IB Maths and GCSE Maths Resources for IB Maths explorations and investigations. Real life maths. Theory of Knowledge (ToK). Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

IB Maths Resources from British International School Phuket:

My name is Andrew Chambers and I am currently working at British International School Phuket. 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.

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.

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:

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:

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 Examssection 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.

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 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.

Henon’s map was created in the 1970s to explore chaotic systems. The general form is created by the iterative formula:

The classic case is when a = 1.4 and b = 0.3 i.e:

To see how points are generated, let’s choose a point near the origin. If we take (0,0) the next x coordinate is given by:

We would then continue this process over several thousands iterations. If we do this then we get the very strange graph at the top of the page – the points are attracted to a flow like structure, which they then circulate round. The graph above was generated when we took our starting coordinate as (0.1,0.1), let’s take a different starting point. This time let’s have (1.1, 1.1):

We can see that exactly the same structure appears. All coordinates close to the origin will get attracted to this strange attractor – except for a couple of fixed points near the origin which remain where they are. Let’s see why. First we can rewrite the iterative formula just in terms of x:

Next we use the fact that when we have a fixed point the x coordinate (and y coordinate) will not change. Therefore we can define the following:

This allows us to then make the following equation:

Which we can then solve using the quadratic formula:

Which also gives y:

So therefore at these 2 fixed points the coordinates do not get drawn to the strange attractor.

Above we can see the not especially interesting graph of the repeated iterations when starting at this point!

But we can also see the chaotic behavior of this system by choosing a point very close to this fixed point. Let’s choose (0.631354477, 0.631354477) which is correct to 9 decimal places as an approximation for the fixed point.

We can see our familiar graph is back. This is an excellent example of chaotic behavior – a very small change in the initial conditions has created a completely different system.

This idea was suggested by the excellent Doing Maths With Python – which is well worth a read if you are interested in computer programing to solve mathematical problems.

This pattern of a fern pictured above was generated by a simple iterative program designed by mathematician Michael Barnsely. I downloaded the Python code from the excellent Tutorialspoint and then modified it slightly to run on repl.it. What we are seeing is the result of 40,000 individual points – each plotted according to a simple algorithm. The algorithm is as follows:

Transformation 1: (0.85 probability of occurrence)

x_{i+1} = 0.85x_{i} +0.04y_{i}

y_{i+1}= -0.04x_{i}+0.85y_{i}+1.6

Transformation 2: (0.07 probability of occurrence)

x_{i+1} = 0.2x_{i} -0.26y_{i}

y_{i+1}= 0.23x_{i}+0.22y_{i}+1.6

Transformation 3: (0.07 probability of occurrence)

x_{i+1} = -0.15x_{i} -0.28y_{i}

y_{i+1}= 0.26x_{i}+0.24y_{i}+0.44

Transformation 4: (0.01 probability of occurrence)

x_{i+1} = 0

y_{i+1}= 0.16y_{i}

So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-100 and assign 1-85 for transformation 1, 86-92 to transformation 2, 93-99 for transformation 3 and 100 for transformation 4. Say I generate the number 36 – therefore I will apply transformation 1.

x_{i+1} = 0.85(0)+0.04(0)

y_{i+1}= -0.04(0)+0.85(0)+1.6

and my new coordinate is (0,1.6). I mark this on my graph.

I then repeat this process – say this time I generate the number 90. This tells me to do transformation 2. So:

x_{i+1} = 0.2(0) -0.26(1.6)

y_{i+1}= 0.23(0)+0.22(1.6)+1.6

and my new coordinate is (-0.416, 1.952). I mark this on my graph and carry on again. The graph above was generated with 40,000 iterations – let’s see how it develops over time:

1000 iterations:

10,000 iterations:

100,000 iterations:

500,000 iterations:

If we want to understand what is happening here we can think of each transformation as responsible for a different part of our fern. Transformation 1 is most likely and therefore this fills in the smaller leaflets. Transformations 2 and 3 fill in the bottom left and right leaflet (respectively) and transformation 4 fills in the stem.

It’s quite amazing to think that a simple computer program can create what looks like art – or indeed that is can replicate what we see in nature so well. This fern is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. You might want to explore the idea of fractals in delving into this topic in more detail.

