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?
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
Find the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr).
James Page of Hethersett High School, Norfolk solved the Holly problem, well done James.
To find this answer I first took a diagram of the shape and put in the angles. I took 45 from 180 degrees to find the interior angle so I could find the inner curve. I then found that this angle is used 8 times in the diagram. Each circle has a radius of 1 cm and a circumference of $2\pi$ cm. To find the length of the edges of these pieces of the holly I did this:
$\begin{eqnarray} \\ 180 - 45 &=& 135 \\ 135/360 &=& 0.375 \\ (135/360) \times 8 \times (\pi \times 2) &=& 6\pi {\rm cm} \end{eqnarray}$
Then there are the two middle ones of which I had to find the circumference and divide by 2 then multiply by 2 for the two of them, giving a total $2\pi$ cm. I then added the two answers to get the perimeter of the holly leaf which is $8\pi = 25.13$ cm (to 2 decimal places).
If you wished to do a 16 spike holly of the same sort the ends would be the same and it would have an extra 3 circles on each side making the perimeter an extra $6\pi$ cm making a total of $14\pi = 43.98$ cm (to 2 decimal places).
The rectangle $ABCD$ has area
$$(2 + 2\sqrt 2)(4 + 2\sqrt 2) = (16 + 12 \sqrt 2).$$
From this we have to subtract the areas of the four triangles at the corners, a total area of 4 sq. cm., and also the areas of 8 sectors of circles with angles of 135 degrees and the areas of the two semicircles, in all a total area of $4\pi$ sq. cm. The area of the 10 spike holly leaf is
$$16 + 12\sqrt 2 - 4 - 4\pi = 12 + 12\sqrt 2 - 4\pi = 16.40 {\rm cm}^2$$