Introduction

Today we will talk about the story of a simple problem that had far-reaching consequences in modern mathematics. It's the reason drops of water are spherical. It's what mathematicians call the "isoperimetric" problem (iso = same, perimeter = length of the boundary).

The problem is this: you are given a rope of a certain length. How would you enclose the maximum area possible with your rope? This is the 2 dimensional isoperimetric problem. Naturally, you can pose the same problem in 3 or more dimensions too. Say you have a surface of a fixed surface area, what's the maximum volume you can enclose?

The problem dates back to around 800 BC and over the millenia, has inspired a lot of developments in various fields of modern analysis and geometry.

The ancient Greeks knew that on the $2d$ plane, the circle would enclose the maximum area. Zenodorus of ancient Greece had a solution, but unbeknowst to him, his solution wasn't complete. 1900 years would pass without any significant progress on the problem. In 1744, the isoperimetric problem inspired the great Euler (read Oiler) to create the foundations of the field of calculus of variations together with Lagrange.

Jacob Steiner (196-1863) put forward many beautiful geometric arguments to solve the isoperimetric problem. Soon after, there was a complete proof of the isoperimetric problem for the 2 dimensional case.

In the late 1800's, Hermann Brunn and Hermann Minkowski discovered a nice little geometric inequality from which the isoperimetric problem could be solved rather elegantly, for all dimensions. This inequality, that would come to be known as the Brunn-Minkowski inequality (BM), and related ideas gave birth to an entire new field in convex geometry. It has many applications throughout geometry and analysis and over the years have made its presence felt in many disparate areas of mathematics.

Our focus for this article would be to take a peek behind the curtain and investigate the inner workings of the BM inequality in the $2d$ plane and see how it follows from simple and natural geometric ideas. We will also see how the isoperimetric problem and the related isoperimetric inequality follow easily from the BM inequality.

Dido and her oxhide rope

Our story begins with the legend of Dido circa 814 BC. Imagine you're the daughter of the king of Tyre (in present day Lebanon). Imagine after your dad's death, your brother who has seized the throne killed your husband in hopes of getting his hidden treasures. What do you do? Well, you throw your husband's treasures in the sea, gather a group of followers and run away as far as possible. At least, according to Virgil (famous ancient Roman poet), that's exactly what Dido did.

In the course of her journey, Dido found herself in the coast of North Africa and asked the local king Iarbas for a bit of help. She asked for a tiny bit of land for temporary refuge for her and her followers. She didn't ask for much, only as much as could be enclosed by an oxhide. The request seemed very modest to the king and he readily agreed.

Dido had a plan. She cut the oxhide into really thin strips, joined them together and decided to enclose the largest tract of land she could with them. And in doing so, she was faced with the so-called isoperimetric problem.

Q. Of all closed curves of length L which is the one that encloses the maximum area ?

As the above figure shows, there are many ways one can enclose an area with a given perimeter $L$. In fact, infinitely many. But according to legend, Dido did manage to find the answer. She figured that the maximum area is enclosed when the curve takes the shape of a circle. And with that land she founded the city of Carthage that would go on to become a powerful and prosperous kingdom. We do not know exactly how Dido solved the problem, it's likely she derived the answer from intuition.

If $L$ happens to be the perimeter of a circle with radius $r$, then we can write $L= 2\pi r $. That would give us the radius to be $r = L/2\pi$ and the area would be $$A_c =\pi r^2 = \frac{L^2}{4\pi} \tag{1}$$ And if we were to believe Dido and accept that yes, the circle is the figure with perimeter $L$ that maximizes the enclosed area, we can then say that for any other figure with perimeter $L$, if it's area is $A$, then $A$ must be less than or equal to $A_c$. That is, we can say, for any figure that has area $A$ and perimeter $L$, $$A \le \frac{L^2}{4\pi} $$ Often written as $$ L^2 - 4\pi A \ge 0 \tag{2}$$ this inequality is called the isoperimetric inequality. And equality holds in the above inequality, that is, $L^2 - 4\pi A = 0$ when the curve takes the form of a circle of perimeter $L$.

