Tilings of Parallelograms with Similar Right Triangles

Abstract

We say that a triangle $T$ tiles the polygon $\mathcal A$ if $\mathcal A$ can be decomposed into finitely many non-overlapping triangles similar to $T$. A tiling is called regular if there are two angles of the triangles, say $\alpha$ and $\beta$, such that at each vertex $V$ of the tiling the number of triangles having $V$ as a vertex and having angle $\alpha$ at $V$ is the same as the number of triangles having angle $\beta$ at $V$. Otherwise the tiling is called irregular. Let $\mathcal P (\delta )$ be a parallelogram with acute angle $\delta$. In this paper we prove that if the parallelogram $\mathcal P (\delta )$ is tiled with similar triangles of angles $(\alpha , \beta , \pi /2)$, then $(\alpha , \beta )=(\delta , \pi /2-\delta )$ or $(\alpha , \beta )=(\delta /2, \pi /2-\delta /2)$, and if the tiling is regular, then only the first case can occur.

We say that a triangle $T$ tiles the polygon $\mathcal A$ if $\mathcal A$ can be decomposed into finitely many non-overlapping triangles similar to $T$. In fact, it is in general very difficult to determine the set of similar triangles which tile a given polygon. Previous papers [1–4, 6] have looked at tiling polygons with similar triangles. In such case one starts with a collection of triangles and are allowed to scale the triangles when placing them. Recently, Laczkovich [5] has looked at tiling polygons, particularly regular triangles, squares, rectangles, and regular $n$-gons with $n$ large enough, with congruent triangles. Clearly, the number of distinct non-similar triangles $T$ such that parallelogram $\mathcal P$ can be dissected into finitely many triangles similar to $T$ is infinite. In this paper we look at tiling parallelograms with similar right triangles.

Let us recall some definitions from [4]. Let $\mathcal A$ be a polygon with vertices $V_{1}, \ldots , V_{N}$, and suppose that $\mathcal A$ is decomposed into non-overlapping similar triangles $\triangle _{1}, \ldots , \triangle _{t}$ of angles $\alpha , \beta , \gamma$. Let $V_{1}, \ldots , V_{m} (m\ge N)$ be an enumeration of the vertices of the triangles $\triangle _{1}, \ldots , \triangle _{t}$. For each $i=1, \ldots , m$ we denote by $p_{i}$ (resp. $q_{i}$ and $r_{i}$) the number of those triangles $\triangle _{j}$ whose angle at the vertex $V_{i}$ is $\alpha$ (resp. $\beta$ and $\gamma$). If $i\le N$ and the angle of $\mathcal A$ at the vertex $V_{i}$ is $\delta _{i}$, then

$$\begin{aligned} p_{i}\alpha +q_{i}\beta +r_{i}\gamma =\delta _{i}. \end{aligned}$$

(1)

If $i>N$ then we have either

$$\begin{aligned} p_{i}\alpha +q_{i}\beta +r_{i}\gamma =2\pi \end{aligned}$$

(2)

or

$$\begin{aligned} p_{i}\alpha +q_{i}\beta +r_{i}\gamma =\pi . \end{aligned}$$

(3)

Namely, (2) holds if $V_{i}$ is in the interior of $\mathcal A$ and whenever $V_{i}$ is on the boundary of a triangle $\triangle _{j}$ then necessarily $V_{i}$ is a vertex of $\triangle _{j}$. In the other cases (3) holds. It is clear that the coefficients $p_{i}, q_{i}, r_{i}$ must satisfy

$$\begin{aligned} \sum \limits _{i=1}^{m} p_{i}=\sum \limits _{i=1}^{m} q_{i}=\sum \limits _{i=1}^{m} r_{i}=t. \end{aligned}$$

(4)

The tiling will be called regular if one of the following statements is true:

• $p_{i}=q_{i}$ for every $i=1,\ldots , m$;

• $p_{i}=r_{i}$ for every $i=1,\ldots , m$;

• $q_{i}=r_{i}$ for every $i=1,\ldots , m$.

Otherwise the tiling is called irregular.

Let $\mathcal P (\delta )$ be a parallelogram with acute angle $\delta$. We first consider the regular tilings of parallelograms with similar right triangles and get the following theorem.

FormalPara Theorem 1

If parallelogram $\mathcal P (\delta )$ has a regular tiling with similar right triangles of angles $(\alpha , \beta , \pi /2)$, then $(\alpha , \beta )=(\delta , \pi /2-\delta )$.

