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Difference between revisions of "Miquel's point"

(→‎Triangle of circumcenters)
(→‎Analogue of Miquel's point)
 
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Similarly, one can prove that <math>\triangle ADE \sim \triangle OO_BO_C, \triangle BDF \sim \triangle OO_AO_C, \triangle CEF \sim \triangle OO_AO_B.</math>  
 
Similarly, one can prove that <math>\triangle ADE \sim \triangle OO_BO_C, \triangle BDF \sim \triangle OO_AO_C, \triangle CEF \sim \triangle OO_AO_B.</math>  
 
*[[Double perspective triangles]]
 
*[[Double perspective triangles]]
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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 +
==Analogue of Miquel's point==
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[[File:5 circles.png|400px|right]]
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Let inscribed quadrilateral <math>ABB'A'</math> and
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points <math>C \in AB', C' \in A'B', D \in A'B</math> be given.
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<cmath>\theta = \odot CC'B', \Theta = \odot BDD', M = \theta \cap \Theta,</cmath>
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<cmath>E = C'D \cap \odot B'CC', D' = AB \cap CE, F = CC' \cap DD'.</cmath>
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Prove that points <math>A, B, B',</math> and <math>M</math> are concyclic.
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<i><b>Proof</b></i>
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<cmath>\angle BMB' =  \angle EMB' - \angle EMB =
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\angle ECB' - \angle ED'B  = \angle BAB' \blacksquare</cmath>
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<i><b>Corollary</b></i>
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The points <math>F, C, D',</math> and <math>M</math> are concyclic.
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The points <math>F, C', D,</math> and <math>M</math> are concyclic.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Six circles crossing point==
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[[File:Fixed Miquel point.png|410px|right]]
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Let <math>\triangle ABC,</math> point <math>P \in BC,</math> point <math>X \in \Omega = \odot ABC</math> be given.
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Denote <math>Y = PX \cap \Omega, D = AX \cap BC, E = AY \cap BC,</math>
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<cmath>\omega = \odot ADE, \theta = \odot PEY, \Theta = \odot PDX,</cmath>
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<math>\sigma = \odot BP</math> tangent to <math>AB, \Sigma = \odot CP</math> tangent to <math>AC.</math>
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Prove that the circles <math>\omega, \Omega, \theta, \Theta, \sigma,</math> and <math>\Sigma</math> have the common point.
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<i><b>Proof</b></i>
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Let <math>M = \omega \cap \Omega \ne A.</math>
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<cmath>\angle YCM = \angle YXM = \angle YAM = \angle EAM = \angle EDM \implies</cmath>
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<math>\angle PDM = \angle PXM \implies</math> points <math>M,P,D,</math> and <math>X</math> are concyclic, <math>M \in \Theta.</math> Similarly <math>M \in \theta, M</math> is the Miquel point of quadrungle <math>EDXY.</math>
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<cmath>\angle CMY = 180^\circ - \angle CAB - \angle BAY.</cmath>
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<cmath>\angle PMY = \angle PEY = \angle ABC - \angle BAY.</cmath>
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<cmath>\angle PMC = \angle CMY - \angle PMY = (180^\circ - \angle CAB - \angle BAY) - (\angle ABC - \angle BAY) = \angle ACB.</cmath>
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<math>\angle PMC = \angle ACB \implies AC</math> is tangent to <math>\Sigma.</math> Similarly, <math>AB</math> is tangent to <math>\sigma.</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Latest revision as of 15:18, 21 October 2024

Miquel and Steiner's quadrilateral theorem

4 Miquel circles.png

Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Prove that the circumcircles of all four triangles meet at a single point.

Proof

Let circumcircle of $\triangle ABC$ circle $\Omega$ cross the circumcircle of $\triangle CEF$ circle $\omega$ at point $M.$

Let $AM$ cross $\omega$ second time in the point $G.$

$CMGF$ is cyclic $\implies \angle BCM = \angle MGF.$

$AMCB$ is cyclic $\implies \angle BCM + \angle BAM = 180^\circ \implies$

$\angle BAG + \angle AGF = 180^\circ \implies AB||GF.$

$CMGF$ is cyclic $\implies \angle AME = \angle EFG.$

$AD||GF \implies \angle ADE + \angle DFG = 180^\circ \implies \angle ADE + \angle AME = 180^\circ \implies$

$ADEM$ is cyclic and circumcircle of $\triangle ADE$ contain the point $M.$

Similarly circumcircle of $\triangle BDF$ contain the point $M$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

Circle of circumcenters

Miquel point.png

Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Prove that the circumcenters of all four triangles and point $M$ are concyclic.

Proof

Let $\Omega, \omega, \Omega',$ and $\omega'$ be the circumcircles of $\triangle ABC, \triangle CEF, \triangle BDF,$ and $\triangle ADE,$ respectively.