Next we will take a look into the ideas behind Brunn and Minkowski's neat little inequality that, for one, proves the isoperimetric inequality not only for the $2d$ case but for all higher dimensions.

We will mainly focus on the $2d$ case where it's easy to see the inner workings of the inequality. But the same arguments can be used without much work for the general case too. We'll also see a sketch of the proof of the Brunn-Minkowski inequality and finally use it to prove the isoperimetric inequality.

Adding two shapes together

Any point on the $2d$ plane, say $u = (u_1,u_2)$ can be thought to represent a vector, namely, the vector joining the origin and the point $u$. That is, it has the direction of the line starting from the origin, $O$ and passing through the point $u$ and has a length equal to the length of the segment $Ou$.

This is usually called a position vector and we can identify the point with the vector and vice-versa.

Say we have another point $v = (v_1, v_2)$. If we define the sum of the two points as the sum of each co-ordinate, that is, $u + v = (u_1 +v_1, u_2 + v_2)$, then $u+v$ is the same as the vector sum of the vectors represented by $u$ and $v$.

Now, take a disc, say $B$ on the $2d$ plane. By disc, we mean, the circle alongwith the region it encloses taken together. Can we define anything meaningful by $B+u$, where $u$ is a vector say $(u_1,u_2)$?

A natural way to define it would be to consider each point $b = (b_1,b_2)$ in $B$ and adding to it the vector $u = (u_1,u_2)$, that is, we are doing $b+u = (b_1+u_1, b_2+u_2)$. That amounts to simply moving the disc $B$ along the direction of $u$ and by a distance equal to the length of the position vector $u$ like so

This makes sense. And similarly, it can be defined for any shape on the plane. Adding a shape by the vector $u$, simply means moving or translating the shape by the vector $u$

Adding a square and a disc

Let's take a square, $K$ and a disc, $B$ centered at the origin. Does it make sense to define the vector sum of $K$ and $B$?

Well, yes. Just like in the previous step, we translated a disc by a specific vector, here $K+B$ is the set of all points of the form $k+b$ where $k$ is any point in $K$ and $b$ is a point in $B$. Basically we take all possible pair of points one in $K$ and the other in $B$ and take their vector sum.

That sounds a bit complicated. How would the shape $K+B$ look like?

The points on the boundary of the square will be translated/moved to the boundary of the new shape.

The new shape would be as if we put discs of the same radius as $B$ at each point of the boundary of the square. The resulting shape would be a square with rounded corners as shown above.

But what if the square and the circle were not placed with their centers at the origin but placed somewhere else like this.

The center of the square is the position vector $u$ and the center of the circle is the position vector $v$. How would $K+B$ look like now? What would the shape be?

Here we can do a nifty little trick:

$1.$ Translate $K$ by $-u$ which takes the center of $K$ to the origin. The resulting shape is $K-u$.
$2.$ Translate $B$ so that it's center now lies at the origin too, that is translate it by $-v$.

And now, we are back to the case where we have a square and a disc centered at the origin and their Minkowski sum would be the familiar square with rounded corners.

So, $K-u + B -v $ is a square with rounded corners. But $K-u + B-v$ is the same as $K+B -(u+v)$. Then to get $K+B$, we simply need to translate the rounded square with center at the origin to the position vector $u+v$. Notationally, it's the trivial cancellation $K+B -(u+v) + u+v = K+B$

And since the translation doesn't change thte shape, $K+B$ is still the same rounded square, just located at $u+v$. Moving a piece of paper lying on your table wouldn't change it's size. Similarly, translating shapes on the plane, won't change its shape or area.

