Difference between revisions of "Spiral similarity"
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The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane. | The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane. | ||
[[File:Spiral center.png|450px|right]] | [[File:Spiral center.png|450px|right]] | ||
+ | [[File:Spiral center 3.png|450px|right]] | ||
Let <math>A' = T(A), B' = T(B),</math> with corresponding complex numbers <math>a', a, b',</math> and <math>b,</math> so | Let <math>A' = T(A), B' = T(B),</math> with corresponding complex numbers <math>a', a, b',</math> and <math>b,</math> so | ||
<cmath>a' = T(a) = x_0 + k (a - x_0), b' = T(b) = x_0+ k (b-x_0) \implies</cmath> | <cmath>a' = T(a) = x_0 + k (a - x_0), b' = T(b) = x_0+ k (b-x_0) \implies</cmath> | ||
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<cmath>x_0=\frac {ab' - ba' }{a-a'+b' -b}, a' - a \ne b' - b.</cmath> | <cmath>x_0=\frac {ab' - ba' }{a-a'+b' -b}, a' - a \ne b' - b.</cmath> | ||
− | Any line segment <math>AB</math> can be mapped into any other <math>A'B'</math> using the spiral similarity. Notation is shown on the diagram. | + | <i><b>Case 1</b></i> Any line segment <math>AB</math> can be mapped into any other <math>A'B'</math> using the spiral similarity. Notation is shown on the diagram. |
<math>P = AB \cap A'B'.</math> | <math>P = AB \cap A'B'.</math> | ||
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<math>\arg(k) =\angle APA'=\angle Ax_0A' =\angle Bx_0B' =\angle Cx_0C'</math> is the angle of rotation. | <math>\arg(k) =\angle APA'=\angle Ax_0A' =\angle Bx_0B' =\angle Cx_0C'</math> is the angle of rotation. | ||
+ | <cmath>\triangle AA'x_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.</cmath> | ||
+ | |||
+ | <i><b>Case 2</b></i> Any line segment <math>AB</math> can be mapped into any other <math>BB'</math> using the spiral similarity. Notation is shown on the diagram. <math>P = B = AB \cap BB'. \Omega </math> is circle <math>ABB,</math> (so it is tangent to <math>BB'), \omega </math> is circle tangent to <math>AB, x_0 = \Omega \cap \omega, x_0 \neq B, C </math> is any point of <math>AB, \theta </math> is circle <math>CBx_0, C' = \theta \cap BB'</math> is the image <math>C</math> under spiral symilarity centered at <math>x_0, |k| = \frac {BB'}{AB}</math> is the dilation factor, <math>\angle Ax_0B = \arg(k)</math> is the angle of rotation. | ||
+ | <cmath>\triangle ABx_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.</cmath> | ||
+ | |||
==Hidden spiral symilarity== | ==Hidden spiral symilarity== | ||
[[File:1932a Pras.png|400px|right]] | [[File:1932a Pras.png|400px|right]] |
Revision as of 13:42, 11 June 2023
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 it 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.
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