In $\Omega' \angle MDF = \angle MBF.$

In $\omega' \angle MDE = \frac {\overset{\Large\frown} {ME}} {2}.$

$ME$ is the common chord of $\omega$ and $\omega' \implies \angle MOE = \overset{\Large\frown} {ME} \implies$

\[\angle MO'o' = \frac {\overset{\Large\frown} {ME}} {2} = \angle MDE.\]

Similarly, $MF$ is the common chord of $\omega$ and $\Omega' \implies \angle MDF = \angle Moo' = \angle MO'o'.$

Similarly, $MC$ is the common chord of $\Omega$ and $\omega' \implies \angle MBC = \angle MOo' \implies$

$\angle MOo' = \angle MO'o' \implies$ points $M, O, O', o,$ and $o'$ are concyclic as desired.

vladimir.shelomovskii@gmail.com, vvsss

Triangle of circumcenters

Miquel perspector.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 $O,O_A, O_B,$ and $O_C$ be the circumcenters of $\triangle ABC, \triangle ADE, \triangle BDF,$ and $\triangle CEF,$ respectively.

Prove that $\triangle O_AO_BO_C \sim \triangle ABC,$ and perspector of these triangles point $X$ is the second (different from $M$) point of intersection $\Omega \cap \Theta,$ where $\Omega$ is circumcircle of $\triangle ABC$ and $\Theta$ is circumcircle of $\triangle O_AO_BO_C.$

Proof

Quadrungle $MECF$ is cyclic $\implies \angle AEM = \angle BFM \implies$ \[\angle AO_AM = 2\angle AEM = 2 \angle BFM = \angle BO_BM.\] \[\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.\] $AO_A = MO_A, BO_B = MO_B, CO_C = MO_C \implies \triangle AO_AM \sim \triangle BO_BM \sim \triangle CO_CM.$

Spiral similarity sentered at point $M$ with rotation angle $\angle AMO_A = \angle BMO_B = CMO_C$ and the coefficient of homothety $\frac {AM}{MO_A} = \frac {BM}{MO_B} =\frac {CM}{MO_C}$ mapping $A$ to $O_A$, $B$ to $O_B$, $C$ to $O_C \implies \triangle O_AO_BO_C \sim \triangle ABC.$

$\triangle AO_AM, \triangle BO_BM, \triangle CO_CM$ are triangles in double perspective at point $M \implies$

These triangles are in triple perspective $\implies AO_A, BO_B, CO_C$ are concurrent at the point $X.$

The rotation angle $\triangle AO_AM$ to $\triangle BO_BM$ is $\angle O_AMO_B$ for sides $O_AM$ and $O_BM$ or angle between $AO_A$ and $BO_B$ which is $\angle AXB \implies M O_AO_BX$ is cyclic $\implies M O_AO_BXO_C$ is cyclic.

Therefore $\angle O_AXO_B = \angle O_AO_CO_B = \angle ACB \implies ABCX$ is cyclic as desired.

Similarly, one can prove that $\triangle ADE \sim \triangle OO_BO_C, \triangle BDF \sim \triangle OO_AO_C, \triangle CEF \sim \triangle OO_AO_B.$

vladimir.shelomovskii@gmail.com, vvsss

Analogue of Miquel's point

5 circles.png

Let inscribed quadrilateral $ABB'A'$ and

points $C \in AB', C' \in A'B', D \in A'B$ be given.

\[\theta = \odot CC'B', \Theta = \odot BDD', M = \theta \cap \Theta,\] \[E = C'D \cap \odot B'CC', D' = AB \cap CE, F = CC' \cap DD'.\] Prove that points $A, B, B',$ and $M$ are concyclic.

Proof

\[\angle BMB' = \angle EMB' - \angle EMB = \angle ECB' - \angle ED'B = \angle BAB' \blacksquare\]

Corollary

The points $F, C, D',$ and $M$ are concyclic.

The points $F, C', D,$ and $M$ are concyclic.

vladimir.shelomovskii@gmail.com, vvsss

Six circles crossing point

Fixed Miquel point.png

Let $\triangle ABC,$ point $P \in BC,$ point $X \in \Omega = \odot ABC$ be given.

Denote $Y = PX \cap \Omega, D = AX \cap BC, E = AY \cap BC,$ \[\omega = \odot ADE, \theta = \odot PEY, \Theta = \odot PDX,\] $\sigma = \odot BP$ tangent to $AB, \Sigma = \odot CP$ tangent to $AC.$

Prove that the circles $\omega, \Omega, \theta, \Theta, \sigma,$ and $\Sigma$ have the common point.

Proof

Let $M = \omega \cap \Omega \ne A.$ \[\angle YCM = \angle YXM = \angle YAM = \angle EAM = \angle EDM \implies\] $\angle PDM = \angle PXM \implies$ points $M,P,D,$ and $X$ are concyclic, $M \in \Theta.$ Similarly $M \in \theta, M$ is the Miquel point of quadrungle $EDXY.$ \[\angle CMY = 180^\circ - \angle CAB - \angle BAY.\] \[\angle PMY = \angle PEY = \angle ABC - \angle BAY.\] \[\angle PMC = \angle CMY - \angle PMY = (180^\circ - \angle CAB - \angle BAY) - (\angle ABC - \angle BAY) = \angle ACB.\] $\angle PMC = \angle ACB \implies AC$ is tangent to $\Sigma.$ Similarly, $AB$ is tangent to $\sigma.$

vladimir.shelomovskii@gmail.com, vvsss

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