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Re: Anti-Steiner Point. New proof?

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  • 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 1 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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