This is a very silly question.

For all regions S contained inside the unit square, what is the infimum of the quantity Perimeter(S)/Area(S)? This ratio being considered is not scale invariant, so it is only the constraint of being contained within the square which implies that this infimum is non-zero.

There are some "obvious" configurations to try, but I do not even know how to use a calculus of variations argument to show that these are local maxima.

Can one do better than $\displaystyle{2 + \sqrt{\pi}}$?

EDIT: One can probably show that an optimal S has $\mathbf{Z}/4\mathbf{Z}$ symmetry. Assume that S consists of line segments along part of boundary, leaving segments of length x at each corner, and then (four quarters of) a shape T in the corner. We can assume that T has volume Ax^2 and perimeter Px. Minimizing the quantity with respect to x, one obtains a minimum of:

$$2 +\sqrt{4 + \frac{(P/2-4)^2}{(A - 4)}},$$

which becomes $2 + \sqrt{\pi}$ when T is the circle, i.e., $A = \pi$ and $P = 2 \pi$.

  • 1
    $\begingroup$ How are you arranging your quarter-discs? One in each corner, sure, but where is the corner of each quarter-disc going, and what's the radius? $\endgroup$ – Yemon Choi Nov 11 '09 at 9:26
  • $\begingroup$ This is not rocket science. It would be easier for you to work it out yourself than for me to write it. Well, easier for me anyway. $\endgroup$ – Anon Nov 11 '09 at 9:41
  • $\begingroup$ Thank you for editing. That's clearer. By the way, AFAIK the purpose of MO is not for you to test people with puzzles. It is to ask people for solutions/pointers. Your original wording gave me the impression you were gluing quarterdiscs onto the midpoint square of the original, hence my inability to meet your standards of thought. $\endgroup$ – Yemon Choi Nov 11 '09 at 10:02
  • $\begingroup$ Should add that I probably misread your original wording, anyway. But the present form is clear, I agree. $\endgroup$ – Yemon Choi Nov 11 '09 at 10:20

Basic observation: the solution will consist of circular arcs joining the sides of the square to each other, and some sides of the square.

Lemma: Consider a line segment of length $\ell$, and real number $p \geq \ell$. Consider all planar regions which contain the line segment in their boundary, and for which the rest of their boundary has length $p$. The largest possible area of such a region is obtained when the nonlinear part of the boundary is a circular arc.

Proof: Otherwise, we could take a circle, draw a chord of length $\ell$ across it cutting off an arc of length $p$, and replace this segment by another shape, to obtain a shape with the same perimeter as the circle and larger area.

Using the basic observation, your problem turns from a problem in the calculus of variations to one in multivariable calculus: mark the 8 points at which your shape leaves the sides of the square, and the radii of the 4 circles which cut the corners. Since, as you observed, the solution has eight-fold symmetry, there are really just two quantities: the length of the four boundary segments and the radius (or the angle) of the arcs cutting the corners.

  • $\begingroup$ That's a nice argument, thx. $\endgroup$ – Anon Nov 11 '09 at 16:04

This is known as the Cheeger problem, you can have a look at the paper "An introduction to the Cheeger problem" (http://www.utgjiu.ro/math/sma/v06/a02.html) and references therein.


The only hand-wavy suggestion I have towards finding extrema is that if S is a subset of the unit square then replacing S with its convex hull ought not to increase the ratio you're trying to minimize, and should strictly decrease it if the original set S was non-convex.

Have you tried feeding |x|^p + |y|^p = 1 into something like Maple/Mathematica, for p > 1, and getting it to compute the desired quantity for the set bounded by that curve?

  • $\begingroup$ I am guessing that S will include some straight line segments along the boundary. $\endgroup$ – Anon Nov 11 '09 at 9:14

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