Difference between revisions of "Spiral similarity"
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===Construction of a similar triangle=== | ===Construction of a similar triangle=== | ||
[[File:1940 Pras.png|380px|right]] | [[File:1940 Pras.png|380px|right]] | ||
− | Let <math>\triangle ABC</math> and point <math>A'</math> on sideline <math>BC</math> be given. Construct <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AC</math> and <math>C'</math> lies on sideline <math>AB.</math> | + | Let triangle <math>\triangle ABC</math> and point <math>A'</math> on sideline <math>BC</math> be given. Construct <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AC</math> and <math>C'</math> lies on sideline <math>AB.</math> |
<i><b>Solution</b></i> | <i><b>Solution</b></i> | ||
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<cmath>B' \in AC \implies B' = A_1B_1 \cap AC.</cmath> | <cmath>B' \in AC \implies B' = A_1B_1 \cap AC.</cmath> | ||
The spiral symilarity <math>T^{-1}(X)</math> centered at <math>A'</math> with the dilation factor <math>k^{-1} = \frac {AC}{AB}</math> and rotation angle <math>-\alpha</math> maps <math>B'</math> into <math>C'</math> and <math>\angle B'A'C' = \alpha, \frac {A'C'}{A'B'} = k^{-1} = \frac {AC}{AB}</math> therefore the found triangle <math>\triangle A'B'C' \sim \triangle ABC</math> is the desired one. | The spiral symilarity <math>T^{-1}(X)</math> centered at <math>A'</math> with the dilation factor <math>k^{-1} = \frac {AC}{AB}</math> and rotation angle <math>-\alpha</math> maps <math>B'</math> into <math>C'</math> and <math>\angle B'A'C' = \alpha, \frac {A'C'}{A'B'} = k^{-1} = \frac {AC}{AB}</math> therefore the found triangle <math>\triangle A'B'C' \sim \triangle ABC</math> is the desired one. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Center of the spiral symilarity for similar triangles=== | ||
+ | [[File:5 133 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> | ||
+ | Prove | ||
+ | |||
+ | a) <math>\angle AB'A' = \angle BC'B' = \angle CA'C'.</math> | ||
+ | |||
+ | b) Center of <math>T</math> is the First Brocard point of triangles <math>\triangle ABC</math> and <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) Let <math>\tau(X)</math> be the spiral symilarity centered at <math>C'</math> with the dilation factor <math>k = \frac {AC}{BC}</math> and rotation angle <math>\gamma = \angle ACB, A_1 = \tau(A), B_1 = \tau(B).</math> | ||
+ | <math>A' = A_1B_1 \cap AC, B' = \tau^{-1}(A').</math> | ||
+ | |||
+ | Denote <math>\varphi = \angle BC'B' \implies</math> <math> \angle CC'A' = 180^\circ - \varphi - \gamma \implies \angle CA'C' = 180^\circ - \gamma - \angle CC'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). | ||
+ | Therefore <math>AB'DA'</math> is cyclic and <math>\angle AB'A' = \angle ADA' =\varphi.</math> | ||
+ | Similarly, <math>\angle BDB' = \angle CDC' = \varphi.</math> | ||
+ | <math>\angle A'DC' = 180^\circ - \gamma, \angle A'DC = 180^\circ - \gamma - \varphi \implies \angle ADC = \angle A'DC' = 180^\circ - \gamma.</math> | ||
+ | <math>\angle CAD + \angle ACD = 180^\circ - \angle ADC = \gamma = \angle ACD + \angle BCD \implies \angle CAD = \angle BCD.</math> | ||
+ | |||
+ | Similarly, <math>\angle CAD = \angle ABD.</math> Therefore, <math>D</math> is the First Brocard point of <math>\triangle ABC.</math> | ||
+ | <math>\triangle A'DC' \sim \triangle ADC \implies \frac {A'C'}{AC} = \frac {A'D}{AD} = \frac {C'D}{CD} \implies \angle DA'C' = \angle DAC \implies \angle DA'C' = \angle DC'B' = \angle DB'A' \implies D</math> is the First Brocard point of <math>\triangle A'B'C'.</math> | ||
+ | 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 04:24, 14 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
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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 the First Brocard point of The spiral symilarity maps into has the center the angle of the rotation
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