Published December 2003,March 2003,December 2011,February 2011.
The Arclets problem set in September 2002 produced some very interesting and inspiring work from Madras College. This short article gives a flavour of the way that Sheila, Shona, Alison Colvin, Sarah, Kathryn and Gordan tackled the problem.
Each of the following shapes is made from arcs of a circle of radius r.
What is the perimeter of a shape with $3$, $4$, $5$ and $n$ "nodes".
What happens when $n$ is very large?
Explain the relationship between your answers and the perimeter of the original circle
Here are arclets with $3$, $4$ and $5$ nodes:
The angles at the centre of the inner circle are $60^{\circ}$
So the angles at the centre of the outer circles are $120^{\circ}$ and $240^{\circ}$ ($120^{\circ}+240^{\circ}= 360^{\circ}$.
We can therefore divide each circle into a $1/3$ ($120^{\circ}$ out of $360^{\circ}$) part and a $2/3$ ($240^{\circ}$ out of $360^{\circ}$) part.
The perimeter of the arclet is made up of 3 "inward" arcs of $1/3$ of the circumference and 3 "outward" arcs of $2/3$ of the circumference.
The inward arcs are $1/4$ of the circumference ($90^{\circ}$  there are 4 inward arcs.
The outward arcs are $1/2$ of the circumference (4 lots of $45^{\circ}$  there are 4 outward arcs.
Because the inner circle is surrounded by five outer circles there are 5 angles  all of $72^{\circ}$ at the centre.
Using the properties of isoseles triangles the outward arcs are $2/5$ of the circumference and the inward arcs are $1/5$ of the circumference.
An image of part of the work Sarah, Kathryn and Gordon did to find the perimeter of the 5node arclet is shown below
Scanned diagrams showing the work of Sheila, Shona and Alison to find the perimeter of a 6node arclet:
We know that these are the equations for the perimeter of 3, 4, 5 and 6 node arclets:
Number of nodes 
3  4  5  6 
Perimeter 
$ \quad 3\times 2 \pi r = 6 \pi r \quad $

$ \quad 3\times 2 \pi r = 6 \pi r \quad$  $ \quad 3\times 2 \pi r = 6 \pi r \quad $  $ \quad 3\times 2 \pi r = 6 \pi r \quad$ 
If we substitute $N$ for the node number we get (in every case):
This is based on the fact that the angles at the centres of the circles will be $1/N$ of a full turn.
If $N$ is very large the node shape begins to look like a circle:
In other words. No matter how many nodes the perimeter will always be 3 circumferences.