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.

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British International School Phuket

Welcome to the British International School Phuket’s maths website. My name is Andrew Chambers and I am currently working at BISP.  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.

There are a huge amount of resources to explore – especially for students doing their IAs and for students looking for revision videos.  You may also like to try our school code breaking site – where you can compete with over 10,000 students from around the world who have made it onto our school leaderboard.

 

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IB Revision with Revision Village

There’s a really great website been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams.  I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL, SL and Studies students.

You choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021) and get the following screen:

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The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions like:

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What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s.

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The Past IB Exams section takes you to full worked solutions to each full paper.

and lastly you can also get a prediction exam for the upcoming year.

For HL students (exam in 2020) you can click here

For SL students (exam in 2020) you can click here

For Studies students (exam in 2020) you can click here

For Analysis students and Applications students (exam in 2021) you can click here

You can also download the Mathematics Studies SL Formula booklet and the Standard Level Formula booklet from here if your teachers haven’t given you a copy.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

Square Triangular Numbers

Square triangular numbers are numbers which are both square numbers and also triangular numbers – i.e they can be arranged in a square or a triangle.  The picture above (source: wikipedia) shows that 36 is both a square number and also a triangular number.  The question is how many other square triangular numbers we can find?

The equation we are trying to solve is:

a2 = 0.5(b2+b)

for some a, b as positive integers. The LHS is the formula to generate square numbers and the RHS is the formula to generate the triangular numbers.

We can start with some simple Python code (which you can run here):

for c in range(1,10001):
 for d in range(1,10001):
  if c**2 == (d**2+d)/2:
   print(c**2, c,d)

This checks the first 10000 square numbers and the first 10000 triangular numbers and returns the following:

1 1 1
36 6 8
1225 35 49
41616 204 288
1413721 1189 1681
48024900 6930 9800

i.e 1225 is the next square triangular number after 36, and can be formed as 352 or as 0.5(492+49). We can see that there are very few square triangular numbers to be found in the first 50 million numbers. The largest we found was 48,024,900 which is made by 69302 or as 0.5(98002+9800).

We can notice that the ratio between each consecutive pair of square triangular numbers looks like it converges as it gives:

36/1 = 36
1225/36 = 34.027778
41616/1225 = 33.972245
1413721/41616 = 33.970612
48024900/1413721 = 33.970564

So, let’s use this to predict that the next square triangular number will be around

48024900 x 33.9706 = 1,631,434,668.

If we square root this answer we get approximately 40391
If we solve 0.5(b2+b) = 1,631,434,668 using Wolfram we get approximately 57120.

Therefore let’s amend our code to look in this region:

for c in range(40380,40400):
 for d in range(57100,57130):
  if c**2 == (d**2+d)/2:
   print(c**2, c,d)

This very quickly finds the next solution as:

1631432881 40391 57121

This is indeed 403912 – so our approximation was very accurate. We can see that this also gives a ratio of 1631432881/48024900 = 33.97056279 which we can then use to predict that the next term will be 33.970563 x 1631432881 = 55,420,693,460. Square rooting this gives a prediction that we will use the 235,416 square number. 235,4162 gives 55,420,693,056 (using Wolfram Alpha) and this is indeed the next square triangular number.

So, using a mixture of computer code and some pattern exploration we have found a method for finding the next square triangular numbers. Clearly we will quickly get some very large numbers – but as long as we have the computational power, this method should continue to work.

Using number theory

The ever industrious Euler actually found a formula for square triangular numbers in 1778 – a very long time before computers and calculators, so let’s have a look at his method:

We start with the initial problem, and our initial goal is to rearrange it into the following form:

Next we make a substitution:

Here, when we get to the equation 1 = x2 – 2y2 we have arrived at a Pell Equation (hence the rearrangement to get to this point).  This particular Pell Equation has the solution quoted above where we can define Pk  as

 

Therefore we have

Therefore for any given k we can find the kth square triangular number.  The a value will give us the square number required and the b value will give us the triangular number required.  For example with k = 3:

This tells us the 3rd square triangular number is the 35th square number or the 49th triangular number.  Both these give us an answer of 1225 – which checking back from our table is the correct answer.

So, we have arrived at 2 possible methods for finding the square triangular numbers – one using modern computational power, and one using the skills of 18th century number theory.

