1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving Area of n-sided Polygons Maximizes When they are Regular?

  1. May 25, 2012 #1

    I'm a math tutor at a community college, and one of the students recently asked me why it is always true that a *regular* polygon (regular meaning equiangular and equilateral) has maximum area for any given perimeter. It makes perfect intuitive sense, but neither I nor any of the other tutors could figure out a rigorous method to prove it (the key word here being "rigorous").

    The textbook skirted around the issue, and Google is turning up fuzzy results at best. Wikipedia states that it is fact, but only cites a book from 1979 that would be difficult to find.

    eCookies for the most elegant proof (::) (::) (::) :)
  2. jcsd
  3. May 25, 2012 #2
    I will have to think more to come up with a proof, but I will make mention of some things now. One difficulty in these problems is you have to know what "shapes" are allowed. If you are only looking at platonic polynomials, a simple proof should be had.

    However, I think these types of proofs can be very difficult if you are looking for full generality. For example, the isoperimetric problem simply states, of all closed curves in ##\mathbb{R}^2## having some fixed length, which one maximizes area? Intuitively the answer is a circle, but its proof is rather difficult.

    In your problem you seem to be only looking at polynomials with fixed perimeter and a fixed number of sides. If you are familiar with calculus of variations, you could exhaust all such polynomials and have a proof. Steiner Symmetrization is also another promising route. However, I think these are some very large guns to for this problem.
  4. May 25, 2012 #3
    It's actually fairly straightforward to prove that, for all *regular* polygons, the area is always less than that of a circle whose radius equals the distance from the center to a vertex of the polygon. You just construct the figure from isosceles triangles, write the area of the triangle in terms of the height of the triangle, and it can be shown that as n-> infinity, the area of the figure -> pi*r^2. The student had no problem accepting this at all.

    The problem is in proving that the regular polygon maximizes the area to begin with for any given perimeter, which seems to be required for the above to be true (in the sense that the circle is not only the maximum area of regular n-gons, but of all n-gons of fixed perimeter).
    Last edited: May 25, 2012
  5. May 25, 2012 #4
    Why not perform induction on the number of triangles? All shapes can be broken up into them, as you have already used. Deform the area of any one triangle in the shapes, calculate area, induct on numberof triangles.
  6. May 26, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You'll also need to fix (or at least limit) the number of edges.

    Not elegant, but here goes:
    Suppose a polygon P has n edges, not all the same length. There must be two adjacent edges, AB, BC, of different length. Drop a perpendicular from B to AC and show that the total area of the two triangles would be increased by replacing B with B', where AB' = B'C = (AB+BC)/2 (easy with a little calculus, probably a non-calculus method too).
    So that shows all the edges must be the same length, and that completes the proof for triangles.

    Suppose all edges are the same but not all angles. There must be two adjacent angles that differ, ABC, BCD. Let AD = H, AB = BC = CD = L, area ABCD = F. Let angle U = (BAD+CDA)/2, V = (BAD-CDA)/2.
    With a bit of working,
    (1) (H2 - L2)/4 = L(H cos V - L cos U)(cos U)
    (2) F = L(H cos V - L cos U)(sin U)
    Thus 4F = (H2 - L2) tan U
    Varying V, F is max when dF/dV = 0, so dU/dV = 0.
    Squaring and adding (1) and (2), then differentiating wrt V:
    (H cos V - L cos U)(-H sin(V) + L sin(U).dU/dV) = 0
    If (H cos V - L cos U) = 0 then F would be 0, so we have sin(V) = 0.
    Thus BAD = CDA, so ABC = BCD.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook