# Neighborly family of coins

Here is a puzzle:

Find 5 identical coins. Can you arrange them so that every coin is touching every other coin?

The solution is here. The hint is: use the third dimension.

My questions are based on this puzzle.

1. Can you prove that six coins can not be arranged to pairwisely touch each other?
2. How much coins can be arranged to pairwisely touch each other in dimension $n$?

Here, "coins" in higher dimensions refer to $2$-dimensional unit disks (or unit $k$-balls for your favorite $k<n$), but the relative interiors of the disks / balls are disjoint.

Here, "coins" may refer to "thin" unit disks, i.e. $B_1^2\times B_\epsilon^{n-2}$ for $\epsilon$ arbitrarily small, or $B_1^k\times B_\epsilon^{n-k}$ for your favorite $k<n$.

• If I were to mint coins in $n$ dimensions, I would make them $B_1^{n-1}\times B_\epsilon^{1}$ instead of $B_1^2\times B_\epsilon^{n-2}$. – Joonas Ilmavirta Mar 2 '15 at 16:38
• @JoonasIlmavirta, as you like. I never saw a higher dimensional coin, so be my guest in higher dimension. – Hao Chen Mar 2 '15 at 17:06
• Are each two coins congruent (isometric)? Perhaps yes? Let radius of the cylinders be $1$. Then the answer should depend on $\epsilon$ -- we shouldn't assume that coins are infinitely/arbitrarily thin, right? The flat variant (non-cylindric) seem very different from cylindric. – Włodzimierz Holsztyński Mar 2 '15 at 18:05
• Of course the arbitrarily thin cylindrical case is extra attractive. – Włodzimierz Holsztyński Mar 2 '15 at 18:07
• @WłodzimierzHolsztyński, I believe that for every $n$ there is $\epsilon_n$ so that the maximum number of coins is the same for all $\epsilon\in(0,\epsilon_n)$. And there probably is an $\epsilon$-independent bound on the number. But it is interesting (and more complicated) to look at all values of $\epsilon$. – Joonas Ilmavirta Mar 2 '15 at 18:25