Filling the plane with squares of harmonic areas

Abstract

Some filling the plane with squares of areas $1, 1/2, 1/3, 1/4, \ldots$ is presented.

A tiling (or a tessellation) of the plane by polygons is a covering of the plane by polygons, so that every point of the plane lies in some polygon, and the polygons have mutually disjoint interiors. Obvious tessellations by equal squares, equilateral triangles, or regular hexagons are well-known (see Fig. 1).

One of the varieties of the tessellation problem is to tile the plane using an infinite number of different tiles. A number set tiles the plane if there exists a tiling of the plane using exactly one square of sidelength n for each n in this set. Golomb (1975) raised the question of whether the set of all natural numbers tiles the plane, calling it the “heterogeneous tiling conjecture”. Henle and Henle (2008) proved that this can be done. Their algorithm of tiling can easily be used to tile the plane using one square of every rational side, i.e., ${\mathbb {Q}}$ tiles the plane. Some subsets of the natural number that tile the plane are also known. Grünbaum and Shephard (1987) presented two algorithms that generate such a subset (see also Herda 1981). Moreover, it has been found (see Henle and Henle 2015) that the even natural numbers tile the plane, however, neither the odd natural numbers nor the prime numbers tile the plane.

The inspiration for this paper is the following geometrical problem presented by Leo Moser [see problem LM6 in Moser (1991)]: can squares of harmonic sidelengths $\ 1/2, 1/3, 1/4, \ldots \$ be packed into a rectangle of area $\ \pi ^2/6-1\$ (the sum of the areas of the squares equals $\ 1/2^2+1/3^2+1/4^2+\cdots = \pi ^2/6-1$)? This problem has not been solved yet [see Ball 1996; Grzegorek and Januszewski 2019; Jennings 1995; Joós 2017; Meir and Moser 1968; Paulhus 1998; Chmielewska et al. 2023 and Tao 2023].

We say that squares $\ Q_1, Q_2, Q_3, \ldots$ fill the subset C of the plane if

• the squares are subsets of C;

• the squares have pairwise disjoint interiors;

• the set of the points of C not covered by any square has Lebesgue measure zero.

Fig. 1

Tiling the plane by equal squares, equilateral triangles, and regular hexagons

Fig. 2

Process of filling the unit square with three squares of sidelength $2^{-n}$ for $n=1,2,\ldots$

FormalPara Example 1

Figure 2 illustrates the process of filling the unit square I with three squares of sidelengths 1/2, three squares of sidelengths 1/4, ... (three squares of sidelength $2^{-n}$ for $n=1,2,\ldots$). The sum of the areas of the squares equals the area of I. It is easy to see that the uncovered part of I after packing 3n squares is the square (in the left upper corner of I) of sidelength $2^{-n}$. Since $\lim _{n \rightarrow \infty } 2^{-n}=0$, it follows that the uncovered part of I (the left upper vertex) has measure zero.

FormalPara Example 2

Eight squares of sidelengths 1/4, 16 squares of sidelengths 1/8, ... (i.e., $2^{n+1}$ squares of sidelengths $2^{-n}$, for $n=2, 3 ,\ldots$) can fill the unit square in a way presented in Fig. 3. It is easy to see that the area of the uncovered part of I after packing all squares of sidelengths $2^{-2}, 2^{-3}, \ldots , 2^{-n}$ equals $2^{-n+1}$. Since $\lim _{n \rightarrow \infty } 2^{-n+1}=0$, it follows that the uncovered part of I after packing all squares (the dotted segment) has measure zero.

Fig. 3

Filling the unit square with $2^{n+1}$ squares of sidelengths $2^{-n}$, for $n=2, 3,\ldots$

Let $S_n$ be a square of area 1/n, for $n=1,2,\ldots$ The sum of the areas of $\ S_1, S_2, S_3, \ldots$ is infinite (the harmonic series $1+1/2+1/3+1/4+\cdots$ is divergent to infinity). We will present some method that permits a filling the plane by these squares. We do not use the word “tiling”, because then each point of ${{\mathbb {R}}}^2$ would have to be covered by a square (some tile). We use the word "filling" allowing for the possibility that the set of points not covered by any square has Lebesgue measure zero.