Rational Approximations to Irrational Numbers

This year two mathematicians (James Maynard and Dimitris Koukoulopoulos) managed to prove a long-standing Number Theory problem called the Duffin Schaeffer Conjecture.  The problem is concerned with the ability to obtain rational approximations to irrational numbers.  For example, a rational approximation to pi is 22/7.  This gives 3.142857 and therefore approximates pi to 2 decimal places.  You can find ever more accurate rational approximations and the conjecture looks at how efficiently we can form these approximation, and to within what error bound.

Finding Rational Approximations for  pi

The general form of the inequality I want to solve is as follows:

Here alpha is an irrational number, p/q is the rational approximation, and f(q)/q can be thought of as the error bound that I need to keep my approximation within.

If I take f(q) = 1/q then I will get the following error bound:

So, the question is, can I find some values of q (where p and q are integers) such that the error bound is less than 1/(q squared)?

Let’s see if we can solve this for when our irrational number is pi, and when we choose q = 6.

We can see that this returns a rational approximation, 19/6 which only 0.02507… away from  pi. This is indeed a smaller error than 1/36.  We won’t be able to find such solutions to our inequality for every value of q that we choose, but we will be able to find an infinite number of solutions, each getting progressively better at approximating pi.

The General Case (Duffin Schaeffer Conjecture)

The general case of this problem states that there will be infinite solutions to the inequality for any given irrational number alpha if and only if the following condition holds:

For:

We will have infinitely many solutions (with p and q as integers in their lowest terms) if and only if:

Here the new symbol represents the Euler totient.  You can read about this at the link if you’re interested, but for the purposes of the post we can transform into something else shortly!

Does f(q) = 1/q provide infinite solutions?

When f(q) = 1/q we have:

Therefore we need to investigate the following sum to infinity:

Now we can make use of an equivalence, which shows that:

 

Where the new symbol on the right is the Zeta function.  The Zeta function is defined as:

So, in our case we have s = 2.  This gives:

But we know the limit of both the top and the bottom sum to infinity.  The top limit is called the Harmonic series, and diverges to infinity. Therefore:

Whereas the bottom limit is a p-series with p=2, this is known to converge. In fact we have:

Therefore because we have a divergent series divided by a convergent one, we will have the following result:

This shows that our error bound 1/(q squared) will be satisfied by infinitely many values of q for any given irrational number.

Does f(q) = 1/(q squared) provide infinite solutions?

With f(q) = 1/(q squared) we follow the same method to get:

But this time we have:

Therefore we have a convergent series divided by a convergent series which means:

So we can conclude that f(q) = 1/(q squared) which generated an error bound of 1/(q cubed) was too ambitious as an error bound – i.e there will not be infinite solutions in p and q for a given irrational number.  There may be solutions out there but they will be rare.

Understanding mathematicians 

You can watch the Numberphile video where James Maynard talks through the background of his investigation and also get an idea what a mathematician feels like when they solve a problem like this!

When do 2 squares equal 2 cubes?

Following on from the hollow square investigation this time I will investigate what numbers can be written as both the sum of 2 squares, 2 cubes and 2 powers of 4. i.e a2+b2 = c3+d3 = e4+f4.

Geometrically we can think of this as trying to find an array of balls such that we can arrange them into 2 squares, or we can rearrange them and stack them to form 2 cubes, or indeed we can arrange them into 2 4-dimensional cubes. I’ll add the constraints that all of a,b,c,d,e,f should be greater than 1 and that the pair of squares or cubes (etc) must be distinct. Therefore we can’t for example have 2 squares the same size.

Infinite solutions

Let’s look at why we can easily find infinite solutions if the squares or cubes (etc) can be the same size.

We want to find solutions to:
a2+b2 = c3+d3 = e4+f4.

so we look at the powers 2,3,4 which have LCM of 12. Therefore if we choose powers with the same base we can find a solution. For example we chose to work with base 2. Therefore we choose

a = 26, b = 26, which gives 212+212
c = 24, d = 24, which gives 212+212
e = 23, f = 23, which gives 212+212

Clearly these will be the same. So we can choose any base we wish, and make the powers into the same multiples of 12 to find infinite solutions.

Writing some code

Here is some code that will find some other solutions:


list1=[]
for a in range(2, 200):
 for b in range(2,200):
  list1.append(a**2+b**2)

list2=[]
for j in list1:
 for c in range(2,200):
  for d in range(2,200):
   if c**3+d**3 == j:
    list2.append(c**3+d**3)
print(list2)

for k in list2:
 for e in range(2,200):
  for f in range(2,200):
   if k == e**4+f**4:
    print(k,e,f)

This returns the following solutions: 8192, 18737, 76832. Of these we reject the first as this is the solution 212+212 which we found earlier and which uses repeated values for the squares, cubes and powers of 4. The 3rd solution we also reject as this is formed by 14 4 + 14 4. Therefore the only solution up to 79202 (we checked every value up to and including 1992 + 1992) is:

18737 = 642+1212 = 173+243 = 114+84.

