A NOTE ON THE ISOPERIMETRIC INEQUALITY

Richard E. Bayne and Myung H. Kwack

In most calculus books students are introduced to an isoperimetric theorem in the following form. Among all rectangles with a given perimeter the square has the largest area [5]. However, this isoperimetric theorem for rectangles can be proved easily in an algebraic class using quadratic functions. The following theorem might be more appropriate for a calculus class. Among all quadrilaterals with a given perimeter and a given side, the trapezoid with the other sides of equal length, and of equal angles between them has the largest area. This isoperimetric theorem for quadrilaterals has a nice application, an isoperimetric theorem for n–polygons, i.e. polygons with n vertices. Among all n–polygons with a given side and a given perimeter, the n–polygon with the maximum area is inscribed in a circle and has all other sides of equal length and of equal angles between them where the existence of such an n–polygon is obtained from the general result that a continuous function on a bounded and closed subset of an Euclidean plane, ℝ^{N}, attains a maximum value.

In this note proofs of the above theorems are presented and the theorems are then utilized along with inscribed polygons to obtain a proof for the following isoperimetric theorem for simple closed curves. Let S be a closed curve formed by a circular arc of length s together with its chord of length ℓ. Then any simple closed curve Σ formed by a curve of length s together with a line segment of length ℓ satisfies the inequality A(Σ) ≤ A(S) where A(σ) denotes the area enclosed by the simple closed curve σ and the equality holds if and only if Σ coincides with S. As a corollary we obtain the isoperimetric theorem for simple closed curves [2]. Any simple closed curve Σ with length s satisfies the inequality 4πA(Σ) ≤ s^{2}, with equality if and only if Σ is a circle.

In his paper [4] “The Isoperimetric Inequality,” Professor Osserman obtains the following inequality for any n–polygon Σ with perimeter s.

\frac{s^{2}}{A(Σ)} ≥ \frac{4}{n} tan \frac{π}{n} > 4π.

He states that one can use inscribed polygons, apply the above inequality and pass to the limit to deduce s^{2} ≥ 4πA(σ) for a simple closed curve σ with length s. “The shortcoming of this proof,” he continues, “is that it does not allow a characterization of the case in which equality holds.” In this paper we show that this shortcoming may be overcome by initially proving the isoperimetric inequality for simple closed curves with a fixed line segment. While we do not know whether the proof of the isoperimetric inequality is new, the methods used should be of interest to teachers and students in a first calculus course. All curves mentioned in this note are rectifiable plane curves.

We begin with an isoperimetric theorem for triangles which may be proved from the study of ellipses in an analytic geometry class. If P_{1}, P_{2}, P_{3} are three different points in ℝ^{2}, then ^{}P̸_{1}P_{2}P_{3} denotes a number in [−π, π] which measures the angle from the ray \overrightarrow{P_{2}P_{1}} to the ray \overrightarrow{P_{2}P_{3}} (a positive number if measured counterclockwise) and A(△P_{1}P_{2}P_{3}) denotes the area of the triangle △P_{1}P_{2}P_{3} if P_{1}, P_{2}, P_{3} are non– colinear and 0 otherwise. The notation |P_{1}P_{2}| denotes the length of P_{1}P_{2}, the line segment from P_{1} to P_{2}.

\underline{Proposition 1}. Among all triangles with a given side and a given perimeter the isosceles triangle has the largest area.

\underline{Proof}. Instead of an algebraic proof, we present a proof where the concept of derivative is used. Any triangle △ABC (see Diagram 1) with a given side BC of length a and a given perimeter a + s is determined by the angle ^{}C̸BA = θ and has an area

f(θ) = \frac{1}{2}ac sin θ

where |AB| = c, |AC| = b and c + b = s. From the identity (s − c)^{2} = c^{2} + a^{2} − 2ac cos θ, we get

c = \frac{s^{2} − a^{2}}{2(s − a cos θ)}.

Diagram 1.

Substituting and differentiating we get

f′(θ) = \frac{1}{4}a(s^{2} − a^{2})\frac{s cos θ − a}{(s − a cos θ)^{2}} .

Thus, f has the maximum value when s cos θ = a, i.e. when ^{}C̸BA = ^{}A̸CB.

