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There are many, many clear and simple proofs of basic but nontrivial facts in high-school mathematics, such as Pythagoras' theorem or the identities $$\sum_{k=1}^nk=\binom{n}{2}\qquad\text{ and }\qquad\sum_{k=1}^n(2k-1)=n^2,$$ that are accessible to the lay person, having no mathematical knowledge beyond perhaps the most basic high-school curriculum, or perhaps even elementary school. See the answers to this question for many such proofs. I'm surprised I can't find such a proof for the fact that the area of a circle of radius $r$ equals $\pi r^2$ that is thoroughly convincing. Let me explain what I find unconvcing about two popular visual proofs.

Parallellogram proof

The proof I see most often is the parallellogram proof, cutting the circle radially into ever smaller wedges and joining them together to approximate a parallellogram:

enter image description here

It leaves the question of convergence open; do these approximations of parallellograms converge to a rectangle with side lengths $r$ and $\pi r$ as we take ever smaller wedges? It turns out that they do and the argument can be made rigorous, but to the critical lay person this need not be clear from the picture.

Triangle proof

Another proof I have seen quite often is the triangle proof, cutting the circle into ever thinner concentric rings and unwrapping them to approximate a right triangle:

enter image description here

Again this leaves the question of convergence; do these approximations of right triangles converge to a right triangle of height $r$ and base $2\pi r$? Again it turns out that they do and the argument can be made rigorous, but again to the critical lay person this need not be clear from the picture.

Other proofs either require techniques beyond the most basic high-school curriculum, such as calculus for the onion proof, or are computationally involved such as Archimedes' proof or variations thereof.

I do assume that the layman is familiar with the fact that the circumference of a circle of radius $r$ equals $2\pi r$.

TL;DR

Given that the circumference of a circle of radius $r$ equals $2\pi r$, is there a simple proof of the fact that its area equals $\pi r^2$, using nothing but the most elementary mathematical techniques (certainly no calculus)?

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    $\begingroup$ Honestly, there is a pretty deep issue here which requires advanced techniques: the circumference of a circle is proportional to the radius, while the area is proportional to the square of the radius. That is, $C = k_1 r$, while $A = k_2 r^2$, where $k_1$ and $k_2$ are constants. It is possible to show, via various techniques, that $k_2 = \frac{1}{2} k_1 = \pi$, but this is a non-trivial fact, and requires some kind of limiting procedure (such as those outlined in your question). $\endgroup$ – Xander Henderson Jan 5 '18 at 17:14
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    $\begingroup$ In order to provide answers we really need to know what is your definition of $\pi$. If $\pi$ is defined as the area of a unit circle, there is very little to prove. $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 17:26
  • $\begingroup$ An elementary proof does not exist. I seem to recall a piece in one of MAA's publications (from the mid 1990s---this may be it) which points out this issue, and I have colleague who has written a pretty detailed paper which tries to patch things together using the techniques available to Archimedes. It is nontrivial. Of course, there are other attacks, but these are through, say, power series representations and differential equations. $\endgroup$ – Xander Henderson Jan 5 '18 at 17:30
  • $\begingroup$ @JackD'Aurizio I believe the question is about the relation between the constants of proportionality for area and circumference, and the fact that they involve the same constant $\pi$. I don't think that it is obvious that the area constant should be half the circumference constant. $\endgroup$ – Xander Henderson Jan 5 '18 at 17:32
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    $\begingroup$ @JackD'Aurizio Define $\pi$ however you like. The problem is showing that, whatever the definition, the value of $\pi$ that appears in the formula $A = \pi r^2$ is the same as the value of $\pi$ that appears in the formula $C = 2\pi r$. While this is not in the question (and the question should probably be edited to correct that), the response to my first comment makes it clear, I think(?), that this is the question being asked. $\endgroup$ – Xander Henderson Jan 5 '18 at 17:36
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I dont know how to draw pictures in the site, so I hope my explanation is clear, but you may have to draw your own. Inscribe a regular $n$-gon into the circle. At the same time exscribe a regular $n$-gon. This is in such a way that the sides of the triangles of the inscribed $n$-gon (which are radii, as in your first picture) extend to the sides of the exscribed $n$-gon. The height $h$ of a triangle of the inscribed $n$-gon is $r\cos \frac{2\pi}{n}$. Thus the area of a triangle of the inscribed $n$-gon is $$\frac{1}{2}br\cos\frac{2\pi}{n},$$ where $b$ is the base. Adding these together we have the area of the inscribed $n$-gon is $$\frac{1}{2}C_i r\cos\frac{2\pi}{n}$$ where $C_i$ is the circumference of the inscribed $n$-gon.

