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