Thursday, 15 December 2011

Infinite Circles Problem

I recently encountered a mathematical problem from NRich Maths about inscribed circles in an equilateral triangle and it really intrigued and after a little bit of intense thought I managed to figure it out.

For those of who wish to know what the problem is without clicking the link:
A circle of radius 1 cm is inscribed in an equilateral triangle. A smaller circle is inscribed at each vertex touching the first circle and tangent to the two 'containing' sides of the triangle. This process is continued ad infinitum...
circles in a triangle 
What is the sum of the circumferences of all the circles?
What is the sum of their areas?
Adding all the circumferences or adding all the areas, which sum grows faster?
Now this might not be immediately obvious as the best way to approach this problem, so we need to think about what we know, what we need to know and the best way to approach this.

What We Know:

  1. The radius of the largest circle is 1cm.
  2. All of the triangles angles are 60° as the triangle is an equilateral.
  3. The area of the first circle is π, the circumference is 2π.
What We Need To Know:
  1. The ratio of the radii from each circle to the next.
  2. The height of the triangle.
  3. The area of all the circles.
  4. The circumference of all the circles.
The height of the triangle may seem like a bit of a strange necessity, but if you know the diameter of the main circle (2cm) then it helps to know what the sum of the diameters of all of the circles will be (height-2).

Now if the the radius were arranged so it was at a right angle to the triangle and a line was drawn from the centre of the largest circle the corner of the side the radius touches the angle would be half of the original angle which is 60° so the new angle is 30°. This is hard to picture but an angle will help that.
Now we have two angles and one side, so we can use the Sin rule to find the size of the line from the centre of the circle to the corner of the triangle.
This means that the radius of the largest circle plus the diameters of all of the other circles is 2. So the height of the triangle 2 plus the radius of the larger circle, which equals 3. It also means that the sum of the diameters of all of the other circles equals 1 too.

Now we can begin to actually tackle the problem of the sum of the circumferences of all the circles. We already know that the circumference of the first circle is 2π, if the sum of all the diameters of the other circles in one line is 1 we can see that the area is then π, but we still have two other sets of circles. So we have the total circumference of 2π+π+π+π which means the total circumference is 5π. Problem one solved.

The second problem is slightly more awkward as the radius is not as easy to find and although the way I am about to explain does work in may not be the most efficient, but it does work and it utilises some very nice Core 2 techniques.

As you may have noticed, there will be an infinite number of circles going into any of the corners (this is caused by the curved shape of a circle against the straight side of the triangle). If we exclude the large circle then the sum of the diameters of into one corner is 1.

The fact that the triangles always get smaller, means that the rate at which they 'increase' is less than 1, we will call this ratio 1/n, the radius of the second triangle will also be 1/n because of the fact that the first radius is 1.

We know that the sum of the diameters equals 1, which means that the sum of 2*radius is also equal to 1. We also know the first term of this series (1/n), the ratio of the series (1/n) and the sum of the series (1). As our ratio is less than 0 we can use the formula covered in C2 for that:
Using that we can rearrange to find what n equals and thus find the ratio. I have included the original equilateral triangle image along with some labelling to help to explain my notation.
So we have that the ratio from radius to radius is 1/3, so to find the sum of the area of all the triangles we must use the sum of an infinite series again. Given that the first term is π (from πr^2 and r = 1), the ratio is r^2, which gives 1/9. We have three of the series so we will times the sum of these by 3, but then we have included the largest triangle three times, so we must subtract this two times (-2π).
This means that the area of all the circles is 11π/8, problem two solved.

The last problem is considerably easy to handle, it simply asks which sum grows faster, this is the one that has a larger ratio. Well the ratio of the area is 1/9, whereas the circumference is 2/3 (r is 1/3, but we want twice this). So this means the circumference increases faster.

This problem really is a lovely one, it combines some relatively simple maths in an advanced form, pieces them all together and leaves you to solve the puzzle. Maths is fun. Maths is really, really fun!

I realise I may have explained fair chunks of this poorly, it is very difficult to convey what is happening and without being in front of you. So if you are left with any questions as to what I have done, or why, simply leave a comment and I will explain or email me at for more information.

Also to let you know, I will be completing a Core 2 revision guide pretty soon (give me a week or so), so keep checking back here for updates on that.

Wednesday, 14 December 2011

Core 2 Numerical Integration: Trapezium Rule

Taken from my upcoming Core 2 Revision Guide. This is a brief preview and an explanation on one of the methods of estimating a definite integral. Others include the mid-ordinate rule, Newton-Cotes formula and the Simpsons rule. However none of these need to be dealt with until later units.

