IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations. Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

British International School Phuket

Welcome to the British International School Phuket’s maths website. I am currently working at BISP and so I am running my site as the school’s maths resources website for both our students and students around the world.

We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.

BISP has a very proud tradition in mathematical excellence. Our students have achieved the top in Thailand awards for the Cambridge Maths IGCSEs 3 years in a row. Pictured above are our world-class maths students for this year, Minjin Kang and Natchongrat (Oy) Terdkiatkhachorn.

Please explore the site – there is a huge amount of content! Some of the most popular includes:

A large “Flipping the classroom” videos section for IB students. This covers the entire IB HL, SL and Studies syllabus.

A new School Code Challenge activity which allows students to practice their code breaking skills – each code hides the password needed to access the next level.

Over 200 ideas to help with students’ Maths Explorations – many with links to additional information to research.

This post is also inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

The challenge is to design a parachute with a big enough area to make sure that someone can land safely on the ground. How can we go about doing this? Let’s start (as in the last post) with some Newtonian maths.

Newton’s Laws:

For an object falling through the air we have:

p_{s}gV – p_{a}gV – F_{D} = p_{s}Va

p_{s} = The density of the falling object
p_{a} = The density of the air it’s falling in
F_{D} = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. p_{a} will be many magnitudes smaller than than p_{s} – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

p_{s}gV – F_{D} ≈ p_{s}Va

We now rewrite things to make it easier to substitute values in later.

p_{s}V = m, where m = mass of an object (as density x volume = mass)
This gives:

mg – F_{D} ≈ ma

and as mg = W (mass x gravitation force = weight) we can rewrite this as:

W – F_{D} ≈ (w/g)a

Now, the key information to know when looking at a parachute design is the terminal velocity that will be reached when the parachute is open – that means the maximum velocity that a parachutist will potentially be hitting the ground traveling.

Now, when a person is traveling at terminal velocity their acceleration is 0, so we can set a = 0 in the equation above to give:

W – F_{D} = 0

Now we need an equation for F_{D} (the drag force).
F_{D} = 0.5p_{a}C_{D}AU^{2}

where
p_{a} = density of the air
C_{D} = the drag coefficient
A = area of parachute
U = velocity

So

when the parachutist is traveling at their terminal velocity with the parachute open we have:

W – F_{D} = 0
W = 0.5p_{a}C_{D}AU^{2}

OK, nearly there. Next thing to consider is what is the maximum velocity we want someone to be traveling when they hit the ground. This is advised to be around 5 m/s – similar to jumping from a 2 metre ladder. Much more than this and you would risk breaking a bone (or worse!)

So we are finally ready to solve our equation. We want to find what value of A (the area of the parachute) will make us land safely.

We have:

p_{a} = 0.6kg/m^{3} (approximate density of air at 3000m)
C_{D} = 1.40 (a calculated drag coefficient for an open parachute)
U = velocity = 5m/s (this is the maximum velocity we want to want to avoid injury)
W = 100kg (we will have this as the combined weight of the parachutist and the parachute)

So,

W = 0.5p_{a}C_{D}AU^{2}
100 = 0.5(0.6)(1.40)A(5)^{2}
A = 9.5m^{2}

So if we had a circular parachute with radius 1.7m it should slow us down sufficiently for us to land safely.

Galileo: Throwing cannonballs off The Leaning Tower of Pisa

This post is inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

Galileo Galilei was an Italian mathematician and astronomer who (reputedly) conducted experiments from the top of the Tower of Pisa. He dropped various objects from in order to measure how long it took for them to reach the bottom, coming to the remarkable conclusion that the objects’ weight did not affect the speed at which it fell. But does that really mean that a feather and a cannonball would fall at the same speed? Well, yes – as long as they were dropped in a vacuum. Let’s have a look at how we can prove that.

