Difference between revisions of "Symmetry"

Revision as of 13:28, 1 September 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

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

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

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

Construction of triangle

Given points $D, E,$ and $F$ at which the segments of the bisectors $AI, BI,$ and $CI,$ respectively intersect the incircle of $\triangle ABC$ centered at $I.$

Construct the triangle $\triangle ABC.$

Construction

We construct the incenter of $\triangle ABC$ as circumcenter of $\odot DEF.$

If these points are collinear or if $\min(\angle DIE, \angle EIF, \angle DIF) \le 90^\circ$ construction is impossible.

We construct bisectors $BEI$ and $CFI.$

We construct the points $D'$ and $D''$ symmetrical to point $D$ with respect to $FI$ and $EI,$ respectively.

We construct the bisector $D'D''$ and choose the point $G$ as the point intersection with the circle $\odot DEF,$ closest to the line $D'D''.$

We construct a tangent to the the circle $\odot DEF,$ at the point $G.$ It intersects the lines $EI$ and $FI$ at points $B$ and $C,$ respectively.

We construct the tangents to $\odot DEF$ which are symmetrical to sideline $BC$ with respect to $BI$ and $CI. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Symmetry with respect angle bisectors 2

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: $\frac {ME}{\sin {EC'M}} : \frac {MF}{\sin {MB'F}} = \frac {EC'}{\sin {EMC'}} : \frac {B'F}{\sin {B'MF}} \implies \frac {ME}{MF} = \frac {EC'}{B'F} = 1. \blacksquare$ vladimir.shelomovskii@gmail.com, vvsss

Symmetry of radical axes

Let triangle $\triangle ABC$ be given. The point $I$ and the circle $\Omega$ are the incenter and the circumcircle of $\triangle ABC.$

Circle $\theta$ centered at $A$ has the radius $AI$ and intersects $\Omega$ at points $D$ and $E.$ Line $\ell$ is the tangent for $\theta$ at the point $I.$

Prove that line $DE$ is symmetry to the line $BC$ with respect axis $\ell.$

Proof

$P = AI \cap \Omega \implies PB = PC = PI \implies$ circle $\Theta$ centered at $P$ contain points $B$ and $C$ and is tangent for $\ell$ and $\theta.$

$DE$ is the radical axis of $\theta$ and $\Omega.$ $BC$ is the radical axis of $\Theta$ and $\Omega.$

$\ell$ is the radical axis of $\theta$ and $\Theta \implies BC, DE,$ and $\ell$ are concurrent (at point $F, \ell = FI.$ ) $\overset{\Large\frown} {AE} + \overset{\Large\frown} {BD} + \overset{\Large\frown} {BP} = 2 \angle DHI, \overset{\Large\frown} {AD} + \overset{\Large\frown} {BD} + \overset{\Large\frown} {CP} = 2 \angle BGI.$ $AE = AD \implies \overset{\Large\frown} {AE} = \overset{\Large\frown} {AD}, BP = CP \implies \overset{\Large\frown} {BP} = \overset{\Large\frown} {CP} \implies \angle DHI = \angle BGI \implies \angle HFI = \angle GFI. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Composition of symmetries

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

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

Composition of symmetries 2

Let triangle $\triangle ABC$ be given. The point $I$ and the circle $\omega$ are the incenter and the incircle of $\triangle ABC.$

Let $S_A$ be the symmetry with respect axis $AI, S_B$ be the symmetry with respect axis $BI, S_C$ the symmetry with respect axis $CI.$ $A' = \omega \cap BC, B' = \omega \cap AC, C' = \omega \cap AB.$ Find the composition of axial symmetries with respect $AI, BI,$ and $CI.$

Solution

It is known that the composition of three axial symmetries whose axes intersect at one point $I$ is an axial symmetry whose axis contains the same point $I.$

Consider the composition of axial symmetries for point $B' : S_A(B') = C', S_B(C') = A', S_C(A') = B' \implies$

$B'$ is a fixed point of transformation.

This means that the desired axis of symmetry contains points $B'$ and $I$ , this is a straight line $B'I. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Symmetry and secant

The circle $\omega$ centered at $O$ and the point $A$ be given. Let $AB$ and $AB'$ be the tangents, $ACD$ be the secant ( $B \in \omega, B' \in \omega, C \in \omega, D \in \omega, AC < AD).$

Segment $AD$ intersects segment $BB'$ at point $E.$ Prove that $\frac {AD}{AC} = \frac{DE}{CE}.$

Proof

Let $AC'D'$ be symmetric to $ACD$ with respect the line $AO.$ $\overset{\Large\frown} {BC} = \overset{\Large\frown} {B'C'}, \overset{\Large\frown} {BD} = \overset{\Large\frown} {B'D'}, M = CD' \cap C'D \implies M \in AO.$

