Tilings of Convex Polygons with Congruent Triangles
Discrete Comput Geom
Tilings of Convex Polygons with Congruent Triangles
M. Laczkovich 0 1
0 M. Laczkovich Department of Mathematics, University College London , Gower Street, London, WC1E 6BT England , UK
1 M. Laczkovich ( ) Department of Analysis, Eötvös Loránd University , Pázmány Péter sétány 1/C, Budapest, 1117 Hungary
By the spectrum of a polygon A we mean the set of triples (α, β, γ ) such that A can be dissected into congruent triangles of angles α, β, γ . We propose a technique for finding the spectrum of every convex polygon. Our method is based on the following classification. A tiling is called regular if there are two angles of the triangles, α and β such that at every vertex of the tiling the number of triangles having angle α equals the number of triangles having angle β. Otherwise the tiling is irregular. We list all pairs (A, T ) such that A is a convex polygon and T is a triangle that tiles A regularly. The list of triangles tiling A irregularly is always finite, and can be obtained, at least in principle, by considering the system of equations satisfied by the angles, examining the conjugate tilings, and comparing the sides and the area of the triangles to those of A. Using this method we characterize the convex polygons with infinite spectrum, and determine the spectrum of the regular triangle, the square, all rectangles, and the regular N gons with N large enough.
Tilings with congruent triangles; Regular and irregular tilings

1 Introduction
This paper is concerned with the following problem. Suppose we are given a convex
polygon A. Decide, whether or not A can be dissected into congruent triangles, and
if there is such a dissection, find all triangles T such that A has a dissection into
congruent triangles similar to T . By a dissection (or tiling) we mean a decomposition
of A into pairwise nonoverlapping polygons. No other conditions are imposed on the
tilings. In particular, it is allowed that two pieces have a common boundary point,
but do not have a common side. By the spectrum of the polygon A we mean the set
of triples (α, β, γ ) such that A can be dissected into congruent triangles of angles
α, β, γ .
A complete solution to this problem would be an algorithm producing the
spectrum of every given convex polygon. Although we cannot solve the problem in this
algorithmic sense of the word, we present a method which seems to work in most
(and possibly, for all) cases. Our method is based on the following classification of
tilings introduced in [3]. Suppose that A is dissected into triangles similar to T (we
do not assume that the triangles are congruent). We say that the tiling is regular if
there are two angles of T , say α and β such that at each vertex V of the tiling, the
number of triangles having V as a vertex and having angle α at V is the same as the
number of triangles having angle β at V . If this condition is not satisfied, then the
tiling is called irregular. Thus the problem of finding the spectrum is divided into two
separate questions: given A, determine those triangles T for which A has a regular
(resp. irregular) tiling with congruent triangles similar to T .
We give a complete solution to the question concerning regular tilings. In
Theorem 2.1 we list all pairs (A, T ) such that A is a convex polygon, T is a triangle, and
there is a regular tiling of A with congruent triangles similar to T . In fact, this is the
main result of the paper, and its proof occupies Sects. 4–9.
As for irregular tilings, our starting point is [3, Theorem 4] stating that for every
polygon A, the number of distinct nonsimilar triangles T such that A has an
irregular tiling with triangles similar to T is finite. More precisely, the number of these
triangles is at most cN 6, where N is the number of vertices of A and c is an absolute
constant. The proof of [3, Theorem 4] is effective, and gives a list of triples (α, β, γ )
such that the angles of every triangle which tiles A irregularly are given by one of the
triples of the list.
As we shall see in Sect. 3, in many cases this list can be reduced considerably
by using the system of equations satisfied by the angles, and considering conjugate
tilings as in [1]. Then, assuming that the tiling consists of congruent triangles, we
may compare the sides and the area of the triangles to those of A in order to obtain
further number theoretical restrictions on the triples (α, β, γ ). Discarding all triples
violating these conditions, we arrive at the list of all triangles that tile A irregularly.
There can be two problems with the application of this argument. The first
problem is that we cannot guarantee that the reduced list obtained by considering the
equations, conjugate tilings and number theoretical restrictions only contains triples
corresponding to a tiling. (Proving this fact would result in an algorithmic solution
of the problem.) However, as we shall see in Sect. 3, in all cases we consider, each
triple contained by the reduced list actually tiles A. One can hope that this happens
to every convex polygon A.
The other problem is that when we arrive at a list of triples from which we cannot
eliminate any item, we have to produce tilings corresponding to these triangles. In
principle, this could be the most difficult step in the procedure. However, in the actual
cases we consider, the existence of these tilings is either trivial (as in the case of
rectangles or regular N gons with N large enough), or relatively easy (as for the
regular triangle). One can hope that this shows a general tendency.
The applications of the method outlined above will be given in Sect. 3. As an
application of Theorem 2.1 we characterize the convex polygons with infinite spectrum
(Theorem 3.1). Then we determine the spectrum of the regular triangle, the square, all
rectangles, and the regular N gons with N large enough (Theorems 3.3, 3.6,
Corollary 3.7 and Theorem 3.4).
2 Regular Tilings of Convex Polygons
We recall the definition of regular tilings. Let A be a polygon with vertices
V1, . . . , VN , and suppose that A is decomposed into nonoverlapping similar
triangles Δ1, . . . , Δt of angles α, β, γ . Let V1, . . . , Vm (m ≥ N ) be an enumeration of the
vertices of the triangles Δ1, . . . , Δt . For every i = 1, . . . , m we shall denote by pi
(resp. qi and ri ) the number of those triangles Δj whose angle at the vertex Vi is α
(resp. β and γ ). If i ≤ N and the angle of A at the vertex Vi is δi , then
If i > N then we have either
or
pi α + qi β + ri γ = δi .
pi α + qi β + ri γ = 2π
pi α + qi β + ri γ = π.
m
i=1
pi =
m
i=1
qi =
m
i=1
ri = t.
(1)
(2)
(3)
(4)
Namely, (2) holds if Vi is in the interior of A and whenever Vi is on the boundary of
a triangle Δj then necessarily Vi is a vertex of Δj . In the other cases (3) will hold. It
is clear that the coefficients pi , qi , ri must satisfy
The tiling will be called regular, if one of the following statements is true:
• pi = qi for every i = 1, . . . , m;
• pi = ri for every i = 1, . . . , m;
• qi = ri for every i = 1, . . . , m.
Otherwise the tiling is called irregular. We call a polygon rational, if its sides are
pairwise commensurable.
Theorem 2.1 Let A be a convex N gon, and suppose that A has a regular tiling with
congruent triangles of angles α, β, γ such that at each vertex of the tiling the number
of angles α is the same as that of β. Then one of the following statements is true.
(i) A is a parallelogram of angles γ and α + β, and the ratio of the sides of A is
a rational multiple of sin α/ sin β.
(ii) A is a rational polygon, N − 2 of its angles equal γ , the other two angles are
integer multiples of α + β, and sin α, sin β, sin γ are pairwise commensurable.
(iii) A is a rational polygon, N − 2 of its angles equal α + β, the other two angles
are integer multiples of γ , and sin α, sin β, sin γ are pairwise commensurable.
(iv) A is a regular triangle, γ = π/3 or γ = 2π/3, and sin α, sin β, sin γ are
pairwise commensurable.
(v) A is a centrally symmetric rational N gon, N = 2k ≥ 8, each angle of A equals
α + β, and the side lengths of A are a1, . . . , ak, a1, . . . , ak in this order, where
a1, . . . , ak−1 constitute a geometric progression of quotient q = sin α/ sin β,
and ak ≥ qk−1 · a1.
(vi) A is a centrally symmetric rational hexagon, each angle of A equals 2π/3,
γ = π/3 or γ = 2π/3, and sin α/ sin β is rational.
(vii) A is a regular N gon, α = β = (π/2) − (π/N ) and γ = 2π/N .
(viii) A is a rational N gon with N ≤ 6, each angle of A equals π/3 or 2π/3, α =
β = π/6 and γ = 2π/3.
(ix) A is a rational hexagon, each angle of A equals 2π/3, γ = π/3, and either
α = π/6, β = π/2 or α = π/2, β = π/6.
We shall prove Theorem 2.1 in Sects. 4–9. In this section first we discuss the most
immediate consequences of Theorem 2.1. Then we show that the cases listed in the
theorem correspond to existing tilings.
Theorem 2.2 The set of those nonsimilar convex polygons which are not
parallelograms and have a regular tiling with congruent triangles is countable.
Proof It is enough to check that the set of those nonsimilar convex polygons which
satisfy the conditions formulated in the cases (ii)–(ix) of Theorem 2.1 is countable.
First note that the set of nonsimilar rational triangles is countable, and thus the
set of those triples (α, β, γ ) for which sin α, sin β, sin γ are pairwise commensurable
is countable. From this observation it is clear that the set of those nonsimilar convex
polygons which satisfy the conditions of cases (ii)–(iv) is countable.
Now consider the case (v). Then the angles of A are equal π − (2π/N ). Since
the sides of A are commensurable, it is clear that the number of these polygons is
countable.
The statement concerning the cases (vi)–(ix) is trivial.
Remark 2.3 The analogous statement about irregular tilings is false. In fact, for every
N , there are continuum many nonsimilar convex N gons that have irregular tilings
with congruent triangles. To see this, let XY Z be a triangle such that XY = XZ and
the angle γ = Y XZ satisfies γ < π/(N − 2). Let Ti denote the triangle obtained
from XY Z by rotating it about the vertex X by the angle i · γ (i = 0, . . . , N − 3).
Then Aγ = iN=−03 Ti is a convex N gon tiled with the triangles T0, . . . , TN−3. It is
clear that different values of γ lead to nonsimilar polygons Aγ .
One can show that the set of nonsimilar convex N gons having irregular tilings
with congruent triangles can be decomposed into countable many families depending
on some continuous parameters. Unfortunately, the description or enumeration of
these families seems to be difficult.
Theorem 2.4 Let A be a convex N gon, and suppose that A has a regular tiling with
congruent triangles of sides a, b, c. Then at least one of the following statements is
true.
(i) N ≤ 6.
(ii) A is a regular N gon.
(iii) A is rational and centrally symmetric.
(iv) The lengths a, b, c are pairwise commensurable.
Proof This is clear from Theorem 2.1.
We close this section by showing that each case listed in Theorem 2.1 occurs. More
precisely, we show that in each case (with the possible exception of (ix)), whenever
the polygon A and the angles α, β, γ satisfy the conditions, then A has a regular tiling
with congruent triangles of angles α, β, γ .
(i) Let A be a parallelogram of sides d1, d2 and angles γ and α + β such that
d1/d2 = (p/q) · (sin α/ sin β), where p and q are positive integers. Then we put
τ = d1/(p sin α). If a = τ sin α and b = τ sin β, then d1 = p · a and d2 = q · b. Thus A
can be decomposed into pq congruent parallelograms of sides a and b. Each of these
parallelograms can be decomposed into two congruent triangles of angles α, β, γ and
of sides a, b and c = τ sin γ . It is clear that the tiling obtained is regular.
Next suppose that A and α, β, γ satisfy the conditions formulated in (ii) or (iii).
Then cos γ is rational, since
cos γ =
sin2 α + sin2 β − sin2 γ
2 sin α sin β
.
Therefore, putting δ = γ if A satisfies (ii) and δ = α + β if A satisfies (iii), we find
that cos δ is rational, N − 2 angles of A equal δ, and the other two angles are integer
multiples of π − δ.
