Correct solutions were submitted by James R Eckert, Jr., Steve Depies, Ray Kremer, Mundia Mubyana.  Further correct solutions came from Jens Voss, Robert McQuaid, Dane Brooke, Khanh Ngo, Edward Lee, Burkart Venzke, Steve Prowse, Sudipta Das, Nancy Schwarzkopf-Schechner, Francesc Suñol, Brian Laughlin, Francisco de Leon-Sotelo y Esteban, Philippe Fondanaiche.

The limit is 3/4. 

Let's first determine how many disks, S_n, are present in the arrangement A_n.   It's obvious that S₁ = 1 and S₂ = 7.  In general, when we go from A_n-1 to A₁ we add n new disks along each of the six sides; the total number of new disks is therefore 6n - 6 since the disks added along each of the corners would otherwise be counted twice.  So S_n = 6(n-1) + S_n-1.  An easy induction gives S_n = 3n² - 3n + 1.

A similar analysis gives the radius of each of the small disks in A_n to be r/(2n-1).

The ratio of the areas is

Note that as n gets larger and larger, the arrangements of disks will approximately form the shape of an inscribed hexagon.  Many people submitted solutions based on the thought that this hexagon would indeed be formed in the limit. Interestingly, the area of this hexagon is about 82.7% of the area of the surrounding circle so even in the limit a good amount of the area remains lost.  Can you explain why that is?

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Page last updated 2 April 2001.

ã 2001 Alberto L Delgado