Difference between revisions of "Steiner line"

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==Dhoner Line==
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A '''Steiner line''' is a line in [[geometry]].
 +
 
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== 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.  
  
Prove that <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>
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<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>
  
<i><b>Proof</b></i>
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=== 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>
*[[Simson line]]
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
==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>
  
<i><b>Proof</b></i>
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=== 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>
  
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
==Ortholine==
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== 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.
  
<i><b>Proof</b></i>
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=== 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>.
  
*[[Miquel's point]]
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=== Proof 2 ===
*[[Simson line]]
 
  
<i><b>Proof 2</b></i>
 
 
[[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.
*[[Complete Quadrilateral]]
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
==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]]
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
==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.  
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
==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.
 
   
 
   
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
==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.
  
'''vladimir.shelomovskii@gmail.com, vvsss'''
+
== See Also ==
 +
 
 +
*[[Miquel's point]]
 +
*[[Simson line]]
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*[[Complete Quadrilateral]]
 +
 
 +
{{stub}}

Latest revision as of 18:52, 18 April 2025

A Steiner line is a line in geometry.

Definition

Steiner and Simson lines.png

Let $ABC$ be a triangle with orthocenter $H. P$ is a point on the circumcircle $\Omega$ of $\triangle ABC.$

Let $P_A, P_B,$ and $P_C$ be the reflections of $P$ in three lines which contains edges $BC, AC,$ and $AB,$ respectively.

$P_A, P_B, P_C,$ and $H$ are collinear. Respective line is known as the Steiner line of point $P$ with respect to $\triangle ABC.$

Proof

Let $D, E,$ and $F$ be the foots of the perpendiculars dropped from $P$ to lines $AB, AC,$ and $BC,$ respectively.

WLOG, Steiner line cross $AB$ at $Y$ and $AC$ at $Z.$

The line $DEF$ is Simson line of point $P$ with respect of $\triangle ABC.$

$D$ is midpoint of segment $PP_C \implies$ homothety centered at $P$ with ratio $2$ sends point $D$ to a point $P_C.$

Similarly, this homothety sends point $E$ to a point $P_B$, point $F$ to a point $P_A,$ therefore this homothety send Simson line to line $P_AP_BP_C.$

Let $\angle ABC = \beta, \angle BFD = \varphi \implies \angle BDF = \beta – \varphi.$ \[P_CP_A||DF \implies \angle P_CYB = \beta – \varphi.\] $P$ is symmetric to $P_C \implies \angle PYD = \beta – \varphi.$

Quadrangle $BDPF$ is cyclic $\implies \angle BPD = \varphi \implies \angle BPY = 90^\circ – \angle BYP – \angle BPD = 90^\circ – \beta.$

$\angle BCH = \angle BPY \implies PY \cap CH$ at point $H_C \in \Omega.$ Similarly, line $BH \cap PZ$ at $H_B \in \Omega.$

According the Collins Claim $YZ$ is $H-line,$ therefore $H \in P_AP_B.$

Collings Clime

Steiner H line.png

Let triangle $ABC$ be the triangle with the orthocenter $H$ and circumcircle $\Omega.$ Denote $H–line$ any line containing point $H.$

Let $l_A, l_B,$ and $l_C$ be the reflections of $H-line$ in the edges $BC, AC,$ and $AB,$ respectively.

Prove that lines $l_A, l_B,$ and $l_C$ are concurrent and the point of concurrence lies on $\Omega.$

Proof

Let $D, E,$ and $F$ be the crosspoints of $H–line$ with $AB, AC,$ and $BC,$ respectively.

WLOG $D \in AB, E \in AC.$ Let $H_A, H_B,$ and $H_C$ be the points symmetric to $H$ with respect $BC, AC,$ and $AB,$ respectively.

Therefore $H_A \in l_A, H_B \in l_B, H_C \in l_C,$ \[AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies\] \[\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.\]

Let $P$ be the crosspoint of $l_B$ and $l_C  \implies BH_CH_BP$ is cyclic $\implies P \in \Omega.$

Similarly $\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP$ is cyclic $\implies P \in \Omega \implies$ the crosspoint of $l_B$ and $l_A$ is point $P.$

Usually the point $P$ is called the anti-Steiner point of the $H-line$ with respect to $\triangle ABC.$


Ortholine

Ortholine.png

Let four lines made four triangles of a complete quadrilateral.

In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Let points $H, H_A, H_B,$ and $H_C$ be the orthocenters of $\triangle ABC, \triangle ADE, \triangle BDF,$ and $\triangle CEF,$ respectively.

Prove that points $H, H_A, H_B,$ and $H_C$ are collinear.

Proof

Let $M$ be Miquel point of a complete quadrilateral.

