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Isosceles?

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  • xpolakis
    Let ABC be a triangle and A B C the cevian triangle of I. If the Euler Lines of ABC, BB C, CC B are concurrent, then is ABC necessarily isosceles? Antreas
    Message 1 of 4 , Jul 6, 2009
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      Let ABC be a triangle and A'B'C' the cevian
      triangle of I.

      If the Euler Lines of ABC, BB'C, CC'B
      are concurrent, then is ABC necessarily isosceles?

      Antreas
    • Nikolaos Dergiades
      Dear Antreas, [APH] ... No. For example construct the triangle ABC with angles A=22.5 B=90 C= 67.5 The triangle B BC is isosceles (BB = BC) and the triangles
      Message 2 of 4 , Jul 6, 2009
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        Dear Antreas,

        [APH]
        > Let ABC be a triangle and A'B'C' the
        > cevian
        > triangle of I.
        >
        > If the Euler Lines of ABC, BB'C, CC'B
        > are concurrent, then is ABC necessarily isosceles?

        No.
        For example construct the triangle ABC with angles
        A=22.5
        B=90
        C= 67.5

        The triangle B'BC is isosceles (BB' = BC)
        and the triangles C'BC, ABC are
        right angled at B.
        Hence their Euler lines are concurrent at B,
        while ABC is not isosceles.

        It would be interesting to see if BC is fixed what is
        the locus of A in order the Euler lines are concurrent.
        Best regards
        Nikos





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      • xpolakis
        Dear Nikos [APH] ... [ND] ... So, if these Euler Lines are concurrent the triangle is either right angled or isosceles or ??? Note that we can replace Euler
        Message 3 of 4 , Jul 6, 2009
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          Dear Nikos

          [APH]
          > > Let ABC be a triangle and A'B'C' the
          > > cevian
          > > triangle of I.
          > >
          > > If the Euler Lines of ABC, BB'C, CC'B
          > > are concurrent, then is ABC necessarily isosceles?

          [ND]
          > No.
          > For example construct the triangle ABC with angles
          > A=22.5
          > B=90
          > C= 67.5
          >
          > The triangle B'BC is isosceles (BB' = BC)
          > and the triangles C'BC, ABC are
          > right angled at B.
          > Hence their Euler lines are concurrent at B,
          > while ABC is not isosceles.
          >
          > It would be interesting to see if BC is fixed what is
          > the locus of A in order the Euler lines are concurrent.

          So, if these Euler Lines are concurrent
          the triangle is either right angled or isosceles or ???

          Note that we can replace Euler lines with other lines
          (OK, OI etc) to have similar problems.

          APH
        • garciacapitan
          Dear Nikos, that was a really nice example. I get that the condition -a^4 b^3 + 2 a^2 b^5 - b^7 + a^5 b c - a^4 b^2 c - a^3 b^3 c + a^2 b^4 c - a^4 b c^2 - a^3
          Message 4 of 4 , Jul 6, 2009
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            Dear Nikos, that was a really nice example.

            I get that the condition

            -a^4 b^3 + 2 a^2 b^5 - b^7 + a^5 b c - a^4 b^2 c - a^3 b^3 c +
            a^2 b^4 c - a^4 b c^2 - a^3 b^2 c^2 - 2 a^2 b^3 c^2 + 2 b^5 c^2 -
            a^4 c^3 - a^3 b c^3 - 2 a^2 b^2 c^3 - b^4 c^3 + a^2 b c^4 -
            b^3 c^4 + 2 a^2 c^5 + 2 b^2 c^5 - c^7 = 0

            implies that Euler Lines of ABC, BB'C, CC'B are concurrent.

            Best regards,

            Francisco Javier.



            --- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:
            >
            > Dear Nikos
            >
            > [APH]
            > > > Let ABC be a triangle and A'B'C' the
            > > > cevian
            > > > triangle of I.
            > > >
            > > > If the Euler Lines of ABC, BB'C, CC'B
            > > > are concurrent, then is ABC necessarily isosceles?
            >
            > [ND]
            > > No.
            > > For example construct the triangle ABC with angles
            > > A=22.5
            > > B=90
            > > C= 67.5
            > >
            > > The triangle B'BC is isosceles (BB' = BC)
            > > and the triangles C'BC, ABC are
            > > right angled at B.
            > > Hence their Euler lines are concurrent at B,
            > > while ABC is not isosceles.
            > >
            > > It would be interesting to see if BC is fixed what is
            > > the locus of A in order the Euler lines are concurrent.
            >
            > So, if these Euler Lines are concurrent
            > the triangle is either right angled or isosceles or ???
            >
            > Note that we can replace Euler lines with other lines
            > (OK, OI etc) to have similar problems.
            >
            > APH
            >
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