How Many Squares Make a Pattern?

A good problem should be more than a mere exercise; it should be challenging and not too easy to solve by the student, and it should require some "dreaming" time. —Howard Eves

Now, we shift our focus to one puzzle of a distinct and unique class of geometric puzzles. In this context, the physical pieces involved may appear ordinary; however, the way they are arranged presents an amusing combinatorial challenge. This particular problem originates from the Mega Test, once considered “the world’s hardest IQ test,” published in April 1985, and I must admit that, prior to encountering it there, I had never seen a puzzle quite like it. Specifically, we must determine which segments can be considered exposed edges, how various stacks must interleave, and the methods needed to demonstrate the optimality of our solution (in this case, we opted for an elementary approach).

To address this intriguing puzzle, we will first present a possible configuration. We will then systematically work through the problem, outlining the steps that lead to a solution. Through a structured approach, we aim to discover a useful approach and achieve a solution that not only satisfies its constraints but also deepens our understanding of similar problems.

How Many Squares Make a Pattern?

What is the minimum number of square sheets of paper sufficient to reproduce the pattern below if the sheets—unfolded, uncut, unmarked, and opaque—are placed flat on top of one another such that each line shown corresponds to the edge of some square insofar as it is not occluded by any overlapping square? Assume the pattern is exclusively formed using squares.

Nine sheets of paper are needed. One large sheet is placed first. Then, three squares of \(\frac{1}{4}\) the size of the largest sheet. One of the squares is placed so that one of its sides is contained in the left side of the largest square, such that a perpendicular bisector can be drawn from the left side of the largest square to the center of the smaller square. The second is placed with one of its sides contained in the bottom side of the largest square, enabling a perpendicular bisector to be drawn from the bottom side of the largest square to the center of this smaller square, covering one-fourth of the area of the first. The third square is placed in the lower right quadrant, covering half of the area of the second. Now add the diamond square inside this last square for a total of five squares. Finally, add four squares of \(\frac{1}{16}\) the size of the largest sheet in the appropriate places for a total of nine (See the figures below).

The diagrams are not drawn at scale or with exact measurements.

Note that this number is the minimum because

• One square sheet is necessary to contain all the others. 
• A second of \(\frac{1}{4}\) the size of the largest sheet is required in the lower right quadrant so that the diamond square rests on it.
• The third is the diamond square inside the aforementioned square sheet.
• At least two additional sheets are required to ensure that at least one of their edges is perpendicular to the edge of the largest square and positioned one-fourth of the distance from one corner of the largest square to another (Two, since there are at least two lines drawn in this fashion, each line being perpendicular to a different side of the largest square and perpendicular to each other)
• At this point, there are only two possibilities: either we can add smaller squares in such a way as to take advantage of the edges already there, or add another larger square of \(\frac{1}{4}\) the size of the largest sheet. In either case, we will need four squares. For if we take the latter approach, we need to place \(3\) larger squares midway between the corners of the largest square (each one having an edge contained in a different side of the largest, none of which is the side that corresponds to the square of the same size that occupies the lower right quadrant), thereby giving the effect of two smaller corner squares at most, and needing other \(3\) smaller squares in the right places to fulfill the pattern. On the other hand, the former approach is described above in the first arrangement provided in the solution.\(^{1}\)

1. Remember that this is a \(3D\) problem, and the order in which the square sheets are arranged does not necessarily reflect the order in which we describe their required quantity to reproduce the pattern.

References

Adapted from Hoeflin, R. K. (1985, April). Problem 27 in The Mega Test. Omni, 7(4), p. 129.