Now it can be shown by induction on N that under these conditions A can be
decomposed into finitely many nonoverlapping rational symmetric trapezoids of angles
δ and π − δ. (See also Lemma 8 of [3], where a more general statement is proved,
except that the rationality of the polygon A is not supposed, and thus the trapezoids
obtained are not necessarily rational either. However, one can check that if A is
rational and cos δ is rational, then the construction of [3, Lemma 8] yields rational
trapezoids.) It follows from Lemma 2.2 of [2] that each of these symmetric trapezoids can
be tiled with congruent copies of a triangle of angles α, β, γ . Then, by Lemma 2.1
of [2], A itself can be tiled with congruent copies of a triangle of angles α, β, γ . It is
easy to check that the tiling obtained this way is regular.
(iv) See Theorem 3.1 of [2].
(v) By assumption, q = a2/a1 is a positive rational number. The conditions imply
that α + β = π − (2π/N ), and γ = 2π/N = π/ k.
Let V1, . . . , VN = V0 be the vertices of A listed counterclockwise such that
Vi−1Vi = ai for every i = 1, . . . , N (see Fig. 1). For every i = 1, . . . , N , let hi be
the halfline starting from Vi such that the angle between the side Vi−1Vi and hi
equals β, and the angle between the side Vi Vi+1 and hi equals α. Let E be the
intersection of hN and h1. Then EV1/EV0 = sin α/ sin β = q, and thus EV1 = q · EV0.
Since V1V2 = q · V0V1 and EV1V2 = α, it follows that the triangle EV1V2 is
similar to the triangle EV0V1, and thus E is the intersection of h1 and h2. Repeating this
argument we find that E is on the halflines hi for every i = 1, . . . , k − 1.
Since ak ≥ qk · a1, there is a point W on the side Vk−1Vk such that Vk−1W = qk−1 ·
a1. (The point W coincides with Vk if ak = qk · a1.) Since W Vk−1 = q · Vk−2Vk−1 and
EVk−1W = α, it follows that the triangle EVk−1W is also similar to the triangle
EV0V1. By kγ = π we find that the points V0, E and W are collinear. Thus the
convex polygon V0, . . . , Vk−1W V0 is tiled by k triangles of angles α, β, γ such that
the similarity ratio of any two of these triangles is rational.
The central symmetry of A implies that there is a point W on the side VN−1V0
such that VN−1W = qk−1 · a1. Also, the convex polygon Vk, . . . , VN−1W Vk is tiled
by k triangles of angles α, β, γ such that the similarity ratio of any two of these
triangles is rational.
The angles of the parallelogram W V0W Vk are α + γ and β. Since W Vk is a
rational multiple of V0V1 and V0W is rational multiple of EV0, it follows that the
ratio W Vk/V0W is a rational multiple of
sin γ sin γ
V0V1/EV0 = sin β = q · sin α .
Therefore, by (i), the parallelogram W V0W Vk can be tiled by congruent triangles of
angles α, β, γ . In this way we decomposed A into nonoverlapping triangles of angles
α, β, γ . It is easy to check that the similarity ratio of any two of these triangles is
rational.
Let Δ1, . . . , Δn be the triangles of this decomposition, where each Δi is similar
to Δ1. Then there are positive rational numbers ri such that Δi is obtained from Δ1
by a similarity transformation with ratio ri . Let pi and q be positive integers such
that ri = pi /q (i = 1, . . . , n). For every i, we can dissect Δi into pi2 congruent
triangles similar to Δ1. In this way we obtain a dissection of A into in=1 pi2 congruent
triangles similar to Δ1. It is easy to see that the tiling obtained is regular.
(vi) Let a1, a2, a3, a1, a2, a3 denote the lengths of the sides of A. It is easy to
see that A can be decomposed into three parallelograms of sides a1, a2; a3, a1; and
a2, a3, respectively. Therefore, these parallelograms are rational, and their angles
equal π/3 and 2π/3. By (i), each of them can be tiled with congruent triangles of
angles α, β, γ . In this way we decomposed A into nonoverlapping triangles of angles
α, β, γ such that the similarity ratio of any two of these triangles is rational. From this
we obtain a tiling of A into congruent triangles of angles α, β, γ as in the previous
case. Clearly, the tiling obtained is regular.
(vii) Connecting the center of A with the vertices we obtain a tiling of A into
congruent triangles of angles α, β, γ . Then we can label the angles (π/2) − (π/N )
of the triangles with α and β in such a way that we obtain a regular tiling.
(viii) Since A is rational and each angle of A equals π/3 or 2π/3, it follows that
A can be decomposed into congruent regular triangles. We can decompose each of
these triangles into three triangles of angles α, β, γ . It is clear that we can label the
acute angles of these triangles with α and β in such a way that we obtain a regular
tiling.
(ix) It is clear that a regular triangle can be tiled with two congruent triangles
having angles α = π/2, β = π/6, γ = π/3. Since every rational polygon with angles
π/3 and 2π/3 can be tiled with congruent regular triangles, it follows that these
polygons can be tiled with congruent triangles having angles α, β, γ . However, we
are looking for regular tilings, and this extra condition excludes the cases when A
is a triangle or a trapezoid or a pentagon. (We shall prove this in Sect. 8.) It is not
clear whether or not every rational hexagon with angles 2π/3 has a regular tiling
with triangles having angles α = π/2, β = π/6, γ = π/3. For some hexagons there
is such a tiling. For example, if A is centrally symmetric, then the existence of such
a tiling follows from (vi), since sin α/ sin β = 2.
3 Applications of Theorem 2.1
For every polygon A we shall denote by c(A) the cardinality of the spectrum of A.
Theorem 3.1 Let A be a convex N gon. Then c(A) = ∞ if and only if A satisfies
one of the following conditions.
(i) A is a regular triangle.
(ii) A is a parallelogram.
(iii) A is rational, and there is a δ such that cos δ is rational, N − 2 angles of A
equal δ, and the other two angles are integer multiples of π − δ.
Proof Let A be a given convex polygon. By [3, Theorem 4], the number of triples
(α, β, γ ) such that A has an irregular tiling with congruent triangles of angles
(α, β, γ ) is finite. Therefore, if c(A) = ∞, then there are infinitely many triples
(α, β, γ ) such that A has a regular tiling with congruent triangles of angles (α, β, γ ).
In particular, there are such triples different from those listed in cases (v), (vii),
(viii) and (ix) of Theorem 2.1. (Note that in (v) of Theorem 2.1 the triple (α, β, γ )
is uniquely determined. Indeed, γ must be equal to 2π/N , and then the condition
sin α/ sin β = q determines α and β as well.) Therefore, A must satisfy one of the
conditions of (i)–(iv) and (vi) of Theorem 2.1. It is clear that in each case, one of
(i)–(iii) of Theorem 3.1 holds. This proves the “only if” part of the theorem.
Now we prove the “if” part. If Δ is a rational triangle and one of its angles equals
π/3 or 2π/3 then, by Theorem 3.1 of [2], the regular triangle can be tiled with
congruent triangles similar to Δ. In Lemma 3.2 of [2] it is shown that there are infinitely
many such triangles Δ, and thus c(A) = ∞ holds for the regular triangle.
If A is a parallelogram of sides a, b and of angles γ and π − γ , then A can be tiled
with congruent triangles of angles α, β, γ whenever sin α/ sin β is a rational multiple
of a/b. It is easy to see that there are infinitely many such triples (α, β, γ ), and thus
c(A) = ∞.
Next suppose that A satisfies (iii). We claim that there are infinitely many triples
(α, β, γ ) such that α, β, γ are the angles of a triangle, γ = δ, and sin α, sin β, sin γ
are pairwise commensurable. For each of these triples, A and (α, β, γ ) satisfy the
conditions of (iii) of Theorem 2.1 and thus, as we saw in the previous section, there
is a tiling of A with congruent triangles of angles α, β, γ . This will prove c(A) = ∞.
Let 0 < γ < π be given such that cos γ is rational. Then d = sin2 γ = 1 − cos2 γ is
a positive rational number. It is wellknown that there are infinitely many points on the
ellipse x2 + d · y2 = 1 having rational coordinates; moreover, the set of these points
is everywhere dense in the ellipse. Indeed, for every s ∈ Q, the point with coordinates
x = ±(s2 − d)/(s2 + d) and y = 2s/(s2 + d) satisfies the equation x2 + d · y2 = 1,
and the set of these points is everywhere dense in the ellipse.
Let (x, y) be such a point with a small positive y. We define α ∈ (0, π/2) by
sin α = y · sin γ . Then sin α/ sin γ = y is rational, and
cos α =
1 − sin2 α =
1 − y2 · sin2 γ =
1 − y2 · d = x.
If y is small enough, then α + γ < π , and we can define β by β = π − α − γ . Then
we have
sin β
sin γ =
sin α cos γ + cos α sin γ
sin γ
= y · cos γ + cos α = y · cos γ + x,
and thus sin α/ sin γ and sin β/ sin γ are both rational. It is clear that this construction
gives infinitely many triples (α, β, γ ) with the required properties.
In the following applications we consider the cases when A is a regular triangle,
a square, a rectangle or a regular N gon with N large enough, and determine the set of
triples (α, β, γ ) such that A can be tiled with congruent triangles of angles (α, β, γ ).
We shall need the following lemma.
Lemma 3.2 Suppose that the convex polygon A has an irregular tiling with
congruent triangles of angles α, β, γ . Then α, β, γ are linear combinations of the angles of
A with rational coefficients.
Proof Consider (1), (2), and (3) at the vertices of the tiling. Since the tiling is
irregular, it follows from [3, Lemma 10] that the determinant
is nonzero for at least one pair of indices (i, j ). Then the corresponding system of
equations
Dij =
1
pi
pj
1
qi
qj
1
ri
rj
α + β + γ = π,
pi α + qi β + ri γ = δi ,
pj α + qj β + rj γ = δj
determines α, β, γ (here δi and δj are either angles of A or equal π or 2π ). Applying
Cramer’s rule, we find that α, β, γ are linear combinations of the angles of A and
of π with rational coefficients. Since π equals the sum of the angles of A divided by
N − 2, we obtain the statement of the lemma.
Theorem 3.3 The regular triangle can be tiled with congruent triangles of angles
α, β, γ if and only if a permutation of (α, β, γ ) satisfies one of the following
conditions:
(i) α = β = π/6 and γ = 2π/3;
(ii) α = π/6, β = π/2, γ = π/3;
(iii) γ ∈ {π/3, 2π/3} and sin α, sin β, sin γ are pairwise commensurable.
Proof We saw already that if (α, β, γ ) is one of the triples listed in the theorem, then
there exists a tiling with the required properties.
Next suppose that the regular triangle is tiled with congruent triangles of angles
α, β, γ . If the tiling is regular then one of (ii)–(iv), (vii) and (viii) of Theorem 2.1
must hold. It is clear that in each of these cases the statement of Theorem 3.3 is true.
Therefore, we may assume that the tiling is irregular. Then, by Lemma 3.2, α, β, γ
are rational multiples of π . Then we can apply Theorem 5.1 of [2], and find that either
a permutation of (α, β, γ ) satisfies (i), (ii) or (iii) of Theorem 3.3, or a permutation
of (α, β, γ ) equals one of the triples
We prove that none of the triples listed in (5) gives an irregular tiling of the regular
triangle. Suppose first (α, β, γ ) = (π/3, π/12, 7π/12). Since the tiling is irregular,
there is an equation pα + qβ + r γ = δ such that q < r and δ ∈ {π/3, π, 2π }.
Multiplying the equation by 12/π we find that 4p + q + 7r is divisible by 4. Then r − q
is also divisible by 4. However, we have 0 < r − q ≤ r ≤ 3, which is impossible.