Newton’s Laws:

For an object falling through the air we have:

p_{s}gV – p_{a}gV – F_{D} = p_{s}Va

p_{s} = The density of the falling object
p_{a} = The density of the air it’s falling in
F_{D} = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

To understand where this equation comes from we note that Newton second law (Force = mass x acceleration) is

F = ma

The LHS of our equation (p_{s}gV – p_{a}gV – F_{D}) represents the forces acting on the object and the RHS (p_{s}Va) represents mass x acceleration.

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. p_{a} will be many magnitudes smaller than than p_{s} – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

p_{s}gV – F_{D} ≈ p_{s}Va

Next, we can note that in a vacuum F_{D} (the drag force) will be 0 – as there is no air resistance. Therefore this can also be ignored to get:

p_{s}gV ≈ p_{s}Va

g ≈ a

But we have a = dU/dt where U = velocity, therefore,

g ≈ a
g ≈ dU/dt
g dt ≈ dU

and integrating both sides will give:

gt ≈ U

Therefore the velocity (U) of the falling object in a vacuum is only dependent on time and the gravitational force. In other words it is independent of the object’s mass. Amazing!

This might be difficult to believe – as it is quite unintuitive. So if you’re not convinced you can watch the video below in which Brian Cox tests this out in the world’s largest vacuum chamber.

If you liked this post you might also like:

War maths – how modeling projectiles plays an essential part in waging wars.

British International School, Phuket (BISP) and Rangsit University, one of Thailand’s largest higher education institutions, are proud to announce the launch of the 2015-16 Thailand Maths Challenge.

This is a competition aimed at students aged between 16 and 18, currently studying at school. Over 500 of Thailand’s top schools have been invited to participate: international schools, bilingual schools, Thai private schools, and state schools in order to find the best young mathematicians in the country.

BISP Headmaster Neil Richards said, “I am immensely proud that the school has initiated the Thailand Maths Challenge, most particularly because of the partnership between the international and Thai communities, and the close collaboration with Rangsit University, who are to be congratulated for their foresight in offering this wonderful opportunity.

“Mathematical ability is just as relevant now as it was before the onset of the computer age, and I hope that the Thailand Maths Challenge will inspire young people to pursue mathematics at the highest possible level.”

The initial round will be conducted at schools where teachers will choose the students who will represent them. The best students from this round will then be invited to a finalists day at Rangsit University, during which the overall winner of the Thailand Maths Challenge will be announced.

Taking part in this competition will offer a very impressive addition to a student’s CV, and the overall prize for the competition will be the incredible offer of a full scholarship to Rangsit University. We look forward to your participation.

You can see further information about the competition here:

For those students studying the IB Maths Higher Level Calculus option, I’ve just finished putting together video playlists to cover the whole option syllabus. These include both videos teaching the course content and also worked past paper solutions. Hopefully this should make what is a demanding unit a little bit more manageable.

Calculus Option Part 1:

Limits, limits at infinity, determining if limits exist, differentiating from first principles, L’Hopital’s Rule, Squeeze theorem, Rolle’s theorem, Mean Value theorem.

Calculus Option Part 2:

differential equations, solving differential equations through separating variables, substitution and integrating factor, sketching slope fields.

Calculus Option Part 3:

Improper integrals, Comparison test, Riemann sums, divergence test, comparison test, limit comparison test, alternating series test, absolute convergence, power series, Ratio test.

Calculus Option Part 4:

Power series, Taylor and Maclaurin series,

Other playlists

You can find all the playlists for the:
HL core maths content here.
SL maths here
Studies maths here.

The coastline paradox arises from the difficulty of measuring shapes with complicated edges such as those of countries like the Britain. As we try and be ever more accurate in our measurement of the British coastline, we get an ever larger answer! We can see this demonstrated below:

This first map represents an approximation of the British coastline with each line representing 200km. With this scale we arrive at an estimation of around 2400km. Yet if we take each line with length 50km we get the following:

This map now has a length of around 3400km. Indeed by choosing ever smaller measuring lengths we can make it much larger still. Coastlines have similar attributes to fractals – which are shapes which exhibit self similarity on ever smaller scales.

