## Squaring the Square## Ross HonsbergerN 1936, FOUR STUDENTS AT TRINITY COLLEGE, Cambridge-Brooks, Smith, Stone, and Tutte--considered the problem of cutting up a rectangle into squares of unequal size (no two alike). It was known, at that time, that a rectangle 32 by 33 could be "squared" as shown in Figure 7.1. ## Figure 7.1Stone became particularly interested in trying to prove that it was impossible to cut a given square into unequal squares. While not able to do this, he did discover a squaring of another rectangle (see Figure 7.2). ## Figure 7.2One way of attacking the problem of finding rectangles that can be squared is to make a sketch of a proposed partition into squares, labelling the edge-length of each square, writing down all the relations that the edge-lengths must satisfy in order to fit into the rectangle, and solving the system of equations thus obtained. We shall carry this out for some examples. Instead of carrying as many unknowns as there are squares in the subdivision, we shall try to label neighbouring squares so that they fit together in the sketch; thus we shall have fewer unknowns to eliminate later. ## Figure 7.3
In Figure 7.3, we label the neighbouring squares x, y and z, as shown. Then it is easy to label the remaining squares in the order x + y, 2x + y, y - z, y - 2z, y - 3z, 2y - 5z. Next, we obtain relations between our unknowns; for example, we can equate the lengths of opposite sides of the containing rectangle The horizontal sides yield 2x + y + x + y = 2y - 5z + y - 2z + y - z, that is, (1) 3z - 2y + 8z = 0; and the vertical sides yield 2y - 5z + 2x + y = y - z + y + x + y, that is, (2) x - 4z = 0. So x = 4z and Y = 10Z. We observe that if we set z = 1, we obtain the tiling shown in Figure 7.1; if we set z equal to any other positive quantity, we obtain the same configuration, blown up (or shrunk) by the factor z. If we prescribe the length of the horizontal edge, say 64", then z would be 2" and the figure would be completely determined. ## Figure 7.4
In Figure 7.4, two unknowns, x and y, suffice to express the edgelengths of all compartments. Equating the lengths of the horizontal sides of the containing rectangle, we obtain
that is, The values x = 16, To show that such schemes do not always yield feasible results, we give another example. ## Figure 7.5
Starting with x and y as shown in Figure X and equating lengths of the vertical borders, we get 2x + y + 3x + y=x + y + y our middle square turns out to be a point and the remaining four squares axe equal. We have merely quartered the given rectangle, which must have been a square! Our examples indicate that
the system of linear equations we get from an arbitrary sketch of a partition
seems to have a unique solution (except for a scaling factor), although the
solution is not necessarily geometrically feasible. It could happen, for example,
that some of the edge-lengths turn out to be negative, and that would not make
sense in the context of our tiling problem.From
our previous experience with systems of linear equations we know that there
may be too many equations (in which case there is no solution) or there may
be too ## Figure 7.6To make things
more precise, consider a given subdivision of a rectangle into squares. With
respect to each horizontal segment of the partition (see Figure a + b = c + d + e telling us that
the sum of the lengths of the edges of squares bordering the segment on one
side is equal to the sum of the lengths of the edges
of squares bordering
the segment on the other side. We shall call all such relations
THEOREM.
Imagine that the rectangle is a plate of uniform thickness made of some conducting
metal. Suppose all points of the upper edge of the rectangle were given the
same electric potential V, and all points on the lower edge had a lower constant
potential V'. (This could be accomplished by coating these edges with some perfect
conductor.) Because of the prescribed potential difference V - V', there will
be a steady flow of current through the rectangle, in the vertical direction.
The rate at which electrons cross over a horizontal interval is proportional
to the length of that interval. Thus, if I is the current crossing over a horizontal
interval of unit length, then the current flowing across an interval of length
I is 11. The Now suppose
that such a current-carrying rectangle has been partitioned into squares. Since
the flow is vertical (there is no flow in the horizontal direction), we may
cut through vertical segments without interfering with the current. Now regard
the cut-up plate as a
The magnitude of the current flowing through each wire is proportional to the length of the side of the square represented by that wire. ## Figure 7.7The law of conservation
of current states that, at each vertex of the graph, the total amount of current
flowing in equals the total amount of current flowing out. For example, the
vertex C (corresponding to the horizontal segment at level c) receives current
which is just one of our horizontal compatibility relations.