It is known that $M \in BB' \implies BM = B'M.$ $2\angle CBB' = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {B'C'} = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {BC} = 2\angle BDC'.$ We use symmetry and get $\angle BMC = \angle B'MC' = \angle BMD \implies \triangle BCM \sim \triangle DBM.$ It is known that $\triangle ABC \sim \triangle ADB \implies$ $\frac {BD^2}{BC^2} = \frac {AD^2}{AB^2} = \frac {AD^2}{AD \cdot AC} = \frac {AD}{AC}.$

Triangles $\triangle BCM$ and $\triangle DBM$ have common side $BM \implies \frac {[DBM]}{[BCM]} = \frac {DE}{CE}.$

Similar triangles $\triangle BCM$ and $\triangle DBM$ have the areas ratio $\frac {[DBM]}{[BCM]} = \frac {BD^2}{BC^2} = \frac {AD}{AC}.$

Therefore $\frac {[DBM]}{[BCM]} = \frac {DE}{CE} = \frac {AD}{AC}. \blacksquare$

According the Cross-ratio criterion the four points $(A,C; E,D)$ are a harmonic range (on the real projective line).

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 203: / Line 203: @@
 [[File:Tangent sym.png|350px|right]]
 The circle <math>\omega</math> centered at <math>O</math> and the point <math>A</math> be given. Let <math>AB</math> and <math>AB'</math> be the tangents, <math>ACD</math> be the secant (<math>B \in \omega, B' \in \omega, C \in \omega, D \in \omega, AC < AD).</math>
 Segment <math>AD</math> intersects segment <math>BB'</math> at point <math>E.</math> Prove that <math>\frac {AD}{AC} = \frac{DE}{CE}.</math>
 <i><b>Proof</b></i>
-Let <math>AC'D'</math> be symmetric to <math>ACD</math> with respect the line <math>AO, \overset{\Large\frown} {BC} = \overset{\Large\frown} {B'C'}, \overset{\Large\frown} {BD} = \overset{\Large\frown} {B'D'}, M = CD' \cap C'D \implies M \in AO.</math>
+Let <math>AC'D'</math> be symmetric to <math>ACD</math> with respect the line <math>AO.</math>
+<cmath>\overset{\Large\frown} {BC} = \overset{\Large\frown} {B'C'}, \overset{\Large\frown} {BD} = \overset{\Large\frown} {B'D'}, M = CD' \cap C'D \implies M \in AO.</cmath>
 It is known that <math>M \in BB' \implies BM = B'M.</math>
-<cmath>2\angle CBB' = overset{\Large\frown} {CC'} + overset{\Large\frown} {B'C'}</cmath>
+<cmath>2\angle CBB' = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {B'C'} = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {BC} = 2\angle BDC'.</cmath>
-<cmath>= overset{\Large\frown} {CC'} + overset{\Large\frown} {BC} = 2\angle BDC'.</cmath>
+We use symmetry and get  <cmath>\angle BMC = \angle B'MC' = \angle BMD \implies \triangle BCM \sim \triangle DBM.</cmath>
-We use symmetry and get  <math>\angle BMC = \angle B'MC' = \angle BMD \implies \triangle BCM \sim \triangle DBM.</math>
+It is known that <math>\triangle ABC \sim \triangle ADB \implies</math>
-It is known that <math>\triangle ABC \sim \triangle ADB \implies \frac {BD^2}{BC^2} = \frac {AD^2}{AB^2} = \frac {AD^2}{AD \cdot AC} = \frac {AD}{AC}.</math>
+<cmath>\frac {BD^2}{BC^2} = \frac {AD^2}{AB^2} = \frac {AD^2}{AD \cdot AC} = \frac {AD}{AC}.</cmath>
 Triangles <math>\triangle BCM</math> and <math>\triangle DBM</math> have common side <math>BM \implies \frac {[DBM]}{[BCM]} = \frac {DE}{CE}.</math>
-Similar triangles <math>\triangle BCM</math> and <math>\triangle DBM</math> have the areas ratio <math>\frac {[DBM]}{[BCM]} = \frac {BD^2}{BC^2} = \frac {AD}{AC}=\frac {DE}{CE}</math> common side <math>BM \implies \frac {[DBM]}{[BCM]} = \frac {DE}{CE} \blacksquare</math>
+Similar triangles <math>\triangle BCM</math> and <math>\triangle DBM</math> have the areas ratio <math>\frac {[DBM]}{[BCM]} = \frac {BD^2}{BC^2} = \frac {AD}{AC}.</math>
+Therefore <math>\frac {[DBM]}{[BCM]} = \frac {DE}{CE} = \frac {AD}{AC}. \blacksquare</math>
 According the Cross-ratio criterion the four points <math>(A,C; E,D)</math> are a harmonic range (on the real projective line).
 '''vladimir.shelomovskii@gmail.com, vvsss'''

Page

Toolbox

Search

Difference between revisions of "Symmetry"

Revision as of 13:28, 1 September 2023

Contents

Hidden symmetry

Symmetry with respect angle bisectors

Symmetry with respect angle bisectors 1

Construction of triangle

Symmetry with respect angle bisectors 2

Symmetry of radical axes

Composition of symmetries

Composition of symmetries 1

Composition of symmetries 2

Symmetry and secant