Therefore if we had 18,737 balls we could arrange them into 2 squares, a 64×64 square and a 121×121 square. Alternatively we could rearrange them into 2 cubes, one 17x17x17 and one 24x24x24. Or we could enter a higher dimensional space and create 2 tesseracts one with sides 11x11x11x11 and the other with 14x14x14x14.

With only 1 solution for around the first 80,000 numbers it looks like these numbers are quite rare – could you find another one? And could you find one that also satisfies g5+h5?

Hollow Cubes investigation

Hollow cubes like the picture above [reference] are an extension of the hollow squares investigation done previously.  This time we can imagine a 3 dimensional stack of soldiers, and so try to work out which numbers of soldiers can be arranged into hollow cubes.

Therefore what we need to find is what numbers can be formed from a3-b3

Python code

We can write some Python3 code to find this out (this can be run here):


for k in range(1,200):

 for a in range(0, 100):
  for b in range(0,100):
   if a**3-b**3 == k :
    print(k,a,b)

This gives the following: (the first number is the number of soldiers and the 2 subsequent numbers are the 2 cubes).

1 1 0
7 2 1
8 2 0
19 3 2
26 3 1
27 3 0
37 4 3
56 4 2
61 5 4
63 4 1
64 4 0
91 6 5
98 5 3
117 5 2
124 5 1
125 5 0
127 7 6
152 6 4
169 8 7
189 6 3

We could perhaps investigate any patterns in these numbers, or explore how we can predict when a hollow cube has more than one solution. I’ll investigate which numbers can be written as both a hollow square and also a hollow cube.

Hollow squares and hollow cubes

list1=[]
for a in range(2, 50):
 for b in range(2,50):
  if a**2-b**2 !=0:
   if a**2-b**2 > 0:
    list1.append(a**2-b**2)

list2=[]
for j in list1:
 for c in range(2,50):
  for d in range(2,50):
   if c**3-d**3 == j:
    list2.append(c**3-d**3)
print(list2)

This returns the following numbers which can all be written as both hollow squares and hollow cubes.

[56, 91, 19, 117, 189, 56, 208, 189, 217, 37, 279, 152, 117, 448, 513, 504, 448, 504, 387, 665, 504, 208, 875, 819, 936, 817, 61, 999, 988, 448, 728, 513, 189, 1216, 936, 784, 335, 469, 1323, 819, 1512, 1352, 1197, 992, 296, 152, 1519, 1512, 1197, 657, 1664, 1323, 1647, 1736, 1701, 1664, 936, 504, 2107, 1387, 1216, 1027, 91, 2015, 279, 2232]

Hollow squares, cubes and hypercubes

Taking this further, can we find any number which can be written as a hollow square, hollow cube and hollow hypercube (4 dimensional cube)? This would require our soldiers to be able to be stretch out into a 4th dimensional space – but let’s see if it’s theoretically possible.

Here’s the extra code to type:

list1=[]
for a in range(2, 200):
 for b in range(2,200):
  if a**2-b**2 !=0:
   if a**2-b**2 > 0:
    list1.append(a**2-b**2)

list2=[]
for j in list1:
 for c in range(2,200):
  for d in range(2,200):
   if c**3-d**3 == j:
    list2.append(c**3-d**3)
print(list2)

for k in list2:
 for e in range(2,200):
  for f in range(2,200):
   if k == e**4-f**4:
    print(k)

Very pleasingly this does indeed find some solutions:

9919: Which can be formed as either 1002-92 or 223-93 or 104-34.

14625: Which can be formed as either 1212-42 or 253-103 or 114-24.

Given that these took some time to find, I think it’ll require a lot of computer power (or a better designed code) to find any number which is a hollow square, hollow cube, hollow hypercube and hollow 5-dimensional cube, but I would expect that there is a number out there that satisfies all criteria. Maybe you can find it?

