Largest area given a fixed perimeter (Part II)

Over the week, I have found a few more things of note to bring to the metaphorical discussion table. Firstly, there is a most welcome patch to the squabble over the proof of necessity of convexity in maximizing area, and secondly, a proposed sketch of a definition of "inside", used when defining area. However, it is still not clear how the definition can be used explicitly in the proof apart from "obvious"-style handwaving at the moment.

Firstly, the patch:

Consider the following 2 propositions:

1) A larger fixed perimeter will yield a larger maximal area.
2) A straight line is the shortest path between 2 fixed points.

For (2), there is not much to say, and is more or less a take-it-or-leave-it axiomatic definition, due in part to the triangle inequality.

For (1), consider 2 fixed perimeters P and P', with P'>P. Consider the shape with the maximal area for P, S. Define S' similarly. Clearly Area(S')>= Area (similar shape to S with perimeter P')* > Area (S), since Area (S)>0 (consider a square, for example). In fact, Area (similar shape to S with perimeter P') = Area (S) * (P'/P) ^2.

Consider a concave (used here possibly loosely to mean non-convex) shape S. Assume it has maximal area.

Then consider of it, two points on its outline A and B such that there exists a point P on line segment AB such that P is not in the shape.

Next, consider the shape S', which is S with arc AB replaced with line segment AB.

Now, consider the shape S'', which is the union of the set of points contained in S and S'.

The area of S'' is clearly greater than that of S, since there is at least one point contained in S'' not contained in S (P), and whatever is contained in S is by definition contained in S''.

3) The perimeter of S'' is also not more than that of S.

<Proof of (3):

Assume otherwise. If so, then there exists an arc CD in between such that its length is less than line CD. This contradicts (2). We are done.>

Hence, S'' has greater area but less than or equal perimeter to S, and hence a shape S''' similar to S'' but with perimeter equal to S with have greater area than S, contradicting the original assumption that S had maximal area for its perimeter.

Finally, on to part 2 of this post, the definition of "inside":

All points at infinite are outside. Consider any line. Each time it intersects the perimeter of the shape indicates a border between "inside" and "outside". The area of a shape is the sum of areas of all "inside" points.

* Similar as in similar figures, used in the same context as congrurent

Largest area given a fixed perimeter

The question is simple: given a fixed perimeter of say P, find the shape that gives the largest possible area. The answer is somewhat equally simple: a circle gives the largest area for a fixed perimeter. The proof, is not so clear. The definition is a trickier problem, which I shall not care to resolve (I define the center of the circle to be the outside of the perimeter...).

Browsing the internet yields more than just one proof and references to proofs that exist. One of which I trust to be rather unambiguously true is referenced to as the proof by calculus of variations, which I suppose works, since calculus is such an overpowered sledgehammer. However, yet another interesting proof presents itself, of which the outline is presented below:

Define X to be a shape that maximizes the area for a given P.

1. X is convex.

Proof: If X is not convex, then let C be a point outside of X, and A and B be points on the perimeter of X such that C lies on AB. Moreover, let there be no points on the outline of X, say Q, such that Q lies on line segment AB. In that case, if we reflect the outline AB about the line AB, the area increases. This contradicts the maximality of area X. Q.E.D..

Attack: What if we cannot pick A and B such that there are no points of the outline of X, Q, such that Q lies between A and B? (a.k.a. fractals)

2. Any polygon with a fixed number of sides N has maximum area when it is regular. (more or less indisputable).

Corollary: A regular polygon with i sides has greater area than one with j sides given a fixed perimeter iff i>j. (Think of degenerate polygons)

3. Therefore, as the number of sides N goes to infinity, the area increases, and we get a circle, and therefore a circle maximizes area for a fixed perimeter P.

Attack: Where did that come from? Can we even apply rule number 2 to "polygons with infinite sides" (a.k.a. shapes with curves)?

Is there a reasonable patch to the leap from part 2 to part 3? I would agree that it is intuitively very probably true that the argument *sort of* works, and just need formalisation, but with infinities I would not be very sure, ever since the last time I saw 1=1+0+0+...=0+1+0+...=0+0+1+...= ... =0+0+0+...=0. Therefore 1=0. Instead I would probably work towards a more geometry-based idea, described handwaving-ly all the way through, since I have not worked out a full "proof" yet:

Lemma 1 remains, despite the attack. Fractals can be dealt with separately, ... some other time... some other person.

2. The outline of X is smooth. (i.e. no line segments anywhere) random rantings.

Imagine the outline as a wire. Now divide the wire into 2 parts of equal length, and call the parts, with direction, AB and BA respectively (i.e. starting with A and going say clockwise to B and starting with B and going clockwise to A respectively). Define the area enclosed by arc AB and line segment BA be called ABA and BAB be defined similarly. if ABA > BAB, then copy ABA over with symmetry on like AB. X clearly does not have this property since otherwise the operation would produce a shape with the same perimeter and larger area. Hence for X, ABA=BAB for every choice of A and B. Now notice that the reflection does not change the area for X, even though it might change the shape. This does not bring us to an immediate conclusion, since X might not be a unique shape. Also, reflecting about AB must result in another convex shape. This fixes X as a circle... somehow? Good night.