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Difference between revisions of "Symmetry"

(→‎Composition of symmetries 1)
(→‎Symmetry with respect angle bisectors 2)
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Symilarly, <cmath>BA' = EB' \implies EB' = DB'. \blacksquare.</cmath>
 
Symilarly, <cmath>BA' = EB' \implies EB' = DB'. \blacksquare.</cmath>
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
−
==Symmetry with respect angle bisectors 2==
+
==Symmetry with respect angle bisectors 1==
 
[[File:Bisectors 3.png|350px|right]]
 
[[File:Bisectors 3.png|350px|right]]
 
The bisector <math>BI</math> intersect the incircle <math>\omega</math> of the triangle <math>\triangle ABC</math> at the point <math>K, B' = \omega \cap AC.</math>
 
The bisector <math>BI</math> intersect the incircle <math>\omega</math> of the triangle <math>\triangle ABC</math> at the point <math>K, B' = \omega \cap AC.</math>
Line 51: Line 51:
 
Similarly <math>\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies </math>
 
Similarly <math>\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies </math>
 
<cmath> \overset{\Large\frown} {DB'} =  \overset{\Large\frown} {EB'} \blacksquare.</cmath>
 
<cmath> \overset{\Large\frown} {DB'} =  \overset{\Large\frown} {EB'} \blacksquare.</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
==Symmetry with respect angle bisectors 2==
 +
[[File:Bisectors 3.png|350px|right]]
 +
Given the triangle <math>\triangle ABC, \omega</math> is the incircle, <math>I</math> is the incenter, <math>B' = \omega \cap AC, C' = \omega \cap AB.</math>
 +
Let <math>D</math> be the point on sideline <math>BC.</math>
 +
Points <math>E</math> and <math>F</math> are symmetrical to point <math>D</math> with respect to the lines <math>BI</math> and <math>CI,</math> respectively. The line <math>\ell \perp EF</math> contains point <math>I.</math>
 +
Prove that <math>M = B'C' \cap \ell</math> is the midpoint <math>EF.</math>
 +
 +
<i><b>Proof</b></i>
 +
 +
The segment <math>EC'</math> is symmetric to <math>DA'</math> with respect to <math>BI,</math> the segment <math>FB'</math> is symmetric to <math>DA'</math> with respect to <math>CI.</math> So <math>EC' = FB'.</math>
 +
Similarly <math>ID = IE = IF \implies \ell \cap EF</math> at midpoint  <math>EF.</math>
 +
 +
<cmath>AB' = AC' \implies \angle AB'C' = \angle AC'B'.</cmath>
 +
<math>\angle MB'F = \angle AB'C</math> or <math>\angle MB'F + \angle AB'C = 180^\circ \implies \sin \angle ECM = \sin \angle MB'F.</math>
 +
<math>\angle EMC' = \angle B'MF</math> or <math>\angle EMC' + \angle B'MF = 180^\circ \implies \sin \angle EMC = \sin \angle B'MF.</math>
 +
We use the Law of Sines and get:
 +
<cmath>\frac {ME}{\sin {EC'M}} : \frac {MF}{\sin {MB'F}} = \frac {EC'}{\sin {EMC'}} : \frac {BF}{\sin {B'MF}} \implies \frac {ME}{MF} = \frac {EC'}{BF} = 1. \blacksquare</cmath>
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
  

Revision as of 13:29, 29 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

Symmetry with respect angle bisectors

Bisectors 1.png
Bisectors 2.png

Given the triangle $\triangle ABC, \omega$ is the incircle, $I$ is the incenter, $B' = \omega \cap AC.$

Points $D$ and $E$ are symmetrical to point $B$ with respect to the lines containing the bisectors $AI$ and $CI,$ respectively.

Prove that $B'$ is the midpoint $DE.$

Proof \[B \in AB \implies D \in AC, B \in CB \implies E \in AC \implies DE \in AC, D \ne E.\] Denote $A' = \omega \cap BC, C' = \omega \cap AB.$

The tangents from point $B$ to $\omega$ are equal $A'B = C'B.$

Point $B'$ is symmetrical to point $C'$ with respect $AI \implies BC'$ is symmetrical to segment $DB' \implies BC' = DB'.$

Symilarly, \[BA' = EB' \implies EB' = DB'. \blacksquare.\] vladimir.shelomovskii@gmail.com, vvsss