This vector addition of shapes is commonly known as Minkowski sums in geometry after the German mathematician Hermann Minkowski of course. The square and the disc are nothing special, the same Minkowski sum can be done for any two set of points. In set notation, one would write this as $$K+ C = \{ k+c \text{ } | k \text{ is in }K, \ c \text{ is in }C\} $$

For a shape (set of points) $K$ on the $2$ dimensional plane, we will denote the area of $K$ by $V(K)$. The area is nothing but the $2$ dimensional volume, and that's what the notation $V$ secretly refers to. At this point, it might be good to recall the very intuitive properties of area.

If a shape $C$ is contained in the shape $K$, $V(K) \ge V(C)$. The total area of two non-overlapping shapes, is the sum of their individual areas. Notationally, if $K$ and $C$ are the shapes under consideration, and they are non overalpping. $V(K \cup C) = V(K) +V(C)$. From this, it also follows, that if the two shapes have some overlap, then $V(K \cup C) < V(K) + V(C)$. The figure above might help see this.

As we noted earlier, if we simply translate a set by a certain vector, the shape remains unchanged and so too does the area. We can write this notationally, for any vector $w$, like so: $$V(K+w) = V(K) $$

What we will do next is try to see how the areas of the Minkowski sums of two shapes relates to the area of the individual shapes. And since the area doesn't vary if we simply translate shapes, whenever we wanna visualize Minkowski sums of shapes that have some symmetry at the center like a square, circle, rectangle etc, We will translate them so that their center lies at the origin $(0,0)$.

Let's go back to our example of figuring out $K+B_r$ where $K$ is a square of side-length $l$ and $B_r$ is a disc of radius $r$. We found that $K+ B_r$ is a larger square with rounded corners.But what about the area $V(K+B_r)$?

From the above figure, we can see that

And here, we notice a curious thing, that $$V(K+B_r)^{\frac{1}{2}} \ge V(K)^{\frac{1}{2}} + V(B_r)^{\frac{1}{2}}$$

Let's go back to the expression for $V(K+B_r)$.

$$V(K+B_r) - V(K) =\pi r^2 + 4lr$$ and $$\frac{V(K+B_r)-V(K)}{r} = \pi r + 4l $$ From here we can see that if we keep decreasing $r$, the expression $\frac{V(K+B_r) - V(K)}{r}$ gets closer and closer to $4l$. If you are familiar with the concept of limits, we can write it concisely as

$s(K)$ here denotes the length of the perimeter of $K$. Here by $r \to 0+$, we are emphasizing the fact that $r$ is non-negative quantity, being the radius of a circle/disc and it approaches $0$ from the right hand side as it's continually decreased.

Turns out this is not only true for the Minkowski sum $K+B_r$ when $K$ is a square, but for for any nice enough set, with a smooth boundary.

Let's take an arbitrary nice shape $K$ on the plane with a smooth boundary. Once again, to see how $K+B_r$ looks like, refer to the pic above. At each point on the boundary of $K$, imagine a disc of radius $r$ and take the resulting figure.

$V(K+B_r)- V(K)$ gives the area of the shaded strip as in the pic above.

If we cut the strip and stretch it out, we get approximately a rectangular strip with one side having the length $L$, that is the perimeter of the shape $K$ and the other side has length $r$. It very closely resembles a thin rectangular strip when $r$ is very small.

So we have approximately, $$ V(K+B_r) - V(K) \approx rL$$ and so, $$\frac{V(K+B_r) - V(K)}{r} \approx L $$ This formula $$\lim_{r\to 0+} \frac{V(K+B_r)-V(K)}{r} = L=s(K)$$ where $s(K)$ denotes the perimeter of $K$ is known as the Minkowski-Steiner formula and relates the area of a figure to its perimeter. It also holds in dimensions $3$ and higher for nice enough objects and relates the $n$ - dimensional volume to the $n-1$ -dimensional volume of the boundary (surface area).

Adding rectangles together, the Minkowski way

Now let's take a look at another example of the Minkowski sum. This time for two rectangles, $K$ and $C$, whose sides are parallel to the $x$ and $y$ axes and so also to each other. Let the corresponding sides of $K$ and $C$, be $a_1,a_2$ and $b_1,b_2$

As before we can translate both the rectangles to the center, since translation won't change the shape or the area or the shape.

