Problem: strips covering a circle -------- >Can one cover a diameter 20 disk (= circle + interior) with >19 strips (= rectangles) each of width 1 ? Solution: --------- Here's a slick and tricky solution. The solution is not due to me; this is an old and well-known brain-teaser. Imagine a spherical surface of radius R. Imagine a pair of parallel planes in space that slice the sphere at an arbitrary position. Consider the "ring" of the sphere trapped between the planes. What's the area of the ring? This is an easy calculus exercise. The answer is: 2*pi*R*a, where "a" is the distance between the planes. The interesting aspect of this solution is that the area is independent of the location of the ring! That is, regardless of where the planes intersect the sphere, the area of the ring remains the same as long as the separation "a" between the planes remains the same. [In particular, if a=2*R, then the "ring" coincides with the sphere itself, and as expected, the area formula above gives 2*pi*R*2*R, that is 4*pi*R^2, for the area of the sphere, as expected.] What has this got to do with your problem? Imagine your circle of radius R together with strip of width "a" overlayed on it. Imagine that what you are seeing represents a projection into two dimensions of a sphere and a ring generated by slicing the sphere by a pair of planes. The area of the ring is 2*pi*R*a, as discussed above. Now, imagine not one strip, but a lot of them, (arbirtary in number and arbitrarily varying thicknesses) overlayed on your circle in a crisscross fashion. Each strip corresponds to the projection of a ring from 3D. Let the sum of the widths of the strips be "b". Therefore the sum of the areas of the rings is 2*pi*R*b. Imagine the crisscrossing strips completely cover the circle. Then the corresponding rings completely cover the sphere. Therefore the sum of the areas of the rings is no less than the area of the sphere, that is: 2*pi*R*b >= 4*pi*R^2, thus: b >= 2*R In other words, if the strips cover the circle, then the sum of their widths is at least equal to the diameter of the circle. Your collection of strips does not have this property, so they won't cover the circle. QED -- Rouben Rostamian From: rouben@math9.math.umbc.edu (Rouben Rostamian) Article: 21335 of sci.math Subject: circles and strips Date: 13 Feb 1993 04:25:33 GMT