# Spiral similarity

## Contents

- 1 Basic information
- 2 Simple problems
- 2.1 Explicit spiral similarity
- 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 Spiral similarity in rectangle
- 2.7 Common point for 6 circles
- 2.8 Three spiral similarities
- 2.9 Superposition of two spiral similarities
- 2.10 Spiral similarity for circles
- 2.11 Remarkable point for spiral similarity
- 2.12 Remarkable point for pair of similar triangles
- 2.13 Remarkable point’s problems
- 2.14 Japan Mathematical Olympiad Finals 2018 Q2

## 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 similarity

Given two similar right triangles and Find and

**Solution**

The spiral similarity centered at with coefficient and the angle of rotation maps point to point and point to point

Therefore this similarity 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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### Spiral similarity in rectangle

Let rectangle be given. Let point

Let points and be the midpoints of segments and respectively.

Prove that

**Proof**

Let be the midpoint

is a parallelogram

and are corresponding medians of and

There is a spiral similarity centered at with rotation angle that maps to Therefore

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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 for 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 parallelogram

Denote

**Corollary**

Let points and be collinear. Then

Therefore is the crosspoint of the bisectors and

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### Remarkable point for pair of similar triangles

Let

Let the points and be the circumcenters of and

Let point be the midpoint of

The point is symmetric to with respect point

Prove:

a) point be the crosspoint of the bisectors and

b)

**Proof**

is parallelogram Denote

Similarly,

The statement that was proved in the previous section.

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### Remarkable point’s problems

**Problem 1**

Let a convex quadrilateral be given,

Let and be the midpoints of and respectively.

Circumcircles and intersect a second time at point

Prove that points and are concyclic.

**Problem 2**

Let triangle be given.

Let point lies on sideline

Denote the circumcircle of the as , the circumcircle of the as .

Let be the circumcenter of

Let circle cross sideline at point

Let the circumcircle of the cross at point

Prove that

**Problem 3**

The circles and are crossed at points and points

Let be the tangent to be the tangent to

Point is symmetric with respect to

Prove that points and are concyclic.

**Problem 4**

The circles and are crossed at points and points

Let be the tangent to be the tangent to

Points and lye on bisector of the angle

Points and lye on external bisector of the angle

Prove that and bisector are tangent to the circle

bisector is tangent to the circle

**Solutions**

Solutons are clear from diagrams.

In each case we use remarcable point as the point of bisectors crossing.

**Solution 1**

We use bisectors and . The points and are concyclic.

* Solution 2*
We use bisectors of and

**Solution 3**

We use bisectors of and .

is the circumcenter of

Circle is symmetric with respect diameter

Point is symmetric to with respect diameter

Therefore

**Solution 4**

Let

Let be midpoint be midpoint .

We need prove that and

Denote

The angle between a chord and a tangent is half the arc belonging to the chord.

is tangent to

is diameter Similarly, is diameter is tangent to

Let be the spiral similarity centered at

Points and are collinear Points and are collinear

Therefore is tangent to

Similarly, is tangent to

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### Japan Mathematical Olympiad Finals 2018 Q2

Given a scalene let and be points on lines and respectively, so that

Let be the circumcircle of and the reflection of across

Lines and meet again at and respectively.

Prove that and intersect on

**Proof**

Let be the orthocenter of

Point is symmetrical to point with respect to height

Point is symmetrical to point with respect to height

is centered at

is symmetrical with respect to heightline

is symmetrical to point with respect to height

is symmetrical to point with respect to height

The isosceles triangles

a) are concyclic.

b) is the spiral center that maps to maps to

Therefore are concyclic and are concyclic.

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