Part III · Geometry Chapter 23 of 60

Symmetry and transformations

An ornament workshop: we draw a motif, and mirrors, rotations and shifts turn it into a rosette, a frieze and a wallpaper. Along the way it turns out that there are only four ways to move the plane without distortion, and seventeen ways to repeat a pattern over the whole plane.

Ages 16–18 55 minutes

Builds on: 22 · Vectors and coordinates

You will learn

recognise the four motions of the plane, decompose any motion into reflections and write a rotation in coordinates
identify the type of a rosette, a frieze and a wallpaper pattern from its symmetries
understand why periodic patterns can have only rotations by 180°, 120°, 90° and 60°, and how a similarity differs from a motion

Line III · Geometry

You are here: 23 · Symmetry

Interchanges 40 · Groups

In the last chapter a boat crossed a river, and every point of it shifted by the same vector. Such a shift distorts nothing: the bow stays the same distance from the stern, the mast the same height. What other ways of moving a figure change none of its distances? A butterfly’s left wing is a mirror copy of its right. A wallpaper border is the same motif repeated along a strip. Floor tiles repeat in two directions at once. It’s all one story: a figure doesn’t change if it’s moved in a certain way.

This chapter is set up as an ornament workshop. We have a motif (a curl, a flag, a footprint) and a set of ways to move it. First we’ll look at the ways themselves and prove there are only four of them. Then we’ll print the motif by rules: around a point we get a rosette, along a strip a frieze, over the whole plane a wallpaper. At the end we’ll go beyond rigid motions: stretch the motif, and turn it inside out with respect to a circle.

Four ways to move a drawing

Let’s start with the key word. If a figure is cut out of paper and put down in a new place, slid, turned or flipped over, the distances between its points don’t change. Geometers describe this as a transformation of the whole plane at once.

A motion (also called an isometry) is a transformation of the plane that preserves distances: for any points $X$ and $Y$, the distance between their images $X'$ and $Y'$ equals $XY$.

The first three examples are familiar to everyone.

A translation by a vector $\vec v$ shifts each point $X$ to the point $X'$ with $\overrightarrow{XX'} = \vec v$. A rotation about a point $O$ by an angle $\varphi$ takes $X$ to the point $X'$ such that $OX' = OX$ and the angle from the ray $OX$ to the ray $OX'$ is $\varphi$ (anticlockwise if $\varphi > 0$). A reflection in a line $\ell$ takes $X$ to the point $X'$ such that $\ell$ is the perpendicular bisector of the segment $XX'$; the points of the line $\ell$ itself stay put.

A rotation by $180^\circ$ comes up so often that it has a name of its own.

A half-turn (point reflection) about a point $O$ is the rotation by $180^\circ$ about $O$: the point $X$ goes to $X'$, and $O$ is the midpoint of the segment $XX'$.

The fourth way is harder to spot, although you’ve seen it on every beach. Walk on wet sand: left footprint, right, left, right. The right footprint is a mirror copy of the left, reflected in the line of your path and moved forward half a step.

A glide reflection is a reflection in a line $\ell$ followed by a translation along the same line by a non-zero vector. The line $\ell$ is called its axis.

Motions come in two kinds. Translation and rotation preserve orientation: the letter R stays an R, though perhaps turned, and going round a triangle anticlockwise stays anticlockwise. Reflection and glide reflection reverse orientation: the loop of the R ends up on the left, a left footprint becomes a right one. However you move a paper figure around on the table, you can’t turn it into its mirror image: you have to turn the sheet over.

Before combining motions, let’s make sure they don’t bend lines or spoil angles.

A motion takes a line to a line, a segment to an equal segment, and a triangle to a congruent triangle. So it preserves angles too.

The idea is that “lying on one line” can be said in the language of distances. A point $B$ lies on the segment $AC$ exactly when $AB + BC = AC$: for any point off the segment, by the triangle inequality, $AB + BC > AC$. A motion preserves all three distances, so $A'B' + B'C' = A'C'$ holds too, and $B'$ lies on the segment $A'C'$. So a segment goes to a segment of the same length, and a line, which is made up of segments, to a line. Triangle $ABC$ goes to triangle $A'B'C'$ with the same side lengths; by the third congruence test they’re congruent, and congruent triangles have equal corresponding angles.

