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.

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 $T(x) = x_0+k (x-x_0),$ where $k$ is a complex number. The magnitude $|k|$ is the dilation factor of the spiral similarity, and the argument $\arg(k)$ 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 $A' = T(A), B' = T(B),$ with corresponding complex numbers $a', a, b',$ and $b,$ so $a' = T(a) = x_0 + k (a - x_0), b' = T(b) = x_0+ k (b-x_0) \implies$ $k = \frac {T(b) - T(a)}{b-a} = \frac {b' - a' }{b - a},$ $x_0=\frac {ab' - ba' }{a-a'+b' -b}, a' - a \ne b' - b.$

Case 1 Any line segment $AB$ can be mapped into any other $A'B'$ using the spiral similarity. Notation is shown on the diagram. $P = AB \cap A'B'.$

$\Omega$ is circle $AA'P, \omega$ is circle $BB'P, x_0 = \Omega \cap \omega, x_0 \neq P,$

$C$ is any point of $AB, \theta$ is circle $CPx_0, C' = \theta \cap A'B'$ is the image $C$ under spiral symilarity centered at $x_0.$ $\triangle AA'x_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.$

$|k| = \frac {A'B'}{AB} = \frac {A'x_0}{Ax_0} = \frac {B'x_0}{Bx_0} = \frac {C'x_0}{Cx_0}$ is the dilation factor,

$\arg(k) =\angle APA'=\angle Ax_0A' =\angle Bx_0B' =\angle Cx_0C'$ is the angle of rotation.

Case 2 Any line segment $AB$ can be mapped into any other $BB'$ using the spiral similarity. Notation is shown on the diagram. $B = AB \cap BB', \Omega$ is circle $ABB$ (so circle is tangent to $BB'), \omega$ is circle tangent to $AB, x_0 = \Omega \cap \omega, x_0 \neq B, C$ is any point of $AB, \theta$ is circle $CBx_0,$ $C' = \theta \cap BB'$ is the image $C$ under spiral symilarity centered at $x_0.$ $\triangle ABx_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.$ $|k| = \frac {BB'}{AB}$ is the dilation factor,

$\angle Ax_0B = \arg(k)$ is the angle of rotation.

Simple problems

Explicit spiral symilarity

Given two similar right triangles $ABC$ and $A'B'C, k = \frac {AC}{BC},$ $\angle ACB = 90^\circ, D = AA' \cap BB'.$ Find $\angle ADB$ and $\frac {AA'}{BB'}.$

Solution

The spiral symilarity centered at $C$ with coefficient $k$ and the angle of rotation $90^\circ$ maps point $B$ to point $A$ and point $B'$ to point $A'.$

Therefore this symilarity maps $BB'$ to $AA' \implies$ $\frac {AA'}{BB'} = k, \angle ADB = 90^\circ.$

vladimir.shelomovskii@gmail.com, vvsss

Hidden spiral symilarity

Let $\triangle ABC$ be an isosceles right triangle $(AC = BC).$ Let $S$ be a point on a circle with diameter $BC.$ The line $\ell$ is symmetrical to $SC$ with respect to $AB$ and intersects $BC$ at $D.$ Prove that $AS \perp DS.$

Proof

Denote $\angle SBC = \alpha, \angle SCB = \beta = 90^\circ - \alpha,$ $\angle SCA = \alpha, \angle BSC = 90^\circ, k = \frac {SC}{SB} = \cot \beta.$ Let $SC$ cross perpendicular to $BC$ in point $B$ at point $D'.$

Then $\frac {BC}{BD'} = \cot \beta.$

Points $D$ and $D'$ are simmetric with respect $AB,$ so $BD = BD' \implies k = \frac {SC}{SB} = \frac {BC}{BD}.$

The spiral symilarity centered at $S$ with coefficient $k$ and the angle of rotation $90^\circ$ maps $B$ to $C$ and $D$ to point $D_0$ such that $k \cdot BD_0 = BC = AC, \angle D_0CS = \angle DBS \implies D_0 = A.$

Therefore $\angle ASC = \angle DSB \implies$ $\angle ASD = \angle ASC - \angle DSC = \angle DSB - \angle DSC = \angle BSC = 90^\circ.$ vladimir.shelomovskii@gmail.com, vvsss

Linearity of the spiral symilarity

$\triangle ABF \sim \triangle BCD \sim \triangle CAE.$ Points $D,E,F$ are outside $\triangle ABC.$

Prove that the centroids of triangles $\triangle ABC$ and $\triangle DEF$ are coinsite.