Sometimes you may be given an equation that seems impossible to integrate given our current known techniques, and that is because they are. So we need a method of determining the area beneath a curve at given limits. Which is actually not too difficult, if we take shapes that are sort of like the curve, find the area of all of them and then add them up we will get roughly the area beneath the curve at those points.

As you may have guessed by now the area will be estimated using trapeziums. The reason for this is that trapeziums naturally slope and can emulate the shape of a curve (given enough of them). The best way to explain how to do this is explain this is to use an example that we can integrate, which I will do as y = x2.

So we have to start by knowing what our limits are (in our case they are at 1 and 4), we also need to know what the height of each of our trapeziums are (the horizontal bit) and what the size of each of the two parallel lines, in our case these are at the points when x = 1, x = 2, x = 3 and x = 4

Once we know each of these we can then find the area of each of the trapeziums, add them together and voila we have an estimate for the area beneath the curve. If you don't remember the area of a trapezium is 0.5(a+b)h where a and b are the parallel sides and h is the height. So the area of each of the trapezium respectively is 0.5*(1+4)*1, 0.5*(4+9)*1 and 0.5*(9+16)*1 which is 2.5, 6.5 and 12.5 so if we add these together (which is 21.5) we get the rough area beneath the graph of y = x2.

If we look at how we have found the area of our three trapeziums [0.5*(1+4)*1 + 0.5*(4+9)*1 + 0.5*(9+16)*1] we can see that there are two things that multiply by each other in each trapezium (0.5 and 1) so if we take them out we get: 0.5*1*(1+4+4+9+9+16), this can be further simplified as there are two 4's and two 9's so we get 0.5*1*[(1+16)+2(4+9)] which is were we derive the formula for the rough area beneath the curve at given limits.

Words you need to know is that an ordinate is the y-coordinate at the values of x we have, so for example when x = 1 the ordinate is also 1, when x = 2 the ordinate is 4, etc. And a strip is a trapezium, so if we have 4 strips we have 4 trapezium, also you should note that if there are n strips there will be n+1 ordinates.

So the word equation for an estimate of the definite integration between limits a and b is:

The formula for this (that you will be given in the exam) is:
Just so you know y0 is the first ordinate, y1 is the second and yn is the last.

This is particularly useful when you need to find the integral of something that our current rules for calculus can not give an answer for it, with √(2x), is an example of where the trapezium rule is useful. In fact using our formula I will solve this between the limits of 3 and 1 with 4 strips.

You will need to know how to make our estimate more accurate, to do this you simply use more strips as the more there are the closer to trapeziums will be to making the actual shape.

If anything I have wrote does not appear to make sense please comment and I will do my best to explain. 

Sunday, 11 December 2011

Do numbers exist?

This seems like a bit of an odd topic for a maths fanatic to discuss, and really it isn't very mathematical to even think about it, as you will you soon find out it is more philosophical.

You'd think it is obvious that the answer is "yes, they do exist", but it really isn't that simple. The fundamental idea of numbers started with counting a number of items, animals, children, etc. and from there it spiralled off. In fact the definition of a number is: "an arithmetical value, expressed as a word or symbol representing a particular quantity".

So numbers 'exist' to serve a purpose for counting, arithmetic and calculations. So if we have 'two' dogs, do we count the number of hairs they have as being them? Do we count the number of bugs on them? Do we need to count the number of atoms in the dog towards the 'two'? So is it really 'two' or is it some uncountable number of atoms? Or quarks? But just because we can not clearly state what two is in real life, it does not mean they do not exist.

Numbers do make interpreting what is happening an awful lot easier, but that doesn't mean they 'exist'. They easily could have been created by man to help with problems, the problem arises when you need to define what 'existing' is. To 'exist' you need to have reality or being, do numbers really have this? They are not a physical, touchable thing and could anyone really argue the case of a number having a reality to it?

You could then argue that nothing really, truly exists, as what is reality? But that is a completely different tangent for a future post. I mean, do thoughts really exist? Do words that are spoke? But anyway, I digress...

Numbers can be thought of as tools developed by us a civilization to help understand the world we live in. But if there was a race of intelligent aliens there is almost no doubt that they would have a form of maths, they may work in a different base (in fact the Aztec's worked in base 6, opposed to our base 10), or some of their basic principles may be different (a negative times a negative could still be a negative, for example). But there is almost no doubt that basic principles of maths will be there. And once they are in place more advanced concepts start to develop like how are these numbers distributed? What about a number between 1 and 2? How do we add numbers? How do we multiply numbers? And so on.