Given a rectangle R, by b(R) denote the length of the bottom of R, and by h(R) the length of the vertical side of R. Moreover, by the width w(R) we mean $\min [ b(R), h(R)]$.

Fig. 4

Packing $S_n$ into the lower left corner of R

Fig. 5

Packing into R

Description of the filling method

• 1. The plane is tiled by congruent squares of sidelength 1 (comp. Fig. 1, left); all the squares are denoted by $K_1, K_2, \ldots$, for example in the ”spiral” manner shown in Animation 1.

• 2. The first square $S_1$ is packed into $K_1$.

• 3. The next square $S_2$ is packed into the lower left corner of $K_2$. After packing, we divide the uncovered part of $K_2$ into $\ U_2 \cup V_2$, where $U_2$ is the rectangle with $b(U_2)=1-1/\sqrt{2}$ and $h(U_2)=1/\sqrt{2}$ and where $b(V_2)=1$ and $h(V_2)=1-1/\sqrt{2}$, as in Fig. 7 or in Fig. 4, if $n=2$ and $b(R)=h(R)=1$. We define the collection ${\mathcal {R}}_{3}$ of rectangles: $\ {\mathcal {R}}_{3} =\{ U_2, V_2\}$.

• 4. Assume that $n\ge 3$, that the squares of the areas $1, 1/2, \ldots , 1/(n-1)\$ have been packed, and that the collection of rectangles ${\mathcal {R}}_{n}$ has been defined.

  • (a) If $S_n$ can be packed into $U_{n-1}$ (from ${\mathcal {R}}_{n}$), then we pack it into the lower left corner of $U_{n-1}$ (see Fig. 6, right). After packing, we divide the uncovered part of $U_{n-1}$ into $\ U_n \cup V_n$, where $b(U_n)=b(U_{n-1})-1/\sqrt{n}$, $h(U_n)=1/\sqrt{n}$, $b(V_n)=b(U_{n-1})$ and $h(V_n)=h(U_{n-1})-1/\sqrt{n}$ (see Fig. 4, if $R=U_{n-1}$). We take $\ {\mathcal {R}}_{n+1} = \bigl ( {\mathcal {R}}_{n}{\setminus } \{U_{n-1} \} \bigr ) \cup \{ U_n, V_n\}$, i.e., we include two new rectangles $U_n$ and $V_n$ in the collection, but we remove the rectangle $U_{n-1}$.

  • (b) If $S_n$ cannot be packed into $U_{n-1}$, i.e., if the width of $U_{n-1}$ is smaller than $1/\sqrt{n}$ (see Fig. 6, left), then we proceed as follows.

    • (i) If there is a rectangle in ${\mathcal {R}}_{n}$ into which the square of area 1/n can be packed, then we choose the one with the smallest width. If there are many rectangles with the same width, then we choose any of them. Denote this rectangle by R. We pack the square into the lower left corner of R (see Fig. 4). After packing, we divide the uncovered part of R into $\ U_n \cup V_n$, where $U_n$ is the rectangle with $b(U_n)=b(R)-1/\sqrt{n}$ and $h(U_n)=1/\sqrt{n}$ and where $b(V_n)=b(R)$ and $h(V_n)=h(R)-1/\sqrt{n}$; it is possible that $U_n=\emptyset$ or that $V_n=\emptyset$ (see Fig. 5). We take $\ {\mathcal {R}}_{n+1} = \bigl ( {\mathcal {R}}_{n}{\setminus } \{R\} \bigr ) \cup \{ U_n, V_n\}$, i.e., we include two new rectangles $U_n$ and $V_n$ in the collection, but we remove the rectangle R.