A chain of footprints in the sand: left, right, left, right. Which motion takes this pattern to itself, moving each footprint to the next?

A left footprint becomes a right one, so the orientation changes: it’s a reflection or a glide reflection. But the next footprint is ahead, not opposite, so a shift of half a step along the line of the path has been added to the reflection in it.

Two mirrors

Motions can be performed one after another. The result is again a motion: if each preserves distances, both in a row do too. Such a transformation is called a composition, as with functions. Let’s start with the simplest material: reflect twice.

Stand two mirrors at an angle, like the leaves of a dressing-table mirror: the mirror reflects the mirror, and you get several copies. Two parallel mirror walls in a lift give an endless corridor in which each next copy is shifted. And two mirrors at a right angle show you without the mirror flip: raise your right hand and the image raises its hand on the same side, like a person facing you. All three effects are explained by one theorem.

Move the mirrors $\ell$ and $m$ by their handles. The black boat is reflected in $\ell$ (dashed), then in $m$ (green). What happens to the angle of rotation if you bring the mirrors together? If you make them parallel?

A reflection in a line $\ell$ followed by a reflection in a line $m$ is:

a rotation about the point where the lines meet, by twice the angle from $\ell$ to $m$, if the lines intersect;
a translation by twice the distance between the lines, perpendicular to them, in the direction from $\ell$ to $m$, if the lines are parallel.

Follow one arbitrary point $X$: its distance to $O$ doesn’t change, and the angle is made up of two pairs of equal angles.

Let $\ell$ and $m$ meet at a point $O$ at an angle $\theta$, and let $X$ be any point. Reflect it in $\ell$. The line $\ell$ is the perpendicular bisector of the segment $XX'$, and every point of a perpendicular bisector is equidistant from the ends of the segment, so $OX' = OX$. The triangles on either side of $\ell$ are congruent, and the angle between the ray $OX$ and the line $\ell$ equals the angle between $\ell$ and the ray $OX'$; call it $\alpha$.

Reflect $X'$ in $m$. In the same way $OX'' = OX'$, and the angles between the rays $OX'$, $OX''$ and the line $m$ are equal; call them $\beta$.

The ray $OX'$ lies between $\ell$ and $m$, so the angle between the mirrors is made up of $\alpha$ and $\beta$: $\theta = \alpha + \beta$.

The ray $OX$ has turned to $OX''$ through the angle $\alpha + \alpha + \beta + \beta = 2\theta$, and the distance to $O$ hasn’t changed. So $X''$ is the image of $X$ under the rotation by $2\theta$ about $O$. The point $X$ was arbitrary, so the two reflections together are this rotation. For other positions of $X$ the angles have to be counted with signs, but the result is the same: check it on the drawing by moving $X$ and the mirror $m$.

Now follow the distances along a perpendicular to the mirrors.

Let $\ell \parallel m$, the distance between them be $d$, and $X$ be any point at distance $a$ from $\ell$. Its reflection $X'$ lies on the perpendicular to $\ell$ through $X$, at the same distance $a$ but on the other side.

The reflection in $m$ again moves the point along the same perpendicular: it’s perpendicular to $m$ too. The point $X'$ is at distance $d - a$ from $m$ (negative if it’s already beyond the mirror), and $X''$ ends up at the same distance on the other side of $m$.

In all, the point has moved along the perpendicular by $a + a + (d - a) + (d - a) = 2d$, from $\ell$ towards $m$, and this doesn’t depend on $X$. So two reflections are a translation by $2d$ perpendicular to the mirrors.

Hence the lift and the dressing-table mirror. Two parallel walls give copies shifted by twice the width of the lift. Two mirrors at an angle $\theta$ give a rotation by $2\theta$, and at $\theta = 90^\circ$ a rotation by $180^\circ$: such an image doesn’t reverse orientation, and so it’s the “right way round”. The order of the reflections matters: first in $m$, then in $\ell$, gives a rotation by the same angle but the other way.