Ramanujan’s Taxi Cabs and the Sum of 2 Cubes

The Indian mathematician Ramanujan (picture cite: Wikipedia) is renowned as one of great self-taught mathematical prodigies.  His correspondence with the renowned mathematician G. H Hardy led him to being invited to study in England, though whilst there he fell sick.  Visiting him in hospital, Hardy remarked that the taxi that had brought him to the hospital had a very “rather dull number” – number 1729.  Ramanujan remarked in reply, ” No Hardy, it’s a very interesting number!  It’s the smallest number expressible as the sum of 2 cubes in 2 different ways!”

Ramanujan was profoundly interested in number theory – the study of integers and patterns inherent within them.  The general problem referenced above is finding integer solutions to the below equation for given values of A:

In the case that A = 1729, we have 2 possible ways of finding distinct integer solutions:

The smallest number which can be formed through 3 distinct (positive) integer solutions to the equation is A = 87, 539, 319.

Although this began as a number theory problem it has close links with both graphs and group theory – and it is from these fields that mathematicians have gained a deeper understanding as to the nature of its solutions.  The modern field of elliptical curve cryptography is closely related to the ideas below and provides a very secure method of encrypting data.

We start by sketching the graph of:

For some given integer value of A. We will notice that the graph has a line of symmetry around y = x and also an asymptote at y = -x.  If we plot:

We can see that both our integer solutions to this problem (1,12) and (9,10) lie on the curve:

Group theory

Groups can be considered as sets which follow a set number of rules with regards to operations like multiplication, addition etc.  Establishing that a set is a group then allows certain properties to be inferred.  If we can establish the following rules hold then we can create an Abelian group.  If we start with a set A and and operation Θ.

1) Identity. For an element e in A, we have a Θ e = a for all a in A.

(for example 0 is the identity element for the addition operation for the set of integers numbers. a+0 = a for all a in the real numbers).

2) Closure.  For all elements a,b in A, a Θ b = c, where c is also in A.

(For example with the addition operation, the addition of 2 integers numbers is still an integer)

3) Associativity. For all elements a,b,c in A, (a Θ b) Θ c = a Θ (b Θ c)

(For example with the addition operation, (1+2) + 3 = 1 + (2+3) )

4) Inverse.  For each a in A there exists a b in A such that a Θ b = b Θ a = e.  Where e is the identity.

(For example with the addition operation, 4+-4 = -4+4 = 0.  0 is the identity element for addition)

5) Commutativity. For all elements a,b in A, a Θ b = b Θ a

(For example with the addition operation 1+2 = 2+1).

As we have seen, the set of integers under the operation addition forms an abelian group.

Establishing a group

So, let’s see if we can establish a Abelian group based around the rational coordinates on our graph.  We can demonstrate with the graph:

We then take 2 coordinate points with rational coordinates (i.e coordinates that can be written as a fraction of integers).  In this case A (1,12) and B (9,10).

We then draw the line through A and B.  This will intersect the graph in a 3rd point, C (except in a special case to be looked at in a minute).

We then reflect this new point C in the line y = x, giving us C’.

In this case C’ is the point  (46/3, -37/3)

We therefore define addition (our operation Θ) in this group as:

A + B = C’.

(1,12) + (9,10) = (46/3, -37/3).

We now need to deal with the special case when a line joining 2 points on the curve does not intersect the curve again.  This will happen whenever the gradient of this line is -1, which will make it parallel to the graph’s asymptote y = -x.

In this case we affix a special point at infinity to the Cartesian (x,y) plane.  We define this point as the point through which all lines with gradient -1 intersect.  Therefore in our expanded geometry, the line through AB will intersect the curve at this point at infinity. Let’s call our special point  Φ.  Now we have a new geometry,  the (x,y) plane affixed with Φ.

We can now create an Abelian group.  For any 2 rational points P(x,y), Q(x,y) we will have:

1) Identity.  P + Φ = Φ + P = P

2) Closure.  P + Q = R. (Where R(x,y) is also a rational point on the curve)

3) Associativity. (P+Q) + R = P+(Q+R)

4) Inverse.  P + (-P) = Φ

5) Commutativity.  P+Q = Q+P

Understanding the identity

Let’s see if we can understand some of these.  For the identity, if we have a point A on the line and the point at infinity then this will contain the line with gradient -1.  Therefore the line between the point at infinity and A will intersect the curve again at B.  Our new point, B’ will be created by reflecting this point in the line y = x.  This gets us back to point A.  Therefore P + Φ = P as required.

Understanding the inverse

With the inverse of our point P(x,y) given as -P = (-x,-y) we can see that this is the reflection in the line y = x.  We can see that we we join up the 2 points reflected in the line y = x we will have a line with slope -1, which will intersect with the curve at our point at infinity.  Therefore P + (-P) = Φ.

