Two unit squares overlap with an area 1/16. The shape of this overlap may be a triangle or a quadrilateral.
Find the minimum distance between their centers.
This is a great problem: easy to pose, hard to answer.
I refined the solution. The best answer is: minimum separation distance
d = 0.933012705 +/- .000000003
It is plotted here.
I had to determine two things via simulations:
(A table from simulations is here.)
1) The minimal separation is made with an overlap shape that
is 3-sided, not 4-sided. (Note that the overlap of two squares
can have between 3-8 sides.)
Here are examples of small overlaps.
Many 4 sided overlaps came close, but the triangle cases win.
2) In the minimal separation the squares centers are connected
by a line perpendicular to the sides. I learned this via
simulation, without a proof. It seems natural however, since
one square is poking a corner onto the other while trying to
stay as close as possible.
The best rotation for the 2nd square around its axis is
about 9.7307 degrees.
Statement 2 does make the problem amenable to calculus. The overlap
area is now determined by only two factors - theta (rotation of 2nd
square on it's axis, and separation d. Since A = f(d , theta) and A is
constant, one can make d a function of theta and differentiate for a
minimum. The math is a bit messy - Euler rotation angles allow the
solution of intersecting lines. Rather than invert and differentiate -
I just ran the math forward and the simulations further.
The parameters listed produce a 1/16 area overlap to an accuracy of 10^(-9).
Edited on July 6, 2023, 3:25 pm
further soln
