Problem
GEO-B1-M06-P031 Different Strips and a Square
There is one grid rectangle of each size \(1\times1,1\times2,1\times3,\ldots,1\times N\), where \(N\ge2\). Can one choose some of them and tile a grid square of area greater than \(1\) without overlaps?
C. Hint 1. Take the longest chosen rectangle.
D. Hint 2. If a \(1\times n\) rectangle is chosen, the square side is at least \(n\), but the total area of all chosen rectangles of length at most \(n\) is small.
E. Full solution.
Assume such a square has been tiled, and let \(1\times n\) be the longest chosen rectangle. Since the square has area greater than \(1\), we have \(n>1\).
The rectangle \(1\times n\) must fit inside the square, even if rotated. Therefore the side length of the square is at least \(n\), and its area is at least \(n^2\).
On the other hand, the chosen rectangles can only be among \(1\times1,1\times2,\ldots,1\times n\). Their total area is at most \(1+2+\cdots+n=\frac{n(n+1)}{2}\).
For \(n>1\), \(\frac{n(n+1)}{2} Therefore a square of area greater than \(1\) cannot be tiled in this way.
A. Source analysis. Main objects: a set of strips, a grid square, and the maximum chosen length. The obvious approach is to try tilings, but the hidden observation is that the longest strip gives a lower bound for the square side. Number of key ideas: 2.
F. Difficulty justification. This is Level 6: a regional-style area problem with an extremal chosen object.
G. Check. This is not a one-step exercise: one must choose the maximal rectangle, get a lower bound for the square area, and compare it with the total area of all possible chosen strips.