18 p^2 q^2 r^4 x^2 y - 6 p q^3 r^4 x^2 y -
                6 p q r^6 x^2 y - 3 q^2 r^6 x^2 y - 18 p^5 q^3 x y^2 +
                12 p^4 q^4 x y^2 + 9 p^3 q^5 x y^2 - 3 p^2 q^6 x y^2 +
                6 p^6 q r x y^2 - 9 p^5 q^2 r x y^2 + 3 p^3 q^4 r x y^2 -
                9 p^2 q^5 r x y^2 + 3 p^6 r^2 x y^2 - 6 p^3 q^3 r^2 x y^2 +
                3 q^6 r^2 x y^2 + 6 p^4 q r^3 x y^2 - 9 p^3 q^2 r^3 x y^2 +
                24 p^2 q^3 r^3 x y^2 - 12 p q^4 r^3 x y^2 - 6 p^3 q r^4 x y^2 +
                18 p^2 q^2 r^4 x y^2 - 3 q^4 r^4 x y^2 - 3 p^2 r^6 x y^2 -
                6 p q r^6 x y^2 + 4 p^6 q^2 y^3 - 5 p^4 q^4 y^3 + p^2 q^6 y^3 +
                4 p^6 q r y^3 + 2 p^4 q^3 r y^3 + 5 p^3 q^4 r y^3 - 2 p q^6 r y^3 +
                p^6 r^2 y^3 - 2 p^3 q^3 r^2 y^3 + q^6 r^2 y^3 - 4 p^4 q r^3 y^3 -
                8 p^3 q^2 r^3 y^3 - 2 p^2 q^3 r^3 y^3 + 5 p q^4 r^3 y^3 -
                2 p^4 r^4 y^3 - 4 p^3 q r^4 y^3 + 2 p q^3 r^4 y^3 - 5 q^4 r^4 y^3 +
                p^2 r^6 y^3 + 4 p q r^6 y^3 + 4 q^2 r^6 y^3 + 3 p^6 q^2 x^2 z -
                3 p^4 q^4 x^2 z - 12 p^4 q^3 r x^2 z - 6 p q^6 r x^2 z -
                3 p^6 r^2 x^2 z - 9 p^5 q r^2 x^2 z + 24 p^3 q^3 r^2 x^2 z +
                18 p^2 q^4 r^2 x^2 z - 3 q^6 r^2 x^2 z + 9 p^5 r^3 x^2 z +
                3 p^4 q r^3 x^2 z - 6 p^3 q^2 r^3 x^2 z - 9 p^2 q^3 r^3 x^2 z -
                6 p q^4 r^3 x^2 z + 12 p^4 r^4 x^2 z + 6 p q^3 r^4 x^2 z -
                18 p^3 r^5 x^2 z - 9 p^2 q r^5 x^2 z + 6 p q r^6 x^2 z +
                3 q^2 r^6 x^2 z - 6 p^6 q^2 x y z + 9 p^5 q^3 x y z -
                6 p^4 q^4 x y z + 9 p^3 q^5 x y z - 6 p^2 q^6 x y z -
                6 p^6 q r x y z + 18 p^5 q^2 r x y z + 6 p^4 q^3 r x y z +
                6 p^3 q^4 r x y z + 18 p^2 q^5 r x y z - 6 p q^6 r x y z -
                6 p^6 r^2 x y z + 18 p^5 q r^2 x y z - 36 p^4 q^2 r^2 x y z -
                6 p^3 q^3 r^2 x y z - 36 p^2 q^4 r^2 x y z + 18 p q^5 r^2 x y z -
                6 q^6 r^2 x y z + 9 p^5 r^3 x y z + 6 p^4 q r^3 x y z -
                6 p^3 q^2 r^3 x y z - 6 p^2 q^3 r^3 x y z + 6 p q^4 r^3 x y z +
                9 q^5 r^3 x y z - 6 p^4 r^4 x y z + 6 p^3 q r^4 x y z -
