Translative Covering a Square with Isosceles Right Triangles

Abstract

A translative covering the rectangle $a\times b$ with homothetic copies of a right isosceles triangle T (of the legs parallel to the sides of $a\times b$) is considered. It is shown that any collection of equal triangles homothetic to T with the total area at least 2 permits a translative covering of $1\times 1$; this bound is tight. It is also demonstrated that any collection of positive homothetic copies of T with the total area at least 3 permits a translative covering of $1\times 1$. Moreover, it is proven that if $a\ge {\displaystyle \frac{5+\sqrt{33}}{4}}b$, then any collection of triangles homothetic to T with the total area at least ${\displaystyle \frac{1}{2}}{(a+b)}^2$ permits a translative covering of $a\times b$; this bound is tight.

1. Introduction

A collection (a sequence) $C_1,C_2,\dots$ of planar convex bodies permits a covering of a planar convex body D, if the sets $C_1,C_2,\dots$ can cover D, i.e., if it is possible to apply translations and rotations to the sets $C_n$ ($n=1,2,\dots$), then the resulting translated and rotated bodies $C_n^{\prime }$ cover D, i.e., $D\subseteq \bigcup C_n^{\prime }$. If we use only translations, then we get a translative covering.

It is known that any collection of squares whose total area is not smaller than 3 permits a translative covering of a unit square (see [1,2,3]). If rotations are allowed, then the lower bound 3 can be reduced to 2 (see [4]). It is also known [5] that any collection of homothetic copies of a triangle $\mathcal{T}$ whose total area is not smaller than two times the area of $\mathcal{T}$ permits a translative covering of $\mathcal{T}$. All these bounds are tight.

Let T be a right isosceles triangle and let I be the unit square with the sides parallel to the legs of T. Song and Su [6] proved that any collection of positive homothetic copies of T with the total area not smaller than 2 permits a covering of I. The generalization of this result for the covering rectangles is given in [7]. The lower bound 2 presented in [6] is tight. However, both translations and 180° rotations are used for the covering. In this note, we do not use rotations. We are not able to generalize this result for the covering I with unequal triangles, but we are able to obtain the tight lower bound for the covering I with equal triangles, as well as for the covering of some rectangles with unequal triangles.

In Section 2 we will show that any collection of congruent triangles homothetic to T with the total area not smaller than 2 permits a translative covering of I; this lower bound is tight.

In Section 3, we will prove that any collection of positive homothetic copies T with the total area not smaller than 3 can translatively cover I; it is very likely that this lower bound is not tight.

Let $R=a\times b$ be a rectangle of sidelengths a and b, with the sides parallel to the legs of T. In Section 4 we will show that if $a\ge {\displaystyle \frac{5+\sqrt{33}}{4}}b$, then any collection of positive homothetic copies of T with the total area not smaller than ${\displaystyle \frac{1}{2}}{(a+b)}^2$ permits a translative covering of R; this lower bound is tight.

2. Translative Covering a Square with Equal Triangles

Lemma 1.

Two congruent right isosceles triangles with legs of length $a>0$ can cover the rectangle $\lambda \times \left({\displaystyle \frac{3}{2}}a-\lambda \right)$ for ${\displaystyle \frac{1}{2}}a\le \lambda \le a$.

Proof. 

Let $\lambda \in \left[{\displaystyle \frac{1}{2}}a,a\right]$ and let $T_a^1,T_a^2$ be right isosceles triangles with legs of length a. Moreover, let $T_a^2$ be the image of $T_a^1$ under the translation by the vector $\left(\lambda -{\displaystyle \frac{1}{2}}a,a-\lambda \right)$. Then the triangles $T_a^1$ and $T_a^2$ cover the rectangle $\lambda \times \left({\displaystyle \frac{3}{2}}a-\lambda \right)$ (see Figure 1).    □

Lemma 2.

The right isosceles triangle with legs of length $a>0$ can cover the rectangle $\mu \times \left(a-\mu \right)$ for $0<\mu <a$.

Proof. 

Let T be a right isosceles triangle with legs of length a. Consider a rectangle R with side lengths $\mu$ and $a-\mu$, where $0<\mu <a$. Place T in such a way that its right-angle vertex coincides with a vertex of R, and the legs of T contain the two sides of R incident with that vertex (see Figure 2).

Two vertices of the rectangle lie on the legs of T at distances $\mu$ and $a-\mu$ from the right-angle vertex; since $0<\mu <a$, these points are not the endpoints of the legs. The fourth vertex of R lies on the hypotenuse of T, since the sum of the lengths of the sides of R adjacent to this vertex is $\mu +(a-\mu )=a,$ which equals the length of the legs of T.