    • (ii) If there is no rectangle in ${\mathcal {R}}_{n}$ into which the square of area 1/n can be packed, i.e., if all rectangles in the collection have widths smaller than $1/\sqrt{n}$, then we find the smallest integer j such that the interior of the square $K_j$ has an empty intersection with any already packed square. We pack $S_n$ into the lower left corner of $K_j$. After packing, we divide the uncovered part of $K_j$ into $\ U_n \cup V_n$, where $U_n$ is the rectangle with $b(U_n)=1-1/\sqrt{n}$ and $v(U_n)=1/\sqrt{n}$ and where $b(V_n)=1$ and $h(V_n)=1-1/\sqrt{n}$. We take $\ {\mathcal {R}}_{n+1} = {\mathcal {R}}_{n} \cup \{ U_n, V_n\}$, i.e., we include two new rectangles $U_n$ and $V_n$ into the collection.

Fig. 6

Rule 4a

Fig. 7

Packing first seven squares

FormalPara Example 3

Figure 7 illustrates the initial process of packing the first seven squares. The first square is packed into $K_1$. The second square $S_2$ is packed into $K_2$. After packing $S_2$, the uncovered part of $K_2$ is divided into two rectangles: $U_2$ and $V_2$. We take ${\mathcal {R}}_3= \{ U_2, V_2\}$. Since $S_3$ cannot be packed into either $U_2$ or into $V_2$ ($1/\sqrt{3} + 1/\sqrt{2}>1$), we pack it into $K_3$ [Rule 4b(ii)]. After packing, the uncovered part of $K_3$ is divided into two rectangles: $U_3$ and $V_3$. We take ${\mathcal {R}}_4= \{ U_2, V_2, U_3, V_3\}$. The next square $S_4$ cannot be packed into any rectangle from ${\mathcal {R}}_4$ ($1/\sqrt{4}+1/\sqrt{3}>1$), so we pack it into $K_4$ [Rule 4b(ii)]. After packing, the uncovered part of $K_4$ is divided into two rectangles: $U_4$ and $V_4$. We take ${\mathcal {R}}_5= \{ U_2, V_2, U_3, V_3, U_4, V_4\}$. Since $1/\sqrt{5}\le 1-1/\sqrt{4}=b(U_4)$ (Rule 4a), we pack $S_5$ into $U_4$. After packing, the uncovered part of $U_4$ is divided into two rectangles: $U_5$ and $V_5$. We take ${\mathcal {R}}_6= \{ U_2, V_2, U_3, V_3, V_4, U_5, V_5\}$ (the rectangle $U_4$ has been removed). Since $S_6$ cannot be packed into $U_5$ we check if there is a rectangle in ${\mathcal {R}}_6$ into which $S_6$ can be packed. There are three such rectangles: $V_3$, $V_4$, and $U_3$. We choose the one with the smallest width, $V_3$ for example. We pack $S_6$ into it. After packing, the uncovered part of $V_3$ is divided into two rectangles: $U_6$ and $V_6$. We take ${\mathcal {R}}_6= \{ U_2, V_2, U_3, V_4, U_5, V_5, U_6, V_6\}$ (the rectangle $V_3$ has been removed). Since $w(U_6)\ge 1/\sqrt{7}$ we pack $S_7$ into $U_6$ (Rule 4a). The next squares are packed similarly.

Animation 1 illustrates the initial process of packing the first 50 squares. Figure 8 illustrates the situation after packing three thousand seven hundred and seventy-six squares. Squares $S_1, S_6, S_{11}, \ldots$ are marked in yellow, $S_2, S_7, S_{12}, \ldots$—in blue, $S_3, S_8, S_{13}, \ldots$—in red, squares $S_4, S_9, S_{14}, \ldots$ are marked in green, and $S_5, S_{10}, S_{15}, \ldots$—in orange.