FormalPara Proof

Suppose that $\mathcal P (\delta )$ can be regularly tiled with similar right triangles of angles $(\alpha , \beta , \pi /2)$. If at each vertex of the tiling the number of angles $\alpha$ is the same as that of $\beta$, then for the angle $\delta$ at a vertex of $\mathcal P (\delta )$ we let $\delta =p(\alpha +\beta )+r\cdot \pi /2$, where $p, r$ are integers. But $0<\delta <\pi /2$, which is impossible. Therefore by symmetry we may assume that at each vertex of the tiling the number of angles $\beta$ is the same as that of right angles $\pi /2$. In the following we shall consider Eqs. (1)–(3).

We first consider Eq. (1) and rewrite it as $p\alpha +r(\beta +\pi /2)=\delta$ or $\pi -\delta$, where $p$ and $r$ are integers. If $p\alpha +r(\beta +\pi /2)=\delta$, then by $0<\delta <\pi /2$ we get $\delta =p\alpha$. This implies that only the angle $\alpha$ occurs at the vertex with angle $\delta$. If $p\alpha +r(\beta +\pi /2)=\pi -\delta$, then by $\pi /2<\pi -\delta <\pi$ we have $r=0$ or 1. When $r=1$, we obtain $p=0$, and thus $\pi -\delta =\beta +\pi /2$. So $\delta =\alpha$ and $(\alpha , \beta )=(\delta , \pi /2-\delta )$, the result is true. When $r=0$, we have $\pi -\delta =p\alpha$. Therefore only the angle $\alpha$ occurs at the vertex with angle $\pi -\delta$.

Secondly, we consider Eq. (2) and rewrite it as $p\alpha +r(\beta +\pi /2)=2\pi$. Clearly, $r\le 3$. If $r=0$, then only $\alpha$ occurs. If $r=1$, then each of $\beta$ and $\pi /2$ occurs once and $\alpha$ occurs $p$ times with $p>1$. If $r=2$, then $p=2$ and each of $\alpha , \beta , \pi /2$ occurs twice. If $r=3$, then $p=0$ and $\beta =\pi /6$, and thus $(\alpha , \beta )=(\pi /3, \pi /6)$. By $\delta =p\alpha$, we get $p=1$ and $(\alpha , \beta )=(\delta , \pi /2-\delta )$, the result is true.

Thirdly, we consider Eq. (3) and denote it by $p\alpha +r(\beta +\pi /2)=\pi$. Clearly, $r\le 1$. If $r=1$, then $p=1$, and thus each of the angles $\alpha , \beta , \pi /2$ occurs once at the vertices with angle $\pi$. If $r=0$, then $\pi =p\alpha$.

From the discussions above we conclude that if $(\alpha , \beta )\ne (\delta , \pi /2-\delta )$ then at the vertices with angles $\pi$ or $2\pi$ the number of angle $\alpha$ occurs is not less than that of $\beta$ and at vertices with angles $\delta$ or $\pi -\delta$ only angle $\alpha$ occurs, which contradicts Eq. (4), and the proof is complete. $\square$

Figure 1 shows that the right triangle $(\delta , \pi /2-\delta , \pi /2)$ can regularly tile the parallelogram $\mathcal P (\delta )$.

Fig. 1

Regular tiling

In order to consider the irregular tiling of parallelograms with similar right triangles, we introduce the following lemma.

FormalPara Lemma 2

The parallelogram $\mathcal P (3\pi /10)$ cannot be tiled with similar right triangles of angles $(\pi /10, 2\pi /5, \pi /2)$.

FormalPara Proof

Suppose that $\mathcal{P }(3\pi /10)$ can be tiled with similar right triangles of angles $(\pi /10, 2\pi /5, \pi /2)$. Then by Theorem 2 of [2], we may assume that the coordinates of each vertex of $\mathcal{P }(3\pi /10)$ and of the triangles of the tiling belong to the field $\mathbb{Q }(\text{ cot }(\pi /10))$. Now, by Lemma 8 of [2], we have $\mathbb{Q }(\text{ cot }(\pi /10))\subset \mathbb{Q }(\zeta )$, where $\zeta =e^{(\pi /20)i}$. Also, since 3 is prime to 20, there is an automorphism $\phi$ of $\mathbb{Q }(\zeta )$ such that

$$\begin{aligned} \phi \big (\text{ cot }{\frac{a}{10}}\pi \big )=\text{ cot }{\frac{3a}{10}}\pi \end{aligned}$$

for every $a$ not divisible by 10. Consider the map

$$\begin{aligned} \Phi (x, y)=(\phi (x), \phi (y)) \quad (x, y\in \mathbb{Q }(\zeta )). \end{aligned}$$

