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Take a long strip of paper. Fold it in half in the same direction a few times. Unfold it and look at the shape the edge of the paper makes. If you folded the paper \(n\) times, then the edge will make an order \(n\) dragon curve, so called because it faintly resembles a dragon. Each of the curves shown on the cover of issue 05 of Chalkdust is an order 10 dragon curve.

The dragon curves on the cover show that it is possible to tile the entire plane with copies of dragon curves of the same order. If any readers are looking for an excellent way to tile a bathroom, I recommend getting some dragon curve-shaped tiles made.

An order \(n\) dragon curve can be made by joining two order \(n-1\) dragon curves with a 90° angle between their tails. Therefore, by taking the cover's tiling of the plane with order 10 dragon curves, we may join them into pairs to get a tiling with order 11 dragon curves. We could repeat this to get tilings with order 12, 13, and so on... If we were to repeat this ad infinitum we would arrive at the conclusion that an order \(\infty\) dragon curve will cover the entire plane without crossing itself. In other words, an order \(\infty\) dragon curve is a space-filling curve.

Like so many other interesting bits of recreational maths, dragon curves were popularised by Martin Gardner in one of his Mathematical Games columns in Scientific American. In this column, it was noted that the endpoints of dragon curves of different orders (all starting at the same point) lie on a logarithmic spiral. This can be seen in the diagram below.

Although many of their properties have been known for a long time and are well studied, dragon curves continue to appear in new and interesting places. At last year's Maths Jam conference, Paul Taylor gave a talk about my favourite surprise occurrence of a dragon.

Normally when we write numbers, we write them in base ten, with the digits in the number representing (from right to left) ones, tens, hundreds, thousands, etc. Many readers will be familiar with binary numbers (base two), where the powers of two are used in the place of powers of ten, so the digits represent ones, twos, fours, eights, etc.

In his talk, Paul suggested looking at numbers in base -1+i (where i is the square root of -1; you can find more adventures of i here) using the digits 0 and 1. From right to left, the columns of numbers in this base have values 1, -1+i, -2i, 2+2i, -4, etc. The first 11 numbers in this base are shown below.

Complex numbers are often drawn on an Argand diagram: the real part of the number is plotted on the horizontal axis and the imaginary part on the vertical axis. The diagram to the left shows the numbers of ten digits or less in base -1+i on an Argand diagram. The points form an order 10 dragon curve! In fact, plotting numbers of \(n\) digits or less will draw an order \(n\) dragon curve.

Brilliantly, we may now use known properties of dragon curves to discover properties of base -1+i. A level \(\infty\) dragon curve covers the entire plane without intersecting itself: therefore every Gaussian integer (a number of the form \(a+\text{i} b\) where \(a\) and \(b\) are integers) has a unique representation in base -1+i. The endpoints of dragon curves lie on a logarithmic spiral: therefore numbers of the form \((-1+\text{i})^n\), where \(n\) is an integer, lie on a logarithmic spiral in the complex plane.

After my talk at EMF 2014, I was sent a copy of MC Escher Kaleidocycles by Doris Schattschneider and Wallace Walker (thanks Bob!). A kaleidocycle is a bit like a 3D flexagon: it can be flexed to reveal different parts of itself.

In this blog post, I will tell you how to make a kaleidocycle from tube maps.

First, fold the cover of a tube map over. This will allow you to have the tube map (and not just its cover) on the faces of your shape.

With the side you want to see facing down, fold the map so that two opposite corners touch.

For this step, there is a choice of which two corners to connect: leading to a right-handed and a left-handed piece. You should make 6 of each type for your kaleidocycle.

Finally, fold the overhanding bits over to complete your module.

The folds you made when connecting opposite corners will need to fold both ways when you flex your shape, so it is worth folding them both ways a few times now before continuing.

Once you have made 12 modules (with 6 of each handedness), you are ready to put the kaleidocycle together.

Take two tube maps of each handedness and tuck them together in a line. Each map is tucked into one of the opposite handedness.

The four triangles across the middle form a net of a tetrahedron. Complete the tetrahedron by putting the last tab into the first triangle. Glue these together.

Take two more tube maps of the opposite handedness to those at the top of the tetrahedron. Fit them into the two triangles poking out of the top of the tetrahedron to make a second tetrahedron.

Repeat this until you have five connected tetrahedra. Finally, connect the triangles poking out of the top and the bottom to make your kaleidocycle.

Take a piece of paper. Fold it in half in the same direction many times. Now unfold it. What pattern will the folds make?

I first found this question in one of Martin Gardner's books. At first, you might that the answer will be simple, but if you look at the shapes made for a few folds, you will see otherwise:

The curves formed are called dragon curves as they allegedly look like dragons with smoke rising from their nostrils. I'm not sure I see the resemblance:

As you increase the order of the curve (the number of times the paper was folded), the dragon curve squiggles across more of the plane, while never crossing itself. In fact, if the process was continued forever, an order infinity dragon curve would cover the whole plane, never crossing itself.

