Professor Erasmus claims that he is able to cut a square into 100 rectangles by making nine horizontal cuts and nine vertical cuts (parallel to the sides of the square), so that

  • exactly 9 of the resulting 100 rectangles are squares, and
  • no two of these 9 resulting squares have the same area.

The professor modestly calls this the "Professor-Erasmus-dissection-of-a-square". Has the professor once again made one of his mathematical blunders, or does such a dissection indeed exist?


3 Answers 3


It is not possible.

Let v be the set of separations between adjacent vertical lines. Let h be the set of separations between adjacent horizontal lines.

  • Every time an element of v matches an element of h, we have a square.
  • No element of one set may match more than one element of the other set, or we would have two equal sized squares.

Thus we must have 9 identical elements in each set, but each set has only 10 elements and the same sum, therefore the 10th element is equal.


No it is not possible.

If we take x as the width of the original cut, and increment it by y each time, then there will be 10 squares with sides length (x+ny) and finally (x-36y) which would have different areas to one another. All other rectangles would be of the form (x+ny) x (x+my). In order for this to work for a square with width 10x, y must be < x/36.

In order to get 9 squares

In order to assume that there are exactly 9 squares we have to move one of the cuts, either horizontal or vertical, from the x+ny pattern. Lets assume vertical, so that we have a square in the penultimate column which is not in the diagonal line of the others. In order to do this, we simply adjust the width of the cut from (x+8y) to (x+my) where m = 0, 1, 2, 3, 4, 5, 6, or 7 creating a final row with height x-36y (with 36y < x) and final column with width x-(36-m)y which obviously doesn't form a square. This produces a square in all but the final column.


In order to produce a square in the penultimate column, it must have for its width the height of one of the rows. This means that the square will HAVE to be the same size as one of the other squares.


all of the cuts as described but with the final vertical cut so that no squares are produced in the penultimate or final column would produce 8 internal squares, and the original square can still be formed. This would produce 9 non-identical squares. But there would then be 101 rectangles produced.


If it needed to be 10 squares it would be quite easy. lets say you cut at distance 1, 3, 6, 10, 15, 21, 28, 36 and 45 of a 55 by 55 square. both in the x and y direction. Then you will have 10 squares along the diagonal with sides of length 1,2,3,4,5,6,7,8,9 and 10.

Now moving any of the cuts would remove 2 squares so I am inclined to say that it is not possible. Not sure if this is proof enough though


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