Difference between revisions of "Steiner line"
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− | == | + | A '''Steiner line''' is a line in [[geometry]]. |
+ | |||
+ | == Definition == | ||
+ | |||
[[File:Steiner and Simson lines.png|500px|right]] | [[File:Steiner and Simson lines.png|500px|right]] | ||
Let <math>ABC</math> be a triangle with orthocenter <math>H. P</math> is a point on the circumcircle <math>\Omega</math> of <math>\triangle ABC.</math> | Let <math>ABC</math> be a triangle with orthocenter <math>H. P</math> is a point on the circumcircle <math>\Omega</math> of <math>\triangle ABC.</math> | ||
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Let <math>P_A, P_B, </math> and <math>P_C</math> be the reflections of <math>P</math> in three lines which contains edges <math>BC, AC,</math> and <math>AB,</math> respectively. | Let <math>P_A, P_B, </math> and <math>P_C</math> be the reflections of <math>P</math> in three lines which contains edges <math>BC, AC,</math> and <math>AB,</math> respectively. | ||
− | + | <math>P_A, P_B, P_C,</math> and <math>H</math> are collinear. Respective line is known as the Steiner line of point <math>P</math> with respect to <math>\triangle ABC.</math> | |
− | + | === Proof === | |
Let <math>D, E,</math> and <math>F</math> be the foots of the perpendiculars dropped from <math>P</math> to lines <math>AB, AC,</math> and <math>BC,</math> respectively. | Let <math>D, E,</math> and <math>F</math> be the foots of the perpendiculars dropped from <math>P</math> to lines <math>AB, AC,</math> and <math>BC,</math> respectively. | ||
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According the Collins Claim <math>YZ</math> is <math>H-line,</math> therefore <math>H \in P_AP_B.</math> | According the Collins Claim <math>YZ</math> is <math>H-line,</math> therefore <math>H \in P_AP_B.</math> | ||
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==Collings Clime== | ==Collings Clime== | ||
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Prove that lines <math>l_A, l_B,</math> and <math>l_C</math> are concurrent and the point of concurrence lies on <math>\Omega.</math> | Prove that lines <math>l_A, l_B,</math> and <math>l_C</math> are concurrent and the point of concurrence lies on <math>\Omega.</math> | ||
− | + | === Proof === | |
Let <math>D, E,</math> and <math>F</math> be the crosspoints of <math>H–line</math> with <math>AB, AC,</math> and <math>BC,</math> respectively. | Let <math>D, E,</math> and <math>F</math> be the crosspoints of <math>H–line</math> with <math>AB, AC,</math> and <math>BC,</math> respectively. | ||
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Usually the point <math>P</math> is called the anti-Steiner point of the <math>H-line</math> with respect to <math>\triangle ABC.</math> | Usually the point <math>P</math> is called the anti-Steiner point of the <math>H-line</math> with respect to <math>\triangle ABC.</math> | ||
− | |||
− | ==Ortholine== | + | == Ortholine == |
[[File:Ortholine.png|400px|right]] | [[File:Ortholine.png|400px|right]] | ||
Let four lines made four triangles of a complete quadrilateral. | Let four lines made four triangles of a complete quadrilateral. | ||
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Prove that points <math>H, H_A, H_B,</math> and <math>H_C</math> are collinear. | Prove that points <math>H, H_A, H_B,</math> and <math>H_C</math> are collinear. | ||
− | + | === Proof === | |
Let <math>M</math> be Miquel point of a complete quadrilateral. | Let <math>M</math> be Miquel point of a complete quadrilateral. | ||
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Using homothety centered at <math>M</math> with ratio <math>2</math> we get <math>4</math> coinciding Stainer lines which contain points <math>H, H_A, H_B,</math> and <math>H_C</math>. | Using homothety centered at <math>M</math> with ratio <math>2</math> we get <math>4</math> coinciding Stainer lines which contain points <math>H, H_A, H_B,</math> and <math>H_C</math>. | ||
− | + | === Proof 2 === | |
− | |||
− | |||
[[File:Steiner 2.png|400px|right]] | [[File:Steiner 2.png|400px|right]] | ||
<math>AH_A \perp DE, CH_C \perp EF \implies AH_A ||CH_C,</math> | <math>AH_A \perp DE, CH_C \perp EF \implies AH_A ||CH_C,</math> | ||
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The segments <math>EF</math> and <math>ED</math> are corresponding segments in similar triangles. | The segments <math>EF</math> and <math>ED</math> are corresponding segments in similar triangles. | ||
Therefore <math>\angle CED = \angle QEF \implies D, E,</math> and <math>F</math> are collinear. | Therefore <math>\angle CED = \angle QEF \implies D, E,</math> and <math>F</math> are collinear. | ||
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==Shatunov-Tokarev line== | ==Shatunov-Tokarev line== | ||
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Of course, it is parallel to Simson line. | Of course, it is parallel to Simson line. | ||
*[[Complete Quadrilateral]] | *[[Complete Quadrilateral]] | ||
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==Shatunov-Tokarev concurrent lines== | ==Shatunov-Tokarev concurrent lines== | ||
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We made simple calculations and get <math>XA''^2 + XC''^2 = XB''^2 + XD''^2,</math> therefore point <math>X</math> lies on <math>P''Q''</math> as desired. | We made simple calculations and get <math>XA''^2 + XC''^2 = XB''^2 + XD''^2,</math> therefore point <math>X</math> lies on <math>P''Q''</math> as desired. | ||
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==Shatunov point== | ==Shatunov point== | ||
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We made simple calculations and get <math>|XA''|^2 + |XC''|^2 = |XB''|^2 + |XD''|^2,</math> therefore point <math>X</math> lies on <math>P''Q''</math> as desired. | We made simple calculations and get <math>|XA''|^2 + |XC''|^2 = |XB''|^2 + |XD''|^2,</math> therefore point <math>X</math> lies on <math>P''Q''</math> as desired. | ||
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==Shatunov chain== | ==Shatunov chain== | ||
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The claim follows from the fact that there are <math>4^2 = 16</math> combinations of quadrilateral vertices, and these 16 quadrilaterals are divided into pairs whose points of intersection with the line <math>PQ</math> coincide. | The claim follows from the fact that there are <math>4^2 = 16</math> combinations of quadrilateral vertices, and these 16 quadrilaterals are divided into pairs whose points of intersection with the line <math>PQ</math> coincide. | ||
− | ' | + | == See Also == |
+ | |||
+ | *[[Miquel's point]] | ||
+ | *[[Simson line]] | ||
+ | *[[Complete Quadrilateral]] | ||
+ | |||
+ | {{stub}} |
Latest revision as of 18:52, 18 April 2025
A Steiner line is a line in geometry.
Contents
[hide]Definition
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.
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
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
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.
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.
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.
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.
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.
See Also
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