Next suppose (α, β, γ ) = (π/3, π/30, 19π/30). Since the tiling is irregular, there
is an equation pα + qβ + r γ = δ such that q < r . Multiplying the equations by 30/π
we find that 10p + q + 19r is divisible by 10. Then r − q is also divisible by 10.
However, we have 0 < r − q ≤ r ≤ 3, which is impossible.
A similar argument works in the case when (α, β, γ ) = (π/3, 7π/30, 13π/30).
Let RN denote the regular N gon. Connecting the center of RN with the vertices,
we obtain a tiling of RN with congruent triangles of angles α = β = (π/2) − (π/N )
and γ = 2π/N . Another tiling is obtained by decomposing each of these isosceles
triangles into two right triangles. The angles of the triangles of the new tiling are
(π/2) − (π/N ), π/2, π/N . This shows that for every N the regular N gon can be
tiled with congruent triangles with angles given by any of the triples
π π π π 2π
2 − N , 2 − N , N
,
π π π π
2 − N , 2 , N
.
For N = 3, 4, 6 there are other tilings of RN . Moreover, in these cases there are
infinitely many other triangles tiling RN since, by Theorem 3.1, we have c(R3) =
c(R4) = c(R6) = ∞. Next we show that this behavior is exceptional among the
regular polygons.
Theorem 3.4
(i) If N = 3, 4, 6, then c(RN ) < ∞.
(ii) If N = 3, 4, 6 and RN has a regular tiling with congruent triangles, then the
angles of the triangles are α = β = (π/2) − (π/N ), γ = 2π/N .
(iii) If N > 420, then c(RN ) = 2.
Lemma 3.5 If N = 3, 4, 6, then RN cannot be tiled with congruent triangles of
angles α = β = π/N , γ = π − (2π/N ).
Proof Suppose there is such a tiling. We may assume that the sides of the triangles
are a, a, c, where a = sin α and c = sin(π − (2π/N )) = sin 2α = 2 sin α cos α. Then
the side of RN equals xa + yc, where x, y are nonnegative integers. If the number of
tiles is t then, comparing the areas we obtain
(6)
Using a = sin α, c = 2 sin α cos α and cot(π/N ) = cos α/ sin α, we obtain
N ·
(xa + yc)2
4
π t 2π
· cot N = 2 · sin2 Nπ · sin N
.
N
4 · (x + 2y cos α)2 = t · sin2 α = t · 1 − cos2 α
and
Ny2 + t cos2 α + N xy cos α +
N x2/4 − t = 0.
(7)
Thus the degree of cos α is at most 2. However, by Theorem 3.9 of [5], the
degree of cos α = cos(π/N ) is φ (2N )/2, and thus φ (2N ) = 2 or φ (2N ) = 4. Since
N = 3, 4, 6, the only possibility is N = 5. Then cos α = cos π/5 = (√5 + 1)/4, and
thus the minimal polynomial of cos α is 4X2 − 2X − 1. Then (7) gives Ny2 + t =
−4 · ((N x2/4) − t ) and N xy = 2((N x2/4) − t ). From the first equation we obtain
t = N (x2 + y2)/3 > N x2/4. Thus the second equation gives N xy < 0, which is
impossible. This contradiction completes the proof.
Proof of Theorem 3.4 (i) Suppose c(RN ) = ∞. The angles of RN equal δ =
π − (2π/N ). Then, by Theorem 3.1, cos δ is rational, and then so is cos 2π/N =
− cos(π − δ). Thus cos 2π/N = 0, ±1/2, ±1 by [5, Corollary 3.12], and hence
N ∈ {3, 4, 6}.
(ii) Let α, β, γ be the angles of the triangles of a regular tiling of RN . If N =
3, 4, 6, then RN has to satisfy one of (ii), (iii), (v) and (vii) of Theorem 2.1. If RN
satisfies (vii), then there is nothing to prove.
Suppose RN satisfies (ii). Then γ = π − (2π/N ) and sin α, sin β, sin γ are
commensurable. Since 2 cos γ = (sin2 α + sin2 β − sin2 γ )/(sin α sin β), it follows that
cos γ is rational. Then, as we saw above, we have N ∈ {3, 4, 6} which is impossible.
If RN satisfies (iii), then γ = 2π/N and the same argument works.
Finally, suppose that RN satisfies (v). Then α + β = π − (2π/N ) and γ = 2π/N .
Since the sides of RN are equal, we have sin α/ sin β = 1, α = β, and α = β =
(π/2) − (π/N ).
(iii) We have to prove that if N > 420 and RN has a tiling with congruent triangles,
then the angles of the triangles are one of given by (6). By (ii), this is true if the tiling
is regular, so we may assume that it is irregular. By Lemma 3.2, this implies that
α, β, γ are rational multiples of π . Now we shall apply the following result proved
in [4]: Let N > 420, and suppose that RN has a tiling with similar triangles of angles
α, β, γ . If α, β, γ are rational multiples of π , then (α, β, γ ) is one of the triples
of (6), or equals (π − (2π/N ), π/N , π/N ). Since, by Lemma 3.5, the latter case is
impossible, this completes the proof.
It is very likely that c(RN ) = 2 for every N = 3, 4, 6. In order to prove this, we
have to show that if N ≤ 420 and N = 3, 4, 6, then there is no irregular tiling of RN .
This amounts to checking a given finite set of triples. Unfortunately, the number of
cases to consider is enormous, and it seems to be hopeless to exclude these triples
without the use of computer. We plan to return to this computation in the forthcoming
paper [4].
Our next aim is to determine those triangles that tile a rectangle.
Theorem 3.6 If the congruent copies of a triangle T tile a rectangle, then T is a
right triangle, and the ratio of the sides of the rectangle is a rational multiple of
the ratio of the perpendicular sides of T . Therefore, a rectangle A can be tiled with
congruent triangles of angles α, β, γ if and only if a permutation of (α, β, γ ) satisfies
the following condition: γ = π/2 and sin α/ sin β is a rational multiple of the ratio
of the sides of A.
We note the following consequence of Theorem 3.6.
Corollary 3.7 The square can be tiled with congruent triangles of angles α, β, γ if
and only if a permutation of (α, β, γ ) satisfies the following condition: γ = π/2 and
sin α/ sin β is rational. Consequently, in every tiling of the square with congruent
triangles, the pieces must be right triangles with commensurable perpendicular sides.
Lemma 3.8 Suppose that a rectangle A is tiled with congruent copies of a right
triangle. Then the ratio of the perpendicular sides of the triangle is a rational multiple
of the ratio of the sides of A.
Proof Let α, β, γ be the angles of the triangle, where γ = π/2 and α ≥ β. If α = β,
then the triangles are isosceles right triangles, and thus the sides of A are
commensurable by [1, Theorem 2]. In this case the statement of the lemma is true. Therefore,
we may assume that α > β, and thus π/4 < α < π/2. If the tiling is regular, then one
of (i), (ii), (iii) and (vii) of Theorem 2.1 must hold. It is clear that in each of these
cases the statement of the lemma is true.
Therefore, we may assume that the tiling is irregular. Then there is an equation
pα + qβ + rγ = δ with p > q and δ ∈ {π/2, π, 2π }. This implies (p − q)α + q(α +
β) + rγ = δ, and thus α + β = γ = π/2 gives mα = sπ/2, where m = p − q and s
are both positive integers. Note that s ≤ 2δ/π ≤ 4. Since π/4 < α < π/2, we have
1/2 < s/m < 1, and thus s/m is one of the fractions
2 3 3 4 4
, , , , .
3 4 5 5 7
Suppose s/m = 2/3. Then α = π/3 and β = π/6. We may assume that the sides of
the triangles are 1, 2, √3. Then the sides of A are x√3 + y and z√3 + u, where
x, y, z, u are nonnegative integers. If the tiling contains t triangles, then comparing
the areas we obtain
This implies x = u = 0 or y = z = 0. In both cases, one side of A is an integer
multiple of sin α = √3/2, and the other side is an integer multiple of cos α = 1/2;
that is, the statement of the lemma is true.
Next suppose s/m = 3/4. Then α = 3π/8 and β = π/8. We may assume that
the sides of the triangles are sin α, cos α and 1. Then the sides of A are x cos α +
y sin α + z and u cos α + v sin α + w, where x, y, z, u, v, w are nonnegative integers.
If the tiling contains t triangles, then, comparing the areas we obtain
t
(x cos α + y sin α + z) · (u cos α + v sin α + w) = 2 · cos α · sin α.
(9)
By Theorem 3.9 of [5], the degree of the algebraic numbers cos α and sin α
equals four. On the other hand, tan α = √2 + 1, and thus each of the numbers
cos2 α, sin2 α, sin α cos α, tan α belongs to the field Q(√2). Thus the left hand side
of (9) equals I + J , where I = (xw + uz) cos √α+ (yw + zv) sin α and J = xu cos2 α +
(xv + yu) cos α sin α + yv sin2 α + z√w ∈ Q( 2). Since the right hand side of (9)
belongs to Q(√2), we find that I ∈ Q( 2), and thus
√
cos α · (xw + uz) + (yw + zv) tan α ∈ Q( 2).
(8)
By cos α ∈/ Q(√2) we obtain I = 0. Now each term of I is nonnegative, and thus
I = 0 implies xw = uz = yw = zv = 0. If z = 0, then we get u = v = 0. Since
u cos α + v sin α + w = w equals a side of A, we have w = 0, and thus x = y = 0.
Then the lefthand side of (9) is zw, while the righthand side of (9) is irrational. This
is a contradiction, and thus z = 0. The same argument shows w = 0. Then, dividing
both sides of (9) by cos2 α we obtain
t
(x + y tan α) · (u + v tan α) = 2 · tan α
or yv tan2 α + (yu + xv − (t /2)) tan α + xu = 0. Since the minimal polynomial of
tan α = √2 + 1 is X2 − 2X − 1, it follows that xu = −yv, and thus xu = yv = 0.
Then we have either x = v = 0 or u = y = 0. In both cases, one side of A is an
integer multiple of sin α, and the other side is an integer multiple of cos α; that is, the
statement of the lemma is true.
Finally, suppose that s/m = 2/3, 3/4. Then s/m is one of the fractions 3/5, 4/5,
4/7. We can see, applying Theorem 3.9 of [5], that in each case the degree of sin α
is greater than the degree of cos α. Consequently, sin α is not an element of the field
Q(cos α). Note also that cos α is irrational.
We may assume that the sides of the triangles are sin α, cos α and 1. Then the
sides of A are x cos α + y sin α + z and u cos α + v sin α + w, where x, y, z, u, v, w
are nonnegative integers. If the tiling contains t triangles, then, comparing the areas
we obtain (9). Since sin2 α ∈ Q(cos α) but sin α ∈/ Q(cos α), (9) implies
and
xu cos2 α + xw cos α + yv sin2 α + uz cos α + zw = 0
(xv + yu) cos α + (yw + zv) · sin α = (t /2) · cos α · sin α.
By (10) we have xu = xw = yv = uz = zw = 0. Also, (11) gives (xv + yu) cos α +
(yw + zv) = (t /2) · cos α. Since cos α is irrational, this implies yw = zv = 0. Since
max(x, y, z) > 0 and max(u, v, w) > 0, these equations imply either x = z = v =
w = 0 or y = z = u = w = 0. In both cases, one side of A is an integer multiple of
sin α, and the other side is an integer multiple of cos α, which completes the proof.
Proof of Theorem 3.6 Suppose that the rectangle A is tiled with congruent copies of
a triangle T . By [1, Theorem 23], one of the following must hold: (i) T is a right
triangle; (ii) the angles of T are (π/6, π/6, 2π/3); or (iii) the angles of T are given
by one of the following triples:
π π 5π
, ,
8 4 8
,
π π 5π
, ,
4 3 12
,
π π 2π
, ,
12 4 3
.