The main consequence sounds almost paradoxical. The result depends only on the meeting point and the angle between the mirrors, or on the direction and distance if they’re parallel. So a pair of mirrors can be rotated about their common point as a single unit, and a parallel pair can be shifted, without changing the result. This trick will soon come in handy.

Two mirrors meet at an angle of $36^\circ$. Through what angle, in degrees, is the image of an object turned after being reflected first in one mirror, then in the other?

By the theorem on two reflections, the angle of rotation is twice the angle between the mirrors: $2 \cdot 36^\circ = 72^\circ$. Such rotations repeat five times before a full turn: $360^\circ : 72^\circ = 5$.

There are no other motions

Translation, rotation, reflection, glide reflection, and nothing else? This can be proved in three steps. First, that a motion is completely determined by where it sends three points.

If $A$, $B$, $C$ don’t lie on one line, the image of any point $X$ under a motion is uniquely determined by the images $A'$, $B'$, $C'$. In particular, a motion that leaves $A$, $B$ and $C$ in place leaves every point in place.

A point can be found from its distances to three reference points, just as a ship is located from its distances to three lighthouses.

The image $X'$ is at the same distance from $A'$ as $X$ is from $A$: a motion preserves distances. So $X'$ lies on the circle with centre $A'$ and radius $AX$.

In the same way $X'$ lies on the circle with centre $B'$ and radius $BX$. Two circles with different centres meet in at most two points, symmetric about the line through the centres. On the drawing these are $X$ and $X^*$, symmetric about $AB$.

The third distance picks one of the two. The point $C$ doesn’t lie on the line $AB$, that is, on the perpendicular bisector of $XX^*$, so $CX \ne CX^*$. Of the two candidates only the one whose distance to $C'$ equals $CX$ will do. If $A' = A$, $B' = B$, $C' = C$, that candidate is the point $X$ itself.

Any motion of the plane can be written as a composition of at most three reflections.

Take a triangle $ABC$ and its image $A'B'C'$ under the given motion. We’ll place mirrors one at a time, each time bringing one more vertex into place without moving the ones already placed.

If $A \ne A'$, reflect everything in the perpendicular bisector of $AA'$: the vertex $A$ lands on $A'$. The other vertices move somewhere; call them $B_1$ and $C_1$.

If $B_1 \ne B'$, reflect in the perpendicular bisector of $B_1B'$. It passes through $A'$, since $A'B_1 = AB = A'B'$: the point $A'$ is equidistant from $B_1$ and $B'$. So $A'$ stays in place, and $B_1$ lands on $B'$. The third vertex moves to $C_2$.

If $C_2 \ne C'$, then $A'C_2 = A'C'$ and $B'C_2 = B'C'$: both points $A'$ and $B'$ lie on the perpendicular bisector of $C_2C'$, that is, this bisector is the line $A'B'$. Reflecting in it doesn’t touch $A'$ and $B'$ and sends $C_2$ to $C'$.

At most three reflections have sent $A$, $B$, $C$ to the same places as the given motion. By the lemma on three points, two motions that act alike on three points not on one line coincide. So our motion is the composition of these reflections.

It remains to understand what three reflections give. Zero reflections give the identity, one gives a reflection, two a rotation or a translation. The trick of rotating a pair of mirrors will deal with three.

Every motion of the plane is the identity, a translation, a rotation, a reflection or a glide reflection.

The cases of zero, one and two reflections are already done; what’s left is a composition of three. We’ll rearrange the mirrors so that the result doesn’t change, until the arrangement becomes simple. The result, the green flag, stays in place the whole time.

Let the motion be reflections first in $\ell$, then in $m$, then in $n$, and let $m$ and $n$ meet at a point $Q$ (the parallel case comes at the end).

Rotate $m$ and $n$ about $Q$ by the same angle so that $m$ becomes perpendicular to $\ell$. The angle between the mirrors and their common point haven’t changed, so by the theorem on two reflections they give the same rotation, and the whole motion is unchanged. Call the new mirrors $m'$ and $n'$.

Now $\ell \perp m'$, and the reflections in them together give a rotation by $2 \cdot 90^\circ = 180^\circ$ about their common point $P$. Any two perpendicular lines through $P$, in either order, give the same half-turn. Rotate the pair about $P$ so that one line, $q$, becomes parallel to $n'$, and the other, $p$, perpendicular to it. The motion is now reflections in $p$, then in $q$, then in $n'$.

