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Difference between revisions of "Miquel's point"

(→‎Miquel and Steiner's quadrilateral theorem)
(→‎Circle of circumcenters)
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<math>\angle MOo' = \angle MO'o' \implies</math> points <math>M, O, O', o,</math> and <math>o'</math> are concyclic as desired.
 
<math>\angle MOo' = \angle MO'o' \implies</math> points <math>M, O, O', o,</math> and <math>o'</math> are concyclic as desired.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Triangle of circumcenters==
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[[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>
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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.
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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>
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<i><b>Proof</b></i>
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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>
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<math>\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.</math>
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<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>
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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>
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</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>
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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.
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 12:19, 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 circumcircles of $\triangle ABC$ and $\triangle O_AO_BO_C.$ Proof Quadrungle $MECF$ is concyclic $\implies \angle AEM = \angle BFM \implies \angle AO_AB = 2\angle AEM = 2 \angle BFM = \angle BO_BM.$ $\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.$ $AO_A = MO_A, BO_B = MO_B, CO_C = MO_C \implies \triangle AO_AM \sim \triangle BO_BM \sim \triangle CO_CM.$ 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.$$ (Error compiling LaTeX. Unknown error_msg)\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$is$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$is cyclic$\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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