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Anti-Steiner Point. New proof?

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  • deocleciomota
    Dear Geometers, Given a line w passing through the orthocenter H of a triangle ABC. We denote by a´, b´, c´the reflections of w in the sidelines BC, CA, AB,
    Message 1 of 2 , Jan 29, 2004
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      Dear Geometers,

      Given a line w passing through the orthocenter H of a triangle ABC.
      We denote by a´, b´, c´the reflections of w in the sidelines BC, CA,
      AB, respectively. Then, the lines a´, b´, c´meet at one point, and
      this point lies on the circumcircle of triangle ABC.

      Solution:
      Let J a point on the circumcircle of the triangle ABC and the points
      P, Q, R the feet of the perpendiculars from J to the sides AB, BC, AC
      of the triangle ABC. The points P, Q and R are collinear (line t),
      (see, Advanced Euclidean Geometry, Roger A. Jonhson, paragraph 192,
      page 137). P is the center of the circle whose diameter is the
      segment JX. Q is the center of the circle whose diameter is the
      segment JZ. R is the center of the circle whose diameter is the
      segment JY. The centers of the three circles are collinear (line t)
      and the circles passes through point J. The three circles form a
      coaxal system of second type or elliptical (see Advanced Euclidean
      Geometry Roger A. Jonhson, paragraph 54 page 36) whose line of
      centers is the line t and the basic points are J and V. The radical
      axis of the coaxal system is perpendicular to line t at point U. The
      Simson line t of the point J bisects, at point D, the line joining
      the point J to the orthocenter H (see Advanced Euclidean Geometry
      Roger A. Jonhson paragraph 327 page 207). The circle of center D and
      radius DH belong to coaxal system. We have
      JX/JP=JV/JU=JH/JD=JZ/JQ=JY/JR=2. Thus the points X, V,H, Z and Y are
      collinear (line w). The central similarity (homothety) with center J
      and ratio 2 carries the line of centers of coaxal system (line t)
      onto line w (see I. M. Yaglom, Geometric Transformation II, page 10).
      The line t is parallel to line w. Under this transformation, the
      points X, Z and Y are the images of the points P, Q and R,
      respectively. Then, all lines passing through the orthocenter H of a
      triangle is parallel to a Simson line of a point on the circumcircle.
      The reflections of the line w in the sidelines AB, CA, BC carries the
      line w onto lines a´, b´, c´, but we have JP=PX; JQ=QZ; JR=RY. When
      the line w is reflected in sidelines AB, CA, BC the images X, Z and Y
      of the points P, Q and R coincide with the point J. Then the anti-
      Steiner point with respect to a triangle is the result of the
      reflections of the images (that are in line w) of the feet of a
      Simson line in sidelines of triangle.
      Is this a new proof?
    • Darij Grinberg
      Dear Deoclecio, ... Yes, of course, it is. But still, showing that every line through H is parallel to some Simson line, and that the Simson line of J bisects
      Message 2 of 2 , Jan 29, 2004
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        Dear Deoclecio,

        In Hyacinthos message #9151, you wrote:

        >> Given a line w passing through the orthocenter H of
        >> a triangle ABC. We denote by a´, b´, c´the
        >> reflections of w in the sidelines BC, CA, AB,
        >> respectively. Then, the lines a´, b´, c´meet at one
        >> point, and this point lies on the circumcircle of
        >> triangle ABC.
        >>
        >> Solution:
        >> Let J a point on the circumcircle of the triangle
        >> ABC and the points P, Q, R the feet of the
        >> perpendiculars from J to the sides AB, BC, AC of
        >> the triangle ABC. The points P, Q and R are
        >> collinear (line t), (see, Advanced Euclidean
        >> Geometry, Roger A. Jonhson, paragraph 192, page
        >> 137). P is the center of the circle whose diameter
        >> is the segment JX. Q is the center of the circle
        >> whose diameter is the segment JZ. R is the center
        >> of the circle whose diameter is the segment JY.
        >> The centers of the three circles are collinear
        >> (line t) and the circles passes through point J.
        >> The three circles form a coaxal system of second
        >> type or elliptical (see Advanced Euclidean
        >> Geometry Roger A. Jonhson, paragraph 54 page 36)
        >> whose line of centers is the line t and the basic
        >> points are J and V. The radical axis of the coaxal
        >> system is perpendicular to line t at point U.
        >> The Simson line t of the point J bisects, at point
        >> D, the line joining the point J to the orthocenter
        >> H (see Advanced Euclidean Geometry Roger A.
        >> Jonhson paragraph 327 page 207). The circle of
        >> center D and radius DH belong to coaxal system. We
        >> have JX/JP=JV/JU=JH/JD=JZ/JQ=JY/JR=2. Thus the
        >> points X, V,H, Z and Y are collinear (line w). The
        >> central similarity (homothety) with center J and
        >> ratio 2 carries the line of centers of coaxal
        >> system (line t) onto line w (see I. M. Yaglom,
        >> Geometric Transformation II, page 10). The line t
        >> is parallel to line w. Under this transformation,
        >> the points X, Z and Y are the images of the points
        >> P, Q and R, respectively. Then, all lines passing
        >> through the orthocenter H of a triangle is
        >> parallel to a Simson line of a point on the
        >> circumcircle.
        >> The reflections of the line w in the sidelines AB,
        >> CA, BC carries the line w onto lines a´, b´, c´,
        >> but we have JP=PX; JQ=QZ; JR=RY. When the line w
        >> is reflected in sidelines AB, CA, BC the images
        >> X, Z and Y of the points P, Q and R coincide with
        >> the point J. Then the anti-Steiner point with
        >> respect to a triangle is the result of the
        >> reflections of the images (that are in line w) of
        >> the feet of a Simson line in sidelines of triangle.
        >> Is this a new proof?

        Yes, of course, it is. But still, showing that every
        line through H is parallel to some Simson line, and
        that the Simson line of J bisects the segment JH is
        some work.

        Note that the line through X, Y, Z and H is called
        STEINER LINE of the point J with respect to triangle
        ABC.

        Sincerely,
        Darij Grinberg
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