1.4 Ordered Sets and Extremal Elements 15Solution.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 byxthe side length of the first (that is, the largest)
square. Arrange the squares inside a square of side
√
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 heightx, and start building the second horizontal
layer of squares by the same rule. Keep going until the squares have been exhausted (see
Figure 4).
Lethbe the total height of the layers. We are to show thath≤
√
2, which in turn
will imply that all the squares lie inside the square of side
√
- To this end, we will find a
lower bound for the total area of the squares in terms ofxandh. Let us mentally transfer
the first square of each layer to the right side of the previous layer. Now each layer exits
the square, as shown in Figure 4.
2 −xh−xFigure 4It follows that the sum of the areas of all squares but the first is greater than or equal
to(
√
2 −x)(h−x). This is because each newly obtained layer includes rectangles of
base
√
2 −xand with the sum of heights equal toh−x. From the fact that the total area
of the squares is 1, it follows that
x^2 +(√
2 −x)(h−x)≤ 1.This implies that
h≤2 x^2 −√
2 x− 1
x−√
2
.
Thath≤
√
2 will follow from2 x^2 −√
2 x− 1
x−