The lines $q$ and $n'$ are parallel, and the two reflections in them are a translation perpendicular to them, that is, along the line $p$, by twice the distance between $q$ and $n'$.

The upshot: a reflection in $p$ followed by a translation along $p$, a glide reflection with axis $p$, or, if the translation is zero, simply a reflection. If $m \parallel n$ but $\ell$ and $m$ meet, the same trick is applied to the pair $\ell$, $m$: it’s rotated until $m$ becomes perpendicular to $n$. And if all three mirrors are parallel, the pair $\ell$, $m$ is shifted, without changing the translation, so that $m$ coincides with $n$; two identical reflections in a row cancel out, and a single reflection is left.

The list of motions is very short, and it’s handy to keep in view as a table.

Motion	Orientation	Fixed points	How many reflections
identity	preserved	all	0
reflection	reversed	a line	1
rotation	preserved	one point	2
translation	preserved	none	2
glide reflection	reversed	none	3

Each reflection reverses orientation, so an even number of reflections preserves it and an odd number reverses it. A rotation can’t be written as three reflections, nor a glide reflection as two.

A rotation in coordinates

The navigator of the last chapter calculated in coordinates, and motions can be translated into them too. A translation is simple: $(x, y) \mapsto (x + v_1,\; y + v_2)$. A rotation about the origin needs trigonometry.

The contribution of the first coordinate: $x$ unit vectors $\vec i$ rotated by $\varphi$, that is, $x\,(\cos\varphi, \sin\varphi)$.

The contribution of the second: $y$ rotated vectors $\vec j$, that is, $y\,(-\sin\varphi, \cos\varphi)$.

Example: at $\varphi = 90^\circ$ we have $\cos\varphi = 0$, $\sin\varphi = 1$, and the point $(3, 1)$ goes to $(-1, 3)$. At $\varphi = 60^\circ$ the point $(2, 0)$ goes to $(1, \sqrt3)$.

Rotate not the point but a rectangle whose diagonal is its position vector. The sides of the rectangle are easy to rotate.

The vector $\vec v = (x, y)$ is the diagonal of a rectangle with sides $x\,\vec i$ and $y\,\vec j$: $\vec v = x\,\vec i + y\,\vec j$.

Rotate the whole rectangle about the origin by the angle $\varphi$. A rotation is a motion, so the rectangle goes to a congruent rectangle, and its sides to $x\,R(\vec i)$ and $y\,R(\vec j)$, where $R(\vec i)$ and $R(\vec j)$ are the rotated unit vectors. The diagonal of the rotated rectangle is $R(\vec v)$, and by the parallelogram rule $R(\vec v) = x\,R(\vec i) + y\,R(\vec j)$.

The vector $R(\vec i)$ has length one and is turned from the $x$-axis by the angle $\varphi$, so by the definition of sine and cosine on the unit circle its coordinates are $(\cos\varphi, \sin\varphi)$. The vector $R(\vec j)$ is $R(\vec i)$ rotated by another $90^\circ$, and such a rotation takes $(a_1, a_2)$ to $(-a_2, a_1)$ (chapter 22). So $R(\vec j) = (-\sin\varphi, \cos\varphi)$.

Add coordinate by coordinate: $R(\vec v) = x\,(\cos\varphi, \sin\varphi) + y\,(-\sin\varphi, \cos\varphi) = (x\cos\varphi - y\sin\varphi,\;\; x\sin\varphi + y\cos\varphi)$.

A reflection in the line through the origin at an angle $\alpha$ to the $x$-axis is just as short to write: $(x, y) \mapsto (x\cos 2\alpha + y\sin 2\alpha,\;\; x\sin 2\alpha - y\cos 2\alpha)$. It’s checked the same way as the rotation: after reflection the vector $\vec i$ makes the angle $2\alpha$ with the $x$-axis and becomes $(\cos 2\alpha, \sin 2\alpha)$, and $\vec j$ makes the angle $2\alpha - 90^\circ$ and becomes $(\sin 2\alpha, -\cos 2\alpha)$. Substituting one such formula into another and using the addition formulas, we get the theorem on two reflections in numbers: a reflection in the line at angle $\alpha$ followed by one at angle $\beta$ is a rotation by $2(\beta - \alpha)$.

