Isoperimetric inequality 1. (1 Viewer)

[seanieg89]

NB: This is a polygonal version of the isoperimetric inequality. The general isoperimetric inequality includes shapes that have curved boundary. I will post a question on this later, but the polygonal version is much easier.

 

[anomalousdecay]

My attempt (I'm a nub at this type of stuff so I may as well give it a go).





















I don't know what to do next that is valid in maths. I don't know how to find x or r.

 

[seanieg89]

My attempt (I'm a nub at this type of stuff so I may as well give it a go).





















I don't know what to do next that is valid in maths. I don't know how to find x or r.

We cannot assume that the polygon is regular or cyclic unfortunately. "Cutting" into triangles is a promising approach though.

 

[anomalousdecay]

We cannot assume that the polygon is regular or cyclic unfortunately. "Cutting" into triangles is a promising approach though.

The regular bit is what stuffed me up I guess.

The only thing I know of is that any polygon's area can be found by cutting them into triangles.

I'll give it a little think about and do some trial and error. Maybe I can get something.

 

[RealiseNothing]

First I will make 2 assumptions:

1) A polygon of perimeter 1 will have it's area maximised when it is a convex polygon.

2) Thus every convex polygon can easily be fit inside a circle. Let's assume that any convex polygon of perimeter 1 fits inside a circle of perimeter 1.

Thus the area of any polygon with perimeter 1 (convex or non-convex) is less than the area of a circle with perimeter 1. Now:







Hence any polygon with perimeter 1 has an area less than and we can make a polygon with area arbitrarily close to this value by making the polygon approach the shape of the circle (i.e. take a cyclic n-gon and let ).

Now I'll have a think about how to prove my assumptions.

 

[RealiseNothing]

Ok my first assumption is easy to prove.

Call a connection of lines a ~confused polygon~ if it is not closed (basically a polygon with a side missing) but still has the potential to be either a convex or non-convex polygon by adding more lines.

Now let's say so far this confused polygon is long and thus we have left to make it either convex or non-convex. Also say that the distance between the end points of this confused polygon is less than .

The shortest distance between two points is their straight line distance. Hence we could connect the two end points and form a convex polygon. We could also add another point theoretically inside this convex polygon, and connect the two end points to this inner point instead. This would form our non-convex polygon.

Now since the shortest distance between two points is their straight line distance, this non-convex polygon has less perimeter left to be used than our convex polygon. However it also has a smaller area, as the non-convex polygon is essentially the convex polygon with a segment (in this case a triangle) cut out of it -> less area.

If this occurred an arbitrary amount of times, it proves that we can construct a convex polygon that has it's area larger than some non-convex polygon of atleast the same perimeter #

 

[RealiseNothing]

We cannot assume that the polygon is regular or cyclic unfortunately.

Hmm I think I may need to prove this in my solution too, to be able to link the 2 assumptions. If this is proven, case closed.

First thoughts is to use the approach:

Let the side lengths of a convex n-gon be such that:



It would suffice to show that where equality holds if which says that the area is maximised if the convex polygon is cyclic and regular.

 

[RealiseNothing]

Ok let's try an induction.

For a triangle with perimeter we have to prove that it's area is maximum when it is an equilateral triangle. Using the AM-GM inequality on the variable part of Heron's formula gives:



Equality holds when and thus the area is maximised when the triangle is equilateral. Thus true for our base case of a 3-gon.

Now assume the maximum area of a k-gon is when it is regular/cyclic/etc.

Editing: Hmm trying to think how to go about this.

 

[RealiseNothing]

Actually, how about this:

Consider a convex polygon with 'n' sides, with perimeter 1. Now suppose we chose three adjacent vertices A, B, and C respectively. Hence if we constructed a triangle out of these three vertices, we could maximise the area of this triangle whilst maintaining the same perimeter. Furthermore the area of the rest of the polygon remains unaffected.

In this case, we must keep the length of constant to maintain the perimeter of the shape. Now we prove that the area of triangle ABC is maximised when .

Since is kept constant, we can consider an ellipse where A and C are the foci, and B a variable point lying on the ellipse (consider ).

Now the length (where 'a' and 'e' assume their usual role in the ellipse). So all we need is to maximise the perpendicular distance from the line AC to the variable point B. This occurs obviously when B lies on the intersection of the perpendicular bisector of AC and the ellipse. Now it is obvious that

Hence we maximise the area of the triangle ABC when the two sides that are common with our n-gon are equal. Rinse and repeat for every set of adjacent sides and QED.

 

[seanieg89]

One of my first ideas was similar to your most recent post Realise, the logic isn't quite right as is though. Think about your last line. If the idea is to get an arbitrary polygon and show that we can make its area bigger by step by step making adjacent sides equal then you are missing the fact that each step you do will modify one of the early sides. (So we will not be able to use this step-by-step construction to compare an arbitrary n-gons area to the regular n-gons area).

Eg/ consider a quadrilateral with sides 2,3,4,4. We can apply your triangle result to move one vertex of this quadrilateral along an ellipse and obtain another quadrilateral with side lengths (2.5,2.5,4,4) with the same perimeter, but slightly larger area. Then doing it to the next pair of vertices, we get: (2.5,3.25,3.25,4). Then (2.5,3.25,3.625,3.625). Etc. We will get a sort of "convergence" towards being a regular polygon, but the triangle result by itself does not prove that the maximal area polygon of a given perimeter is regular.

