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Galileo’s Inclined Planes

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!

Finding focus with Archimedes

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.

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:

Screen Shot 2021-01-09 at 6.51.05 PM

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.


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!

Generating e through probability and hypercubes

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).


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 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.



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 download here.  Mark-scheme download here.]

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.]

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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.]

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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.]

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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.]

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 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.]

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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.]

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 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.]

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 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.]

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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.]

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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.]

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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.]

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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.]

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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.]

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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]

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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]

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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.


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.


The IB Maths Exploration Guide and the IB Maths Modelling and Statistics Exploration Guide are suitable for all IB students.  

They are both written by an IB teacher with an MSc. in Mathematics, 10 years experience teaching IB Standard and Higher Level and who has worked as an IB examiner (IA moderation).

Resource Number 1

The Exploration Guide talks you through:

  1. An introduction to the essentials about the investigation,
  2. The new marking criteria,
  3. How to choose a topic,
  4. Examples of around 70 topics that could be investigated,
  5. Useful websites for use in the exploration,
  6. A student checklist for completing a good investigation,
  7. Common mistakes that students make and how to avoid them,
  8. General stats projects advice,
  9. A selection of some interesting exploration topics explored in more depth,
  10. Teacher advice for marking,
  11. Templates for draft submissions,
  12. Advice on how to use Geogebra, Desmos and Tracker in your exploration,
  13. Some examples of beautiful maths using Geogebra and Desmos.

[If you don’t have a PayPal account you can just click on the relevant credit card icon].

Exploration Guide

A comprehensive 63 page pdf guide to help you get excellent marks on your maths investigation. [This is not an automatic download but will be emailed the same day].


Resource Number 2

The Modelling and Statistics Guide talks you through various techniques useful for statistical and modelling explorations. Topics included are:

  1. Linear regression
  2. Quadratic regression
  3. Cubic regression
  4. Exponential regression
  5. Linearisation using log scales
  6. Trigonometric regression
  7. Pearson’s Product investigations: Height and arm span
  8. Binomial investigations: ESP powers
  9. Poisson investigations: Customers in a shop
  10. 2 sample t tests: Reaction times
  11. Paired t tests: Reaction times
  12. Chi Squared: Efficiency of vaccines
  13. Bernoulli trials: Polling confidence intervals
  14. Spearman’s rank: Taste preference of cola
  15. Sampling techniques and experiment design.

[If you don’t have a PayPal account you can just click on the relevant credit card icon].

Modelling and Statistics Guide

A 60 page pdf guide full of advice to help with modelling and statistics explorations. [This is not an automatic download but will be emailed the same day].


Resource Number 3

The Exploration Guide and the Modelling and Statistics Guide can be purchased together for a discount.

Exploration Guide AND the Modelling and Statistics Guide

Both guides included together for a discount. [This is not an automatic download but will be emailed the same day].



The Martingale system

The Martingale system was first used in France in 1700s gambling halls and remains used today in some trading strategies.  I’ll look at some of the mathematical ideas behind this and why it has remained popular over several centuries despite having a long term expected return of zero.

The scenario

You go to a fair ground and play a simple heads-or-tails game.  The probability of heads is 1/2 and tails is also 1/2.  You place a stake of counters on heads.  If you guess correctly you win that number of counters.  If you lose, you double your stake of counters and then the coin is tossed again.  Every time you lose you double up your stake of counters and stop when you finally win.

Infinitely deep pockets model:

You can see that in the example above we always have a 0.5 chance of getting heads on the first go, which gives a profit of 1 counter.  But we also have a 0.5 chance of a profit of 1 counter as long as we keep doubling up our stake, and as long as we do indeed eventually throw heads.  In the example here you can see that the string of losing throws don’t matter [when we win is arbitrary, we could win on the 2nd, 3rd, 4th etc throw].  By doubling up, when you do finally win you wipe out your cumulative losses and end up with a 1 counter profit.

This leads to something of a paradoxical situation, despite only having a 1/2 chance of guessing heads we end up with an expected value of 1 counter profit for every 1 counter that we initially stake in this system.

So what’s happening?  This will always work but it requires that you have access to infinitely deep pockets (to keep your infinite number of counters) and also the assumption that if you keep throwing long enough you will indeed finally get a head (i.e you don’t throw an infinite number of tails!)

Finite pockets model:

Real life intrudes on the infinite pockets model – because in reality there will be a limit to how many counters you have which means you will need to bail out after a given number of tosses.  Even if the probability of this string of tails is very small, the losses if it does occur will be catastrophic –  and so the expected value for this system is still 0.

