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A square is partitioned into some rectangles. For each rectangle with side lengths $a,b$, its score is $a^2+b^2$. It turns out that the sum of scores of all rectangles is equal to the score of the big square. Is it true that all the rectangles in the partition are squares?

The converse is true: if all the rectangles in the partition are squares, then the score is simply two times the area of the square. Since the areas of the little squares add up to the area of the big square, the scores also do.

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Suppose we have rectangle with length $a$ and width $b$. Area of rectangle is $ab$. When $a=b$, the score $a^2+b^2 = 2ab$ (i.e. twice the area of the square is the score).

So we can think of the big square as having a total score twice as large as its area.

Think of ways to show that any non-square rectangles must produce a scorer greater than twice their area. That is show that:

$a^2+b^2 > 2ab$ whenever $a$ is not equal $b$.

If that is the case, then any sum of rectangle scores that includes a non-square would then sum to something above the big squares score.


Further hint: in showing $a^2+b^2 > 2ab$, it is the same to show that:

$a^2-2ab+b^2>0$

which is the same as showing

$(a-b)^2 > 0$.

Is the above line true whenever a not equal to b?

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Suppose that $a_i$ is the longer side length, so the sum of areas is $$\sum a_ib_i=a^2$$

and $$\sum a_i^2+b_i^2-2a_ib_i=2a^2-2a^2$$

$$\sum(a_i-b_i)^2=0$$

We know that all the terms of the last sum are positive, so $a_i-b_i$ must equal zero for all $i$.

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