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Proving that regular shapes have maximum area proportional to the circumference
By Hyukjin Yoon on Tuesday, December 11, 2001 - 06:36 pm:

How can you prove that regular shapes have the maximum area proportional to the circumference for all shapes?
A proof applicable to all n-gons will be helpful.

By Dan Goodman on Wednesday, December 12, 2001 - 01:27 am:

I have a sketch argument which I think proves it, but needs to be fleshed out a bit. The proof assumes that if you have a triangle with one side of length L fixed, and circumference C fixed, and the other sides length M and N then the area A is maximised if you have that M=N. I'm pretty sure this is true, but haven't checked it.

Right, now suppose you had an n-gon which had 2 sides joined with different lengths. If we can prove that we can find another polygon with larger area and same circumference when this happens then we're done, because the only time this won't occur is when all the sides have the same length, i.e. it is regular.

If we had such a polygon, let's call the vertices X, Y and Z so that XY and YZ are two adjacent sides of different length. We also require that the polygon is convex (i.e. any line between two points inside the polygon are contained in the polygon). There is an easy argument to prove that this is true which I won't write out unless you want me to go into more detail. So, the triangle with vertices X,Y,Z is contained in the polygon at the very least and there are no vertices of the polygon in the interior of the triangle. Moreover, the rest of the polygon (not including the bit in the triangle) is entirely on the other side of the line between X and Z (not the side that the triangle is in), because otherwise it wouldn't be convex. Again, I can make this bit more explicit if you like? Now we can make the area of this triangle larger by moving Y so that XY and YZ are the same length (and also so that XY+YZ stays the same). We can be sure that this will be OK because the rest of the polygon is kept safely away from the area in which we are moving things (that's why we need the convexity bit). However, this increases the area of the entire polygon and keeps the circumference the same. Tada!

By Dan Goodman on Wednesday, December 12, 2001 - 01:30 am:

Hmm, reading over that I realise it was very sketchy. I can write more and even draw a picture if you'd like? Let me know...

By Michael Doré on Wednesday, December 12, 2001 - 01:55 am:

I think this is a nice way of showing that if there exists a shape with maximal area (for fixed perimeter) then this shape must have all its sides are equal. However it is not entirely clear to me that the set of possible areas (with fixed perimeter) attains its sup, and even if we can show this we have to also show all the polygon's angles are the same (which is not guarenteed by having equal lengths).

By the way, the initial statement about the triangle is true. If a triangle has side lengths a,b,c then by Heron's formula its area A is given by:

A = sqrt(s(s-a)(s-b)(s-c))

where s = (a + b + c)/2. So if s and a are fixed then using the AM-GM inequality:

A = sqrt(s(s-a))sqrt(s-b)(s-c)) £ sqrt(s(s-a))((s-b) + (s-c))/2

with equality iff s - b = s - c, i.e. b = c.

So A £ sqrt(s(s-a))*a/2 with equality iff b = c = s - a/2.

By Hyukjin Yoon on Wednesday, December 12, 2001 - 07:26 am:

A drawing would be very helpful if you do, because i do have some difficulties understanding this.
Thanks very much for your help so far.

By Dan Goodman on Wednesday, December 12, 2001 - 07:21 pm:

OK, here's a picture or three:

Polygon diagrams

So, the first diagram should illustrate what's going on in my first message. The rest of the polygon has to be contained in the blue area, the green triangle has to be part of the polygon. The area of the green triangle is maximised (for x+z constant) when x=z. The second diagram I'll explain below. The third diagram shows why the polygon would have to be convex. If it weren't, you could increase the area by getting rid of one of the nonconvex bits. Obviously, you'd also decrease the total length by doing this, but that's even better. Now just scale the whole shape up a bit so that the total length is the same, this will further increase the area.

The second diagram is an attempt to deal with the problem Michael rightly pointed out, that you need all the angles to be the same too. Again, if we could prove that the area of the quadrilateral ABDC (where all the of the lengths AB, BD and CA are the same but DC need not be the same) with points C and D fixed is maximised when the angles a and b are the same, then we're done. Because if the polygon of maximum area had two different angles, we'd be in the situation in the second diagram and we could increase the area by making the angles the same. I imagine there is a similar way of showing that this is true like the method Michael described above for the triangle.

By Kerwin Hui on Wednesday, December 12, 2001 - 11:08 pm:

Indeed there is a way to show all angles are the same when the area is maximised. It relies on the fact that given 4 sides of a quadrangle, the maximum area occurs when the quadrangle is cyclic. I will leave the proof as an exercise to Hyukjin. (It requires no more than 4 lines of algebra)

Let A,B,C,D be consecutive vertices of our n-gon. (n is at least 4, for a triangle with all sides equal implies all angle equals.) Now we know AB=BC=CD, and AD is fixed. The maximum area occurs if and only if ABCD is cyclic. It follows that when the area is maximised, angle ABD=angle ACD and angle CBD=angle BCA. Hence all angles of our n-gon are equal.

Kerwin

By Hyukjin Yoon on Thursday, December 13, 2001 - 04:34 pm:

Thanks for your help. I am very thankful.