Is there any way to arrive at $\pi$ without mentioning the circle's radius or diameter?

Question

Given a circle of arbitrary size, is there any way to arrive at $\pi$ or $\tau$ (if you will) without any reference to the circle's radius or diameter?

Answer 1

Let $C$ be the circumference, $A$ the area and $r$ the radius of the given circle. We have $C=2\pi r$ and $A=\pi r^{2}$. Therefore $A=C^{2}/4\pi$, i.e., $\pi = C^{2}/4A$, and this is an expression for $\pi$ that does not (explicitly, at least) involve the radius or the diameter.

To be specific, then, the answer to your original question is "yes".

Answer 2

Or, try this: circumscribe a square around your circle. Throw a lot of darts at the square and let $p$ be the probability that a dart lies in the circle. Then $$\pi=4p$$.

Answer 3

Draw a smallest square that encloses the circle. To obtain $\pi$, divide the area of the circle by the area of the square and multiply the result by $4$.

Answer 4

Toss a coin $3600 = 60^2$ times. The expected number of heads is $1800$. The probability that the observed number of heads is at least $1788$ but not more than $1809$ is $$ \int_{1788-1/2}^{1809+1/2} \varphi\left( \frac{x-1800}{\sqrt{3600/4\,{}}} \right) \, \frac{dx}{\sqrt{3600/4\,{}}} $$ where $$ \varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}. $$ In the first half of the 18th century Abraham de Moivre discovered this, except that he found the constant $1/\sqrt{2\pi\,{}}$ only numerically. It's in his book The Doctrine of Chances. James Stirling found it in this closed form.

On a closely related note (almost the same thing, really), we have $$ \lim_{n\to\infty} \frac{n!}{n^n\sqrt n} = \sqrt{2\pi\,{}}. $$

Answer 5

I'm not sure what we're supposed to do with the circle if we're not allowed to measure it. Assuming you know some calculus, there are other definitions of $\pi$, though, that don't mention circles and geometry. I give an example of such a definition below.

Assume there is a function $f(x) : \mathbb{R} \to \mathbb{R}$ with the following properties: $f(0) = 0$, $f'(0) = 1$, and, for every $x$, $f''(x) = -f(x)$. It turns out there is such a function: the sine function $f(x) = \sin(x)$. Furthermore, it's not too hard to show it's the only function with those properties (though I won't do so here). The sine function is periodic (i.e., it repeats), and its period (i.e., how far you have to go in $x$ before you see it repeat) is $2\pi$. This defines $\pi$ purely from calculus in terms of properties of a solution to a simple equation.

Answer 6

$\pi = 2\inf \{ x \geq 0 : \cos(x)=0\} $.

Answer 7

Without any reference to radius/diameter is a little confusing.

But you can show that an $n$-sided polygon that is inscribed into a circle of radius $r$ is composed of $n$ isoceles triangles with side lengths $r$, $r$, and $2r\sin{180° \over n}$. Finding the area of one isoceles triangle and multiplying it by $n$, you can show that the area of the polygon is $r^2×n\sin{180° \over n}\cos{180° \over n}$. If the number of sides that the polygon has approaches $\infty$, the term $n\sin{180° \over n}\cos{180° \over n}$ approaches $\pi$.

Doesn't directly relate to the radius but still includes it, so I am not sure if this answered your question.