# Smallest square to wrap a cylinder

Suppose you need to gift-wrap a cylinder (e.g., a can of tennis balls, or a large candle) of height $h$ and radius $r$. Here wrap is the natural sense of covering the surface area of the cylinder completely, without cutting the square, creasing however needed. What is the smallest square that suffices for a given $h$ and $r$? For example, a rectangle of dimensions $(h+2r) \times (2 \pi r)$ suffices for how one might wrap a can of tennis balls or a stout candle:

In this $h=3$ and $r=1$ case, the rectangle has dimenions $5 \times 6.28$, and so a square of side $2 \pi$ suffices. But is that optimal?

Merry Christmas!

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@Alexander E.'s edit: I did mean square, for the smallest-area rectangle, can, I believe, approach the surface area of the cylinder, by using a long, thin rectangle. So I took the liberty of rolling back. – Joseph O'Rourke Dec 25 '12 at 21:24
@Joseph: sorry, now I understand. I removed my edit. – Alexandre Eremenko Dec 25 '12 at 21:29
@Alexandre: No problem; and sorry for misspelling your name! – Joseph O'Rourke Dec 25 '12 at 21:46
If the paper has pictures of, say, reindeer, do we need to have them come out entire on either the top or bottom or cylindrical part, or can we bend them? – Will Jagy Dec 25 '12 at 22:16
@Will: That would greatly complicate the question, to require matching patterns! Let's pursue that after solving the easier question I posed. – Joseph O'Rourke Dec 25 '12 at 22:20

This is in response to Vectornaut's question about wrapping with a thin rectangle. In Geometric Folding Algorithms: Linkages, Origami, Polyhedra, it is argued (Theorem 15.2.1, p.236) that any polyhedron can be covered with a thin strip with arbitrarily small surface area beyond that of the polyhedron. A polyhedral approximation to a cylinder then yields the claim. This is Figure 52.2 (p.234), which gives some idea of switchback turns of the strip used in the argument:

Here is a link to the original 1999 paper by Demaine, Demaine, and Mitchell, "Folding Flat Silhouettes and Wrapping Polyhedral Packages: New Results in Computational Origami": link.

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Super cool—thanks! – Vectornaut Dec 26 '12 at 23:53

You can wrap the cylinder with a square whose diagonal has length $2(h + 2r)$: put the cylinder in the center of the square and fold the corners up to meet at the top of the cylinder. When $h/r$ is less than $\pi - 2$, this method uses less paper than the OP's method. If $h/r$ is very small (if you're wrapping a CD, for example), it uses just a little over $2/\pi$ as much paper as the OP's method.

I haven't done a comparison with Gerhard Paseman's method...

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Very clean and simple---Thanks! – Joseph O'Rourke Dec 26 '12 at 13:49

Perhaps Joseph can fill in this description with a picture.

In square with vertices A, B, C, and D in clockwise order, draw a line from A to a point P on BC. If the length of AP is longer than the circumference R of the inscribed circle of ABP, then a cylinder with circumference R and height which I leave you to determine can be wrapped by ABCD.