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Difference between revisions of "Miquel's point"

(→‎Circle of circumcenters)
(→‎Triangle of circumcenters)
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==Triangle of circumcenters==
 
==Triangle of circumcenters==
 
[[File:Miquel perspector.png|500px|right]]
 
[[File:Miquel perspector.png|500px|right]]
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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 four lines made four triangles of a complete quadrilateral.
−
Let points <math>O,O_A, O_B,</math> and <math>O_C</math> be the circumcenters of <math>\triangle ABC, \triangle ADE, \triangle BDF,</math> and <math>\triangle CEF,</math> respectively.  
+
 
−
Prove that  <math>\triangle O_AO_BO_C \sim \triangle ABC,</math> and perspector of these triangles point <math>X</math> is the second (different from <math>M</math>) point of intersection circumcircles of <math>\triangle ABC</math> and <math>\triangle O_AO_BO_C.</math>
+
In the diagram these are <math>\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.</math>
 +
 
 +
Let points <math>O,O_A, O_B,</math> and <math>O_C</math> be the circumcenters of <math>\triangle ABC, \triangle ADE, \triangle BDF,</math> and <math>\triangle CEF,</math> respectively.
 +
 +
Prove that  <math>\triangle O_AO_BO_C \sim \triangle ABC,</math> and perspector of these triangles point <math>X</math> is the second (different from <math>M</math>) point of intersection <math>\Omega \cap \Theta,</math> where <math>\Omega</math> is circumcircle of <math>\triangle ABC</math> and <math>\Theta</math> is circumcircle of <math>\triangle O_AO_BO_C.</math>
 +
 
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
−
Quadrungle <math>MECF</math> is concyclic <math>\implies \angle AEM = \angle BFM \implies \angle AO_AB = 2\angle AEM = 2 \angle BFM = \angle BO_BM.</math>
+
 
−
<math>\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.</math>  
+
Quadrungle <math>MECF</math> is cyclic <math>\implies \angle AEM = \angle BFM \implies</math>
 +
<cmath>\angle AO_AB = 2\angle AEM = 2 \angle BFM = \angle BO_BM.</cmath>
 +
<cmath>\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.</cmath>  
 
<math>AO_A = MO_A, BO_B = MO_B, CO_C = MO_C \implies \triangle AO_AM \sim \triangle BO_BM \sim \triangle CO_CM.</math>
 
<math>AO_A = MO_A, BO_B = MO_B, CO_C = MO_C \implies \triangle AO_AM \sim \triangle BO_BM \sim \triangle CO_CM.</math>
−
Spiral similarity sentered at point <math>M</math> with rotation angle <math>\angle AMO_A = \angle BMO_B = CMO_C</math> and the coefficient of homothety <math>\frac {AM}{MO_A} = \frac {BM}{MO_B} =\frac {CM}{MO_C}</math> mapping <math>A</math> to <math>O_A</math>, <math>B</math> to <math>O_B</math>, <math>C</math> to <math>O_C \implies </math>\triangle O_AO_BO_C \sim \triangle ABC.<math>
+
 
−
</math>\triangle AO_AM, \triangle BO_BM, \triangle CO_CM<math> are triangles in double perspective at point </math>M \implies<math> these triangles are in triple perspective </math>\implies AO_A, BO_B, CO_C<math> are concurrent at the point </math>X.<math>
+
Spiral similarity sentered at point <math>M</math> with rotation angle <math>\angle AMO_A = \angle BMO_B = CMO_C</math> and the coefficient of homothety <math>\frac {AM}{MO_A} = \frac {BM}{MO_B} =\frac {CM}{MO_C}</math> mapping <math>A</math> to <math>O_A</math>, <math>B</math> to <math>O_B</math>, <math>C</math> to <math>O_C \implies \triangle O_AO_BO_C \sim \triangle ABC.</math>
−
The rotation angle </math>\triangle AO_AM<math> to </math>\triangle BO_BM<math> is </math>O_AMO_B<math> for sides </math>O_AM<math> and </math>O_BM<math> or angle between </math>AO_A<math> and </math>BO_B<math> which is </math>\angle AXB \implies M O_AO_BX<math> is cyclic </math>implies M O_AO_BXO_C<math> is cyclic </math>\implies \angle O_AXO_B = \angle  O_AO_CO_B = \angle ACB \implies ABCX$ is cyclic as desired.
+
 
 +
<math>\triangle AO_AM, \triangle BO_BM, \triangle CO_CM</math> are triangles in double perspective at point <math>M \implies</math>
 +
 
 +
these triangles are in triple perspective <math>\implies AO_A, BO_B, CO_C</math> are concurrent at the point <math>X.</math>
 +
 
 +
The rotation angle <math>\triangle AO_AM</math> to <math>\triangle BO_BM</math> is <math>\angle O_AMO_B</math> for sides <math>O_AM</math> and <math>O_BM</math> or angle between <math>AO_A</math> and <math>BO_B</math> which is <math>\angle AXB \implies M O_AO_BX</math> is cyclic <math>\implies M O_AO_BXO_C</math> is cyclic.
 +
 
 +
Therefore  <math>\implies \angle O_AXO_B = \angle  O_AO_CO_B = \angle ACB \implies ABCX</math> is cyclic as desired.
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 11:32, 6 December 2022

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

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

vladimir.shelomovskii@gmail.com, vvsss

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