Symmetry with respect angle bisectors 1

Bisectors 3.png

The bisector $BI$ intersect the incircle $\omega$ of the triangle $\triangle ABC$ at the point $K, B' = \omega \cap AC.$ The point $D$ is symmetric to $K$ with respect to $AI,$ the point $E$ is symmetric to $K$ with respect to $CI.$ Prove that $B'I$ is the bisector of the segment $DE.$

Proof

The point $A'= \omega \cap BC$ is symmetric to $C' = \omega \cap BA$ with respect to $AI \implies \overset{\Large\frown} {KA'} = \overset{\Large\frown} {KC'}.$

The point $C'$ is symmetric to $B'$ with respect to $AI \implies \overset{\Large\frown} {KC'} = \overset{\Large\frown} {DB'}.$ Similarly $\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies$ \[\overset{\Large\frown} {DB'} = \overset{\Large\frown} {EB'} \blacksquare.\] vladimir.shelomovskii@gmail.com, vvsss

Symmetry with respect angle bisectors 2

Bisectors 3.png

Given the triangle $\triangle ABC, \omega$ is the incircle, $I$ is the incenter, $B' = \omega \cap AC, C' = \omega \cap AB.$ Let $D$ be the point on sideline $BC.$ Points $E$ and $F$ are symmetrical to point $D$ with respect to the lines $BI$ and $CI,$ respectively. The line $\ell \perp EF$ contains point $I.$ Prove that $M = B'C' \cap \ell$ is the midpoint $EF.$

Proof

The segment $EC'$ is symmetric to $DA'$ with respect to $BI,$ the segment $FB'$ is symmetric to $DA'$ with respect to $CI.$ So $EC' = FB'.$ Similarly $ID = IE = IF \implies \ell \cap EF$ at midpoint $EF.$

\[AB' = AC' \implies \angle AB'C' = \angle AC'B'.\] $\angle MB'F = \angle AB'C$ or $\angle MB'F + \angle AB'C = 180^\circ \implies \sin \angle ECM = \sin \angle MB'F.$ $\angle EMC' = \angle B'MF$ or $\angle EMC' + \angle B'MF = 180^\circ \implies \sin \angle EMC = \sin \angle B'MF.$ We use the Law of Sines and get: \[\frac {ME}{\sin {EC'M}} : \frac {MF}{\sin {MB'F}} = \frac {EC'}{\sin {EMC'}} : \frac {BF}{\sin {B'MF}} \implies \frac {ME}{MF} = \frac {EC'}{BF} = 1. \blacksquare\] 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 $\ell, S_m$ 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.

\[B = S_l(A), C = S_m(B) = S_m(S_l(A)) \implies \overset{\Large\frown} {AC} = 2 \alpha.\] \[F = S_l(C), E = S_m(F) = S_m(S_l(C)) \implies \overset{\Large\frown} {CE} = 2 \alpha.\] \[D = S_l(E), A = S_m(D) = S_m(S_l(E)) \implies \overset{\Large\frown} {EA} = 2 \alpha.\] 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

Composition of symmetries 1

Bisectors 4.png

Let the triangle $\triangle ABC$ be given.

$\omega$ is the incircle, $I$ is the incenter, $O$ is the circumcenter of $\triangle ABC.$ \[A' = \omega \cap BC, B' = \omega \cap AC, C' = \omega \cap AB.\] The point $A''$ is symmetric to $A'$ with respect to $AI, B''$ is symmetric to $B'$ with respect to $BI, C''$ is symmetric to $C'$ with respect to $CI.$

Prove: a)$A''C'' || AC;$

b) $P = AA'' \cap BB'' \cap CC'' \in IO.$

Proof

a) Denote $\varphi$ the smaller angle between $AI$ and $CI.$

$S_A$ is the symmetry with respect axis $AI, S_C$ is the symmetry with respect axis $CI.$

$A' = S_C(B'), A'' = S_A(A') = S_A(S_C(B')) \implies \overset{\Large\frown} {B'A''} = 2 \varphi$ counterclockwise direction.

$C' = S_A(B'), C'' = S_C(C') = S_C(S_A(B')) \implies \overset{\Large\frown} {B'C''} = 2 \varphi$ clockwise direction.

Therefore $\overset{\Large\frown} {B'A''} = \overset{\Large\frown} {C''B'} \implies A''C''$ is parallel to tangent line for $\omega$ at point $B' \implies A''C'' || AC.$

b) $A''C'' || AC, A''B'' || AB, B''C'' || BC \implies \triangle ABC$ is homothetic to $\triangle A''B''C''.$

$\omega$ is the circumcenter of $\triangle A''B''C'' \implies$

The center of the homothety lies on the line passing through the circumcenters of the triangles. $\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

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