The isoperimetric inequality

Our story begins at around 800 BCE in ancient Greece with Dido of Tyre. As many stories from that time, this too started with a murder, when Dido’s brother Pygmalion killed her husband. Fleeing her brother, Dido eventually reached north Africa (where Tunisia is today), and there a local ruler offered her the following deal. For her fortune, she will get an ox hide, and what ever land she can encompass with it, she can have. Since Dido was quite a clever woman, she cut out the hide into thin long strips, and joined them together for one long strip. She then faced with quite an interesting problem – what is the largest area that she could encompass with that given strip?

Eventually Dido used this strip to form a half circle, where the other side was bounded by the Mediterranean sea, and there she established the city of Carthage where she was the queen. And now, some years later, we can marvel at her choice and see why this is actually the best that Dido could have done.

For a more formal discussion, we have some close simple curve with bounded area $A$ and perimeter $P$ . If we fix the perimeter $P$ , what is the largest possible bounded area?

Clearly, once we fix $P$ , the area $A$ cannot be as large as we want, so it should be bounded from above by a function of $P$ . What should be this function?

For the Hebrew listeners out there, I also made a video about this here: https://youtu.be/z6T718F1NaE.

Trying to find some intuition, think about taking your curve and filling it with as much material as you can (or in dimension 3 – fill a balloon with air). This material will “push out” the perimeter, and we expect it to become more and more like a circle. For a circle with radius $R$ the perimeter is $P=2\pi R$ while the area is $A=\pi R^2$ , so that $4\pi A=P^2$ . If we actually expect this to be the best possible, then for other curves we should have $4\pi A\leq P^2$ . Indeed, this is the case, and this inequality is called the isoperimetric inequality (for the plane).

As proving it for general curves seems quite complicated, since general curves can be complicated, lets start with simpler curves – the polygons.

The isoperimetric inequality for polygons

Let us assume now that our curve is a polygon with $N$ vertices, and we want to show that the inequality $\frac{4\pi A}{P^2}\leq 1$ holds for it as well. In other words, we are now looking for the polygons where this ratio is maximal, and want to show that it is less than $1$ there.

If you want to play around with the polygons by yourselves, you can also do it here: https://openprocessing.org/sketch/1299967

Step 1:

By our intuition, increasing this ratio should mean that the polygon is getting “closer” to to a circle. The question is how do we formalize this “closeness”? For example, let look on the following “new moon” like hexagon:

While the edges at the part ABCD is a sort of approximation of a part of a circle, the part at DEFA is the opposite of what we expect from a circle – it “curves” in the opposite direction. To fix this, we can simply take the part DEFA and straighten it to a line:

Since the straight line between D and A is the shortest curve between these two points, then certainly we decreased the perimeter. Also, we clearly increased the area of the polygon, so the ratio $\frac{4\pi A}{P^2}$ is larger for the new polygon.

We can do this operation for any two vertices in the polygon where the segment between them is outside the polygon. After finitely many such operation we get a more “rounded” polygon, where the segment between any two vertices is inside (including boundary) of the polygon, and it is not hard to show that actually for any two points in the polygon, their connecting segment is in the polygon. This property, that any two points in a set has their connecting segment in the set also is exactly what convexity means. In other words, we have just proved that

Claim: If $\frac{4\pi \cdot A}{P^2}$ is maximal, then the polygon is convex.

Step 2:

Now that our polygon is a bit more “rounded” how can we “round” it even more?

Lets think about it like queen Dido. Can we increase the area without changing the perimeter? We can start with something easy – can we change the location of a single vertex that satisfies this? Since the only change of the perimeter is for the edges touching this vertex, the only thing that we need to check is that the sum of the distances from this vertex to its neighbors is fixed. Luckily for us, this has a very simple geometric interpretation – the vertex must be on an ellipse where its neighbors are the focus points:

What can we say about the area? Well, we already know that our polygon needs to be convex, so the vertex needs to be on the “outer” part of the segment connecting its neighbors (above the purple line in the image). In this case, the area of the polygon is the area of the triangle formed by the vertex and its neighbors, and the fixed area from the rest of the polygon (below the purple line). Thus, to increase the area we only need to find the triangle with maximal area.

This last problem is quite simple to solve (try it before reading!). If we have any ellipse (which we might as well assume is aligned to the axes) $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$ with $a > b > 0$ , then its focus points are at $(\pm \sqrt{a^2-b^2}, 0)$ . Their actual values are not really important, other than the fact that they are both on the $X$ -axis and at the same distance from the origin. If we connect these two points with a point on the ellipse $(x,y)$ to form triangle, then its area is $\sqrt{a^2-b^2}*|y|$ , so to maximize it, we simply need to maximize $|y|$ . But we are on the ellipse, so that the two points which maximize it are $(0, \pm b)$ .