The point $(3, 1)$ is rotated about the origin by $90^\circ$ anticlockwise. Find the coordinates of its image.

With $\cos 90^\circ = 0$ and $\sin 90^\circ = 1$ the formula gives $x' = 3 \cdot 0 - 1 \cdot 1 = -1$, $y' = 3 \cdot 1 + 1 \cdot 0 = 3$. We get $(-1, 3)$. Check: the length $\sqrt{10}$ hasn’t changed, and the dot product $(3, 1) \cdot (-1, 3) = 0$: the angle is a right angle.

A butterfly, a snowflake, a triskelion

Now let’s start printing. Take a figure and ask which motions take it to itself.

A symmetry of a figure is a motion that takes the figure to itself. The identity is a symmetry of any figure.

A butterfly has one non-trivial symmetry: the reflection in the axis of its body. The letter S has the half-turn. The triskelion, the three running legs on the flag of the Isle of Man, has rotations by $120^\circ$ and $240^\circ$, but no mirror symmetry: a reflection would make the legs run the other way. A square has eight symmetries: four rotations and four reflections. A snowflake has twelve: six rotations and six reflections.

The set of symmetries of a figure follows rules. The composition of two symmetries is again a symmetry: if each motion puts the figure back in place, so do both in a row. The inverse of a symmetry is also a symmetry. Sets with these properties will be called groups in chapter 40, and the symmetries of the square will be the first example there. For now let’s learn to manufacture such sets.

The kaleidoscope was patented by the Scottish physicist David Brewster in 1817. The simplest kaleidoscope has, inside a tube, two mirrors at an angle of $\frac{180^\circ}{n}$ and coloured bits of glass between them. By the theorem on two reflections the mirrors give a rotation by $\frac{360^\circ}{n}$, and all the reflections together a rosette with $n$ rotations and $n$ axes.

Draw inside the circle: each stroke is repeated by all the symmetries. Turn off the mirrors: only the rotations are left, and the pattern swirls in one direction.

However much you draw, a rosette always has one of two sets: either just $n$ rotations about the centre (this set is written $C_n$) or these rotations together with $n$ reflections in axes through the centre (the set $D_n$). This isn’t a property of the kaleidoscope but a theorem.

If a figure has finitely many symmetries, they all leave one and the same point $G$ in place and form one of two sets: $C_n$, the rotations about $G$ by multiples of $\frac{360^\circ}{n}$, or $D_n$, the same rotations plus $n$ reflections in lines through $G$.

The trick is to find a common fixed point: the centre of mass of the images of one point. After that the smallest angle of rotation decides everything.

Take any point $X$ and apply all the symmetries of the figure to it. We get a finite set of points $X_1, \dots, X_k$, the images of $X$.

Let $G$ be their centre of mass, the point with the average coordinates. Any symmetry $g$ permutes these points: $g(X_i) = g(h(X))$ for some symmetry $h$, and the composition $g \circ h$ is again a symmetry, so $g(X_i)$ is in the set too. A motion takes the average of points to the average of their images: in coordinates it’s a rotation or reflection plus a translation, $g(x, y) = x\,\vec a + y\,\vec b + \vec c$, and such a formula commutes with averaging. So $g(G)$ is the centre of mass of the same points, that is, $g(G) = G$.

All the symmetries leave $G$ in place. By the classification, a motion with a fixed point is the identity, a rotation about it or a reflection in a line through it: translations and glide reflections have no fixed points. There are finitely many rotations; let $\varphi$ be the smallest positive angle among them.

Every angle of rotation $\psi$ is a multiple of $\varphi$. Otherwise $\psi = k\varphi + r$ with $0 < r < \varphi$, and the rotation by $\psi$ followed by one by $-k\varphi$ would give a symmetry, a rotation by $r$, smaller than $\varphi$. For the same reason $360^\circ$ is a multiple of $\varphi$: $\varphi = \frac{360^\circ}{n}$, and there are exactly $n$ rotations. That’s the set $C_n$.