Line $KLMN$ is the line which contain $4$ Simson lines of $4$ triangles.

Using homothety centered at $M$ with ratio $2$ we get $4$ coinciding Stainer lines which contain points $H, H_A, H_B,$ and $H_C$.

Proof 2

Steiner 2.png

$AH_A \perp DE, CH_C \perp EF \implies AH_A ||CH_C,$

$AH \perp BC, EH_C \perp CF \implies AH ||EH_C,$

$EH_A \perp AD, CH \perp AB \implies EH_A ||CH.$

Points $A, E,$ and $C$ are collinear.

According the Claim of parallel lines, points $H, H_A,$ and $H_C$ are collinear.

Similarly points $H, H_B,$ and $H_C$ are collinear as desired.

Claim of parallel lines

Let points $A, B,$ and $C$ be collinear.

Let points $D, E, F$ be such that $AF||CD, BF||CE, AE||BD.$

Prove that points $D, E,$ and $F$ are collinear.

Proof

Pras 1 12.png

Let $P = AE \cap CD, Q = AF \cap CE.$

\[\angle CEP = \angle QEA, AQ||CP \implies \angle QAE = \angle CPE \implies\] \[\triangle AEQ \sim \triangle PEC.\]

\[AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},\]

\[CQ||BF \implies \frac {AF}{QF} = \frac {AB}{BC} = \frac {PD}{CD}.\]

The segments $EF$ and $ED$ are corresponding segments in similar triangles. Therefore $\angle CED = \angle QEF \implies D, E,$ and $F$ are collinear.

Shatunov-Tokarev line

Shatunov line.png

Let the quadrilateral $ABCD$ be given ($ABCD$ is not cyclic). Let points $E$ and $F$ be the midpoints of $BD$ and $AC,$ respectively. Let points $P$ and $Q$ be such points that $PA = PB, PC = PD, QA = QD, QB = QC.$

a) Prove that $PQ \perp EF.$

b) Prove that the point $X$ lies on the line $PQ$ iff $XA^2 + XC^2 = XB^2 + XD^2.$

Proof

a) Let $\omega$ be the circle centered at $F$ with radius $BE.$ Let $\Omega$ be the circle centered at $E$ with radius $AF.$ $PE$ is the median of $\triangle PBD \implies PE^2 = \frac {PB^2 + PD^2}{2} – BE^2.$

The power of the point $P$ with respect to the circle $\Omega$ is $Pow_{\Omega}(P) = PE^2 – AF^2 = \frac {PB^2 + PD^2}{2} – BE^2 – AF^2.$

$PF$ is the median of $\triangle PAC \implies PF^2 = \frac {PA^2 + PC^2}{2} – AF^2.$

The power of the point $P$ with respect to the circle $\omega$ is $Pow_{\omega}(P) = PF^2 – BE^2 = \frac {PA^2 + PC^2}{2} – BE^2 – AF^2 =  \frac {PB^2 + PD^2}{2} – BE^2 – AF^2 = Pow_{\Omega}(P).$

Therefore $P$ lies on the radical axis of $\Omega$ and $\omega.$ Similarly, $Q$ lies on these line. So the line $PQ$ is the radical axes of $\Omega$ and $\omega.$

This line is perpendicular to Gauss line $EF$ which is the line of centers of two circles $\Omega$ and $\omega$ as desired.

b) $XE$ is the median of $\triangle XBD \implies XE^2 = \frac {XB^2 + XD^2}{2} – BE^2.$

$XF$ is the median of $\triangle XAC \implies XF^2 = \frac {XA^2 + XC^2}{2} – AF^2.$

$X$ lies on the radical axes of $\Omega$ and $\omega \implies XE^2 – XF^2 = AF^2 – BE^2 \implies$ \[\frac {XB^2 + XD^2}{2} – BE^2 – ( \frac {XA^2 + XC^2}{2} – AF^2) = AF^2 – BE^2 \implies XB^2 + XD^2 = XA^2 + XC^2.\]

If the point $X$ satisfies the equation $XB^2 + XD^2 = XA^2 + XC^2$ then locus of $X$ is the straight line (one can prove it using method of coordinates).

The points $P$ and $Q$ 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 $E$ and $F$ with radii $BE$ and $AF,$ respectively.

Of course, it is parallel to Simson line.

Shatunov-Tokarev concurrent lines

Shatunov 3 concurrent lines.png

Let the quadrilateral $ABCD$ be given ($ABCD$ is not cyclic).

