Isoperimetric Pairs

Liam is asleep, but I am not sure how long that will last.  That means time is of the essence here, and so I am going to forego any pretense of an introduction and jump right into some math.

Definition: Two natural numbers (m, n) are said to be an isoperimetric pair if the  number of 1\times 1 bricks needed to enclose mn units of area (arranged as an m \times n rectangle) is as small as possible.  That is, if

\displaystyle m+n = \left\lceil\frac{mn}{\lceil\sqrt{mn}\rceil}\right\rceil + \left\lceil\sqrt{mn}\right\rceil.

They may not word it exactly like this, but the REU students have been working hard (and playing hard!) on answering this question: Given a fixed number m for what values of n is (m,n) an isoperimetric pair?  (In my previous post, I called the product mn a hydropronic number.)

A brute force approach will confirm that when m = 1 the only possible values for n are 1, 2, or 3.  Similarly the only pairs for which m = 2 are (2, 2), (2, 3), (2, 4), and (2, 5).  Of course these students have been busy pattern-hunting, which allowed them/us to make a number of conjectures.  I list some them below, labeling all of them as claims, and placing a check-mark next to one to indicate that it has been proved.  Unchecked claims are open conjectures that might be fun to spend a friday afternoon exploring.

  • Claim 1: (\checkmark) All pairs of the form (n, n) and (n, n+1) are isoperimetric pairs.
  • Claim 2: All pairs of the form (n, n+2) are isoperimetric pairs.
  • Claim 3: If (m, n) is an isoperimetric pair, then so is (m+1, n+1).
  • Claim 4: Fix m \in \mathbb{N} and let m^* = \max\{n \in \mathbb{N} : (m, n) \text{ is an isoperimetric pair }\}.  Then (m, k) is an isoperimetric pair for all natural numbers k where m \leq k \leq m^*.
  • Claim 5: Using the same notation as in Claim 4, m^* +1 \leq (m+1)^* \leq m^*+2
  • Claim 6: Given m \in \mathbb{N} the number m^* = f(m) where f(m) is … not yet known.  In other words, this conjecture is complete; the pattern for m^* is not yet clear ….

What do you think?  Interestingly, pictures can be very helpful in deciding how best to approach or verify any of these claims.  Also, Liam’s still asleep!

Update (11:39 am): Someone really should write a program that finds and plots these pairs.  Something like this:  IF n \geq m and \displaystyle m+n = \left\lceil\frac{mn}{\lceil\sqrt{mn}\rceil}\right\rceil + \left\lceil\sqrt{mn}\right\rceil THEN print (m, n) and you need to repeat this, increasing n until it fails.  THEN this entire thing needs to be repeated, changing m to m+1 — some sort of nested if-then thingy, I suspect.

Update (12:57 pm): I am not too certain why this only now occurred to me, but if we take our condition for (m, n) being an isoperimetric pair and remove all of the ceilings from the expression (I know, but stick with me for just a second) one obtains m + n = \frac{mn}{\sqrt{mn}} + \sqrt{mn} which of course simplifies to

m+n = 2\sqrt{mn}.

If you divide both sides by 2 the left hand side becomes the mean of m and n while the right hand side becomes what is called the geometric mean.  So (very) roughly speaking, our pairs (m, n) consist of numbers whose arithmetic and geometric means coincide.  This is a well studied situation; in fact, an inequality between the two means exists, with equality holding if and only if m = n.

Again, this is a very rough approximation to our situation, since our equation holds for various choices of m and n.  Still, I’m curious that there may be some more geometry lurking in the background here.  Any thoughts?

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