Through our graphical understanding the commutativity rule also follows immediately, It doesn’t matter which of the 2 points come first when we draw a line that connects them, therefore P+Q = Q+P.

Understanding associativity and closure

Neither associativity nor closure are obvious from our graph.  We could check individual points to show that (P+Q) + R = P+(Q+R), but it would be harder to explain why this always held.  Equally whilst it’s clear that P+Q will always create a point on the curve it’s not obvious that this will be a rational point.

In fact we do have both associativity and closure for our group as we have the following algebraic definition for our addition operation:

For a given point:

On a given curve of the form:

The addition of 2 points is given by:

In the case of our curve:

If we take P = (1,12).  P + P will be given by:

We can check this result graphically.  If P and Q are the same point, then the line that passes through both P and Q has to be the tangent to the curve at that point.  Therefore we would have:

Here the tangent at A does indeed meet the curve again – at point C, which does reflect in y = x to give us the coordinates above.

We could also find this intersection point algebraically.  If we differentiate the original curve to find the gradient when x = 1 we can find the equation of the tangent when x=1 and then substitute this back into the equation of the curve to find the intersection point.  This would give us:

We would then reverse the x and y coordinates to reflect in the line y = x.  This also gives us the same coordinates.

More generally if we have the 2 rational coordinates on the curve:

We have the algebraic formula for addition as:

 

If P = (1,12) and Q = (9,10), P + Q would give (after much tedious substitution!):

This agrees with the coordinates we found earlier using the much easier geometrical approach. As we can see from this formula, both coordinate points will always be rational – as they will be composed of combinations of our original rational coordinates.  For any given curve there will be a generator set of coordinates through which we can generate all other rational coordinates on the curve through our addition operation.

So, we seem to have come a long way from our original goal – finding integer solutions to an algebraic equation. Instead  we seem to have got sidetracked into studying graphs and establishing groups.  However by reinterpreting this problem as one in group theory then this then opens up many new mathematical techniques to help us understand the solutions to this problem.

A fuller introduction to this topic is the very readable, “Taxicabs and the Sum of Two Cubes” by Joseph Silverman (from which the 2 general equations were taken) .

Waging war with maths: Hollow squares

The picture above [US National Archives, Wikipedia] shows an example of the hollow square infantry formation which was used in wars over several hundred years.  The idea was to have an outer square of men, with an inner empty square.  This then allowed the men in the formation to be tightly packed, facing the enemy in all 4 directions, whilst the hollow centre allowed the men flexibility to rotate (and also was a place to hold supplies).  It was one of the infantry formations of choice against charging cavalry.

So, the question is, what groupings of men can be arranged into a hollow square?  This is a current Nrich investigation, so I thought I’d do a mini-investigation on this.

We can rethink this question as asking which numbers can be written as the difference between 2 squares. For example in the following diagram (from the Nrich task Hollow Squares)

We can see that the hollow square formation contains a larger square of 20 by 20 and a smaller hollow square of 8 by 8.  Therefore the number of men in this formation is:

202-82 = 336.

The first question we might ask therefore is how many numbers from 1-100 can be written as the difference between 2 squares?  These will all be potential formations for our army.

I wrote a quick code on Python to find all these combinations.  I included 0 as a square number (though this no longer creates a hollow square, rather just a square!). You can copy this and run it in a Python editor like Repl.it.


for k in range(1,50):

  for a in range(0, 100):
   for b in range(0,100):
    if a**2-b**2 == k :
     print(k,a,b)

This returned the following results:

1 1 0
3 2 1
4 2 0
5 3 2
7 4 3
8 3 1
9 3 0
9 5 4
11 6 5
12 4 2
13 7 6
15 4 1
15 8 7
16 4 0
16 5 3
17 9 8
19 10 9
20 6 4
21 5 2
21 11 10
23 12 11
24 5 1
24 7 5
25 5 0
25 13 12
27 6 3
27 14 13
28 8 6
29 15 14
31 16 15
32 6 2
32 9 7
33 7 4
33 17 16
35 6 1
35 18 17
36 6 0
36 10 8
37 19 18
39 8 5
39 20 19
40 7 3
40 11 9
41 21 20
43 22 21
44 12 10
45 7 2
45 9 6
45 23 22
47 24 23
48 7 1
48 8 4
48 13 11
49 7 0
49 25 24

Therefore we can see that the numbers with no solutions found are:

2,6,10,14,18,22,26,30,34,38,42,46,50

which are all clearly in the sequence 4n-2.