Hence, the entire rectangle R is contained in T. Therefore the triangle T covers the rectangle $\mu \times (a-\mu )$.    □

For covering with one triangle, we take $\mu =1$ and $a=2$ in Lemma 2.

Remark 1.

A single triangle of area 2 (i.e., with $a=2$) can cover I.

It is clear that the number 2 cannot be replaced here by any smaller one. Therefore, the lower bound 2 presented in Theorem 1 is tight.

For covering with two equal triangles, we take $\lambda ={\displaystyle \frac{3}{4}}a$ and $a={\displaystyle \frac{4}{3}}$ in Lemma 1.

Corollary 1.

Two congruent right isosceles triangles with legs of length $a>0$ can cover the square with the sidelength ${\displaystyle \frac{3}{4}}a$.

Corollary 2.

Two congruent right isosceles triangles with legs of length ${\displaystyle \frac{4}{3}}$ (with the total area $2·{\displaystyle \frac{1}{2}}·{\left({\displaystyle \frac{4}{3}}\right)}^2={\displaystyle \frac{16}{9}}$) can cover the unit square I.

The covering with three and four equal triangles is presented in Figure 3 and Figure 4.

Remark 2.

Three congruent right isosceles triangles with legs of length ${\displaystyle \frac{8}{7}}$ (with the total area $3·{\displaystyle \frac{1}{2}}·{\left({\displaystyle \frac{8}{7}}\right)}^2={\displaystyle \frac{96}{49}}$) can cover the unit square I.

Remark 3.

Four congruent right isosceles triangles with legs of length 1 (with the total area $4·{\displaystyle \frac{1}{2}}·1^2=2$) can cover the unit square I.

Theorem 1.

Any collection of n congruent triangles ($n=1,2,\dots$) homothetic to T with the total area not smaller than 2 permits a translative covering of I.

Proof. 

Consider a collection of n congruent triangles $T_a$, each homothetic to T, with the total area not smaller than 2 and let a denote the length of the legs of $T_a$. Since the area of one triangle is equal to ${\displaystyle \frac{1}{2}}{a}^2$, the condition $n·{\displaystyle \frac{1}{2}}{a}^2\ge 2$ implies that

$$a\ge {\displaystyle \frac{2}{\sqrt{n}}}.$$

If $n\in \{1,2,3,4\}$, a collection of n triangles with the total area 2 can cover I (see Remark 1, Corollary 2 and Remarks 2 and 3).

Assume that $n>4$.

In the covering process, we will first cover the square with an appropriate family of rectangles. Then, the triangles from the collection will be placed in such a way that they cover these rectangles.

In Figure 5, Figure 6, Figure 7, Figure 8, Figure 9, Figure 10, Figure 11 and Figure 12, any rectangle with a marked diagonal will be covered with two triangles (as in Lemma 1), and any rectangle without a marked diagonal will be covered with a single triangle (as in Lemma 2). For example, in Figure 5, left, the square of sidelength ${\displaystyle \frac{3}{4}}a$ will be covered with two triangles, while the square of sidelength $2-{\displaystyle \frac{7}{4}}a$ will be covered with one triangle.

In many cases, we will use one of the following two methods.

In the method $M_1\left(k\right)$ (for example, see Figure 6, middle, where $k=2$) the unit square is covered with $k^2$ rectangles of sidelengths ${\displaystyle \frac{1}{k}}$ and ${\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}}$ and with $n-2k^2$ rectangles of sidelengths $1-k({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}})=2-{\displaystyle \frac{3}{2}}ka$ and ${\displaystyle \frac{1}{n-2k^2}}$. By Lemma 1 (we take $\lambda ={\displaystyle \frac{1}{k}}$), each rectangle ${\displaystyle \frac{1}{k}}\times \left({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}}\right)$ can be covered with two triangles $T_a$. Moreover, by Lemma 2 (we take $\mu =2-{\displaystyle \frac{3}{2}}ka$), the rectangle $(2-{\displaystyle \frac{3}{2}}ka)\times \left({\displaystyle \frac{1}{n-2k^2}}\right)$ can be covered with $T_a$, if

$$a-(2-{\displaystyle \frac{3}{2}}ka)\ge {\displaystyle \frac{1}{n-2k^2}}.$$

Obviously, $a\ge {\displaystyle \frac{2}{\sqrt{n}}}$. Consequently, to check that the triangles can cover I it suffices to show that

$$\begin{array}{c}\hfill {\lambda }_1(n,k)\ge 0,\end{array}$$

where

$$\begin{array}{c}\hfill {\lambda }_1(n,k)={\displaystyle \frac{2}{\sqrt{n}}}(1+{\displaystyle \frac{3}{2}}k)-2-{\displaystyle \frac{1}{n-2k^2}}.\end{array}$$

We will verify this inequality in Cases 4 and 9.

In the method $M_2\left(k\right)$ (for example, see Figure 5, middle and right, where $k=2$) the unit square is covered with $k·(k-1)$ rectangles of sidelengths ${\displaystyle \frac{1}{k}}$ and ${\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}}$ and with $n-2k(k-1)$ rectangles of sidelengths $1-(k-1)({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}})$ and ${\displaystyle \frac{1}{n-2k(k-1)}}$. By Lemma 1, each rectangle ${\displaystyle \frac{1}{k}}\times \left({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}}\right)$ can be covered with two triangles $T_a$ (we take $\lambda ={\displaystyle \frac{1}{k}}$). Moreover, by Lemma 2 (we take $\mu =1-(k-1)({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}})$), the rectangle $\left(1-(k-1)({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}})\right)\times {\displaystyle \frac{1}{n-2k(k-1)}}$ can be covered with $T_a$, provided that

$$\begin{array}{c}\hfill a-(1-(k-1)({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{k}}))\ge {\displaystyle \frac{1}{n-2k(k-1)}}.\end{array}$$

Consequently, to check that the triangles can cover I, it suffices to show that

$$\begin{array}{c}\hfill {\lambda }_2(n,k)\ge 0,\end{array}$$

where

$$\begin{array}{c}\hfill {\lambda }_2(n,k)={\displaystyle \frac{2}{\sqrt{n}}}({\displaystyle \frac{3}{2}}k-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{k}}-2-{\displaystyle \frac{1}{n-2k(k-1)}}.\end{array}$$

We will verify this inequality in Cases 2, 6, and 9.

Consider ten cases depending on the number of triangles in the collection.

• Case 1: $n=5$. We cover I with four rectangles as shown in Figure 5, left. Now, we check that five triangles $T_a$, for $a\ge {\displaystyle \frac{2}{\sqrt{5}}}$, can cover these rectangles. According to Corollary 1, two triangles $T_a$ can cover a square of sidelength ${\displaystyle \frac{3}{4}}a$. Since ${\displaystyle \frac{7}{4}}a-1=a-\left(1-{\displaystyle \frac{3}{4}}a\right),$ by Lemma 2, a rectangle of size $\left(1-{\displaystyle \frac{3}{4}}a\right)\times \left({\displaystyle \frac{7}{4}}a-1\right)$ can be covered with a single triangle $T_a$. Thus, we cover two such rectangles with two triangles $T_a$. Moreover, by taking $\mu ={\displaystyle \frac{1}{2}}a$ in Lemma 2, we conclude that $T_a$ can cover a square of sidelength ${\displaystyle \frac{1}{2}}a$. It is easy to verify that $2-{\displaystyle \frac{7}{2}}a\le {\displaystyle \frac{1}{2}}a\mathrm{for}a\ge {\displaystyle \frac{2}{\sqrt{5}}}.$ Therefore, a square of sidelength $2-{\displaystyle \frac{7}{2}}a$ can be covered with $T_a$.
• Case 2: $n\in \{6,7\}$. We use the method $M_2\left(2\right)$ (see Figure 5, middle and right). It is easy to verify that

  $$\begin{array}{c}\hfill {\lambda }_2(n,2)={\displaystyle \frac{2}{\sqrt{n}}}(3-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{2}}-2-{\displaystyle \frac{1}{n-4}}>0.\end{array}$$
• Case 3: $n\in \{8,9\}$. We cover the square with four equal squares, as shown in Figure 6, left. Clearly,

  $$\begin{array}{c}\hfill 8·{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{4}}a\right)}^2\ge 4·{\displaystyle \frac{9}{16}}·{\left({\displaystyle \frac{2}{\sqrt{n}}}\right)}^2={\displaystyle \frac{9}{n}}\ge 1,\mathrm{for}n\in \{8,9\}.\end{array}$$

  By Corollary 1 these rectangles can be covered with eight triangles $T_a$.
• Case 4: $10\le n\le 13$. We use the method $M_1\left(2\right)$ (see Figure 6, middle, where $n=13$). It is easy to verify that

  $$\begin{array}{c}\hfill {\lambda }_1(n,2)={\displaystyle \frac{2}{\sqrt{n}}}(1+3)-2-{\displaystyle \frac{1}{n-8}}>0.\end{array}$$
• Case 5: $n=14$ (see Figure 6, right). Since $(2-3a)<a-{\displaystyle \frac{1}{3}}$ for $a\ge {\displaystyle \frac{2}{\sqrt{14}}}$, the rectangle of size ${\displaystyle \frac{1}{3}}\times (2-3a)$ can be covered with two triangles $T_a$.
• Case 6: $15\le n\le 19$. We use the method $M_2\left(3\right)$ (see Figure 7, left). It is easy to verify that

  $$\begin{array}{c}\hfill {\lambda }_2(n,3)={\displaystyle \frac{2}{\sqrt{n}}}\left({\displaystyle \frac{9}{2}}-{\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{3}}-2-{\displaystyle \frac{1}{n-12}}>0.\end{array}$$
• Case 7: $20\le n\le 22$. See Figure 7, middle. By

  $$\begin{array}{c}\hfill {\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{4}}\ge {\displaystyle \frac{5}{3}}-3a,\mathrm{for}a\ge {\displaystyle \frac{2}{\sqrt{n}}}\ge {\displaystyle \frac{2}{\sqrt{22}}},\end{array}$$

  six rectangles ${\displaystyle \frac{1}{2}}\times \left({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{2}}\right)$ and four rectangles ${\displaystyle \frac{1}{4}}\times ({\displaystyle \frac{5}{3}}-3a)$ can be covered with 20 triangles $T_a$.
• Case 8: $23\le n\le 29$, see Figure 7, right. By

  $$\begin{array}{c}\hfill a-\left({\displaystyle \frac{23}{12}}-{\displaystyle \frac{9}{2}}a\right)={\displaystyle \frac{11}{2}}a-{\displaystyle \frac{23}{12}}\ge {\displaystyle \frac{11}{\sqrt{n}}}-{\displaystyle \frac{23}{12}}\ge {\displaystyle \frac{1}{n-20}},\mathrm{for}23\le n\le 29,\end{array}$$

  I can be covered with n triangles $T_a$.
• Case 9: $30\le n\le 205$.
  If $30\le n\le 35$ (see Figure 8, left), then we use method $M_2\left(4\right)$:

  $$\begin{array}{c}\hfill {\lambda }_2(n,4)={\displaystyle \frac{2}{\sqrt{n}}}(6-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{4}}-2-{\displaystyle \frac{1}{n-24}}\ge 0.\end{array}$$

  If $36\le n\le 45$ (see Figure 8, middle), then we use $M_1\left(4\right)$:

  $$\begin{array}{c}\hfill {\lambda }_1(n,4)={\displaystyle \frac{2}{\sqrt{n}}}(1+6)-2-{\displaystyle \frac{1}{n-32}}>0.\end{array}$$

  If $46\le n\le 56$ (see Figure 8, right), then we use $M_2\left(5\right)$:

  $$\begin{array}{c}\hfill {\lambda }_2(n,5)={\displaystyle \frac{2}{\sqrt{n}}}({\displaystyle \frac{15}{2}}-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{5}}-2-{\displaystyle \frac{1}{n-40}}>0.\end{array}$$

  If $57\le n\le 68$ (see Figure 9, left), we use $M_1\left(5\right)$:

  $$\begin{array}{c}\hfill {\lambda }_1(n,5)={\displaystyle \frac{2}{\sqrt{n}}}(1+{\displaystyle \frac{15}{2}})-2-{\displaystyle \frac{1}{n-50}}>0.\end{array}$$

  If $69\le n\le 81$ (see Figure 9, middle), we use $M_2\left(6\right)$:

  $$\begin{array}{c}\hfill {\lambda }_2(n,6)={\displaystyle \frac{2}{\sqrt{n}}}(9-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{6}}-2-{\displaystyle \frac{1}{n-60}}>0.\end{array}$$

  If $82\le n\le 95$ (see Figure 9, right), we use $M_1\left(6\right)$:

  $$\begin{array}{c}\hfill {\lambda }_1(n,6)={\displaystyle \frac{2}{\sqrt{n}}}(1+9)-2-{\displaystyle \frac{1}{n-72}}>0.\end{array}$$

  If $96\le n\le 111$ (see Figure 10, left), we use $M_2\left(7\right)$:

  $$\begin{array}{c}\hfill {\lambda }_2(n,7)={\displaystyle \frac{2}{\sqrt{n}}}({\displaystyle \frac{21}{2}}-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{7}}-2-{\displaystyle \frac{1}{n-84}}>0.\end{array}$$

  If $112\le n\le 127$ (see Figure 10, middle), we use the method $M_1\left(7\right)$:

  $$\begin{array}{c}\hfill {\lambda }_1(n,7)={\displaystyle \frac{2}{\sqrt{n}}}(1+{\displaystyle \frac{21}{2}})-2-{\displaystyle \frac{1}{n-98}}>0.\end{array}$$

