I was told this puzzle last friday by Peter Winkler (who had mentioned that it was told to him by a japanese fellow who is perhaps the one you are referring to).

The solution in the $n \leq 10 $ case is to consider the tiling of the plane by unit height hexagons. Inscribe within each of these hexagons a unit circle. This grid of circles has density > 0.90 on the plane, and so if you randomly place this grid on the plane you accordingly have expected number of points covered > 9 (out of the 10), and this implies exists an arrangement that covers 10. (theres a few details missing from this probabilistic method argument, but you get the basic idea).

I believe for the $n>10$ case we have some way of computing an upper bound on the density of a sphere packing on the plane that rules it out in general. (or something to that extent)

I was told this puzzle last Friday by Peter Winkler (who had mentioned that it was told to him by a Japanese fellow who is perhaps the one you are referring to).

The solution in the $n \leq 10 $ case is to consider the tiling of the plane by unit height hexagons. Inscribe within each of these hexagons a unit circle. This grid of circles has density > 0.90 on the plane, and so if you randomly place this grid on the plane you accordingly have expected number of points covered > 9 (out of the 10), and this implies exists an arrangement that covers 10. (theres a few details missing from this probabilistic method argument, but you get the basic idea).

I believe for the $n>10$ case we have some way of computing an upper bound on the density of a sphere packing on the plane that rules it out in general. (or something to that extent)

I was told this puzzle last friday by Peter Winkler (who had mentioned that it was told to him by a japanese fellow who is perhaps the one you are referring to).

The solution in the $n \leq 10 $ case is to consider the tiling of the plane by unit height hexagons. Inscribe within each of these hexagons a unit circle. This grid of circles has density > 0.90 on the plane, and so if you randomly place this grid on the plane you accordingly have expected number of points covered > 9, and this implies exists an arrangement that covers 10. (theres a few details missing from this probabilistic method argument, but you get the basic idea).

I believe for the $n>10$ case we have some way of computer an upper bound on the density of a sphere packing on the plane that rules it out in general. (or something to that extent)

I was told this puzzle last friday by Peter Winkler (who had mentioned that it was told to him by a japanese fellow who is perhaps the one you are referring to).

The solution in the $n \leq 10 $ case is to consider the tiling of the plane by unit height hexagons. Inscribe within each of these hexagons a unit circle. This grid of circles has density > 0.90 on the plane, and so if you randomly place this grid on the plane you accordingly have expected number of points covered > 9 (out of the 10), and this implies exists an arrangement that covers 10. (theres a few details missing from this probabilistic method argument, but you get the basic idea).

I believe for the $n>10$ case we have some way of computing an upper bound on the density of a sphere packing on the plane that rules it out in general. (or something to that extent)