### 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?

# Contact

##### Stage: 4 Challenge Level:

Well done Sam from Kings Wimbledon for your solution:

1) Coin radius 1: How long would the box need to be for every point on the coin's circumference to touch the box as I roll it along?

The left bottom quarter of the disc is not in contact, which is a radius' width. Then the whole circumference must roll along the base of the box, until another quarter of circumference is not in contact in the bottom right and corner.
So the disc must be two radii (from when the disc is at the sides) plus one circumference long. This is $l=1+1+(2\times\pi\times1)=2(1+\pi)=8.28$.
For any radius $r$, the length of the box is $l=r+r+(2\times\pi\times r)=2(r+\pi)$

2) If the two edges are the same length, how long do they need to be to ensure every point on the circumference touches? What if they're not the same length?

Call the length of the edges $l$. Then the amount of the circumference that touches each edge is $l-r$, as at the corner there will be a quarter of the circumference that will not be touched.
Obviously if $l-r> c$ then the whole circumference will touch both edges, so we want to find the minimum length an edge can have to touch the whole circumference. So you need
$2\times(l-r)+\frac{2\pi r}{4}=2\times(l-r)+\frac{\pi r}{2}=2\times(2\pi r)$
$2\times(l-r)=\frac{7}{2} \times \pi r$
$l=\frac{7}{4} \times \pi r +r$
$l=r\times(\frac{7}{4}\pi + 1)$
That is, $\frac{7}{8}$ of the circumference, and then plus the extra radius that is missed in the corner.

For edges of different lengths, obviously if both lengths are at least as long as $r\times(\frac{7}{4}\pi + 1)$ then it will work, and the whole circumference will be covered.
But the more interesting question is, if one of the edges is shorter than this figure, how long will the other need to be to make up for it, and ensure the whole circumference is still touched.
The answer is that if one length is less than $r\times(\frac{7}{4}\pi + 1)$, then the other length must be the entire circumference plus the extra radius lost in the corner. This can be seen from labeling four equidistant points on the disc and studying their movement as the disc rotates around the edges.

3) Rectangular tray: What proportion of the circumference touches the tray as it makes one complete circuit of a 4 by 3 tray?

Initially just looking at one side of the tray we can see how much of the circumference touches this. We can find the arc length that touches on any side by looking at the length of the side minus two radii from the corners. By noting the formula $c\times f=a$, the formula relating the cicumference, fraction of the circumference and the arc length of that fraction, we find $f=\frac{l-2r}{2\pi r}$.
So for a disc of radius $1$ in the $4 \times 3$ box, the proportion touched on the first side is $\frac{4-2r}{2\pi r}=\frac{1}{\pi}$, and then a quarter is missed, then $\frac{3-2r}{2\pi r}=\frac{1}{2\pi}$ and miss a quarter again. Then repeat for the last two sides. So after two sides $\frac{1}{\pi}+\frac{1}{2\pi}=\frac{3}{2\pi}$ of the circumference has been touched, and we have moved $\frac{3}{2\pi}+\frac{1}{2}$ of the circumference. Which is a distance of $3+\pi$, just less than the actual circumference of $2\pi$. So then the new bits that will be touched are $2\pi-(3+\pi)=\pi-3$ on each side from now on. So on the last two sides $2(\pi-3)$ of the circumfernce is touched, which is $\frac{2(\pi-3)}{2\pi}=\frac{\pi-3}{\pi}$ of the circumference. Adding this to the proportion touched on the first two sdies, the total proportion touched is $\frac{3}{2\pi}+\frac{\pi-3}{\pi}=\frac{2\pi-3}{2\pi}$ of the circumfernce. Which is a total arc length of $2\pi \times \frac{2\pi-3}{2\pi}=2\pi-3$.

Then as for every side the disc touches it touches anew $\pi-3$ of the circumference it will eventually touch the entire circumfernce.