On Short Cuts - or - Fencing in Rectangular Strips

The isoperimetric inequality for shapes in R 2 states that the area enclosed by a simple closed curve is at most that of a circle of the same length, and that equality occurs only for circles. This immediately implies that among all simple closed curves enclosing a given area, a circle is the shortest.

Several variations on the isoperimetric inequality were considered in the literature (see e.g. [7]). In this paper we shall discuss inequalities involving the notion of “free perimeter " for a shape S, located inside a simple, bounded, planar domain D. We may assume that there is a border or wall surrounding this domain, or alternatively that this domain is an island surrounded by water. A simple shape S inside this domain will be defined by a boundary curve, some portion of which may touch and even follow the border (wall / shoreline) of the domain / island. The free perimeter of the shape will be defined as the length of the boundary curve of S that does not overlap with, or trace, the border of the enclosing domain D.

The problem that we can pose with these definitions is the following : given the domain D, determine the shape with the shortest free-perimeter that has a given area A. This problem is, of course, that of determining the way to cut out a shape of a total area A from D with the least effort of cutting, i.e. with the shortest cut. Equivalently, this is the problem of determining the shortest length “fence” that can separate a contiguous region of area A inside the domain D.

This interpretation clearly explains the totally misleading title of our paper, in which we do not take any short-cuts and of course we do not discuss fencing as a sport that happens to be played on a rectangular, strip-shaped, “ring”.

A related problem is that of finding the connected shape of largest area that can be “lifted out” of D with a total length of “cuts” or “fences” less than or equal to L.

In this paper we solve the problem raised above when the region D is a rectangle. We prove that the shortest cut, i.e. the minimum free perimeter, that separates a shape with half of the area of D has, as expected, the length of the shorter side of the rectangle. We then provide the shortest free perimeter for all Area(S) Area(D) ratios from 0 to 1.

We note that the problem we discuss is closely related to the problem A26, “Dividing up a piece of land by a short fence”, discussed in the book “Unsolved Problems in Geometry” [4]. The challenge posed there is that of dividing a convex shape into two equal-area parts. We refer the interested reader to [4] and to some recent follow-up papers [5,6].

Fig. 1. An illustration of the notion of “free perimeter “. The area of the shape S equals A, while its free perimeter equals l1 + l2.

Let D[X, Y ] be a bounding rectangle of dimensions X and Y , with X ≤ Y . Let A be the area we want to enclose with a region of shape S, and denote by l F P (S) the length of the free perimeter of the shape S. Let us denote by l (A) the length of the free perimeter of a shape S with area A, such that S has the smallest value of l F P (S) out of all the shapes of area A. Namely :

We shall be interested in determining the value of l (A) for A ∈ [0, XY ]. For this, we shall first prove the following result :

Proof. To prove the above stated, and rather natural and hardly surprising result, we shall need to combine several simple facts.

Given any shape of area A in the plane, and perimeter of length l we have :

with equality achieved for a circle.

Given any shape S of area A in a half plain domain, with free perimeter of l F P (S) we have :

Proof. If S touches the boundary of the half-plane, let us reflect it along the boundary line, thereby generating a (symmetric) shape of area 2A in the plane. For this “double shape” S we have :

and with the classical isoperimetric inequality of Fact 1 we obtain :

hence :

Fact 3 The Quarter-Plane Isoperimetric Inequality Given any shape S of area A in a quarter plain domain, with free perimeter of l F P (S) we have :

Proof. If S touches the two orthogonal boundaries of the quarter-plane, let us reflect it symmetrically into the three quarters plane domain boundary, generating a shape S in the plane, of area 4A. For S we have :

and with the classical isoperimetric inequality of Fact 1 we obtain :

yielding :

] may touch the sides of the boundary of the rectangle D(X, Y ) in several ways. We may have S that touches 0,1,2,3 or 4 sides. Let us consider these cases separately : Case 0 : S touches 0 sides of D[X, Y ]. In this case, the classical isoperimetric inequality of Fact 1 yields :

In this case, Fact 2 yields :

In this case we have either S touches two opposite sides, yielding l F P (S) ≥ 2 min{X, Y } ≥ 2X, or S touches two adjacent sides, in which case Fact 3 provides :

We also have that :

Notice that for all i, S C i cannot touch more than 2 sides of the rectangle D[X, Y ], since this would imply that S is disconnected.

By Facts 1,2 and 3 we the

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