This map defines a collineation that maps every triangle of the tiling into a triangle of angles $(3\pi /10, \pi /5, \pi /2)$. Also, $\Phi$ does not change the orientation of these triangles (see pp. 290–291 of [2]). Now $\Phi$ maps $\mathcal{P }(3\pi /10)$ into a parallelogram of angles $9\pi /10$ and $\pi /10$. Therefore, the conjugate tiling induced by $\Phi$ gives a tiling of parallelogram of angles $9\pi /10$ and $\pi /10$ with triangles of angles $(3\pi /10, \pi /5, \pi /2)$. But this is clearly impossible, since the angles of the triangles are greater than $\pi /10$. This proves the lemma. $\square$

FormalPara Theorem 3

If the parallelogram $\mathcal P (\delta )$ has an irregular tiling with similar right triangles of angles $(\alpha , \beta , \pi /2)$, then $(\alpha , \beta )=(\delta , \pi /2-\delta )$ or $(\alpha , \beta )=(\delta /2, \pi /2-\delta /2)$.

FormalPara Proof

Suppose that $\mathcal P (\delta )$ has an irregular tiling with similar right triangles of angles $(\alpha , \beta , \pi /2)$. If $\alpha =\beta$, then $(\alpha , \beta , \pi /2)=(\pi /4, \pi /4, \pi /2)$ and this is the case of $(\delta , \pi /2-\delta , \pi /2)$. Therefore, we may assume that $\alpha >\beta$, and thus $\pi /4<\alpha <\pi /2$. Since the tiling is irregular, there is an equation $p\alpha +q\beta +r\cdot \pi /2=\tau$ with $p>q$ and $\tau \in \{\delta , \pi -\delta , \pi , 2\pi \}$, where $p, q, r$ are integers. That is, $(p-q)\alpha =\tau -(q+r)\cdot \pi /2$. We need to consider the following cases.

Case 1: $\tau =\delta$. In this case we have $(p-q)\alpha =\delta -(q+r)\cdot \pi /2$. Since $p>q$ and $\delta <\pi /2$, we get $q+r=0$, and thus $p\alpha =\delta$. Since $\alpha >\pi /4$, this implies $p=1$, $\alpha =\delta$, and $(\alpha , \beta )=(\delta , \pi /2-\delta )$.

Case 2: $\tau =\pi -\delta$. Suppose $(p-q)\alpha =\pi -\delta -(q+r)\cdot \pi /2$. This implies $q+r\le 1$. If $q+r=0$, then $p\alpha =\pi -\delta$. Since $\pi /4<\alpha <\pi /2$, we get $p=2$ or $p=3$. If $p=2$, then $\alpha =\pi /2-\delta /2$ and we have $(\alpha , \beta )=(\pi /2-\delta /2, \delta /2)$.

If $p=3$, then $\alpha =(\pi -\delta )/3$ and $\beta =\pi /6+\delta /3$. Since $\alpha >\beta$, this implies $\delta <\pi /4$. Thus the angle $\delta$ cannot be tiled with angles $\alpha$. Therefore, we have $\delta =n\beta$. Thus $n=1$ or $n=2$, which give $\delta =\pi /4$ or $\delta =\pi$, both are impossible.

If $q+r=1$, then $(p-q)\alpha =\pi /2-\delta$. Since $\alpha >\pi /4$, this implies $\alpha =\pi /2-\delta$ and $(\alpha , \beta )=(\pi /2-\delta , \delta )$.

Case 3: $\tau =\pi$. In this case we have $(p-q)\alpha =\pi -(q+r)\cdot \pi /2$. This implies $q+r\le 1$. If $q+r=0$, then $p\alpha =\pi$. Since $\pi /4<\alpha <\pi /2$, we get $p=3$, $\alpha =\pi /3$, and $(\alpha , \beta )=(\pi /3, \pi /6)$. If $\delta =\pi /3$, then $\delta =\alpha$, and thus $(\alpha , \beta )=(\delta , \pi /2-\delta )$. If $\delta =n\cdot \pi /6$, then $n=1$ or $n=2$. When $n=2$, this is the above case. When $n=1$, then $\delta =\pi /6$, and thus $(\alpha , \beta )=(\pi /2-\delta , \delta )$.