We can attempt to classify the dimension of fractals by using decimals. Just as 1 dimension represents a straight line and 2 dimensions represents a surface, we can have a pattern with dimension (say) 1.32. These dimensions make sense in terms of classifying fractal. A fractal with dimension close to 1 will be close to a straight line, one with a dimension close to 2 will be very “crinkly” indeed.

We can use the graph above, which was used by one of the founding fathers of fractal mathematics – Mandelbrot – to help expand his early ideas on the subject. The x axis is a log base 10 scale of the length chosen to measure the coastline in. The y axis is a log base 10 scale of the subsequent coastline length. So for example if we take our first estimate of the British coastline, i.e measurements of 200km, which achieved an estimate of 2400km – then we would plot the coordinate ( log(200), log(2400) ) For our second estimate this achieves the point (log(50), log(3400) ).

We can see that countries with steeper slopes (i.e those whose coastline greatly increases with ever smaller measuring scales) will have a more jagged coastline and so can be regarded as having a higher dimension. Mandelbrot assigned the coastline dimension as related to the gradient of the slope.

Finding the gradient of a log-log graph

However to find the gradient of the lines above is slightly complicated by the fact that we have a log-log plot. There is a formula we should use:

In the formula above, m is the gradient and F_{1} and F_{2} are the corresponding y values to x_{1} and x_{2}. So using our coordinate values ( log(200), log(2400) ) and (log(50), log(3400) ) we would get a slope of:

log(2400/3400)/log(200/50) = -0.251

We then take the absolute value of this and add 1 – which gives a coastline dimension of 1.251 for Britain’s West coast.

We can also read off the approximate values from the graph. If we take the points (1.5, 3.3) and (2.7, 3) then we have a slope of:

log(3/3.3)/log(2.7/1.5) = -0.162 which gives a coastline dimension of 1.162.

Actually, with a more accurate reading of this scale Mandelbrot arrived at a coastline dimension of 1.25 for Britain – agreeing with our previous working out.

The coastline dimensions of other countries

The coastline of the German land frontier was assigned a dimension of 1.15 – i.e it is not as jagged as that of Britain. Meanwhile below we can see the South African coast:

This has a very smooth coastline – and as such the log-log graph looks to have an almost flat gradient. As such it has a dimension of 1.02.

You might think that winning at rock, paper, scissors was purely a matter of chance – after all mathematically each outcome has the same probability. We can express the likelihood of winning in terms of a game theory grid:

It is clear that in theory you would expect to win, draw and lose with probability 1/3. However you can actually exploit human psychology to give yourself a significant edge at this game. Below is a report of a Chinese study on the psychology of game players:

Zhijian and co carried out their experiments with 360 students recruited from Zhejiang University and divided into 60 groups of six players. In each group, the players played 300 rounds of Rock-Paper-Scissors against each other with their actions carefully recorded.

As an incentive, the winners were paid in local currency in proportion to the number of their victories. To test how this incentive influenced the strategy, Zhijian and co varied the payout for different groups. If a loss is worth nothing and a tie worth 1, the winning payout varied from 1.1 to 100.

The results reveal a surprising pattern of behavior. On average, the players in all the groups chose each action about a third of the time, which is exactly as expected if their choices were random.

But a closer inspection of their behavior reveals something else. Zhijian and co say that players who win tend to stick with the same action while those who lose switch to the next action in a clockwise direction (where R → P → S is clockwise).

So, for example if person A chooses Rock and person B chooses Paper, then person B wins. Human nature therefore seems to mean that person B is more likely to stick to a winning strategy and choose Paper again, whilst person A is more likely to copy that previous winning behaviour and also choose Paper. A draw.

So you can exploit this by always moving anticlockwise i.e R → S → P. To look at our example again, person A chooses Rock and person B chooses Paper, then person B wins. This time person A follows his previous pattern and still chooses Paper, but person B exploits this new knowledge to choose Rock. Player B wins.

You can play against a Wolfram Alpha AI player here. This program will track your win percentage, and will also adapt its behavior to exploit any non-random behavior that you exhibit. Even though you may not be conscious of your biases, they probably will still be there – and the designers of this simulator are confident that the program will be beating you after about 50 games. Have a go!

