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Let $D$ be a disc of diameter 20, and suppose you are given 19 rectangles, each of which is $1 \times 20$. Can $D$ be covered completely by the rectangles? Note that the rectangles can be arranged "any which-way" for the covering.

Note: this was a popular poser in my grad student days; some of us came up with an answer, but had a hard time nailing down the proof. I'd like to hear from anyone with a good approach, or even a good reference, to this question.

Added note: the rectangles are not to be cut up. And the question may be posed differently in terms of 19 width 1 "strips", each infinitely long (i.e. copies of $[0,1] \times R$). In this form the question becomes: Can we cover the diameter-20 disk using 19 of these width 1 strips? [If one could cover via strips, then each strip could be trimmed to a 1X20 rectangle without uncovering any covered points in the disk.]

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    Then you would have 20 rectangles, but problem says 19. :-) I$f$ one desires, the rectangles can be viewed as 19 "strips", all of width 1 and infinite length (copies of $[$0$,1] \times R$). This is because given any configuration of 19 such strips which covered. one could trim each strip to be a $1 \times 2$0$$ recta$n$gle without uncovering anything.2012-11-08

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I remember that problem from university! (But I didn't solve it, somebody gave me a big clue)

The answer is that it is impossible, you can't cover the circle of diameter 20 using only 19 strips $1\times 20$.

To see why, suppose you have such cover and imagine a sphere of diameter 20 cut in half by our circle. Find the orthogonal projection of each strip on the sphere, it is a ring, and it is easy to compute it's area $2\pi R x$, where $R$ is the radius of the sphere and $x$ the width of the strip.

But between all the rings, they cover the whole sphere so the total area must be at least $4\pi R^2$, so $ N \times 2 \pi R x \ge 4 \pi R^2 $ where $N$ is the number of strips. Letting $N=19, R=10$ and $x=1$ we obtain a contradiction.

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    If the disk is partitioned into 20 width one parts of the form $[x_i,x_i+1]$ for $x_i=-10..,9$, the projections on the sphere cover it exactly. And since the surface areas of the slices only depend on their width, this shows you need 20 to cover the surface area, but only have 19. [The surface area of a slice on the sphere being the same as that on a cylinder...] So even without a calculation I should have noticed this, if only I'd thought about the 3-d aspect!2012-11-09