If there’s at least one reflection $s$, then for each rotation $r$ the composition $r \circ s$ reverses orientation and leaves $G$ in place: it’s a reflection, and different rotations give different ones. Conversely, any reflection $s'$ gives a rotation $s' \circ s$ (two reflections with a common point), and then $s' = (s' \circ s) \circ s$, since $s \circ s$ is the identity. There are exactly as many reflections as rotations: $n$. That’s the set $D_n$.

How many symmetries does a regular hexagon have?

Six rotations about the centre by multiples of $60^\circ$ (including zero), and six reflections: three axes through opposite vertices and three through the midpoints of opposite sides. $12$ in all: that’s the set $D_6$, like a snowflake.

Friezes: seven ways to walk

Now let’s repeat the motif along a strip. Such band patterns are on Greek vases, embroidery on a sleeve, balcony railings and rolls of decorative ribbon.

A frieze is a pattern on an infinite strip that goes to itself under a translation along the strip. Among these translations there’s a shortest one, and all the others are its multiples.

Besides translations, a frieze can have other symmetries, and there turn out to be exactly seven combinations. John Conway suggested remembering them by footprints: each type is the gait of a person walking along the strip.

The set of symmetries of any frieze is one of seven: translations only (“hop”); a glide reflection (“step”); vertical mirrors (“sidle”); half-turns (“spinning hop”); vertical mirrors, half-turns and a glide reflection (“spinning sidle”); a horizontal mirror (“jump”); horizontal and vertical mirrors with half-turns (“spinning jump”).

Let’s find out which motions take the strip to itself at all, and then which of them can live together.

A motion that takes the strip to itself also takes its midline, the only line equidistant from the edges, to itself. A line is taken to itself only by translations along it, the reflection in the line itself (a horizontal mirror), reflections in lines perpendicular to it (vertical mirrors), half-turns about its points, and glide reflections with their axis on it. Rotations by other angles tilt the line, and reflections and glide reflections with other axes move it.

Divide the symmetries of the frieze into two heaps: the first preserves the direction along the strip (translations, the horizontal mirror, glide reflections), the second reverses it (vertical mirrors and half-turns). The composition of two motions from the second heap reverses the direction twice and lands in the first; the composition of motions from different heaps lands in the second.

The first heap comes in three kinds. Let $t$ be the length of the shortest translation. (a) Translations only. (b) There’s a horizontal mirror. (c) There’s no mirror but there’s a glide reflection; then its shift is half of $t$, up to a multiple of $t$. Indeed, a glide reflection done twice is a translation by twice the shift, which is a multiple of $t$; and if the shift itself were a multiple of $t$, adding the reverse translation would give a horizontal mirror.

The second heap is either empty or entirely recovered from one of its elements $g$: any other element $h$ of it equals $g \circ (g^{-1} \circ h)$, and $g^{-1} \circ h$ lies in the first heap. In case (a) the element $g$ is a vertical mirror or a half-turn, but not both at once: a vertical mirror and a half-turn together would give a motion from the first heap that reverses orientation, that is, a horizontal mirror or a glide reflection, and in case (a) there are none. In cases (b) and (c) a vertical mirror brings a half-turn with it and vice versa: their composition with the horizontal mirror or glide reflection gives the missing element.

Case (a) gives three sets: translations only, translations with vertical mirrors, translations with half-turns. Cases (b) and (c) give two each: without the second heap and with it. In all $3 + 2 + 2 = 7$, and all seven occur: the footprints on the buttons in the experiment below draw them.

To identify the type of a frieze, four questions asked in order are enough, like a key in a field guide to plants:

Is there a horizontal mirror? If so, with vertical mirrors it’s “spinning jump”, without them “jump”.
Are there vertical mirrors? If so, with half-turns it’s “spinning sidle”, without them “sidle”.
Are there half-turns? If so, “spinning hop”.
Is there a glide reflection? If so, “step”; if not, “hop”.

In “draw” mode choose a type and draw in the strip: the motif repeats by the type’s rules. In “identify” mode the frieze picks a random motif and type; work out the type and check against the footprints on the buttons.