This is not the only way to cover a plane with dragon curves: the curves tessellate.

Dragon curves of different orders can also fit together:

To generate digital dragon curves, first notice that an order \(n\) curve can be made from two order \(n-1\) curves:

This can easily be seen to be true if you consider folding paper: If you fold a strip of paper in half once, then \(n-1\) times, each half of the strip will have made an order \(n-1\) dragon curve. But the whole strip has been folded \(n\) times, so is an order \(n\) dragon curve.

Because of this, higher order dragons can be thought of as lots of lower order dragons tiled together. An the infinite dragon curve is actually equivalent to tiling the plane with a infinite number of dragons.

If you would like to create your own dragon curves, you can download the Python code I used to draw them from GitHub. If you are more of a thinker, then you might like to ponder what difference it would make if the folds used to make the dragon were in different directions.

Last weekend, I attended Electromagnetic Field, a camp for hackers, geeks, makers and the interested. On the Sunday, I gave a talk on four mathematical ideas/tasks which I have encountered over the past few years: Flexagons, Folding Tube Maps, Braiding and Sine Curves. I'd love to see photos, hear stories, etc from anyone who tries these activities: either comment on here or tweet @mscroggs.

It's probably best to start by showing you what a flexagon is...

What you saw there is called a trihexaflexagon. Tri- because it has three faces; -hexa- because it is a hexagon; and -flexagon because it can be flexed to reveal the other faces.

The story goes that, in 1939, Arthur H. Stone, who was an Englishman studying mathematics at Princeton, was trimming the edges off his American paper to fit in his English folder. He was fiddling with the offcuts and found that if he folded the paper under itself in a loop, he could make a hexagon; and when this hexagon was folded up as we saw, it would open out to reveal a different face.

The way it flexes can be shown on a diagram: In the circles, the colour on either side of the flexagon is shown and the lines show flexes which can be made.

When Stone showed his flexagon to other students at Harvard, they were equally amazed by it, and they formed what they called 'The Flexagon Committee'. Members of the committee included Richard Feynman, who was then still a graduate student. The committee could meet regularly and soon discovered other flexagons, the first of which was the hexahexaflexagon: Again shaped like a hexagon, but this time with six faces.

A hexahexaflexagon is created by taking a longer strip of paper and rolling it around itself like this. The shorter strip at the end is then folded and glued in the same way the trihexaflexagon was. Once made, the hexahexaflexagon can be flexed. From some positions, the flexagon can be flexed in different ways to reveal different faces. Due to this, finding some of the faces can be quite difficult.

The committee went on to find other flexagons which could be made, again made by first folding into a shorter strip, then folding up like the trihexaflexagon.

The committee later found that hexaflexagons with any number of faces could be made by starting with a certain shaped strip, rolling it up then folding it like a trihexaflexagon.

Our second story starts with me sitting on the tube reading Alex's Adventures in Numberland by Alex Bellos on the tube. In his book, Alex describes how to fold a tetrahedron, or triangle-based pyramid, from two business cards. With no business cards to hand, I picked up two tube maps and followed the steps: first, I folded it corner to corner; then I folded the overlaps over.I made another one of these, but the second a mirror image of the first, slotted them together and I had my tetrahedron.

Then I made a tube map cube by making six squares like so and slotting them together.

While making these shapes, I discovered an advantage of tube maps over business cards: Due to the pages, folded tube maps have slots to tuck the tabs into, so the solids are pretty sturdy.

Making these shapes got me wondering: what other Platonic solids could I make?

In 2D, we have regular shapes: shapes with all the sides of the same length and all angles equal. Platonic solids are sort of the 3D equivalent of this: they are 3D shapes where every face is the same regular shape and at each vertex the same number of faces meet.

For example, our tetrahedron is a Platonic solid because every side is a regular triangle, and three triangles meet at every vertex. Our cube is a Platonic solid because every side is a square (which is a regular shape) and three squares meet at every vertex.

In order to fold all the Platonic solids, we must first find out how many there are.

To do this, we're going to start with a triangle, as it is the 2D shape with the smallest number of sides, and make Platonic solids.

If we try to put two triangles at each vertex, then they'll squash flat; so that's no good. We've seen that three triangles at each vertex makes a tetrahedron. If we put four triangles at each vertex then we get an octahedron.

Five triangles at each vertex gives us an icosahedron.

Each angles in an equilateral triangle is 60°. So if we put six triangles at each vertex the angles add up to 360°, a full turn. This means that the triangles will lie flat, giving us a nice pattern for a kitchen floor, but not a solid. Any more than 6 triangles will add up to more than 360 and also not give a solid. So we have found all the Platonic solids whose faces are triangles.

Next, four sided faces. Three squares at each vertex gives us a cube. Four squares at each vertex will add up to 4 times 90°... 360° again, so another kitchen floor and as before we have all the Platonic solids whose faces are squares.