  If $128\le n\le 145$ (see Figure 10, right), we use $M_2\left(8\right)$:

  $$\begin{array}{c}\hfill {\lambda }_2(n,8)={\displaystyle \frac{2}{\sqrt{n}}}(12-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{8}}-2-{\displaystyle \frac{1}{n-112}}>0.\end{array}$$

  If $146\le n\le 164$ (see Figure 11, left), we use $M_1\left(8\right)$:

  $$\begin{array}{c}\hfill {\lambda }_1(n,8)={\displaystyle \frac{2}{\sqrt{n}}}(1+12)-2-{\displaystyle \frac{1}{n-128}}>0.\end{array}$$

  If $165\le n\le 184$ (see Figure 11, middle), we use $M_2\left(9\right)$:

  $$\begin{array}{c}\hfill {\lambda }_2(n,9)={\displaystyle \frac{2}{\sqrt{n}}}({\displaystyle \frac{27}{2}}-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{9}}-2-{\displaystyle \frac{1}{n-144}}>0.\end{array}$$

  Finally, if $185\le n\le 205$ (see Figure 11, right), we use $M_1\left(9\right)$:

  $$\begin{array}{c}\hfill {\lambda }_1(n,9)={\displaystyle \frac{2}{\sqrt{n}}}(1+{\displaystyle \frac{27}{2}})-2-{\displaystyle \frac{1}{n-162}}>0.\end{array}$$
• Case 10: $n\ge 206$. Clearly, for any $a>0$ there is an integer m such that

  $$\begin{array}{c}\hfill {\displaystyle \frac{1}{m+1}}<{\displaystyle \frac{3}{4}}a\le {\displaystyle \frac{1}{m}},\mathrm{i}.\mathrm{e}.,{\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m+1}}<a\le {\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m}}.\end{array}$$

  Note that

  $$\begin{array}{c}\hfill {\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m+1}}<{\displaystyle \frac{2}{3}}·{\displaystyle \frac{2m+1}{m(m+1)}}\le {\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m}}.\end{array}$$

Subcase 10.1: ${\displaystyle \frac{2}{3}}·{\displaystyle \frac{2m+1}{m(m+1)}}<a\le {\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m}}$. It is easy to check that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{m}}·\left({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{m}}\right)>1,\mathrm{for}a>{\displaystyle \frac{2}{3}}·{\displaystyle \frac{2m+1}{m(m+1)}}.\end{array}$$

This implies that I can be covered with $m·(m+1)$ rectangles of size ${\displaystyle \frac{1}{m}}\times \left({\displaystyle \frac{3}{2}}a-{\displaystyle \frac{1}{m}}\right)$ (as shown in Figure 12, left, where $m=10$ and $a={\displaystyle \frac{2}{\sqrt{235}}}$). Each such rectangle can be covered with two triangles $T_a$ (we take $\lambda ={\displaystyle \frac{1}{m}}$ in Lemma 1). Consequently, I can be covered with $2m(m+1)$ triangles $T_a$ and to prove that I can be covered with n triangles $T_a$ it suffices to check that $n\ge 2m(m+1)$. Observe that

$$\begin{array}{c}\hfill m(m+1)a^2<m(m+1)·{\left({\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m}}\right)}^2={\displaystyle \frac{16}{9}}·{\displaystyle \frac{m+1}{m}}\le 2\mathrm{for}m\ge 8.\end{array}$$

Since $n·{\displaystyle \frac{1}{2}}{a}^2\ge 2$ (the total area of the triangles is not smaller than 2), it follows that $n>2m(m+1)$.

Subcase 10.2: ${\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{m+1}}<a\le {\displaystyle \frac{2}{3}}·{\displaystyle \frac{2m+1}{m(m+1)}}$. We cover I with $(m+1)·(m+1)$ squares of sidelength ${\displaystyle \frac{1}{m+1}}$ (as illustrated in Figure 12, right, where $m=9$). By Corollary 1, each such square can be covered with two triangles $T_a$ (in our subcase ${\displaystyle \frac{3}{4}}a>{\displaystyle \frac{1}{m+1}}$). Consequently, I can be covered with $2{(m+1)}^2$ triangles $T_a$. Observe that

$$\begin{array}{c}\hfill {(m+1)}^2{a}^2\le {(m+1)}^2·{\displaystyle \frac{4}{9}}·{\displaystyle \frac{{(2m+1)}^2}{m^2{(m+1)}^2}}={\displaystyle \frac{4}{9}}·{\left({\displaystyle \frac{2m+1}{m}}\right)}^2\le 2\mathrm{for}m\ge 9.\end{array}$$

Since $n·{\displaystyle \frac{1}{2}}{a}^2\ge 2$, it follows that $n\ge 2{(m+1)}^2$. Consequently, I can be covered with n triangles $T_a$.   □

3. Translative Covering a Square with Unequal Triangles

By an s-rectangle we mean a rectangle $c\times d$, where $d\in \{1,1/2,1/4,1/8,1/16,\dots ,\}$.