If T is a right triangle, then the statement of the theorem is true by Lemma 3.8.
Suppose that the angles of T are (π/6, π/6, 2π/3). We may assume that the sides
of the triangles are 1, 1, √3. Then the sides of A are x√3 + y and z√3 + u, where
(10)
(11)
(12)
x, y, z, u are nonnegative integers. If the tiling contains t triangles, then, comparing
the areas we obtain
(x√3 + y) · (z√3 + u) = 2t · √3.
This implies x = u = 0 or y = z = 0. In both cases, there is a side XY of A such
that XY is an integer. Let X = U0, U1, . . . , Uk = Y be a division of XY such that
each subinterval Ui−1Ui (i = 1, . . . , k) is a side of a triangle Ti of the tiling. Since
Ui−1Ui = 1 for every i, it follows that the angle of Ti at one of the vertices Ui−1 and
Ui equals 2π/3. However, the angle of T1 at U0 = X and the angle of Tk at Uk = Y
must be π/6, and thus there is a 0 < i < k such that the angle of both Ti−1 and Ti at
Ui equals 2π/3. Then the triangles Ti−1 and Ti overlap, which is impossible.
In order to complete the proof of Theorem 3.6, we have to prove the following.
Lemma 3.9 Suppose that the angles of T are given by one of the triples of (12).
Then no rectangle can be tiled with congruent copies of T .
Proof First we suppose that a rectangle A of vertices V1, V2, V3, V4 is tiled with
congruent triangles with angles α = π/8, β = 5π/8, γ = π/4. We may assume
that V1 is the origin and V2 is the point (1, 0). Since each of the numbers cot π/8 =
√2 + 1, cot 5π/8 = √2 − 1 and cot π/4 = 1 belongs to Q(√2), it follows from [1,
Theorem 2] that the coordinates of the vertices of the triangles belong to Q(√2).
Let a, b, c denote the sides of the triangles. Then a/c = (sin π/8)/(sin π/4) and
b/c = (sin 5π/8)/(sin π/4). By Theorem 3.9 of [5], the degree of sin π/8 and of
sin 5π/8 equals 4, and thus the ratios a/c and b/c do not belong to Q(√2).
There is a division V1 = U0, . . . , Uk = V2 of the side V1V2 such that Ui−1Ui is
the side of a triangle Ti of the tiling for every i = 1, . . . , k. Let xi denote the first
coordinate of Ui . Since xi − xi−1 ∈ Q(√2) for every i = 1, . . . , k and a/c, b/c ∈/
√
Q( 2), it follows that either xi − xi−1 = Ui−1Ui ∈ {a, b} for every i, or xi − xi−1 =
Ui−1Ui = c for every i.
Suppose Ui−1Ui = c for every i. Then the angles of Ti at the points Ui−1 and Ui
are π/8 and 5π/8 in some order. Since 5π/8 > π/2, it follows that the angle of T1 at
U0 = V1 and the angle of Tk at Uk = V2 must be π/8, and thus there is a 0 < i < k
such that the angle of both Ti−1 and Ti at Ui equals 5π/8. Then the triangles Ti−1
and Ti overlap, which is impossible. This proves that Ui−1Ui ∈ {a, b} for every i.
Next we prove that either Ui−1Ui = a for every i, or Ui−1Ui = b for every i. In
order to prove this we shall need to consider a conjugate tiling as described in [1].
x −Lyet√φ2 d(xen,oyte∈ tQhe). aTuhtoemnoΦrp(xh1is,mx2)of=th(φe (fixe1l)d, φQ((x√2)2))ddefiefinneseda bcyollφin(exa+tioyn√on2)th=e
set of vertices of the tiling. Let X denote the image of X under Φ. Then V1 = V1 and
V2 = V2. The points V1, V2, V3, V4 are the vertices of a rectangle A and, according
to [1], the images of the triangles are nonoverlapping and constitute a tiling of A .
Since the images of the triangles Ti are nonoverlapping, it follows that the
points U0, . . . , Uk constitute a division of the segment V1, V2 in this order. Since
Ui = (φ (xi ), 0) for every i, it follows that the sequence (φ (xi ))ik=0 is strictly
increasing.
Suppose that there are indices 1 ≤ i, j ≤ k such that xi − xi−1 = a and xj −
xj−1 = b. Then φ (xi ) = φ (xi−1) + φ (a) and φ (xj ) = φ (xj−1) + φ (b). Therefore, the
numbers φ (a), φ (b) are positive. On the other hand, a/b = (sin π/8)/(sin 5π/8) =
√2 − 1, and thus
φ (a)/φ (b) = φ (a/b) = −
which is a contradiction. This proves that either Ui−1Ui = a for every i, or Ui−1Ui =
b for every i.
Thus V1V2 is an integer multiple of either a or b. The same is true for the side
V2V3, and thus the area of A is an integer multiple of one of the numbers a2, b2, ab.
On the other hand, the area of any of the triangles is ab(cos γ )/2 = ab√2/4, and
thus the area of A is an integer multiple of ab√2/4. Therefore, one of the numbers
(ab√2)/a2, (ab√2)/b2, (ab√2)/(ab) is rational. However, a/b = √2 − 1, and thus
each of these numbers is irrational, a contradiction. This proves that no rectangle can
be tiled with congruent triangles with angles α = π/8, β = 5π/8, γ = π/4.
Next suppose that a rectangle A of vertices V1 = (0, 0), V2 = (1, 0) and V3, V4
is tiled with congruent triangles with angles α = π/4, β = 5π/12, γ = π/3. Since
each of the numbers cot π/4 = 1, cot 5π/12 = 2 − √3, cot π/3 = √3/3 belongs to
Q(√3), it follows from [1, Theorem 2] that the coordinates of each vertex of A and
of each triangle belong to Q(√3).
Let a, b, c denote the sides of the triangles. Then
a/c = (sin π/4)/(sin π/3) =
√ √ √
2/ 3 ∈/ Q( 3).
Since sin 5π/12 = (√3 + 1)/(2√2), we have
√ √
b/c = (sin 5π/12)/(sin π/3) = ( 3 + 1)/( 6),
and thus b/c does not belong to Q(√3) either.
There is a division V1 = U0, . . . , Uk = V2 of the side V1V2 such that Ui−1Ui is the
side of a triangle Ti of the tiling for every i = 1, . . . , k. Since a/c, b/c ∈/ Q(√3), it
follows that either Ui−1Ui ∈ {a, b} for every i, or Ui−1Ui = c for every i.
Suppose Ui−1Ui = c for every i. Then the angles of Ti at the points Ui−1 and
Ui are π/4 and 5π/12. It is clear that the angle of T1 at U0 = V1 must be π/4 and,
similarly, the angle of Tk at Uk = V2 must be π/4. Therefore, there exists an index
0 < i < k such that the angle of Ti−1 at Ui is 5π/12 and the angle of Ti at Ui is also
5π/12. Since π − 2 · (5π/12) = π/6 < min(α, β, γ ), this is clearly impossible. Thus
Ui−1Ui ∈ {a, b} for every i.
Next we prove that either Ui−1Ui = a for every i, or Ui−1Ui = b for every i. Let
ψ denote the automorphism of the field Q(√3) defined by ψ (x + y√3) = x − y√3
(x, y ∈ Q). Then Ψ (x1, x2) = (ψ (x1), ψ (x2)) defines a collineation on the set of
vertices of the tiling. Let X denote the image of X under Ψ . Then V1 = V1 and
V2 = V2. The points V1, V2, V3, V4 are the vertices of a rectangle A and the images
of the triangles are nonoverlapping.
Therefore, the points U0, . . . , Uk constitute a division of the segment V1, V2 in this
order. Let xi denote the first coordinate of Ui . Then Ui = (ψ (xi ), 0) for every i, and
hence the sequence (ψ (xi ))ik=0 is strictly increasing.
Suppose that there are indices 1 ≤ i, j ≤ k such that xi − xi−1 = a and xj − xj−1 =
b. Then ψ (xi ) = ψ (xi−1) + ψ (a) and ψ (xj ) = ψ (xj−1) + ψ (b). Therefore, the
numbers ψ (a), ψ (b) are positive. On the other hand, a/b = (sin π/4)/(sin 5π/12) =
√3 − 1, and thus
ψ (a)/ψ (b) = ψ (a/b) = −
which is a contradiction. This proves that either Ui−1Ui = a for every i, or Ui−1Ui =
b for every i.
Thus V1V2 is an integer multiple of either a or b. The same is true for the side
V2V3, and thus the area of A is an integer multiple of one of the numbers a2, b2, ab.
On the other hand, the area of any of the triangles is ab(cos γ )/2 = ab√3/4, and
thus the area of A is an integer multiple of ab√3/4. Therefore, one of the numbers
(ab√3)/a2, (ab√3)/b2, (ab√3)/(ab) is rational. However, a/b = √3 − 1, and thus
each of these numbers is irrational, a contradiction. This proves that no rectangle can
be tiled with congruent triangles with angles α = π/4, β = 5π/12, γ = π/3.
Finally, suppose that a rectangle is tiled with congruent triangles with angles α =
π/4, β = π/12, γ = 2π/3. It is easy to check that the conjugate tiling corresponding
to the automorphism of the field Q(√3) defined by ψ (x + y√3) = x − y√3 (x, y ∈
Q) is a tiling of a rectangle with congruent triangles with angles α = π/4, β =
5π/12, γ = π/3 (see [1], p. 291). As we proved above, this is impossible.
4 Proof of Theorem 2.1: Some Preliminary Results
We start with the following simple observation.
Lemma 4.1 Let A be a parallelogram of angles γ and π − γ . Suppose that A is tiled
with congruent copies of a triangle of angles α, β, γ and of sides a, b, c. If one of the
sides of A is an integer multiple of a, then the ratio of the sides of A is a rational
multiple of sin α/ sin β, and thus (i) of Theorem 2.1 holds.
Proof If the other side of A is of length d, then the area of A equals kad · sin γ
with an integer k. Since the area of each triangle of the tiling equals (ab sin γ )/2,
we obtain 2kad sin γ = t · ab sin γ and d = (t /2k) · b, where t is the number of
triangles of the tiling. Thus the ratio of the sides of A equals ka/d = (2k2/t ) · (a/b) =
(2k2/t ) · (sin α/ sin β).
In the sequel we fix a regular tiling of the convex polygon A with the congruent
triangles Δ1, . . . , Δt of angles α, β, γ such that equations (1)–(3) are of the form
pi (α + β) + ri γ = δi , pi (α + β) + ri γ = 2π and pi (α + β) + ri γ = π , respectively.
The vertex Vi of the tiling is called normal, if pi = ri . If pi = ri and Vi is not a vertex
of A, then we say that the vertex Vi is exceptional. Whenever
p(α + β) + rγ = vπ
(13)
is an equation satisfied by α, β, γ , then we shall call (p − s)(α + β) + (r − s)γ =
(v − s)π the reduced form of (13), where s = min(p, r). Thus the reduced form of
the equation at a normal vertex is 0 · (α + β) + 0 · γ = 0 · π .
Let a, b, c denote the sides of the triangles Δi opposite to the angles α, β, γ . Then
a/b = sin α/ sin β and b/c = sin β/ sin γ .
We shall say that a triangle Δ is supported by a segment U V , if one of the sides
of Δ is a subset of U V .
