Spiral Similarity

Lesson · Advanced

Geometry

A spiral similarity is a geometric transformation combining a rotation and a dilation (scaling) about the same fixed point. The fixed point is called the center of spiral similarity.

A spiral similarity preserves shape but may change size and orientation. A direct spiral similarity preserves orientation; an indirect spiral similarity also involves reflection and reverses orientation.

We use the following signed similarity notation when orientation matters:

\triangle ABC\overset{+}{\sim}\triangle DEF\quad\text{(direct)},\qquad\triangle ABC\overset{-}{\sim}\triangle DEF\quad\text{(indirect)}.

Direct Spiral Similarity

A direct spiral similarity centered at $P$ maps

A\longrightarrow A',\qquad B\longrightarrow B',\qquad C\longrightarrow C'.

All three vertices are rotated through the same directed angle, and their distances from $P$ are multiplied by the same positive factor.

1. Equal ratios

\frac{PA'}{PA}=\frac{PB'}{PB}=\frac{PC'}{PC}=k,\qquad k>0.

Here $k$ is the scaling factor.

2. Equal directed rotation angles

\angle APA'\equiv\angle BPB'\equiv\angle CPC'\pmod{360^\circ}.

All three directed angles are measured with the same orientation.

3. Similar triangles

The transformation preserves angles and multiplies every length by the same factor. Hence

\triangle ABC\overset{+}{\sim}\triangle A'B'C',

with corresponding sides satisfying

\frac{A'B'}{AB}=\frac{B'C'}{BC}=\frac{C'A'}{CA}=k.

Important Observation

The fact that $\triangle ABC\sim\triangle A'B'C'$ alone does not prove that a given point $P$ is their spiral similarity center. We must verify that the corresponding vertices are related by one common rotation angle and one common scaling factor about $P$ .

Indirect Spiral Similarity

Consider triangles $ABC$ , $A'B'C'$ , and $A''B''C''$ , with $P$ as the fixed point. First, a direct spiral similarity centered at $P$ maps

A\longrightarrow A',\qquad B\longrightarrow B',\qquad C\longrightarrow C'.

Next, reflection across the dotted vertical line through $P$ maps

A'\longrightarrow A'',\qquad B'\longrightarrow B'',\qquad C'\longrightarrow C''.

The composition is an indirect similarity fixing $P$ and mapping

A\longrightarrow A'',\qquad B\longrightarrow B'',\qquad C\longrightarrow C''.

This transformation reverses orientation while preserving shape.

1. Equal ratios

\frac{PA''}{PA}=\frac{PB''}{PB}=\frac{PC''}{PC}=k,\qquad k>0.

2. Reversed directed angles

Unlike direct similarity, indirect similarity reverses directed angles:

\angle A''B''C''\equiv-\angle ABC\pmod{360^\circ}.

Corresponding angles have equal magnitudes but opposite orientations.

3. Similar triangles

\triangle ABC\overset{-}{\sim}\triangle A''B''C'',

with corresponding sides satisfying

\frac{A''B''}{AB}=\frac{B''C''}{BC}=\frac{C''A''}{CA}=k.

Important Observation

An indirect similarity does not generally rotate all points through the same angle. Instead, it combines scaling, rotation, and reflection.

The composition of two direct spiral similarities is another direct similarity. It need not have a fixed center: it can be a nontrivial translation.

Suppose the first similarity has rotation angle $\theta_1$ and scaling factor $k_1>0$ , while the second has rotation angle $\theta_2$ and scaling factor $k_2>0$ . Their composition has:

1. Rotation angle

\boxed{\theta\equiv\theta_1+\theta_2\pmod{360^\circ}}

2. Scaling factor

\boxed{k=k_1k_2}

Important: The center of the resulting similarity is not necessarily the center of either original transformation. If the composition is a nontrivial translation, it has no fixed center.

Take any two distinct points $A,B$ . Suppose the first similarity maps

A\longrightarrow A',\qquad B\longrightarrow B',

and the second maps

A'\longrightarrow A'',\qquad B'\longrightarrow B''.

