How do we know the ratio between circumference and diameter is the same for all circles?

Solution 1:

This is not a very rigorous proof, but it is how I was taught the fact that the circumference of a circle is proportional to its radius.

Consider two concentric circles as in the diagram above. The radius of the smaller one is $r$, while that of the larger one, $R$; their circumferences are $c$ and $C$ respectively.

We draw two lines through the center to meet each circle, forming two triangles as shown. The ratio of their sides $r/R = r/R$, and they have a common angle $\alpha$, so they are similar. Thus $k/K = r/R$. Also note that if $\beta$ denotes the full (360 degree) angle of a circle, then $\beta/\alpha \cdot k \approx c$ and $\beta/\alpha \cdot K \approx C$.

We can say that $\frac{c}{C} \approx \frac{\beta/\alpha \cdot k}{\beta/\alpha \cdot K} = \frac{r}{R}$. As the angle $\alpha$ becomes smaller and smaller (tending towards zero, to make a limiting argument) the approximations $\beta/\alpha \cdot k \approx c$ and $\beta/\alpha \cdot K \approx C$ grow more accurate. In the limiting case -- and this is where the 'proof' is slightly nonrigorous -- we get that $\frac{c}{C} = \frac{r}{R}$.

Thus $c/r = C/R$ or equivalently $c/(2r) = C/(2R)$: the circumference divided by the diameter is always a constant for any two circles since any two circles can be made concentric by a trivial translation. We call this magic constant $\pi$.

Solution 2:

Here is a rigorous proof.

First, we need a definition for the distance of two points along a section of a curve. This is called arc length and is defined as the limit of the sum of line-segments of the curve.

The proper definition may be found here. Using this definition, it follows that the arc length $s$ between two points $a$ and $b$ for a continuously differentiable function $f: \mathbb{R} \to \mathbb{R}$ may be calculated by

$$ s = \int_a^b \sqrt{1 + f'(x)^2} dx$$

Lets compute with this formula the arc length $s$ of the upper half of a circle with radius $r$. Using Pythagoras the function describing the upper circle goes from $-r$ to $r$ and is defined as $f(x) = \sqrt{r^2-x^2}$

Also by the chain-rule we have

$$ f'(x)= \frac{-x}{\sqrt{r^2-x^2}}$$

thus

$$ s = \int_{-r}^r \sqrt{1 + \frac{x^2}{r^2-x^2} } dx = \int_{-r}^r \sqrt{ \frac{r^2}{r^2-x^2} } dx= \int_{-r}^r \sqrt{ \frac{1}{1-(x/r)^2} } dx$$

which is if you substitute* with $\phi(x)=rx$:

$$ s =r \int_{-1}^1 \frac{1}{\sqrt{1-x^2}}dx $$

Now $$\int_{-1}^1 \frac{1}{\sqrt{1-(x)^2}} dx:= c_0$$ is clearly a constant that does not depend on the circle (it does not depend on $r$). Since $2s=C$ is the circumference and $2r =d$ is the diameter we have

$$\frac{C}{d} = \frac{s}{r} = c_0$$ for any circle. Which shows, that the ratio between the circumference and diameter of any circle is always $c_0$. This constant is known by the symbol $\pi$.

*To be precise, the substitution rule is

$$ \int_{a}^b h(\phi(x)) \phi'(x) dx = \int_{\phi(a)}^{\phi(b)} h(u)du $$

in this case we have $$ h(x) = \sqrt{\frac{1}{1-(\frac{x}{r})^2} } \text{ and }\phi(x) = rx$$

Solution 3:

To show that $\pi$ is constant we must show that given two circles of diameters $d_1$ and $d_2$ and circumferences $c_1$ and $c_2$, respectively, that $\frac{c_1}{d_1}=\frac{c_2}{d_2}$.

If $d_1=d_2$ then the two circles are congruent because one can be placed upon the other and they will line up. Without loss of generality we can assume $d_1\lt d_2$. Draw the circles concentrically. Then $d_2=kd_1$ for some $k$. If we can show that $c_1=kc_2$ then $\frac{c_2}{d_2}=\frac{kc_1}{kd_1}=\frac{c_1}{d_1}$ and we will be done.

Label the common center of the two circles $O$. Construct two rays emanating from $O$ outwards that divide each of the circles' circumferences into $n$ equal parts where $n\gt 2$. Label the points where the two rays intersect the inner circle $A$ and $B$ and the points where they intersect the outer circle $A'$ and $B'$. Choose the $A'$ so that it lies on the same ray as $A$.

Now consider the triangles $AOB$ and $A'OB'$. These two triangles are similar and moreover, the ratio $\frac{\overline {A'B'}}{\overline{AB}}=\frac{A'}{A}=\frac{\frac{d_2}{2}}{\frac{d_1}{2}}=\frac{d_2}{d_1}=\frac{kd_1}{d_1}=k$. Therefore, $\overline{A'B'}=k\overline{AB}$.

The perimeter of the regular $n$-gon inscribed inside the inner circle is given by $n$ copies of $\overline{AB}$ and the perimeter of the regular $n$-gon inscribed inside the outer circle is $n$ times the length of $\overline{A'B'}$.

Therefore, the perimeter of the inner $n$-gon is $n|\overline{AB}|$ and the perimeter of the outer $n$-gon is given by $n|\overline{A'B'}|=kn|\overline{AB}|$. The ratio of the two perimeters is $k$ and this is independent of $n$.

Taking $n$ arbitrarily large gives that the ratio of the two circumferences are also equal to $k$ and the result is proved.

Solution 4:

Very interesting question. I reckon that any answer will have to do with what the perimeter is even defined as. If we take it as limit of n-glons, then it is easy to see that ratio of perimeters of polygons (at least $2n$-polygons) to their diameter is constant. Thus in limit, too, the ratio will remain constant.

Note: Early Greek mathematicians did have an idea of circles (and many other curved objects) as limits of polygons.