This solution is hard to explain so if you don't understand my explanation please forgive me.

First consider the minimum and maximum possible areas the rectangles can cover, not neccessarily forming a square. The maximum area would be (10*9)+(8*7)+(6*5)+(4*3)+(2*1)=190. The minimum area would be (10*1)+(9*2)+(8*3)+(7*4)+(6*5)=110. The three possible squares between these two bounds are 121 (11 by 11), 144 (12 by 12), and 169 (13 by 13).

Try to tile a 12 by 12 square. It makes common sense that one rectangle should fit in each corner of the square with the fifth one in the middle. There is no other possibility since the sides of the rectangles are all different lengths. Each side of the square shall consist of the sides of two different rectangles. Thus the sides will consist of these sums: 10+2, 9+3, 8+4, 7+5. The middle rectangle will be the leftover two numbers, 6 by 1.

It also stands to reason that each side of the square will consist of the long side of one rectangle and the short side of another. To form a middle rectangle of side 6 we must subtract the side of one rectangle from the side of another, the possibilities being 10-4=6 and 8-2=6. 9-3=6 is not possible since these two dimensions must form a side of the square. In the 10-4 case we would have the 10+2 side of the square opposite of the 8+4 side. The other two opposite sides being 9+3 and 7+5, all odd numbers. Note that it is impossible to subtract one odd number from another and obtain 1. Thus the 10-4 case is impossible. Under the 8-2 case we still have the 10+2 and the 8+4 on opposite sides of the square. Again it will be impossible to form a 1 remainder with odd numbered sides only the other way. Thus the 12 by 12 square is impossible to tile.

The 13 by 13 square will have sides of lengths 10+3, 9+4, 8+5, and 7+6, with a 2 by 1 rectangle in the middle. Through a process of elimination as in the 12 by 12 case we can't help but stumple upon the solutions.

The 11 by 11 square can have sides of lengths 10+1, 9+2, 8+3, 7+4, and 6+5. Four of these will go along the sides and the fifth shall form the middle rectangle. Again it is a process of trial and error to find the solutions.

This is a picture of just one solution:

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Go back to Problem 98

Go back to Shack's Math Problems

Michael Shackleford, A.S.A