There are some additional tips for winning at rock paper scissors – if you are in a single game competition then choose paper. This is because men are most likely to choose rock, and scissors are the least popular choice. Also you should try some reverse psychology and announce what you will throw. Most opponents will not believe you and modify their throw as a result.

Rock, Paper, Scissors, Lizard, Spock

You can of course make the game as complicated as you wish – the version above was popularised (though not invented by) The Big Bang Theory. The grid below shows the possible outcomes for this game:

And of course, why stop there? Below is a 15 throw version of the game

If you’ve honed your strategy then maybe you could compete in the a professional rock, paper, scissors tournament – here you can watch the final of a $50,000 Las Vegas competition.

If you liked this post you might also like:

Game Theory and Tic Tac Toe – Tic Tac Toe has already been solved using Game Theory – this topic also brings in an introduction to Group Theory.

This post builds on some of the ideas in the previous post on elliptical curves. This blog originally appeared in a Plus Maths article I wrote here. The excellent Numberphile video above expands on some of the ideas below.

On a (slightly simplified) level elliptical curves they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the curve below is an elliptical curve. This curve also has an added point at infinity though we don’t need to worry about that here. Elliptical curve cryptography is based on the difficulty in solving arithmetic problems on these curves. If you remember from the last post, we have a special way of defining the addition of 2 points.

Let’s say take 2 points A and B on y² = x³ -4x + 1. In the example we have A = (2,1) and B = (-2,1). We now want to find an answer for A + B which also is on the elliptical curve. If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve. So, we define the addition A + B through the following geometric steps.

We join up the points A and B. This line intersects the curve in one more place, C.

We then reflect the point C in the x axis. We then define this new point C’ = A + B. In this case this means that (2,1) + (-2,1) = (1/4, -1/8).

Trying another example, y² = x³ -5x + 4 (below), with A = (1,0) and B = (0,2) we have C = (3,-4) and C’ = (3,4). Therefore (1,0) + (0,2) = (3,4).

We also need to have a definition when A and B define the same point on the curve. This will give us the definition of 2A. In this case we take the tangent to the curve at that point, and then as before find the intersection of this line and the curve, before reflecting the point. This probably is easier to understand with another graph:

Here we used the graph y² = x³ -5x + 4 again. This time point A = B = (-1.2, 2.88) and we have drawn the tangent to the curve at this point, which gives point D, which is then reflected in the x axis to give D’. D’ = (2.41, -2.43). Therefore we can say 2A = D’, or 2(-1.2, 2.88) = (2.41, -2.43).

Now addition of points is defined we can see how elliptical curve cryptography works. The basic idea is that given 2 points on the curve, say A and B, it takes a huge amount of computing power to work out the value a such that aA = B. For example, say I use the curve y² = x³ -25x to encrypt, and the 2 points on the curve are A = (-4,6) and B = (1681/144 , -62279/1728). Someone who wanted to break my encryption would need to find the value a such that a(-4,6) = (1681/144 , -62279/1728). The actual answer is a =2 which we can show graphically. As we want to show that 2(-4,6) = (1681/144 , -62279/1728) , we can use the previous method of finding the tangent at the point (-4,6):

We can then check with Geogebra which shows that B’ is indeed (1681/144 , -62279/1728). When a is chosen so that it is very large, this calculation becomes very difficult to attack using brute force methods – which would require checking 2(4,-6), 3(4,-6), 4(4,-6)… until the solution (1681/144 , -62279/1728) was found.

NSA and hacking data

Elliptical curve cryptography has some advantages over RSA cryptography – which is based on the difficulty of factorising large primes – as less digits are required to create a problem of equal difficulty. Therefore data can be encoded more efficiently (and thus more rapidly) than using RSA encryption. Currently the digital currency Bitcoin uses elliptical curve cryptography, and it is likely that its use will become more widespread as more and more data is digitalised. However, it’s worth noting that as yet no-one has proved that it has to be difficult to crack elliptical curves – there may be a novel approach which is able to solve the problem in a much shorter time. Indeed many mathematicians and computer scientists are working in this field.