What is important in this whole argument, is not actually finding the area of the triangle, but noting that it is a very symmetric triangle – the edges connecting our vertex on the ellipse to the focus points of the ellipse have the same length. In our original problem, this means that we maximize the area exactly when the two edges touching the vertex have the same length. As we can do it for any vertex, we conclude that :

Claim: If $\frac{4\pi \cdot A}{P^2}$ is maximal, then all the edges of the polygon have the same length.

Step 3:

We are already in a good place – our polygon must be convex, and all of its edges have the same length. To really determine the shape of the polygon we also need to say something about the angles. We expect them to all be the same, so that the polygon would be as “rounded” as it can be, but the two properties from above are not enough, for example, we can have a polygon like that:

We now need a third trick to round it even more. We can no longer move a single vertex, since this will change the lengths of the edges, so instead we can move a whole edge as follows:

Here too we have a similar problem to the one from the previous step. The perimeter doesn’t change, and the area is the area of the quadrilateral that we change + the rest of the area of the polygon, so we only need to maximize the area of the quadrilateral, where three of its edges have the same length.

The entire idea of this post, is that maximizing the area means that our shape is “rounded” in a sense. As we shall see below, this claim is true for this small problem as well. If we are given 4 edge lengths, and we want to construct a quadrilateral with maximal area, then its vertices must be on a circle (how more “rounded” can you get here?). What does it tell us about the angles? We have 3 edges which have the same length and bounded in a circle, and I claim that this means that the angles between them must be the same. One way to see it is by taking the triangles defined by these edges and the center of the circle as follows:

By construction, all of these triangles are congruent (their edges have the same lengths – two green and an orange edge in the picture above), so the angle between the edges must be the same (twice the angle between the orange and green edges).

Another, more visual, way, is to rotate our circle and see that one angle must fall exactly on the second, so that they must be the same.

In any case, we conclude that the angles on both side of the edge must be the same and therefore:

Claim: If $\frac{4\pi \cdot A}{P^2}$ is maximal, then all the angles of the polygon have the same size.

Rounding up quadrilaterals

Consider now the problem of a quadrilateral with edges of length $a,b,c,d$ as follows:

We want to show that from this family of quadrilaterals, the one with the largest area is bounded in a circle. It is not hard to show that the area is a one parameter function – if we know even one angle, we can find out all the rest and compute the area. However, this function is going to be quite complicated, so instead we will use another approach which utilizes some symmetry of the problem. For the Hebrew listeners among you, I made a video about this problem here https://youtu.be/mgX1Z3CZ7to . You can also try to play with these quadrilaterals in the interactive program here https://openprocessing.org/sketch/1304517

Consider the two opposite angles $t,s$ in the quadrilateral, and break it to two triangles using the diagonal separating them. Given these angles, we can compute the area of these triangles to get that the total area is

$A = f(s,t) = (ab\sin(t) + cd\sin(s))/2$ .

However, not every two angles create triangles that we can glue together to form a quadrilateral. The condition should be that the length of the 3rd edge in both triangles (the diagonal) must have the same length. In order to get that length $x$ we can use the cosine theorem (which is a generalization of the famous Pythagorean theorem) to get that

$a^2+b^2 - 2ab\cos(t) = x^2 = c^2 + d^2 -2cd\cos(s)$ .

In other words, we want to find the maximum for $f(s,t)$ given the condition

$g(s,t) = [a^2+b^2 - 2ab\cos(t)] - [c^2 + d^2 -2cd\cos(s)] = 0$ .

While now we have a two parameter function with a condition, both of $f$ and $g$ are very simple function which have nice symmetry in $t$ and $s$ . The main tool to deal with this sort of problems is the Lagrange multipliers. Since this is a post about the isoperimetric inequality, I will leave the more visual explanation for another time, and go straight to the formulation which simply tells us that a maximum can be reached only if $\nabla f = \lambda \cdot \nabla g$ for some constant (multiplier) $\lambda$ . A simple calculation shows us that

$\tan(t) = -4\lambda=-\tan(s)=\tan(-s)$ .

Since $\tan(\cdot)$ is $\pi$ periodic, we conclude that $t+s=k\cdot \pi$ for some integer $k$ . Furthermore, it should not be too hard to convince yourselves that $0<t,s<\pi$ , so that the only possibility is that $k=1$ . This means that the sum of opposite angles in the quadrilateral is $t+s=\pi$ , which is exactly the condition for the quadrilateral to be bounded in a circle.