Then $K + C$ would be the shape obtained by placing the rectangle $C$ at each point of the boundary of $K$ and taking the composite shape so obtained as in the figure above. Turns out $K+C$ is simply a larger rectangle with side lengths $a_1+b_1$, $a_2+b_2$. That is, $$V(K+L)^{\frac{1}{2}} \ge V(K)^{\frac{1}{2}} + V(L)^{\frac{1}{2}} \tag{$\circ$}$$ We encountered this inequality in the other example too. And it's no coincidence. In fact, it turns out this inequality holds for all "nice enough" sets/shapes. By "nice enough", we mean sets for which the area is defined.

This inequality is known as the Brunn-Minkowski inequality. Just as it holds here in $2$-dimensions it also holds in higher dimensions. That is, if $K$ and $L$ are $n$-dimensional objects and $V(\cdot)$ denotes the $n$-dimensional volume, then $$V(K+L)^{\frac{1}{n}} \ge V(K)^{\frac{1}{n}} + V(L)^{\frac{1}{n}} $$ It was found that equality holds when $K$ and $L$ are the same after possible translations or/and scaling.

Brunn-Minkowski Inequality -sketch of proof

We already saw that the Brunn-Mminkowski inequality holds for two rectangles with sides parallel to each other and the $x$ and $y$ - axes. As we will see, this is about all the information we need to say that it holds for all $2$ dimensional sets (well, rechnically all of those for which the concept of area is defined, mathematicians call them "measurable sets")

Minkowski sum of a collection of rectangles in a grid

Suppose $A$ is a collection of $2$ non-overlapping rectangles in a grid like shown in the figure above and $B$ is $1$ rectangle also from another rectangular grid. What can we say about the Minkowski sum of $A$ and $B$ and the area, $V(A+B)$?

For starters, note that $A$ is basically a composite shape made up of two grid rectangles $A_1$ and $A_2$ taken together. In set notation, we can write that as a union: $$A = A_1 \cup A_2 $$ Recall that $$A +B = \{ a+b | a \text{ is in A}, \text{ $b$ is in B} \} $$ Since $A_1$ and $A_2$ do not overlap, if we take any point $a$ in $A$, it must be either in $A_1$ or in $A_2$, can't be in both.

So $A+B$ is actually just $A_1 + B$ and $A_2 +B$ taken together (their union, that is).

Remember that translating $A$ and $B$ won't change the shape of $A +B$ as we saw before. So, now, we can safely shift $A$ horizontally till we have the $y$-axis as one of the grid-lines and $A_2$ just to the right of it. That is, all points in $A_2$ have their $x$-coordinates non-negative.

We also shift $B$ so that the $y$- axis passes somewhere through $B$ dividing it into $2$ smaller rectangles, $B_1$ to the left and $B_2$ to the right such that $$\frac{V(A_1)}{V(A_2)} = \frac{V(B_1)}{V(B_2)} $$

The motivation behind shifting and having the $y$-axis ($x=0$) dividing $A$ and $B$ in such a way is to ensure that $A_1 + B_1$ will always lie to the left of the $y$ - axis and $A_2$ + $B_2$ will always lie to the right of the $y$- axis. And because of that they will not overlap (except possibly only on a line segment that is part of the $y$-axis, but the area of a line, a 1 dimensional figure is $0$, so we can disregard that overlap).

To see this note that the $x$-coordinates of all points in $A_1$ and $B_1$ are always negative or zero (non-positive) and when they are added together, the result will also be non-positive. Similarly, the $x$-coordinates of all points in $A_2 + B_2$ are all non-negative.

