**Euler’s 9 Point Circle**

This is a nice introduction to some of the beautiful constructions of geometry. This branch of mathematics goes in and out of favour – back in the days of Euclid, constructions using lines and circles were a cornerstone of mathematical proof, interest was later revived in the 1800s through Poncelot’s projective geometry – later leading to the new field of non Euclidean geometry. It’s once again somewhat out of fashion – but more accessible than ever due to programs like Geogebra (on which the below diagrams were plotted). The 9 point circle (or at least the 6 point circle was discovered by the German Karl Wilhelm von Feuerbach in the 1820s. Unfortunately for Feuerbach it’s often instead called the Euler Circle – after one of the greatest mathematicians of all time, Leonhard Euler.

So, how do you draw Euler’s 9 Point Circle? It’s a bit involved, so don’t give up!

Step 1: Draw a triangle:

Step 2: Draw the perpendicular bisectors of the 3 sides, and mark the point where they all intersect (D).

Step 3: Draw the circle through the point D.

Step 4: From each line of the triangle, draw the perpendicular line through its third angle. For example, for the line AC, draw the perpendicular line that goes through both AC and angle B. (The altitudes of the triangle). Join up the 3 altitudes which will meet at a point (E).

Step 5: Join up the mid points of each side of the triangle with the remaining angle. For example, find the mid point of AC and join this point with angle B. (The median lines of the triangle). Label the point where the 3 lines meet as F.

Step 6: Remove all the construction lines. You can now see we have 3 points in a line. D is the centre of the circle through the points ABC, E is where the altitudes of the triangle meet (the orthoocentre of ABC) and F is where the median lines meet (the centroid of ABC).

Step 7: Join up the 3 points – they are collinear (on the same line).

Step 8: Enlarge the circle through points A B C by a scale factor of -1/2 centered on point F.

Step 9: We now have the 9 point circle. Look at the points where the inner circle intersects the triangle ABC. You can see that the points M N O show the points where the feet of the altitudes (from step 4) meet the triangle.

The points P Q R show the points where the perpendicular bisectors of the lines start (i.e the midpoints of the lines AB, AC, BC)

We also have the points S T U on the circle which show the midpoints of the lines between E and the vertices A, B, C.

Step 10: We can drag the vertices of the triangle and the above relationships will still hold.

In the second case we have both E and D outside the triangle.

In the third case we have E and F at the same point.

In the fourth case we have D and E on opposite sides of the triangle.

So there we go – who says maths isn’t beautiful?

]]>**Log Graphs to Plot Planetary Patterns**

This post is inspired by the excellent Professor Stewart’s latest book, Calculating the Cosmos. In it he looks at some of the mathematics behind our astronomical knowledge.

**Astronomical investigations**

In the late 1760s and early 1770s, 2 astronomers Titius and Bode both noticed something quite strange – there seemed to be a relationship in the distances between the planets. There was no obvious reason as to why there would be – but nevertheless it appeared to be true. Here are the orbital distances from the Sun of the 6 planets known about in the 1760s:

Mercury: 0.39 AU

Venus: 0.72 AU

Earth: 1.00 AU

Mars: 1.52 AU

Jupiter: 5.20 AU

Saturn: 9.54 AU

In astronomy, 1 astronomical unit (AU) is defined as the mean distance from the center of the Earth to the centre of the Sun (149.6 million kilometers).

Now, at first glance there does not appear to be any obvious relationship here – it’s definitely not linear, but how about geometric? Well dividing the term above by the term below we get r values of:

1.8, 1.4, 1.5, 3.4, 1.8

4 of the numbers are broadly similar – and then we have an outlier of 3.4. So either there was no real pattern – or there was an undetected planet somewhere between Mars and Jupiter? And was there another planet beyond Saturn?

**Planet X**

Mercury: 0.39 AU

Venus: 0.72 AU

Earth: 1.00 AU

Mars: 1.52 AU

Planet X: x AU

Jupiter: 5.20 AU

Saturn: 9.54 AU

Planet Y: y AU

For a geometric sequence we would therefore want x/1.52 = 5.20/x. This gives x = 2.8 AU – so a missing planet should be 2.8 AU away from the Sun. This would give us r values of 1.8, 1.4, 1.5, 1.8, 1.9, 1.8. Let’s take r = 1.8, which would give Planet Y a distance of 17 AU.

