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Here is an answer for the square case: Following what Henry did, using matlab/octave calculations, one gets that 8 is the maximum.

Synopsis: you can lose a bit more than 11% of the area and still have an area that divides into 8 less than 9 times. Now note that we don't lose that much area from the intersection of the tipped square and the square annulus unless the tip of the square is between approx 17 and 73 degrees. We are also guaranteed to have at least one side of the tipped square at least .5 units from a corner of the central square. (see the figure below -- the angle bounds are approximate)

(source)

Now, a bit of thought and playing around will convince you that the area red rectangle is a lower bound on the inaccessible area to other squares in the annulus, thus that area can be added back to the lost area.

(I did the calculations twice, found a simple arithmetic error that did not effect conclusion -- I am pretty sure this is right.)

Here is an answer for the square case: Following what Henry did, using matlab/octave calculations, one gets that 8 is the maximum.

Synopsis: you can lose a bit more than 11% of the area and still have an area that divides into 8 less than 9 times. Now note that we don't lose that much area from the intersection of the tipped square and the square annulus unless the tip of the square is between approx 17 and 73 degrees. We are also guaranteed to have at least one side of the tipped square at least .5 units from a corner of the central square. (see the figure below -- the angle bounds are approximate)

     (source)

Now, a bit of thought and playing around will convince you that the area red rectangle is a lower bound on the inaccessible area to other squares in the annulus, thus that area can be added back to the lost area.

(I did the calculations twice, found a simple arithmetic error that did not effect conclusion -- I am pretty sure this is right.)

Here is an answer for the square case: Following what Henry did, using matlab/octave calculations, one gets that 8 is the maximum.

Synopsis: you can lose a bit more than 11% of the area and still have an area that divides into 8 less than 9 times. Now note that we don't lose that much area from the intersection of the tipped square and the square annulus unless the tip of the square is between approx 17 and 73 degrees. We are also guaranteed to have at least one side of the tipped square at least .5 units from a corner of the central square. (see the figure below -- the angle bounds are approximate)

alt text http://s20.postimage.org/qrxof1f31/square_kissing_number.jpg

Now, a bit of thought and playing around will convince you that the area red rectangle is a lower bound on the inaccessible area to other squares in the annulus, thus that area can be added back to the lost area.

(I did the calculations twice, found a simple arithmetic error that did not effect conclusion -- I am pretty sure this is right.)

Here is an answer for the square case: Following what Henry did, using matlab/octave calculations, one gets that 8 is the maximum.

Synopsis: you can lose a bit more than 11% of the area and still have an area that divides into 8 less than 9 times. Now note that we don't lose that much area from the intersection of the tipped square and the square annulus unless the tip of the square is between approx 17 and 73 degrees. We are also guaranteed to have at least one side of the tipped square at least .5 units from a corner of the central square. (see the figure below -- the angle bounds are approximate)

(source)

Now, a bit of thought and playing around will convince you that the area red rectangle is a lower bound on the inaccessible area to other squares in the annulus, thus that area can be added back to the lost area.

(I did the calculations twice, found a simple arithmetic error that did not effect conclusion -- I am pretty sure this is right.)

Here is an answer with the square case: Following what Henry did, using matlab/octave calculations, one gets that 8 is the maximum.

Synopsis: you can lose a bit more than 11% of the area and still have an area that divides into 8 less than 9 times. Now note that we don't lose that much area from the intersection of the tipped square and the square annulus unless the tip of the square is between approx 17 and 73 degrees. We are also guaranteed to have at least one side of the tipped square at least .5 units from a corner of the central square. (see the figure below -- the angle bounds are approximate)

alt text http://s20.postimage.org/qrxof1f31/square_kissing_number.jpg

Now, a bit of thought and playing around will convince you that the area red rectangle is a lower bound on the inaccessible area to other squares in the annulus, thus that area can be added back to the lost area.

(I did the calculations twice, found a simple arithmetic error that did not effect conclusion -- I am pretty sure this is right.)

Here is an answer for the square case: Following what Henry did, using matlab/octave calculations, one gets that 8 is the maximum.

Synopsis: you can lose a bit more than 11% of the area and still have an area that divides into 8 less than 9 times. Now note that we don't lose that much area from the intersection of the tipped square and the square annulus unless the tip of the square is between approx 17 and 73 degrees. We are also guaranteed to have at least one side of the tipped square at least .5 units from a corner of the central square. (see the figure below -- the angle bounds are approximate)

alt text http://s20.postimage.org/qrxof1f31/square_kissing_number.jpg

Now, a bit of thought and playing around will convince you that the area red rectangle is a lower bound on the inaccessible area to other squares in the annulus, thus that area can be added back to the lost area.

(I did the calculations twice, found a simple arithmetic error that did not effect conclusion -- I am pretty sure this is right.)