It doesn't seem like a very difficult task
to find the closed curve in the plane that, for a given perimeter, encloses the largest area.
To find this curve, we can start by calculating the areas of a few figures with
a perimeter of 1. The triangle, the square, the pentagon... decagon, hectagon,
chiliagon, myriagon. No matter how many sides you add, the area of any regular
polygon with a given perimeter is always less than that of the circle.
You can try with ellipses, ovals, ovoids, or devilish polygons. You can
integrate the weirdest curve you can think of, but the result will always be less
than the area of the circle. Given that, the circle stands out
as the great candidate for the solution to the isoperimetric problem. But, is it?
This simple problem is one of the oldest in mathematics, THE ISOPERIMETRIC
PROBLEM, and its history begins with a legend and continues with wonderful episodes.
Its simplicity should not deceive us because it took mathematicians more than 2,000 years
to prove it, and for that, we had to invent new areas of mathematics.
Additionally, the isoperimetric problem is one of the few mathematical problems that made it into
literature. Poets and writers from the ancient and medieval worlds incorporated it into their works.
The most famous of all is Virgil, who in the Aeneid tells us the story of the ingenious Queen Dido.
In the 9th century B.C., King Matan ruled in the Phoenician city of Tyre, located in what
is now Lebanon. This king had three children: Pygmalion the eldest, Elissa
our protagonist, and the young Anna. After King Matan's death, Pygmalion
proclaimed himself king of Tyre. Our next character is Sychaeus, priest of the temple
of Melqart, who possessed great riches that Pygmalion secretly coveted.
Pygmalion arranged the marriage of his sister Elissa to the much older priest Sychaeus.
He waited some time and then tried to convince his sister to reveal where Sychaeus hid his riches.
Elissa realized at that moment that she had been used. So, she told her brother a wrong place
for the treasure, under the altar, hiding the real location, the garden. That same night, Pygmalion sent
assassins to kill Sychaeus. When Elissa found her husband dead,
she unearthed the treasure and fled from Tyre, taking her sister Anna and a small entourage with her.
In her desperate flight, Elissa did not want to settle in any Phoenician colony
for fear of her brother, and thus she reached the coasts of Africa, where a tribe of Libyans lived, whose king was Iarbas. Elissa asked King Iarbas for a piece of land
along the coastline where she could settle. The king, as a sign of hospitality, offered her
all the land she could encompass with the hide of a bull. The sovereign made this "generous" offer
with the intention of giving her the minimum land. But Elissa maximized it with cunning, cutting
the animal's hide into very thin strips to make a very long thread with which she
drew a semicircle. In that space, she founded the famous city of Carthage, and thus she began to be called QUEEN DIDO.
Dido solved the problem using her intuition. Greek mathematicians were
the first to try to prove that this intuition was correct. The first to tackle the
problem was Zenodorus around 200 B.C. His treatise "On Isoperimetric Figures" is lost, and
we only know it by references. Since then, many have tried to prove the isoperimetric problem. Some of these proofs were incomplete,
if not erroneous, but they have led to the development of interesting and fruitful ideas.
The list of mathematicians who have worked on this problem includes Galileo, Euler, the
Bernoulli brothers, Gauss, Legendre, Steiner, Weierstrass, Schmidt, and many others.
Today, the isoperimetric problem remains an active field of
research in areas such as differential geometry, discrete and convex geometry, probability, and
Banach space theory... But today, by the isoperimetric problem, we
understand something more complex, a problem in which we try to find a surface that minimizes the perimeter (or hypersurface) under one or more volume constraints and possibly with additional boundary and symmetry conditions.
Although the problem solved by Queen Dido deals with a half-plane and has
as a solution the semicircle, we will pose the problem for the plane and soon
we will see that both problems are equivalent. The ISOPERIMETRIC THEOREM states that given a
fixed length L, the plane curve with perimeter L that encloses the largest area is the circle.
Probably the Swiss Jakob Steiner is the mathematician most strongly associated with the isoperimetric problem.
In this video, we will follow his purely geometric reasoning, although first, allow me to introduce
a couple of necessary preliminary lemmas. To begin with, let's PROVE a similar problem in its approach. We start with a FAMILY of mathematical objects. In this case, triangles of any size and shape. We impose a CONDITION that
they must meet: Two sides of the triangle must have fixed lengths a and b.