Fig. 8

The initial process of filling the plane with squares

Observation 1. Each square $S_i$ has been packed into $K_1 \cup \ldots \cup K_i$ by the method presented above, for $i=1,2,\ldots$ (the squares are packed one by one).

Observation 2. Into each $K_j$ ($j=1,2,\ldots$) a square has been packed (the sum of the areas of the squares is infinite while the area of $K_1 \cup \ldots \cup K_j$ is finite).

Observation 3. If there is no rectangle in ${\mathcal {R}}_{n}$ into which $S_n$ can be packed for some integer n (we pack $S_n$ into the new $K_j$), then all rectangles in ${\mathcal {R}}_{n}$ have widths smaller than $1/\sqrt{n}$ [see Rule 4b(ii)].

FormalPara Lemma 1

Assume that the squares $S_1, S_2, \ldots$ have been packed by our method. Let $n \ge 3$ be an integer. If R is an arbitrary rectangle from ${\mathcal {R}}_{n}$, then the sum of the areas of squares from the sequence $S_n, S_{n+1}, \ldots$ packed in R is greater than 1/4 times the area of R.

FormalPara Proof

Let $S_{n_1}$ be the first square packed into R (in the lower left corner). Denote by $L_{n_1}$ the rectangle with $b(L_{n_1})=b(R)$ and $h(L_{n_1})=1/\sqrt{n_1}$ such that $S_{n_1 } \subset L_{n_1} \subset R$ (see Fig. 9). Let $l_1$ be the greatest integer such that

$$\begin{aligned} \frac{1}{\sqrt{n_1}} + \frac{1}{\sqrt{n_1+1}} + \cdots + \frac{1}{\sqrt{n_1+l_1}} \le b(R). \end{aligned}$$

By Rule 4a, the squares $S_{n_1}, \ldots , S_{n_1+l_1}$ are packed into $L_{n_1}$. It is easy to see that

$$\begin{aligned} \frac{1}{\sqrt{n_1}} + \frac{1}{\sqrt{n_1+1}} + \cdots + \frac{1}{\sqrt{n_1+n_1}} > n_1 \cdot \frac{1}{\sqrt{2n_1}} \ge 1 \ge b(R), \end{aligned}$$

provided that $n_1\ge 2$. Consequently, $l_1 \le n_1$ and $\frac{1}{\sqrt{n_1+l_1}}\ge \frac{1}{\sqrt{2n_1}} > \frac{1}{2\sqrt{n_1}}$, i.e., the sidelength of $S_{n_1+l_1}$ is greater than half the sidelength of $S_{n_1}$. Since the area of $S_{n_1}$ equals $h(L_{n_1})\cdot \frac{1}{\sqrt{n_1}}$, it follows that the sum of the areas of squares $S_{n_1}, \ldots , S_{n_1+l_1}$ contained in $L_{n_1}$ is greater than

$$\begin{aligned}{} & {} h(L_{n_1})\cdot \frac{1}{\sqrt{n_1}} + \frac{1}{2}\Bigl [ b(R) - \frac{1}{\sqrt{n_1}} - \frac{1}{\sqrt{n_1+l_1}} \Bigr ] h(L_{n_1}) \\{} & {} \quad = \frac{1}{2}h(L_{n_1})\Bigr [ b(R) + \frac{1}{\sqrt{n_1}} - \frac{1}{\sqrt{n_1+l_1}}\Bigr ] \ge \frac{1}{2} h(L_{n_1})b(R), \end{aligned}$$

i.e., is greater than 1/2 times the area of $L_{n_1}$.

Let $S_{n_2}$ be the first square packed into $V_{n_1}$ (the lower left vertex of $S_{n_2}$ is the upper left vertex of $S_{n_1}$). Denote by $L_{n_2}$ the rectangle with $b(L_{n_2})=b(R)$ and $h(L_{n_2})=1/\sqrt{n_2}$ such that $S_{n_2 } \subset L_{n_2} \subset R$ (see Fig. 9). Similarly, one can check that the sum of the areas of squares packed in $L_{n_2}$ is greater than 1/2 times the area of $L_{n_2}$.