Lemma 3.

Any right isosceles triangle $T_r$ of legs of length not greater than 2 contains an s-rectangle of the area equal to $4/9$ times the area of $T_r$.

Proof. 

Let k be a non-negative integer such that $2^{-k}\le t<2·2^{-k}$, where t denotes the length of the leg of $T_r$.

• Case 1: $t\ge {\displaystyle \frac{3}{2}}·2^{-k}$. It is easy to check that ${\displaystyle \frac{2}{9}}{t}^2-2^{-k}t+4^{-k}\le 0$. Thus

  $${\displaystyle \frac{4}{9}}·{\displaystyle \frac{1}{2}}{t}^2\le (t-2^{-k})2^{-k}.$$

  This means that $T_r$ contains the s-rectangle $(t-2^{-k})\times 2^{-k}$ of the area not smaller than $4/9$ times the area of $T_r$ (see Figure 13, left). Clearly, $t-2^{-k}<2·2^{-k}-2^{-k}=2^{-k}$.
• Case 2: $t<{\displaystyle \frac{3}{2}}·2^{-k}$. It is easy to verify that ${\displaystyle \frac{2}{9}}{t}^2-{\displaystyle \frac{1}{2}}·2^{-k}t+{\displaystyle \frac{1}{4}}·4^{-k}<0$. Thus

  $$\begin{array}{c}\hfill {\displaystyle \frac{4}{9}}·{\displaystyle \frac{1}{2}}{t}^2<(t-{\displaystyle \frac{1}{2}}·2^{-k})·{\displaystyle \frac{1}{2}}·2^{-k}.\end{array}$$

  This means that $T_r$ contains the s-rectangle $(t-2^{-k-1})\times 2^{-k-1}$ of the area greater than $4/9$ times the area of $T_r$ (see Figure 13, right). Clearly, $t-2^{-k-1}<{\displaystyle \frac{3}{2}}·2^{-k}-2^{-k-1}=2·2^{-k-1}$.    □

Figure 13. s-rectangle contained in $T_r$.

Figure 13. s-rectangle contained in $T_r$.

Lemma 4.

If a right isosceles triangle $T_r$ contains a rectangle $\lambda \times {\displaystyle \frac{1}{2^j}}$ for $j\ge 1$ and $\lambda >1/2^j$, then $T_r$ also contains a rectangle $(\lambda -{\displaystyle \frac{1}{2^j}})\times {\displaystyle \frac{2}{2^j}}$.

Proof. 

See Figure 14, where $j=1$.    □

Now we describe the main covering method. Let $R_1=a\times 1$ (for $a>0$) and let $T_1,T_2,\dots$ be a collection of triangles homothetic to T such that $t_1\ge t_2\ge \dots$, where $t_i$ is the length of the leg of $T_i$, $i=1,2,3,\dots$. The rectangle $R_1$ is divided into $2^k$ horizontal layers of height $2^{-k}$, for each positive integer k (see Figure 15, left, where $k=1$ and Figure 15, right, where $k=2$).

By Lemma 3 we know that any triangle $T_i$ contains an s-rectangle $P_i=a_i\times b_i$ such that $a_i·b_i\ge {\displaystyle \frac{2}{9}}{t}_i^2$ and that $a_i\le 2b_i$. We will use these rectangles $P_i\subset T_i$ for the covering. The rectangles, except perhaps the last one, will be placed in layers of the proper height.

• The covering algorithm.

[1]

The first triangle $T_1$ is placed so that the corresponding rectangle $P_1$ is placed into the lowest layer of height $b_1$ at the left-bottom corner of the layer, i.e., $P_1$ is packed at the left-bottom corner of $R_1$ (see Figure 16 left and right).

• Assume that $n>1$ and that $T_1,\dots ,T_{n-1}$ have been placed. Let $k=k\left(n\right)$ be an integer such that $b_n=1/2^k$. Denote by $L_1,\dots ,L_{2^k}$ all the layers in $R_1$ of height $b_n$. If each point of a layer $L_i$ is covered by a placed triangle, then $L_i$ is full.

[2]

If there is enough empty space in $R_1$ to pack $P_n$, then we pack $P_n$ into the lowest possible layer of height $b_n$ as far to the left as possible (see $P_2$ in Figure 16 left and right).

[3]

Otherwise, we find the smallest integer j such that the layer $L_j$ is not full.