\underline{Proposition 2}. Among the quadrilaterals ABCD with a given side AD and a given perimeter, the quadrilateral with the maximum area is the trapezoid satisfying the following (see Diagram 2):

(i) |AB| = |BC| = |CD|. (ii) ^{}D̸AB = 2^{}C̸AB = ^{}C̸DA = 2^{}D̸BC and ^{}A̸BC = ^{}B̸CD. (iii) The quadrilateral ABCD is symmetric with respect to the perpendicular bi-

sector of AD. (iv) There is a point O such that |OA| = |OB| = |OC| = |OD|.

Diagram 2.

\underline{Proof}. Since the quadrilateral with the largest area is convex, (i) follows from Proposition 1. Any convex quadrilateral ABCD such that |AD| = ℓ and |AB| = |BC| = |CD| = a with a and ℓ fixed is determined by the angle ^{}D̸AB = θ (see Diagram 2) and has an area

f(θ) = \frac{1}{2}(ℓa sin θ + 2a^{2} sin α cos α) = \frac{a}{2}(ℓ sin θ + a sin 2α)

where α = ^{}D̸BC. So f′(θ) = 0 when

\frac{dα}{dθ} = −\frac{ℓ cos θ}{2a cos 2α}.

Differentiating the identity (2a cos α)^{2} = a^{2} + ℓ^{2} − 2aℓ cos θ, we also get

\frac{dα}{dθ} = −\frac{ℓ sin θ}{2a sin 2α}.

It follows that f has the maximum value when

\frac{ℓ cos θ}{2a cos 2α} = \frac{ℓ sin θ}{2a sin 2α}, i.e. tan θ = tan 2α.

Consequently the quadrilateral ABCD has the maximum area when θ = 2α. Now suppose θ = 2α. Since a sin 2α = 2a cos α sin γ where γ = ^{}B̸DA, we get

sin α = sin γ and α = γ.

It follows that

AD∥BC and ^{}A̸BC = ^{}B̸CD = π − 2α.

The statements (ii) and (iii) now follow from equivalence of triangles. Finally the intersection point O of the bisectors of the angles ^{}A̸BC and ^{}B̸CD satisfies the conditions of (iv).

Now we will consider the case of n–polygons. First we show the existence of a polygon with the maximum area among the set of n–polygons with a given perimeter and a given side.

\underline{Proposition 3}. Among all n–polygons with a given perimeter s and a given side of length ℓ there is a polygon with the maximum area.

\underline{Proof}. Let Γ be the closed and bounded set of points (a_{11}, …, a_{1n}, …, a_{n1}, …, a_{nn}) in ℝ^{n^{2}} satisfying the following conditions.

1. A_{1} = (0, …, 0), A_{n} = (ℓ, 0, …, 0),

2. ∑_{i=1}^{n−1} |A_{i}A_{i+1}| + |A_{n}A_{1}| = s, where A_{i} = (a_{i1}, …, a_{in}) for i = 1, …, n,

3. 0 ≤ ^{}A̸_{i}A_{1}A_{i−1} for i = 3, …, n, and ∑_{i=3}^{n} ^{}A̸_{i}A_{1}A_{i−1} ≤ π.

The function f: Γ → ℝ given by

f(a_{11}, …, a_{1n}, …, a_{n1}, …, a_{nn}) = ∑_{i=2}^{n−1} A(△A_{i}A_{1}A_{i+1})

is continuous, and so f attains the maximum value in Γ. Since a non–convex polygon cannot have the maximum area, the proof is complete.

Proposition 2, Proposition 3, mathematical induction and equivalence of triangles lead to the following theorem.

\underline{Theorem 4}. Among the n–polygons with a given side and a given perimeter the n–polygon R_{n} which has the maximum area satisfies the following conditions.

(i) All sides with the possible exception of the given are of equal length.

(ii) All angles not touching the given side are of equal measure and so are the

angles touching the given side. (iii) The polygon R_{n} is symmetric with respect to the perpendicular bisector of the

given side. (iv) The polygon R_{n} is inscribed in a circle.

In particular, among all n–polygons with a given perimeter the regular n– polygon has the maximum area.