1. Scaling factor

Since the first transformation has scaling factor $k_1$ ,

A'B'=k_1AB.

Similarly, the second transformation has scaling factor $k_2$ , so

A''B''=k_2A'B'.

Therefore,

A''B''=k_2k_1AB.

Hence the composition has scaling factor

\boxed{k=k_1k_2}.

2. Rotation angle

The first transformation rotates every directed segment through angle $\theta_1$ :

\angle(AB,A'B')\equiv\theta_1\pmod{360^\circ}.

Similarly,

\angle(A'B',A''B'')\equiv\theta_2\pmod{360^\circ}.

Adding these directed angles gives

\begin{aligned}\angle(AB,A''B'')&\equiv\angle(AB,A'B')+\angle(A'B',A''B'')\\&\equiv\theta_1+\theta_2\pmod{360^\circ}.\end{aligned}

Therefore, the composition has rotation angle

\boxed{\theta\equiv\theta_1+\theta_2\pmod{360^\circ}}.

Both transformations preserve orientation, so their composition also preserves orientation. Thus, the composition is a direct similarity with the required scaling factor and rotation angle. $\square$

Important Observation

If $k_1k_2=1$ and $\theta_1+\theta_2\equiv0^\circ\pmod{360^\circ}$ , the resulting transformation is either a translation or the identity. Otherwise, it has a unique fixed point and is a direct spiral similarity centered at that point.

On sides $AC$ and $AB$ of triangle $ABC$ , construct externally the isosceles right triangles $DAC$ and $EAB$ , respectively, such that

\angle ADC=\angle AEB=90^\circ.

If $K$ is the midpoint of $BC$ , prove that triangle $KDE$ is an isosceles right triangle.

Triangle ABC with external right isosceles triangles DAC and EAB, and midpoint K

Construct an isosceles right triangle $BCK'$ externally to triangle $ABC$ , such that

\angle BK'C=90^\circ.

Additional right isosceles triangle BCK'

Consider a counterclockwise spiral similarity centered at $C$ , with rotation angle $45^\circ$ and scaling factor $\sqrt2$ . Under this transformation,

D\longrightarrow A,\qquad K'\longrightarrow B.

Next, consider a counterclockwise spiral similarity centered at $B$ , with rotation angle $45^\circ$ and scaling factor $1/\sqrt2$ . Under this transformation,

A\longrightarrow E,\qquad C\longrightarrow K'.

The composition of these two spiral similarities is itself a direct similarity. Its rotation angle is

45^\circ+45^\circ=90^\circ,

and its scaling factor is

\sqrt2\cdot\frac1{\sqrt2}=1.

Thus, the composition is a counterclockwise rotation through $90^\circ$ mapping

D\longrightarrow E.

To identify its center, observe that the same composition sends $K'$ first to $B$ and then to $B$ , so it sends $K'$ to $B$ , not to $K$ . Instead, use the explicit center formula for the two rotations/dilations, or equivalently the midpoint-vector relation below.

Let $J$ denote counterclockwise rotation through $90^\circ$ about the origin, and write points as vectors. Because the triangles are external and isosceles right triangles,

d=\frac{a+c}{2}+\frac12J(c-a),\qquad e=\frac{a+b}{2}+\frac12J(a-b).

The signs depend on the orientation of the figure; in the displayed configuration these are the correct outward choices. With $k=(b+c)/2$ , direct subtraction gives

e-k=J(d-k).

Therefore the composition rotating $D$ to $E$ is the counterclockwise quarter-turn about $K$ , and consequently

\boxed{KD=KE},\qquad\boxed{\angle DKE=90^\circ}.\quad\checkmark

Hence triangle $KDE$ is an isosceles right triangle.

Spiral Similarity

Direct Spiral Similarity

1. Equal ratios

2. Equal directed rotation angles

3. Similar triangles

Important Observation

Indirect Spiral Similarity

1. Equal ratios

2. Reversed directed angles

3. Similar triangles

Important Observation

1. Direct similarity (orientation-preserving)

2. Indirect similarity (orientation-reversing)

1. Rotation angle

2. Scaling factor

1. Scaling factor

2. Rotation angle

Important Observation