Now moving up again to five sided faces. Three pentagons at each vertex will gives us a dodecahedron, which looks like this.

This is the best I could do.

(After the talk, I was shown a few better ways to fold pentagons. Watch this space for my attempts...) Now if we try four pentagons around a vertex: the internal angle in a pentagon is 108°. 4 times 108° is 432°. This is more than a full turn, so we don't get a solid.

Moving up again, if we take three hexagons we get another tessellation. Shapes with more than six sides will all have larger angles than this so three make more than a full turn. Therefore, we have found and folded all the Platonic solids.

In 2012, I posted this on my blog and got the following comment:

I'm pretty sure this was a joke, but one hour, 48 tube maps and a lot of glue later:

A few months ago, my mother showed my a way to make braids using a cardboard octagon with a slot cut on each side and a hole in the middle.

To make a braid, seven strands of wool are tied together, fed through the hole, then one tucked into each slot.

Now, we jump over two strands, pick the third strand and move it to the vacant slot. So first, we jump over the orange and green and move the red strand.

Then we jump the light blue and yellow and move the dark blue.

And so on..

After a while, the braid looks like this:

Once I'd made a few braids, I began to wonder which other numbers of threads could be used to make braids like this. To investigate this I found it useful to represent braids by drawing connections to show where a thread is moved. This shows the first move:

Then the second move:

And so on until you get:

After the octagon, I tried braiding on a hexagon, moving the second thread each time. Here's what happened:

I only moved the yellow and green threads and nothing interesting happened. When I drew this out as before, it demonstrated what had gone wrong: three slots are missed so three threads are never moved.

So we need to find out when slots are missed and when all the slots are hit. To do this, let's call the number of slots \(a\), and let \(b\) be the number thread we pick each time. For example, in the first braid that worked \(a\) was 8 and \(b\) was 3.

First we'll label the slots. Label the slot which starts empty 0, then number anti-clockwise. This numbering puts all the multiples of \(a\) at the bottom slot.

Now let's look at which slots we visit. We start at0, then visit \(b\), then \(2b\), then \(3b\) and so on. We visit all the multiples of \(b\).

Therefore we will reach the bottom slot again and finish our loop when we reach a common multiple of \(a\) and \(b\). The first time this happens will be at the lowest common multiple, or:

On our way to this slot, we visited one slot for every \(a\) we passed, so the number of slots we have visited is

and we will visit every slot if

or, equivalently if

This is true when, \(a\) and \(b\) have no common factors, or in other words are coprime; which can be written

So we've found that if \(a\) is the number of slots and \(b\) is the jump then the braid will not work unless \(a\) and \(b\) are coprime.

For example, if \(a\) is 6 and \(b\) is 2 then 2 is a common factor so the braid fails. And, if \(a\) is 8 and \(b\) is 3 then there are no common factors and the braid works. And, if \(a\) is 12 and \(b\) is 5 then there are no common factors and the braid works.

But, if \(a\) is 5 and \(b\) is 2 then there are no common factors but the braid fails.

The rule I've explained is still correct, and explains why some braids fail. But if \(a\) and \(b\) are coprime, we need more rules to decide whether or not the braid works.

And that's as far as I've got, so I'm going to finish braiding with two open questions: Why does the 5 and 2 braid fail? And for which numbers \(a\) and \(b\) does the braid work?

For the last part of the talk, I did a practical demonstration of how to draw a sine curve using five people.

I told the first person to stand on the spot and the second person to stand one step away, hold a length of string and walk.

The third person was instructed to stay in line with the second person, while staying on a vertical line.

The fourth person was told to walk in a straight line at a constant speed.

And the fifth person had to stay in line with both the third and fourth people. This led them to trace a sine curve.

To explain why this is a sine curve, consider the following triangle:

As our first two people are one step apart, the hypotenuse of this triangle is 1. And so the opposite (vertical) side is equal to the sine of the angle.

I like to finish with a challenge, and this task leads nicely into two challenge questions:

1. How could you draw a cosine curve with five people?

2. How could you draw a tan(gent) curve with five people?

Here is a collection of Christmas relates mathematical activities.

I first encountered flexagons sometime around October 2012. Soon after, we used this template to make them at school with year 11 classes who had just taken GCSE papers as a fun but mathematical activity. The students loved them. This lead me to adapt the template for Christmas:

And here is an uncoloured version of the template on that site if you'd like to colour it yourself and a blank one if you'd like to make your own patterns:

The excitement of flexagons does not end there. There are templates around for six faced flexagons and while writing this piece, I found this page with templates for a great number of flexagons. In addition, there is a fantastic article by Martin Gardner and a two part video by Vi Hart.

I discovered the Fröbel star while searching for a picture to be the Wikipedia Maths Portal picture of the month for December 2013. I quickly found these very good instructions for making the star, although it proved very fiddly to make with paper I had cut myself. I bought some 5mm quilling paper which made their construction much easier. With a piece of thread through the middle, Fröbel starts make brilliant tree decorations.