9 Lattice-Point Polygons

In this chapter, we explore the nature of regular polygonal figures that might be found in various kinds of lattices, including the integer lattice, the triangular lattice, the hexagonal lattice, and other more general lattices.

9.1 Which regular polygons can be found using vertices of the 2 × 1 brick tiling? Prove your answer.

Theorem. The figures that can be formed with lattices in the infinite 2 × 1 regular brick lattice are exactly the same as those in the 1 × 1 integer lattice, because these are in fact the same lattice. Consequently, the only regular polygons to be found in the brick lattice are squares.

Proof. Consider the 2 × 1 regular brick tiling, where by regular I mean that on each subsequent row, the midpoint of each brick lines up with the edge between two bricks on the next row. If you consider carefully the lattice points of the brick lattice, the main observation to be made is that they are exactly the same as the lattice points of the 1 × 1 integer lattice.

The point is that the long-side midpoints as well as the corners are lattice points in the brick lattice, because each such midpoint joins the corners of bricks in the next row. For the purpose of finding those vertex points, therefore, we are in effect tiling with the bricks cut in half, with 1 × 1 tiles, and this gives us the integer lattice. The figures to be found in the brick tiling, consequently, are the same as those in the integer lattice. For regular polygons, theorem 70 in the main text shows that only squares are to be found.

9.3 Prove that in the square lattice, any line segment joining two lattice points is the side of a square whose vertices are lattice points.

Theorem. In the square lattice, any line segment joining two lattice points is the side of a square whose vertices are lattice points.

Proof. We may consider the integer lattice, consisting of points (x, y) where x and y are integers. Suppose that A and B are lattice points. Notice that if we rotate the entire lattice by 90^° around any lattice point, the lattice lands directly upon itself, with the rows landing on columns and columns on rows. Consider a counterclockwise such rotation around the point A. This rotation takes the point B to a point B′.

A similar rotation centered at B will bring A to a lattice point A′, and these four points form a square, since the distances are the same and the angles are 90^°. So AB is the side of a lattice-point square.

9.5 Prove that the square lattice is not necessarily invariant under the reflection swapping two lattice points.

It is clear that the reflection swapping any two vertices on the same horizontal row or the same vertical column will preserve the lattice. And also, the reflection swapping any two vertices on a 45^° diagonal from each other will preserve the lattice. But it turns out that these reflections are the only lattice-preserving reflections.

Theorem. The square lattice is not invariant under all the reflections swapping any two lattice points.

Proof. It suffices to find such a reflection that doesn’t take the lattice to itself. Consider the reflection swapping two opposite corners of a 2 × 1 rectangle. We have oriented the lattice here on an angle, so that the two points A and B lie horizontally, and so the reflection is through the dotted vertical line between them.

Thus, the original lattice is reflected to the new lattice, and clearly, the transformation does not take lattice points to lattice points.

9.7 Can some nonsquare regular polygons arise from lattice points in some rectangular lattice, not necessarily square? Exactly which regular polygons can arise in such a lattice?

First, it is easy to see that some nonsquare regular polygons can be made in a rectangular lattice. For example, consider a unit rectangle of size . This is exactly the right distance to form an equilateral triangle as follows:

Let us attempt to provide a completely general answer to the question, by classifying the possibilities for regular polygons in a rectangular lattice.

Theorem. Every rectangular lattice is one of the following types:

1. No regular polygons of any kind can be formed using lattice points.

2. Squares can be formed from lattice points, but no other regular polygons.

3. Equilateral triangles and regular hexagons can be formed from lattice points, but no other regular polygons.

And furthermore, there are rectangular lattices of each of these types.

Proof. First of all, observe that by theorem 76 in the main text, we can never find any regular polygon in any kind of lattice, including any rectangular lattice, except possibly triangles, squares, and hexagons. So we need to consider only these three cases. Also, equilateral triangles and regular hexagons are to be found together, by the arguments in the main text, since from any equilateral triangle we may form a regular hexagon by translating along lattice line segments. And if we may form a hexagon, we may also form one twice as big (by side length), thereby forming a hexagon whose center is also a lattice point. But this will make six equilateral triangles. So the triangles and hexagons come together.

Next, let us argue that squares and equilateral triangles can never arise in the same rectangular lattice. Suppose that a rectangular lattice contains a square. Take two adjacent vertices of the square, not on the same vertical line. So their horizontal displacement d is an integer multiple of the base width w of the unit rectangle, d = nw. The adjacent edge of the square has exactly the same vertical displacement d, which must be an integer multiple of the base height h of the unit rectangle d = mh. So nw = mh and therefore are commensurable; both w and h are integer multiples of the common unit length h/n. So we can refine the rectangular lattice to a square lattice, using that common unit. Theorem 70 in the main text shows that there are no equilateral triangles to be found in that integer lattice, and so none can be found in this rectangular lattice either.

