How can I improve my explanation of why the ratio $\pi=\frac{C}{d}$ holds for all circles?

Question

I'm trying to informally explain why $\pi$ holds for all circles. I would like to know if there is anything pertinent that I can add, or that is wrong with this explanation. It's an explanation, not an attempt at a rigorous proof, so go easy.

$\pi = \frac{C}{d}$ is the ratio of $C$, the circumference of the circle, to $d$, the diameter of the circle. The ratio holds for all circles because all circles are "similar". That is to say they are all the same essential shape and that they only differ in size, thus for "similarity" to hold this ratio must extend to all circles.

It is easier to see why $\pi$ holds for all circles if we rewrite it in terms of the radius $r$ as $\pi=\frac{2\pi r }{2r}$. According to this, any change in the radius $r$ (and thus circumference and ultimately the size of the circle) will affect an equivalent change in the numerator and denominator of the equation. So in essence any changes in the size of the radius $r$ and subsequently the circle are negated, so the ratio $\pi=\frac{C}{d}$ holds for all circle.

Answer 1

A slightly better explanation (I believe): consider a regular $n$-gon inscribed in a circle, and decompose it into $n$ isoceles triangles with the two equal sides equal to the radius $r$ of the circle. The polygon's perimeter $P_n$ and the circle's radius are in a certain ratio, say $r_n:1$. (So $P_n/r=r_n)$. Then, if you vary the circle's radius, the polygon's perimeter changes proportionally, with proportionality factor $r_n$, because of similar triangles. So, if the $r_n$ approach a limit as $n$ increases (which you can make plausible numerically), then this limit must be the ratio of any circle's circumference to its radius. (Assuming that a circle may be viewed as an "$\infty$-gon".)

Answer 2

Your explanation does little more than point out that he fact that $\frac Cd=\pi$ for all circles follows from the fact that all circles are similar.

Indeed, there's little else that you can do without diving into some pretty complicated analysis, so if that's all you're interested in saying, then that's fine.

A more succinct explanation is given by:

Choose some circle $V$ with circumference $C$ and diameter $d$, and define $\pi=\frac Cd$. Now choose some other circle $V'$ with circumference $C'$ and diameter $d'$. Any two circles are similar; in particular, $V$ and $V'$ are similar, and we have:

$$ \frac C{C'}=\frac d{d'} $$

Therefore, we can conclude that

$$ \frac{C'}{d'}=\frac Cd=\pi $$

Since the circle $V'$ was arbitrary, we conclude that the ratio of the circumference to the diameter of a circle is $\pi$ for all circles.

To get a rigorous proof, you ha to define what you mean by the length of a curve. Once you have done that, there is a good formula for finding the length of a curve parametrized by smooth functions:

If $\gamma=(\gamma_1,\gamma_2):[a,b]\to\mathbb R^2$ is a smooth curve, then the length of $\gamma$ is given by $$ L(\gamma)=\int_a^b \|\dot\gamma(t)\|dt=\int_a^b\sqrt{\gamma_1(t)^2+\gamma_2(t)^2}dt $$

In particular, if $\gamma$ is a circle of radius $r$ about a point $(X,Y)$, we can write it as

$$ \gamma:[0,2\pi]\to\mathbb R^2:t\mapsto(X+r\cos t,Y+r\sin t) $$

Then

$$ \dot\gamma(t)=(-r\sin t,r\cos t) $$

and

$$ C=L(\gamma)=\int_0^{2\pi} \sqrt{r^2\sin^2(t)+r^2\cos^2(t)}dt=2\pi r $$

Then we have the diameter $d=2r$, and so $\frac Cd=\pi$. Throughout, $\pi$ is defined to be the first positive zero of the (analytically defined) $\sin$ function. Of course, there's lots of analysis to do to check that $\cos$ and $\sin$ really do parametrize a circle, and that they behave the way they do with respect to differentiation, but this is the easiest way to get to a rigorous proof.

Update: having read Kim's answer, I've now changed my mind: inscribing a regular $n$-gon is the best way of proving this rigorously. It uses the definition of the length of a curve directly; that is:

$$ L(\gamma)=\sup\left\{\sum_{i=1}^n d\left(\gamma(t_i),\gamma(t_{i-1})\right)\;\middle|\;n\in\mathbb N,0=t_0<t_1<\dots<t_n=1\right\} $$

where $d(p,q)$ is the Euclidean distance. In addition, no calculus is required for that approach (just some care regarding limits).