So we predict a planet around 2.8 AU from the Sun, and another one around 17 AU from the Sun. In 1781, Uranus was discovered – 19.2 AU from the Sun, and in 1801 Ceres was discovered at 2.8 AU. Ceres is what is now classified as a dwarf planet – the largest object in the asteroid belt between Jupiter and Mars.

**Log Plots**

Using graphs is a good way to graphically see relationships. Given that we have a geometrical relationship in the form d = ab^n with a and b as constants, we can use the laws of logs to rearrange to give log d = log a + n log b.

Therefore we can plot log d on the y axis, and n on the x axis. If there is a geometrical relationship we will see us a linear relationship on the graph, with log a being the y intercept and the gradient being log b.

(n=1) Mercury: d = 0.39 AU. log d = -0.41

(n=2) Venus: d = 0.72 AU. log d = -0.14

(n=3) Earth: d = 1.00 AU. log d = 0

(n=4) Mars: d = 1.52 AU. log d = 0.18

(n=5) Ceres (dwarf): d = 2.8 AU. log d = 0.45

(n=6) Jupiter: d = 5.20 AU. log d = 0.72

(n=7) Saturn: d = 9.54 AU. log d = 0.98

(n=8) Uranus: d = 19.2 AU. log d = 1.28

We can use Desmos’ regression tool to find a very strong linear correlation – with y intercept as -0.68 and gradient as 0.24. Given that log a is the y intercept, this gives:

log a = -0.68

a = 0.21

and given that log b is the gradient this gives:

log b = 0.24

b = 1.74

So our final formula for the relationship for the spacing of the n ordered planets is:

d = ab^n

distance = 0.21 x (1.74)^n.

**Testing the formula**

So, using this formula we can predict what the next planetary distance would be. When n = 9 we would expect a distance of 30.7 AU. Indeed we find that Neptune is 30.1 AU – success! How about Pluto? Given that Pluto has a very eccentric (elliptical) orbit we might not expect this to be as accurate. When n = 10 we get a prediction of 53.4 AU. The average AU for Pluto is 39.5 – so our formula does not work well for Pluto. But looking a little more closely, we notice that Pluto’s distance from the Sun varies wildly – from 29.7 AU to 49.3 AU, so perhaps it is not surprising that this doesn’t follow our formula well.

**Other log relationships**

Interestingly other distances in the solar system show this same relationship. Plotting the ordered number of the planets against the log of their orbital period produces a linear graph, as does plotting the ordered moons of Uranus against their log distance from the planet. Why these relationships exist is still debated. Perhaps they are a coincidence, perhaps they are a consequence of resonance in orbital periods. Do some research and see what you find!

]]>This is a quick example of how using Tracker software can generate a nice physics-related exploration. I took a spring, and attached it to a stand with a weight hanging from the end. I then took a video of the movement of the spring, and then uploaded this to Tracker.

**Height against time**

The first graph I generated was for the height of the spring against time. I started the graph when the spring was released from the low point. To be more accurate here you can calibrate the y axis scale with the actual distance. I left it with the default settings.

You can see we have a very good fit for a sine/cosine curve. This gives the approximate equation:

y = -65cos10.5(t-3.4) – 195

(remembering that the y axis scale is x 100).

This oscillating behavior is what we would expect from a spring system – in this case we have a period of around 0.6 seconds.

**Momentum against velocity**

For this graph I first set the mass as 0.3kg – which was the weight used – and plotted the y direction momentum against the y direction velocity. It then produces the above linear relationship, which has a gradient of around 0.3. Therefore we have the equation:

p = 0.3v

If we look at the theoretical equation linking momentum:

p = mv

(Where m = mass). We can see that we have almost perfectly replicated this theoretical equation.

**Height against velocity**

I generated this graph with the mass set to the default 1kg. It plots the y direction against the y component velocity. You can see from the this graph that the velocity is 0 when the spring is at the top and bottom of its cycle. We can then also see that it reaches its maximum velocity when halfway through its cycle. If we were to model this we could use an ellipse (remembering that both scales are x100 and using x for vy):

If we then wanted to develop this as an investigation, we could look at how changing the weight or the spring extension affected the results and look for some general conclusions for this. So there we go – a nice example of how tracker can quickly generate some nice personalised investigations!

