Difference between revisions of "Symmetry"
(→Symmetry of radical axes) |
(→Symmetry with respect angle bisectors 1) |
||
Line 49: | Line 49: | ||
The point <math>C'</math> is symmetric to <math>B'</math> with respect to <math>AI \implies \overset{\Large\frown} {KC'} = \overset{\Large\frown} {DB'}.</math> | The point <math>C'</math> is symmetric to <math>B'</math> with respect to <math>AI \implies \overset{\Large\frown} {KC'} = \overset{\Large\frown} {DB'}.</math> | ||
− | Similarly <math>\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies | + | |
− | + | Similarly <math>\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies \overset{\Large\frown} {DB'} = \overset{\Large\frown} {EB'} \blacksquare.</math> | |
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Construction of triangle== | ||
+ | [[File:Construction tr.png|350px|right]] | ||
+ | Given points <math>D, E,</math> and <math>F</math> at which the segments of the bisectors <math>AI, BI,</math> and <math>CI,</math> respectively intersect the incircle of <math>\triangle ABC</math> centered at <math>I.</math> Construct the triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Construction</b></i> | ||
+ | |||
+ | We construct the incenter of <math>\triangle ABC</math> as circumcenter of <math>\odot DEF.</math> | ||
+ | |||
+ | If these points are collinear or if <math>\min(\angle DIE, \angle EIF, \angle DIF) \le 90^\circ</math> construction is impossible. | ||
+ | |||
+ | We construct bisectors <math>BEI</math> and <math>CFI.</math> | ||
+ | |||
+ | We construct the points <math>D'</math> and <math>D''</math> symmetrical to point <math>D</math> with respect to <math>FI</math> and <math>EI.</math> | ||
+ | |||
+ | We construct the bisector <math>D'D''</math> and choose the point <math>G</math> as the point intersection with the circle <math>\odot DEF,</math> closest to the line <math>D'D''.</math> | ||
+ | We construct a tangent to the the circle <math>\odot DEF,</math> at the point <math>G.</math> | ||
+ | It intersects the lines <math>EI</math> and <math>FI</math> at points <math>B</math> and <math>C,</math> respectively. | ||
+ | We construct the tangents to <math>\odot DEF</math> which are symmetrical to sideline <math>BC</math> with respect to <math>BI</math> and <math>CI. \blacksquare</math> | ||
+ | |||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 12:46, 31 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.
Contents
[hide]Hidden symmetry
Let the convex quadrilateral be given.
Prove that
Proof
Let be bisector
Let point be symmetric with respect
is isosceles.
Therefore vladimir.shelomovskii@gmail.com, vvsss
Symmetry with respect angle bisectors
Given the triangle is the incircle, is the incenter,
Points and are symmetrical to point with respect to the lines containing the bisectors and respectively.
Prove that is the midpoint
Proof Denote
The tangents from point to are equal
Point is symmetrical to point with respect is symmetrical to segment
Symilarly, vladimir.shelomovskii@gmail.com, vvsss
Symmetry with respect angle bisectors 1
The bisector intersect the incircle of the triangle at the point The point is symmetric to with respect to the point is symmetric to with respect to Prove that is the bisector of the segment
Proof
The point is symmetric to with respect to
The point is symmetric to with respect to
Similarly
vladimir.shelomovskii@gmail.com, vvsss
Construction of triangle
Given points and at which the segments of the bisectors and respectively intersect the incircle of centered at Construct the triangle
Construction
We construct the incenter of as circumcenter of
If these points are collinear or if construction is impossible.
We construct bisectors and
We construct the points and symmetrical to point with respect to and
We construct the bisector and choose the point as the point intersection with the circle closest to the line We construct a tangent to the the circle at the point It intersects the lines and at points and respectively. We construct the tangents to which are symmetrical to sideline with respect to and
vladimir.shelomovskii@gmail.com, vvsss
Symmetry with respect angle bisectors 2
Given the triangle is the incircle, is the incenter,
Let be the point on sideline
Points and are symmetrical to point with respect to the lines and respectively. The line contains point
Prove that is the midpoint
Proof
The segment is symmetric to with respect to the segment is symmetric to with respect to So
Similarly at midpoint
or or We use the Law of Sines and get: vladimir.shelomovskii@gmail.com, vvsss
Symmetry of radical axes
Let triangle be given. The point and the circle are the incenter and the circumcircle of
Circle centered at has the radius and intersects at points and Line is the tangent for at the point
Prove that line is symmetry to the line with respect axis
Proof
circle centered at contain points and and is tangent for and
is the radical axis of and is the radical axis of and
is the radical axis of and and are concurrent (at point )
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
is the symmetry with respect axis is 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.
Therefore vladimir.shelomovskii@gmail.com, vvsss
Composition of symmetries 1
Let the triangle be given.
is the incircle, is the incenter, is the circumcenter of The point is symmetric to with respect to is symmetric to with respect to is symmetric to with respect to
Prove: a)
b)
Proof
a) Denote the smaller angle between and
is the symmetry with respect axis is the symmetry with respect axis
counterclockwise direction.
clockwise direction.
Therefore is parallel to tangent line for at point
b) is homothetic to
is the circumcenter of
The center of the homothety lies on the line passing through the circumcenters of the triangles.
vladimir.shelomovskii@gmail.com, vvsss
Composition of symmetries 2
Let triangle be given. The point and the circle are the incenter and the incircle of
Let be the symmetry with respect axis be the symmetry with respect axis the symmetry with respect axis Find the composition of axial symmetries with respect and
Solution
It is known that the composition of three axial symmetries whose axes intersect at one point is an axial symmetry whose axis contains the same point
Consider the composition of axial symmetries for point
is a fixed point of transformation.
This means that the desired axis of symmetry contains points and , this is a straight line
vladimir.shelomovskii@gmail.com, vvsss