Changing the iterations

We can explore what happens when we change the iterations very slightly.

Christmas tree

Crazy spiral

Modern art

You can modify the code to run this here. Have a go!

This post is based on the maths and ideas of Hahn’s Calculus in Context – which is probably the best mathematics book I’ve read in 20 years of studying and teaching mathematics. Highly recommended for both students and teachers!

Hahn talks us though the mathematics, experiments and thought process of Galileo as he formulates his momentous theory that in free fall (ignoring air resistance) an object falling for t seconds will fall a distance of ct² where c is a constant. This is counter-intuitive as we would expect the mass of an object to be an important factor in how far an object falls (i.e that a heavier object would fall faster). Galileo also helped to overturned Aristotle’s ideas on motion. Aristotle had argued that any object in motion would eventually stop, Galileo instead argued that with no friction a perfectly spherical ball once started rolling would roll forever. Galileo’s genius was to combine thought experiments and real data to arrive at results that defy “common sense” – to truly understand the universe humans had to first escape from our limited anthropocentric perspective, and mathematics provided an opportunity to do this.

Inclined Planes

Galileo conducted experiments on inclined planes where he placed balls at different heights and then measured their projectile motion when they left the ramp, briefly ran past the edge of a flat surface and then fell to the ground. We can see the set up of one ramp above. The ball starts at O, and we mark as h this height. At an arbitrary point P we can see that there are 2 forces acting on the ball, F which is responsible for the ball rolling down the slope, and f, which is a friction force in the opposite direction. At point P we can mark the downwards force mg acting on the ball. We can then use some basic rules of parallel lines to note that the angles in triangle PCD are equal to triangle AOB.

Galileo’s times squared law of fall

We have the following equation for the total force acting on the ball at point P:

We also have the following relationship from physics, where m is the mass and a(t) the acceleration:

This therefore gives:

Next we can use trigonometry on triangle PCD to get an equation for F:

and so:

Next we can use another equation from physics which gives us the frictional force on a perfectly spherical, homogenous body rolling down a plane is:

So this gives:

We can then integrate to get velocity (our constant of integration is 0 because the velocity is 0 when t = 0)

and integrate again to get the distance travelled of the ball (again our constant of integration is 0):

When Galileo was conducting his experiments he did not know g, instead he noted that the relationship was of the form;

where c is a constant related to a specific incline. This is a famous result called the times squared law of fall. It shows that the distance travelled is independent of the mass and is instead related to the time of motion squared.

Velocity also independent of the angle of incline

Above we have shown that the distance travelled is independent of the mass – but in the equation it is still dependent on the angle of the incline. We can go further and then show that the velocity of the ball is also independent of the angle of incline, and is only dependent on the height at which the ball starts from.

If we denote as t_b as the time when the ball reaches point A in our triangle we have:

This is equal to the distance from AO, so we can use trigonometry to define:

This can then be rearranged to give:

this is the time taken to travel from O to A. We can the substitute this into the velocity equation we derived earlier to give the velocity at point A. This is:

This shows that the velocity of the ball at point A is only dependent on the height and not the angle of incline or mass. The logical extension of this is that if the angle of incline has no effect on the velocity, that this result would still hold as the angle of incline approaches and then reaches 90 degrees – i.e when the ball is in free fall.

Galileo used a mixture of practical experiments on inclined planes, mathematical calculations and thought experiments to arrive at his truly radical conclusion – the sign of a real genius!

This post is based on the maths and ideas of Hahn’s Calculus in Context – which is probably the best mathematics book I’ve read in 20 years of studying and teaching mathematics. Highly recommended for both students and teachers!

Hard as it is to imagine now, for most of the history of mathematics there was no coordinate geometry system and therefore graphs were not drawn using algebraic equations but instead were constructed. The ancient Greeks such as Archimedes made detailed studies of conic sections (parabolas, ellipses and hyperbola) using ideas of relationships in constructions. The nice approach to this method is that it makes clear the link between conic sections and their properties in reflecting light – a property which can then be utilized when making lenses. A parabolic telescope for example uses the property that all light collected through the scope will pass through a single focus point.

Let’s see how we can construct a parabola without any algebra – simply using the constructions of the Greeks. We start with a line and a focus point F not on the line. This now defines a unique parabola.