]]>**Predicting the UK election using linear regression**

The above data is the latest opinion poll data from the Guardian. The UK will have (another) general election on June 8th. So can we use the current opinion poll data to predict the outcome?

**Longer term data trends**

Let’s start by looking at the longer term trend following the aftermath of the Brexit vote on June 23rd 2016. I’ll plot some points for Labour and the Conservatives and see what kind of linear regression we get. To keep things simple I’ve looked at randomly chosen poll data approximately every 2 weeks – assigning 0 to July 1st 2016, 1 to mid July, 2 to August 1st etc. This has then been plotted using the fantastic Desmos.

**Labour**

You can see that this is not a very good fit – it’s a very weak correlation. Nevertheless let’s see what we would get if we used this regression line to predict the outcome in June. With the x axis scale I’ve chosen, mid June 2017 equates to 23 on the x axis. Therefore we predict the percentage as

y = -0.130(23) + 30.2

y = 27%

Clearly this would be a disaster for Labour – but our model is not especially accurate so perhaps nothing to worry about just yet.

**Conservatives**

As with Labour we have a weak correlation – though this time we have a positive rather than negative correlation. If we use our regression model we get a prediction of:

y = 0.242(23) + 38.7

y = 44%

So, we are predicting a crushing victory for the Conservatives – but could we get some more accurate models to base this prediction on?

**Using moving averages**

The Guardian’s poll tracker at the top of the page uses moving averages to smooth out poll fluctuations between different polls and to arrive at an averaged poll figure. Using this provides a stronger correlation:

**Labour**

This model doesn’t take into account a (possible) late surge in support for Labour but does fir better than our last graph. Using the equation we get:

y = -0.0764(23) + 28.8

y = 27%

**Conservatives**

We can have more confidence in using this regression line to predict the election. Putting in the numbers we get:

y = 0.411(23) + 36.48

y = 46%

**Conclusion**

Our more accurate models merely confirm what we found earlier – and indeed what all the pollsters are predicting – a massive win for the Conservatives. Even allowing for a late narrowing of the polls the Conservatives could be on target for winning by over 10% points – which would result in a very large majority. Let’s see what happens!

]]>A farmer has 40m of fencing. What is the maximum area he can enclose?

**Case 1: The rectangle:**

Reflection – the rectangle turns out to be a square, with sides 10m by 10m. Therefore the area enclosed is 100 metres squared.

**Case 2: The circle:**

Reflection: The area enclosed is greater than that of the square – this time we have around 127 metres squared enclosed.

**Case 3: The isosceles triangle:**

Reflection – our isosceles triangle turns out to be an equilateral triangle, and it only encloses an area of around 77 metres squared.

**Case 4, the n sided regular polygon**

Reflection: Given that we found the cases for a 3 sided and 4 sided shape gave us the regular shapes, it made sense to look for the n-sided regular polygon case. If we try to plot the graph of the area against n we can see that for n ≥3 the graph has no maximum but gets gets closer to an asymptote. By looking at the limit of this area (using Wolfram Alpha) as n gets large we can see that the limiting case is the circle. This makes sense as regular polygons become closer to circles the more sides they have.

**Proof of the limit using L’Hospital’s Rule**

Here we can prove that the limit is indeed 400/pi by using L’Hospital’s rule. We have to use it twice and also use a trig identity for sin(2x) – but pleasingly it agrees with Wolfram Alpha.

So, a simple example of how an investigation can develop – from a simple case, getting progressively more complex and finishing with some HL Calculus Option mathematics.

]]>**Cracking ISBN and Credit Card Codes**

ISBN codes are used on all books published worldwide. It’s a very powerful and useful code, because it has been designed so that if you enter the wrong ISBN code the computer will immediately know – so that you don’t end up with the wrong book. There is lots of information stored in this number. The first numbers tell you which country published it, the next the identity of the publisher, then the book reference.

**Here is how it works:**

Look at the 10 digit ISBN number. The first digit is 1 so do 1×1. The second digit is 9 so do 2×9. The third digit is 3 so do 3×3. We do this all the way until 10×3. We then add all the totals together. If we have a proper ISBN number then we can divide this final number by 11. If we have made a mistake we can’t. This is a very important branch of coding called error detection and error correction. We can use it to still interpret codes even if there have been errors made.

If we do this for the barcode above we should get 286. 286/11 = 26 so we have a genuine barcode.

