Difference between revisions of "Steiner line"
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Let <math>\angle ABC = \beta, \angle BFD = \varphi \implies \angle BDF = \beta – \varphi.</math> | Let <math>\angle ABC = \beta, \angle BFD = \varphi \implies \angle BDF = \beta – \varphi.</math> | ||
<cmath>P_CP_A||DF \implies \angle P_CYB = \beta – \varphi.</cmath> | <cmath>P_CP_A||DF \implies \angle P_CYB = \beta – \varphi.</cmath> | ||
− | <math>P</math> is | + | <math>P</math> is symmetric to <math>P_C \implies \angle PYD = \beta – \varphi.</math> |
− | + | Quadrangle <math>BDPF</math> is cyclic <math>\implies \angle BPD = \varphi \implies \angle BPY = 90^\circ – \angle BYP – \angle BPD = 90^\circ – \beta.</math> | |
<math>\angle BCH = \angle BPY \implies PY \cap CH</math> at point <math>H_C \in \Omega.</math> | <math>\angle BCH = \angle BPY \implies PY \cap CH</math> at point <math>H_C \in \Omega.</math> | ||
Line 56: | Line 56: | ||
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> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Ortholine== | ||
+ | [[File:Ortholine.png|400px|right]] | ||
+ | Let four lines made four triangles of a complete quadrilateral. | ||
+ | |||
+ | In the diagram these are <math>\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.</math> | ||
+ | |||
+ | Let points <math>H, H_A, H_B,</math> and <math>H_C</math> be the orthocenters of <math>\triangle ABC, \triangle ADE, \triangle BDF,</math> and <math>\triangle CEF,</math> respectively. | ||
+ | |||
+ | Prove that points <math>H, H_A, H_B,</math> and <math>H_C</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>M</math> be Miquel point of a complete quadrilateral. | ||
+ | |||
+ | Line <math>KLMN</math> is the line which contain <math>4</math> Simson lines of <math>4</math> triangles. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | *[[Miquel's point]] | ||
+ | *[[Simson line]] | ||
+ | |||
+ | <i><b>Proof 2</b></i> | ||
+ | [[File:Steiner 2.png|400px|right]] | ||
+ | <math>AH_A \perp DE, CH_C \perp EF \implies AH_A ||CH_C,</math> | ||
+ | |||
+ | <math>AH \perp BC, EH_C \perp CF \implies AH ||EH_C,</math> | ||
+ | |||
+ | <math>EH_A \perp AD, CH \perp AB \implies EH_A ||CH.</math> | ||
+ | |||
+ | Points <math>A, E,</math> and <math>C</math> are collinear. | ||
+ | |||
+ | According the <i><b>Claim of parallel lines,</b></i> points <math>H, H_A,</math> and <math>H_C</math> are collinear. | ||
+ | |||
+ | Similarly points <math>H, H_B,</math> and <math>H_C</math> are collinear as desired. | ||
+ | |||
+ | <i><b>Claim of parallel lines</b></i> | ||
+ | |||
+ | Let points <math>A, B,</math> and <math>C</math> be collinear. | ||
+ | |||
+ | Let points <math>D, E, F</math> be such that <math>AF||CD, BF||CE, AE||BD.</math> | ||
+ | |||
+ | Prove that points <math>D, E,</math> and <math>F</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | [[File:Pras 1 12.png|400px|right]] | ||
+ | Let <math>P = AE \cap CD, Q = AF \cap CE.</math> | ||
+ | |||
+ | <cmath>\angle CEP = \angle QEA, AQ||CP \implies \angle QAE = \angle CPE \implies</cmath> | ||
+ | <cmath>\triangle AEQ \sim \triangle PEC.</cmath> | ||
+ | |||
+ | <cmath>AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},</cmath> | ||
+ | |||
+ | <cmath>CQ||BF \implies \frac {AF}{QF} = \frac {AB}{BC} = \frac {PD}{CD}.</cmath> | ||
+ | |||
+ | 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. | ||
+ | *[[Complete Quadrilateral]] | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Shatunov-Tokarev line== | ||
+ | [[File:Shatunov line.png|500px|right]] | ||
+ | Let the quadrilateral <math>ABCD</math> be given (<math>ABCD</math> is not cyclic). Let points <math>E</math> and <math>F</math> be the midpoints of <math>BD</math> and <math>AC,</math> respectively. Let points <math>P</math> and <math>Q</math> be such points that <math>PA = PB, PC = PD, QA = QD, QB = QC.</math> | ||
