Problem: strips covering a circle
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>Can one cover a diameter 20 disk (= circle + interior) with
>19 strips (= rectangles) each of width 1 ?
Solution:
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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