This unique parabola is defined as all the points A such that the distance from A to F is equal as the perpendicular distance from A to the line.

We can see above that point A must be on our parabola because the distance AB is the same as the distance AF.

We can also see that point C must be on our parabola because the length CD is the same as CF. Following this same method we could eventually construct every point on our parabola. This would finally create the following parabola:

Focus point of a parabolic mirror

We can now see how this parabola construction gives us an intrinsic understanding of reflective properties. If we have a light source entering parallel to the perpendicular though the focus then we can use the fact that this light will pass through the focus to find the path the light traces before it is reflected out.

Newton made use of this property when designing his parabolic telescope. It’s interesting to note how a different method leads to a completely different appreciation of the properties of a curve.

Finding the area under a quadratic curve without calculus

Amazingly a method for finding the area under a quadratic curve was also discovered by the Greek scientist and mathematician Archimedes around 2200 years ago – and nearly 2000 years before calculus. Archimedes’ method was as follows.

Choose 2 points on the curve, join them to make 2 sides of a triangle. Choose the 3rd point of the triangle as the point on the quadratic with the same gradient as the chord. This is best illustrated as below. Here I generated a parabola with focus at (0,1) and line with the x axis.

Here I chose points B and C, joined these with a line and then looked for the point on the triangle with the same gradient. This then gives a triangle with area 4. Archimedes then discovered that the area of the parabolic segment (i.e the total area enclosed by the line BC and the parabola) is 4/3 the area of the triangle. This gives 4/3 of 4 which is 5 1/3. Once we have this we can find the area under the curve (i.e the integral) using simple areas of geometric shapes.

Using calculus

We can check that Archimedes’ method does indeed work. We want to find the area enclosed by the 2 following equations:

This is given by:

It works! Now we can try a slightly more difficult example. This time I won’t choose 2 points parallel to the x-axis.

This time I find the gradient of the line joining B and C and then find the point on the parabola with the same gradient. This forms my 3rd point of the triangle. The area of this triangle is approximately 1.68. Therefore Archimedes’ method tells us the area enclosed between the line and the curve will be approximately 4/3 (1.68) = 2.24. Let’s check this with calculus:

Again we can see that this method works – our only error was in calculating an approximate area for the triangle rather than a more precise answer.

So, nearly 2000 years before the invention of calculus the ancient Greeks were already able to find areas bounded by line and parabolic curves – and indeed Archimedes was already exploring the ideas of the limit of sums of areas upon which calculus in based.

Finding the average distance between 2 points on a hypercube

This is the natural extension from this previous post which looked at the average distance of 2 randomly chosen points in a square – this time let’s explore the average distance in n dimensions. I’m going to investigate what dimensional hypercube is required to have an average distance of more than one, and then also what happens to the average distance as n approaches infinity.

Monte Carlo method

The Monte Carlo method is a very powerful technique which utilizes computational power. Basically we use the fact that the average of a very large number of trials will serve as an approximation to an exact result. In this case I will run a Python program 10 million times – each time it will select 2 coordinate points and then work out the distance between them. It will then find the average of these 10 million trials. The code above generates 2 coordinates in 3 dimensional space inside a unit cube. We can modify this for n-dimensional space by remembering that Pythagoras still works in higher dimensions.

Results

Running this code helps to generate the above results. This answers our first question – we need a 7 dimensional unit hypercube until the average distance between two randomly chosen points is greater than 1. We can also see that the difference between the average distances is reducing – but it’s not clear if this will approach a limit or if it will continue growing to infinity. So let’s do some more trials.

Further trials

This takes us up to a 22-dimensional hypercube. At this point it’s probably useful to plot a graph to see the trend.

Reciprocal model

This reciprocal model is of the form:

We can see that this is a pretty good fit (R squared 0.9994). If this model is accurate then this would suggest that the average distance approaches a limit as n approaches infinity.

Polynomial model

This polynomial model is of the form:

We can see that this is also a very good fit (R squared 0.9997). If this model is accurate then as b is greater than 0, this would suggest that the average distance approaches infinity as n approaches infinity.