Thinking about this, we can see that this can be written as 2(2n-1) which is the product of an even number and an odd number. This means that all numbers in this sequence will require an odd factor in each of their factor pairs:

eg. 50 can be written as 10 (even) x 5 (odd) or 2 (even) x 25 (odd) etc.

But with a2-b2 = (a+b)(a-b), due to symmetries we will always end up with (a+b) and (a-b) being both even or both odd, so we can’t create a number with a factor pair of one odd and one even number. Therefore numbers in the sequence 4n-2 can’t be formed as the difference of 2 squares. There are some nicer (more formal) proofs of this here.

A battalion with 960 soldiers

Next we are asked to find how many different ways of arranging 960 soldiers in a hollow square. So let’s modify the code first:


for a in range(0, 1000):
 for b in range(0,1000):
  if a**2-b**2 == 960 :
   print(a,b)

Which gives us the following solutions:

31 1
32 8
34 14
38 22
46 34
53 43
64 56
83 77
122 118
241 239

General patterns

We can notice that when the number of soldiers is 1,3,5,7,9,11 (2n-1) we can always find a solution with the pair n and n-1. For example, 21 can be written as 2n-1 with n = 11. Therefore we have 10 and 11 as our pair of squares. This works because 112-102 = (11+10)(11-10) returns the factor pair 21 and 1. In general it always returns the factor pair, 2n-1 and 1.

We can also notice that when the number of soldiers is 4,8,12,16,20 (4n) we can always find a solution with the pair n+1 and n-1. For example, 20 can be written as 4n with n = 5. Therefore we have 6 and 4 as our pair of squares. This works because 62-42 = (6+4)(6-4) returns the factor pair 10 and 2. In general it always returns the factor pair, 2n and 2.

And we have already shown that numbers 2,6,10,14,18,22 (4n-2) will have no solution. These 3 sequences account for all the natural numbers (as 2n-1 incorporates the 2 sequences 4n-3 and 4n-1).

So, we have found a method of always finding a hollow square formation (if one exists) as well as being able to use some computer code to find other possible solutions. There are lots of other avenues to explore here – could you find a method for finding all possible combinations for a given number of men? What happens when the hollow squares become rectangles?

Finding the volume of a rugby ball (prolate spheroid)

With the rugby union World Cup currently underway I thought I’d try and work out the volume of a rugby ball using some calculus.  This method works similarly for American football and Australian rules football.   The approach is to consider the rugby ball as an ellipse rotated 360 degrees around the x axis to create a volume of revolution.  We can find the equation of an ellipse centered at (0,0) by simply looking at the x and y intercepts.  An ellipse with y-intercept (0,b) and x intercept (a,0) will have equation:

Therefore for our rugby ball with a horizontal “radius” (vertex) of 14.2cm and a vertical “radius” (co-vertex) of 8.67cm will have equation:

We can see that when we plot this ellipse we get an equation which very closely resembles our rugby ball shape:

Therefore we can now find the volume of revolution by using the following formula:

But we can simplify matters by starting the rotation at x = 0 to find half the volume, before doubling our answer.  Therefore:

Rearranging our equation of the ellipse formula we get:

Therefore we have the following integration:

Therefore our rugby ball has a volume of around 4.5 litres.  We can compare this with the volume of a football (soccer ball) – which has a radius of around 10.5cm, therefore a volume of around 4800 cubic centimeters.

We can find the general volume of any rugby ball (mathematically defined as a prolate spheroid) by the following generalization:

We can see that this is very closely related to the formula for the volume of a sphere, which makes sense as the prolate spheroid behaves like a sphere deformed across its axes. Our prolate spheroid has “radii” b, b and a – therefore r cubed in the sphere formula becomes b squared a.

Prolate spheroids in nature

The image above [wiki image NASA] is of the Crab Nebula – a distant Supernova remnant around 6500 light years away.  The shape of Crab Nebula is described as a prolate spheroid.

The Shoelace Algorithm to find areas of polygons

This is a nice algorithm, formally known as Gauss’s Area formula, which allows you to work out the area of any polygon as long as you know the Cartesian coordinates of the vertices.  The case can be shown to work for all triangles, and then can be extended to all polygons by first splitting them into triangles and following the same approach.