Last, let us show that all three types of rectangular lattices exist. First, the square lattice is a rectangular lattice having squares but no other regular polygons, as proved in the main text. Next, we observed above that in the rectangular lattice with unit rectangle of size , one may find equilateral triangles. The arguments above show that this lattice will therefore also exhibit regular hexagons, but not squares. Finally, consider a rectangular lattice using a base rectangle 1 × h, where h is irrational and not a rational multiple of . For example, let h be transcendental. In this case, there can be no square, since as previously we saw that if a rectangular lattice has a square at lattice points, then the base and height of the unit rectangle will be commensurable, which they are not here. If there is an equilateral triangle, then we may assume one vertex is at the origin. Let A be another vertex of the triangle not on the same vertical line as the first point. So it is at coordinate (n, mh) for some integers n and m. The third vertex B is obtained by rotating A by 60^°, and so using the rotation matrix we see that it has x-coordinate . This must be an integer, in order for the point to be on the lattice, and so also must be an integer k, which means , contrary to the choice of h. So for such a value of h, there can be no equilateral triangles and hence no regular hexagons, and also no squares. So all three types of rectangular lattices occur.

9.8 Prove that in any rectangular lattice, using a rectangle whose side lengths are commensurable, the only regular polygon to be found using lattice points is a square.

Theorem. The only regular polygon to be found at lattice points in a rectangular lattice, based on a unit rectangle having commensurable side lengths, is a square.

Proof. Consider a rectangular lattice based on a unit rectangle having commensurable side lengths. Commensurability means that the sides have a common unit measure; each of them is an integer multiple of some common unit length, perhaps very small. Let us imagine constructing a square lattice using that common unit length. Because of commensurability, our rectangular lattice points will arise as points in that fine square lattice. In other words, the rectangular lattice is refined by the square lattice based on the common unit. And so any regular polygon to be found in the rectangular lattice can also be found in the square lattice. By theorem 70, the only such figures are squares.

9.9 Using colored chalk on the tiled plaza in the town’s market square, a child connects the centers of some of the hexagons like this:

Has she made any squares?

Well, since this is a child playing with mathematical figures, I am likely to encourage her regardless and ask her also to draw other kinds of figures; the “squares” are surely square enough for a child’s purpose. But if she is mathematically curious and serious about the question, wanting to know whether in fact any of them are exactly square, as a mathematical fact, then I should be delighted to discuss it further with her. The answer is that one cannot actually make a perfect square by joining the centers in a hexagonal lattice. In the image, the child’s rectangles are aligned with the hexagonal sides, but even if she had drawn them with an inclined orientation, she still will not have made any squares.

Theorem. No square can be made by joining the centers of hexagons in the hexagonal lattice.

Proof. We refer here, of course, to nondegenerate squares. By including the center points of the hexagons into the lattice we would form exactly the triangular lattice, because each hexagon consists of six equilateral triangles formed by joining its center to its outer vertices. If we could form a square using the hexagonal centers in the hexagonal lattice, therefore, then we would be able to form a square in the triangular lattice. But theorem 73 shows that there is no square to be found in the triangular lattice. And so we cannot have found such a square using the centers of hexagons in the hexagonal lattice.

Alternative proof. An alternative argument could proceed without the triangular lattice observation by observing that the centers of the hexagons are all uniformly offset from a hexagonal vertex by the same vector. So by translating the figure, we would find a square using lattice points in the hexagonal lattice, which we know is impossible by theorem 73.

Meanwhile, let us also observe that by making the attempted square large enough or, equivalently, by using a fine enough hexagonal lattice, we can make a rectangle that is as close to being square as desired. For example, we can calculate the aspect ratios of the nearly square figures the child has already drawn. If the hexagons have unit side length, then each unit hexagon has width 2 and height . The small dark rectangle therefore has size , for an aspect ratio of 1.15. The medium-sized rectangle is , for a slightly better aspect ratio of 0.96. And the large rectangle is , for an aspect ratio of 1.01, which is very nearly square.

9.11 Prove that there are no regular polygons, except squares, in the rational plane ℚ × ℚ.

Theorem. No regular polygons, except squares, can be formed by vertices in the rational plane ℚ × ℚ.

