So now it'seems I have shifted my focus from the Pythagorean Theorem to something a tad bit different. I do an activity with high school Geometry classes after teaching them about the trigonometric functions and eventually the area formula for a triangle involving the sine of the included angle. The activity is a guided exploration that leads students to the fact that pi = limit (n->infinity) (n/2)sin(360/n) which comes from increasing the number of sides n of an inscribed regular polygon in a unit circle... (9th grade geometry students do not know about radians yet...). I also use this in pre-calculus courses to hint at the idea of limits and Riemann Sums, etc...

Well in toying with this idea and also the circumscribed polygons, I have managed to come up with the following statement:

n sin(180/n) < pi < 2n (csc(180/n) - cot(180/n))

I then happened to notice that the expression sin(180/n) is in each term on the outer ends of this inequality.

In doing a tad bit of reading on Archimedes (I am no expert on this at all), I noticed that he was able to show that pi was between 3 10/70 and 3 10/71 by using regular 96-gons. I imagine this was quite a daunting task in his day.

So then pi is between 2 rational numbers.

What I would *like* to do with the aforementioned inequality is to use it to prove that pi is IRrational. Here is my thinking:

As n gets infinitely large, all three sides of the inequality approach the same value. Starting with a convenient value, say n = 6, we have:

3 < pi < 12(2 - sqrt(3)), the rightmost value being irrational.

Now here's the $64 million question:

Multiplying n = 6 by successive powers of 2 gives constant use of the half angle identity sin(n/2) = sqrt(( 1 - cos(n) )/2).

I believe (but do not know how to 'prove') that every such value of n all the way up to infinity gives irrationals on both sides. (The trouble being that I can't show that 1 - cos(n) is never a rational multiple of sqrt(2) for such values of n). If that is true, then by the squeeze theorem, pi must also be irrational.

Nevertheless, if the proof is a bust, I still like the inequality.