Only seventeen wallpapers
Tonight's idea
Look at wallpaper, a tiled floor or printed fabric, and imagine it going on forever. A wallpaper pattern repeats in two different directions: slide the whole thing by the right step and it lands exactly on itself. There must also be a shortest slide that works, which rules out a blank wall or plain stripes.
A symmetry is any move of the whole plane that lands the pattern on itself. Besides slides there are turns about a point, mirror flips across a line, and glides (a flip across a line, then a slide along that line). Together they form a group, as in Corner 32.
Sorted by their symmetry groups, wallpaper patterns come in exactly 17 kinds. And a turn centre of a wallpaper can only have order 2, 3, 4 or 6.
A turn centre has order \(n\) when the smallest turn about it that works is \(\frac{1}{n}\) of a full turn. Evgraf Fedorov proved the count of 17 in 1891. Tonight you hunt for turn centres and mirrors, and see why order 5 never appears.
Watch one: a floor of square tiles
The pattern is the lines of a square grid. Turns. About a corner where four tiles meet, a quarter turn lands the grid on itself, so that corner has order 4. So does the middle of every tile. About the midpoint of a side, a quarter turn fails (it stands that side upright, where there is no line), but a half turn works: order 2. There are no other turn centres.
Mirrors. Every grid line is a mirror. So is every line through tile middles parallel to the grid lines, and every diagonal line through corners. Four mirrors cross at each tile middle, four at each corner and two at each side midpoint.
Fold it up. The mirrors cut each tile into 8 equal triangles. Fold the plane along every mirror, like a paper fan, and the whole pattern stacks onto one triangle. Its corners are a tile middle, a tile corner and a side midpoint, with angles \(45^\circ\), \(45^\circ\) and \(90^\circ\). That is \(180^\circ \div 4\), \(180^\circ \div 4\) and \(180^\circ \div 2\): the folded-up shape remembers the orders 4, 4 and 2.
Your turn
- Pattern A is a honeycomb of regular hexagons. (a) Find the order of the turn centre at the middle of a hexagon, at a corner where three hexagons meet, and at the midpoint of a side. Mark one of each on the picture. (b) How many mirror lines pass through each of those three points? (c) The mirrors cut each hexagon into equal triangles. How many triangles, and what are the angles of each?
- Pattern B is a brick wall. Each brick is 2 long and 1 tall, and each row is shifted along by 1. (a) A half turn about the middle of a brick works. Find two other kinds of point where a half turn works, and mark them. One kind lies on the long sides. (b) Explain why no quarter turn, about any point, is a symmetry of this wall. (c) Which lines are mirrors? Is the long straight line between two rows a mirror?
- Suppose the mirrors of a wallpaper cut the plane into triangles. Where \(n\) mirrors cross at a corner of a triangle, the angle there is \(180^\circ \div n\), and the corner is a turn centre of order \(n\). (Check both against the square tiles.) Say the three corners have \(a\), \(b\) and \(c\) mirrors. (a) Use the angle sum of a triangle to show that \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\). (b) Find every solution in whole numbers with \(2 \le a \le b \le c\). (c) Which solutions are the square tiles and the honeycomb? Can a corner of such a triangle have order 5?
- Pattern C starts as a square grid. In each grid square, go round clockwise and mark the point a quarter of the way along each side, then join the four marks into a tilted square. (a) Does a quarter turn about the middle of a grid square still work? About a grid corner? (b) Any symmetry of C must land the long straight lines on each other, so it is also a symmetry of the plain grid. Explain why no mirror of the plain grid is a symmetry of C. (c) Why does this pattern show that exercise 3 cannot be the whole reason order 5 never appears?
