From: erick@sfu.ca (Erick Wong) Subject: Re: Rectangles Divided Into Squares Date: Tue, 14 Sep 1999 23:48:06 -0700 Newsgroups: sci.math Keywords: Using Kirchhoff's laws to solve combinatorial problems jonl@cs.pdx.edu (Jon P LeVitre) wrote: >I've been thinking that you can express the width and height of the >rectangle as a system of linear equations and solve for the size >of any square in terms of another. This is easy enough with >examples I've looked at, but I'm not sure how to make a solid >sci.math-postable proof out of it. > >I know that there must be a simple indirect proof, but it hasn't >come to me yet. Here's a proof that appeals to (electro-)physical intuition. I first read it in Ross Honsberger's MAA booklet, "Ingenuity in Mathematics". As an aside, the relevant essay is actually about a really neat problem about dividing a square into smaller squares of distinct sizes (finitely of course). What's really surprising is that it's actually possible! Anyway, here's the argument: picture the rectangle as a metal plate with uniform resistance. Now run a current from the top edge to the bottom edge (so that the potential at any point on the plate is independent of its horizontal position). By elementary physics, the resistance of a square subplate is _independent_ of the size of the square. So we can assume that all squares have unit resistance (current = voltage). The key observation is that the current and potential difference across one of our squares is directly proportional to its side length. We define a circuit by placing a node at every horizontal line that meets the top or bottom of a square. This is finite, of course :). To complete the abstraction, we place a unit resistor across any two nodes that share the same square. Now we ask a very natural question: how do Kirchoff's Laws in our circuit translate into facts about our squares (really vice versa)? 1) The total current going into a node is equal to the current going out of a node. In other words the sum of currents (read: square sides) across any horizontal line is constant. Well of course, that's the width of the rectangle! 2) The total potential difference along any cycle is zero. Since the voltage drop across any square is proportional to its height, this too is clear as the cycle must return to its original vertical position :). What has this given us? It shows the physical constraints of the squares "fitting together" are sufficient to imply Kirchoff's Laws for the related circuit, and therefore the voltage/current through each resistor is uniquely determined up to a scalar multiple (the current applied). Translated back, the side of each square is uniquely determined up to a scalar multiple (the width of the square). Since we're dealing with linear equations with integer coefficients, the unique width/height ratio of the square must be rational. Has anyone seen any other elegant proof of this result? -- Erick