Reflection

Quite annoyingly we have 2 model which both fit the data very accurately – but predict completely different results! Logically we could probably say that we would expect the average distance to approach infinity as n approaches infinity – and also we could possibly justify this by the fact that the polynomial model is a slightly better fit. Given the similarity between the 2 models it probably time to find out the actual results for this.

Average n-dimensional distance bounds

Not surprisingly the mathematics required to work this out is exceptionally difficult – and ends up with non-solvable integrals which require analytic solutions. The Monte Carlo method with very large numbers of trials is a reasonably good approach to approximating this answer. There is however a very useful lower and upper bound for the average distance in n dimensional space given by:

This shows immediately that the average distance will approach infinity as n grows large – as the lower bound will grow to infinity. Quite pleasingly we can see that the polynomial model we derived is similar to the lower bound. We can plot both upper and lower bound along with our polynomial model to see how these all compare. We have lower bound (green), polynomial model (black) and upper bound (green):

We can see that our polynomial model very closely follows the upper bound in our domain. As we extend the domain this polynomial approximation remains above the lower and tracks the upper bounds before gradually growing less accurate. When n is 50 our model predicts a distance of 2.94, whereas the upper bound is 2.88. This is quite a nice result – we have used the Monte Carlo method to derive a polynomial approximation to the average distance in n-dimensional hypercubes and it both closely follows the upper bound over a reasonable domain and also is of a very similar form to the lower bound. We can use this lower bound to see that a 36 dimensional hypercube (and higher) would be guaranteed to have an average distance of more than 2.

Conclusion

This was a nice example of the power of the Monte Carlo method in these kind of problems – we were able to use it quite successfully to get a polynomial approximation which turned out to be reasonably accurate. We could have significantly improved this accuracy by running 100 million (or 1 billion etc) trials each time – though this would have probably required a more powerful computer!

Find the average distance between 2 points on a square

This is another excellent mathematical puzzle from the MindYourDecisions youtube channel. I like to try these without looking at the answer – and then to see how far I get. This one is pretty difficult (and the actual solution exceptionally difficult!) The problem is to take a square and randomly choose 2 points somewhere inside. If you calculate the distance between the 2 points, then do this trial approaching an infinite number of times what will the average distance be? Here is what I did.

Simplify the situation: 1×1 square

This is one of the most important strategies in tackling difficult maths problems. You simplify in order to gain an understanding of the underlying problem and possibly either develop strategies or notice patterns. So, I started with a unit square and only considered the vertices. We can then list all the possible lengths:

We can then find the average length by simply doing:

2×2 square

We can then follow the same method for a 2×2 square. This gives:

Which gives an average of:

Back to a 1×1 square

Now, we can imagine that we have a 1 x 1 square with dots at every 0.5. This is simply a scaled version of the 2×2 square, so we can divide our answer by 2 to give:

3×3 square

Following the same method we have:

This gives an average of:

Back to a 1×1 square

and if we imagine a 1×1 square with dots at every 1/3. This is simply a scaled version of the 3×3 square, so we can divide our answer by 3 to give:

We can then investigate what happens as we consider more and more dots inside our 1×1 square. When we have considered an infinite number then we will have our average distance – so we are looking the limit to infinity. This suggests using a graph. First I calculated a few more terms in the sequence:

Then I plotted this on Desmos. The points looked like they fit either an exponential or a reciprocal function – both which have asymptotes, so I tried both. The reciprocal function fit with an R squared value of 1. This is a perfect fit so I will use that.

This was plotted using the regression line:

And we can find the equation of the horizontal asymptote by seeing what happens when x approaches infinity. This will give a/c. Using the values provided by Desmos’ regression I got 0.515004887. Because I have been using approximate answers throughout I’ll take this as 0.52 (2sf). Therefore I predict that the average distance between 2 points in a 1×1 square will be approximately 0.52. And more generally, the average distance in an n x n square will be 0.52(n). This is somewhat surprising as a result – it’s not obvious why it would be a little over half the distance from 0 to 1.

