Mathagony Aunt

9th September 2005 at 01:00
I am a technology teacher, trying to create an interactive artwork based on circles. How do you work out what fraction of a circle another circle is? Do you always have to work from the area formula?

With different size circles, one is actually an enlargement of the other, so they can be defined as similar shapes. The properties of similar shapes can therefore be used to compare the areas: the fraction is given by comparing the squares of the radii.

We can deduce this from the formula for the area of a circle. Suppose that one circle has a radius rcm and a larger circle has a radius Rcm. The area of the small circle is p X r X r = pr2 and the area of the larger circle is p X R X R = pR2.

The fraction is: ("cancelling" by p).

So if we have two circles, one of radius 2cm and one of 8cm then As both 4 and 16 are in the 4 times table we can "cancel" (divide top and bottom of the fraction) by 4. So we can say that the area of the smaller circle is 116th of the area of the bigger circle.

But the smaller radius is only a quarter of the length of the radius in the larger circle. Comparing the lengths as fractions we have The fractional comparison of length, area and volume of similar shapes is probably best understood by looking at similar shapes when they are squares and cubes, as we measure area in "square units" and volume in "cubic units". Imagine two squares, A and B, with side lengths of 2cm and 8cm respectively. Making a fractional comparison of the sides, we have as before: Area of square A = 22 = 2 X 2 = 4cm2. Area of square B = 82 = 8 X 8 = 64cm2. The area of square A is therefore 464 = 116 that of square B.

Now take two cubes with sides 2cm and 8cm respectively. Volume of cube A = 23 = 2 X 2 X 2 = 8cm3. Volume of cube B = 83 = 8 X 8 X 8 = 512cm3. Writing this as a fraction, we have: so the volume of cube A is 164 that of cube B.

Provided the shapes you are trying to compare are similar (ie, one is an enlargement of the other), a fractional comparison can be made very easily.

For length we have length A divided by length B = lL; for area we have the square of length A divided by the square of length B = l2L2; for volume we have the cube of length A divided by the cube of length B = l3L3.

Orange box Orange layers touching in hexagonal array More densely filling the space this way.

Market oranges neatly stacked in a pile.

But greengrocers now can smugly smile.

In 1998, Thomas Hales proved, my brothers, That packing oranges, each touching twelve others, An arrangement filling the space most densely, a fact.

The research papers confirm this in the abstract.

Scheduled to be published later, There are about 3 gigs of data, With proof now close to 300 pages Carefully detailing the stages.

The humble shopkeepers more spatially aware Already knew the best way to display their fare!

This poem was inspired by a report about the work of Professor Thomas Hales from the University of Pittsburgh, who proved Johannes Kepler's conjecture of 1661, that the best arrangement for packing spheres is to begin with a hexagonal lattice (the shape formed when 7 coins of the same size are laid on a table, with one surrounded by 6 others), and then to place each successive layer as another hexagonal lattice with the spheres in the lowest point of the previous layer. As the container is filled, each sphere will touch 12 others. This maximises the number of spheres in the container, just like oranges are stacked in a shop. After 300 pages of notes and 3 gigabytes of computer programming, data and results, Thomas Hales announced the proof of Kepler's conjecture complete.


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