That said, the idea is definitely of value and you could probably fashion a proof out of it

 

[seanieg89]

Had another look at this problem. There are some things that are a little tricky to rigorously prove, but here is the outline.

0. Use a compactness argument to show that there must exist a maxima of A/L^2 over the space of n-gons.

1. Show that the max of A/L^2 over all n-gons increases as n gets bigger.

2. Show that a non-convex polygon cannot have an optimal A/L^2, (by showing that it's convex hull is a polygon of fewer sides that has greater area and smaller perimeter).

3. Show that a convex polygon must had all sides equal. An argument similar to Realise's attempt works here. If not all sides are equal there must exist a pair of adjacent and unequal sides, which we can then peturb slightly along the ellipse (only slightly, otherwise we would have to prove we don't get self intersection) so as to increase the area of our poly while keeping perim fixed.

4. Use a similar argument to 3 but using quadrilaterals rather than triangles to show that if all internal angles are not equal then out polygon is non-optimal.

5. Conclude that the unique optimal n-gon up to scaling is the regular n-gon.

 

[cineti970128]

Haha... y all these very complicated ideas

I think I'm putting forward a very stupid idea but pls don't flame me
Maximum area of n polygon made possible with a fixed perimeter is a circle- n approaching infinity(Assumption which can be logically deduced)
Then this infinitely sided polygon ie circle with a perimeter 1 has radius 1/2pi
Putting this into the area formula gives you1/4pi.
Hence since the maximum possible polygon has this area the rest n sided polygon must have an area less than this regardless whether they are cyclic or regular or whatsoever.

I think this method is easier.
Pls correct me if I'm wrong

 

[seanieg89]

Maximum area of n polygon made possible with a fixed perimeter is a circle- n approaching infinity(Assumption which can be logically deduced)

1. A circle is not a polygon, it doesn't make sense to talk about an infinitely sided polygon.

2. How can you "logically deduce" that the circle of circumference 1 is the unique curve that maximises A ? This is in fact a harder problem than the one you are trying to prove!

 

[cineti970128]

But you see, the area of a polygon with a fixed perimeter gets maximised as the number of sides increases which gets restricted as a circle with the 1 unit perimeter. A circle is not a polygon yes but it's like the restricting space. The maximum space possible by a fixed perimeter.
Ok imagine a1 unit long string. Form any shape with pins to shape it. How can you maximise an area formed with this! The polygon approaching an unit CIRCLE!

It's maximum because it's so obvious. Maybe there is a mathematically way to show this. BUT our intuition tells us that it works sorta
Sorry for this messy explanation.

 

[Carrotsticks]

But you see, the area of a polygon with a fixed perimeter gets maximised as the number of sides increases which gets restricted as a circle with the 1 unit perimeter. A circle is not a polygon yes but it's like the restricting space. The maximum space possible by a fixed perimeter.
Ok imagine a1 unit long string. Form any shape with pins to shape it. How can you maximise an area formed with this! The polygon approaching an unit CIRCLE!

It's maximum because it's so obvious. Maybe there is a mathematically way to show this. BUT our intuition tells us that it works sorta
Sorry for this messy explanation.

- You are stating facts, but the idea here is to PROVE them, as 'intuitive' as they may be. We cannot always rely on our intuition (case in point Harmonic Series)

- The bolded part is in essence the isoperimetric inequality. You just reworded it.

 

[seanieg89]

It's maximum because it's so obvious. Maybe there is a mathematically way to show this. BUT our intuition tells us that it works sorta
Sorry for this messy explanation.

I agree that it is quite an intuitive fact that the circle is optimal in the isoperimetric sense, but the point is to find a mathematical way to PROVE this. The whole point of proof is that it gives us absolute certainty. I could easily list about 50 counterintuitive facts in mathematics.

When things are not so "clear cut", what happens if two people's intuition say something different? A question is only resolved with proof, until then it is nothing but a conjecture (guess).

 

[cineti970128]

I agree that it is quite an intuitive fact that the circle is optimal in the isoperimetric sense, but the point is to find a mathematical way to PROVE this. The whole point of proof is that it gives us absolute certainty. I could easily list about 50 counterintuitive facts in mathematics.

When things are not so "clear cut", what happens if two people's intuition say something different? A question is only resolved with proof, until then it is nothing but a conjecture (guess).

That's true. That makes this very hard then

 

[RealiseNothing]

Haha... y all these very complicated ideas

I think I'm putting forward a very stupid idea but pls don't flame me
Maximum area of n polygon made possible with a fixed perimeter is a circle- n approaching infinity(Assumption which can be logically deduced)
Then this infinitely sided polygon ie circle with a perimeter 1 has radius 1/2pi
Putting this into the area formula gives you1/4pi.
Hence since the maximum possible polygon has this area the rest n sided polygon must have an area less than this regardless whether they are cyclic or regular or whatsoever.

I think this method is easier.
Pls correct me if I'm wrong

uhhh did you even read the thread?