So whether they are invented or they exist, they are a necessity. It takes away the "what if?" factor from so many elements and provides concrete and quantifies otherwise incomprehensible things and data. Numbers are so useful for everything, so really it doesn't matter if they exist or not because what they accomplish is real, the data, the facts and the information they provide about the world we live in is real, they have a real impact.

So rather than getting bogged down in the semantics of whether numbers 'exist' or not should not detract from the joy and beauty of mathematics. Maths is fun, do not let your philosophical stand point alter any of that.

Wednesday, 7 December 2011

Riemann Hypothesis

Now I am definitely not an expert in this field, and in fact even the experts aren't really experts in the conventional sense. No one is an expert on it in the conventional sense, it is still unsolved. Over 150 years old and it still remains unsolved not for the lack of trying! In fact it is so important to mathematicians that the Clay Mathematics Institute has put a $1,000,000 bounty on its head (that is, you get $1,000,000 if you manage to solve it).

But what actually is the Riemann Hypothesis? It is a conjecture about the location of the non-trivial zeros of the Riemann Zeta function, it states that all the zeros should lie on the critical strip 0.5+it. "Oh yeah!", I hear you cry, now you get it, obviously. I will explain what this means properly later on in this post. But first I will state what it means. If true it implies a lot of things about the distribution of prime numbers, and as you may or may not know they are very irregular and very difficult to find as the numbers get very, very big.

To track back to my earlier point, what is the Riemann Zeta Function ( it is denoted as ζ(s), ζ being the Greek lower case from which z was derived)?
Where s is an imaginary number, a+ib.
This requires that you understand sequences and seriesimaginary numbers and imaginary exponents. The real intrigue of this comes from the fact that it can be represented by Euler's product.
As you may or may not notice this is comprised of the prime numbers, this means that there is a sort of subliminal link between the natural numbers and the prime numbers. This showed that the prime numbers were not just positioned randomly and are not merely the building blocks to numbers but there is an actual link between them and the natural numbers.

The Riemann Zeta Function on its face doesn't look too difficult, I mean it is just an infinite sequence, even with a complex power you'd expect this to be possible and even pretty easy. But that is not the case at all, part of the reason is how sporadic complex exponents can be, and although it is not too difficult to find solutions (using a high powered computer thousands can be found each hour) it is incredibly, incredibly hard to find a proof for all the solutions.
The plot of the Riemann Zeta Function, the red line is the
real part, the blue part is the imaginary part.
You can see, this function seems to have little to no consistency to it, but a fair amount is known about the function. A lot of the zeros do actually satisfy the hypothesis, over 10 trillion of them in fact. And you'd think that is a proof alone, but as it often involves an iterated log (a log of a log, log(log(x)) and this increases very, very, very slowly in fact log(log(10,000,000,000)) = 1, so 10 trillion really isn't anything. If it is still holding true for log(log(x))>40 there may be a greater unanimous opinion on the truth of the hypothesis.

Every mathematician worth his salt has had an encounter with the Riemann Hypothesis and it has withheld every single attempt thus far. The maths used to try and tackle the problem is so complex that entirely new branches of mathematics have been created to deal with it, this maths to laymen has literally nothing, at all, to do with the prime numbers. It is so complex and far away from the problem that it almost boggles mathematicians minds, but it consumes them, it is their passion and life.

Prime numbers are the passion for many and the Riemann Hypothesis is merely an extension of that, and hopefully it will be solved in my life time.

If you have caught the prime number bug I suggest you read the excellent book by Karl Sabbagh called Dr Riemann's Zeros.

Sunday, 4 December 2011

A-Level Maths Revision: AQA Core 1

Download the revision guide now!

Now this took an awful lot of work, like so much, it is 23 pages long and I really hope that it will help you a lot. Tell your friends about it and just spread the word. There are over 50 questions in the revision guide and I meticulously detail over every aspect of Core 1. I use both graphics and words to explain all terminologies and chapters.
If you need any help on anything in this revision guide or anything else comment here or email me at

Download the revision guide here: AS Maths Core 1 Revision Guide.

Current Corrections:
1.) d.) (-16-11 sqrt(2))/14
"knew" incorrectly in place of "new"
General formatting errors