In the same way we choose $S_{n_3}$, $S_{n_4}, \ldots$ and we define rectangles $L_{n_3}$, $L_{n_4}, \ldots$ contained in R. Clearly, $\ h(L_{n_1})+h(L_{n_2})+\ldots > \frac{1}{2} h(R)$ (otherwise at least one more rectangle $L \subset R$ can be created). Thus the sum of the areas of $L_{n_1}$, $L_{n_2}, \ldots$ is greater than 1/2 times the area of R. As a consequence, the sum of the areas of squares in R is greater than $(1/2) \cdot (1/2)=1/4$ times the area of R. $\square$

Fig. 9

Density of packing squares in $L_{m_i}$

FormalPara Theorem 1

The plane can be filled with the squares $S_1, S_2, \ldots$.

FormalPara Proof

We use our filling method. We will show that the uncovered part of the plane has Lebesque measure zero.

Since the plane is tiled into countable many squares $\ K_1, K_2, \ldots$, it suffices to check that the uncovered part of any $K_j$ ($j=1,2,\ldots )$ has Lebesque measure zero (the union of countably many sets of measure zero has measure zero).

Let j be a positive integer.

Moreover, let $S_l$ be the first square packed into $K_{j+1}$. Denote by $\sigma _n$ the area of the uncovered part of $K_j$ after packing $S_{n-1}$ (for $n=l+1, l+2,\ldots$) and let ${\mathcal {K}}_n$ be the collection of rectangles from ${\mathcal {R}}_n$ that are contained in $K_j$. Obviously, $\sigma _n$ is equal to the sum of the areas of rectangles from ${\mathcal {K}}_n$. Clearly, the sequence $(\sigma _n)$ is decreasing, and $\sigma _n \ge 0$ for all n. Denote by $\sigma$ the limit of this sequence. To show that the uncovered part of $K_j$ has Lebesque measure zero it suffices to verify that $\sigma =0$. Assume the opposite, that $\sigma >0$. This means that $\sigma _n \ge \sigma$ for all $n\ge l$ and that there exists an integer $m>l$ such that $\sigma _m < \frac{4}{3} \sigma$. By Lemma 1, the total area of the next squares packed into rectangles from ${\mathcal {K}}_m$ is greater than 1/4 times the area of rectangles from ${\mathcal {K}}_m$. Hence $\ \lim _{n \rightarrow \infty } \sigma _n \le \sigma _m - \frac{1}{4}\sigma _m = \frac{3}{4} \sigma _m < \frac{3}{4} \cdot \frac{4}{3}\sigma = \sigma$, which is a contradiction. This means that the uncovered part of $K_j$ has Lebesque measure zero. $\square$

Similarly to Henle and Henle (2015), the squares can fill more than one plane.

FormalPara Remark 1

The squares $\ S_1, S_2, S_3, \ldots$ can fill countably many planes.

FormalPara Proof

It is enough to determine well the squares $K_i$. Each plane is tiled by the squares of sidelength 1. All tiles are numbered. The squares that tile the first plane are denoted $\ K_1, K_2, K_4, K_7, K_{11}, K_{16}, \ldots$, i.e., are denoted by integers $\ 1+\frac{1}{2}j(j+1)$, for $j=0,1,2,\ldots$. The squares that tile the second plane are denoted $\ K_3, K_5, K_8, K_{12}, K_{17}, \ldots$, i.e., are denoted by integers $\ 2+\frac{1}{2}j(j+1)$ for $j=1,2, \ldots$. The squares that tile the n-th plane ($n=1,2,3,\ldots$) are denoted by integers $\ n+\frac{1}{2}j(j+1)$ for $j=n-1, n ,n+1, \ldots$ $\square$