Arclets Explained

Age 11 to 16
Article by NRICH team

Published March 2003,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.

Arclets Image

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:

3-node solution

3 Node Solution 3 Node 2nd Diagram

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.

\begin{eqnarray} \mbox{Perimeter} &=&((3 \times \frac{1}{3}) + (3 \times \frac{2}{3}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

4-node solution

For the 4 node arclets the angles at the centre are $90^{\circ}$ 4 Node Solution

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.

\begin{eqnarray} \mbox{Perimeter} &=&((4 \times \frac{1}{4}) + (4 \times \frac{2}{4}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

5-node solution

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.


5 Arc 5 arc2 5 arc 3
\begin{eqnarray} \mbox{Perimeter} &=&((5 \times \frac{1}{5}) + (5 \times \frac{2}{5}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

An image of part of the work Sarah, Kathryn and Gordon did to find the perimeter of the 5-node arclet is shown below

5 Node Arclet by Sarah, Kathryn and Gordon

6-node solution

Scanned diagrams showing the work of Sheila, Shona and Alison to find the perimeter of a 6-node arclet:

Scanned diagrams showing the work of Sheila, Shona and Alison to find the perimeter of a 6-node arclet:

 

\begin{eqnarray} \mbox{Perimeter for 6 nodes} &=&((6 \times \frac{1}{6}) + (6 \times \frac{2}{6}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

Finding the perimeter of the N - node 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):

\begin{eqnarray} \mbox{Perimeter for N nodes} &=&((N \times \frac {1}{N}) + (N \times \frac{2}{N}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

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:

Many Circles

In other words. No matter how many nodes the perimeter will always be 3 circumferences.