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**Sphere packing problem: Pyramid design**

Sphere packing problems are a maths problems which have been considered over many centuries – they concern the optimal way of packing spheres so that the wasted space is minimised. You can achieve an average packing density of around 74% when you stack many spheres together, but today I want to explore the packing density of 4 spheres (pictured above) enclosed in a pyramid.

**Considering 2 dimensions**

First I’m going to consider the 2D cross section of the base 3 spheres. Each sphere will have a radius of 1. I will choose A so that it is at the origin. Using some basic Pythagoras this will give the following coordinates:

**Finding the centre**

Next I will stack my single sphere on top of these 3, with the centre of this sphere directly in the middle. Therefore I need to find the coordinate of D. I can use the fact that ABC is an equilateral triangle and so:

**3D coordinates**

Next I can convert my 2D coordinates into 3D coordinates. I define the centre of the 3 base circles to have 0 height, therefore I can add z coordinates of 0. E will be the coordinate point with the same x and y coordinates as D, but with a height, *a*, which I don’t yet know:

In order to find *a *I do a quick sketch, seen below:

Here I can see that I can find the length AD using trig, and then the height DE (which is my *a* value) using Pythagoras:

**Drawing spheres**

The general equation for spheres with centre coordinate (a,b,c) and radius 1 is:

Therefore the equation of my spheres are:

Plotting these on Geogebra gives:

**Drawing a pyramid**

Next I want to try to draw a pyramid such that it encloses the spheres. This is quite difficult to do algebraically – so I’ll use some technology and a bit of trial and error.

First I look at creating a base for my pyramid. I’ll try and construct an equilateral triangle which is a tangent to the spheres:

This gives me an equilateral triangle with lengths 5.54. I can then find the coordinate points of F,G,H and plot them in 3D. I’ll choose point E so that it remains in the middle of the shape, and also has a height of 5.54 from the base. This gives the following:

As we can see, this pyramid does not enclose the spheres fully. So, let’s try again, this time making the base a little bit larger than the 3 spheres:

This gives me an equilateral triangle with lengths 6.6. Taking the height of the pyramid to also be 6.6 gives the following shape:

This time we can see that it fully encloses the spheres. So, let’s find the density of this packing. We have:

Therefore this gives:

and we also have:

Therefore the density of our packaging is:

Given our diagram this looks about right – we are only filling less than half of the available volume with our spheres.

**Comparison with real data**

[Source: Minimizing the object dimensions in circle and sphere packing problems]

We can see that this task has been attempted before using computational power – the table above shows the average density for a variety of 2D and 3D shapes. The pyramid here was found to have a density of 46% – so our result of 44% looks pretty close to what we should be able to achieve. We could tweak our measurements to see if we could improve this density.

So, a nice mixture of geometry, graphical software, and trial and error gives us a nice result. You could explore the densities for other 2D and 3D shapes and see how close you get to the results in the table.

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**The Mathematics of Cons – Pyramid Selling**

Pyramid schemes are a very old con – but whilst illegal, still exist in various forms. Understanding the maths behind them therefore is a good way to avoid losing your savings!

The most basic version of the fraud starts with an individual making the following proposition, “pay me $1000 to join the club, all you then need to do is recruit 6 more people to the club (paying $1000 each) and you will have made a $5000 profit.”

There are lots of variations – and now that most people are aware of pyramid selling, now normally revolve around multi-level-marketing (MLM). These are often still pyramid schemes, but encourage participants to believe it is a genuine business by actually having a sales product which members have to sell. However the main focus of the business is still the same – taking money off people who then make their money back after having signed up a set number of new recruits.

The following graphic from Consumer Fraud Reporting is a clear mathematical demonstration why these frauds only end up enriching those at the top of the pyramid:

You can see that if the requirement was to recruit 8 new members, that by the 9th level you would need to have 1 billion people already signed up. Even with the need to recruit just 4 new members you still have rapid exponential growth which very quickly means you will run out of new potential members. For pyramid schemes it is only those in the first 3-4 levels (the white cells) that stand any real chance of making money – and these levels are usually filled by those in on the scam.

Ponzi schemes (like that run by Bernie Madoff) use a similar method. A conman takes money from investors promising (say) 10% annual returns. Lots of investors sign up. The conman then is able to use the lump sum investments to pay the 10% annual returns. This scam can last for years, with people thinking that they are getting a good rate of return, only to find out eventually that actually their lump sum investment has gone.

This is a good topic to look at with graphs (plotting exponential growth), interest rates, or exponential sequences – and shows why understanding maths is an important financial skill.

If you like this topic you might also like:

Benford’s Law – Using Maths to Catch Fraudsters – the surprising mathematical law that helps catch criminals.

Amanda Knox and Bad Maths in Courts – when misunderstanding mathematics can have huge consequences .

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1) Exploration Guides and Paper 3 Resources

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