+ | |||
+ | a) Prove that <math>PQ \perp EF.</math> | ||
+ | |||
+ | b) Prove that the point <math>X</math> lies on the line <math>PQ</math> iff <math>XA^2 + XC^2 = XB^2 + XD^2.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) Let <math>\omega</math> be the circle centered at <math>F</math> with radius <math>BE.</math> Let <math>\Omega</math> be the circle centered at <math>E</math> with radius <math>AF.</math> | ||
+ | <math>PE</math> is the median of <math>\triangle PBD \implies PE^2 = \frac {PB^2 + PD^2}{2} – BE^2.</math> | ||
+ | |||
+ | The power of the point <math>P</math> with respect to the circle <math>\Omega</math> is <math>Pow_{\Omega}(P) = PE^2 – AF^2 = \frac {PB^2 + PD^2}{2} – BE^2 – AF^2.</math> | ||
+ | |||
+ | <math>PF</math> is the median of <math>\triangle PAC \implies PF^2 = \frac {PA^2 + PC^2}{2} – AF^2.</math> | ||
+ | |||
+ | The power of the point <math>P</math> with respect to the circle <math>\omega</math> is <math>Pow_{\omega}(P) = PF^2 – BE^2 = \frac {PA^2 + PC^2}{2} – BE^2 – AF^2 = \frac {PB^2 + PD^2}{2} – BE^2 – AF^2 = Pow_{\Omega}(P).</math> | ||
+ | |||
+ | Therefore <math>P</math> lies on the radical axis of <math>\Omega</math> and <math>\omega.</math> Similarly, <math>Q</math> lies on these line. | ||
+ | So the line <math>PQ</math> is the radical axes of <math>\Omega</math> and <math>\omega.</math> | ||
+ | |||
+ | This line is perpendicular to Gauss line <math>EF</math> which is the line of centers of two circles <math>\Omega</math> and <math>\omega</math> as desired. | ||
+ | |||
+ | b) <math>XE</math> is the median of <math>\triangle XBD \implies XE^2 = \frac {XB^2 + XD^2}{2} – BE^2.</math> | ||
+ | |||
+ | <math>XF</math> is the median of <math>\triangle XAC \implies XF^2 = \frac {XA^2 + XC^2}{2} – AF^2.</math> | ||
+ | |||
+ | <math>X</math> lies on the radical axes of <math>\Omega</math> and <math>\omega \implies XE^2 – XF^2 = AF^2 – BE^2 \implies</math> | ||
+ | <cmath>\frac {XB^2 + XD^2}{2} – BE^2 – ( \frac {XA^2 + XC^2}{2} – AF^2) = AF^2 – BE^2 \implies XB^2 + XD^2 = XA^2 + XC^2.</cmath> | ||
+ | |||
+ | If the point <math>X</math> satisfies the equation <math>XB^2 + XD^2 = XA^2 + XC^2</math> then locus of <math>X</math> is the straight line (one can prove it using method of coordinates). | ||
+ | |||
+ | The points <math>P</math> and <math>Q</math> 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 <math>E</math> and <math>F</math> with radii <math>BE</math> and <math>AF,</math> respectively. | ||
+ | |||
+ | Of course, it is parallel to Simson line. | ||
+ | *[[Complete Quadrilateral]] | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Shatunov-Tokarev concurrent lines== | ||
+ | [[File:Shatunov 3 concurrent lines.png|500px|right]] | ||
+ | Let the quadrilateral <math>ABCD</math> be given (<math>ABCD</math> is not cyclic). | ||
+ | |||
+ | Let points <math>A'</math> and <math>A''</math> be on the line <math>AB</math> such that <math>AA' = AA''</math>. Similarly | ||
+ | <cmath>B' \in BC, B'' \in BC, C' \in CD, C'' \in CD,</cmath> | ||
+ | <cmath>D' \in AD, D'' \in AD,</cmath> | ||
+ | <cmath>BB' = BB'' = CC' = CC'' = DD' = DD'' = AA'.</cmath> | ||
+ | |||
+ | Let points <math>Q, Q',</math> and <math>Q''</math> be the crosspoints of the bisectors <math>AD \cap BC, A'D' \cap B'C', A''D'' \cap B''C''.</math> | ||
+ | |||
+ | Similarly points <math>P, P',</math> and <math>P''</math> are the crosspoints of the bisectors <math>AB \cap CD, A'B' \cap C'D', A''B'' \cap C''D''.</math> | ||
+ | |||
+ | Prove that lines <math>PQ, P'Q',</math> and <math>P''Q''</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Segment <math>XA</math> is the median of the <math>\triangle XA'A'' \implies 2(XA^2 + AA'^2) = XA'^2 + XA''^2.</math> | ||
+ | |||