Finite pockets model capped at 4 tosses:

In the example above we only have a 1/16 chance of losing – but when we do we lose 15 counters.  This gives an expected value of:

Finite pockets model capped at n tosses:

If we start with a 1 counter stake then we can represent the pattern we can see above for E(X) as follows:

Here we use the fact that the losses from n throws are the sum of the first (n-1) powers of 2. We can then notice that both of these are geometric series, and use the relevant formula to give:

Therefore the expected value for the finite pockets model is indeed always still 0.

So why does this system remain popular?

So, given that the real world version of this has an expected value of 0, why has it retained popularity over the past few centuries?  Well, the system will on average return constant linear growth – up until a catastrophic loss.  Let’s say you have 100,000 counters and stake 1 counter initially.  You can afford a total of 16 consecutive losses.  The probability of this is only:

but when you do lose, you’ll lose a total of:

So, the system creates a model that mimics linear growth, but really the small risk of catastrophic loss means that the system still has E(X) = 0.  In the short term you would expect to see the following very simple linear relationship for profit:

With 100,000 counters and a base trading stake of 1 counter, if you made 1000 initial 1 counter trades a day you would expect a return of 1000 counters a day (i.e 1% return on your total counters per day).  However the longer you continue this strategy the more likely you are to see a run of 16 tails – and see all your counters wiped out.

Computer model

I wrote a short Python code to give an idea as to what is happening. Here I started 9 people off with 1000 counters each.  They have a loss limit of 10 consecutive losses.  They made starting stakes of 1 counter each time, and then I recorded how long before they made a loss of 10 tosses in a row.

For anyone interested in the code here it is:


The program returned the following results.  The first number is the number of starting trades until they tossed 10 tails in a row.  The second number was their new account value (given that they had started with 1000 counters, every previous trade had increased their account by 1 counter and that they had then just lost 1023 counters).

1338, 1315
1159, 1136
243, 220
1676, 1653
432, 409
1023, 1000
976, 953
990, 967
60, 37

This was then plotted on Desmos. The red line is the trajectory their accounts were following before their loss.  The horizontal dotted line is at y = 1000 which represents the initial account value.  As you can see 6 people are now on or below their initial starting account value.  You can also see that all these new account values are themselves on a line parallel to the red line but translated vertically down.

From this very simple simulation, we can see that on average a person was left with 884 counters following hitting 10 tails.  i.e below initial starting account.  Running this again with 99 players gave an average of 869.

999 players

I ran this again with 999 players – counting what their account value would be after their first loss.  All players started with 1000 counters.  The results were:

31 players bankrupt: 3%

385 players left with less than half their account value (less than 500): 39%

600 players with less than their original account value (less than 1000): 60%

51 players at least tripled their account (more than 3000): 5%

The top player ended up with 6903 counters after their first loss.

The average account this time was above starting value (1044.68).  You can see clearly that the median is below 1000 – but that a small number of very lucky players at the top end skewed the mean above 1000.

Second iteration

I then ran the simulation again – with players continuing with their current stake.  This would have been slightly off because my model allowed players who were bankrupt from the first round to carry on [in effect being loaned 1 counter to start again].  Nevertheless it now gave:

264 players bankrupt: 26%

453 players left with less than half their account value (less than 500): 45%

573 players with less than their original account value (less than 1000): 57%

95 players at least tripled their account (more than 3000): 10%

The top player ended up with 9583 counters after their second loss.

We can see a dramatic rise in bankruptcies – now over a quarter of all players.  This would suggest the long term trend is towards a majority of players being bankrupted, though the lucky few at the top end may be able to escape this fate.

Screen Shot 2020-04-08 at 1.09.39 PM

This carries on our exploration of projectile motion – this time we will explore what happens if gravity is not fixed, but is instead a function of time.  (This idea was suggested by and worked through by fellow IB teachers Daniel Hwang and Ferenc Beleznay).   In our universe we have a gravitational constant – i.e gravity is not dependent on time.  If gravity changed with respect to time then the gravitational force exerted by the Sun on Earth would lessen (or increase) over time with all other factors remaining the same.

Interestingly time-dependent gravity was first explored by Dirac and some physicists have tried to incorporate time dependent gravity into cosmological models.  As yet we have no proof that gravity is not constant, but let’s imagine a university where it is dependent on time.

