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 2017-03-08 
This post appeared in issue 05 of Chalkdust. I strongly recommend reading the rest of Chalkdust.
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.
Top: Folding a strip of paper in half four times leads to an order four dragon curve (after rounding the corners). Bottom: A level 10 dragon curve resembling a dragon.
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.
The endpoints of dragon curves of order 1 to 10 with a logarithmic spiral passing through them.
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.
Number in base -1+iComplex number
00
11
10-1+i
11(-1+i)+(1)=i
100-2i
101(-2i)+(1)=1-2i
110(-2i)+(-1+i)=-1-i
111(-2i)+(-1+i)+(1)=-i
10002+2i
1001(2+2i)+(1)=3+2i
1010(2+2i)+(-1+i)=1+3i
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.
Numbers in base -1+i of ten digits or less plotted on an Argand diagram.
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.
If you'd like to play with some dragon curves, you can download the Python code used to make the pictures here.
                        
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 2017-01-13 
I wrote this post with, and after much discussion with Adam Townsend. It also appeared on the Chalkdust Magazine blog.
Recently, Colin "IceCol" Beveridge blogged about something that's been irking him for a while: those annoying social media posts that tell you to work out a sum, such as \(3-3\times6+2\), and state that only $n$% of people will get it right (where \(n\) is quite small). Or as he calls it "fake maths".
A classic example of "fake maths".
This got me thinking about everyone's least favourite primary school acronym: BODMAS (sometimes known as BIDMAS, or PEMDAS if you're American). As I'm sure you've been trying to forget, BODMAS stands for "Brackets, (to the power) Of, Division, Multiplication, Addition, Subtraction" and tells you in which order the operations should be performed.
Now, I agree that we all need to do operations in the same order (just imagine trying to explain your working out to someone who uses BADSOM!) but BODMAS isn't the order mathematicians use. It's simply wrong. Take the sum \(4-3+1\) as an example. Anyone can tell you that the answer is 2. But BODMAS begs to differ: addition comes first, giving 0!
The problem here is that in reality, we treat addition and subtraction as equally important, so sums involving just these two operations are calculated from left-to-right. This caveat is quite a lot more to remember on top of BODMAS, but there's actually no need: Doing all the subtractions before additions will always give you the same answer as going from left-to-right. The same applies to division and multiplication, but luckily these two are in the correct order already in BODMAS (but no luck if you're using PEMDAS).
So instead of BODMAS, we should be using BODMSA. But that's unpronounceable, so instead we suggest that from now on you use MEDUSA. That's right, MEDUSA:
This is big news. MEDUSA vs BODMAS could be this year's pi vs tau... Although it's not actually the biggest issue when considering sums like \(3-3\times6+2\).
The real problem with \(3-3\times6+2\) is that it is written in a purposefully confusing and ambiguous order. Compare the following sums:
$$3-3\times6+2$$ $$3+2-3\times6$$ $$3+2-(3\times6)$$
In the latter two, it is much harder to make a mistake in the order of operations, because the correct order is much closer to normal left-to-right reading order, helping the reader to avoid common mistakes. Good mathematics is about good communication, not tricking people. This is why questions like this are "fake maths": real mathematicians would never ask them. If we take the time to write clearly, then I bet more than \(n\)% of people will be able get the correct answer.
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We use BEDMAS in Canada (Brackets, Exponents, Division, Multiplication, Addition, Subtraction) But we are taught that you do whichever comes first from left to right if they are the addition/ subtraction or multiplication/division. So it could also be BEMDAS, or BEMDSA, or BEDMSA. It just uses the order the that rolls off the tongue more.
Brodaha
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we use BOMAL - Brackets, Overs, Multiplication/Division, Addition/Subtraction, Left to Right. I agree they need to know negative numbers to fully understand and use BODMAS, BIDMAS, BEDMAS, PODMAS, PIDMAS, PEDMAS, BOMAL or MEDUSA
tiny
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If we could just teach young children about positive and negative numbers, then this wouldn't be a problem. Subtraction is just the addition of negative numbers. Division is also the multiplication of fractions. This is why BOMA/PEMA is the optimal method. I think MEDUSA is very creative, though.
Blan
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 2016-12-20 
Last week, I posted about the Christmas card I designed on the Chalkdust blog.
The card looks boring at first glance, but contains 12 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s green and 2s red will reveal a Christmassy picture.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically converted to base 3 and coloured...
#Answer (base 10)Answer (base 3)
1000000000
2000000000
3000000000
4000000000
5000000000
6000000000
7000000000
8000000000
9000000000
10000000000
11000000000
12000000000
  1. The square number larger than 1 whose square root is equal to the sum of its digits.
  2. The smallest square number whose factors add up to a different square number.
  3. The largest number that cannot be written in the form \(23n+17m\), where \(n\) and \(m\) are positive integers (or 0).
  4. Write down a three-digit number whose digits are decreasing. Write down the reverse of this number and find the difference. Add this difference to its reverse. What is the result?
  5. The number of numbers between 0 and 10,000,000 that do not contain the digits 0, 1, 2, 3, 4, 5 or 6.
  6. The lowest common multiple of 57 and 249.
  7. The sum of all the odd numbers between 0 and 66.
  8. One less than four times the 40th triangle number.
  9. The number of factors of the number \(2^{756}\)×\(3^{12}\).
  10. In a book with 13,204 pages, what do the page numbers of the middle two pages add up to?
  11. The number of off-diagonal elements in a 27×27 matrix.
  12. The largest number, \(k\), such that \(27k/(27+k)\) is an integer.
                        