                36 p^2 q^2 r^4 x y z + 6 p q^3 r^4 x y z - 6 q^4 r^4 x y z +
                9 p^3 r^5 x y z + 18 p^2 q r^5 x y z + 18 p q^2 r^5 x y z +
                9 q^3 r^5 x y z - 6 p^2 r^6 x y z - 6 p q r^6 x y z -
                6 q^2 r^6 x y z - 3 p^4 q^4 y^2 z + 3 p^2 q^6 y^2 z -
                6 p^6 q r y^2 z - 12 p^3 q^4 r y^2 z - 3 p^6 r^2 y^2 z +
                18 p^4 q^2 r^2 y^2 z + 24 p^3 q^3 r^2 y^2 z - 9 p q^5 r^2 y^2 z -
                3 q^6 r^2 y^2 z - 6 p^4 q r^3 y^2 z - 9 p^3 q^2 r^3 y^2 z -
                6 p^2 q^3 r^3 y^2 z + 3 p q^4 r^3 y^2 z + 9 q^5 r^3 y^2 z +
                6 p^3 q r^4 y^2 z + 12 q^4 r^4 y^2 z - 9 p q^2 r^5 y^2 z -
                18 q^3 r^5 y^2 z + 3 p^2 r^6 y^2 z + 6 p q r^6 y^2 z +
                3 p^6 q^2 x z^2 - 3 p^2 q^6 x z^2 + 6 p^6 q r x z^2 +
                6 p^4 q^3 r x z^2 - 6 p^3 q^4 r x z^2 - 6 p q^6 r x z^2 -
                9 p^5 q r^2 x z^2 - 9 p^3 q^3 r^2 x z^2 + 18 p^2 q^4 r^2 x z^2 -
                18 p^5 r^3 x z^2 - 6 p^3 q^2 r^3 x z^2 + 24 p^2 q^3 r^3 x z^2 +
                12 p^4 r^4 x z^2 + 3 p^3 q r^4 x z^2 - 12 p q^3 r^4 x z^2 -
                3 q^4 r^4 x z^2 + 9 p^3 r^5 x z^2 - 9 p^2 q r^5 x z^2 -
                3 p^2 r^6 x z^2 + 3 q^2 r^6 x z^2 - 3 p^6 q^2 y z^2 +
                3 p^2 q^6 y z^2 - 6 p^6 q r y z^2 - 6 p^4 q^3 r y z^2 +
                6 p^3 q^4 r y z^2 + 6 p q^6 r y z^2 + 18 p^4 q^2 r^2 y z^2 -
                9 p^3 q^3 r^2 y z^2 - 9 p q^5 r^2 y z^2 + 24 p^3 q^2 r^3 y z^2 -
                6 p^2 q^3 r^3 y z^2 - 18 q^5 r^3 y z^2 - 3 p^4 r^4 y z^2 -
                12 p^3 q r^4 y z^2 + 3 p q^3 r^4 y z^2 + 12 q^4 r^4 y z^2 -
                9 p q^2 r^5 y z^2 + 9 q^3 r^5 y z^2 + 3 p^2 r^6 y z^2 -
                3 q^2 r^6 y z^2 + p^6 q^2 z^3 - 2 p^4 q^4 z^3 + p^2 q^6 z^3 +
                4 p^6 q r z^3 - 4 p^4 q^3 r z^3 - 4 p^3 q^4 r z^3 +
                4 p q^6 r z^3 + 4 p^6 r^2 z^3 - 8 p^3 q^3 r^2 z^3 +
                4 q^6 r^2 z^3 + 2 p^4 q r^3 z^3 - 2 p^3 q^2 r^3 z^3 -
                2 p^2 q^3 r^3 z^3 + 2 p q^4 r^3 z^3 - 5 p^4 r^4 z^3 +
                5 p^3 q r^4 z^3 + 5 p q^3 r^4 z^3 - 5 q^4 r^4 z^3 + p^2 r^6 z^3 -
                2 p q r^6 z^3 + q^2 r^6 z^3)

