The overlap of two unit squares is a rectangle with area 1/16.
Find the minimum distance between their centers.
The distance between the centers is the same as the distance between any two corresponding points of the two squares.
So then I drew the problem as a single unit square with a x by y rectangle in the corner.
Then distance from the inside corner of the rectangle to the far opposing corner of the square is the same as the original distance between the two centers.
So given the small rectangle is x by y then the distance can be expressed as D^2 = (1-x)^2 + (1-y)^2, such that x*y=1/16.
Expand and simplify:
D^2 = 2 - 2x +x^2 - 1/(8x) + 1/(256x^2)
Differentiate:
2D * D' = -2 + 2x + 1/(8x^2) - 1/(128x^3)
Now equate to zero and multiply through to clear fractions:
0 = 256x^4 - 256x^3 + 16x - 1
This factors pretty easily into
(4x-1) * (4x+1) * (16x^2+16x+1) = 0
Then x = 1/4, x=-1/4, x = 1/2 +/- sqrt(3)/4
The x=-1/4 must be rejected.
x=1/4 corresponds to the symmetric solution of a 1/4 by 1/4 overlap.
x = 1/2 +/- sqrt(3)/4 are actually the two lengths of the non-symmetric solution (if x is one, then y is the other)
If x=1/4 then D^2 = 9/8, or D = (3/4)*sqrt(2) = 1.0607
If x=1/2 +/- sqrt(3)/4 then D^2 = 7/8, or D = sqrt(14)/4 = 0.935414
And the boundary case x=1/16, y=1 has D^2 = 15/16, or D=sqrt(15)/4 = 0.9682.
The smallest of these extrema is a distance of sqrt(14)/4, which occurs when the 1/16 area overlap rectangle has dimensions of 1/2 + sqrt(3)/4 by 1/2 - sqrt(3)/4.
Analytic Solution