Finally, we introduce a FUNCTION TO OPTIMIZE. In this case (as in the isoperimetric problem), we want the area of the triangle to be MAXIMUM. Among all possible triangles with two sides of lengths 'a' and 'b,' WHICH ONE HAS THE LARGEST AREA?
To answer this question, we place the side of length 'b' as the base
of the triangle and remember that the area of a triangle is precisely base x height / 2.
We are interested in making the red point such that the height 'h' is as large as possible. But the red point
is one of the ends of the segment of length 'a' with the other end fixed to the base. Therefore,
that point must belong to the circumference centered at that end and with radius 'a.'
Clearly, 'h' reaches its maximum when it coincides with the side of length 'a,' and this
is the case when the sides of the triangle form a right angle. Thus, the triangle with two
sides of lengths 'a' and 'b' with the largest area is the RIGHT TRIANGLE. The area is A = a x b / 2.
Our second PRELIMINARY LEMMA is related to a theorem by the great Thales on
circumferences and inscribed triangles. As we saw in the video linked. Any triangle
inscribed in a circumference such that one of its sides is a diameter is a right triangle.
What is interesting about this result is that the RECIPROCAL is also true: given a
segment, the geometric locus of the points from which the segment subtends
an angle of 90° is a circumference. You can find this characterization of the circumference in the book by Rademacher and Toeplitz “Numbers and Shapes.” We leave you a link
to this book in the video description. Let's prove the isoperimetric theorem
using these two lemmas! To do this, we will begin by distinguishing between
two types of plane curves. A curve that verifies that given any two points
on its perimeter, the segment that joins them is contained within it is said to be CONVEX. If there
are points on the perimeter such that the segment joining them is not contained within, then the curve is NON-CONVEX. Let's see that if a curve is NON-CONVEX, it cannot be the curve with the maximum area and given perimeter, because we will construct a new
curve with the same perimeter and larger area. Let's take a segment that lies outside our figure.
This segment divides the perimeter into two curves of lengths R and Q whose sum is L.
Now, using the segment as an axis of symmetry, we can consider a new curve
Q’ symmetric to Q, and thus both have the same length. The curves Q, Q’ and the segment,
determine two regions I and II of identical area. Well, the curves R and Q’ determine a new
curve whose perimeter is also L, but whose area is larger than the area enclosed by the original curve. We can do this with any non-convex curve, and therefore NON-CONVEX curves
are ruled out as solutions. The solution to the isoperimetric problem must be CONVEX.
Let's consider a convex curve. Any chord divides the perimeter into arcs R and Q that
sum to L. By fixing one of the ends and moving the other, we can make the two arcs have the same
length and divide the figure into two hemispheres of areas B and C, whose sum is the total.
If these two hemispheres did not have the same area, for example, B > C, let's see that we can construct a curve with the same perimeter and larger area than the previous one. We consider the segment
as the axis of symmetry and project the arc R, obtaining a figure with a perimeter R + R = L.
The area of this new figure will be A’ = B + B, and since B is greater than C, we have that
the area A’ is greater than the area A = B + C. The curve with the maximum area and
fixed perimeter L must satisfy that a segment dividing its perimeter into equal parts
also divides its area into equal parts. To find the solution, it is enough
to find the curve with perimeter L/2 and ends on a line that encloses the largest
area. This was precisely the problem that Queen Dido intuitively solved.
Let's consider any point on the perimeter. If we draw the segments that join this point
with the ends, we obtain two regions S_1 and S_2 and a triangle T, so that area B
is the sum of the areas of these three regions. Let's consider the angle of triangle
T at the chosen point on the perimeter. Let's see that if the angle is less than
90°, then we can construct a new curve with the same perimeter but larger area.
To do this, we take the triangle T out of the figure as if it were a puzzle piece. We can
separate the end points of the curve so that regions S_1 and S_2 do not change
until the angle they form is 90°. In this way, triangle T’ has two sides in
common with triangle T but with a right angle. We saw at the beginning that the triangle with
two fixed sides and the largest area was the right triangle, so T’ > T, and the new figure has the same
perimeter L/2 but larger area. Let's reassemble the original figure and
choose another point on its perimeter. By drawing segments from this point to the
ends, we have three regions, but in this case, the angle formed is greater than 90°.