Stretch: why never a fifth
Suppose a wallpaper had turn centres of order \(n\). Slides carry them to more centres of order \(n\), and in a wallpaper these cannot crowd as close together as you like (take that on trust). So pick two, \(A\) and \(B\), as close as any two are, at distance \(d\). Turn the pattern about \(A\) by \(\frac{1}{n}\) of a turn: \(B\) lands on a centre \(B'\). Turn about \(B\) by \(\frac{1}{n}\) of a turn the other way: \(A\) lands on a centre \(A'\). Measuring a careful drawing gives the distance from \(A'\) to \(B'\): for \(n = 3\) it is \(2d\); for \(n = 4\) it is \(d\); for \(n = 5\) about \(0.38d\); for \(n = 6\) it is 0; for \(n = 8\) about \(0.41d\). (a) Which of these \(n\) are impossible, and why? (b) Why is 0 not a problem? (c) Some tilings found by Roger Penrose in the 1970s land on themselves after a fifth of a turn about one special point. Which step of the argument fails for them?
Answers
- (a) Middle of a hexagon: order 6. Corner: order 3, because three hexagons meet there and a third of a turn swaps them round. Side midpoint: order 2. (b) Six through a middle (three through opposite corners, three through opposite side midpoints), three through a corner (each runs along one side and on through the middle of the hexagon opposite), two through a side midpoint (the line along the side and the line across it). (c) 12 triangles, with angles \(30^\circ\), \(60^\circ\), \(90^\circ\). Those are \(180^\circ \div 6\), \(180^\circ \div 3\), \(180^\circ \div 2\), matching the orders 6, 3, 2.
- (a) The midpoint of a short side, and the points on a long side a quarter of the way in from either end. Each quarter-way point is halfway between an upright joint in the row above and one in the row below. All of these have order 2. (b) A quarter turn would carry a brick lying flat to a brick standing on end, 1 wide and 2 tall, and the wall has none. (c) The upright lines through brick middles, which also run along joints in the rows above and below, and the level lines through brick middles. The line between two rows is not a mirror: a flip across it would carry each joint above the line to the spot directly below, but the joints below are shifted along by 1. The quarter-way points in (a) lie on no mirror at all.
- (a) The angles add to \(180^\circ\), so \(\frac{180}{a} + \frac{180}{b} + \frac{180}{c} = 180\). Divide by 180. (b) If \(a\) were 4 or more, the sum would be at most \(\frac{3}{4}\), so \(a\) is 2 or 3. With \(a = 2\) you need \(\frac{1}{b} + \frac{1}{c} = \frac{1}{2}\): \(b = 2\) leaves nothing for \(\frac{1}{c}\), \(b = 3\) gives \(c = 6\), \(b = 4\) gives \(c = 4\), and a larger \(b\) would need \(c \lt b\). With \(a = 3\) you need \(\frac{1}{b} + \frac{1}{c} = \frac{2}{3}\), so \(b = c = 3\). The solutions are \((2, 3, 6)\), \((2, 4, 4)\) and \((3, 3, 3)\). (c) The square tiles are \((2, 4, 4)\) and the honeycomb is \((2, 3, 6)\). The third, with three angles of \(60^\circ\), is a kaleidoscope of three mirrors in a triangle. No solution contains a 5, so no.
- (a) Yes to both. A quarter turn about the middle of a grid square turns its tilted square onto itself. A quarter turn about a grid corner carries each grid square to a neighbour, and the marks stay a quarter of the way along going clockwise, because a turn does not change clockwise into anticlockwise. (b) A flip does change clockwise into anticlockwise. It would carry a tilted square to one whose marks are a quarter of the way along going anticlockwise, and C has none. (c) C has turn centres of order 4 and order 2, but no mirrors and so no mirror triangles. Exercise 3 says nothing about wallpapers like this one, so a full proof needs another idea.
Stretch. (a) \(n = 5\) and \(n = 8\). There \(A'\) and \(B'\) are two different centres of order \(n\) that are closer together than \(d\), but \(A\) and \(B\) were chosen as close as any two are. For \(n = 3\) and \(n = 4\) the new pair is no closer than \(d\), so nothing goes wrong. (b) Distance 0 means \(A'\) and \(B'\) are the same centre, not two centres that are too close. (c) A Penrose tiling does not repeat: no slide lands it on itself. So there is no second centre \(B\) and the argument never starts. The rule is only for patterns that repeat in two directions. For every \(n\) from 7 upwards the same distance comes out between 0 and \(d\), so those orders are impossible too. Only 2, 3, 4 and 6 are left, and you have now seen each of them.