First we assume that a is not a linear combination of b and c with nonnegative
rational coefficients. We define a directed graph Γa on the set of vertices of the tiling
as follows. Let XY be a maximal segment belonging to the union of the
boundaries of the triangles Δi (i = 1, . . . , t ), and suppose that the segment XY lies in
the interior of A, except perhaps the endpoints X and Y . Then there are divisions
X = U0, U1, . . . , Uk = Y and X = V0, V1, . . . , V = Y of the segment XY such that
each subinterval Ui−1Ui (i = 1, . . . , k) is a side of a triangle Ti of the tiling supported
by XY and lying on the same side of the segment XY , and each subinterval Vj−1Vj
(j = 1, . . . , ) is a side of a triangle Tj of the tiling supported by XY and lying on the
other side of XY . Suppose that exactly one of the lengths U0U1 and V0V1 equals a.
By symmetry, we may assume that U0U1 = a = V0V1. Since a is not a linear
combination of b and c with nonnegative rational coefficients, it follows that there is a
unique index 1 ≤ i0 < k such that Ui−1Ui = a for every i ≤ i0 and Ui0 Ui0+1 = a. If
all these conditions are satisfied, then we connect the vertices X and Ui0 by a directed
edge X−−U→i0 . Note that Ui0 belongs to the interior of A and is different from each of the
points Vj (j = 0, . . . , ). Thus Ui0 is in the interior of one of the sides of the triangle
Tj for a suitable j .
Let Γa denote the set of all directed edges defined as above. It is clear that the
indegree of any vertex is zero or one. As we saw above, a vertex V can have an
incoming edge only if V belongs to the interior of A and if V is in the interior of one
of the sides of a triangle of the tiling.
Lemma 4.2 Suppose that a is not a linear combination of b and c with nonnegative
rational coefficients. If V is a normal vertex and V is the endpoint of an edge of Γa ,
then at least one edge of Γa starts from V .
Proof Let X−→V be an edge, and let V = Ui0 , where XY is a maximal segment
belonging to the union of the boundaries of the triangles Δi , and X = U0, U1, . . . , Uk = Y
and T1, . . . , Tk are as in the definition of the graph. Put T = Ti0 and T = Ti0+1. Then
the side of T lying on the segment XY equals a, and the side of T lying on the
segment XY is different from a, and thus the angle of T off the line XY is α, and the
angle of T off the line XY is different from α. Since V is normal, the equation at V
must be α + β + γ = π . Therefore, V is the common vertex of three triangles.
Considering all possible positions of the angle α in these triangles, we can see that
in each case at least one edge starts from V (see Fig. 2). Note that a = b and a = c.
Lemma 4.3 Suppose that a is not a linear combination of b and c with nonnegative
rational coefficients. Let XY be a side of A, and suppose that every vertex V lying
in the interior of the segment XY is normal and is such that no edge of the graph Γa
starts from V . Then each of the following statements is true.
(i) At least one of the angles of A at X and Y is such that the corresponding
equation is of the form p(α + β) = δ with a positive integer p.
(ii) If there is a triangle Δ supported by XY such that its angles on the side XY
are β and γ , then every triangle supported by XY is a translated copy of Δ. In
particular, the length of XY equals k · a, where k is a positive integer.
(iii) If no edge of the graph Γa starts from X or from Y then at least one of the
angles of A at X and Y is such that the corresponding equation is different from
α + β = δ.
Proof There is a division X = U0, U1, . . . , Uk = Y of the segment XY such that each
subinterval Ui−1Ui (i = 1, . . . , k) is a side of a triangle Ti of the tiling supported by
XY . By assumption, Ui is normal, and no edge starts from Ui for every 1 ≤ i ≤ k − 1.
It is easy to check, by inspecting the possible cases, that either Ti+1 is a translated
copy of Ti , or one of the four cases presented in Fig. 3 holds.
It follows from the convexity of A that the equation at X is of the form p(α + β) =
δ or rγ = δ. Suppose it is rγ = δ. Then the angle of T1 at the vertex U0 = X is γ .
Then, considering the possible cases according to Fig. 3, we can see that each triangle
Ti is either a translated copy of T1, or its angle at the vertex Ui is α. This implies that
the angle of Tk at Y is α or β, and thus the equation at Y is p(α + β) = δ. This
proves (i).
If Ti has angles β and γ on XY then, considering again the possible cases
according to Fig. 3, we can see that each triangle Tj is a translated copy of Ti . This
proves (ii).
Suppose that the equation at the vertex X is of the form α + β = δ. If no edge
starts from X, then either the angle of T1 at X is β and at U1 is γ , or its angle at
the vertex X is α. In the first case each triangle Ti must be a translated copy of T1.
Then the angle of Tk at Y equals γ , and thus the equation at Y cannot be of the form
α + β = δ. Suppose that the angle of T1 at X is α. Then, as Fig. 3 shows, the angle
of Ti at Ui−1 is α for every i. In particular, the angle of Tk at Uk−1 is α. If the angle
of Tk at Y equals γ , then the equation at Y cannot be of the form α + β = δ. On the
other hand, if the angle of Tk at Y equals β and the equation at Y is α + β = δ, then
there is an edge starting from the vertex Y . This proves (iii).
Next we consider the case when c is not a linear combination of a and b with
nonnegative rational coefficients. We define the directed graph Γc the same way as
we defined Γa , except that we replace a by c in the definition.
Lemma 4.4 Suppose that c is not a linear combination of a and b with nonnegative
rational coefficients. Let XY be a side of A, and suppose that every vertex V lying
in the interior of the segment XY is normal and is such that no edge of the graph Γc
starts from V . Then
(i) at least one of the angles of A at X and Y is such that the corresponding equation
is of the from p(α + β) = δ with a positive integer p; and
(ii) if there is a triangle Δ supported by XY such its angles on the side XY are α and
β, then every triangle supported by XY is a translated copy of Δ. In particular,
the length of XY equals k · c, where k is a positive integer.
Proof There is a division X = U0, U1, . . . , Uk = Y of the segment XY such that each
subinterval Ui−1Ui (i = 1, . . . , k) is a side of a triangle Ti of the tiling supported
by XY . It is easy to check, by inspecting the possible cases, that either Ti+1 is a
translated copy of Ti , or one of the presented in Fig. 4 four cases holds. Then we can
repeat the argument of the proof of Lemma 4.3.
Lemma 4.5 Suppose that c is not a linear combination of a and b with nonnegative
rational coefficients. Let the segment XY belong to the union of the boundaries of the
triangles, and let T and T be two triangles supported by XY and lying on the same
side of XY . Suppose that T and T have a common vertex V on the segment XY and
that V is either on the boundary of A, or is an inner point of a side of a triangle. If
the side of T lying on XY equals c and the side of T lying on XY is different from
c, then there is an edge of Γc starting from V .
Proof If V is normal, then we can easily check that an edge starts from V (see Fig. 2
with α replaced by γ ).
Suppose the vertex V is exceptional. Then the equation at V is of the form rγ = π
or p(α + β) = π . If the equation is rγ = π , then neither of the triangles T and T
can have a side of length c lying on XY , which is impossible. Thus the equation is
p(α + β) = π . Let T1, . . . , T2p denote the triangles having V as a vertex listed
clockwise. We may assume, by symmetry, that T1 = T and T2p = T . There are halflines
hi (i = 0, . . . , 2p) starting from V such that Ti lies in the angular domain bounded
by hi−1 and hi for every i = 1, . . . , 2p. Let αi and βi denote the angles of Ti at its
vertices lying on the halflines hi−1 and hi , respectively. Then α1 = γ and β2p = γ
by the conditions on T and T . Since the angle of Ti at V is different from γ for
every i, it follows that there exists an index i < 2p such that αi = γ and βi+1 = γ or
αi = γ and βi+1 = γ . Then an edge of Γc starts from V along the halfline hi , which
completes the proof of the lemma.
Lemma 4.6 If a and b are commensurable and c is not a rational multiple of a, then
the graph Γc is empty: it has no edge.
Proof By Lemma 4.5, the indegree is not greater than the outdegree at each vertex.
Therefore, the indegree equals the outdegree at each vertex. Since no edge arrives
at the boundary of A, it follows that no edge of Γc starts from the boundary of A.
Suppose that Γc is not empty, and let G denote the set of vertices of nonzero
outdegree. Then each vertex V ∈ G is the starting point of an edge and also the endpoint
of another edge. Let V be a vertex of the convex hull of G. Then V ∈ G, and thus V is
the endpoint of an edge X−→V . Let V = Ui0 , where XY is a maximal segment
belonging to the union of the boundaries of the triangles, and let X = U0, U1, . . . , Uk = Y
and T1, . . . , Tk be as in the definition of the graph. Since X ∈ G, it follows that the
segment V Y is outside the convex hull of G, and thus no edge starts or arrives at any
point of V Y except at V .
Suppose that an inner point W of the segment XY is the endpoint of an edge Z−−→W.
Then W is an inner point of a side of a triangle which must be supported by XY . Thus
the starting point Z of the edge must be either X or Y . It cannot be Y , as Y is outside
the convex hull of G, and thus Y ∈/ G. Thus Z = X, and then W = V . Therefore,
among the vertices that are inner points of XY only V belongs to G.
Let be the smallest positive index for which U is a common vertex of triangles
lying on different sides of XY . Then i0 < ≤ k. It follows from Lemma 4.5 that
for every i0 < i ≤ , the length of the segment Ui−1Ui is either a or b. Therefore,
XU = i0c + ra with a positive rational r .
Let d1, . . . , dν be the side lengths of the triangles supported by XU and lying on
the side opposite to the triangles Ti . Then either each di equals c or each is different
from c. Indeed, otherwise an edge would start from an inner point of XY different
from V which is impossible. Thus the length XU is either k c with a positive integer
k or sa with a positive rational s. Since both of the equation i0c + ra = k c and
i0c + ra = sa implies that c is a rational multiple of a, both cases are impossible.
Lemma 4.7 If a and b are commensurable and c is not a rational multiple of a, then
the sides of A are pairwise commensurable.
and let θi denote the directed angle between the positive xaxis and hi (i = 1, . . . , k).
Then we define
k
i=1
Φ(P ) =
χ (θi ) · Xi−1Xi .
(14)
The definition makes sense, as the function χ is periodic mod 2π . Using the fact
that χ (θ + π ) = −χ (θ ) for every θ ∈ G it is easy to see that the function Φ is
additive in the following sense: if P is decomposed into the nonoverlapping
polygons P1, . . . , Pt ∈ PG, then Φ(P ) = ti=1 Φ(Pi ). It is clear that Φ is invariant under
translations. Also, it follows from the multiplicative property of χ that if P ∈ PG and
P is obtained from P by a rotation of angle θ ∈ G, then Φ(P ) = χ (θ ) · Φ(P ).
Let αi be the angle of P at the vertex Xi (i = 1, . . . , k), and put αi = π − αi . Since
the halfline hi+1 is obtained from hi by a rotation of angle αi , it follows that
θi ≡ θ1 + α1 + · · · + αi−1
(mod 2π )
for every 1 ≤ i ≤ k. Thus χ (θi ) = χ (θ1) · χ (α1) · · · χ (αi−1) for every 1 ≤ i ≤ k, and
Φ(P ) = χ (θ1) · X0X1 + χ α1 · X1X2 + · · · + χ α1 · · · χ αk−1 · Xk−1Xk . (15)
Now we turn to the proof of Theorem 2.1. We shall consider the following five
cases separately: γ > π/2 and γ = 2π/3; γ < π/2 and γ = π/3; γ = 2π/3; γ =
π/3; γ = π/2.