\underline{Proof}. Let R_{n} = A_{1}A_{2} ⋯ A_{n−1}A_{n}A_{1} be the n–polygon with the given segment A_{1}A_{n} and with the maximum area (see Diagram 3). From Proposition 2, we conclude

\begin{matrix}|A_{1}A_{2}| & = |A_{2}A_{3}| = ⋯ = |A_{n−1}A_{n}| and \\ ^{}A̸_{1}A_{2}A_{3} = & ^{}A̸_{2}A_{3}A_{4} = ⋯ = ^{}A̸_{n−2}A_{n−1}A_{n} = π − 2α \end{matrix}

where α = ^{}A̸_{3}A_{1}A_{2}. From Proposition 2 and equivalence of triangles we get

α = ^{}A̸_{i+1}A_{1}A_{i} = ^{}A̸_{j+1}A_{n}A_{j} for i = 2, …, n − 1, j = 1, …, n − 2.

Diagram 3.

Therefore,

A̸_{n}A_{1}A_{2} = ^{}A̸_{n−1}A_{n}A_{1} = (n − 2)α

from which follow (ii) and (iii).

Let O be the intersection point of the bisectors of the angles ^{}A̸_{1}A_{2}A_{3} and A̸_{2}A_{3}A_{4}. Then again from equivalence of triangles (iv) follows, i.e. all the vertices of the polygon R_{n} lie on the circle with center O and the radius equal to |OA_{1}|.

As an application of the theorem above, we present an isoperimetric theorem for simple closed curves containing a given segment and having a given length.

\underline{Theorem 5}. Let Σ be a simple closed curve formed by a curve of length s together with a line segment of length ℓ (s > ℓ). Let 0 < θ < π be defined by

\frac{ℓ}{s} = \frac{sin(π − θ)}{π − θ}.

Then

A(Σ) ≤ A(S) = \frac{s}{4(π − θ)}(s + ℓ cos θ)

with equality if and only if Σ coincides with the simple closed curve S formed by a circular arc of length s and its chord of length ℓ.

\underline{Remark}. The function

g(x) = \frac{sin x}{x}

is strictly decreasing on the interval (0, π) and thus θ in Theorem 5 is uniquely determined. See [1] for a simple proof which avoids the use of a formula of an area of a sector of a circle for

lim_{x→0} \frac{sin x}{x} = 1

which is used in the proof below.

\underline{Proof}. Let P Q be the given segment. For each positive integer n ≥ 4, let P_{n} be an n–polygon with perimeter s_{n} + ℓ with vertices in Σ which include P, Q and which divide Σ into n pieces of equal length. By Theorem 4, A(P_{n}) ≤ A(R_{n}) where R_{n} = A_{1}A_{2} ⋯ A_{n}A_{1} is the n–polygons with perimeter s_{n} + ℓ such that

(i) R_{n} is inscribed in a circle with center O and radius r_{n} and (ii) |A_{1}A_{n}| = |P Q| and |A_{1}A_{2}| = |A_{2}A_{3}| = ⋯ = |A_{n−1}A_{n−2}|.

Diagram 4.

Let γ_{n} = (1/2)^{}A̸_{2}OA_{1} and θ_{n} = π − (n − 1)γ_{n} (see Diagram 4). Then since

r_{n} = \frac{ℓ}{2 sin θ_{n}} = \frac{s_{n}}{2(n − 1) sin γ_{n}}

we have

\frac{ℓ}{s_{n}} = \frac{sin θ_{n}}{(n − 1) sin(\frac{π−θ_{n}}{n−1} )}.

Taking limits as n → ∞ we get

\frac{ℓ}{s} = lim_{n→∞} \frac{sin(π − θ_{n})}{π − θ_{n}}

and thus,

lim_{n→∞} θ_{n} = θ.

The area A(R_{n}) is given by

A(R_{n}) = (n − 1)r_{n}^{2} cos γ_{n} sin γ_{n} + r_{n}^{2} cos θ_{n} sin θ_{n}

= \frac{r_{n}}{2} (s_{n} cos γ_{n} + ℓ cos θ_{n})

= \frac{r_{n}}{2} (s_{n} cos \frac{π − θ_{n}}{n − 1} + ℓ cos θ_{n}).

