There were a number of correct solutions and part solutions to this problem.
Most of you picked up the increase of $ \frac{\pi}{3}$ as the number of sides of the polygon increased by one, although I am not sure that you could see "why" this was happening. A common omission in your solutions is an attempt to look back at the problem and try to see how the rule you have found fits with the mathematical context itself. I hope this makes sense.

Another small but important point is to remember that describing $ \frac{1}{3}$ as 0.3 recurring and $ \frac{2}{3}$ as 0.6 recurring lacks elegance and makes it look as if you used a calculator and did not think about what the mathematics was!!

Well done to:
William and Edward, Josh, Nick and Jeffy, who began to think about the "why" of the pattern (all from King's College School, Cambridge). Katherine and Rosie (The Mount School, York), Andrei (School 205, Bucharest) and Clement. A partial solution was also recieved from Vivienne (Akadmisches Gymnasium, Vienna).

The angle at the centre of the arc of each petal can be calculated as:

$360^{\circ} - 2 \times60^{\circ}$

The interior angle of the polygon $ = 240^{\circ} - 180(1-2/n)^{\circ}$

Giving the length of the side of the equilateral triangle as 1 unit

The length of each arc (petal) can be calculated as:

$ 2\pi(240 - \frac{180(1-\frac{2}{n})}{360}) $

As there as many arcs as there are sides of the polygon, the total length of all the arcs is:

$ n \times 2\pi (240 - \frac{180\left(1-\frac{2}{n}\right)}{360}) = \pi(\frac{n}{3} + 2) $

The perimeter of the petals increases by $ \frac {\pi}{3} $ when the polygon increases its number of sides by one.