5 Case I: γ > π/2, γ = 2π/3
In this case, in each of equations (1), if ri > 0, then the equation must be of the
form γ = δi . We claim that in each of equations (2) and (3) we have ri ≤ pi . Indeed,
otherwise the reduced form of the equation in question would be rγ = vπ , where r is
a positive integer and v = 1 or 2. However, as π/2 < γ < π and γ = 2π/3, no such
equation is possible.
Let P (α + β) + Rγ = (N − 2)π be the sum of equations (1). Then, by (4), we
have P ≤ R. If N = 3 then P (α + β) + Rγ = π , and thus P ≤ R gives P = 0 or
P = R = 1. If P = 0 then Rγ = π which is impossible. If P = R = 1, then at one
of the angles of A the equation is α = δ which contradicts the regularity of the tiling.
Thus N = 3 is impossible, and we have N ≥ 4.
By P (α + β) + Rγ = (N − 2)π and P ≤ R we have R ≥ N − 2, and thus the
possible values of R are N − 2, N − 1 and N . If R = N − 2, then necessarily P =
N − 2, and in each of the equations (2) and (3) we have pi = ri .
If R = N − 1, then N − 1 of the angles of A equal γ . The equation P (α + β) +
(N − 1)γ = (N − 2)π gives P (α + β) + (N − 1)(π − α − β) = (N − 2)π and
π = (N − 1 − P )(α + β). There must exist an equation pi (α + β) + ri γ = vπ with
ri < pi and v = 1, 2. The reduced form of this equation is p(α + β) = vπ , where
we have p = (N − 1 − P )v. Since pi = ri , it follows that p + P ≤ R; that is,
(N − 1 − P )v + P ≤ N − 1. Thus v = 1, and there is no other vertex with rj < pj and
j > N . Therefore, apart from the vertices of A, there is only one vertex of the tiling
with pi = ri , and the reduced equation at this vertex equals (N − 1 − P )(α + β) = π .
Finally, if R = N , then each angle of A equals γ . Thus α + β = π − γ = 2π/N .
There must exist an equation pi (α + β) + ri γ = vπ with ri < pi and v = 1, 2. The
reduced form of this equation is p(α + β) = vπ , where we have 2p = N v. Since
pi = ri , it follows that p + P ≤ R; that is, N · (v/2) ≤ N . If v = 2, then there
is no other vertex with rj < pj and j > N . If v = 1, then there is one more such
vertex, where the reduced equation is (N /2) · (α + β) = π . Therefore, apart from the
vertices of A, there are at most two vertices with pi = ri . If there is one such vertex,
then the equation at this vertex is N (α + β) = 2π , and if there are two such vertices,
then N is even and the corresponding reduced equations are (N /2) · (α + β) = π .
Summing up: there are three cases. In the first case P = R = N − 2, and pi = ri
for every i > N ; that is, every vertex other than the vertices of A is normal. In this
case N − 2 of the vertices of A equals γ , and the other two vertices equal p(α + β)
and p (α + β), where p + p = N − 2.
In the second case R = N − 1, every vertex other than the vertices of A is normal
with one exception. In the exceptional vertex the reduced equation is (N − 1 − P )(α +
β) = π . In this case N − 1 of the vertices of A equals γ , and one equals P (α + β).
In the third case R = N , every vertex other than the vertices of A is normal with at
most two exceptions. If there is only one exceptional vertex, then the corresponding
equation is N (α + β) = 2π ; and if there are two exceptional vertices, then their
reduced equations are (N /2) · (α + β) = π . In this case each angle of A equals γ .
5.1 Subcase Ia
First we assume that a is not a linear combination of b and c with nonnegative rational
coefficients. Then we may consider the directed graph Γa .
Suppose that P = R = N − 2. Then there is no exceptional point other than the
vertices of A and thus, by Lemma 4.2, the outdegree is not smaller than the
indegree at each vertex. Therefore, they are equal everywhere, and thus the graph Γa
is the union of disjoint directed cycles. Since the indegree is zero at each vertex
belonging to the boundary of A, it follows that no edge starts from the boundary
of A. Therefore, by (i) of Lemma 4.3, A does not have two consecutive vertices with
angle γ . However, we know that A has N − 2 vertices where its angle is γ . Since
N − 2 ≥ 2, it follows that N = 4. Then P = 2, and thus two equations at the vertices
of A are of the form α + β = δ. In other words, A is a quadrilateral with angles
γ , α + β, γ , α + β in this order. Therefore, A is a parallelogram.
Let X be a vertex of A with angle γ , and let T be the unique triangle of the tiling
having X as a vertex. Let U be the vertex of T with angle β, and suppose that U is
on the side XY . Then, by (ii) of Lemma 4.3, the length of the side XY equals ka,
where k is a positive integer. Therefore, by Lemma 4.1, (i) of Theorem 2.1 holds in
this case.
Next suppose that R = N − 1. Let V denote the exceptional vertex where the
reduced equation is (N − 1 − P )(α + β) = π . If, at the vertex V , the outdegree
is not smaller than the indegree then, as this is also true at every other vertex, it
follows that the outdegree is equal to the indegree everywhere, and Γa is the union
of disjoint directed cycles. Since the indegree is zero at each vertex belonging to the
boundary of A, it follows that no edge starts from the boundary of A. We know that
N − 1 ≥ 3 vertices of A equals γ , and thus there are at least N − 2 ≥ 2 sides of A
such that A has angle γ at each endpoint. However, the exceptional point V can be an
inner point of at most one of these sides, so there is a side without exceptional point
and outgoing edge such that A has angle γ at each endpoint. By (i) of Lemma 4.3,
this is impossible.
Therefore, the outdegree is smaller than the indegree at the vertex V . This means
that there is an edge arriving at V , but no edge starts from V . Then V is an inner point
of A, and thus there is no exceptional point on the boundary of A. Since the
outdegree is not smaller than the indegree at the vertices different from V , it follows
that Γa is the union of disjoint directed cycles and one path arriving at V . Therefore,
on the boundary of A there is at most one vertex with positive outdegree. Then we
can find again a side of A without exceptional points and without outgoing edge such
that A has angle γ at each endpoint, which is impossible.
Next we consider the case R = N . Then all angles of A equal γ , and then there
are N ≥ 4 sides such that the angles of A at the endpoints equal γ . We know that
there are at most two exceptional points, where the outdegree can be smaller than
the indegree. The deficit can be at most two, and thus Γa is the union of disjoint
directed cycles and at most two paths arriving at the exceptional point(s). Considering
the possible cases according to the number of exceptional points on the boundary of
A, we can check that in each case there is a side of A without exceptional points
and without outgoing edge such that A has angle γ at each endpoint. But this is a
contradiction again. This completes the proof in the subcase when a is not a linear
combination of b and c with nonnegative rational coefficients.
5.2 Subcase Ib
Suppose that b is not a linear combination of a and c with nonnegative rational
coefficients. Since the roles of a and b are symmetric in the conditions as well as in the
statements of the theorem, this subcase can be treated similarly to Subcase Ia.
Therefore, we may assume that a is a linear combination of b and c with
nonnegative rational coefficients, and b is a linear combination of a and c with nonnegative
rational coefficients. This implies that either a, b, c are pairwise commensurable, or
a and b are commensurable and c is not a rational multiple of a.
If a, b, c are pairwise commensurable, then sin α, sin β, sin γ are pairwise
commensurable, and so are the sides of A. We show that this implies P = R = N − 2,
and thus in this case statement (ii) of the theorem holds. Indeed, if R = N − 1 or
R = N , then α and β satisfy an equation of the form p(α + β) = vπ , and thus γ is
a rational multiple of π . Since cos γ = (c2 − a2 − b2)/(2ab) is rational as well, it
follows that γ = π/2, π/3 or 2π/3 (see [5, Corollary 3.12]), which is not the case.
5.3 Subcase Ic
Thus we are left with the case when a and b are commensurable and c is not a rational
multiple of a. Then the graph Γc is empty by Lemma 4.6.
If P = R = N − 2, then using (i) of Lemma 4.4 and the corresponding argument
above, we can see that A is a parallelogram. Since A is rational by Lemma 4.7, we
find that the case (i) of Theorem 2.1 holds.
The cases R = N − 1 and R = N are impossible. Indeed, in these cases there is
a side XY of A such that the angle of A at the vertices X and Y equals γ , and XY
does not contain exceptional points. This, together with the fact that Γc is empty,
contradicts (i) of Lemma 4.4.
6 Case II: γ < π/2, γ = π/3
In this case, in each of equations (1), if pi > 0, then the equation must be of the form
α + β = δi . We claim that in each of equations (2) and (3) we have ri ≥ pi . Indeed,
otherwise the reduced form of the equation in question would be p(α + β) = vπ ,
where p is a positive integer and v = 1 or 2. However, as π/2 < α + β < π and
α + β = 2π/3, no such equation is possible.
Let P (α + β) + Rγ = (N − 2)π be the sum of equations (1). Then, by (4), we
have P ≥ R. If N = 3 then P (α + β) + Rγ = π , and thus P ≥ R gives R = 0 or
P = R = 1. If R = 0 then P (α + β) = π which is impossible. If P = R = 1, then at
one of the angles of A the equation is α = δi which contradicts the regularity of the
tiling. Thus N = 3 is impossible, and we have N ≥ 4.
By P (α + β) + Rγ = (N − 2)π and P ≥ R we have P ≥ N − 2, and thus the
possible values of P are N − 2, N − 1 and N . If P = N − 2, then necessarily
R = N − 2, and in each of the equations (2) and (3) we have pi = ri .
If P = N − 1, then N − 1 of the angles of A equal α + β. The equation
(N − 1)(α + β) + Rγ = (N − 2)π gives (N − 1 − R)γ = π . There must exist an
equation pi (α + β) + ri γ = vπ with ri > pi and v = 1, 2. The reduced form of this
equation is rγ = vπ , where we have r = (N − 1 − R)v. Since pi = ri , it
follows that, apart from the vertices of A, there is only one vertex of the tiling with
pi = ri , and that the reduced equation at this vertex is (N − 1 − R)γ = π .
Finally, if P = N , then each angle of A equals α + β. Thus γ = π − (α + β) =
2π/N . There must exist an equation pi (α + β) + ri γ = vπ with ri > pi and v = 1, 2.
The reduced form of this equation is rγ = vπ , where we have 2r = N v. Since
pi = ri , it follows that, apart from the vertices of A, there are at most two
vertices with pi = ri . If there is one such vertex, then the reduced equation at this vertex
is N γ = 2π , and if there are two such vertices, then N is even and the corresponding
reduced equations are (N /2)γ = π .
Summing up: there are three cases. In the first case P = R = N − 2, and pi = ri
for every i > N . In this case N − 2 of the vertices of A equals α + β, and the other
two vertices equal rγ and r γ , where r + r = N − 2.
In the second case P = N − 1, and pi = ri for every i > N with exactly one
exception, where the reduced equation is (N − 1 − R)γ = π . In this case N − 1 of
the vertices of A equals α + β, and one equals Rγ .
In the third case P = N , and pi = ri for every i > N with at most two
exceptions. If there is only one exceptional vertex, then the corresponding equation is
N γ = 2π , and if there are two exceptional vertices, then their reduced equations
are (N /2)γ = π . In this case each angle of A equals α + β.
6.1 Subcase IIa
First we assume that a is not a linear combination of b and c with nonnegative rational
coefficients, and consider the graph Γa . If R = N − 2, then there are no exceptional
points apart from the vertices of A, and the indegree equals the outdegree at every
vertex. Thus no vertex starts from the boundary of A. By (iii) of Lemma 4.3 this
implies that there are no adjacent vertices of A with angle α + β. As in case I, we can
infer that A is a parallelogram, and using an analogous argument, we can check that
in this case the statement (i) of the theorem holds. Repeating the analogous argument
of case I, we can also see that the cases R = N − 1 and R = N are impossible.
