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Find all $n\geq2$ such that we can have n points $p_i$, and n circles $C_t$, such that if $j\neq k$, then $p_j$ is contained in $C_k$, and each $C_q$ has a point $p_q$ at its center.

Could you please help me? I don't even understand how am I supposed to create such a circle...

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Am maybe missing something, but for circles with large radii, is there any limit to the number whose centers could be placed in a small area so they were all contained in the collection of circles? – daniel Dec 3 '11 at 13:07
up vote 2 down vote accepted

I am assuming that you are asking for the centre $p_i$ of the circle $C_i$ to lie on the circle $C_k$ for all $k \neq i$. (Otherwise, as daniel suggests above, the answer is (uncountably) infinite.)

First note that if we do have a collection of such circles, then their radii must all correspond. (To see this, note that since $p_i$ lies on $C_k$, then the distance from $p_i$ to the centre $p_k$ of $C_k$ must equal the radius of $C_k$. Similarly, since $p_k$ lies on $C_i$ then the distance from $p_k$ to $p_i$ must be the radius of $C_i$. But these two distances are the same, so the radii of the circles correspond.)

The above also shows that the points must all be the same distance apart. So, without loss of generality, we are looking for the number of points which are all distance $1$ apart. Clearly we can find $2$ points which are distance $1$ apart. For $n=3$ these three points must define an equilateral triangle, and there's no problem with that. However, once you start to look for $4$ such points, problems arise. The basic idea is that given two points $p_1, p_2$ of distance $1$ apart there are only two points are are distance $1$ from each of these (and they lie on opposite sides of the line defined by $p_1, p_2$), however these two points are $\sqrt{3} > 1$ apart from each other.

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Thank you. Do I understand correctly that only n=2 and n=3 work? – Tom Reilly Dec 3 '11 at 17:55
@Tom Reilly: That's my thinking, anyway. – arjafi Dec 5 '11 at 11:21

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