### Just Rolling Round

P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?

### Coke Machine

The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design. Coins inserted into the machine slide down a chute into the machine and a drink is duly released. How many more revolutions does the foreign coin make over the 50 pence piece going down the chute? N.B. A 50 pence piece is a 7 sided polygon ABCDEFG with rounded edges, obtained by replacing AB with arc centred at E and radius EA; replacing BC with arc centred at F radius FB ...etc..

### Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

# Efficient Packing

##### Stage: 4 Challenge Level:

First we need to find how many discs we can stack $1 \ \mathrm{m}$ in the $y$-direction, the centres of all the discs lie on a line at 60 degrees to the horizontal.
The total number stacked vertically $= \frac{1}{d\sin 60^\circ}$ (where $d$ is the diameter of a disc measured in m)

Case 1: $d = 10 \ \mathrm{cm} =0.1 \ \mathrm{m}$

$\textrm{Number of discs stacked vertically} = 11$
$\textrm{Total number of discs} = (6 \times 10) + (5 \times 9) = 105$
$\textrm{Packing fraction} = \frac{\textrm{No. Discs} \times \pi r^2}{1} = 0.825$

Case 2: d=1cm = 0.01m

Number of discs stacked vertically = 115
Total number of discs = (58x100) + (57x99)= 11443
Packing fraction =$\frac{No. Discs x \pi r^2}{1}$= 0.899

Case 3: d = 1mm =0.001m

Number of discs stacked vertically = 1154
Total number of discs = (577x1000) + (577x999)= 1153423
Packing fraction =$\frac{No. Discs x \pi r^2}{1}$= 0.906

Extension

The most efficent method of packing spheres is face centred cubic packing, FCC packing will provide us with an upper limit to the number of spheres we can pack.

In the Face Centered Cubic (FCC) unit cell there is one host sphere at each corner and one host sphere in each face. Since each corner sphere contributes one eighth of its volume to the cell interior, and each face sphere contributes one half of its volume to the cell interior (and there are six faces), then there are a total of $\frac{1}{8}x8 + \frac{1}{2} x 6 = 4$ spheres in the unit cell.

If we define
a = length of one side of the unit cell
r = radius of one sphere

we can see that 4rsin(45) = a.

The volume fraction of such a unit cell is the number of spheres in the cell multiplied by the volume of a sphere and then divided by volume of cube.

Volume fraction = $\frac{4x \frac{4}{3}\pi r^3}{(2\sqrt(2)r)^3}$ = 0.74

The number of spheres in a volume of $1m^3$ is therefore:

$\frac{0.74}{\frac{4}{3}\pi (0.005)^3}= 1413295$

This is the upper limit to the number of spheres we may pack. It is likely we will not actually be able to pack quite so many. If we did adopt this method we would have fractions of spheres along the sides of the container, as shown in the unit cell above.