**Check whether the following are ISBNs**

1) 0-13165332-6

2) 0-1392-4191-4

3) 07-028761-4

**Challenge (harder!) :**The following ISBN code has a number missing, what is it?

1) 0-13-1?9139-9

Answers in white text at the bottom, highlight to reveal!

Credit cards use a different algorithm – but one based on the same principle – that if someone enters a digit incorrectly the computer can immediately know that this credit card does not exist. This is obviously very important to prevent bank errors. The method is a little more complicated than for the ISBN code and is given below from computing site Hacktrix:

You can download a worksheet for this method here. Try and use this algorithm to validate which of the following 3 numbers are genuine credit cards:

1) 5184 8204 5526 6425

2) 5184 8204 5526 6427

3) 5184 8204 5526 6424

Answers in white text at the bottom, highlight to reveal!

ISBN:

1) Yes

2) Yes

3) No

1) 3 – using x as the missing number we end up with 5x + 7 = 0 mod 11. So 5x = 4 mod 11. When x = 3 this is solved.

Credit Card: The second one is genuine

If you liked this post you may also like:

NASA, Aliens and Binary Codes from the Stars – a discussion about how pictures can be transmitted across millions of miles using binary strings.

Cracking Codes Lesson – an example of 2 double period lessons on code breaking

]]>**Modelling a Nuclear War **

With the current saber rattling from Donald Trump in the Korean peninsula and the instability of North Korea under Kim Jong Un (incidentally a former IB student!) the threat of nuclear war is once again in the headlines. Post Cold War we’ve got somewhat used to the peace afforded by the idea of mutually assured destruction – but this peace only holds with rational actors in charge of pushing the buttons. The closest we have got to a nuclear war between 2 nuclear powers was in the 1962 Cuban missile crisis – and given the enormous nuclear arsenals of the US and the then USSR this could have pretty much ended civilization as we know it. In that period, modelling of the effects of nuclear war was a real priority. So let’s have a look at some current modelling predictions for the effects of a nuclear war. Those of a nervous disposition may wish to look away!

**Nuclear blast radius**

The picture at the top of the post is the nuclear blast radius calculated from this site. It shows the effects of a 100 megaton airburst (equivalent to 100 million tonnes of TNT explosive). This is the biggest nuclear bomb that the USSR ever tested. If dropped on London it would have a fireball radius of 6km, an air blast radius of 33 km (destroying most buildings) and a thermal radiation radius of 74km. The site estimates that this single bomb would cause 6 million deaths and another 6 million injuries. And remember this is a single bomb – there are collectively around 15,000 nuclear weapons in the world (the majority shared between the US and Russia).

**Nuclear Winter**

Whilst the effects of a single bomb would be absolutely catastrophic for both a country and also for the global economy, it would not be an extinction event for humanity – however scientists have modelled the consequences of a *nuclear war *which would effect the climate to such an extent that it could lead to global mass extinctions.

Let’s have a look at one of those papers – the pessimistically titled:

Nuclear winter revisited with a modern climate model and current nuclear arsenals: Still catastrophic consequences.

In this paper the authors look at 2 scenarios – the long term climate effect of (a) the detonation of 1/3 of the world’s arsenal of nuclear weapons and (b) the detonation of the full arsenal of the world’s nuclear weapons. Let’s leave to the side that this would almost certainly end civilisation as we know it – but what would be in store for those lucky (?) enough to survive such an event?

**Changes to global temperature and rainfall**

This above graphic is a double line graph – with the red lines relating to changes in temperature and the black line corresponding to the changes in precipitation. The middle 2 lines relate to case (a) and the bottom 2 lines relate to case (b). The y axis relates to years. You can see from this graph that a large nuclear war where 1/3 of the nuclear arsenal was released would have a significant effect on both global temperatures and rainfall. 5 years after the detonations you would have a global temperature 3-4 degrees lower than normal, and even a decade later it would still be a degree lower than normal. For the full nuclear arsenal case the effects would be catastrophic – a average global drop in temperature of close to 9 degrees 2-3 years after the event. To put this in context – the last ice age had global temperatures around 5 degrees lower than present. Meanwhile the average rainfall would drop by around 1.6mm/day equivalent to a 45% global drop in rainfall.

