Spiral similarity
Contents
[hide]- 1 Basic information
- 2 Simple problems
- 2.1 Explicit spiral symilarity
- 2.2 Hidden spiral symilarity
- 2.3 Linearity of the spiral symilarity
- 2.4 Construction of a similar triangle
- 2.5 Center of the spiral symilarity for similar triangles
- 2.6 Common point for 6 circles
- 2.7 Three spiral similarities
- 2.8 Superposition of two spiral similarities
- 2.9 Spiral similarity for circles
- 2.10 Remarkable point of spiral similarity
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.
Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation).
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 For any points and the center of the spiral similarity taking to point is also the center of a spiral similarity taking to This fact explain existance of Miquel point.
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 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 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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Common point for 6 circles
Let and point on sideline be given. where lies on sideline and lies on sideline
Denote
Prove that circumcircles of triangles have the common point.
Proof
so there is the spiral symilarity taking to Denote the center of the center of is the secont crosspoint of circumcircles of and but this center is point so these circles contain point . Similarly for another circles.
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Three spiral similarities
Let triangle be given. The triangle is constructed using a spiral similarity of with center , angle of rotation and coefficient
A point is centrally symmetrical to a point with respect to
Prove that the spiral similarity with center , angle of rotation and coefficient taking to
Proof
Corollary
Three spiral similarities centered on the images of the vertices of the given triangle and with rotation angles equal to the angles of take to centrally symmetric to with respect to
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Superposition of two spiral similarities
Let be the spiral similarity centered at with angle and coefficient
Let be spiral similarity centered at with angle and coefficient
Let
Prove: a) is the crosspoint of bisectors and
b)
Algebraic proof
We use the complex plane Let Then Geometric proof
Denote
Then
Let be the midpoint be the point on bisector such that be the point on bisector such that Then
is the crosspoint of bisectors and
Corollary
There is another pare of the spiral similarities centered at and with angle coefficients and
In this case
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Spiral similarity for circles
1. Let circle cross circle at points and Point lies on
Spiral similarity centered at maps into
Prove that points and are collinear.
Proof
Arcs
Corollary
Let points and be collinear.
Then exist the spiral similarity centered at such that
2. Let circle cross circle at points and
Points and lie on
Let be the tangent to be the tangent to
Prove that angle between tangents is equal angle between lines and
Proof
There is the spiral similarity centered at such that Therefore angles between these lines are the same.
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Remarkable point of spiral similarity
Circles and centered at points and respectively intersect at points and Points and are collinear. Point is symmetrical to with respect to the midpoint point Prove:
a)
b)
Proof
a) cross in midpoint
b) is parallelogramm
Denote
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