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2016-03-31
Pythagoras's Theorem is perhaps the most famous theorem in maths. It is also very old, and for over 2500 years mathematicians have been explaining why it is true.
This has led to hundreds of different proofs of the theorem. Many of them were collected in the 1920s in The pythagorean proposition by Elisha Scott Loomis [1]. Let's have a look at some of them.
Using similar triangles
For our first proof, start with a right angled triangle, \(ABC\), with sides of lengths \(a\), \(b\) and \(c\).
![](javascript:showlimage("pyth0.png"))
Add a point \(D\) on the hypotenuse such that the line \(AD\) is perpendicular to \(BC\). Name the lengths as shown in the second diagram.
![](javascript:showlimage("pyth1.png"))
![](javascript:showlimage("pyth2.png"))
\(ABC\) and \(DBA\) are similar triangles, so:
$$\frac{b}{x}=\frac{c}{b}$$
$$b^2=xc$$
\(ABC\) and \(DAC\) are similar triangles, so:
$$\frac{a}{c-x}=\frac{c}{a}$$
$$a^2=c^2-cx$$
Adding the two equations gives:
$$a^2+b^2=c^2$$
Constructing a quadrilateral
This proof shows the theorem is true by using extra lines and points added to the triangle. Start with \(ABC\) as before then add a point \(D\) such that \(AD\) and \(BC\) are perpendicular and of equal length. Add points \(E\) on \(AC\) and \(F\) on \(AB\) (extended) such that \(DE\) and \(AC\) are perpendicular and \(DF\) and \(AB\) are perpendicular.
![](javascript:showlimage("pyth3.png"))
By similar triangles, it can be seen that \(DF=b\) and \(DE=a\).
As the two diagonals of \(BACD\) are perpendicular, its area is \(\tfrac12c^2\).
![](javascript:showlimage("pyth4.png"))
The quadrilateral \(BACD\).
The area of \(BACD\) is also equal to the sum of the areas of \(ABD\) and \(ACD\). The area of \(ABD\) is \(\tfrac12b^2\). The area of \(ACD\) is \(\tfrac12a^2\).
![](javascript:showlimage("pyth5.png"))
The triangles \(ABD\) and \(ACD\).
Therefore, \(\tfrac12a^2+\tfrac12b^2=\tfrac12c^2\), which implies that \(a^2+b^2=c^2\).
Using a circle
This proof again uses extra stuff: this time using a circle. Draw a circle of radius \(c\) centred at \(C\). Extend \(AC\) to \(G\) and \(H\) and extend \(AB\) to \(I\).
![](javascript:showlimage("pyth6.png"))
![](javascript:showlimage("pyth6-zoomed.png"))
By the intersecting chord theorem, \(AH\times AG = AB\times AI\). Using the facts that \(AI=AB\) and \(CH\) and \(CG\) are radii, the following can be obtained from this:
$$(c-a)\times(c+a)=b\times b$$
$$c^2-a^2=b^2$$
$$a^2+b^2=c^2$$
Rearrangement proofs
A popular method of proof is dissecting the smaller squares and rearranging the pieces to make the larger square. In both the following, the pieces are coloured to show which are the same:
![](javascript:showlimage("pyth-re1.png"))
![](javascript:showlimage("pyth-re2.png"))
Alternatively, the theorem could be proved by making copies of the triangle and moving them around. This proof was presented in The pythagorean proposition simply with the caption "LOOK":
![](javascript:showlimage("pyth-re3.png"))
Moving proof
This next proof uses the fact that two parallelograms with the same base and height have the same area: sliding the top side horizontally does not change the area. This allows us to move the smaller squares to fill the large square:
![](javascript:showlimage("pyth-moving1-squashed.gif"))
Using vectors
For this proof, start by labelling the sides of the triangle as vectors \(\alpha\), \(\beta\) and \(\gamma\).
![](javascript:showlimage("pyth7.png"))
Clearly, \(\gamma = \alpha+\beta\). Taking the dot product of each side with itself gives:
$$\gamma\cdot\gamma = \alpha\cdot\alpha+2\alpha\cdot\beta+\beta\cdot\beta$$
\(\alpha\) and \(\beta\) are perpendicular, so \(\alpha\cdot\beta=0\); and dotting a vector with itself gives the size of the vector squared, so:
$$|\gamma|^2=|\alpha|^2+|\beta|^2$$
If you don't like any of these proofs, there are of course many, many more. Why don't you tweet me your favourite.
References
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2016-03-30
Take a piece of paper. Fold it in half in the same direction many times. Now unfold it. What pattern will the folds make?
![](javascript:showlimage("folding.png"))
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:
![](javascript:showlimage("young-dragons.png"))
Dragon curves of orders 1 to 6.
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:
![](javascript:showlimage("drawn-dragon.png"))
An order 10 dragon curve.