**Localised effects of changes in precipitation **

This above graphic shows the distribution of the effects of precipitation following the detonation of the full nuclear arsenal one year on. You can see that not all parts of the globe are equally effected. The countries near the equator see a massive drop in rainfall (more than 3.5mm/day) along with large parts of North America and Western Europe

**Localised effects in the change in temperatures:**

This above graphic shows the distribution of the effects of temperature following the detonation of the full nuclear arsenal one year on. As with the rainfall, you can see startling changes – parts of North America would be 20-30 degrees colder than average, parts of Russia 30-35 degrees colder. You can see the misleading nature of global temperature averages here. The global average temperature drop after 1 year was “only” 5 degree – but the parts where the majority of people live see temperature drops many times this. The global average is brought up by the relatively small change in global ocean temperatures.

**Results of a nuclear winter**

These changes to the climate alone would be sufficient to destroy agricultural production for the global food chain for a number of years. One gloomy assessment in 1986 referenced in the paper is that the majority of people who had somehow survived the nuclear bombs and radiation would in any case die in the following years of starvation as crops failed across the globe. So in short given have the ability to cause our own extinction event, let’s hope those with their fingers on the nuclear buttons are rational enough never to press them.

]]>Topics included are:

1. Number

2. Circle Theorems and angles

3. Algebra

4. Volume

5. Statistics

6. Solving equations using graphs

7. Trigonometry

8. Linear graphs and inequalities

9. Transformations

10. Probability and Venn Diagrams

11. Vectors

12. Functions

13. Sequences

For each topic I have chosen a few past paper questions to talk through. Hopefully this should be useful for students sitting their exams in the coming weeks.

]]>The Drake Equation was intended by astronomer Frank Drake to spark a dialogue about the odds of intelligent life on other planets. He was one of the founding members of SETI – the Search for Extra Terrestrial Intelligence – which has spent the past 50 years scanning the stars looking for signals that could be messages from other civilisations.

In the following video, Carl Sagan explains about the Drake Equation:

where:

*N = the number of civilizations in our galaxy with which communication might be possible (i.e. which are on our current past light cone);
R* = the average number of star formation per year in our galaxy
fp = the fraction of those stars that have planets
ne = the average number of planets that can potentially support life per star that has planets
fl = the fraction of planets that could support life that actually develop life at some point
fi = the fraction of planets with life that actually go on to develop intelligent life (civilizations)
fc = the fraction of civilizations that develop a technology that releases detectable signs of their existence into space
L = the length of time for which such civilizations release detectable signals into space*

The desire to encode and decode messages is a very important branch of mathematics – with direct application to all digital communications – from mobile phones to TVs and the internet.

All data content can be encoded using binary strings. A very simple code could be to have 1 signify “black” and 0 to signify “white” – and then this could then be used to send a picture. Data strings can be sent which are the product of 2 primes – so that the recipient can know the dimensions of the rectangle in which to fill in the colours.

If this sounds complicated, an example from the excellent Maths Illuminated handout on codes:

If this mystery message was received from space, how could we interpret it? Well, we would start by noticing that it is 77 digits long – which is the product of 2 prime numbers, 7 and 11. Prime numbers are universal and so we would expect any advanced civilisation to know about their properties. This gives us either a 7×11 or 11×7 rectangular grid to fill in. By trying both possibilities we see that an 11×7 grid gives the message below.

More examples can be downloaded from the Maths Illuminated section on Primes (go to the facilitator pdf).

A puzzle to try:

“If the following message was received from outer space, what would we conjecture that the aliens sending it looked like?”

0011000 0011000 1111111 1011001 0011001 0111100 0100100 0100100 0100100 1100110

Hint: also 77 digits long.

This is an excellent example of the universality of mathematics in communicating across all languages and indeed species. Prime strings and binary represent an excellent means of communicating data that all advanced civilisations would easily understand.

Answer in white text below (highlight to read)

Arrange the code into a rectangular array – ie a 11 rows by 7 columns rectangle. The first 7 numbers represent the 7 boxes in the first row etc. A 0 represents white and 1 represents black. Filling in the boxes and we end up with an alien with 2 arms and 2 legs – though with one arm longer than the other!

