# How many unit squares can overlap a given unit square without overlapping each other?

How many unit squares can overlap a given unit square without overlapping each other?

@calculus has managed to arrange 7 squares (see this GeogebraTube page). This seems like the maximum possible, but how to prove it formally?

• Have you tried putting the seventh square in the middle of the configuration, rather than round the outside? Dec 30, 2014 at 15:23
• You can put the seventh in the middle, since the diagonal is of greater length than the sides. There is space enough. The optimal configuration of overlapping is to be tangent. Dec 30, 2014 at 15:29
• 7 squares are possible: directupload.net/file/d/3852/a4wm5vz6_jpg.htm Dec 30, 2014 at 15:55
• @calculus Thanks! Now this seems absolutely full. But how can you prove that 8 are not possible? Dec 30, 2014 at 19:30
• @ErelSegalHalevi That is the difficulty of this problem. At the moment I have no idea how to prove it. But there is a great chance, that someone else have an idea. Dec 30, 2014 at 20:09

Apparently $8$ squares are possible. Start with the example suggested by Peter Woolfitt and try to pack everything as tightly as possible.

In the following picture, each white square is tilted with respect to the gray square by a multiple of $\frac{\pi}{8}$ and the intersections are the obvious ones, i.e. each two squares intersect either in a whole side or in a single point.

Now perturb the angles and positions slightly so that the overlapping conditions are satisfied.

In case someone wants to play with this example, here is the Mathematica code used to produce the pictures. First, some useful functions:

ClearAll[A, T, sq]
A[phi_] := {{Cos[phi], -Sin[phi]}, {Sin[phi], Cos[phi]}}
T = {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {0, 0}};
sq[v_, phi_] := Line[Table[A[phi].t + v, {t, T}]]


The first picture:

Graphics[{Opacity[0.2], Polygon[T], Opacity[1], Gray, sq[{0, 0}, 0],
Black, sq[{0, 0}, 3 Pi/4], sq[{0, 0}, 5 Pi/4], sq[{0, 1}, 5 Pi/8],
sq[{0, 1}, Pi/8], sq[{1, 0}, -Pi/8], sq[{1, 0}, -5 Pi/8],
sq[{1, 1}, 0], sq[{1/2, (1 - Sqrt[2])/2} - {0.1, 0.1}, Pi/4]}]


The perturbation:

Graphics[{Opacity[0.2], Polygon[T], Opacity[1], Gray, sq[{0, 0}, 0],
Black, sq[{0, 0} + {0.01, 0.02}, 3 Pi/4], sq[{0, 0} + {0.02, 0.01}, 5 Pi/4],
sq[{0, 1} + {0.01, -0.01}, 5 Pi/8 - 0.05], sq[{0, 1} + {0.03, -0.01}, Pi/8 - 0.05],
sq[{1, 0} + {-0.01, 0.03}, -Pi/8 + 0.05], sq[{1, 0} + {-0.01, 0.01}, -5 Pi/8 + 0.05],
sq[{1, 1} - {0.01, 0.01}, 0], sq[{1/2, (1 - Sqrt[2])/2} - {0.12, 0.12}, Pi/4]}]

• Is it possible to prove that more than 8 squares are impossible? Aug 12, 2015 at 11:45
• @ErelSegal-Halevi: I'll think about it and if I find a proof, I'll expand the answer. Aug 12, 2015 at 11:51
• Very nicely done! Aug 12, 2015 at 19:16

This is a comment, but too large to fit in the comments section

Pure area arguments are unlikely to work as shown by this near miss for $8$ squares: