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In this New York Times article, Steven Strogatz offers the following argument for why the area of a circle is $\pi r^2$. Suppose you divide the circle into an even number of pizza slices of equal arc length, and wedge them together in such a way that half of the slices have an arc at the bottom, and half of the slices have an arc at the top:

Area of circle

Then, the base of the shape created has length $\pi r$, and its height is $r$. As the number of slices tends to infinity, the limiting case is that of a rectangle:

Limiting case

Hence, the area of the circle is $\pi r^2$. Although this argument is very geometrically appealing, it also seems fairly difficult to make rigorous. I suppose the most challenging part is showing that the base of the shape really does become arbitrarily flat, and its height becomes arbitrarily vertical, if that makes sense. How might we convert this intuitive argument into a rigorous proof?

Joe
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    Arbitrarily vertical can be explained by $\sin\theta\sim\theta$ for small $\theta$. – Andrew Chin May 29 '21 at 21:13
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    More slices= less at length per piece= less deviation from a line segment ... – Roddy MacPhee May 29 '21 at 21:15
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    You can approximate the area of the shape from above and from below using parallelograms, whose base and height you can calculate with a bit of trigonometry. You can then show that as the number of slices increases, the difference between the approximation from above and below goes to $0$, so the approximations have to converge to the actual area. – Vercassivelaunos May 29 '21 at 21:33
  • Very similiar to my answer about the volume of a torus. – TonyK May 29 '21 at 21:55
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    Archimedes used inscribed and circumscribed polygons to approximate $\pi$ since he could calculate the size of polygonal wedges exactly. You could use the same idea and the squeeze theorem to prove this. – CyclotomicField May 29 '21 at 22:09
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    There's a subtlety here that a lot of answers are glossing over: the fact that the circumference of a circle is less than the perimeter of a circumscribed polygon is not trivial. – eyeballfrog May 30 '21 at 16:47
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    I just want to point out that while this proof gives you the area of the circle as function of its circumference, it doesn't give you a numerical value for π. In order to get that, you need to put in a little more work to select a sequence of polygons where you can compute the length of the short side. – tobi_s May 31 '21 at 08:28
  • " it doesn't give you a numerical value for π." I don't think anywhere in the question that it was implied it would. However the question does assume as a given without any consideration or doubt that for all circles there is a constant value so that Circumference = Constant x radius. Such an assumption should be examined and critiqued. @tobi_s – fleablood Jun 03 '21 at 17:40
  • @fleablood: I understand that the 'classical' approach to circles is quite difficult to make $100%$ rigorous, as to define $\pi$ as the ratio of the circumference to the diameter requires us to first be able to define what an arc length is, and also show that all circles are similar as you have mentioned. However, I'm willing to disregard these issues because it is possible to make the classical approach rigorous, even if this is quite a challenge compared to more modern approaches. – Joe Jun 03 '21 at 17:47
  • The other thing you must watch out for (as @eyeballfrog points out) is if a line segment is the limit of many curved arcs, that does not mean the length of the line is the limit of the lengths of the arcs. In this case it is but in the diagonal step paradox that $\sqrt 2 = 2$ it is not (Imagine a series of zig-zags a step up and a step down and $\sqrt 2$ steps across, as the zig-zags get small they "approach" the shape of a line but their lengths (total: $2$) do not approach the length of the line $\sqrt 2$.) – fleablood Jun 03 '21 at 18:07
  • @fleablood thanks for rephrasing what I said. My experience is that if you present something like this image to math-curious people who've never experienced math outside of high school then they would end up either telling me that the proof is wrong / useless / whatnot because it doesn't give a value for π or they'd keep asking me how that gives the value of π, so I'd end up amending it with the circumference proof. Generally, people seem to finish school believing that numbers are what's important where in fact ratios are. It's for this reason that have this urge to point out such things. – tobi_s Jun 04 '21 at 02:21

3 Answers3

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For simplicity I will assume the number of slices $n$ is even.

If you connect the four "corners" of the wedge figure, you obtain an "inner" parallelogram with one pair of sides having length exactly $r$, and another pair of sides having length approximately $\pi r$. The height of the parallelogram is $r \cos \frac{\pi}{n}$ (which tends to $r$); the length is $n r \sin \frac{\pi}{n}$ (which tends to $\pi r$). So the area of the "inner" parallelogram tends to $\pi r^2$.

You can also consider inscribing the wedge figure inside a slightly larger parallelogram. The height is again $r \cos \frac{\pi}{n}$ but with two additional "crusts" each having thickness $r(1 - \cos \frac{\pi}{n})$, so the final height is $r(2-\cos \frac{\pi}{n})$ (which also tends to $r$). I think the length is the same as the that of the inner parallelogram: $nr \sin \frac{\pi}{n}$ (which tends to $\pi r$).