Projectile motion when gravity is time dependent

Screen Shot 2020-04-10 at 2.12.03 PM

We can start off with the standard parametric equations for projectile motion. Here v is the initial velocity, theta is the angle of launch, t can be a time parameter and g is the gravitational constant (9.81 on Earth).  We can see that the value for the vertical acceleration is the negative of the gravitational constant.  So the question to explore is, what if the gravitational constant was time dependent?  Another way to think about this is that gravity varies with respect to time.

Linear relationship

If we have the simplest time dependent relationship we can say that:

Screen Shot 2020-04-08 at 1.01.41 PM

where a is a constant.  If a is greater than 0 then gravity linearly increases as time increases, if a is less than 0 than gravity linearly decreases as time increases.  For matters of slight convenience I’ll define gravity (or the vertical acceleration) as -3at.  The following can then be arrived at by integration:

Screen Shot 2020-04-10 at 2.17.28 PM

This will produce the following graph when we fix v = 10, a = 2 and vary theta:

Screen Shot 2020-04-08 at 1.09.39 PM

Now we can use the same method as in our Projectile Motion Investigation II to explore whether these maximum points lie in a curve.  (You might wish to read that post first for a step by step approach to the method).

Screen Shot 2020-04-08 at 1.44.34 PM

therefore we can substitute back into our original parametric equations for x and y to get:

Screen Shot 2020-04-08 at 1.46.02 PM

We can plot this with theta as a parameter.  If we fix v = 4 and a =2 we get the following graph:

Screen Shot 2020-04-08 at 1.33.06 PM

Compare this to the graph from Projectile Motion Investigation II, where we did this with gravity constant (and with v fixed as 10):

Screen Shot 2020-04-06 at 9.34.04 PM

The Projectile Motion Investigation II formed a perfect ellipse, but this time it’s more of a kind of egg shaped elliptical curve – with a flat base.  But it’s interesting to see that even with time dependent gravity we still have a similar relationship to before!

Inverse relationship

Let’s also look at what would happen if gravity was inversely related to time.  (This is what has been explored by some physicists).

In this case we get the following results when we launch projectiles (Notice here we had to use the integration by parts trick to integrate ln(t)).  As the velocity function doesn’t exist when t = 0, we can define v and theta in this case as the velocity and theta value when t = 1.

Screen Shot 2020-04-10 at 2.34.42 PM

Now we use the same trick as earlier to find when the gradient is 0:

Screen Shot 2020-04-10 at 2.37.22 PM

Substituting this back into the parametric equations gives:

Screen Shot 2020-04-10 at 2.42.57 PM

The ratio v/a will therefore have the greatest effect on the maximum points.

v/a ratio negative and close to zero:

v = 40, a = -2000, v/a = -0.02

Screen Shot 2020-04-10 at 2.52.57 PM

This gives us close to a circle, radius v, centred at (0,a).

v = 1, a = -10, v/a = -0.1

Screen Shot 2020-04-10 at 2.59.20 PM

Here we can also see that the boundary condition for the maximum horizontal distance thrown is given by x = v(e).

v/a ratio negative and large:

v = 40, a = -2, v/a = -20.

Screen Shot 2020-04-10 at 2.48.30 PM

We can see that we get an egg shape back – but this time with a flatter bulge at the top and the point at the bottom.  Also notice how quickly the scale of the shape has increased.

v/a ratio n/a (i.e a = 0)

Screen Shot 2020-04-10 at 3.07.17 PM

Here there is no gravitational force, and so projectiles travel in linear motion – with no maximum.

Envelope of projectiles for the inverse relationship

This is just included for completeness, don’t worry if you don’t follow the maths behind this bit!

Screen Shot 2020-04-15 at 10.09.59 AM

Screen Shot 2020-04-15 at 10.10.11 AM

Therefore when we plot the parametric equations for x and y in terms of theta we get the envelope of projectile motion when we are in a universe where gravity varies inversely to time.  The following graph is generated when we take v = 300 and a = -10.  The red line is the envelope of projectiles.

Screen Shot 2020-04-15 at 10.11.19 AM

A generalized power relationship

Lastly, let’s look at what happens when we have a general power relationship i.e gravity is related to (a)tn.  Again for matters of slight convenience I’ll look at the similar relationship -0.5(n+1)(n+2)atn.

