Problem
Given finitely many squares whose areas add up to $1$, show that they can be arranged without overlaps inside a square of area $2$.
[Taken from the book Putnam and Beyond by Razvan Gelca and Titu Andreescu ]
Given Proof
The guess is that a tight way of arranging the small squares inside the big square is by placing the squares in order of decreasing side length.
To prove that this works, denote by $x$ the side length of the first (that is, the largest) square. Arrange the squares inside a square of side $\sqrt 2$ in the following way.
- Place the first in the lower-left corner, the next to its right, and so on, until obstructed by the right side of the big square.
- Then jump to height $x$, and start building the second horizontal layer of squares by the same rule. Keep going until the squares have been exhausted.
Let $h$ be the total height of the layers. We are to show that $h \le \sqrt 2$, which in turn will imply that all the squares lie inside the square of side $\sqrt 2$. To this end, we will find a lower bound for the total area of the squares in terms of $x$ and $h$.
Let us mentally transfer the first square of each layer to the right side of the previous layer. Now each layer exits the square.
$\color\red{\text{It follows that the sum of the areas of all squares but the first is greater than or equal to } (\sqrt 2 − x)(h − x). \text{ This is because each newly obtained layer includes rectangles of base } \sqrt 2−x \text{ and with the sum of heights equal to } h−x}$.
From the fact that the total area
of the squares is $1$, it follows that $$x^2 + (\sqrt 2 − x)(h − x) \le 1$$
This implies that
$h \le 2x^2 − \sqrt 2x − 1x −\sqrt 2$
That is $h \le \sqrt 2$ will follow from $2x^2 −\sqrt 2x − 1x −\sqrt 2\le \sqrt 2$
This is equivalent to
$2x2 − 2\sqrt 2x + 1 \ge 0$,
or $(x\sqrt 2 − 1)2 \ge 0$,
which is obvious and we are done.
My Doubt
What I don't understand is the $\color{red}{red}$ part when the author says
It follows that the sum of the areas of all squares but the first is greater than or equal to $(\sqrt 2 − x)(h − x)$. This is because each newly obtained layer includes rectangles of base $\sqrt 2−x$ and with the sum of heights equal to $h−x$
I wonder if someone could explain why the layer includes such rectangles?
In my opinion when the square is stacked, there is always space beneath each layer since the left square of the previous layer is larger than the squares on the right (beneath this layer).
Maybe this extended image helps:
The dotted squares on the right are the copied first squares of the next row. The red rectangles are covered completely by squares (after moving the first squares down ant to the right); each red rectangle has a width of $\sqrt2-x$; the sum of the heights of the red rectangles is the sum of the green bars on the right, which equals the sum of the same green bars on the left; as the green bars on the left connect to one long bar of length $h-x$, we conclude that the total red area is $(\sqrt2-x)(h-x)$, and of course is $\le 1-x^2$.