Difference between revisions of "Symmetry"
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Therefore <cmath>\angle ABC = \angle AEC = \angle ADC \blacksquare.</cmath> | Therefore <cmath>\angle ABC = \angle AEC = \angle ADC \blacksquare.</cmath> | ||
| + | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
| + | |||
| + | ==Composition of symmetries== | ||
| + | [[File:Combination S.png|200px|right]] | ||
| + | Let the inscribed convex hexagon <math>ABCDEF</math> be given, <math>AB || CF || DE, BC ||AD || EF.</math> | ||
| + | |||
| + | Prove that <math>\angle ABC = 120^\circ.</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | 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. | ||
| + | |||
| + | <cmath>B = T_l(A), C = T_m(B) = T_m(T_l(A)) \implies \overset{\Large\frown} {AC} = 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> | ||
| + | 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> | ||
| + | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 13:41, 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 
then 
the other cases hold by symmetry because the steps are the same.
Hidden symmetry
Let the convex quadrilateral 
be given.
![\[AC = DE, \angle CAD + \angle ACB = 180^\circ.\]](http://latex.artofproblemsolving.com/b/f/9/bf98317d3bd9b5a53145b71a03b6af4ca6d32dab.png)
Prove that 
Proof
Let 
be bisector 
Let point 
be symmetric 
with respect 
![\[\angle CAD = \angle ACE \implies \angle CAD + \angle ACB = 180^\circ \implies E \in BC.\]](http://latex.artofproblemsolving.com/3/5/d/35d0b1b58bd30670c2cc74850e1c49519971a05f.png)
is isosceles.
Therefore ![\[\angle ABC = \angle AEC = \angle ADC \blacksquare.\]](http://latex.artofproblemsolving.com/9/1/a/91a86a1b1c0f771c5ca09990db9e4b4dd2939c6a.png)
vladimir.shelomovskii@gmail.com, vvsss
Composition of symmetries
Let the inscribed convex hexagon 
be given, 
Prove that 
Proof
Denote 
the circumcenter of 
the common bisector 
the common bisector 
the smaller angle between lines and 
the symmetry with respect axis 
the symmetry with respect axis 
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.
![\[B = T_l(A), C = T_m(B) = T_m(T_l(A)) \implies \overset{\Large\frown} {AC} = 2 \alpha.\]](http://latex.artofproblemsolving.com/8/8/0/8803332be7c06a053c066cc81f5eee82eb70d2a5.png)
![\[F = T_l(C), E = T_m(F) = T_m(T_l(C)) \implies \overset{\Large\frown} {CE} = 2 \alpha.\]](http://latex.artofproblemsolving.com/8/3/6/8368e822b0e3b3e2b5e74acc0c53bda9f2bf88f2.png)
![\[D = T_l(E), A = T_m(D) = T_m(T_l(E)) \implies \overset{\Large\frown} {EA} = 2 \alpha.\]](http://latex.artofproblemsolving.com/b/1/6/b1601e9b7d1c53ffb6c2a4f21f957ae2cbe6799a.png)
Therefore ![\[\overset{\Large\frown} {AC} + \overset{\Large\frown} {CE} + \overset{\Large\frown} {EA} = 6 \alpha = 360^\circ \implies \alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.\]](http://latex.artofproblemsolving.com/e/3/b/e3b35b29543e67a43c306bdd03c847b7029d5b98.png)
vladimir.shelomovskii@gmail.com, vvsss