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@Matthew: Thank you for the prompt response! It makes sense now and perhaps I should have read a little closer!
Dan Whitman
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@Dan Whitman: Find the difference between the original number and the reverse of the original. Call this difference \(a\). Next add \(a\) to the reverse of \(a\)...
Matthew
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In number 4 what are we to take the difference between? Do you mean the difference between the original number and its reverse? If so when you add the difference back to the reverse you simply get the original number, which is ambiguous. I am not sure what you are asking us to do here.
Dan Whitman
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 2016-03-15 
This article first appeared in issue 03 of Chalkdust. I highly recommend reading the rest of the magazine (and trying to solve the crossnumber I wrote for the issue).
It all began in December 1956, when an article about hexaflexagons was published in Scientific American. A hexaflexagon is a hexagonal paper toy which can be folded and then opened out to reveal hidden faces. If you have never made a hexaflexagon, then you should stop reading and make one right now. Once you've done so, you will understand why the article led to a craze in New York; you will probably even create your own mini-craze because you will just need to show it to everyone you know.
The author of the article was, of course, Martin Gardner.
A Christmas flexagon.
Martin Gardner was born in 1914 and grew up in Tulsa, Oklahoma. He earned a bachelor's degree in philosophy from the University of Chicago and after four years serving in the US Navy during the Second World War, he returned to Chicago and began writing. After a few years working on children's magazines and the occasional article for adults, Gardner was introduced to John Tukey, one of the students who had been involved in the creation of hexaflexagons.
Soon after the impact of the hexaflexagons article became clear, Gardner was asked if he had enough material to maintain a monthly column. This column, Mathematical Games, was written by Gardner every month from January 1956 for 26 years until December 1981. Throughout its run, the column introduced the world to a great number of mathematical ideas, including Penrose tiling, the Game of Life, public key encryption, the art of MC Escher, polyominoes and a matchbox machine learning robot called MENACE.

Life

Gardner regularly received topics for the column directly from their inventors. His collaborators included Roger Penrose, Raymond Smullyan, Douglas Hofstadter, John Conway and many, many others. His closeness to researchers allowed him to write about ideas that the general public were previously unaware of and share newly researched ideas with the world.
In 1970, for example, John Conway invented the Game of Life, often simply referred to as Life. A few weeks later, Conway showed the game to Gardner, allowing him to write the first ever article about the now-popular game.
In Life, cells on a square lattice are either alive (black) or dead (white). The status of the cells in the next generation of the game is given by the following three rules:
For example, here is a starting configuration and its next two generations:
The first three generations of a game of Life.
The collection of blocks on the right of this game is called a glider, as it will glide to the right and upwards as the generations advance. If we start Life with a single glider, then the glider will glide across the board forever, always covering five squares: this starting position will not lead to the sad ending where everything is dead. It is not obvious, however, whether there is a starting configuration that will lead the number of occupied squares to increase without bound.
Gosper's glider gun.
Originally, Conway and Gardner thought that this was impossible, but after the article was published, a reader and mathematician called Bill Gosper discovered the glider gun: a starting arrangement in Life that fires a glider every 30 generations. As each of these gliders will go on to live forever, this starting configuration results in the number of live cells perpetually increasing!
This discovery allowed Conway to prove that any Turing machine can be built within Life: starting arrangements exist that can calculate the digits of pi, solve equations, or do any other calculation a computer is capable of (although very slowly)!

Encrypting with RSA

To encode the message \(809\), we will use the public key:
$$s=19\quad\text{and}\quad r=1769$$
The encoded message is the remainder when the message to the power of \(s\) is divided by \(r$:
$$809^{19}\equiv\mathbf{388}\mod1769$$

Decrypting with RSA

To decode the message, we need the two prime factors of \(r\) (\(29\) and \(61\)). We multiply one less than each of these together:
\begin{align*} a&=(29-1)\times(61-1)\\[-2pt] &=1680. \end{align*}
We now need to find a number \(t\) such that \(st\equiv1\mod a\). Or in other words:
$$19t\equiv1\mod 1680$$
One solution of this equation is \(t=619\) (calculated via the extended Euclidean algorithm).
Then we calculate the remainder when the encoded message to the power of \(t\) is divided by \(r\):
$$388^{619}\equiv\mathbf{809}\mod1769$$

RSA

Another concept that made it into Mathematical Games shortly after its discovery was public key cryptography. In mid-1977, mathematicians Ron Rivest, Adi Shamir and Leonard Adleman invented the method of encryption now known as RSA (the initials of their surnames). Here, messages are encoded using two publicly shared numbers, or keys. These numbers and the method used to encrypt messages can be publicly shared as knowing this information does not reveal how to decrypt the message. Rather, decryption of the message requires knowing the prime factors of one of the keys. If this key is the product of two very large prime numbers, then this is a very difficult task.

