## Comparing two geometric problems ### Concentric circles

When the circles are concentric it is easy to see that whether the rings close or not depends only on the ration of the radii. Elementary trigonometry allows us to find the corresponding ratios for which the rings close. Figure 10 shows the aproximate values of the first 25 obtained using the following formula3 $\mbox{ration of the radii }=\frac{\sin\left(\frac{\pi}{n}\right)+1}{1-\sin\left(\frac{\pi}{n}\right)}.$ We have, then, an answer to our problem, as it was posed: there does not always exist a ring that closes and it is enough to choose, as a counterexample, the pair of circles represented in figures 6 and 7. What we have done till now suggests another interesting question: is there any pair of initial circles for which the ring closes for a particular choice of the first circle but does not close for another choice of that circle? This is a question that does not seem easy to tackle. However, from what we wrote before, if such a pair of initial circles exists, they will necessarily be non-concentric. It would be great if we could reduce the general case to the particular case of concentric circunferences, for which the answer is known. How can we try to do that?

Closed rings
Number of circle Ratio of the raddi
3 $$13,92820323$$
4 $$5,828427125$$
5 $$3,851839996$$
6 $$3$$
7 $$2,532843231$$
8 $$2,239828809$$
9 $$2,039606729$$
10 $$1,894427191$$
11 $$1,784478148$$
12 $$1,698396372$$
13 $$1,629211496$$
14 $$1,572416528$$
15 $$1,524970987$$
16 $$1,484750842$$
17 $$1,450228262$$
18 $$1,420276625$$
19 $$1,394047222$$
20 $$1,370888706$$
21 $$1,350292994$$
22 $$1,331857993$$
23 $$1,315261385$$
24 $$1,300241804$$
25 $$1,286585105$$

Figure 10

3 We have seen the formula for rings with $$n$$ circles that close after one turn. If we want rings with $$p$$ circles that close after $$q$$ turns ($$p$$ and $$q$$ prime to each other) the corresponding expression for the ratio of the raddi is $\frac{\sin\left(\frac{\pi}{p/q}\right)+1}{1-\sin\left(\frac{\pi}{p/q}\right)}=\frac{\sin\left(\frac{q \pi}{p}\right)+1}{1-\sin\left(\frac{q \pi}{p}\right)}.$

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