Wallpaper: seventeen

Finally, let’s repeat the motif in two directions at once, like floor tiles or the pattern on wallpaper.

A wallpaper pattern (periodic pattern) is a pattern on the plane that goes to itself under translations by two non-parallel vectors. All its translations are made up of whole multiples of two shortest ones and form a lattice.

Which rotations can such a pattern have? A rotation by $90^\circ$ occurs in square tiling, by $60^\circ$ in a honeycomb, by $120^\circ$ in a triangular grid. But you can’t tile a floor with regular pentagons, and that’s no accident.

If a wallpaper pattern goes to itself under a rotation by $\frac{360^\circ}{n}$ about some point, then $n$ is $1$, $2$, $3$, $4$ or $6$.

Take the two closest centres of rotation and show that for the “forbidden” $n$ there would be even closer ones.

Call a point about which the pattern goes to itself under a rotation by $\theta = \frac{360^\circ}{n}$ a centre of order $n$. Under any translation of the pattern such a centre goes to another one like it, so there are many centres. They can’t be arbitrarily close: a rotation about one centre and the reverse rotation about another together give a translation no longer than twice the distance between them, and the pattern has no translations shorter than the shortest. So there are two closest centres of order $n$, $A$ and $B$.

Rotate the point $B$ about $A$ by $\theta$ to get $B'$. A rotation about $A$ is a symmetry of the pattern, and a symmetry takes a centre of order $n$ to a centre of the same order. So $B'$ is also a centre of order $n$.

Likewise rotate $A$ about $B$ by $-\theta$ to get a centre $A'$.

Let $AB = 1$, $A = (0, 0)$, $B = (1, 0)$. Then $B' = (\cos\theta, \sin\theta)$ and $A' = (1 - \cos\theta, \sin\theta)$, so $A'B' = |1 - 2\cos\theta|$. For $n = 5$ this is $|1 - 2\cos 72^\circ| \approx 0.38$, and for $n \ge 7$ the angle $\theta$ is less than $60^\circ$ and $0 < 2\cos\theta - 1 < 1$. In both cases $A'$ and $B'$ turn out to be different centres closer together than $A$ and $B$, which contradicts the choice of $A$ and $B$. That leaves $n = 2$, $3$, $4$, for which $A'B' \ge AB$, and $n = 6$, for which $A' = B'$.

Fivefold symmetry does occur in nature, in the starfish, the wild rose flower, many viruses. The ban applies only to patterns that repeat by translations. In 1974 Roger Penrose devised tilings made of two kinds of rhombus, with fivefold symmetry, that never repeat. And in 1982 Dan Shechtman obtained an alloy whose diffraction pattern had tenfold symmetry, impossible for a crystal. Such substances were called quasicrystals, and in 2011 Shechtman received the Nobel Prize in Chemistry for them. Penrose tilings can be built in Patterns, in the “Quasiperiodic tilings” layer.

The lattice, rotations of orders $2$, $3$, $4$, $6$, and the possible mirrors and glide reflections give exactly seventeen types of wallpaper pattern. This classification was published in 1891 by the crystallographer Evgraf Fedorov and, independently, Arthur Schoenflies; in 1924 George Pólya derived all seventeen again and drew an example of each. The proof is a long but elementary case analysis of the same kind as for friezes: you work out what the largest rotation can be, which lattice is compatible with it, and how mirrors and glide reflections are added to the rotations. In full it would take a chapter of its own, so we don’t give it here.

Choose a group and draw: the stroke is repeated by all its symmetries. The names are the international notation: the digit is the highest order of rotation, $m$ a mirror, $g$ a glide reflection, $c$ a centred lattice. The “symmetries” switch shows the mirrors, glide axes and rotation centres.

You can play further in the Patterns utility. The “Wallpaper symmetry” layer repeats any picture by one of the seventeen groups, “Kaleidoscope” builds $C_n$ and $D_n$ rosettes, and Islamic star patterns show how girih patterns grow out of a simple lattice.

Can a wallpaper pattern go to itself under a rotation by $72^\circ$?

$72^\circ = \frac{360^\circ}{5}$, and by the crystallographic restriction $n$ can only be $1$, $2$, $3$, $4$ or $6$. Regular pentagons don’t cover the plane without gaps: their angle, $108^\circ$, doesn’t divide $360^\circ$.