For the exscribed $n$-gon, the height of a triangle is $r$, and so the area of the exscribed $n$-gon is $$\frac{1}{2}C_e r$$ where $C_e$ is the circumference of the exscribes $n$-gon. Thus we have

$$\frac{1}{2}C_i r\cos\frac{2\pi}{n}\leq A\leq \frac{1}{2}C_e r$$

What is the relation between $C_i$ and $C_e$ ? It follows from similar triangles that

$$\frac{C_e}{C_i}=\frac{r}{h}=\frac{1}{\cos\frac{2\pi}{n}}$$

Thus we have

$$\frac{1}{2}C_i r\cos\frac{2\pi}{n}\leq A\leq \frac{1}{2}C_i r\frac{1}{\cos\frac{2\pi}{n}}$$ Now as $n\to\infty$, $$\cos\frac{2\pi}{n}\to 1$$ and $$C_i\to C$$ where $C$ is the circumference of the circle. This is in fact the definition of the circumference. Further, $$C=2\pi r$$ is the definition of $\pi$. So we have

$$A=\frac{1}{2}Cr=\pi r^2.$$

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  • $\begingroup$ @Servaes Thank-you. What dont you like about that claim ? The definition of the length of a curve is the limit of the lengths of all polygon approximations. So actually, I think its not a claim at all, but a definition. $\endgroup$ – Rene Schipperus Jan 8 '18 at 17:28
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This is a calculus-based proof, but it is very elementary. As a reference, see Keith Ball - An Elementary Introduction to Modern Convex Geometry.

For any $r>0$, let $L(r)$ be the length of a circle with radius $r$ and let $A(r)$ be the enclosed area.
By convexity (which also ensures that $A$ and $L$ are well-defined) $$ L(r)=\lim_{\varepsilon\to 0^+}\frac{A(r+\varepsilon)-A(r)}{\varepsilon}.\tag{$\delta$}$$ Since concentric circles with any radius are homothetic, $A(r) = r^2 A(1)$ and $L(r)=r L(1)$.
On the other hand $(\delta)$ gives $L(r)=\frac{d}{dr}A(r)$, hence $L(1)=\color{red}{2}\,A(1)$.

So we may equivalently define $\pi$ as $A(1)$ or as $\frac{1}{2}L(1)$.
And we have that in $\mathbb{R}^n$ the surface area of the Euclidean unit ball is just $n$ times its volume.

$(\delta)$ also follows from this pictorial argument, where the outer "polygon" is obtained by summing (in the set-sense) a small disk to the inner polygon:

enter image description here

It is visually clear what happens if we let the number of sides go to $+\infty$ while preserving the circum-radii of the inner and outer polygon.

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    $\begingroup$ @Servaes: consider two concentric circles with radii $r,r+\varepsilon$ and a set of orthogonal vectors to the boundary of the inner circle. The area of the annulus exactly equals $\varepsilon L(r) + A(\varepsilon)$. Have you ever seen the proof of the isoperimetric inequality through the Brunn-Minkowski inequality? See pages $111+$ of my notes. $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 18:36
  • $\begingroup$ I like the picture. Can it be used to argue Michael Behrend's premise? Regardless, interesting answer. (+1) $\endgroup$ – CopyPasteIt Jan 7 '18 at 20:49
  • $\begingroup$ @MikeMathMan: convex boundaries are always well-approximated by polygons, hence yes, it can be used to prove Micheal's premise, too. $\endgroup$ – Jack D'Aurizio Jan 7 '18 at 20:55
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The onion proof can be done without calculus, provided one assumes the following: the area of a circular ring is greater than width $\times$ inner circumference, and less than width $\times$ outer circumference. For a positive integer $n$, divide the circle into concentric rings of equal width $r/n$. If the above assumption is accepted, then the area of the $k$th ring ($k = 1, 2, \ldots, n$) lies between $(r/n) \times 2\pi r(k - 1)/n = 2\pi r^2(k - 1)/n^2$ and $2\pi r^2k/n^2$. From the formulae $$0 + 1 + 2 + \cdots + (n - 1) = (n^2 - n)/2,\quad 1 + 2 + 3 + \cdots + n = (n^2 + n)/2$$ it follows that the area of the whole circle lies between $\pi r^2(1 - 1/n)$ and $\pi r^2(1 + 1/n)$. Since this is true for any $n$, the area is $\pi r^2$.

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We can explain to the layman that thousands of years ago Euclid knew that there was a fixed number $K$ such that the area of a circle of radius $r$ is given by $K r^2$, and that this result was obtained without calculus.

Now, we tell the layman to look at $y = \sqrt{1-x^2}$ in the first quadrant and the region $G$ bounded by the graph and $x = 0$ and $y = 0$. Without calculus, in the obvious way (uh, with the help of a computer) they can approximate the area of $G$ and then multiply by $4$ to get the area of the unit circle = $\pi$.

We explain that this brute force method certainly seems convincing that we are talking about the same number $\pi$ in both the formula for the circumference and the formula for the area of a circle, but it is not a proof. Without exploring Archimedes proof or calculus, they will need to accept some amount of 'hand waving' in any purported proof. But this quick video seems accessible,

Archimedes' Proof that the Area of a Circle is $ \pi r^2$

If the layman needs to see a Cartesian brute force method, they can inspect the google spreadsheet Calculate $\pi$. At the bottom of the spreadsheet ($\text{row } = 1027$) they will find that

$\quad 3.139603642 \lt \pi \lt 3.143509892$

with the average of the 'squeeze' = $3.141556767$.

If the layman is with us so far, they might become so interested that they begin to study calculus in their spare time.

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  • $\begingroup$ "in the obvious way" - the idea is that our hypothetical layman is not a mathematician (not that interested in proofs) but is good with computers and ideas - so really close to 'discovering integration/calculus' on their own. $\endgroup$ – CopyPasteIt Jan 9 '18 at 1:02

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