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7859Re: CONCURRENT EULER LINES (was: Thebault point)

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  • Darij Grinberg
    Sep 11, 2003
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      Dear Antreas,

      In Hyacinthos message #7858, you wrote (partly):

      >> Let ABC be a triangle and AHa, BHb, CHc its altitudes.
      >>
      >> Special case #1:
      >>
      >> Let Ab, Ac be the reflections of A in Hc, Hb, resp.
      >> Let Bc, Ba be the reflections of B in Ha, Hc, resp.
      >> Let Ca, Cb be the reflections of C in Hb, Ha, resp.
      >>
      >> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
      ^^^^^^^^^^^^^^^^^^^
      I guess you mean AAbAc, BBcBa, CCaCb.
      The Euler lines of these triangles concur since the
      orthocenter H of triangle ABC is their common
      circumcenter.

      >> concur (at H of ABC).

      >> Problem: Are their Nine Point Circles concurrent?

      Yes. In fact, the nine-point circle of a triangle
      ABC is the reflection of the circumcircle of triangle
      AB'C' in the line B'C', where A', B', C' are the
      midpoints of the sides BC, CA, AB. This yields that
      the nine-point circle of triangle AAbAc is the
      reflection of the circumcircle of triangle AHbHc in
      the line HbHc. Similarly for the other two nine-point
      circles. Since the circles AHbHc, BHcHa, CHaHb concur
      at H, the three nine-point circles concur at the
      antigonal conjugate of H with respect to triangle
      HaHbHc.

      NOTE. Antigonal conjugates? If P is a point in the
      plane of a triangle ABC, then the reflections of the
      circles PBC, PCA, PAB in the sidelines BC, CA, AB
      have a common point. This point is called antigonal
      conjugate of P with respect to triangle ABC. But I
      see (Hyacinthos message #7826) that you know this,
      maybe with another name.

      The point where your nine-point circles concur is
      not in the ETC.

      >> Generalization:
      >>
      >> Let Pac, Pab be points on AB, AC resp. such that:
      >>
      >> APac / AHc = APab / AHb = t
      >>
      >> Similarly we define the points Pba, Pbc; Pcb, Pca
      >>
      >> The Euler Lines of the Triangles APabPac, BPbcPba,
      >> CPcaPcb are concurrent

      Yes, they are: the proof is simple. Moreover, if
      J = X(125) is the intersection of the Euler lines of
      triangles AHbHc, BHcHa, CHaHb, then the Euler lines
      of triangles APabPac, BPbcPba, CPcaPcb meet at the
      point W lying on the line HJ and dividing the segment
      HJ in the ratio

      HW 2-t
      -- = ---.
      WJ t-1

      Remark that the line HJ passes through X(74).

      Sincerely,
      Darij Grinberg
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