Really Hard Circle Question

Question

A circle of radius r is centered at every lattice point in the plane (that is, every point whose coordinates are both integers). In every lattice rectangle (that is, a rectangle all of whose vertices are at integer coordinates), exactly half of the area is inside these circles. Compute r.

Answer 1

The answer is r = 1/√(2π) ≈ 0.3989, and the problem is much friendlier than it looks once you notice what the phrase “every lattice rectangle” is really telling you.

The key idea: one unit square is the whole problem

Every lattice rectangle is built out of 1×1 lattice squares. So if the covered fraction is 1/2 in every rectangle, it has to be 1/2 in a single unit square — and conversely, if it is 1/2 in the unit square, it is automatically 1/2 in every rectangle, because you are just adding up identical cells. The word “every” is a hint, not extra work: it is telling you the pattern is periodic, so you only ever need one cell.

Counting the covered area in that square

Take the square with corners (0,0), (1,0), (0,1), (1,1). Circles are centered at all four corners, and each contributes a quarter of itself to the square. Four quarters make one whole circle, so the covered area inside the unit square is exactly

πr²

This is the step that makes the problem collapse. You are not computing a messy union — the four corner pieces reassemble into one full circle, and each unit square carries exactly one circle’s worth of area.

Solve

The square has area 1, so “exactly half the area is inside the circles” means

πr² = 1/2

r² = 1/(2π)

r = 1/√(2π) = √(2π)/(2π) ≈ 0.39894

The check you must not skip

The four-quarters argument quietly assumes the circles do not overlap each other. If they did, the pieces would double-count and the covered area would be less than πr².

The closest pair of lattice points are distance 1 apart, so the circles stay disjoint exactly when 2r ≤ 1, that is r ≤ 0.5. Our answer is r ≈ 0.3989, comfortably under. So the assumption holds and the answer stands.

Always run that check on this type of problem. If the required coverage had been, say, 90% instead of 50%, you would have needed r ≈ 0.535, the circles would overlap, and the whole quarter-circle shortcut would collapse — you would be stuck computing lens-shaped intersections instead. The clean answer here is a consequence of 1/2 being small enough, not a general fact.

I checked this numerically as well: sampling a fine grid over the unit square with r = 1/√(2π) gives a covered fraction of 0.50002, converging to 1/2 as the grid refines.

A pleasant coincidence

1/√(2π) is the normalizing constant of the standard normal distribution — the number out front of the bell curve. There is no deep connection here; both just come from the same integral showing up in different places. But it makes the answer easy to recognize and remember.