Let points $A'$ and $A''$ be on the line $AB$ such that $AA' = AA''$. Similarly \[B' \in BC, B'' \in BC, C' \in CD, C'' \in CD,\] \[D' \in AD, D'' \in AD,\] \[BB' = BB'' = CC' = CC'' = DD' = DD'' = AA'.\]

Let points $Q, Q',$ and $Q''$ be the crosspoints of the bisectors $AD \cap BC, A'D' \cap B'C', A''D'' \cap B''C''.$

Similarly points $P, P',$ and $P''$ are the crosspoints of the bisectors $AB \cap CD, A'B' \cap C'D', A''B'' \cap C''D''.$

Prove that lines $PQ, P'Q',$ and $P''Q''$ are concurrent.

Proof

Segment $XA$ is the median of the $\triangle XA'A'' \implies 2(XA^2 + AA'^2) = XA'^2 + XA''^2.$

Similarly $2(XB^2 + BB'^2) = XB'^2 + XB''^2, 2(XC^2 + CC'^2) = XC'^2 + XC''^2, 2(XD^2 + DD'^2) = XD'^2 + XD''^2.$

Let $PQ$ cross $P'Q'$ at point $X \implies XA^2 + XC^2 = XB^2 + XD^2, XA'^2 + XC'^2 = XB'^2 + XD'^2.$

We made simple calculations and get $XA''^2 + XC''^2 = XB''^2 + XD''^2,$ therefore point $X$ lies on $P''Q''$ as desired.

Shatunov point

Shatunov point.png

Let the quadrilateral $ABCD$ be given ($ABCD$ is not cyclic).

Let points $A', B', C',$ and $D'$ be on the lines $AD, AB, BC,$ and $CD,$ respectively such that $|AA'| = |BB'| = |CC'| = |DD'| = d.$

Let points $A'', B'', C'',$ and $D''$ be on the segments $AA', BB', CC',$ and $DD',$ respectively such that $\frac {|AA''|}{|A'A''|} = \frac {|BB''|}{|B'B''|} = \frac {|CC''|}{|C'C''|} = \frac {|DD''|}{|D'D''|} = \frac {m}{n},$ where $m + n = 1.$

Let points $Q, Q',$ and $Q''$ be the crosspoints of the bisectors $AD \cap BC, A'D' \cap B'C', A''D'' \cap B''C''.$

Similarly points $P, P',$ and $P''$ are the crosspoints of the bisectors $AB \cap CD, A'B' \cap C'D', A''B'' \cap C''D''.$

Prove that lines $PQ, P'Q',$ and $P''Q''$ are concurrent.

Proof

Segment $XA''$ is the cevian to the side AA' of the $\triangle XAA'.$

We use the Stewart's theorem and get: \[m \cdot|XA'|^2 + n \cdot |XA|^2) = |XA''|^2 + mn \cdot d^2.\] Similarly $m \cdot|XB'|^2 + n \cdot |XB|^2) = |XB''|^2 + mn \cdot d^2,$ \[m \cdot|XC'|^2 + n \cdot |XC|^2) = |XC''|^2 + mn \cdot d^2,\] \[m \cdot|XD'|^2 + n \cdot |XD|^2) = |XD''|^2 + mn \cdot d^2.\] Let $PQ$ cross $P'Q'$ at point $X \implies |XA|^2 + |XC|^2 = |XB|^2 + |XD|^2, |XA'|^2 + |XC'|^2 = |XB'|^2 + |XD'|^2.$

We made simple calculations and get $|XA''|^2 + |XC''|^2 = |XB''|^2 + |XD''|^2,$ therefore point $X$ lies on $P''Q''$ as desired.


Shatunov chain

Shatunov 8 points chain.png

Let the quadrilateral $ABCD$ be given ($ABCD$ is not cyclic).

Let points $A'$ and $A''$ be on the line $AB$ such that $|AA'| = |AA''|$. Similarly $B' \in BC, B'' \in BC, C' \in CD, C'' \in CD, D' \in AD, D'' \in AD,$ $|BB'| = |BB''| = |CC'| = |CC''| = |DD'| = |DD''| = |AA'|.$

Let points $Q$ and $P$ be the crosspoints of the bisectors $AD \cap BC$ and $AB \cap CD.$

We made quadrilateral $KLMN$ using one point from the pare ${A',A''},$ one point from the pare ${B',B''},$ one point from the pare ${C',C''},$ one point from the pare ${D',D''}.$ For each quadrilateral we find the crosspoints of the bisectors $KL \cap MN$ and $KN \cap LM$ and named these points as ${Q_i,P_i}, i = 1..16.$

Prove that lines $P_iQ_i$ cross line $PQ$ in 8 points and positions of these points are fixed for given $ABCD$ (not depend from the length of $AA'.)$

Proof

The claim follows from the fact that there are $4^2 = 16$ combinations of quadrilateral vertices, and these 16 quadrilaterals are divided into pairs whose points of intersection with the line $PQ$ coincide.

See Also

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