Let’s see if we can work out the algorithm ourselves using the construction at the top of the page.  We want the area of the triangle (4), and we can see that this will be equivalent to the area of the rectangle minus the area of the 3 triangles (1) (2) (3).

Let’s start by adding some other coordinate points for the rectangle:

Therefore the area of the rectangle will be:

(1) + (2) +(3) +(4): (x3-x2)(y1-y3)

And the area of triangles will be:

(1): 0.5(x3-x2)(y2-y3)

(2): 0.5(x1-x2)(y1-y2)

(3): 0.5(x3-x1)(y1-y3)

Therefore the area of triangle (4) will be:

Area = (x3-x2)(y1-y3) – 0.5(x3-x2)(y2-y3) – 0.5(x1-x2)(y1-y2) – 0.5(x3-x1)(y1-y3)

Therefore we have our algorithm!  Let’s see if it works with the following coordinates added:

x1 = 2  x = 1  x = 3
y1 = 3 y = 2  y = 1

Area = (x3-x2)(y1-y3) – 0.5(x3-x2)(y2-y3) – 0.5(x1-x2)(y1-y2) – 0.5(x3-x1)(y1-y3)

Area = (3-1)(3-1) – 0.5(3-1)(2-1) – 0.5(2-1)(3-2) – 0.5(3-2)(3-1)

Area = 4 – 1 – 0.5 – 1 = 1.5 units squared

We could check this using Pythagoras to find all 3 sides of the triangle, followed by the Cosine rule to find an angle, followed by the Sine area of triangle formula, but let’s take an easier route and ask Wolfram Alpha (simply type “area of a triangle with coordinates (1,2) (2,3) (3,1)).  This does indeed confirm an area of 1.5 units squared.  Our algorithm works.  We can of course simplify the area formula by expanding brackets and simplifying.  If we were to do this we would get the commonly used version of the area formula for triangles.

The general case for finding areas of polygons

The general formula for the area of an n-sided polygon is given above.

For a triangle this gives:

For a quadrilateral this gives:

For a pentagon this gives:

You might notice a nice shoelace like pattern (hence the name) where x coordinates criss cross with the next y coordinate along.  To finish off let’s see if it works for an irregular pentagon.

If we arbitrarily assign our (x1, y1) as (1,1) and then (x2, y2) as (3,2), and continue in a clockwise direction we will get the following:

area = absolute of 0.5( 1×2 + 3×4 + 3×1 + 4×0 + 2×1 – 3×1 – 3×2 – 4×4 – 2×1 – 1×0)

area = 4.

Let’s check again with Wolfram Alpha – and yes it does indeed have an area of 4.

It could be a nice exploration task to take this further and to explore how many different methods there are to find the area of polygons – and compare their ease of use, level of mathematics required and aesthetic appeal.

Soap Bubbles and Catenoids

Soap bubbles form such that they create a shape with the minimum surface area for the given constraints.  For a fixed volume the minimum surface area is a sphere, which is why soap bubbles will form spheres where possible.  We can also investigate what happens when a soap film is formed between 2 parallel circular lines like in the picture below: [Credit Wikimedia Commons, Blinking spirit]


In this case the shape formed is a catenoid – which provides the minimum surface area (for a fixed volume) for a 3D shape connecting the two circles.  The catenoid can be defined in terms of parametric equations:

Where cosh() is the hyperbolic cosine function which can be defined as:

For our parametric equation, t and u are parameters which we vary, and c is a constant that we can change to create different catenoids.  We can use Geogebra to plot different catenoids.  Below is the code which will plot parametric curves when c =2 and t varies between -20pi and 20 pi.

 

We then need to create a slider for u, and turn on the trace button – and for every given value of u (between 0 and 2 pi) it will plot a curve.  When we trace through all the values of u it will create a 3D shape – our catenoid.

Individual curve (catenary)


Catenoid when c = 0.1

Catenoid when c = 0.5

Catenoid when c = 1

Catenoid when c = 2

Wormholes

For those of you who know your science fiction, the catenoids above may look similar to a wormhole.  That’s because the catenoid is a solution to the hypothesized mathematics of wormholes.  These can be thought of as a “bridge” either through curved space-time to another part of the universe (potentially therefore allowing for faster than light travel) or a bridge connecting 2 distinct universes.

Screen Shot 2019-09-29 at 7.03.34 PM

Above is the Morris-Thorne bridge wormhole [Credit The Image of a Wormhole].

