LAST EDITED ON Jun-22-02 AT 10:29 AM (EST)
 
>Well, it certainly can't be done in a plane. I've seen some
>tricks to do it involving folding the paper as you draw the
>line and drawing on the back of the paper. Of course, that's
>not really helpful mathematically ;)

It is in fact, because there's all mathematics one needs to solve that problem. However, the problem itself is a little different from what you discuss. You seem to think that the task is to draw that shape of five rectangles, but it's not. The figure is given. Think of it as a house blueprint with five rooms such that every wall segment has a door. The task is to walk through all the doors, not to draw a blueprint.

>The proof I was taught involved the T junctions of the
>figure. Notice that whenever three segments meet in a T
>intersection, then when you are drawing the figure, one
>segment must begin or end at that intersection. If you do
>not start drawing at that point, then when you visit it for
>the first time, you obviously have to leave that point
>through a different segment. Thus, you have drawn two of the
>three segments, and there is only one segment remaining
>connected to that vertex. Obviously, when you draw that
>segment, there will be no way to leave that vertex again.
>For these kinds of problems, it follows that for every
>intersection of an odd number of segments, you must begin or
>end drawing a line there. Since the line you are drawing
>begins and ends somewhere, it can handle only two of the
>intersections with an odd number of segments. The figure you
>give has 8 T intersections, so it will take at least four
>seperate lines to draw.
>
>This method can certainly disprove whether you can draw a
>figure in so many seperate, non-overlapping lines, but I am
>not sure if it can give the number of lines needed. It gives
>4 lines as a floor, but I am not sure if it proves that you
>can actually draw the figure in 4 lines. It seems like it
>should, so I wonder if any figure can be drawn in so many
>lines as half the number of intersections with an odd number
>of segments. Does anyone know any proofs or counterexamples?

The problem (regardless of the interpretation) is one of those resolved by Euler's theorem, the theorem that started graph theory. Assume, e.g. as you think, there's a graph with 8 T vertices, other vertices being of even degree, i.e. being incident with an even number of edges.

Connect the 8 T (or, in general, odd degree) vertices with 4 new edges making all of them even. Now, Euler's theorem says that, for such a graph, there exists an Euler cycle, a closed path on the graph - a sequence of adjacent edges - that contains all the edges and each only once.

Now, remove the 4 add-on edges from the cycle. The cycle is bound to get split into four pieces, since the number of loose ends is 8.