Since current flows in at the upper edge of the plate, moves down vertically and flows out at the lower edge, we see that every point at the same horizontal level is at the same potential, and points at higher horizontal levels have higher potentials than points at lower horizontal levels. According to Ohm's law,
network, say
This is one of our vertical compatibility relations.
The law of conservation of current and Ohm's law yield a system of linear equations equivalent to the system S of compatibility relations. Now it seems plausible that if a difference in potential is prescribed for a network such as ours (say the potentials at vertices
Since these physical laws axe equivalent with the geometrical compatibility relations (the prescribed potential difference corresponding to the vertical dimension of the rectangle), the assertion that the system S has a unique solution (see p. 52) is seen to be just a corollary of Kirchhoff's theorem. A rigorous proof of Kirchhoff's theorem is beyond the scope of this essay.$ Let us now re-examine the problem of partitioning a rectangle into squares. First we note that a solution of the linear system S may include
non-positive quantities. These cannot be made to correspond to lengths of sides
of subdivision squares, so such solutions of the system S are not solutions
of the tiling problem.
We might try to attack the rectangle-tiling problem by first devising a systematic method of enumerating all possible partitions into
A squaring or tiling in which no two squares are the same size is said to be "perfect". Tutte and his friends were chiefly after perfect tilings. They wanted to find a "perfect square", that is, a square with a perfect tiling. All their results involved non-square rectangles. And so they began to think that a perfect square did not exist. However, in 1939 Roland Sprague of Berlin found one, and, since then, many others have been discovered. Attention was then turned to finding the perfect square with the smallest number of squares in it (i.e., of lowest order). The record to date is one of order 24, found by an amateur mathematician, T. H. Willcocks of Bristol, England. (See Figure 7.8.) ## Figure 7.8 (Compound) perfect square of order 24.
A tiling tiling is said to be "simple" if the arrangement of the squares does
not form any smaller squared rectangle inside the original. Willcocks' perfect
square is compound, not simple. Accordingly, attention was focussed on the
It has been shown, using computers, that there is no perfect square of order less than 20, simple or otherwise. Consequently, there is not much room for improvement in the above records. So many perfect squares have been found that experts in the field suspect that tiled rectangles of unequal sides are the harder to come by.
We close this essay with the following easy-to-prove theorem:
## Figure 7.9. Simple perfect square of order 25
## Figure 7.10. View of bottom rectangle of the box.This smallest cube on the bottom, being out in the middle part of the bottom, must be bordered on every side by a bigger cube (it is the smallest). Its upper surface, then, is completely walled in; see Figure 7.11. In order to cover this upper surface, even smaller cubes must be used. ## Figure 7.11Among the cubes on the upper surface of S, the smallest, again occurring in the middle part, is surrounded by bigger ones. Consequently, even smaller ones still must occur in a third layer on the top of this walled-in cube. This argument continues without end, implying that there is no end to the number of cubes that must be employed. EXERCISES ## Figure 7.12Prove that
there exists some rectangle which can be tiled into
In tiling an equilateral triangle with unequal equilateral triangles, prove: (b) the smallest
tile, T, touching the top of S meets the top of S in just ## Figure 7.14(c) Does this lead to an endless tower of little triangles, providing a proof of the theorem "it is impossible to tile an equilateral triangle with unequal equilateral triangles (no two alike)", which was first proved by Tutte around 1948? ## FOOTNOTES+ Some headway in this direction has been made, for example, by applying Euler's formula to the graphs of our networks. + A path in our
network leads from a vertex along an edge
$The
interested reader may look it up, for example, in R, E. Scott,
## REFERENCESJ. Bouwkamp, Ak. van Wet.,
1946, Vol. 49, pp. 1176-1188; 1947, Vol. 50, pp. 1296-1299. |