If $q+r=1$, then $(p-q)\alpha =\pi /2$. From $\pi /4<\alpha <\pi /2$ we know that this case is impossible.

Case 4: $\tau =2\pi$. In this case we have $(p-q)\alpha =2\pi -(q+r)\cdot \pi /2$. Since $p>q$ and $\pi /4<\alpha <\pi /2$, we have $q+r\le 3$.

If $q+r=3$, then $(p-q)\alpha =\pi /2$. From $\pi /4<\alpha <\pi /2$ we know this is impossible.

If $q+r=2$, then $(p-q)\alpha =\pi$. Since $\pi /4<\alpha <\pi /2$, we get $\alpha =\pi /3$, which is the result in Case 3.

If $q+r=1$, then $(p-q)\alpha =3\pi /2$. This implies $\alpha =3\pi /8$ or $\alpha =3\pi /10$.

When $\alpha =3\pi /8$, then $(\alpha , \beta )=(3\pi /8, \pi /8)$. If $\delta =3\pi /8$ or $\delta =\pi /8$, our result is true. Otherwise, we have $\delta =m\cdot \pi /8$. It is clear that $m\ne 1, 3$. If $m=2$, then $\delta =\pi /4$, and this is the case of $(\alpha , \beta )=(\pi /2-\delta /2, \delta /2)$. If $m\ge 4$, then $\delta =m\cdot \pi /8\ge \pi /2$, a contradiction.

When $\alpha =3\pi /10$, then $(\alpha , \beta )=(3\pi /10, \pi /5)$. If $\delta =3\pi /10$, or $\delta =\pi /5$, our result is true. Otherwise, we have $\delta =n\cdot \pi /5$. It is clear that $n\ne 1$. If $n=2$, then $\delta =2\pi /5$, and this is the case of $(\pi /2-\delta /2, \delta /2)$. If $n\ge 3$, then $\delta =n\cdot \pi /5>\pi /2$, a contradiction.

If $q+r=0$, then $p\alpha =2\pi$. Since $\pi /4<\alpha <\pi /2$, we have $\alpha =2\pi /5$, $\alpha =\pi /3$, or $\alpha =2\pi /7$.

When $\alpha =2\pi /7$, then $(\alpha , \beta )=(2\pi /7, 3\pi /14)$. If $\delta =2\pi /7$, or $\delta =3\pi /14$, then the result is true. If $\delta \ne 2\pi /7$ and $\delta \ne 3\pi /14$, then by $0<\delta <\pi /2$ we may assume that $\delta =m\cdot 3\pi /14$. If $m=2$, then $\delta =3\pi /7$, and this is the case of $(\pi /2-\delta /2, \delta /2)$. If $m\ge 3$, then $\delta =m\cdot 3\pi /14>\pi /2$, a contradiction.

When $\alpha =\pi /3$, this is a result of Case 3.

When $\alpha =2\pi /5$, then $(\alpha , \beta )=(2\pi /5, \pi /10)$. If $\delta =2\pi /5$ or $\delta =\pi /10$, the result is true. Otherwise, we have $\delta =n\cdot \tfrac{\pi }{10}$. If $n=2$, this is the case of $(\pi /2-\delta /2, \delta /2)$. If $n=3$, then $\delta =3\pi /10$, and by Lemma 2 we conclude a contradiction. If $n\ge 5$, then $\delta =n\cdot \pi /10\ge \pi /2$, a contradiction. The proof is complete. $\square$

Figure 2 shows that the right triangles $(\delta , \pi /2-\delta , \pi /2)$ and $(\delta /2, \pi /2-\delta /2, \pi /2)$ can irregularly tile the parallelogram $\mathcal P (\delta )$, where the numbers 1, 2, 3 in the left figure denote the angles $\delta , \pi /2-\delta , \pi /2$, respectively; the numbers 1, 2, 3 in the right figure denote the angles $\delta /2, \pi /2-\delta /2, \pi /2$, respectively.

Fig. 2

Irregular tiling

FormalPara Remark 4

Since the above proofs never used that $\mathcal P (\delta )$ is a parallelogram, only that two of its angles equal $\delta$ and two equal $\pi -\delta$, for the trapezoid $\mathcal T (\delta )$ with acute angle $\delta$, we have the same results as that of parallelogram $\mathcal P (\delta )$.