It turns out that the largest ellipse in any rhombus always occupies π/4 of the area of the rhombus. (For, mapping the rhombus linearly onto a square multiplies all areas by the same factor, so the inverse mapping carries the maximum ellipse in the square — namely, a circle — onto the maximum ellipse in the rhombus. The fraction of area covered is the same in both cases.) So the N=3 case, the three 60º-120º rhombi of side = 1 in the unit disk D can cover the fraction π/4 of the area of the inscribed hexagon, itself having area = 6 sqrt(3)/4 = sqrt(27/4), for a total area of π*sqrt(27/64) and a total ratio of (area of 3 ellipses / area of circle) = sqrt(27/64) = 0.6495190+. This *might* be the maximum for N = 3. Meanwhile as n —> oo, the fraction —> π/sqrt(12) = 0.9068996+ (approaching e.g. a triangular arrangement of circular disks). So the fraction must increase eventually. —Dan I wrote: ----- I'm still not sure, but it looks as though for N = 3 ellipses in a unit disk if you chop the inscribed hexagon into 3 60º - 120º rhombi of side = 1, then possibly the largest ellipses in such a rhombus add up to more area than the largest congruent circular disks that can all 3 exist disjointly in D. But that *might* be the only case... -----