Now, the Minkowski sum, $A + B$, of course contains $A_1 + B_1$ and $A_2 + B_2$ in it, so the area of $A+B$ is greater than the sum of the areas of $A_1 +B_1$ and $A_2+B_2$. That is, $$V(A+B) \ge V(A_1+B_1) + V(A_2+B_2) $$ $A_1, B_1, A_2, B_2$ are rectangles and we already know from before that the Brunn-Minkowski inequality is true for rectangles. So, of course, we can write
$$V(A_1+B_1)^{\frac{1}{2}} \ge V(A_1)^{\frac{1}{2}} +V(B_1)^{\frac{1}{2}} $$ $$V(A_2+B_2)^{\frac{1}{2}} \ge V(A_2)^{\frac{1}{2}} +V(B_2)^{\frac{1}{2}} $$

And in turn, we have

Remember that we had chosen $B_1$ and $B_2$ such that $$\frac{V(A_1)}{V(A_2)} = \frac{V(B_1)}{V(B_2)} $$ That is, for some number $c$, $$\frac{V(B_1)}{V(A_1)} = \frac{V(B_2)}{V(A_2)} = c $$

$A_1$ and $A_2$ are non-overlapping and so are $B_1$ and $B_2$, so we can write $$V(A) = V(A_1) + V(A_2) $$ $$V(B) = V(B_1) + V(B_2) $$ Then

Using this in the inequality in $(\ast)$, we get

That is, $$V(A+B)^{\frac{1}{2}} \ge V(A)^{\frac{1}{2}} + V(B)^{ \frac{1}{2}} $$ and we have Brunn-Minkowski inequality for the sum of a shape composed of $2$ grid rectangles and another shape composed of $1$ grid rectangle.

An exactly similar argument, along with induction can be used to show the Brunn-Minkowski inequality for two collections of any number of grid rectangles. It could be a straightforward exercise to try it out yourself.

Suppose we try to cover two arbitrary shapes $K$ and $C$ with rectangular strips with sides parallel to the $x$ and $y$ axes. Let's denote the collection of rectangular strips in $C$ by $A$ and $B$ denote the collection of rectangles in $C$. See figure above.

We have already seen that the Brunn-Minkowski Inequality holds for collection of rectangles with sides parallel to the $x$ and $y$ axes. Here that would impy $$V(A+B)^{\frac{1}{2}} \ge V(A)^{\frac{1}{2}} + V(B)^{\frac{1}{2}} $$

The areas of $K$ and $C$ can be approximated increasingly better by finer and finer rectangular strips that are more in number. And at each step, the Brunn-Minkowski inequality is true for those collections of rectangular strips.

Passing onto the limit, we would have that the Brunn-Minkowski inequality holds for $K$ and $C$ as well.

An almost similar argument can be made in dimensions $3$ and higher, dealing with $n$- dimensional boxes instead of rectangles and it can be shown that the Brunn-Minkowski inequality holds in higher dimensions too.

Isoperimetric inequality revisited

With the Brunn-Minkowski inequality in our toolbox, tackling the isoperimetric inequality becomes a straightforward application.

Say, we have a shape $K$ that has perimeter $L$ and area $A$, i.e., $V(K) = A$. Then if we take the Minkowski sum of K and a disc of radius $r$, say $B_r$ $$\frac{V(K+B_r) -V(K)}{r} \to L$$ as $r \to 0+$ (from the Minkowski-Steiner formula we saw before).

Also, the Brunn-Minkowski inequality gives us

That is,

And then $$$$

That is, we have $$L \ge 2\sqrt{A}\sqrt{\pi} $$ And squaring it, we have our beloved isoperimetric inequality $$L^2 \ge 4\pi A $$

Equality holds in the Brunn-Minkowski inequality if the two objects are the same up to translation and scaling, which is to say that they have the same shape. For example if one of them is a circle centered at the origin, for equality to hold, the other must be a circle as well but maybe of a different radius or maybe with their center located away from the origin.

And equality in $L^2 \ge 4\pi A$ implies that we also have equality in the Brunn-Minkowski inequality $$$$

Then $K$ must be a disc since possibly scaling and translating the disc $B_r$ would give us a disc.

That is, equality holds in the isoperimetric inequality when the shape is a disc of perimeter $L$ and area $A$.

$$ $$
$\blacksquare $