Taking limits we arrive at the inequality.

Diagram 5.

To prove the last assertion suppose Σ is convex and A(Σ) = A(S). Let S = S_{0} ∪ P Q where S_{0} is the circular part of S with the end points P, Q. Let S_{1} be the circular arc with the end points P, Q such that Ŝ = S_{1} ∪ S_{0} is a circle with circumference ŝ. Let Σ̂ = (Σ − P Q) ∪ S_{1} be the simple closed curve enclosing P Q (see Diagram 5). Since A(Σ) = A(S) we have A(Σ̂) = A(Ŝ). It suffices to show that Σ̂ and Ŝ coincide. Let A, B be points on Σ̂ which divide Σ̂ into two arcs of equal length. We may assume that AB divides the region enclosed by the curve Σ̂ into two regions of equal area. We claim |AB| = ŝ/π. To prove this let Σ_{1} be the simple closed curve which is the union of the circular arc of length ŝ/2 with the end points A, B and its reflection about AB. If |AB| < ŝ/π, the curve Σ_{1} is not convex (see Diagram 6) and thus, by the first part of Theorem 5,

A(Σ̂) ≤ A(Σ_{1}) < A(Ŝ) = \frac{ŝ^{2}}{4π} ,

a contradiction. If |AB| = ŝ/π then the intersection points D, E of the bisector of AB with Σ_{1} divide Σ_{1} into two arcs of equal length (see Diagram 7) and |DE| < ŝ/π and as before we get a contradiction.

Diagram 6. Diagram 7.

Diagram 8.

Finally suppose |AB| = ŝ/π and Σ̂ is not a circle. Let M be the midpoint of AB. There is a point N on Σ̂ such that |MN| < ŝ/2π. Let Σ_{2} be the union of the arc ANB of Σ̂ and its reflection about M (see Diagram 8). Then N and its reflection point N′ about M divide Σ_{2} into two arcs of equal length and |NN′| < ŝ/π. As before then A(Σ̂) ≤ A(Σ_{2}) < A(Ŝ), a contradiction and so Σ̂ must be a circle and thus, Σ = S.

The following isoperimetric theorem for simple closed curves which has been given many proofs [2,4] follows from Theorem 5 with ℓ = 0, i.e. when the given segment is a single point.

\underline{Corollary 6}. If Σ is a simple closed curve with length s, then 4πA(Σ) ≤ s^{2} with equality if and only if Σ is the circle with circumference s.

Among all simple closed curves which are formed by curves of length s together with line segments joining its end points, the one that bounds the maximum area must be formed by, from Theorem 5, a circular arc. It is known that among the circular arcs of a fixed length the smaller the curvature (i.e. the larger the diameter), the longer the chord [3]. The following corollary states that the diameter of the one which bounds the maximum area coincides with the length of its chord and can be proved by considering the closed curve formed with its mirror image.

\underline{Corollary 7}. Let Σ be a simple closed curve formed by a circular arc of length s together with its chord. Then 2πA(Σ) ≤ s^{2} with equality if and only if Σ is the curve formed by the semicircle of length s with its chord.

\underline{References}

1. R. Bayne, J. Joseph, and M. Kwack, “On the Length of a Circular Arc,” Missouri Journal of Mathematical Sciences, 11 (1999), 84–86.

2. S. S. Chern, Curves and Surfaces in Euclidean Space, Global Differential Geometry, Studies in Mathematics, Vol. 27, 99–139, Mathematical Association of America, 1989.

3. M. P. Do Carmo, Differential Geometry of Curves and Surfaces, Prentice–Hall, Englewood Cliffs, NJ, 1976.

4. R. Osserman, “The Isoperimetric Inequality,” Bulletin of the American Mathematical Society, Vol. 84, No. 6, 1978.

5. M. Spivak, Calculus, 2nd ed. Publish or Perish, Inc., Houston, Texas, 1980.

Richard E. Bayne Department of Mathematics Howard University Washington, D.C. 20059 email: bayne@scs.howard.edu

Myung H. Kwack Department of Mathematics Howard University Washington, D.C. 20059 email: mkwack@fac.howard.edu