Difference between revisions of "Spiral similarity"
(→Center of the spiral symilarity for similar triangles) |
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[[File:5 133 Pras.png|400px|right]] | [[File:5 133 Pras.png|400px|right]] | ||
[[File:5 133a Pras.png|400px|right]] | [[File:5 133a Pras.png|400px|right]] | ||
− | Let triangle <math>\triangle ABC</math> and point <math>C' \ne C, C' \ne B</math> on sideline <math>BC</math> be given. <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AB</math> and <math>A'</math> lies on sideline <math>AC.</math> The spiral symilarity <math>T</math> maps <math>\triangle ABC</math> into <math>\triangle A'B'C'.</math> | + | Let triangle <math>\triangle ABC</math> and point <math>C' (C' \ne C, C' \ne B)</math> on sideline <math>BC</math> be given. <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AB</math> and <math>A'</math> lies on sideline <math>AC.</math> The spiral symilarity <math>T</math> maps <math>\triangle ABC</math> into <math>\triangle A'B'C'.</math> |
Prove | Prove | ||
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<math>A' = A_1B_1 \cap AC, B' = \tau^{-1}(A').</math> | <math>A' = A_1B_1 \cap AC, B' = \tau^{-1}(A').</math> | ||
− | Denote <math>\varphi = \angle BC'B' \implies | + | Denote <math>\varphi = \angle BC'B' \implies \angle CC'A' = 180^\circ - \varphi - \gamma \implies </math> |
− | + | <math>\angle CA'C' = 180^\circ - \gamma - \angle CC'A' = \varphi.</math> Similarly <math>\angle AB'A' = \varphi.</math> | |
− | Similarly <math>\angle AB'A' = \varphi.</math> | ||
b) It is well known that the three circumcircles <math>AA'B', BB'C',</math> and <math> CC'A' </math> have the common point (it is <math>D</math> in the diagram). | b) It is well known that the three circumcircles <math>AA'B', BB'C',</math> and <math> CC'A' </math> have the common point (it is <math>D</math> in the diagram). | ||
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<cmath>\angle A'DC' = 180^\circ - \gamma, \angle A'DC = 180^\circ - \gamma - \varphi \implies </cmath> | <cmath>\angle A'DC' = 180^\circ - \gamma, \angle A'DC = 180^\circ - \gamma - \varphi \implies </cmath> | ||
<cmath>\angle ADC = \angle A'DC' = 180^\circ - \gamma.</cmath> | <cmath>\angle ADC = \angle A'DC' = 180^\circ - \gamma.</cmath> | ||
− | <math>\angle CAD + \angle ACD = 180^\circ - \angle ADC = \gamma = \angle ACD + \angle BCD \implies \angle CAD = \angle BCD.</math> | + | <math>\angle CAD + \angle ACD = 180^\circ - \angle ADC = \gamma = \angle ACD + \angle BCD \implies \angle CAD = \angle BCD = \psi.</math> |
− | Similarly, <math>\angle CAD = \angle ABD.</math> | + | Similarly, <math>\angle CAD = \angle ABD = \angle BCD = \psi.</math> |
Therefore, <math>D</math> is the First Brocard point of <math>\triangle ABC.</math> | Therefore, <math>D</math> is the First Brocard point of <math>\triangle ABC.</math> | ||
− | <math> | + | <math>AB'DA'</math> is cyclic <math>\implies \angle A'B'D = \angle A'AD = \psi.</math> Similarly, <math>\angle B'C'D = \angle C'A'D = \psi.</math> |
− | Therefore <math>D</math> is the First Brocard point of <math>\triangle A'B'C'.</math> | + | |
+ | Therefore <math>D</math> is the First Brocard point of <math>\triangle A'B'C',</math> and <math>\triangle A'DC' \sim \triangle ADC, \triangle A'DB' \sim \triangle ADB.</math> | ||
− | + | Therefore the spiral symilarity <math>T</math> maps <math>\triangle ABC</math> into <math>\triangle A'B'C'</math> has the center <math>D,</math> the angle of the rotation <math>\varphi.</math> | |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 06:14, 14 June 2023
Contents
Basic information
A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important.
The transformation is linear and transforms any given object into an object homothetic to given.
On the complex plane, any spiral similarity can be expressed in the form where
is a complex number. The magnitude
is the dilation factor of the spiral similarity, and the argument
is the angle of rotation.
The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane.
Let with corresponding complex numbers
and
so
Case 1 Any line segment can be mapped into any other
using the spiral similarity. Notation is shown on the diagram.
is circle
is circle
is any point of
is circle
is the image
under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Case 2 Any line segment can be mapped into any other
using the spiral similarity. Notation is shown on the diagram.
is circle
(so circle is tangent to
is circle tangent to
is any point of
is circle
is the image
under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Simple problems
Explicit spiral symilarity
Given two similar right triangles and
Find
and
Solution
The spiral symilarity centered at with coefficient
and the angle of rotation
maps point
to point
and point
to point
Therefore this symilarity maps to
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Hidden spiral symilarity
Let be an isosceles right triangle
Let
be a point on a circle with diameter
The line
is symmetrical to
with respect to
and intersects
at
Prove that
Proof
Denote
Let
cross perpendicular to
in point
at point
Then
Points and
are simmetric with respect
so
The spiral symilarity centered at with coefficient
and the angle of rotation
maps
to
and
to point
such that
Therefore
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Linearity of the spiral symilarity
Points
are outside
Prove that the centroids of triangles and
are coinsite.
Proof
Let where
be the spiral similarity with the rotation angle
and
A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore
We use the property of linearity and get
Let
be the centroid of
so
is the centroid of the
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Construction of a similar triangle
Let triangle and point
on sideline
be given. Construct
where
lies on sideline
and
lies on sideline
Solution
Let be the spiral symilarity centered at
with the dilation factor
and rotation angle
so image of any point
lies on
The spiral symilarity
centered at
with the dilation factor
and rotation angle
maps
into
and
therefore the found triangle
is the desired one.
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Center of the spiral symilarity for similar triangles
Let triangle and point
on sideline
be given.
where
lies on sideline
and
lies on sideline
The spiral symilarity
maps
into
Prove
a)
b) Center of is the First Brocard point of triangles
and
Proof
a) Let be the spiral symilarity centered at
with the dilation factor
and rotation angle
Denote
Similarly
b) It is well known that the three circumcircles and
have the common point (it is
in the diagram).
Therefore is cyclic and
Similarly,
Similarly,
Therefore, is the First Brocard point of
is cyclic
Similarly,
Therefore is the First Brocard point of
and
Therefore the spiral symilarity maps
into
has the center
the angle of the rotation
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