[3a]

If either $j\le 2^k-2$ or $j=2^k$, then we place $P_n$ along the bottom of $L_j$ so that the left side of $P_n$ is the right side of the last rectangle packed into this layer.

[3b]

If $j=2^k-1$ and if $L_{2^k}$ is full, then we place $P_n$ along the bottom of $L_j$ so that the left side of $P_n$ is the right side of the last rectangle packed into this layer.

[3c]

If $j=2^k-1$ and if $L_{2^k}$ is not full, then we check which layer from among $L_{2^k-1}$ and $L_{2^k}$ is the least filled. Initially, we place $P_n$ along the bottom of this last filled layer so that the left side of $P_n$ is the right side of the last rectangle packed into this layer.

[3c1]

If, after placing $P_n$, the distance between the right side of $P_n$ and the right side of $R_1$ is not greater than $b_n$, then we do not change anything; in that case the area of the part of $P_n$ lying outside $R_1$ is not greater than $b_n^2$.

[3c2]

If after placing $P_n$ (together with $T_n\supset P_n$) the distance between the right side of $P_n$ and the right side of $R_1$ is greater than $b_n$ (as $P_3$ on Figure 16 right), then we will use another part of $T_n$ for the covering; by Lemma 4 we can cover with $T_n$ the uncovered part of $L_{2^k-1}\cup L_{2k}$ (the uncovered part of $R_1$), which is contained in a rectangle $P_n^{\prime }$ of sides of length $(a_n-b_n)\times 2b_n$ (as $P_3^{\prime }$ on Figure 16, right); as a consequence, $R_1$ is covered by triangles $T_1,\dots ,T_n$.

Example 1.

On Figure 16, left the rectangle $P_1$ is packed by Rule 1, the rectangles $P_2, P_3, P_4, P_8,$ and $P_9$ by Rule 2, the rectangles $P_5,P_6$ and $P_7$ are placed by Rule 3a, the rectangle $P_{10}$, of height 1/8, by Rule 3c1.

Example 2.

On Figure 16, to the right of the rectangle $P_1$ is packed by Rule 1 and the rectangle $P_2$ by Rule 2. Since there is not enough space to put $P_3$ in $L_{2^k-1}$ or in $L_{2^k}$, we use Rule 3c2. The whole $R_1$ is covered.

Lemma 5.

Any collection of positive homothetic copies T of homothety ratio not greater than 2, with the total area not smaller than ${\displaystyle \frac{9}{4}}(a+{\displaystyle \frac{1}{3}})$ permits a translative covering of the rectangle $R_1=a\times 1$, where $a>0$.

Proof. 

Let $T_1,T_2,\dots$ be a collection of triangles homothetic to T and let $t_i$ denote the length of the leg of $T_i$, $i=1,2,3,\dots$. Without loss of generality, we can assume that $t_1\ge t_2\ge \dots$.

We cover $R_1$ by the method presented above. Assume that $R_1$ is not covered by $T_1,T_2,\dots$. We will show that this leads to a contradiction.

Since $R_1$ is not covered, no rectangle has been placed by Rule 3c2. Observe that all placed rectangles $P_i$ have disjoint interiors. Some rectangles (placed by Rule 1 or 2) are contained in $R_1$. If a rectangle $P_m$ has been placed by Rule 3b, then the area of the part of $P_m$ lying outside $R_1$ is not greater than $b_m·2b_m$. If a rectangle $P_m$ has been placed by Rule 3c1, then the area of the part of $P_m$ lying outside $R_1$ is not greater than $b_m^2$.

The sum of the areas of parts of rectangles lying outside $R_1$ is smaller than ${\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{4^2}}+\dots ={\displaystyle \frac{1}{3}}$. This means that the total area of the rectangles $P_1$, $P_2,\dots$ is smaller than $a·1+{\displaystyle \frac{1}{3}}$. Consequently, by Lemma 3, the sum of the areas of triangles $T_1,T_2,\dots$ is smaller than ${\displaystyle \frac{9}{4}}(a+{\displaystyle \frac{1}{3}})$, which is a contradiction.    □

Taking $a=1$ in Lemma 5, we get the following result.

Theorem 2.

Any collection of positive homothetic copies T with the total area not smaller than 3 permits a translative covering of I.

Conjecture 1.

Any collection of positive homothetic copies T with the total area not smaller than 2 permits a translative covering of I.

4. Translative Covering a Rectangle with Triangles

Lemma 6.

Any collection $T_1,T_2,\dots$ of positive homothetic copies T of homothety ratio not greater than 2, with the total area not smaller than ${\displaystyle \frac{9}{4}}(ab+{\displaystyle \frac{1}{3}}{b}^2)$ permits a translative covering of the rectangle $R=a\times b$.