+ | Similarly <math>2(XB^2 + BB'^2) = XB'^2 + XB''^2, 2(XC^2 + CC'^2) = XC'^2 + XC''^2, 2(XD^2 + DD'^2) = XD'^2 + XD''^2.</math> | ||
+ | |||
+ | Let <math>PQ</math> cross <math>P'Q'</math> at point <math>X \implies XA^2 + XC^2 = XB^2 + XD^2, XA'^2 + XC'^2 = XB'^2 + XD'^2.</math> | ||
+ | |||
+ | 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. | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Shatunov point== | ||
+ | [[File:Shatunov point.png|450px|right]] | ||
+ | Let the quadrilateral <math>ABCD</math> be given (<math>ABCD</math> is not cyclic). | ||
+ | |||
+ | Let points <math>A', B', C',</math> and <math>D'</math> be on the lines <math>AD, AB, BC,</math> and <math>CD,</math> respectively such that <math>|AA'| = |BB'| = |CC'| = |DD'| = d.</math> | ||
+ | |||
+ | Let points <math>A'', B'', C'',</math> and <math>D''</math> be on the segments <math>AA', BB', CC',</math> and <math>DD',</math> respectively such that <math>\frac {|AA''|}{|A'A''|} = \frac {|BB''|}{|B'B''|} = \frac {|CC''|}{|C'C''|} = \frac {|DD''|}{|D'D''|} = \frac {m}{n},</math> where <math>m + n = 1.</math> | ||
+ | |||
+ | Let points <math>Q, Q',</math> and <math>Q''</math> be the crosspoints of the bisectors <math>AD \cap BC, A'D' \cap B'C', A''D'' \cap B''C''.</math> | ||
+ | |||
+ | Similarly points <math>P, P',</math> and <math>P''</math> are the crosspoints of the bisectors <math>AB \cap CD, A'B' \cap C'D', A''B'' \cap C''D''.</math> | ||
+ | |||
+ | Prove that lines <math>PQ, P'Q',</math> and <math>P''Q''</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Segment <math>XA''</math> is the cevian to the side AA' of the <math>\triangle XAA'.</math> | ||
+ | |||
+ | We use the Stewart's theorem and get: | ||
+ | <cmath>m \cdot|XA'|^2 + n \cdot |XA|^2) = |XA''|^2 + mn \cdot d^2.</cmath> | ||
+ | Similarly <math> m \cdot|XB'|^2 + n \cdot |XB|^2) = |XB''|^2 + mn \cdot d^2,</math> | ||
+ | <cmath>m \cdot|XC'|^2 + n \cdot |XC|^2) = |XC''|^2 + mn \cdot d^2,</cmath> | ||
+ | <cmath>m \cdot|XD'|^2 + n \cdot |XD|^2) = |XD''|^2 + mn \cdot d^2.</cmath> | ||
+ | Let <math>PQ</math> cross <math>P'Q'</math> at point <math>X \implies |XA|^2 + |XC|^2 = |XB|^2 + |XD|^2, |XA'|^2 + |XC'|^2 = |XB'|^2 + |XD'|^2.</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Shatunov chain== | ||
+ | [[File:Shatunov 8 points chain.png|450px|right]] | ||
+ | Let the quadrilateral <math>ABCD</math> be given (<math>ABCD</math> is not cyclic). | ||
+ | |||
+ | Let points <math>A'</math> and <math>A''</math> be on the line <math>AB</math> such that <math>|AA'| = |AA''|</math>. Similarly <math>B' \in BC, B'' \in BC, C' \in CD, C'' \in CD, D' \in AD, D'' \in AD,</math> | ||
+ | <math>|BB'| = |BB''| = |CC'| = |CC''| = |DD'| = |DD''| = |AA'|.</math> | ||
+ | |||
+ | Let points <math>Q</math> and <math>P</math> be the crosspoints of the bisectors <math>AD \cap BC</math> and <math>AB \cap CD.</math> | ||
+ | |||
+ | We made quadrilateral <math>KLMN</math> using one point from the pare <math>{A',A''},</math> one point from the pare <math>{B',B''},</math> one point from the pare <math>{C',C''},</math> one point from the pare <math>{D',D''}.</math> | ||
+ | For each quadrilateral we find the crosspoints of the bisectors <math>KL \cap MN</math> and <math>KN \cap LM</math> and named these points as <math>{Q_i,P_i}, i = 1..16.</math> | ||
+ | |||
+ | Prove that lines <math>P_iQ_i</math> cross line <math>PQ</math> in 8 points and positions of these points are fixed for given <math>ABCD</math> (not depend from the length of <math>AA'.)</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 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. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 12:07, 12 May 2024
Contents
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
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss