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This is probably very simple and I am sure I can hack something together. However, I am looking for an elegant solution to my problem.

I have a hollow circle with, let us say, an outer radius of 47.3 cm and an inner radius of 39.5 cm. Hence, the thickness of the hollow circle is 7.8 cm. Btw, is there a better name than hollow circle?

I would like to know the area of this hollow circle (now problem) minus the area of the maximum number of mini circles that can fit into the hollow circle. Here the mini circles have a radius of 2.5 cm and are snugly placed at the outer circumference of the hollow circle.

I hope I have expressed myself comprehensibly. Thanks a lot.

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I think the word you want is "annulus". This is Latin for "ring". –  MJD Sep 19 '12 at 14:12

1 Answer 1

up vote 2 down vote accepted

The area of such an annulus is simply $\pi(r_1^2-r_2^2)$. For the minicircles of radius $r$ each, you need to know their number $n$. As $r_1-r_2>2r$, the circles fit comfortably inside the annulus.. The $n$ lines from the center of the annulus to the centers of the small circles have length $r_1-r$ and an angle $\alpha$ such that circles don't overlap, i.e. $\sin\frac\alpha2\ge\frac r{r_1-r}$. Thus you should calculate $\alpha_\min=2\arcsin\frac r{r_1-r}$, let $n=\left\lfloor\frac{2\pi}{\alpha_\min}\right\rfloor$ and then have the toal area $$ A = \pi(r_1^2 -r_2^2-n r^2).$$

In your example, $r_1=47.3$, $r_2=39.5$, $r=2.5$, hence $\alpha_\min\approx0.111665149$, $n=\lfloor56.28\rfloor=56$, finally $$ A\approx 1027.426$$

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Just manually computed it (using a slightly more cumbersome approach) and it is correct. Excellent, thank you. –  csetzkorn Sep 19 '12 at 14:51

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