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RE: [EMHL] POINT ON THE EULER LINE ?

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  • Paul Yiu
    Dear Antreas, [APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC, ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections
    Message 1 of 6 , Feb 6, 2013
      Dear Antreas,

      [APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC, ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections of BC, CA, AB, in La, Lb, Lc, resp.

      The circumcenter Q of the triangle bounded by the lines (L1,L2,L3) is on the line L.

      True?

      *** No.

      Best regards
      Sincerely
      Paul
    • Antreas Hatzipolakis
      Dear Paul This time there are too many points P on the Euler line..... ! Let ABC be a triangle, P a point, A ,B ,C the reflections of A,B,C in the
      Message 2 of 6 , Feb 7, 2013
        Dear Paul

        This time there are too many points P on the Euler line..... !

        Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in the
        perpendicular
        bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian triangle of P
        wrt A'B'C'.

        Which is the locus of P such that ABC, A"B"C" are perspective?

        APH


        On Thu, Feb 7, 2013 at 1:50 AM, Paul Yiu <yiu@...> wrote:

        > **
        >
        >
        > Dear Antreas,
        >
        > [APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC,
        > ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections of
        > BC, CA, AB, in La, Lb, Lc, resp.
        >
        >
        > The circumcenter Q of the triangle bounded by the lines (L1,L2,L3) is on
        > the line L.
        >
        > True?
        >
        > *** No.
        >
        > Best regards
        > Sincerely
        > Paul
        >
        > _
        >


        [Non-text portions of this message have been removed]
      • Paul Yiu
        Dear Antreas, This is wonderful. [APH]: Let ABC be a triangle, P a point, A ,B ,C the reflections of A,B,C in the perpendicular bisectors of BC, CA, AB,
        Message 3 of 6 , Feb 7, 2013
          Dear Antreas,

          This is wonderful.

          [APH]: Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in the
          perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian triangle of P
          wrt A'B'C'. Which is the locus of P such that ABC, A"B"C" are perspective?

          A ' = (a^2:-(b^2-c^2):b^2-c^2),

          If P = (u:v:w), then
          A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) : -c^2(v+w)((b^2-c^2)u-a^2w))
          etc.

          A''B''C'' and ABC are perspective if and only if P lies on the Euler line.
          The perspector Q also lies on the Euler line.

          If OP P PH = t : 1-t, then
          OQ : QH = (1+t) : -8t\cos A\cos B\cos C

          Here are some examples:

          P Q
          ------------
          G X(25)
          O H
          H X(24)
          N X(3518)
          L O
          X(21) X (28)
          X(22) G
          X(23) X(468)
          X(186 X(403)

          Best regards
          Sincerely
          Paul
        • Antreas Hatzipolakis
          Dear Paul Note that the triangle A B C , where A ,B ,C are the reflections of A,B,C in the perpendicular bisectors, is the triangle which is the antipodal
          Message 4 of 6 , Feb 7, 2013
            Dear Paul

            Note that the triangle A'B'C', where A',B',C' are the reflections of A,B,C
            in the perpendicular bisectors,
            is the triangle which is the antipodal triangle of the circumcevian
            triangle of H.

            So we can replace H with an other fixed point T on the Euler line.

            That is:

            Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the
            circumcevian
            triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the
            antipodes of At, Bt,Ct).

            Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

            Which is the locus of P such that ABC, A"B"C" are perspective?

            I think that the locus is the Euler Line.

            We can furthermore generalize it by taking T as an arbitrary fixed point.
            The locus will be some line, I guess.

            Greetings

            Antreas

            On Thu, Feb 7, 2013 at 5:14 PM, Paul Yiu <yiu@...> wrote:

            > **
            >
            >
            > Dear Antreas,
            >
            > This is wonderful.
            >
            > [APH]: Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C
            > in the
            >
            > perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian
            > triangle of P
            > wrt A'B'C'. Which is the locus of P such that ABC, A"B"C" are perspective?
            >
            > A ' = (a^2:-(b^2-c^2):b^2-c^2),
            >
            > If P = (u:v:w), then
            > A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) :
            > -c^2(v+w)((b^2-c^2)u-a^2w))
            > etc.
            >
            > A''B''C'' and ABC are perspective if and only if P lies on the Euler line.
            > The perspector Q also lies on the Euler line.
            >
            > If OP P PH = t : 1-t, then
            > OQ : QH = (1+t) : -8t\cos A\cos B\cos C
            >
            > Here are some examples:
            >
            > P Q
            > ------------
            > G X(25)
            > O H
            > H X(24)
            > N X(3518)
            > L O
            > X(21) X (28)
            > X(22) G
            > X(23) X(468)
            > X(186 X(403)
            >
            > Best regards
            > Sincerely
            > Paul
            >
            > _
            >


            [Non-text portions of this message have been removed]
          • Paul Yiu
            Dear Antreas, You are right. [APH]: Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the circumcevian triangle of T, A B C the antipodal
            Message 5 of 6 , Feb 7, 2013
              Dear Antreas,

              You are right.

              [APH]: Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the
              circumcevian triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the
              antipodes of At, Bt,Ct).

              Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

              Which is the locus of P such that ABC, A"B"C" are perspective?

              *** The locus is the line OT, and the perspector Q lies on OT as well.

              If OP : PT = t: 1-t, then
              OQ : QT = R^2(1+t) : - (R^2-OT^2) t.

              Best regards
              Sincerely
              Paul
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