Brute forcing using Python

We can also write a quick code to approximate this answer using Python (This is a Monte Carlo method). I generate 4 random numbers to represent the 2 x-coordinates and 2-y coordinates of 2 random points. I then work out the distance between them and repeat this 10 million times, then calculate the average distance. This gives:

Checking with the actual answer

Now for the moment of truth – and we watch the video to find out how accurate this is. The correct answer is indeed 0.52 (2sf) – which is great – our method worked! The exact answer is given by:

Our graphical answer is not quite accurate enough to 3 sf – probably because we relied on rounded values to plot our regression line. Our Python method with 10 million trials was accurate to 4 sf. Just to keep my computer on its toes I also calculated this with 100 million trials. This gave 0.5214126210834646 (now accurate to 5 sf).

We can also find the percentage error when using our graphical method. This is only:

Overall this is a decent result! If you are feeling extremely brave you might want to look at the video to see how to do this using calculus.

Extension: The average distance between 2 points in a unit circle

I modified the Python code slightly to now calculate the average distance between 2 points in a unit circle. This code is:

which returns an answer of 0.9054134561871364. I then looked up what the exact answer is. For the unit circle it is 128/(45 pi). This is approximately 0.9054147874. We can see that our computer method was accurate to 5 sf here. Again, the actual mathematical proof is extremely difficult.

Reflection

This is a nice example of important skills and techniques useful in mathematics – simplification of a problem, noticing patterns, graphical methods, computational power and perseverance!

This is a really beautiful solution to an interesting probability problem posed by fellow IB teacher Daniel Hwang, for which I’ve outlined a method for solving suggested by Ferenc Beleznay. The problem is as follows:

On average, how many random real numbers from 0 to 1 (inclusive) are required for the sum to exceed 1?

1 number

Clearly if we choose only 1 number then we can’t exceed 1.

2 numbers

Here we imagine the 2 numbers we pick as x and y and therefore we can represent them as a coordinate pair. The smallest pair (0,0) and the largest pair (1,1). This means that the possible coordinates fit inside the unit square shown above. We want to know for what coordinate pairs we have the inequality x + y > 1. This can be rearrange to give y > 1-x. The line y = 1-x is plotted and we can see that any coordinate points in the triangle BCD satisfy this inequality. Therefore the probability of a random coordinate pair being in this triangle is 1/2.

3 numbers

This time we want to find the probability that we exceed 1 with our third number. We can consider the numbers as x, y, z and therefore as 3D coordinates (x,y,z). From the fact that we are choosing a third number we must already have x +y <1. We draw the line x+y = 1, which in 3D gives us a plane. The volume in which our coordinate point must lie is the prism ABDEFG.

We now also add the constraint x+y+z >1. This creates the plane as shown. If our coordinate lies inside the pyramid ABDE then our coordinates will add to less than 1, outside this they will add to more than 1.

The volume of the pyramid ABDE = 1/3 (base area)(perpendicular height).

The volume of the prism ABDEFG = (base area)(perpendicular height).

Given that they share the same perpendicular height and base area then precisely 1/3 of the available volume would give a coordinate point that adds to less than 1, and 2/3 of the available volume would give a coordinate point that adds to more than 1.

Therefore we have the following tree diagram:

Exceeds 1 with 2 numbers = 1/2

Does not exceed 1 with 2 numbers, exceeds 1 with 3 numbers = 1/2 x 2/3 = 1/3.

Does not exceed 1 with 2 numbers, does not exceed 1 with 3 numbers = 1/2 x 1/3 = 1/6.

4 numbers

If you been following so far this is where things get interesting! We can now imagine a 4 dimensional unit cube (image above from Wikipedia) and a 4D coordinate point (x,y,z,a).

Luckily all we care about is the ratio of the 4-D pyramid and the 4-D prim formed by our constraints x+y+z <1 and x+y+z+a >1.

We have the following formula to help:

The n-D volume of a n-D pyramid = 1/n (base)(perpendicular height).

Therefore:

The 4-D volume of a 4-D pyramid = 1/4 (base 3D volume)(perpendicular height).

The 4-D volume of the prism ABDEFG = (base 3D volume)(perpendicular height).

Given that the 2 shapes share the same base and perpendicular height, the hyper-pyramid occupies exactly 1/4 of the 4-D space of the hyper-prism. So the probability of being in this space is 1/4 and 3/4 of being outside this space.