Similarity and homothety

Motions preserve size. Let’s drop that requirement and allow stretching, equally in all directions.

A homothety (dilation) with centre $O$ and ratio $k \ne 0$ takes each point $X$ to the point $X'$ with $\overrightarrow{OX'} = k\,\overrightarrow{OX}$.

Move the centre $O$ and change $k$. What happens at $k = 1$, at $k = -1$, and for negative $k$?

A homothety with ratio $k$ takes any segment $XY$ to a segment $X'Y'$ with $\overrightarrow{X'Y'} = k\,\overrightarrow{XY}$: it’s parallel to $XY$ (or lies on the same line) and $|k|$ times as long. So a homothety preserves angles and multiplies areas by $k^2$.

By definition $\overrightarrow{OX'} = k\,\overrightarrow{OX}$ and $\overrightarrow{OY'} = k\,\overrightarrow{OY}$. Subtract: $\overrightarrow{X'Y'} = \overrightarrow{OY'} - \overrightarrow{OX'} = k\,(\overrightarrow{OY} - \overrightarrow{OX}) = k\,\overrightarrow{XY}$; the distributive law is at work here. So the vectors $\overrightarrow{X'Y'}$ and $\overrightarrow{XY}$ are collinear, and their lengths differ by a factor of $|k|$. A triangle goes to a triangle with sides $|k|$ times as long, that is, to a similar one, and their angles are equal. A square of side $a$ goes to a square of side $|k|a$ and area $k^2a^2$; for other figures the factor $k^2$ comes out the same way as in the chapter on area, where any figure is approximated by little squares.

A similarity transformation is a transformation that multiplies all distances by the same number $k > 0$. Every such transformation is a homothety with ratio $k$ followed by a motion: if a homothety with ratio $\frac1k$ is applied after it, the distances return to what they were. Two figures are similar if one can be taken to the other by a similarity transformation; for triangles this agrees with the definition from chapter 17.

Homothety has already worked for us, though without a name. In chapter 17 the Euler line was proved by shrinking the plane by half towards the centroid $G$ and turning it half a turn: that’s the homothety with centre $G$ and ratio $-\frac12$, taking the vertices to the midpoints of the sides and the altitudes to the perpendicular bisectors. The pantograph, used since the early seventeenth century to copy drawings at another scale, is based on homothety; it was invented by the Jesuit astronomer Christoph Scheiner. So is the slide projector: the picture on the screen is the image of the slide under a homothety centred at the lens with a negative ratio, which is why slides are put in upside down.

A preview: inversion

The workshop’s last tool bends straight lines, so it’s no longer a motion. But it preserves angles and is responsible for Escher’s strangest prints.

Inversion in the circle with centre $O$ and radius $R$ takes a point $X \ne O$ to the point $X'$ on the ray $OX$ with $OX \cdot OX' = R^2$.

The points of the circle itself stay put, the inside of the disc is turned outwards and vice versa: a point at distance $\frac R2$ from the centre goes out to distance $2R$. Inversion has astonishing properties, which we’ll prove when they’re needed: it turns lines not through $O$ into circles through $O$ (minus the point $O$ itself), circles into circles or lines, and it preserves the angles between curves. In chapter 58 inversions will become the “reflections” of Lobachevsky’s geometry in the Poincaré model, inside a disc where the lines are arcs of circles perpendicular to its edge. That’s how Escher’s prints Circle Limit (1958–1960) are drawn, with fish, angels and devils shrinking towards the edge of the disc. The “Hyperbolic tilings” layer in Patterns builds similar designs.

Where next

The most symmetric figure in the plane is the circle: all the rotations about its centre and all the reflections in its diameters take it to itself, infinitely many symmetries. Put a circle at the base of a cone and cut the cone with a slanting plane. The section is an ellipse, and of the infinite set of symmetries four are left; tilt the plane more and you get a parabola, which has only one reflection. The Greeks studied these curves in order to double the cube, and two thousand years later Kepler discovered that every planet moves along one of them. Which curves come from cutting a cone, and why they matter so much: chapter 24.