Wikipedia:Reference desk/Archives/Mathematics/2012 March 19

= March 19 =

Partitioning a square into five subsquares
It is a nice exercise to show that any square can be partitioned into n subsquares, for any n greater than or equal to six. However, even though it is "obvious" that it can't be done for exactly five subsquares, I haven't been able to think of a proof. Does anyone have any suggestions?

Thanks.50.103.231.24 (talk) 03:14, 19 March 2012 (UTC)


 * It may not be elegant but you can reach a formal proof by considering cases. For example, first show that at least one side of the large square has to be split between exactly two squares. Then show this cannot give a solution if the two squares are the same size, and also not if they are different sizes. PrimeHunter (talk) 03:37, 19 March 2012 (UTC)


 * The trick is the following. If you partition a square to more than one subsquares, then every corner of the original square must be contained by a different square in the partition.  This accounts for four of the five squares, so the fifth square could lie on only at most one side of the big square.  Thus, for three of the four sides of the big square, that side must be contained by the two corner subsquares.  This means the side length of those two corner squares must add up to the side length of the big square.  If this is true for three sides of the big square, that must imply either that all four corner squares are exactly half the size of the big square, in which case you have a partition to four squares, or if two of the corner squares have size greater than half of the big squares, which is impossible as those squares would overlap.  (The idea of the proof is from Reiman István, A geometria és határterületei)  &#x2013; b_jonas 08:30, 19 March 2012 (UTC)

Is the formula for the edge largest and smallest possible subsquare in a n-partition of the unit square the following? While I have a fairly good feeling for this, the formula for the edge of the smallest possible subsquare seems to be uglier and based on the value of n mod 3. Naraht (talk) 20:58, 19 March 2012 (UTC)
 * If n is even, (n-2)/n (split into an n-1/n square and n-1 1/n squares)
 * If n is odd, (n-5)/n (take the split for n-3 and then split one of the small squares into 4 pieces)
 * If n mod 3 =1, 1/(2^((n-1)/3)) (start with quarters and keep breaking down one of the smallest squares into quarters
 * If n mod 3 =2, 1/(2^((n-2)/3)) (start with a 3/4x3/4 and 7*1/4x1/4 and then keep ...)
 * If n mod 3 = 0, 1/(3*2^((n-6)/3)) (start with a 2/3x2/3 and 5*1/3x1/3 and then keep ...)
 * I don't think the minimum lengths are right. There is a partition for 8 with a 3/5x3/5, 3 2/5x2/5, and 4 1/5x1/5.  I don't know if 1/5 is the smallest possible for 8 or not.  In general it seems like a hard problem to account for all possible partitions for a given n. Rckrone (talk) 04:34, 20 March 2012 (UTC)
 * There exists a partitioning with 24 squares where the smallest subsquare has a side of 1/175. See also Squaring the square 84.197.178.75 (talk) 10:05, 20 March 2012 (UTC)

Thanks, everyone, for your replies. 97.64.192.126 (talk) 12:47, 21 March 2012 (UTC)