We can now extend our tree diagram:

Does not exceed 1 with 2 numbers, does not exceed 1 with 3 numbers, exceeds with 4 numbers = 1/2 x 1/3 x 3/4 = 1/8

Does not exceed 1 with 2 numbers, does not exceed 1 with 3 numbers, does not exceed with 4 numbers = 1/2 x 1/3 x 1/4 = 1/24.

In general a hyper-pyramid in n dimensional space occupies exactly 1/n of the space of the hyper-prism – so we can now continue this tree diagram.

Expected value

We can make a table of probabilities to find how many numbers we expect to use in order to exceed one.

Which gives us the following expected value calculation:

Which we can rewrite as:

But we have:

Therefore this gives:

So on average we would need to pick e numbers for the sum to exceed one! This is quite a remarkable result – e, one of the fundamental mathematical constants has appeared as if by magic on a probability question utilizing hyper-dimensional shapes.

Demonstrating this with Python

Running the Python code shown above will simulate doing this experiment. The computer generates a “random” number, then another and carries on until the sum is greater than 1. It then records how many numbers were required. It then does this again 1 million times and finds the average from all the trials.

1 million simulations gives 2.7177797177797176. When we compare this with the real answer for e, 2.7182818284590452353602874713527, we can see it has taken 1 million simulations to only be correct to 4sf.

Even 5 million simulations only gives 2.7182589436517888, so whilst we can clearly see that we will eventually get e, it’s converging very slowly. This may be because we are reliant on a random number generator which is not truly random (and only chooses numbers to a maximum number of decimal places rather than choosing from all values between 0 and 1).

I think this is a beautiful example of the unexpected nature of mathematics – we started out with a probability problem and ended up with e, via a detour into higher dimensional space! We can also see the power of computers in doing these kinds of brute force calculations.

Adapting and exploring maths challenge problems is an excellent way of finding ideas for IB maths explorations and extended essays. This problem is taken from the book: The first 25 years of the Superbrain challenges. I’m going to see how many different ways I can solve it.

The problem is to find all the integer solutions to the equation above. Finding only integer solutions is a fundamental part of number theory – a branch of mathematics that only deals with integers.

Method number 1: Brute force

This is a problem that computers can make short work of. Above I wrote a very simple Python program which checked all values of x and y between -99 and 99. This returned the only solution pairs as:

Clearly we have not proved these are the only solutions – but even by modifying the code to check more numbers, no more pairs were found.

Method number 2: Solving a linear equation

We can notice that the equation is linear in terms of y, and so rearrange to make y the subject.

We can then use either polynomial long division or the method of partial fractions to rewrite this. I’ll use partial fractions. The general form for this fraction can be written as follows:

Next I multiply by the denominator and the compare coefficients of terms.

This therefore gives:

I can now see that there will only be an integer solution for y when the denominator of the fraction is a factor of 6. This then gives (ignoring non integer solutions):

I can then substitute these back to find my y values, which give me the same 4 coordinate pairs as before:

Method number 3: Solving a quadratic equation

I start by making a quadratic in x:

I can then use the quadratic formula to find solutions:

Which I can simplify to give:

Next I can note that x will only be an integer solution if the expression inside the square root is a square number. Therefore I have:

Next I can solve a new quadratic as follows:

As before I notice that the expression inside my square root must be a square number. Now I can see that I need to find m and n such that I have 2 square numbers with a difference of 24. I can look at the first 13 square numbers to see that from the 12th and 13th square numbers onwards there will also be a difference of more than 24. Checking this list I can find that m = 1 and m = 5 will satisfy this equation.

This then gives:

which when I solve for integer solutions and then sub back into find x gives the same four solutions:

Method number 4: Graphical understanding

Without rearranging I could imagine this as a 3D problem by plotting the 2 equations:

This gives the following graph:

We can see that the plane intersects the curve in infinite places. I’ve marked A, B on the graph to illustrate 2 of the coordinate pairs which we have found. This is a nice visualization but doesn’t help find our coordinates, so lets switch to 2D.

In 2D we can use our rearranged equation:

This gives the following graph:

Here I have marked on the solution pairs that we found. The oblique asymptote (red) is y = 2x-1 because as x gets large the fraction gets very small and so the graph gets closer and closer to y = 2x -1.

