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Published 2003 Revised 2009
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 5-node arclet is shown below
Scanned diagrams showing the work of Sheila, Shona and Alison to find the perimeter of a 6-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):
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.
The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?
Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.