Something to think about

Gardner had no education in maths beyond high school, and at times had difficulty understanding the material he was writing about. He believed, however, that this was a strength and not a weakness: his struggle to understand led him to write in a way that other non-mathematicians could follow. This goes a long way to explaining the popularity of his column.
After Gardner finished working on the column, it was continued by Douglas Hofstadter and then AK Dewney before being passed down to Ian Stewart.
Gardner died in May 2010, leaving behind hundreds of books and articles. There could be no better way to end than with something for you to go away and think about. These of course all come from Martin Gardner's Mathematical Games:
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 2015-10-21 
This post also appeared on the Chalkdust Magazine blog.
If you're like me, then you will be disappointed that all of the home nations have been knocked out of the Rugby World Cup. If you're really like me, doing some maths related to rugby will cheer you up...
The scoring system in rugby awards points in packets of 3, 5 and 7. This leads a number of interesting questions that you can find in my guest puzzle on Alex Bellos's Guardian blog. In this blog post, we will focus on another area of rugby: conversion kicking.

Conversion kicks

When a try is scored by putting the ball down behind the line, the scoring team gets to take a conversion kick. This kick must be taken in line with where the try was scored but it is up to the kicker how far away the kick should be taken. But how far back should the ball be taken to make the kick easiest?
Too close (red) and too far away (blue) will give small angles to aim at. Somewhere in the middle is needed (green).
One way to answer this question is to look to maximise the angle between the posts which the kicker will have to aim at: if the kick is taken too close to or too far from the goal line there will be a very thin angle to aim at. Somewhere between these extremes there will be a maximum angle to aim at.
When looking to maximise this angle, we can use one of the 'circle theorems' which have tormented many generations of GCSE maths students: 'angles subtended by the same arc at the circumference are equal'. This means that if a circle is drawn going through both posts, then the angle made at any point on this circle will be the same.
The angles made by the red and blue lines are equal because 'angles subtended by the same arc at the circumference are equal'.
A larger circle drawn through the posts will give a smaller angle. If a vertical line is drawn which just touches the right of the circle, then the point at which it touches the circle will be the best place on this line to take a kick. This is because any other point on the line will be on a larger circle and so make a smaller angle.
Using this method for circles of different sizes leads to the following diagram, which shows where the kick should be taken for every position a try could be scored:
The best place to take a kick?
This, however, is not the best place to take the kick.

Taking account of height

When a try is scored near the posts, the above method recommends a position from where the ball must be kicked at an impossibly steep angle to go over. To deal with this problem, we are going to have to look at the situation from the side.
When kicked, the ball will travel along a parabola (ignoring air resistance and wind as their effects will be small[citation needed]). Given a distance from the posts, there will be two angles which the ball can be kicked at and just make it over the bar. Kicking at any angle between these two will lead to a successful conversion. Again, we have an angle which we would like to maximise.
The highest (blue) and lowest (red) the ball can be kicked while still going over the bar.
However, the position where this angle is maximised is very unlikely to also maximise the angle we looked at earlier. To find the best place to kick from, we need to find a compromise point where both angles are quite big.
To do this, imagine that the kicker is standing inside a large sphere. For each point on the sphere, kicking the ball at the point will either lead to it going over or missing. We can draw a shape on the sphere so that aiming inside the shape will lead to scoring. Our sensible kicker will aim at the centre of this shape.
But our kicker will not be able to aim perfectly: there will be some random variation. We can predict that this variation will follow a Kent distribution, which is like a normal distribution but on the surface of a sphere. We can use this distribution to calculate the probability that our kicker will score. We would like to maximise this probability.
The Kent distribution can be adjusted to reflect the accuracy of the kicker. Below are the optimal kicking positions for an inaccurate, an average and a very accurate kicker.
The best place to take a kick for a bad kicker (top), an average kicker (middle) and a good kicker (bottom). All the kickers kick the ball at 30m/s.
As you might expect, the less accurate kicker should stand slightly further forwards to make it easier to aim. Perhaps surprisingly, the good kicker should stand further back when between the posts than when in line with the posts.
The model used to create these results could be further refined. Random variation in the speed of the kick could be introduced. Or the kick could be made to have more variation horizontally than vertically: there are parameters in the Kent distribution which allow this to be easily adjusted. In fact, data from players could be used to determine the best position for each player to kick from.
In addition to analysing conversions, this method could be used to determine the probability of scoring 3 points from any point on the pitch. This could be used in conjunction with the probability of scoring a try from a line-out to decide whether kicking a penalty for the posts or into touch is likely to lead to the most points.
Although estimating the probability of scoring from a line-out is a difficult task. Perhaps this will give you something to think about during the remaining matches of the tournament.
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