All points on this curve are solutions to the equation – but we can see that the only integer solution pairs will be when x is small. When x is a large integer then the curve will be close to the asymptote and hence will return a number slightly bigger than an integer.

So, using this approach we would check all possible integer solutions when x is small, and again should be able to arrive at our coordinate pairs.

So, 4 different approaches that would be able to solve this problem. Can you find any others?

You can download all 17 of the Paper 3 questions for free here: [PDF].

The full typed mark scheme is available to download at the bottom of the page.

Seventeen IB Higher Level Paper 3 Practice Questions

With the new syllabus just started for IB Mathematics we currently don’t have many practice papers to properly prepare for the Paper 3 Higher Level exam. As a result I’ve put together 17 full investigation questions – each one designed to last around 1 hour, and totaling around 40 pages of questions and 600 marks worth of content. This has been specifically written for the Analysis and Approaches syllabus – though some parts would also be suitable for Applications.

Below I have split the questions into individual pdfs, with more detail about each one. For each investigation question I have combined several areas of the syllabus in order to create some level of discovery – and in many cases I have introduced some new mathematics (as will be the case on the real Paper 3).

Topics explored:

Paper 1: Rotating curves: [Individual question downloadhere. Mark-scheme downloadhere.]

Students explore the use of parametric and Cartesian equations to rotate a curve around the origin. You can see a tutorial video on this above. The mathematics used here is trigonometry (identities and triangles), functions and transformations.

Paper 12: Circumscribed and inscribed polygons [Individual question download here].

Students explore different methods for achieving an upper and lower bound for pi using circumscribed and inscribed polygons. You can see a video solution to this investigation above. The mathematics used here is trigonometry and calculus (differentiation and L’Hopital’s rule).

Paper 2: Who killed Mr. Potato? [Individual question download here.]

Students explore Newton’s Law of Cooling to predict when a potato was removed from an oven. The mathematics used here is logs laws, linear regression and solving differential equations.

Paper 3: Graphically understanding complex roots [Individual question download here.]

Students explore graphical methods for finding complex roots of quadratics and cubics. The mathematics used here is complex numbers (finding roots), the sum and product of roots, factor and remainder theorems, equations of tangents.

Paper 4: Avoiding a magical barrier [Individual question download here.]

Students explore a scenario that requires them to solve increasingly difficult optimization problems to find the best way of avoiding a barrier. The mathematics used here is creating equations, optimization and probability.

Paper 5 : Circle packing density [Individual question download here.]

Students explore different methods of filling a space with circles to find different circle packing densities. The mathematics used here is trigonometry and using equations of tangents to find intersection points.

Paper 6: A sliding ladder investigation [Individual question download here.]

Students find the general equation of the midpoint of a slipping ladder and calculate the length of the astroid formed. The mathematics used here is trigonometry and differentiation (including implicit differentiation). Students are introduced to the ideas of parametric equations.

Paper 7: Exploring the Si(x) function [Individual question download here.]

Students explore methods for approximating non-integrable functions and conclude by approximating pi squared. The mathematics used here is Maclaurin series, integration, summation notation, sketching graphs.

Paper 8: Volume optimization of a cuboid [Individual question download here.]

Students start with a simple volume optimization problem but extend this to a general case of an m by n rectangular paper folded to make an open box. The mathematics used here is optimization, graph sketching, extended binomial series, limits to infinity.

Paper 9: Exploring Riemann sums [Individual question download here.]

Students explore the use of Riemann sums to find upper and lower bounds of functions – finding both an approximation for pi and also for ln(1.1). The mathematics used here is integration, logs, differentiation and functions

Paper 10 : Optimisation of area [Individual question download here.]

Students start with a simple optimisation problem for a farmer’s field then generalise to regular shapes. The mathematics used here is trigonometry and calculus (differentiation and L’Hopital’s rule)

Paper 11: Quadruple Proof [Individual question download here.]

Students explore 4 different ways of proving the same geometrical relationship. The mathematics used here is trigonometry (identities) and complex numbers.

Paper 13: Using the binomial expansion for bounds of accuracy [Individual question download here.]

Students explore methods of achieving lower and upper bounds for and non-calculator methods for calculating logs. The mathematics used here is the extended binomial expansion for fractional and negative powers and integration.

