Steiner line
Contents
[hide]Steiner line
Let be a triangle with orthocenter is a point on the circumcircle of
Let and be the reflections of in three lines which contains edges and respectively.
Prove that and are collinear. Respective line is known as the Steiner line of point with respect to
Proof
Let and be the foots of the perpendiculars dropped from to lines and respectively.
WLOG, Steiner line cross at and at
The line is Simson line of point with respect of
is midpoint of segment homothety centered at with ratio sends point to a point
Similarly, this homothety sends point to a point , point to a point therefore this homothety send Simson line to line
Let is symmetric to
Quadrangle is cyclic
at point Similarly, line at
According the Collins Claim is therefore
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Collings Clime
Let triangle be the triangle with the orthocenter and circumcircle Denote any line containing point
Let and be the reflections of in the edges and respectively.
Prove that lines and are concurrent and the point of concurrence lies on
Proof
Let and be the crosspoints of with and respectively.
WLOG Let and be the points symmetric to with respect and respectively.
Therefore
Let be the crosspoint of and is cyclic
Similarly is cyclic the crosspoint of and is point
Usually the point is called the anti-Steiner point of the with respect to
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Ortholine
Let four lines made four triangles of a complete quadrilateral.
In the diagram these are
Let points and be the orthocenters of and respectively.
Prove that points and are collinear.
Proof
Let be Miquel point of a complete quadrilateral.
Line is the line which contain Simson lines of triangles.
Using homothety centered at with ratio we get coinciding Stainer lines which contain points and .
Proof 2
Points and are collinear.
According the Claim of parallel lines, points and are collinear.
Similarly points and are collinear as desired.
Claim of parallel lines
Let points and be collinear.
Let points be such that
Prove that points and are collinear.
Proof
Let
The segments and are corresponding segments in similar triangles. Therefore and are collinear.
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Shatunov-Tokarev line
Let the quadrilateral be given ( is not cyclic). Let points and be the midpoints of and respectively. Let points and be such points that
a) Prove that
b) Prove that the point lies on the line iff
Proof
a) Let be the circle centered at with radius Let be the circle centered at with radius is the median of
The power of the point with respect to the circle is
is the median of
The power of the point with respect to the circle is
Therefore lies on the radical axis of and Similarly, lies on these line. So the line is the radical axes of and
This line is perpendicular to Gauss line which is the line of centers of two circles and as desired.
b) is the median of
is the median of
lies on the radical axes of and
If the point satisfies the equation then locus of is the straight line (one can prove it using method of coordinates).
The points and are satisfies this equation, so this line contain these points as desired.
It is easy to understand that this line is parallel to Steiner line which is the radical axis of the circles centered at and with radii and respectively.
Of course, it is parallel to Simson line.
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Shatunov-Tokarev concurrent lines
Let the quadrilateral be given ( is not cyclic).
Let points and be on the line such that . Similarly
Let points and be the crosspoints of the bisectors
Similarly points and are the crosspoints of the bisectors
Prove that lines and are concurrent.
Proof
Segment is the median of the
Similarly
Let cross at point
We made simple calculations and get therefore point lies on as desired. vladimir.shelomovskii@gmail.com, vvsss
Shatunov point
Let the quadrilateral be given ( is not cyclic).
Let points and be on the lines and respectively such that
Let points and be on the segments and respectively such that where
Let points and be the crosspoints of the bisectors
Similarly points and are the crosspoints of the bisectors
Prove that lines and are concurrent.
Proof
Segment is the cevian to the side AA' of the
We use the Stewart's theorem and get: Similarly Let cross at point
We made simple calculations and get therefore point lies on as desired.
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Shatunov chain
Let the quadrilateral be given ( is not cyclic).
Let points and be on the line such that . Similarly
Let points and be the crosspoints of the bisectors and
We made quadrilateral using one point from the pare one point from the pare one point from the pare one point from the pare For each quadrilateral we find the crosspoints of the bisectors and and named these points as
Prove that lines cross line in 8 points and positions of these points are fixed for given (not depend from the length of
Proof
The claim follows from the fact that there are combinations of quadrilateral vertices, and these 16 quadrilaterals are divided into pairs whose points of intersection with the line coincide.
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