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Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle


Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?


M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.


Age 14 to 16 Challenge Level:

James Page of Hethersett High School, Norfolk solved the Holly problem, well done James.

Length $AB = 2 + 2\sqrt2$ cm and length $BC = 4 + 2\sqrt 2$ cm.

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

$$16 + 12\sqrt 2 - 4 - 4\pi = 12 + 12\sqrt 2 - 4\pi = 16.40 {\rm cm}^2$$