How many circles (radius $r$) are needed to cover circle whose radius is $2$ times bigger (radius $2r$).
I think we need to use area which is $S=\pi R^2$ but I don't really know what to do.
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2$\begingroup$ This is a difficult problem: First you have to devise a covering that you assume to be optimal. Then comes the difficult part: If your covering uses, say, 10 circles you have to prove that one definitely cannot do with 9 circles. The area estimate you propose could be an idea, but it only gives that you need at least 5 small circles: 4 for covering the area and at least one for the unavoidable overlap. $\endgroup$– Christian BlatterCommented Mar 8, 2011 at 13:53
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2$\begingroup$ This page (www2.stetson.edu/~efriedma/circovcir) comes at it the other way: "How big a circle can I cover with n circles?". It claims the "I can cover a circle twice as big" case is trivial. $\endgroup$– RawlingCommented Mar 8, 2011 at 14:19
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1$\begingroup$ @Rawling: That link is not working now. A replacement is this page. $\endgroup$– hardmathCommented Nov 18, 2022 at 18:32
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1 Answer
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This is a disk covering problem. As you have stated it, it is not quite as difficult as some others.
The first task is to find the minimum number of small circles which cover the circumference of the bigger circle rather than the whole area. If this is $m$ then it will be impossible to have a regular $m$-gon with edges of length $2r$ fitting strictly inside the circle of radius $2r$. This implies $m \ge 6$.
It turns out to be just possible to cover the circumference of the bigger circle with 6 smaller circles, and ignoring symmetries there is only one way to do it (you get a regular hexagon whose vertices lie on the bigger circle and whose edges are diameters of the smaller circles). But this leaves an uncovered area in the middle, which needs at least one (and in fact exactly one) more.
So the answer is seven smaller circles.
