Difference between revisions of "Spiral similarity"
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Therefore <math>\angle ASC = \angle DSB \implies</math> | Therefore <math>\angle ASC = \angle DSB \implies</math> | ||
<cmath>\angle ASD = \angle ASC - \angle DSC = \angle DSB - \angle DSC = \angle BSC = 90^\circ.</cmath> | <cmath>\angle ASD = \angle ASC - \angle DSC = \angle DSB - \angle DSC = \angle BSC = 90^\circ.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Linearity of the spiral symilarity=== | ||
+ | [[File:1933 Pras.png|400px|right]] | ||
+ | <math>\triangle ABF \sim \triangle BCD \sim \triangle CAE.</math> Points <math>D,E,F</math> are outside <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that the centroids of triangles <math>\triangle ABC</math> and <math>\triangle DEF</math> are coinsite. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\vec y = T(\vec x),</math> where <math>T</math> be the spiral similarity with <math>\angle BAF= \angle CBD = \angle ACE</math> and <math>k = \frac {|AF|}{|AB|} = \frac {|DB|}{|BC|} = \frac {|EC|}{|CA|}.</math> | ||
+ | A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore <math>\vec AF = T(\vec AB), \vec BD = T(\vec BC), \vec CE = T(\vec CA).</math> | ||
+ | <math>\vec AB + \vec BC + \vec CA = \vec 0.</math> | ||
+ | We use the property of linearity and get <math>\vec AF + \vec BD + \vec CE = k(\vec AB + \vec BC + \vec CA) = k \cdot \vec 0 = \vec 0.</math> | ||
+ | Let <math>G</math> be the centroid of <math>\triangle ABC</math> so <math>\vec GA + \vec GB + \vec GC = vec 0 \implies \vec GD + \vec GE + \vec GF = vec 0 \implies G</math> is the centroid of the <math>\triangle DEF.</math> | ||
+ | |||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 13:07, 12 June 2023
Contents
[hide]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
vladimir.shelomovskii@gmail.com, vvsss
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 vladimir.shelomovskii@gmail.com, vvsss
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 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
vladimir.shelomovskii@gmail.com, vvsss