Screen Shot 2020-04-10 at 2.21.52 PM

This gives (following the same method as above:

Screen Shot 2020-04-08 at 6.48.11 PM

Screen Shot 2020-04-08 at 6.48.46 PM

As we vary n we will find the plot of the maximum points.  Let’s take the velocity as 4 and a as 2.  Then we get the following:

When n = 0:

Screen Shot 2020-04-08 at 8.18.07 PM

When n = 1:

Screen Shot 2020-04-08 at 8.16.42 PM

When n =2:

Screen Shot 2020-04-08 at 8.18.21 PM

When n = 10:

Screen Shot 2020-04-08 at 8.18.44 PM

We can see the general elliptical shape remains at the top, but we have a flattening at the bottom of the curve.

When n approaches infinity:

Screen Shot 2020-04-08 at 8.26.45 PM

We get this beautiful result when we let n tend towards infinity – now we will have all the maximum points bounded on a circle (with the radius the same as the value chosen as the initial velocity.  In the graph above we have a radius of 4 as the initial velocity is 4. Notice too we have projectiles traveling in straight lines – and then seemingly “bouncing” off the boundary!

If we want to understand this, there is only going to be a very short window (t less than 1) when the particle can upwards – when t is between 0 and 1 the effect of gravity is effectively 0 and so the particle would travel in a straight line (i.e if the initial velocity is 5 m/s it will travel 5 meters. Then as soon as t = 1, the gravity becomes crushingly heavy and the particle falls effectively vertically down.

Classical Geometry Puzzle: Finding the Radius

This is another look at a puzzle from Mind Your Decisions.  The problem is to find the radius of the following circle:

We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more!  First I’ll look at my attempt to solve this.  It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.

Method 1, extra construction lines:

These are the extra construction lines required to solve this problem.  Here is the step by step thought process:

  1. Find the hypotenuse of triangle AGC.
  2. Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
  3. Therefore triangle AGC and GBD are similar, so length BG = 4.  We can now use Pythagoras to find length BD.
  4. We can find length CD by Pythagoras.
  5. Now we have 3 sides of a triangle, CDB.  This allows use to find angle BDC using the cosine rule.
  6. Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
  7. Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
  8. Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
  9. Therefore the radius is approximately 4.03.

Method 2, creating a coordinate system

This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above).  We start by creating a coordinate system based around point G at (0,0).  Because we have perpendicular lines we can therefore create coordinates for A, B and C.  We also mark the centre of the circle as (p,q).

First we start with the equation of a circle centre (p.q):

Next we create 3 equations by substituting in our coordinates:

Next we can do equation (3) – equation (1) to give:

Next we can substitute this value for p into equations (1) and (3) and equate to get:

Lastly we can substitute both values for p and q into equation (1) to find r:

We get the same answer as before – though this definitely feels like a “cleaner” solution.  There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius).  Perhaps explore any other methods for solving this – what are the relative merits of each approach?

The Mordell Equation [Fermat’s proof]

Let’s have a look at a special case of the Mordell Equation, which looks at the difference between an integer cube and an integer square.  In this case we want to find all the integers x,y such that the difference between the cube and the square gives 2.  These sorts of problems are called Diophantine problems and have been studied by mathematicians for around 2000 years.  We want to find integer solution to:

First we can rearrange and factorise, using the property of imaginary numbers.

Next we define alpha and beta such that:

For completeness we can say that alpha and beta are part of an algebraic number field:

Next we use an extension of the Coprime Power Trick, which ensures that the following 2 equations have solutions (if our original equation also has a solution). Therefore we define:

We can then substitute our definition for alpha into the first equation directly above and expand:

Next we equate real and imaginary coefficients to give:

This last equation therefore requires that either one of the following equations must be true:

If we take the case when b = 1 we get:


If we take the case when b = -1 we get

Therefore our solution set is (a,b): (1,1), (1,-1), (-1,1), (-1,-1.  We substitute these possible answers into our definition for y to give the following:

We can then substitute these 2 values for y into the definition for x to get:

These therefore are the only solutions to our original equation.  We can check they both work:

We can see this result illustrated graphically by plotting the graph:

and then seeing that we have our integer solutions (3,5) and (3,-5) as coordinate on this curve.

This curve also clearly illustrates why we have a symmetrical set of solutions, as our graph is symmetrical about the x axis.

This particular proof was first derived by Fermat (of Fermat’s Last Theorem fame) in the 1600s and is an elegant example of a proof in number theory.  You can read more about the Mordell Equation in this paper (the proof above is based on that given in the paper, but there is a small mistake in factorization so that y = 7 and y = -7 is erroneously obtained)


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IB Maths Exploration Guide

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

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IB HL Paper 3 Practice Questions 

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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:

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