### 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?

# Steel Cables

##### Stage: 4 Challenge Level:

It was great to see so many different ways of approaching this! Here are some of the responses that we received from students.

Calvin, from Ethiopia, gave the following interesting solution:

At first I noticed a pattern: the number of strands per cable was going up by multiples of 6 every time. Then I plotted the data for sizes 1-10:

 Size 1 2 3 4 5 6 7 8 9 10 Strands 1 7 19 37 61 91 127 169 217 271

and graphed them. The result was a parabola, so I found its equation and that was the answer. The answer is s = 3n2 - 3n + 1, where s is the number of strands required and n is the size of the cable.

Great! But how did Calvin find that formula? Patrick, from Otterbourne, suggests a way:

The sequence starts off: 1, 7, 19, 37, 61, ...
If we take the difference between successive terms, we get the sequence 6, 12, 18, 24, ...
If we take successive differences again, we get 6, 6, 6, ...

This is constant, so we must have a quadratic equation s = an$^2$ + bn + c.
In fact, half of 6 is 3, so we know that a = 3. Therefore

s = 3n$^2$ + bn + c

In order to find b and c, we can substitute values in:

(n = 1, s = 1) 1 = 3 + b + c

(n = 2, s = 7) 7 = 12 + 2b + c

and we can solve these simultaneous equations to get b = -3 and c = 1.

Szymon from Priory Academy LSST sent us a similar solution:

We know (similar to the above) that s = 3n$^2$ + bn + c. We can tabulate some values:

 Strands (s) 1 7 19 37 61 91 3n$^2$ 3 12 27 48 75 108 Difference (s - 3n$^2$) -2 -5 -8 -11 -14 -17

Now we can work out the successive differences between these terms as before.

These are very nice solutions if you've seen this sort of method before. If not, don't worry: there are other ways of going about it! Matthew, from Saint Peter and Paul's, sent us this interesting 'counting argument':

We can view the size 5 hexagon as being made up from 6 triangles of side length 5. Each of these triangles contains 15 strands, so we get 6$\times$15 = 90. But wait - these triangles overlap, so we've overcounted! Adjacent triangles overlap in 5 strands, one of which is the centre, and all of the triangles overlap in this centre strand; so we need to subtract 1 from our total for each strand that we've double-counted, and we've counted the centre 6 times so we must subtract 5 from our total for that:

90 - 6$\times$4 $-$5 = 61

For the size n hexagon, the same thing happens:

The hexagon is made from 6 triangles of side length n, so each triangle contains n(n+1)/2 strands (the nth triangular number), so 6 triangles give 6n(n+1)/2 strands, or 3n(n+1) strands.

Adjacent triangles overlap in (n-1) strands plus the centre strand, so we need to subtract (n-1) for each pair of adjacent triangles (and there are 6 of them), and then we need to subtract 5 for the centre.

So we get 3n(n+1) - 6(n-1) - 5 strands, which is 3n$^2$ - 3n + 1.

We can break a size n hexagon into 3 big triangles (of side length n) and 3 small triangles (of side length n-2). Since the 3 big triangles meet in the middle, when we add up the number of strands in each we have to subtract 2 because we have overcounted the middle strand by 2:

$\frac{3n(n+1)}{2}$ + $\frac{3(n-2)(n-1)}{2}$ - 2 = 3n$^2$ - 3n + 1

Karl from King's School, Grantham, sent us this slick solution:

The middle row contains 2n-1 strands. The top part can be seen as a large triangle (of side length 2n-2) minus a small triangle (of side length n-1), giving in total:

$\frac{(2n-2)(2n-1)}{2}$  - $\frac{(n-1)(n)}{2}$ = $\frac{3n^2 - 5n + 2}{2}$

The bottom part is the same as the top.

This gives us a total of (3n$^2$ - 5n + 2) + (2n - 1) = 3n$^2$ - 3n + 1.

Crayola, from the British School in Tokyo, knew a short-cut:

We have to work out n + (n+1) + (n+2) + ... + (n+x) + (n+x-1) + ... + (n+1) + n,
where n+x is the longest row of the hexagon.

What is x?

Well, there are 2x + 1 terms in this sum, one for each row, and the hexagon has 2n-1 rows, so x = n-1.

That is, we have to work out n + (n+1) + ... + (2n - 2) + (2n - 1) + (2n - 2) + ... + (n+1) + n. We can regroup these as follows:

2[n + (n+1) + ... + (2n-2)] + (2n-1).

Now, a theorem of Gauss tells us that, to work out n + (n+1) + ... + (2n-2), we simply have to add the first and last number together, multiply the result by the number of terms in the sum, and divide by 2:

2$\times$$\frac{(3n-2)(n-1)}{2}$ + (2n-1) = 3n$^2$ - 3n + 1

Well spotted! Of course, if you didn't know this result, no problem - you can work it out simply by one of the other solutions above. Gauss himself probably figured it out by drawing pictures like Karl's triangles above!