Paper 14: Radioactive Decay [Individual question download here.]

Students explore discrete decay models, using probability density functions to investigate the decay of Carbon-14 and then explore the use of Euler’s method to approximate more complex decay chains. The mathematics used here is integration, probability density functions and Euler’s method of approximation

Paper 15: Probability generating functions [Individual question download here.]

Students explore the use of probability generating functions to find probabilities, expected values and variance for the binomial distribution and Poisson distribution for predicting the eruption of a volcano.

Paper 16: Finding the Steiner inellipse using complex numbers [Individual question download here]

Students use a beautiful relationship between complex numbers and an ellipse tangent to the midpoints of a triangle. This relationship allows you to find the equation of an ellipse from coordinate points of a triangle.

Paper 17: Elliptical curves [Individual question download here]

Students explore a method for adding points on an elliptical curve. This has links with elliptical curve cryptography.

Mark-scheme download

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IB HL Paper 3 Practice Questions and markscheme.

100 pages of preparatory questions with answers for the IB HL Analysis P3 exam. Please note this is not an automatic download and will be sent the same day.

$9.00

IB HL Paper 3 Practice Questions and markscheme AND Exploration Guide

All the Paper 3 questions and mark scheme AND the 63 page Exploration Guide. The Exploration Guide includes: Investigation essentials, Marking criteria guidance, 70 hand picked interesting topics, Useful websites for use in the exploration, A student checklist for top marks, Avoiding common student mistakes, A selection of detailed exploration ideas, Advice on using Geogebra, Desmos and Tracker. And more! Please note this is not an automatic download and will be sent the same day.

Using Desmos or Geogebra to design a picture or pattern is quite a nice exploration topic – but here’s an idea to make your investigation stand out from the crowd – how about converting your image to a 3D printed design?

Step 1

Create an image on Desmos or Geogebra. Remove the axes and grid pattern. This image is a pre-drawn image already on Desmos available here.

Step 2

Take a screen capture image of your picture (jpeg, gif, png). We need to convert this to a SVG file. You can convert these for free at sites like picsvg.

Step 3

Lastly we need to use a 3D editing site . You can join up with a site like Tinkercad for free.

Step 4

Making our 3D model. We import our SVG file and we get the image above. We can then resize this to whatever dimensions we wish – and also add 3D depth.

Lastly I would then save this file and send it to a 3D printer. You can see the finished file below:

So, if we printed this we’d get something like this:

3D printing the Eiffel Tower

Let’s use another Desmos art work. The Eiffel Tower above was a finalist in their annual art competition drawn by Jerry Yang from the USA.

This is then converted to the SVG file above.

And this is the result on Tinkercad when I add some depth and change the colour scheme. Let’s see what that would look like printed:

Pretty good- we’ve created a cheap tourist souvenir in about 5 minutes!

Mathematical art

I thought I’d have a go at making my own mathematical art. I started with using some polar coordinates to create this nice pattern:

Which then creates the following 3D shape:

This topic has a lot of scope for exploration and links with art, design technology and engineering. Thanks to our ever resourceful ICT wizz at school Jon for assistance, and also thanks for this excellent method which was posted by Ryan on Thingiverse. You can also explore huge numbers of ready made 3D templates on the site.

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IB Maths Exploration Guide

IB Maths Exploration Guide

A comprehensive 63 page pdf guide to help you get excellent marks on your maths investigation. Includes:

Full revision notes for SL Analysis (60 pages), HL Analysis (112 pages) and SL Applications (53 pages). Beautifully written by an experienced IB Mathematics teacher, and of an exceptionally high quality. Fully updated for the new syllabus. A must for all Analysis and Applications students!

Seventeen full investigation questions – each one designed to last around 1 hour, and totaling around 40 pages and 600 marks worth of content. There is also a fully typed up mark scheme. Together this is around 120 pages of content.

A 60 page pdf guide full of advice to help with modelling and statistics explorations – focusing in on non-calculator methods in order to show good understanding. Includes:

Pearson’s Product: Height and arm span

How to calculate standard deviation by hand

Binomial investigation: ESP powers

Paired t tests and 2 sample t tests: Reaction times