In this case, we can also take out the triangle T and in this case, bring the ends
together to form a right angle and find a new figure with the same perimeter L/2 but
larger area since triangle T’ is right. It is clear that the original figure cannot
be the one with the largest area and fixed perimeter unless the segment between its ends
subtends a right angle from EVERY point on the perimeter. This was precisely a
characterization of the semicircle! Having ruled out all figures except the
circle, the only possibility left is the circle, and the theorem is proven... or is it? Steiner's proof shows that if there is a
solution to the isoperimetric problem, then that solution is the circle, but are we sure
that the problem has a solution? Although we intuitively see that such
a curve must exist, in mathematics, intuition cannot
substitute for rigor. Not without some cruelty, Perron, in an article published in 1913, said
that using an argument similar to Steiner's, one could prove something as absurd
as what I am about to tell you... Let's prove the following Theorem
(in quotes). The number 1 is the greatest element of the set of natural numbers.
Let's begin with the proof (in quotes). The number 1 is characterized as the
only natural number that, when squared, the result is itself.
So, given any natural number n other than 1, its square is greater than n. In this way, we have constructed a new natural number
greater than n and, therefore, discard n as the maximum of the natural numbers.
Once we have discarded all numbers other than 1 as the maximum of the set of
natural numbers, only 1 remains, and therefore, we conclude that it is
the maximum of that set. Obviously, this is not a proof, and the set of
natural numbers has no maximum. In fact, the proof is flawed
precisely because it assumes the existence of such a maximum. If the maximum
existed, once all naturals except 1 were discarded, we could indeed conclude that 1 is the maximum.
That is why it is so important to prove that a solution to the isoperimetric problem exists
to conclude that it is the circle. To rigorously prove the isoperimetric theorem,
a new field of mathematics had to be developed: the calculus of variations.
Another classic problem lies at the birth of this theory: the brachistochrone problem,
the curve of fastest descent. In 1696, Johann Bernoulli proposed a problem
as a challenge to the mathematical community. Given two points A and B in a vertical plane,
what is the curve that a particle subject only to gravity should follow to go from
A to B in the shortest possible time? The solution, which is none other than the cycloid, was
found by Bernoulli himself and by other great mathematicians, including the brilliant Newton, who is the protagonist of one of the most amusing episodes in the history of mathematics known as
"The Lion's Paw." But this would require a whole video, so let's get back to our topic.
The tools developed for solving Bernoulli's challenge gave rise to the calculus of variations. Karl
Weierstrass used this branch of mathematics in 1879 to formulate the first complete and rigorous
proof of the isoperimetric theorem. Mathematicians always like to generalize
or dualize or seek analogs of theorems. Thus, the isoperimetric theorem
has versions in higher dimensions. For example, in dimension 3, the isoperimetric
theorem states that, given a fixed area, the three-dimensional figure with that area
that encloses the largest volume is the sphere. We can also dualize the isoperimetric
theorem by swapping the roles of perimeter and area. The dualized theorem
would state that given a fixed area, the plane curve enclosing that area with
the SMALLEST perimeter is the circle. This dual result is equivalent to the
original isoperimetric theorem, as shown in this wonderful book
by Rademacher and Toeplitz, "Numbers and Shapes," which we have already mentioned in this video.
Having said this, we can do both procedures: generalize to dimension 3 and dualize. The
resulting theorem would state: Given a fixed volume, the three-dimensional
figure with the SMALLEST area is the sphere. Do you know who knows this theorem
well and applies it every winter? Our dog Emmy, who by the way, is
a lover of math books. Have they reorganized the
bookshelf again? Where have they put Sáenz de Cabezón's books? I'm in one of them!
Emmy has a fixed volume. A bit larger than it should be to be a half-greyhound according to
our vet. That can't be changed in the short term, but when she is cold,
she would like her exterior surface to have the smallest area
possible to avoid losing heat. So what she does is curl up in such a way that
she forms a shape as close to a sphere as possible. By the isoperimetric theorem, her area
will be the smallest possible, and she will be warmer. In our next video, I want to show you
a math magic trick related to isoperimetric problems, which is the best
math magic trick I've ever seen. Don't miss it!
