9153Re: Anti-Steiner Point. New proof?

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• Jan 29, 2004
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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