Whole Rectangles
May 25th, 2007
A short introduction to Graph Theory is needed for this one. If you already are familiar with Graph Theoretic constructs feel free to skip it.
A graph G, is a pair (V,E) where V is a set of vertices and E is a set of edges. Each edge is an unordered pair of the form (u, v) where u and v are vertices (i.e. they belong to V). The degree of a vertex t (denoted deg(t)) is the number of edges containing it:
deg(t) = #{ e | e = (s, t) ^ s belongs to V }
In this post I only consider finite and simple graphs. A finite graph is a graph whose vertex-set V is a finite set. A simple graph is a graph in which there are no loops (i.e. edges of the form (u, u)) and no multiple edges (i.e. E is a proper set so it cannot contain the same element twice).
The following is a trivial claim:
The sum of the degrees in a graph equals twice the number of edges.
It is trivial to prove this claim by induction on the number of edges (on a graph with no edges it is clear, and by adding an edge to the edge set of the graph the sum of degrees increases by two).
Use this claim to solve the following riddle:
A rectangle is called whole if at least one of its sides is an integer. For example, a rectangle of 2 by 3/5 is whole as well as a rectangle of sqrt(5) by 3. A rectangle of 1/2 by 1/2 is not whole. Examples:

A set T of rectangles constitues a tiling of a rectangle R if the rectangles in T are disjoint and for every point p in R there is a rectangle S in T such that p belongs to S. Example:

Prove that if R is a rectangle and T is a tiling of R consisting only of whole rectangles (i.e. every rectangle S in T is whole) then R is whole.
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May 26th, 2007 at 4:13 pm
[no spoilers, but points to spoilers]
This is indeed a great riddle, because one can solve it in so many ways: using graph-theory, using complex integration (the ‘standard’ proof), etc.
For a few simple proofs, see http://www.inference.phy.cam.ac.uk/mackay/rectangles/ (also inside: a link to an article from 1987, that proves the theorem in 14 *different* ways).
By the way, the same theorem holds even if you define a ‘whole’ rectangle as one having at least one *rational* side, or even one *algebraic* side.