6.2 Subcase IIb
Since the roles of a and b are symmetric, we have the same conclusion if b is not a
linear combination of a and c with nonnegative rational coefficients.
Therefore, we may assume that a is a linear combination of b and c with
nonnegative rational coefficients, and b is a linear combination of a and c with nonnegative
rational coefficients. This implies that either a, b, c are pairwise commensurable, or a
and b are commensurable and c is not a rational multiple of a. If a, b, c are pairwise
commensurable, then (iii) of the theorem is true. Indeed, if R = N − 1 or R = N ,
then γ is a rational multiple of π , which is impossible (see the analogous argument
in case I).
6.3 Subcase IIc
In this subcase we assume that a and b are commensurable and c is not a rational
multiple of a. Then the graph Γc is empty by Lemma 4.6.
Lemma 6.1 If a and b are commensurable and c is not a rational multiple of a, then
either (i) of Theorem 2.1 holds, or P = N ; that is, each angle of A equals α + β.
Proof Let X, Y, Z be consecutive vertices of A, and suppose that the equation at X
is rγ = δ. We show that if there are no exceptional points on the sides XY and Y Z,
then the equation at Z is also of the form rγ = δ. Indeed, by (i) of Lemma 4.4, the
equation at Y must be α + β = δ. The proof of Lemma 4.4 also shows that there is a
triangle Tk supported by the side XY , and there is a point Uk−1 in the interior of XY
such that Uk−1Y is a side of Tk , and the angle of Tk at Uk−1 is γ .
Let T1, . . . , Tm be the triangles supported by the side Y Z, and let Y = V0, . . . , Vm =
Z be a division of Y Z such that Vi−1Vi is a side of Ti for every i = 1, . . . , m. Then
the angle of T1 at the vertex Y equals α or β. Since no edge of Γc starts from Y and
the angle of Tk at Uk−1 equals γ , it follows that the angle of T1 at V1 must be γ .
Now the vertices Vi are normal by assumption, and no edge of Γc starts from any of
them. Therefore, each Ti has angle γ at the vertex Vi (see Fig. 4). In particular, Tm
has angle γ at Vm = Z, and thus the equation at Z is of the form rγ = δ.
Suppose R = N − 2. Then there are no exceptional points, and thus it follows from
what we proved above that the equation at every second vertex of A is of the form
rγ = δ. Then N = 4 and A is a parallelogram. By Lemma 4.7, the sides of A are
commensurable. Since a/b is rational, it follows that (i) of Theorem 2.1 holds.
Next suppose P = N − 1. Then there is one single exceptional point. Let
Z1, Y1, X, Y2, Z2 be consecutive vertices of A such that the equation at the vertex
X is of the form rγ = δ. (The vertices Z1 and Z2 may coincide.) Since only one of
the sides Z1Y1, Y1X, XY2, Y2Z2 can contain the exceptional point, it follows that
the equation at either Z1 or Z2 is of the form rγ = δ. This, however, contradicts the
assumption P = N − 1. Therefore, the only remaining possibility is P = N , which
completes the proof.
Therefore, we are left with the case when P = N . If N = 4, then A is a rectangle.
Since α + β = π/2, sin α/ sin β = a/b is rational and A is rational by Lemma 4.7,
it follows that (i) of Theorem 2.1 holds. Thus we may assume that N ≥ 5. Our next
aim is to show that in this case (v), (vi) or (vii) of Theorem 2.1 holds. The rest of the
section is devoted to the proof of this statement.
In Lemma 4.8 we proved that those triangles of the tiling that have their side of
length c in the interior of A come in pairs, and each pair forms a quadrilateral which
is either a parallelogram of sides a, b and of angles γ , α + β, or a kite of sides a, b
and of angles 2β, γ , 2α, γ .
Lemma 6.2 Suppose that a and b are commensurable, c is not a rational multiple of
a, and P = N . Then among the pairs described above only parallelograms can occur.
More precisely, there are no kites, unless α = β when each kite is a parallelogram.
Proof Suppose this is not true; that is, α = β and there exists at least one kite. We
shall denote by Γkite the set of those directed segments X−→Y for which the segment
XY is the common side of two triangles having angles α at X and β at Y . Then the
line going through the edge X−→Y of Γkite is the axis of symmetry of a kite described
in Lemma 4.8.
If X−→Y is an edge of Γkite, then Y is in the interior of A. Indeed, the left hand side
of the equation at Y is of the form p(α + β) + rγ with p ≥ 2. Since r ≥ p ≥ 2, we
have p = r = 2, and the equation at Y is 2α + 2β + 2γ = 2π .
By Lemma 4.8, the triangles having Y as a vertex and having angle α or β at Y
come in pairs. Each pair forms a quadrilateral which is either a parallelogram having
angle α + β or γ at Y , or a kite having angle 2α or 2β at Y .
Since p = r = 2, it follows that the arrangement of the triangles around Y must be
one of two cases presented in Fig. 5.
Since there is exactly one kite having angle 2α at Y , it follows that there is exactly
one edge of Γkite starting from Y . Therefore, Γkite is the union of disjoint cycles, and
every vertex of Γkite is in the interior of A. Note also that each cycle of Γkite is a
simple polygon of angles α + β or α + 2γ + β.
We shall need another graph on the set of vertices. Let XZ be a maximal segment
contained by the union of the boundaries of the triangles Δi and such that XZ belongs
to the interior of A except perhaps the endpoints X and Z. There are divisions X =
U0, U1, . . . , Uk = Z and X = V0, V1, . . . , V = Z of the segment XZ such that each
subinterval Ui−1Ui (i = 1, . . . , k) is a side of a triangle Ti of the tiling supported by
XZ and lying on the same side of the segment XZ, and each subinterval Vj−1Vj
(j = 1, . . . , ) is a side of a triangle Tj of the tiling supported by XZ and lying on
the other side of XZ. Suppose that the angle of T1 at the vertex U1 and the angle
of T1 at the vertex V1 both equal γ . Then there is a maximal index 1 ≤ i0 ≤ k such
that the angle of Ti at the vertex Ui equals γ for every i = 1, . . . , i0. Similarly, there
is a maximal index 1 ≤ j0 ≤ such that the angle of Tj at the vertex Vj equals γ
for every j = 1, . . . , j0. Then we connect the vertices X and Y by a directed edge
X−→Y where Y = Ui0 if Ui0 is closer to X than Vj0 , Y = Vj0 if Vj0 is closer to X than
Ui0 , and Y = Ui0 = Vj0 otherwise. We denote by Γexcept the set of these edges. (The
notation will be justified later, when we show that each edge of Γexcept starts from a
vertex of A and arrives at an exceptional point.)
Let X−→Y be an edge of Γexcept. We show that (i) no inner point of the segment XY
can be a vertex of Γkite, and (ii) Y is a vertex of Γkite or Y is an exceptional point.
In the proof of these statements we shall use the notation of the definition of the
graph Γexcept. By symmetry, we may assume that Y = Ui0 . Let V be an inner point
of XY , and suppose that V is a vertex of Γkite. Then the arrangement of the triangles
around V cannot be as shown by (B) of Fig. 5 since 2α + γ = π and 2β + γ = π . But
it cannot be as shown by (A) of Fig. 5 either. Indeed, if i and j are such that Ti , Tj
have V as a vertex, then they are supported by the segment XZ and have angles γ
at V . However, those two triangles in (A) of Fig. 5 which have angle γ at the given
point are not supported by a common segment. This proves (i).
Next suppose that Y is not a vertex of Γkite and that Y is normal. Since Y = Ui0 ,
we have either U0 = Vj0 or U0 is closer to X than Vj0 . This implies that either Y is in
the interior of a side of one of the triangles Tj , or Y is the vertex of a triangle Tj such
that the angle of Tj at Y equals γ . In the first case the equation at Y is α + β + γ = π ,
and the angle of Ti0 at Y equals γ . By Lemma 4.6, no edge of Γc starts from Y . This
implies that the angle of the triangle Ti0+1 at the point Ui0+1 must be γ . (Indeed,
by considering the possible arrangements of the angles around Y we can see that
otherwise an edge of Γc would start from Y .) This, however, contradicts the choice
of i0, since i0 was the largest index such that Ti has angle γ at Ui for every i ≤ i0.
In the second case there are two triangles supported by the continuation of the
segment XY and having angle γ at Y . Since Y is normal, there are four other triangles
having Y as a vertex, and their angles at Y are α, α, β, β. The two triangles with angle
α at V cannot be adjacent, because in that case, depending on the location of their
angle γ , either there would be an edge of Γc starting from Y , or Y would be a vertex
of Γkite. Thus the angles of the four triangles at V are either α, β, α, β or β, α, β, α in
this order. Thus Y is an inner point of the segment XZ. Then we can check, the same
way as in the previous case, that the angle of Ti0+1 at the point Ui0+1 must be γ . This,
again, contradicts the maximality of i0, which completes the proof of (ii).
If P is a simple polygon, then the bounded component of R2 \ P will be denoted
by P °.
Let C be a cycle of Γkite. We shall also denote by C the polygon formed by the
edges belonging to C. Let X be a vertex of C such that the angle of C at X is convex;
that is, equals α + β. (For short, in this case we shall say that X is a convex vertex
of C.) Let h denote the halfline starting from X as in (B) of Fig. 5. Then there is a
point Y on h such that X−→Y is an edge of Γexcept. Since h intersects C°, and no interior
point of XY can be a vertex of Γkite, it follows that one of the following must hold:
(i) Y is a vertex of C; (ii) Y ∈ C° and Y is a vertex of Γkite; or (iii) Y ∈ C° and Y is
exceptional.
Note that in case (i) the angle of C at Y is concave; that is, equals α + 2γ + β.
Indeed, Y must be a vertex of C as in (B) of Fig. 5. Since the halfline h arrives at Y
from C°, it follows that the angle of C at Y equals α + 2γ + β.
Our next aim is to show that for every cycle C of Γkite, C° contains all exceptional
points. In order to prove this we may assume that C° does not contain any vertex of
Γkite. Indeed, if C◦ ∩ Γkite = ∅, then we take a cycle C1 of Γkite belonging to C° and
having minimal area. Then C1◦ does not contain any vertex of Γkite, and if C1◦ contains
all exceptional points then so does C°.
As we proved above, if V is a convex vertex of C, then there is an edge V−−→W of
Γexcept such that either W is a concave vertex of C or W ∈ C° and W is exceptional.
(Here we used the fact that C° does not contain any vertex of Γkite.) If u and v denote
the number of convex and concave vertices of C, respectively, then we have
u(α + β) + v(α + β + 2γ ) = (u + v − 2)π.
Since γ = 2π/N and α + β = π(1 − (2/N )), this implies u = v + N . Let V and
V be distinct convex vertices of C, and let V−−→W, V−−−W→ be the corresponding edges
of Γexcept. It is clear that if W and W are concave vertices of C then they must be
distinct (see (B) of Fig. 5). Since u = v + N , it follows that there are at least N convex
vertices of C such that the corresponding edge aV−−→W arrives at an exceptional point
belonging to C°. This proves the claim if there is only one exceptional point. Suppose
there are two exceptional points: E and F . Then N is even, and the equation at E
is either (N /2)γ = π or α + β + ((N /2) + 1)γ = 2π . In both cases, the number
of edges of Γexcept ending at E is at most N /2; this follows from the fact that if an
edge arrives at E then there are two triangles having E as a vertex and such that they
are supported by the same segment, and their angles at E equal γ . The same is true
for F . Since there are at least N edges arriving at exceptional points belonging to C°,
it follows that both E and F must be in C°. This proves our claim.