So the area of the wedge figure is between the areas of the two parallelograms $\pi r^2 cos \frac{\pi}{n} \frac{\sin(\pi/n)}{\pi/n}$ and $\pi r^2 (2-\cos \frac{\pi}{n}) \frac{\sin(\pi/n)}{\pi/n}$. Since both of these areas converge to $\pi r^2$, so does the area of the circle.

Picture of the inner and outer parallelograms:

enter image description here

angryavian
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  • The length of each side is only $nr\sin(\pi/n)$, because it's made up of only half the triangle bases (the other half are on the other side of the parallelogram). Your answer is off by a factor of 2. – eyeballfrog May 30 '21 at 14:08
  • @eyeballfrog Thanks, I've now corrected this. – angryavian May 30 '21 at 16:59
  • I don't get the larger parallelogram part. Can you make a drawing of it, or perhaps a more rigorous definition of how you make a "slightly larger parallelogram"? – justhalf May 30 '21 at 18:57
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    @justhalf Look at each pizza slice. You can "cut off the crust" to get a smaller triangle with height $r \cos (\pi/n)$. (This builds the inner parallelogram.) You can also "add more crust" to get a larger triangle with height $r$. (This builds the larger parallelogram.) More generally you can think of my answer as sandwiching a circle between an inscribed polygon and a circumscribed polygon. – angryavian May 30 '21 at 19:05
  • @justhalf I've added a picture. Also, in my previous comment I was wrong; the argument using the outer parallelogram is slightly different from the circumscribed polygon approach. – angryavian May 31 '21 at 04:39
  • Thanks for the image! That's what I imagined as well. However, the description of the crusts having "length" instead of "height" confused me. – justhalf May 31 '21 at 04:45
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    @justhalf Ah, I see, sorry for the unclear wording. I've changed it to hopefully be clearer. – angryavian May 31 '21 at 04:46
  • Thanks for the prompt response! It's definitely clearer now =D – justhalf May 31 '21 at 04:47
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The circle and the rectangle must have the same area if it is the case that no matter how small you make a positive number $\varepsilon,$ the difference between their areas is less than $\varepsilon.$ As you make $\varepsilon$ smaller, the areas of the circle and the rectangle are not changing, yet they always continue to differ in area by less than $\varepsilon.$

So the idea is this: First observe that the sum of the areas of the wedges is equal to the area of the circle. Then, given the small positive number $\varepsilon,$ you can make the number of wedges so large that the difference between the area of the rectangle and the sum of the areas of the wedges is less than $\varepsilon.$ That establishes the inequality discussed in my first paragraph above.

Michael Hardy
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  • Hi Michael. I don't fully understand your answer: the 'limiting' rectangle shown in the diagram has $\pi r^2$, whereas the sum of the areas of the wedges is also $\pi r^2$. So it doesn't seem necessary to 'make the numbers of wedges so large that the difference between the area of the rectangle and the sum of the areas of the wedges is less than $\epsilon$'—if the two shapes have the same area, then any number of wedges will suffice. What am I missing? – Joe Jun 03 '21 at 21:35
  • @Joe : What you're missing is just about the whole point. You DO NOT KNOW that the sum of the areas of the wedges is $\pi r^2$ because that is just the thing that you are trying to prove. – Michael Hardy Jun 04 '21 at 03:04
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Whether this is rigorous will depend on your definitions of area and length, and whether this is a proof will depend on what we are allowed to presuppose, but I will try to show how control the error.

We only need to show that for small $x$, one of those segments with arc length $rx$ (making $x$ the angle) will be close enough in area to a rectangle with area $\frac{rx\cdot r}2$. Now from your usual diagram used to introduce trigonometric functions, we see that the area is between $\frac{r\cos x\cdot r\sin x}2$ and $\frac{r\cdot r\tan x}2$. Since the circumference of the circle is $2\pi r$, we will need $\frac{2\pi}x$ of these segments. (You can choose $x$ to make this an integer if you like, i.e. set $x=\frac{2\pi}n$ and let $n\to\infty$.) This sandwiches the area of the circle between $$\lim_{x\to0} \frac{2\pi}x \cdot \frac{r\cos x\cdot r\sin x}2 = \lim_{x\to0} \pi r^2\cdot \cos x\cdot\frac{\sin x}x = \pi r^2$$ and $$\lim_{x\to0} \frac{2\pi}x \cdot \frac{r\cdot r\tan x}2 = \lim_{x\to0} \pi r^2\cdot \frac1{\cos x}\cdot\frac{\sin x}x = \pi r^2.$$ So the area must be $\pi r^2$. Of course, this assumes that we already know that $\lim_{x\to0}\frac{\sin x}x=1$.

Carsten S
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