If you enjoyed this post you may also like:

Cracking Codes Lesson – a double period lesson on using and breaking codes

Cracking ISBN and Credit Card Codes– the mathematics behind ISBN codes and credit card codes

]]>**Sequence Investigation**

This is a nice investigation idea from Nrich. The above screen capture is from their Picture Story puzzle. We have successive cubes – a 1x1x1 cube, a 2x2x2 cube etc.

The cubes are then rearranged to give the following shape. The puzzle is then to use this information to discover a mathematical relationship. This was my first attempt at this:

1^{3} = 1^{2}

2^{3} = (1+2)^{2} – 1^{2}

3^{3} = (1+2+3)^{2} – (1+2)^{2}

4^{3} = (1+2+3+4)^{2} – (1+2+3)^{2}

n^{3} = (1+2+3+4+…+n)^{2} – (1+2+3+…+ (n-1))^{2}

This is not an especially attractive relationship – but nevertheless we have discovered a mathematical relationship using the geometrical figures above. Next let’s see why the RHS is the same as the LHS.

RHS:

(1+2+3+4+…+n)^{2} – (1+2+3+…+ (n-1))^{2}

= ([1+2+3+4+…+ (n-1)] + n)^{2} – (1+2+3+…+ (n-1))^{2}

= (1+2+3+…+ (n-1))^{2} + n^{2} + 2n(1+2+3+4+…+ (n-1)) – (1+2+3+…+ (n-1))^{2}

= n^{2} + 2n(1+2+3+4+…+ (n-1))

next we notice that 1+2+3+4+…+ (n-1) is the sum of an arithmetic sequence first term 1, common difference 1 so we have:

1+2+3+4+…+ (n-1) = (n-1)/2 (1 + (n-1) )

1+2+3+4+…+ (n-1) = (n-1)/2 + (n-1)^{2}/2

1+2+3+4+…+ (n-1) = (n-1)/2 + (n^{2} – 2n + 1)/2

Therefore:

2n(1+2+3+4+…+ (n-1)) = 2n ( (n-1)/2 + (n^{2} – 2n + 1)/2 )

2n(1+2+3+4+…+ (n-1)) = n^{2} -n + n^{3} – 2n^{2} + n

Therefore

n^{2} + 2n(1+2+3+4+…+ (n-1)) = n^{2} + n^{2} -n + n^{3} – 2n^{2} + n

n^{2} + 2n(1+2+3+4+…+ (n-1)) = n^{3}

and we have shown that the RHS does indeed simplify to the LHS – as we would expect.

**An alternative relationship**

1^{3} = 1^{2}

1^{3}+2^{3} = (1+2)^{2}

1^{3}+2^{3}+3^{3} = (1+2+3)^{2}

1^{3}+2^{3}+3^{3}+…n^{3} = (1+2+3+…+n)^{2}

This looks a bit nicer – and this is a well known relationship between cubes and squares. Could we prove this using induction? Well we can show it’s true for n =1. Then we can assume true for n=k:

1^{3}+2^{3}+3^{3}+…k^{3} = (1+2+3+…+k)^{2}

Then we want to show true for n = k+1

ie.

1^{3}+2^{3}+3^{3}+… k^{3} + (k+1)^{3}= (1+2+3+…+k + (k+1))^{2}

LHS:

1^{3}+2^{3}+3^{3}+… k^{3} + (k+1)^{3}

= (1+2+3+…+k)^{2} + (k+1)^{3}

RHS:

(1+2+3+…+k + (k+1))^{2}

= ([1+2+3+…+k] + (k+1) )^{2}

= [1+2+3+…+k]^{2} + (k+1)^{2} + 2(k+1)[1+2+3+…+k]

= [1+2+3+…+k]^{2 }+ (k+1)^{2} + 2(k+1)(k/2 (1+k)) (sum of a geometric formula)

= [1+2+3+…+k]^{2 }+ (k+1)^{2} + 2(k+1)(k/2 (1+k))

= [1+2+3+…+k]^{2 } + k^{3}+ 3k^{2} + 3k + 1

= (1+2+3+…+k)^{2} + (k+1)^{3}

Therefore we have shown that the LHS = RHS and using our induction steps have shown it’s true for all n. (Write this more formally for a real proof question in IB!)

So there we go – a couple of different mathematical relationships derived from a simple geometric pattern – and been able to prove the second one (the first one would proceed in a similar manner). This sort of free-style pattern investigation where you see what maths you can find in a pattern could make an interesting maths IA topic.

]]>