Proof. 

Assume that $T_1,T_2,\dots$ is a sequence of positive homothetic copies T of homothety ratio not greater than 2, with the total area not smaller than ${\displaystyle \frac{9}{4}}(ab+{\displaystyle \frac{1}{3}}{b}^2)$. Let $\mathcal{H}$ be a homothety with the ratio $h=1/b$. By Lemma 5 we know that the triangles $\mathcal{H}\left(T_1\right),\mathcal{H}\left(T_2\right),\dots$ can translatively cover $\mathcal{H}\left(R\right)={\displaystyle \frac{a}{b}}\times 1$. This means that $T_1,T_2,\dots$ can translatively cover R.    □

Theorem 3.

Any collection of positive homothetic copies T with the total area not smaller than ${\displaystyle \frac{1}{2}}{(a+b)}^2$ permits a translative covering of the rectangle $a\times b$, provided that $a\ge {\displaystyle \frac{5+\sqrt{33}}{4}}b$.

Proof. 

Let $R=a\times b$, where $a\ge {\displaystyle \frac{5+\sqrt{33}}{4}}b\approx 2.686b.$ Assume that $T_1,T_2,\dots$ is a sequence of positive homothetic copies T, with the total area not smaller than ${\displaystyle \frac{1}{2}}{(a+b)}^2$. We lose no generality in assuming that $t_1\ge t_2\ge \dots$, where $t_i$ denotes the length of the leg of $T_i$.

• Case 1: $t_1\le 2b$. Since ${\displaystyle \frac{1}{2}}{(a+b)}^2\ge {\displaystyle \frac{9}{4}}(ab+{\displaystyle \frac{1}{3}}{b}^2)$ for $a\ge {\displaystyle \frac{5+\sqrt{33}}{4}}b$, by Lemma 6 we deduce that the triangles can cover R.
• Case 2: $t_1\ge a$ and $t_2\le 2b$. The first triangle is placed so that the part of R not covered by this triangle is contained in a square Q of sidelength $a+b-t_1$ (see Figure 17, left). It is easy to verify that

  $$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}{(a+b)}^2-{\displaystyle \frac{1}{2}}{t}_1^2\ge 3{(a+b-t_1)}^2.\end{array}$$

  By Theorem 2, we deduce that the remaining triangles $T_2,T_3,\dots$, of the total area greater than ${\displaystyle \frac{1}{2}}{(a+b)}^2-{\displaystyle \frac{1}{2}}{t}_1^2$, can cover Q (the total area of the triangles is greater than three times the area of Q).
• Case 3: $2b<t_1<a$ and $t_2\le 2b$. The first triangle is placed so that the part of R not covered by this triangle is contained in a rectangle $R_{new}=(a+b-t_1)\times b$ (see Figure 17, right). It is easy to verify that

  $$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}{(a+b)}^2-{\displaystyle \frac{1}{2}}{t}_1^2\ge {\displaystyle \frac{9}{4}}[(a+b-t_1)b+{\displaystyle \frac{1}{3}}{b}^2].\end{array}$$

  This implies, by Lemma 6, that the remaining triangles $T_2,T_3,\dots$, of the total area greater than ${\displaystyle \frac{1}{2}}{(a+b)}^2-{\displaystyle \frac{1}{2}}{t}_1^2$, can cover $R_{new}$.
• Case 4: there is an integer $m\ge 2$ such that $t_m>2b$ and $t_{m+1}\le 2b$. Any $T_i$ can cover the rectangle $(a_i-b)\times b$. Consequently, $T_1,\dots ,T_m$ can cover $(t_1+\dots +t_m-mb)\times b$. It is easy to see (Figure 18) that the sum of the areas of $T_1,\dots ,T_m$ is smaller than the area of a triangle $T_{new}$ of legs of length $t_1+\dots +t_m-(m-1)b$. Consider the following sequence of triangles: $T_{new},T_{m+1},T_{m+2},\dots$. By Case 2 or Case 3, we know that this sequence permits a translative covering of R. Consequently, $T_1,T_2,\dots$ can cover R.
• Case 5: $t_i>2b$ for each i. We place the triangles as in Case 4. If R is not covered, then the sum of the areas of the triangles is not greater than the area of the triangle $T_{new}$ of legs of length smaller than $a+b$ (see Figure 18), which is a contradiction.    □

Clearly, any copy of T with the homothety ratio smaller than ${\displaystyle \frac{1}{2}}{(a+b)}^2$ can not translatively cover the rectangle $a\times b$. This means that the lower bound ${\displaystyle \frac{1}{2}}{(a+b)}^2$ given in Theorem 3 is tight.