Further exploration:

This is a topic with lots of interesting areas to explore – the individual curves (catenary) look similar to, but are distinct from parabola.  These curves appear in bridge building and in many other objects with free hanging cables.  Proving that catenoids form shapes with minimum surface areas requires some quite complicated undergraduate maths (variational calculus), but it would be interesting to explore some other features of catenoids or indeed to explore why the sphere is a minimum surface area for a given volume.

If you want to explore further you can generate your own Catenoids with the Geogebra animation I’ve made here.

For students taking their exams in 2021 there is a big change to the IB syllabus – there will now be 4 possible strands: IB HL Analysis and Approaches, IB SL Analysis and Approaches, IB HL Applications and Interpretations, IB SL Applications and Interpretations.

IB Applications and Interpretations

Screen Shot 2019-07-27 at 8.57.12 PM

There is a reasonable cross-over between the current Studies course and the new Applications SL course.  However there is quite a lot of new content – and as such the expectation is that this course could be quite a bit more challenging that the current SL course.   The HL Applications course is a rather odd mix of former Maths Studies topics, former SL topics and former IB HL Statistics topics.

Some key points:

  1. The SL Applications course will be a complete sub-set of the HL Applications course, and the HL exam will include some of the same questions as the SL exam.
  2. Both SL and HL will only have calculator papers (and have no non-calculator papers like Analysis)
  3. Both SL and HL will have Paper 1 consisting of short questions and Paper 2 with longer style questions (similar to the current Maths Studies course).
  4. HL students will do an investigation style Paper 3 – potentially with the use of technology.  This will lead students through an investigation on any topic on the syllabus.
  5. The Exploration coursework will remain – however the guidance is now that it should be 12-20 pages (rather than 6-12 previously).

What does this all mean for Applications SL?

If the IB follow through with their stated plan to have both new SL courses (Applications and Analysis) the same difficulty then either:

(a) The Analysis course will remain at the same level of difficulty as the current Maths SL and therefore many students who up until now have taken Maths Studies will find the new Applications course extremely challenging.

(b) The Analysis course will be made easier than the current Maths SL course, so that the new Applications course is also a little more accessible – though still harder than the current Maths Studies course.

I would predict that (b) is the more likely of these two – otherwise there will be a significant cohort of IB students (around 30%) who fail to get even a Level 4 in their maths.  At the moment I would advise that all weaker students should definitely take this course (IGCSE Grade C and below), but it may be the case that it is a good option for more stronger students who have traditionally taken SL rather than Studies.

What does this all mean for Applications HL?

This is really hard to work out – if Applications SL remains accessible for students with low IGCSE grades, and Applications HL contains a subset of these questions, then that would suggest that the Applications HL would be significantly easier than the current HL course.  However, there are a number of challenging topics on the Applications HL syllabus which could well be used to stretch top students.  Again the stated aim of the IB is for the two HL courses to be the same difficulty – so this is one we will really have to wait and see with.

Based on the current information I would advice only students with an A* in IGCSE to take this course. It would appear to be aimed at students who need some mathematical skills for their university courses (such as biology, medicine or business) but who do not want to study mathematics or a field with substantial mathematics in it (such as engineering, physics, computer science etc).

Resources for teachers and students

This will be a work in progress – but to get started we have:

General resources

  1. A very useful condensed pdf of the Applications and Interpretations formula book for both SL and HL.
  2. An excellent overview of the changes to the new syllabus – including more detailed information as to the syllabus changes, differences between the two courses and also what 10 of the leading universities have said with regards to course preferences.
  3. University acceptance.  Information collated by a group of IB teachers on university requirements as to which course they will require for different subjects (this may be not be up to date, so please check).

Investigation resources for Paper 3 [Higher Level]

  1. Old IA investigations

Standard Level

[Links removed – hopefully the IB will provide these resources elsewhere]

(a) All SL IA investigations from 1998 to 2009 : This is an excellent collection to start preparations for the new Paper 3.

(b) Specimen investigations: These are 8 specimen examples of IA investigations from 2006 with student answers and annotations.

(c) SL IA investigations 2011-2012: Some more investigations with teacher guidance.

(d) SL IA investigations 2012-2013: Some more investigations with teacher guidance.

(e) Koch snowflakes: This is a nice investigation into fractals.

Higher Level 

(a) All HL IA investigations from 1998 to 2009:  Lots more excellent investigations – with some more difficult mathematics.

(b) HL IA investigations 2011-2012: Some more investigations with teacher guidance.

(c) HL IA investigations 2012-2013: Some more investigations with teacher guidance.

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