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**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:

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!

**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****.**]

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**

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

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.

$14.00

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:

- An introduction to the essentials about the investigation,
- The new marking criteria,
- How to choose a topic,
- Examples of around 70 topics that could be investigated,
- Useful websites for use in the exploration,
- A student checklist for completing a good investigation,
- Common mistakes that students make and how to avoid them,
- General stats projects advice,
- A selection of some interesting exploration topics explored in more depth,
- Teacher advice for marking,
- Templates for draft submissions,
- Advice on how to use Geogebra, Desmos and Tracker in your exploration,
- 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].

$9.00

**Resource Number 2**

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

- Linear regression
- Quadratic regression
- Cubic regression
- Exponential regression
- Linearisation using log scales
- Trigonometric regression
- Pearson’s Product investigations: Height and arm span
- Binomial investigations: ESP powers
- Poisson investigations: Customers in a shop
- 2 sample t tests: Reaction times
- Paired t tests: Reaction times
- Chi Squared: Efficiency of vaccines
- Bernoulli trials: Polling confidence intervals
- Spearman’s rank: Taste preference of cola
- 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].

$9.00

**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].

$13.50

**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.

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**

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:

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:

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

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).

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

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

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

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.

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

Substituting this back into the parametric equations gives:

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

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

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

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.

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)**

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!

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.

**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)t^{n}. Again for matters of slight convenience I’ll look at the similar relationship -0.5(n+1)(n+2)at^{n}.

This gives (following the same method as above:

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:

When n = 1:

When n =2:

When n = 10:

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:**

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:

- Find the hypotenuse of triangle AGC.
- Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
- Therefore triangle AGC and GBD are similar, so length BG = 4. We can now use Pythagoras to find length BD.
- We can find length CD by Pythagoras.
- Now we have 3 sides of a triangle, CDB. This allows use to find angle BDC using the cosine rule.
- Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
- Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
- 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.
- 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)

**Can you solve Oxford University’s Interview Question?**

The excellent Youtube channel Mind Your Decisions is a gold mine for potential IB maths exploration topics. I’m going to follow through my own approach to problem posed in the video. The problem is to be able to trace the movement of the midpoint of a ladder as it slides down a wall. This has apparently been used as an Oxford interview question to test the ability to investigate novel problems.

It’s normally a good idea to start with a specific case with some nice numbers, to see what happens. So, I’ll choose a 3,4,5 triangle, where the ladder has a fixed length of 5 and has endpoints with coordinates (0,4) and (3,0). The midpoint is given by ((0+3)/2, (4+0)/2) = (1.5, 2).

Next I imagine what would happen to the point (3,0) if the ladder slipped down the wall. (3,0) would become (3+t,0) where t is a parameter. Given that the length is fixed as 5, I can now find the new height of the ladder up the wall using Pythagoras:

The new height is given by:

Therefore the new midpoint is given by:

We can now define our curve parametrically:

Therefore we can make t the subject in the first equation to get an equation just in terms of x and y.

Therefore we can rearrange to get:

This is the equation of a circle centred at (0,0) with radius 2.5:

This graph therefore traces the movement of the midpoint of the ladder (note that when the ladder was vertical against the wall the midpoint would be 2.5 high hence the graph starts at (0,2.5).

**The general case**

Now we have worked through the maths for a specific case, the general case isn’t too much extra work. For a triangle with base a and height b we would have the following midpoint coordinates:

This would lead to the following equation:

Which would rearrange to give the equation of the circle:

This is a circle centered at (0,0) with radius:

**Another approach**

This method is an alternative to the version above – this time using trigonometry. We start with the triangle below:

and then let the ladder slide to get the following (as the angle will get smaller t will be negative):

We can then define the midpoint coordinates as:

We can then rearrange and square both sides to get the following:

We can then use the trig identity for cosine squared theta + sine squared theta = 1:

Which rearranges to give the same result as before:

So, there we go – we’ve passed an Oxford interview question with a couple of different methods! The approach of first exploring the topic with a simple case is often a good starting point for these sorts of problems – as it allows you to gain an understanding of what is happening without getting too bogged down with variables. You can watch the video for a quicker solution – are there any other ways of approaching this problem you can find? How could this problem be modified?

**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!