The most rounded polygons

At this point we know that for $\frac{4\pi \cdot A}{P^2}$ to be maximal, all the edges of our polygon must be the same, as well as all of the angles, or in other words our polygon is regular.

Computing the area and perimeter is now a simple task. Without loss of generality, we can assume that the polygon is bounded in a circle of radius 1 (changing the radius doesn’t change the ratio from above!). If the polygon has degree $N$ , then connecting the vertices to the center of the circle, divides the polygon into $N$ triangles, each have two edges of length 1, and the angle between them is $\frac{2\pi}{N}$ . This means that the last edge has size $2\cdot \sin(\frac{\pi}{N})$ and the area of each triangle is $1\cdot 1\cdot \sin(\frac{2\pi}{N})/2$ . It follows that the area and perimeter are

$A = N\cdot \sin(\frac{2\pi}{N})/2=N\cdot \sin(\frac{pi}{N})\cos(\frac{\pi}{N})$

$P = N\cdot 2\cdot \sin(\frac{\pi}{N})$ .

Computing the ratio we get

$\displaystyle \frac{4\pi A}{P^2} = \frac{4\pi N\cdot \sin(\frac{pi}{N})\cos(\frac{\pi}{N})}{(N\cdot 2\cdot \sin(\frac{\pi}{N}))^2} = \frac{\pi \cdot \cos(\frac{\pi}{N})}{N\cdot \sin(\frac{\pi}{N})} = \frac{\pi/N}{\tan(\pi/N)}$ .

As a simple exercise, show that the last expression is always smaller than 1 where $N\geq 3$ , and this completes the proof for the polygon case.

The general curve

In general we have some close curve on the plane, e.g. a continuous injective map $\varphi:S^1\to \mathbb{R}^2$ , and we want to look at its perimeter and the area of the set it encloses. For simplicity, you may assume that the curve is piecewise smooth function. What can we say about this much more complicated version?

We worked hard for the polygon case, so lets use the ideas from there. First, we already know that we want convex polygons, but the same argument works for general curve. If we look at the convex hull of the area bounded by the curve, then it is of course larger than the bounded area itself. More over, the boundary of this convex hull is similar to the original curve, but we “straighten” out several parts, and just like before, the straight line between two points is the shortest curve between them, so we also decreased the perimeter. In other word:

Claim: If $\frac{4\pi \cdot A}{P^2}$ is maximal, then the area bounded by the curve is convex.

If the convex hull is in itself a polygon, then we can solve it directly using the previous section. But even if not, we can at least approximate the curve using a polygon. By choosing enough points on the curve and taking the polygon defined by them, we can assume that both the area and perimeter of this polygon are as close as we want to the area and perimeter respectively of the original curve.

We can actually say something stronger since the curve bounded a convex set. Each edge on our polygon is (1) inside the original bounded set , and (2) it is shorter then the part of the curve between its endpoints. In other words, both the area and perimeter of the new polygon are smaller than the area and perimeter of the original curve, or formally

$A-\varepsilon \leq A_{poly} \leq A, \; \; P-\varepsilon \leq P_{poly} \leq P$ .

Using these inequalities, and the fact that we know that the isoperimetric inequality holds for polygons, we get that

$4\pi A \leq 4\pi(A_{poly}+\varepsilon)\leq P_{poly}^2 + 4\pi \varepsilon \leq P^2 + 4\pi\varepsilon$ .

Since $\varepsilon$ can be as small as we want, we conclude that $4\pi A\leq P^2$ which is exactly what we wanted to prove.

Some more isoperimetric inequalities

The isoperimetric inequality is not unique to the plane and there are analogues in many other systems. We can go to higher dimensions spaces (e.g. in the 3-dimensional world the best possible ratio between volume and surface is attained in the sphere), we could have other variations (e.g. close the perimeter with a given straight line, like in Dido’s case), even go to more discrete versions (like with graphs) etc.

The attempt to maximize the area while minimizing the perimeter and its generalizations, appear in many problems, and in particular we can see how nature has “solved” it many time.

Probably the best example is a soap bubble. While the soapy surface part itself “wants” to be as small as possible, the air inside it cannot be pressured to be too small. Hence, looking to minimize its surface area, the bubble gets it familiar look – the sphere.

Another place that we see it, and actually practice it, is when it is really cold outside. While our body generates heat, we always lose some of it through our surface area. This is why, when we want to keep ourselves warm, we try to curl into a ball as much as possible. We will probably won’t increase our volume just to get warm (at least we can’t do it quick), but this way we can reduce our surface area, and the best way to do it, is to be as “rounded” as possible.

There are of course many other variations and applications for the isoperimetric inequality, but this will be left for another time.

The isoperimetric inequality