I have a polygon with $n$-vertices and of fixed length, I need to find a condition for which the polygon gives the maximum area without using isoperimetric inequality.

  1. I tried with the simple polygon "Triangle" ($n=3$) and find that it gives maximum area when it's 3 sides are equal. (See here for proof)

  2. Next, I break one side of the triangle to form a Quadrilateral ($n=4$). And find that it gives maximum area when it is a Square. (See proof here)

My intuition tells me that if I increase $n$ then the corresponding polygon will give maximum if their sides are all equal, I assumed this is true for $n=k$ but I am unable to show this for $n=k+1$. Or is there is any other way to prove it? Any help will be appreciated. Thanks in advance.

  • $\begingroup$ All sides and all angles are equal, not just all sides. But yes, that's the right answer. I don't know if induction is the way to prove it, though; a square and a regular pentagon don't have much in common. $\endgroup$ – Arthur Mar 21 '18 at 6:47
  • $\begingroup$ See math.stackexchange.com/questions/868026/… $\endgroup$ – lhf Jun 8 '18 at 15:22
  • $\begingroup$ Any generalization to 3D will be very messy, as there are not regular polygons with $n$ faces for every $n$. $\endgroup$ – Connor Harris Jun 8 '18 at 15:33

Suppose you have a plane $n$-gon with perimeter $L$ and maximum area.

If there are two sides whose lengths differ then there are two adjacent sides $AB$ and $BC$ with different lengths. If you replace the triangle $ABC$ with an isosceles triangle $AB'C$on base $AC$ with $|AB|+|BC| = |AB'|+|B'C|$ you will increase the area of the polygon, contradicting the area maximality.

I suspect a similar argument can prove the angles are all equal. I haven't thought that through.

Edit. In fact both my argument for the equality of the side lengths and the argument for angles is the core of the answer at this question, linked from the comments: Given a polygon of n-sides, why does the regular one (i.e. all sides equal) enclose the greatest area given a constant perimeter?

To extend the question to three dimensions requires a definition of the area of a polygon that might not be planar. You will have to clarify the question.

  • $\begingroup$ What about the existence of such a polygon attaining the maximum (not supremum) area with given perimeter? $\endgroup$ – Saad Jun 8 '18 at 15:26
  • $\begingroup$ @AlexFrancisco That does indeed require some further argument, as in the linked question. This subtlety may or may not matter to the OP. $\endgroup$ – Ethan Bolker Jun 8 '18 at 15:30
  • $\begingroup$ I want to mean "The area of the polygon in 3D space" by "the volume of a closed surface in that space". My question is can this approximation procedure be extended so that we can prove that a simple closed surface with fixed surface area gives maximum volume iff it is a spere? $\endgroup$ – Sujit Bhattacharyya Jun 8 '18 at 15:33
  • $\begingroup$ @SujitBhattacharyya It is true that the sphere maximizes volume for given surface area, but I haven't studied a proof. For sure there are not enough regular ones like the regular $n$-gons to do the job. The wikipedia page may answer all your questions. en.wikipedia.org/wiki/Isoperimetric_inequality $\endgroup$ – Ethan Bolker Jun 8 '18 at 15:40

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