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.
![](javascript:showlimage("dragon-tile.png"))
When tiled, this picture demonstrates how dragon curves tessellate. For a demonstration, try obtaining infinite lives...
Dragon curves of different orders can also fit together:
![](javascript:showlimage("fd1.png"))
![](javascript:showlimage("fd2.png"))
![](javascript:showlimage("fd3.png"))
Drawing dragon curves
To generate digital dragon curves, first notice that an order \(n\) curve can be made from two order \(n-1\) curves:
![](javascript:showlimage("dragon-recursion.png"))
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.
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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.
![](javascript:showlimage("flexE.jpg"))
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:
- Any live cell with one or no live neighbours dies of loneliness;
- Any live cell with four or more live neighbours dies of overcrowding;
- Any dead cell with exactly three live neighbours becomes alive.
For example, here is a starting configuration and its next two generations:
![](javascript:showlimage("life1.jpg"))
![](javascript:showlimage("life2.jpg"))
![](javascript:showlimage("life3.jpg"))
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.
![](javascript:showlimage("glider_gun.jpg"))
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:
- Find a number base other than 10 in which 121 is a perfect square.
- Why do mirrors reverse left and right, but not up and down?
- Every square of a 5-by-5 chessboard is occupied by a knight.
- Is it possible for all 25 knights to move simultaneously in such a way that at the finish all cells are still occupied as before?
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2016-01-01
The deadline for entering the 2015 Advent calendar competition has now passed. You can see all the puzzles and their answers here.
Thank-you to everyone who took part in the competition! In total, 23 people submitted answers to the puzzles. Every one of these had correct answers to at least 18 of the puzzles. The winners are as follows:
| # | Name | Details |
| 1 | Scott | All correct at 5:00:18 GMT on Christmas Day |
| 2 | Louis de Mendonca | All correct at 5:00:32 GMT on Christmas Day |
| 3 | Alex Bolton | All correct at 5:00:34 GMT on Christmas Day |
| 4 | Martin Harris | All correct at 6:15 GMT on Christmas Day |
| 5 | Linus Hamilton | All correct at 14:12 GMT on Christmas Day |
| 6 | Zephi | All correct at 20:40 GMT on Christmas Day |
| 7 | Daniel Chiverton | All but one (5 December) correct at 5:00:24 GMT on Christmas Day |
| 8 | Jon Palin | All but one (12 December) correct at 5:00:34 GMT on Christmas Day |
| 9 | Kathryn Coffin | All but one (5 December) correct at 6:28 GMT on Christmas Day |
| 10 | Félix Breton | All but one (15 December) correct at 9:05 GMT on Christmas Day |
I will be in touch will all the entrants in the next few days and I will post pictures of prizes here once they are on their way!
I have already started working on puzzles for next year's (in fact this year's) calendar, so make sure you're back here in December...
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@stephan: It looks like the code I wrote to check the solutions were unique contained errors. This might explain why first question was very difficult to solve with logic alone.×1
Matthew
×1 Are you sure http://www.mscroggs.co.uk/puzzles/126 have unique solutions. See e.g. http://cryptarithms.awardspace.us/solv... which gives lots of solutions
stephan
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2015-11-24
This year, the front page of mscroggs.co.uk will feature an advent calendar. Behind each door, there will be a puzzle with a three digit solution.
As you solve the puzzles, your answers will be stored in a cookie. Behind the door on Christmas Day, there will be a form allowing you to submit your answers. The first person to submit the correct answers will win this array of prizes:
![](javascript:showlimage("prizes.jpg"))
The prizes include an mscroggs.co.uk t-shirt, a DVD of Full Frontal Nerdity and a beautiful mscroggs.co.uk 2015 winner's medal:
![](javascript:showlimage("medal.jpg"))
I will be adding to the pile of prizes throughout December.
The next nine people to submit the correct answers will win a winner's medal plus a smaller goody bag of prizes. If less than ten correct entries are submitted, prizes will be given to those with one incorrect answer, then those with two incorrect answers and so on.
Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT.
To win a prize, you must submit your entry before the start of 2016. Only one entry will be accepted per person. Once ten correct entries have been submitted, I will add a note here and below the calendar. If you have any questions, ask them in the comments below or on Twitter.
So once December is here, get solving! Good luck and have a very merry Christmas!
Edit: The winners and answers can now be found here.
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How many wrong answers do I have ? And when will you publish the solutions ?×3 ×3 ×2 ×3 ×3
Anonymous2
×3 ×3 ×2 ×3 ×3 I've submitted my answer! On Christmas too! It's just that I sorta crammed all of the answers today :/
Are you allowed to tell me if I am correct?×3 ×3 ×3 ×3 ×2
Are you allowed to tell me if I am correct?
Anonymous1
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