18
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ Have you tried putting the seventh square in the middle of the configuration, rather than round the outside? $\endgroup$
    – Mark Bennet
    Dec 30, 2014 at 15:23
  • $\begingroup$ 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. $\endgroup$
    – Martigan
    Dec 30, 2014 at 15:29
  • 5
    $\begingroup$ 7 squares are possible: directupload.net/file/d/3852/a4wm5vz6_jpg.htm $\endgroup$
    – callculus42
    Dec 30, 2014 at 15:55
  • $\begingroup$ @calculus Thanks! Now this seems absolutely full. But how can you prove that 8 are not possible? $\endgroup$
    – Erel Segal-Halevi
    Dec 30, 2014 at 19:30
  • 1
    $\begingroup$ @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. $\endgroup$
    – callculus42
    Dec 30, 2014 at 20:09

2 Answers 2

Reset to default
11
+50
$\begingroup$

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.

enter image description here

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

enter image description here

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]}]
$\endgroup$
3
  • 1
    $\begingroup$ Is it possible to prove that more than 8 squares are impossible? $\endgroup$
    – Erel Segal-Halevi
    Aug 12, 2015 at 11:45
  • 1
    $\begingroup$ @ErelSegal-Halevi: I'll think about it and if I find a proof, I'll expand the answer. $\endgroup$
    – Dejan Govc
    Aug 12, 2015 at 11:51
  • 1
    $\begingroup$ Very nicely done! $\endgroup$
    – Peter Woolfitt
    Aug 12, 2015 at 19:16
9
$\begingroup$

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:

squares

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .