Isoperimetric Theorem and InequalityIsoperimetric Theorem has been known from the time of antiquity. Elegant all by itself, the Theorem also has a fascinating history. The most essential contribution towards its rigorous proof was given in 1841 and is due to Jacob Steiner (1796-1863). At the time it stood at the center of controversy between adherents of analytic (i.e. using Calculus) and synthetic (pure Geometric) methods. Although accepting validity of the analytic methods, Steiner only used synthetic approach. His proof contained a flaw that later was fixed by analytic approach. To better understand the nature of the discussion let me first formulate two equivalent statements. Statement 1Among all planar shapes with the same perimeter the circle has the largest area. Statement 2Among all planar shapes with the same area the circle has the shortest perimeter. Theorem
ProofAs a shorthand, denote Statement 1 as A and Statement 2 as B. Assume A holds and let us prove B. Suppose, on the contrary, that B is false. Then for a given circle C there exists a figure F with the same area but with a perimeter shorter than that of C. Shrink C into a smaller circle C' whose perimeter equals that of F. The area of C' will clearly be smaller than that of C and, consequently, it will be smaller than the area of F. Now, this contradicts our assumption that A holds: C' and F have the same perimeter but the area of the circle C' is less than the area of F. Thus A=>B. The implication B=>A is proven in precisely the same fashion. Q.E.D.  Note that we did not prove either A or B but rather that they are either both true or both false. Mathematical theorems in general have premises and conclusions and assert that the latter follow from the former. Human reasoning often resorts to the well known modus ponens syllogism: if A=>B and A are true then so is B. Oftentimes, the premise A seems to be so obviously true that a proof of A=>B is mistakenly substituted for a proof of B. See if you can detect an inconsistency in the following Proof of Statement 1
The proof will proceed in several steps. First of all, let us note that the figure that solves
Statement 1 (we hope it's a circle) must be convex. Indeed, assume it's not (see Fig 1).
On the second step we proof the following Lemma 1
Proof of Lemma 1Assume, on the contrary, the area S1T is larger than T2S (Fig 3). Then reflecting S1T in ST will get the shape S1T3 with the same perimeter as S1T2 but a larger area.
From this proof it appears that it's sufficient to solve the following "half" of the problem: Among all arcs with a given length and endpoints on a line ST find the one that along with the line ST encloses the largest area. This question was previously posed elsewhere on this site. Lemma 2
Proof of Lemma 2
Suffice it to show that every angle inscribed into the arc is right. If it's not, slide points
S until the angle becomes right. Let pieces of the arc move along. As an exercise prove that
among all triangles with two given sides the one whose sides form a right angle has the largest
area. Elegant as it is this proof of Statement 1 contains a flaw. On each of the three steps we assumed that the shape answering conclusions of the steps existed and the Lemmas have only been proved under this assumption. Ultimately, we assumed that there exists a figure having a maximum area among all the shapes with the same perimeter. Under this assumption we proved that such a shape is bound to be a circle. Denote the existence hypothesis as H. What we have actually shown is an implication H=>A. In order to prove A we still have to demonstrate that H holds true. Existence of the optimal shape in the sense of Statement 1 is not at all obvious. For example, if Statement 1 required us to determine a shape with the smallest area for a given perimeter, such shape would not exist at all. Once we understood this point it's less important to actually complete the proof. H is proven with a limiting procedure which is quite simple but requires some basic elements of Calculus. Both Statements 1 and 2 can be expressed in a more algebraic form which indeed underlines the fact that they are equivalent. Denote the perimeter and area of a planar shape L and A, respectively. Then
which is known as the Isoperimetric Inequality. The equality only holds for a circle. Fixing L we get Statement 1. Fixing A Statement 2. Isoperimetric Inequality can be generalized in several ways. For one, it extends into higher dimensional spaces. For example, if S is a surface area while V a volume of a 3d body then
The inequality translates into two equivalent statements of which one claims that among all three dimensional solid bodies with a given surface area the sphere has the largest volume. What is the other one? Another generalization pertains to the Geometric Theory of Fuzzy Sets. Regular sets for which it's assumed to be known whether a point belongs to a set or not are called crisp among the fuzzy sets. For a fuzzy set a point may belong to a set with a certain degree that changes from 0 (does not belong at all) to 1 (belongs unequivocally). The theory defines fuzzy circles and fuzzy spheres. In Reference [1] I have shown that the general Isoperimetric Inequality holds for fuzzy sets. The equality is only reached for crisp circles, spheres and their multidimensional analogues. Pappus of Alexandria (ca 300 A.D.) wrote
In a sense it is very satisfying to have the circle as a solution to Statement 1. The circle is unique in its roundness. It's symmetric in every straight line passing through its center. The circle is perfect and, thus, is a good candidate to satisfy the Isoperimetric Theorem. Polygons are less smooth than a circle, less perfect one can say. Still in every family of polygons there is one which is less imperfect than the rest. Interestingly, in their respective families these "less imperfect shapes" satisfy isoperimetric conditions.
References
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This would imply
existence of two points from the figure such that the connecting segment would not itself lie inside
our shape. Then reflecting a region between the segment and the boundary of the shape
as on Fig. 2, it would be possible to increase its area without modifying its perimeter.
Since the area of the two red regions
did not change but the area of the triangle (Fig 5) grew, the whole area between the new arc and the
line ST has increased. This shows that unless the curve is a semicircle we can always increase
the area in question by moving points S and T. This proves Lemma 2 and with it Statement 1.| 37225077 |
