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Difference between revisions of "Symmetry"

(→‎Hidden symmetry)
(→‎Composition of symmetries)
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==Composition of symmetries==
 
==Composition of symmetries==
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[[File:Combination S.png|200px|right]]
+
[[File:Combination S.png|300px|right]]
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Let the inscribed convex hexagon <math>ABCDEF</math> be given, <math>AB || CF || DE, BC ||AD || EF.</math>  
+
[[File:Combination Sy.png|300px|right]]
−
 
+
Let the inscribed convex hexagon <math>ABCDEF</math> be given,
 +
<cmath>AB || CF || DE, BC ||AD || EF.</cmath>  
 
Prove that <math>\angle ABC = 120^\circ.</math>
 
Prove that <math>\angle ABC = 120^\circ.</math>
  
 
<i><b>Proof</b></i>  
 
<i><b>Proof</b></i>  
  
−
Denote <math>O</math> the circumcenter of <math>ABCDEF, \ell</math> the common bisector <math>AB || CF || DE, m</math> the common bisector <math>BC ||AD || EF, \ell \cap m = O, \alpha</math> the smaller angle between lines <math>\ell</math> and <math>m, S_l.</math> the symmetry with respect axis <math>\ell, T_m</math> the symmetry with respect axis <math>m.</math>
+
Denote <math>O</math> the circumcenter of <math>ABCDEF,</math>
 +
 
 +
<math>\ell</math> the common bisector <math>AB || CF || DE, m</math> the common bisector <math>BC ||AD || EF,</math>
 +
 
 +
<math>\ell \cap m = O, \alpha</math> the smaller angle between lines <math>\ell</math> and <math>m,</math>
 +
 
 +
<math>S_l</math> is the symmetry with respect axis <math>\ell, T_m</math> is the symmetry with respect axis <math>m.</math>
  
 
It is known that the composition of two axial symmetries with non-parallel axes is a rotation centered at
 
It is known that the composition of two axial symmetries with non-parallel axes is a rotation centered at
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<cmath>F = T_l(C), E = T_m(F) = T_m(T_l(C)) \implies \overset{\Large\frown} {CE} = 2 \alpha.</cmath>
 
<cmath>F = T_l(C), E = T_m(F) = T_m(T_l(C)) \implies \overset{\Large\frown} {CE} = 2 \alpha.</cmath>
 
<cmath>D = T_l(E), A = T_m(D) = T_m(T_l(E)) \implies \overset{\Large\frown} {EA} = 2 \alpha.</cmath>
 
<cmath>D = T_l(E), A = T_m(D) = T_m(T_l(E)) \implies \overset{\Large\frown} {EA} = 2 \alpha.</cmath>
−
Therefore <cmath>\overset{\Large\frown} {AC} + \overset{\Large\frown} {CE} + \overset{\Large\frown} {EA} = 6 \alpha = 360^\circ \implies \alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.</cmath>
+
Therefore <cmath>\overset{\Large\frown} {AC} + \overset{\Large\frown} {CE} + \overset{\Large\frown} {EA} = 6 \alpha = 360^\circ \implies</cmath>
 +
<cmath>\alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.</cmath>
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 14:49, 28 August 2023

A proof utilizes symmetry if the steps to prove one thing is identical to those steps of another. For example, to prove that in triangle ABC with all three sides congruent to each other that all three angles are equal, you only need to prove that if $AB = AC,$ then $\angle C = \angle B;$ the other cases hold by symmetry because the steps are the same.

Hidden symmetry

Hidden S.png
Hidden Sy.png

Let the convex quadrilateral $ABCD$ be given. \[AC = DE, \angle CAD + \angle ACB = 180^\circ.\]

Prove that $\angle ABC = \angle ADC.$

Proof

Let $\ell$ be bisector $AC.$

Let point $E$ be symmetric $D$ with respect $\ell.$

\[\angle CAD = \angle ACE \implies \angle CAD + \angle ACB = 180^\circ \implies E \in BC.\] $AE = CD = AB \implies \triangle ABE$ is isosceles.

Therefore \[\angle ABC = \angle AEC = \angle ADC \blacksquare.\] vladimir.shelomovskii@gmail.com, vvsss

Composition of symmetries

Combination S.png
Combination Sy.png

Let the inscribed convex hexagon $ABCDEF$ be given, \[AB || CF || DE, BC ||AD || EF.\] Prove that $\angle ABC = 120^\circ.$

Proof

Denote $O$ the circumcenter of $ABCDEF,$

$\ell$ the common bisector $AB || CF || DE, m$ the common bisector $BC ||AD || EF,$

$\ell \cap m = O, \alpha$ the smaller angle between lines $\ell$ and $m,$

$S_l$ is the symmetry with respect axis $\ell, T_m$ is the symmetry with respect axis $m.$

It is known that the composition of two axial symmetries with non-parallel axes is a rotation centered at point of intersection of the axes at twice the angle from the axis of the first symmetry to the axis of the second symmetry.

\[B = T_l(A), C = T_m(B) = T_m(T_l(A)) \implies \overset{\Large\frown} {AC} = 2 \alpha.\] \[F = T_l(C), E = T_m(F) = T_m(T_l(C)) \implies \overset{\Large\frown} {CE} = 2 \alpha.\] \[D = T_l(E), A = T_m(D) = T_m(T_l(E)) \implies \overset{\Large\frown} {EA} = 2 \alpha.\] Therefore \[\overset{\Large\frown} {AC} + \overset{\Large\frown} {CE} + \overset{\Large\frown} {EA} = 6 \alpha = 360^\circ \implies\] \[\alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.\] vladimir.shelomovskii@gmail.com, vvsss

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