Tuesday, 4 January 2011

Mathematical Problems, 3B - A lower value of Pi

Returning to pi, and this time using geometry to calculate a minimum value.  Instead of using a square around the outside of a circle, this method will use a square within a circle.  Last time we looked at square ABCD, this time, it's WXYZ.


Now the circumference of the circle is greater than the perimeter of WXYZ.  We know this for sure because the shortest distance between two corners of a square is the straight line that connects them, and the circle is a curved line and therefore must be longer.

If we call the centre of the circle O, then we can see that WOX is a right-angled triangle, and WO = OX = radius of the circle.

Using Pythagoras, we can determine the length WX:

WX 2 = WO2 + OX2

And since WO = OX = radius of the circle, r, then WX2 = 2 r2
And WX = sqrt (2 r2)  = sqrt 2 x r

Now, the full perimeter of the square is four times WX (since the square has four sides),
4 WX = 4 sqrt (2 r2)  = 4 x sqrt 2 x r

We want to describe pi in terms of the diameter of the circle, not the radius, so substitute d = 2r and this gives

Perimeter of square (4 WX) = 4 x sqrt 2 x d/2
Perimeter of square = 2 x sqrt 2 x d

We know the circumference of the circle is pi x d
The perimeter of the square is 2 x sqrt 2 x d
Therefore, pi is greater than 2 x sqrt 2  (approximately 2.82)

Combining this with the result from the previous post gives us the approximation

2 sqrt 2< pi < 4

In my next post, I'll work out how to use superscripts and square root signs in HTML (I hope) and I'll show a way of approximating pi using statistics rather than geometry.  In a future post, I'll also look at close-packed circles, and calculate how much of the available area they can fill.

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