Proof

Let $\vec y = T(\vec x),$ where $T$ be the spiral similarity with the rotation angle $\angle BAF= \angle CBD = \angle ACE$ and $k = \frac {|AF|}{|AB|} = \frac {|DB|}{|BC|} = \frac {|EC|}{|CA|}.$

A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore $\vec AF = T(\vec AB), \vec BD = T(\vec BC), \vec CE = T(\vec CA).$ $\vec AB + \vec BC + \vec CA = \vec 0.$ We use the property of linearity and get $\vec AF + \vec BD + \vec CE = k(\vec AB + \vec BC + \vec CA) = k \cdot \vec 0 = \vec 0.$ Let $G$ be the centroid of $\triangle ABC$ so $\vec GA + \vec GB + \vec GC = \vec 0 \implies$

$\vec GD + \vec GE + \vec GF = \vec 0 \implies G$ is the centroid of the $\triangle DEF.$

vladimir.shelomovskii@gmail.com, vvsss

Construction of a similar triangle

Let triangle $\triangle ABC$ and point $A'$ on sideline $BC$ be given. Construct $\triangle A'B'C' \sim \triangle ABC$ where $B'$ lies on sideline $AC$ and $C'$ lies on sideline $AB.$

Solution

Let $T(X)$ be the spiral symilarity centered at $A'$ with the dilation factor $k = \frac {AB}{AC}$ and rotation angle $\alpha = \angle BAC, A_1 = T(A), B_1 = T(B).$

$A_1B_1 = T(AB)$ so image of any point $C' \in AB$ lies on $A_1B_1.$ $B' \in AC \implies B' = A_1B_1 \cap AC.$ The spiral symilarity $T^{-1}(X)$ centered at $A'$ with the dilation factor $k^{-1} = \frac {AC}{AB}$ and rotation angle $-\alpha$ maps $B'$ into $C'$ and $\angle B'A'C' = \alpha, \frac {A'C'}{A'B'} = k^{-1} = \frac {AC}{AB}$ therefore the found triangle $\triangle A'B'C' \sim \triangle ABC$ is the desired one.

vladimir.shelomovskii@gmail.com, vvsss

Center of the spiral symilarity for similar triangles

Let triangle $\triangle ABC$ and point $C' (C' \ne C, C' \ne B)$ on sideline $BC$ be given. $\triangle A'B'C' \sim \triangle ABC$ where $B'$ lies on sideline $AB$ and $A'$ lies on sideline $AC.$ The spiral symilarity $T$ maps $\triangle ABC$ into $\triangle A'B'C'.$ Prove

a) $\angle AB'A' = \angle BC'B' = \angle CA'C'.$

b) Center of $T$ is the First Brocard point of triangles $\triangle ABC$ and $\triangle A'B'C'.$

Proof

a) Let $\tau(X)$ be the spiral symilarity centered at $C'$ with the dilation factor $k = \frac {AC}{BC}$ and rotation angle $\gamma = \angle ACB, A_1 = \tau(A), B_1 = \tau(B).$ $A' = A_1B_1 \cap AC, B' = \tau^{-1}(A').$

Denote $\varphi = \angle BC'B' \implies \angle CC'A' = 180^\circ - \varphi - \gamma \implies$ $\angle CA'C' = 180^\circ - \gamma - \angle CC'A' = \varphi.$ Similarly $\angle AB'A' = \varphi.$

b) It is well known that the three circumcircles $AA'B', BB'C',$ and $CC'A'$ have the common point (it is $D$ in the diagram).

Therefore $AB'DA'$ is cyclic and $\angle AB'A' = \angle ADA' =\varphi.$

Similarly, $\angle BDB' = \angle CDC' = \varphi.$

$\angle A'DC' = 180^\circ - \gamma, \angle A'DC = 180^\circ - \gamma - \varphi \implies$ $\angle ADC = \angle A'DC' = 180^\circ - \gamma.$ $\angle CAD + \angle ACD = 180^\circ - \angle ADC = \gamma = \angle ACD + \angle BCD \implies \angle CAD = \angle BCD = \psi.$

Similarly, $\angle CAD = \angle ABD = \angle BCD = \psi.$

Therefore, $D$ is the First Brocard point of $\triangle ABC.$

$AB'DA'$ is cyclic $\implies \angle A'B'D = \angle A'AD = \psi.$ Similarly, $\angle B'C'D = \angle C'A'D = \psi.$

Therefore $D$ is the First Brocard point of $\triangle A'B'C',$ and $\triangle A'DC' \sim \triangle ADC, \triangle A'DB' \sim \triangle ADB.$

Therefore the spiral symilarity $T$ maps $\triangle ABC$ into $\triangle A'B'C'$ has the center $D,$ the angle of the rotation $\varphi.$

vladimir.shelomovskii@gmail.com, vvsss

Common point for 6 circles

Let $\triangle ABC$ and point $A'$ on sideline $BC$ be given. $\triangle A'B'C' \sim \triangle ABC$ where $B'$ lies on sideline $AC$ and $C'$ lies on sideline $AB.$

Denote $D = BB' \cap CC', E = AA' \cap CC', F = BB' \cap AA'.$

Prove that circumcircles of triangles $\triangle ABF, \triangle A'B'F, \triangle BCD,$ $\triangle B'C'D, \triangle ACE, \triangle A'C'E$ have the common point.

Proof

$\triangle A'B'C' \sim \triangle ABC$ so there is the spiral symilarity $T$ taking $\triangle ABC$ to $\triangle A'B'C', T(ABC) = A'B'C'.$ Denote $O$ the center of $T.$ $A'B' = T(AB), A'B = \tau(AB'), F = AA' \cap BB' \implies$ the center of $\tau$ is the secont crosspoint of circumcircles of $\triangle ABF$ and $\triangle A'B',$ but this center is point $O,$ so these circles contain point $O$ . Similarly for another circles.