Government digital spy agencies like the NSA and GCHQ are also very interested in such encryption techniques. If there was a method of solving this problem quickly then overnight large amounts of encrypted data would be accessible – and for example Bitcoin currency exchange would no longer be secure. It also recently transpired that the NSA has built “backdoor” entries into some elliptical curve cryptography algorithms which have allowed them to access data that the people sending it thought was secure. Mathematics is at the heart of this new digital arms race.

If you enjoyed this post you might also like:

RSA Encryption – the encryption system which secures the internet.

Circular inversion – learn about some other geometrical transformations used in university level mathematics.

Elliptical curves are a very important new area of mathematics which have been greatly explored over the past few decades. They have shown tremendous potential as a tool for solving complicated number problems and also for use in cryptography. (This blog is based on the article I wrote for Plus Maths here).

Andrew Wiles, who solved one of the most famous maths problems of the last 400 years, Fermat’s Last Theorem, using elliptical curves. In the last few decades there has also been a lot of research into using elliptical curves instead of RSA encryption to keep data transfer safe online. So, what are elliptical curves? On a simple level they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the following is an elliptical curve:

If we’re being a bit more accurate, we also need 4a³ + 27b² ≠ 0. This stops the graph having “singular points” which cause problems with the calculations. We also have a “point at infinity” which can be thought of as an extra point added on to the usual x,y plane representing infinity. This also helps with calculations – though we don’t need to go into this in any more detail here!

What makes elliptical curves so useful is that we can create a group structure using them. Groups are very important mathematical structures – again because of their usefulness in being applied to problem solving. A group needs to have certain properties. For example, we need to be able to combine 2 members of the group to create a 3rd member which is also in the group. This is how it is done with elliptical curves:

Take 2 points A and B on y² = x³ -4x + 1. In the example we have A = (2,1) and B = (-2,1). We now want to find an answer for A + B which also is on the elliptical curve. If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve. So, we define the addition A + B through the following geometric steps.

We join up the points A and B. This line intersects the curve in one more place, C.

We then reflect the point C in the x axis. We then define this new point C’ = A + B. In this case this means that (2,1) + (-2,1) = (1/4, -1/8).

Trying another example, y² = x³ -5x + 4 (above), with A = (1,0) and B = (0,2) we have C = (3,-4) and C’ = (3,4). Therefore (1,0) + (0,2) = (3,4).

We can also have elliptical curves as a collection of discrete points. We start with the curve y² = x³ +x+1, above, and just look at the positive integer solutions mod 7. For example, when x = 1,

y² = 1³ +1 + 1

y² = 3

So this has no integer solution.

Next, when x = 2 we have:

y² = 2³ +2 +1 = 11.

However when we are working mod 7 we look at the remainder when 11 is divided by 7 (which is 4). So:

y² = 4 (mod 7)

y = 2

When x = 3 we have:

y² = 3³ +3 +1 = 31

y² = 3 (mod 7)

which has no integer solutions.

In fact, all the following coordinate points satisfy the equation (mod 7):

(2,2), (0,1), (0,6), (2,5).

Even though all this might seem very abstract, these methods of calculating points on elliptical curves form the basis of elliptical cryptography. The basic idea is that it takes computers a very long time to make these sorts of calculations – and so they can be used very effectively to encrypt data.

If you enjoyed this post you might also like:

RSA Encryption – the encryption system which secures the internet.

Circular inversion – learn about some other geometrical transformations used in university level mathematics.

The one time pad is a very secure way of encoding messages – so secure in fact that it is unbreakable as long as the agents using it don’t make any mistakes. It was a very popular method of sending coded messages during the Second World War – and persisted well into the Cold War, when secret agents on the ground in a foreign country could encode their reports without fear that they would be broken by the enemy.