Proof. Suppose a figure is formed by finitely many points in the rational plane ℚ × ℚ. Since the rational numbers are closed under integer multiples, we may find a scaled-up instance of this figure by multiplying through by a common multiple of the denominators of the rational numbers arising as coordinates of points in the figure. Thus, we find a scaled-up version of the figure in the integer lattice. Since there are no regular polygons except squares in the integer lattice, it therefore follows that there can be none in the rational plane ℚ × ℚ.

9.13 Prove that the ratio of any two parallel line segments joining points in the integer lattice is a rational number. Is this true without the “parallel” qualifier?

Let us begin with the first part of the question.

Theorem. The ratio of the lengths of any two parallel line segments joining points in the integer lattice is rational.

Proof. Let us first give a soft proof. Consider a line segment in the integer lattice, such as the segment AB pictured here. The length of the line segment will be an integer multiple of the length of the most basic segment having that slope. In the instance here, for example, AB has slope 2/3 and consists of three segments of the 3 × 2 diagonal.

Since the lengths of any two parallel segments will thus be multiples of the same unit length in that direction, the ratio of their lengths will be rational.

We can also give a messier but more explicit proof using coordinates.

Alternative explicit proof. Suppose that segment AB joins the point A = (a₀, a₁) to the point B = (b₀, b₁) and that it is parallel to segment CD joining C = (c₀, c₁) to D = (d₀, d₁). Since the segments are parallel, they have the common slope

The respective lengths are

and

Putting this together, we may compute the ratio of the lengths, and then factor out a term:

The compound radical expression in the middle drops out, because it is 1 by the common slope identity. Therefore, the length ratio is a rational number.

Observation. The previous theorem is not true without the “parallel” qualifier.

Proof. The ratio of the diagonal of the unit square to the side is , which is not rational.

9.15 Explain what happens precisely with the parallelogram argument proving theorem 76 when it is applied to triangles, squares, or hexagons.

Theorem 76 in the main text asserts that there are no regular polygons to be found in any lattice other than triangles, squares, and hexagons. The proof of this argument is concerned with regular polygons other than these three. One should form all the parallelograms from two adjacent edges of the regular polygon, taking in each case the resulting fourth point of the parallelogram, which we know must also be a lattice point; one thereby produces a smaller instance of the regular polygon. This is a contradiction, if one already had the smallest possible instance. And so for these kinds of regular polygons, there are none to be found in any lattice.

The argument breaks down with triangles, squares, and hexagons, however, as it must, since triangles, squares, and hexagons can be found in certain lattices. But let us see exactly how the argument breaks down; let us see what happens with the construction when it is applied to triangles, squares, and hexagons.

With an equilateral triangle, the vertices formed from the parallelograms constructed from adjacent edges will give you a much larger triangle, as indicated at the left. This breaks the minimality part of the argument, since the new triangle is not smaller than the original triangle, and so there is no contradiction.

With the square, the parallelograms formed by adjacent edges rebuild the very same square again, since the fourth vertex is already on the square. This also breaks the minimality argument, since the new square is not smaller than the original; it is indeed exactly the same as the original. With the hexagon, the fourth point of each parallelogram is the center point of the hexagon. Thus, the new vertices do not form another smaller hexagon at all, only a single point (a degenerate hexagon). So in each of the three cases, the argument toward contradiction breaks down, and no contradiction is made. This is of course to be expected, since we already know how these three kinds of regular polygons can occur in lattices.

9.17 For each integer n ≥ 3, let d(n) be the smallest dimension such that there is a regular planar n-gon realized at points of rational coordinates in dimension d(n), if possible. For which values of n is d(n) defined, and what is the complete list of values of the function d(n)?

Theorem. For each integer n ≥ 3, let d(n) be the smallest dimension for which a regular n-gon can be realized at rational coordinates in that dimension, if possible. Then the following is the complete list of values where d is defined:

Proof. By theorem 76, we know that the only regular polygons to be found in any lattice whatsoever are triangles, squares, and hexagons, and so only these types of regular polygons will be found in the rational lattice of any dimension. So d(n) is defined at most on n = 3, 4, and 6. Clearly, the square can be found in dimension 2, but not smaller, so d(4) = 2. Meanwhile, there are no triangles or hexagons in the two-dimensional integer lattice and therefore also not in the two-dimensional rational lattice. So d(3) and d(6) are at least 3. But in fact, they are both exactly 3, because one can form an equilateral triangle from three vertices of a cube, by slicing off one corner, and one can form a regular hexagon by joining the midpoints of certain sides in the cube.

In fact, we have already argued in exercise 9.7 that once you have formed an equilateral triangle in a lattice, then you may also form a hexagon and conversely, and so we knew that d(3) = d(6), and now we know furthermore that d(3) = d(6) = 3.