                and the Euler lines of AB'C', BC'A', CA'B' are concurrent at

                {2 p^3 q^2 + 2 p^3 r^2 + 3 p q^2 r^2 + q^3 r^2 + q^2 r^3 + 2 p^4 q t -
                p^3 q^2 t + 3 p^2 q^3 t + 2 p^4 r t + 4 p^3 q r t +
                6 p^2 q^2 r t + 4 p q^3 r t + q^4 r t - p^3 r^2 t + 6 p^2 q r^2 t +
                q^3 r^2 t + 3 p^2 r^3 t + 4 p q r^3 t + q^2 r^3 t + q r^4 t -
                p^4 q t^2 + 3 p^3 q^2 t^2 - p^2 q^3 t^2 + p q^4 t^2 - p^4 r t^2 +
                7 p^3 q r t^2 + 7 p^2 q^2 r t^2 + p q^3 r t^2 + 3 p^3 r^2 t^2 +
                7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 + q^3 r^2 t^2 - p^2 r^3 t^2 +
                p q r^3 t^2 + q^2 r^3 t^2 + p r^4 t^2 - p^3 q^2 t^3 + p^2 q^3 t^3 -
                4 p^3 q r t^3 - p^2 q^2 r t^3 + 2 p q^3 r t^3 - p^3 r^2 t^3 -
                p^2 q r^2 t^3 + 2 p q^2 r^2 t^3 + p^2 r^3 t^3 + 2 p q r^3 t^3,
                2 p^2 q^3 + p^3 r^2 + 3 p^2 q r^2 + 2 q^3 r^2 + p^2 r^3 +
                3 p^3 q^2 t - p^2 q^3 t + 2 p q^4 t + p^4 r t + 4 p^3 q r t +
                6 p^2 q^2 r t + 4 p q^3 r t + 2 q^4 r t + p^3 r^2 t +
                6 p q^2 r^2 t - q^3 r^2 t + p^2 r^3 t + 4 p q r^3 t + 3 q^2 r^3 t +
                p r^4 t + p^4 q t^2 - p^3 q^2 t^2 + 3 p^2 q^3 t^2 - p q^4 t^2 +
                p^3 q r t^2 + 7 p^2 q^2 r t^2 + 7 p q^3 r t^2 - q^4 r t^2 +
                p^3 r^2 t^2 + 7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 + 3 q^3 r^2 t^2 +
                p^2 r^3 t^2 + p q r^3 t^2 - q^2 r^3 t^2 + q r^4 t^2 + p^3 q^2 t^3 -
                p^2 q^3 t^3 + 2 p^3 q r t^3 - p^2 q^2 r t^3 - 4 p q^3 r t^3 +
                2 p^2 q r^2 t^3 - p q^2 r^2 t^3 - q^3 r^2 t^3 + 2 p q r^3 t^3 +
                q^2 r^3 t^3,
                p^3 q^2 + p^2 q^3 + 3 p^2 q^2 r + 2 p^2 r^3 + 2 q^2 r^3 + p^4 q t +
                p^3 q^2 t + p^2 q^3 t + p q^4 t + 4 p^3 q r t + 4 p q^3 r t +
                3 p^3 r^2 t + 6 p^2 q r^2 t + 6 p q^2 r^2 t + 3 q^3 r^2 t -
                p^2 r^3 t + 4 p q r^3 t - q^2 r^3 t + 2 p r^4 t + 2 q r^4 t +
                p^3 q^2 t^2 + p^2 q^3 t^2 + p^4 r t^2 + p^3 q r t^2 +
                7 p^2 q^2 r t^2 + p q^3 r t^2 + q^4 r t^2 - p^3 r^2 t^2 +
                7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 - q^3 r^2 t^2 + 3 p^2 r^3 t^2 +
                7 p q r^3 t^2 + 3 q^2 r^3 t^2 - p r^4 t^2 - q r^4 t^2 +
                2 p^3 q r t^3 + 2 p^2 q^2 r t^3 + 2 p q^3 r t^3 + p^3 r^2 t^3 -
                p^2 q r^2 t^3 - p q^2 r^2 t^3 + q^3 r^2 t^3 - p^2 r^3 t^3 -
                4 p q r^3 t^3 - q^2 r^3 t^3}

                ND

                > Let A'B'C', A"B"C" be two homothetic
                > Equilateral triangles.
                >
                > Conjecture:
                >
                > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
                > (and of A"B'C', B"C'A', C"A'B')
                >
                > True?
                >
                > APH
              • Antreas
                Dear Nikos Thanks!! Using this Therorem we can find new Triangle centers, but with very complicated coordinates, I guess! A family of triangle equilateral
                Message 7 of 9 , Jan 7, 2013
                • 0 Attachment
                  Dear Nikos

                  Thanks!!

                  Using this Therorem we can find new Triangle centers, but with very
                  complicated coordinates, I guess!
                  A family of triangle equilateral triangles is Morley triangle and its homothetics. We have already discussed the case of the 1st Morley and Russel triangles. It will be interesting if the two centers we get are on the Euler line of the reference triangle (remarked by Frank Jackson, Hyacinthos #21354)

                  APH


                  --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades wrote:
                  >
                  > Dear Antreas,
                  > If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics
                  > and A'B'C' is the homothetic of ABC in the homothety PA' = t.PA
                  > then the Euler lines of A'BC, B'CA, C'AB are concurrent at
                  > Q = ......

                  >
                  > The locus of Q is ...........

                  >
                  > and the Euler lines of AB'C', BC'A', CA'B' are concurrent at......
                  >
                  > ND
                  >
                  > > Let A'B'C', A"B"C" be two homothetic
                  > > Equilateral triangles.
                  > >
                  > > Conjecture:
                  > >
                  > > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
                  > > (and of A"B'C', B"C'A', C"A'B')
                  > >
                  > > True?
                  > >
                  > > APH
                  >
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