The picture above is from a one time pad code sheet which was used in the Second World War. What would happen is that the secret agent would be issued with a code book with lots of pages such as the page shown above. The person who was to receive the coded message would have an identical copy of this book kept securely in the home country. Now, the secret agent would infiltrate a foreign country and start to gather data. To encode it he would choose a page from the codebook, encode his message using the method will look at in a minute, and then the receiver would once he knew which page of the codebook had been used, would be able to decipher very quickly.

The one time pad is uncrackable as long as the codebook uses truly randomly generated numbers (not easy!) and as long as each page from the codebook is only used once – and is never discovered by the enemy. Sometimes this code was cracked because agents became lazy and reused the same pages, other times because the agent was captured and their codebook discovered. Nevertheless, in the age before digital communications it was a very secure method of encoding data.

How does the one time pad work?

It’s all based on Modulo arithmetic (mod 26). This sounds complicated, but basically we just work out the remainder when we divide by 26. For example 4 is still 4 mod 26, because 4 divided by 26 is 0 remainder 4. 30 is also 4 mod 26 because 30 divided by 26 is 1 remainder 4. For negatives we just keep adding 26 until we get a positive number. eg. -10 = 16 mod 26.

Now, say I want to encode the message ” Attack”. I assign each letter in the alphabet a value between 0 and 25 (A = 0, B = 1 etc).

So Attack = 0 19 0 2 10

Next I use my one time pad. Let’s use the one pictured at the top of the page. The first 5 random numbers are 54048, so I add these in turn working in mod 26 if required.

0 + 5 = 5

19 + 4 = 23

0 + 0 = 0

2 + 4 = 6

10 + 8 = 18

Now I convert these new number back into the alphabet using A = 0, B = 1 etc as before. This gives my code word as:

FXAGS

I can then send this back home using my chosen method (e.g. in a letter perhaps hidden as the first word of every sentence etc). The person receiving the code simply has to work in reverse using the same code pad. He converts FXAGS to numbers:

5 23 0 6 18

and he notes the numbers in his code pad are 54048 so he takes away to reveal the message:

5 – 5 = 0

23 – 4 = 19

0 – 0 = 0

6 – 4 = 2

18 – 8 = 10

This reveals the hidden message, Attack!

The beauty of this code is that it can even be done without a code pad – indeed deep cover secret agents were often too scared of being caught with a code book that they used ordinary books. This works in the same way, as long as both parties have agreed in advance on what copy book to use. Say I choose to use John Le Carre’s spy classic The Honourable Schoolboy, page 1 paragraph 1. This can sit innocuously on my bookshelf and yet I can create an equivalent to a code pad by taking each word in turn to make a number. In this case the first word is Afterwards – which would give me the code pad numbers 0 5 19 4 17 22 0 17 3 18. I could then use this to encode in the same manner as above.

We’re now running a huge school code breaking competition. There are a total of 7 competitions to enter – each one with a number of codes to crack. Each time a code is cracked this gives the password to access the next clue. Students who break all codes will be entered onto the leaderboard of fame. Levels of codes range from upper KS2 to post 16 – so there should be something for everyone.

Choose your level, and good luck!

Level 1: Easy, Age 11-16. This is intended to be a gentle introduction to code breaking. Hints provided throughout.

Level 2: Easy, Age 11-16. This builds on some of the skills from the Level 1 code and includes a mixture of common codes, from Morse Codes to substitution ciphers.

Level 3: Murder mystery, Medium, Age 11-16. A murder has been committed in the Maths Department! Solve the clues to uncover the murderer, the weapon and the room. Level 4: Murder mystery, Medium, Age 11-16. Another murder has been committed! Who can solve it first?

Level 5: Medium, Age 11-16. This level combines knowledge of a variety of codes and methods of solving them. Caesar shifts, transposition codes and more.

Level 6: Hard, Age 15-18. This is a real tough challenge – not for the faint hearted! It is primarily aimed at post 16 students and you can expect to look at binary codes, RSA codes and modulo arithmetic.

Level Extra: Easy, Age 11-16. A bonus code breaking challenge in case you solve all the others!