Let C1 and C2 be disjoint cycles of Γkite. Since both of C1◦ and C2◦ contain the
exceptional points, we have C1◦ ∩ C2◦ = ∅, and thus we have either C2 ⊂ C1◦ or C1 ⊂ C◦.
2
Therefore, we may list the cycles of Γkite as C1, . . . , Cs , where Ci+1 ⊂ Ci◦ for every
Fig. 6
i = 1, . . . , s − 1, and Cs◦ contains all exceptional points. In particular, there are no
exceptional points in A \ C◦.
1
We prove that C1 is convex. Suppose this is not true, and let V be a concave vertex
of C1. Then there is an edge V−−→W of Γexcept starting from V (see (B) of Fig. 5). Since
the edge starts from A \ C1◦ and A \ C1◦ contains no vertex of Γkite and no exceptional
points, it follows that W belongs to the boundary of A \ C1◦. That is, either W is in the
boundary of A, or W is a vertex of C1. But W cannot be a boundary point of A, since
W is either a vertex of Γkite or is exceptional. Hence W is a vertex of C1. Clearly,
the segment V W is in the exterior of C1 (except the points V and W ). The argument
above proves that for every concave vertex V of C1 there is another vertex W of C1
such that the segment V W is in the exterior of C1. These segments V W are pairwise
disjoint, since any intersection would be an exceptional point.
For every concave vertex V , the vertices V and W divide C1 into the subarcs σ1V
and σ2V . Let PiV denote the simple polygon (V W ) ∪ σiV (i = 1, 2). We may choose
the indices in such a way that (P1V )◦ is disjoint from C◦, and (P2V )◦ contains C◦.
1 1
Let V be a concave vertex of C1 for which the area of (P1V )◦ is minimal. Since
(P1V )◦ ∩ C1◦ = ∅, it follows that there exists a concave vertex V in the subarc σ1V .
If V−−−W→ is the edge of Γexcept starting from V then, as the segments V W and V W
disjoint, W belongs to σ1V . Then the area of (P1V )◦ is smaller than that of (P1V )◦,
contradicting the choice of V . This contradiction proves that C1 is convex. Then
every angle of C1 equals α + β = (1 − (2/N ))π , and hence C1 has N vertices.
Let X be a vertex of A. Since no edge of Γc starts from X, the triangles having X
as a vertex must be arranged as in Fig. 6.
Suppose that they are arranged as in (B) of Fig. 6. Let XY be a side of A. Let
X = U0, . . . , Uk = Y be a division of XY such that each Ui−1Ui is the side of a
triangle Ti of the tiling. Since the angle of T1 at the vertex U1 is γ and every vertex
on XY is normal, it follows that the angle of Ti at the vertex Ui is γ for every i (see
Fig. 4). In particular, the angle of Tk at the vertex Uk = Y is γ , which is impossible.
Therefore, the triangles T and T having X as a vertex must be arranged as in (A)
of Fig. 6. Then the angles of T and of T opposite to the sides on the boundary of
A equal γ , which means that there exists an edge X−−→W of Γexcept starting from X.
Fig. 7
Since A \ C1◦ contains no vertex of Γkite and no exceptional points, it follows that W
belongs to the boundary of A \ C◦. We can check, using a previous argument, that
1
W is a vertex of C1. Clearly, the segment XW is in the exterior of C1 (except the
point W ).
Let V1, . . . , VN = V0 denote the vertices of A such that Vi−1 and Vi are adjacent
for every i = 1, . . . , N . As we saw above, there are vertices Wi of C1 such that V−−iW→i
is a vertex of Γexcept for every i = 1, . . . , N . The segments Vi Wi are pairwise disjoint,
since an intersection of two of them would be an exceptional point.
Since C1 is an N gon, it has no vertices other than W1, . . . , WN . Put W0 = WN .
Then Wi−1 and Wi are adjacent vertices of C1 for every i = 1, . . . , N . Indeed,
suppose that W1, Wi , Wj , . . . , Wk is a list of the consecutive vertices of C1, where 2 <
i < N . Then P = {V1V2 . . . Vi Wi W1V1} and Q = {Vi Vi+1 . . . VN , V1, W1, Wi , Vi } are
simple polygons. If Wj ∈ P °, then all vertices of C1 other than W1 and Wi belong to
P °, since the sides of C1, with the exception of WkW1, W1Wi and Wi Wj , are disjoint
from the boundary of P . Since WN ∈/ P °, this is impossible. If Wj ∈ Q°, then we
obtain a contradiction by W2 ∈ P °. This shows that W1 and Wi can be adjacent vertices
of C1 only if i = 0 or i = 2. We find, in the same way, that Wi−1 and Wi are
adjacent for every i = 1, . . . , N . Therefore, the quadrilaterals Ri = {Vi−1, Vi , Wi , Wi−1}
(i = 1, . . . , N ) form a ring along the boundary of A as shown in Fig. 7.
Now we prove that this is impossible (assuming α = β). Since the equations at
the vertices of A are of the form α + β = δ, the angles of Ri at the vertices Vi−1
and Ri are equal to either α or β. We shall assume that the angle of R1 at the vertex
V0 equals α. (The same argument applies if it equals β.) Let T denote the triangle
having V0 as a vertex and lying in R1. Since V−−0−W→0 is an edge of Γexcept, the angle
of T opposite to the side supported by V0V1 equals γ . By (ii) of Lemma 4.4, each
triangle supported by V0V1 is a translated copy of T . This implies that the angle of
R1 at the vertex V1 equals β. Then the angle of R2 at the vertex V1 equals α, and (ii)
of Lemma 4.4 gives that the angle of R1 at the vertex V2 equals β. Continuing the
argument we find that for every i, the angle of Ri at the vertex Vi−1 equals α, and its
angle at the vertex Vi equals β.
The vertex W0 is a vertex of Γkite, and it is the endpoint of the edge V−−0−W→0 of
Γexcept. Therefore, the angle of R1 at the vertex W0 equals either γ + α or γ + β
(see (B) of Fig. 5). Suppose it is γ + α. Then the angle of R1 at the vertex W1 equals
γ + β, since 2π − α − β − (γ + α) = γ + β. Thus the angle of R2 at the vertex W1
Fig. 8
equals γ + α, and its angle at W2 equals γ + β. Continuing the argument we find that
for every i, the angles of Ri are as in (A) of Fig. 8. If the angle of R1 at the vertex W0
equals γ + β, then we find that for every i, the angles of Ri are as in (B) of Fig. 8.
Now we prove that both cases are impossible if α = β. First suppose that each
Ri looks like (A) of Fig. 8. There is a point Y on the segment Vi−1Vi such that
Vi−1Wi−1Y = γ and Wi Wi−1Y = α. Then the segments Y Wi−1 and Vi Wi are
parallel to each other. Thus the quadrilateral Y Vi Wi Wi−1 is a trapezoid. If α > β, then
(γ + β) + β < π , and hence Vi Wi > Y Wi−1. The angles of the triangle Vi−1Y Wi−1
are α, β, γ , and thus the condition α > β implies Y Wi−1 > Vi−1Wi−1. Therefore,
α > β implies Vi Wi > Vi−1Wi−1 for every i. Thus
which is impossible. If α < β, then a similar argument gives
V1W1 < V2W2 < · · · < VN WN < V1W1,
V1W1 > V2W2 > · · · > VN WN > V1W1,
(16)
(17)
also impossible. Next suppose that the quadrilaterals Ri look like (B) of Fig. 8. Then
the sides Wi−1Wi and Vi−1Vi are parallel to each other. There is a point Y on the
segment Vi−1Vi such that Vi−1Wi−1Y = γ and Wi Wi−1Y = β. Then Y Vi Wi Wi−1
is a parallelogram, and thus Vi Wi = Y Wi−1. Suppose α > β. Then, as the angles of
the triangle Vi−1Y Wi−1 are α, β, γ , we have Y Wi−1 > Vi−1Wi−1. Therefore, α > β
implies Vi Wi > Vi−1Wi−1 for every i, which implies (16). If α < β, then a similar
argument gives (17). Since both are impossible, the proof is complete.
Lemma 6.3 Suppose that a and b are commensurable, c is not a rational multiple
of a, and P = N ; that is, each angle of A equals α + β. Then one of (v), (vi) or (vii)
of Theorem 2.1 holds.
Proof We shall use the notation introduced in the proof of Lemma 6.2. We proved
there that if X−→Y is an edge of Γexcept, then Y is either a vertex of Γkite, or is an
exceptional point. (Note that the argument proving this statement did not use the condition
α = β.) If α = β, then Γkite is empty, and thus Y must be an exceptional point. The
same conclusion holds if α = β. Indeed, suppose that α = β and the endpoint Y is
normal. Then Y is a vertex of Γkite, and the triangles around Y are arranged as in
Fig. 5. It is clear that the case of (A) is impossible. In the case of (B), the halfline h is
a continuation of the segment XY , since α + α + γ = α + β + γ = π . Now the two
triangles supported by h have angle γ at their vertices lying on h, which contradicts
the fact that Y is an endpoint of an edge. Indeed, in the definition of Γexcept Y is
defined as Ui0 , where i0 is the maximal index such that the triangle Ti has angle γ at
Ui for every i ≤ i0, while, in (B) of Fig. 5, i0 + 1 also has this property. This proves
that Y must be exceptional.
In the proof of Lemma 6.2 it was also shown that each vertex of A is a
starting point of an edge of Γexcept. (This argument was also independent of the
condition α = β.) Let V−−iW→i be an edge of Γexcept for every i = 0, . . . , N , where
W1, . . . , WN = W0 are exceptional points.
As we saw earlier, we have either Wi−1Vi−1Vi = α and Vi−1Vi Wi = β for
every i = 1, . . . , N , or Wi−1Vi−1Vi = β and Vi−1Vi Wi = α for every i = 1, . . . , N .
Since the roles of α and β are symmetric, we may assume the former.
Suppose that there is one exceptional point, E. Then Wi = E for every i, and the
edges V−−i→E (i = 1, . . . , N ) decompose A into the nonoverlapping triangles Vi−1Vi E.
Since EVi−1Vi = α and Vi−1Vi E = β for every i = 1, . . . , N , the triangles
Vi−1Vi E are similar to each other. If V0E > V1E, then Vi−1E > Vi E for every i,
and we obtain
V0E > · · · > VN−1E > V0E,
which is impossible. We get a similar contradiction if V0E < V1E. Thus V0E = V1E,
α = β, A is a regular N gon, and we obtain (vii) of Theorem 2.1.
Next suppose that there are two exceptional points, E and F . Then N is even,
N = 2k, and the equation at E is either kγ = π or α + β + (k + 1)γ = 2π . It is clear
that in the first case the number of edges of Γexcept with endpoint E is at most k − 1,
and in the second case this number is at most k. The same is true for F . Since there
are (at least) N = 2k edges arriving at E and F , it follows that the equations at E
and F must be α + β + (k + 1)γ = 2π . In particular, E and F are in the interior of
A, and both of them are the endpoints of k edges. By shifting the indices we may
assume that there is an index 1 ≤ j < N such that Wi = E for every i = 0, . . . , j − 1,
and WN−1 = Wj = F .
The edges starting from the vertices of A are pairwise disjoint, except the
endpoints. This implies that E is in the interior of the simple polygon {VN−1, . . . , Vj , F },
and thus Wi = F for every i = j, . . . , N − 1. Therefore, we have j = k, W0 = · · · =
Wk−1 = E and Wk, . . . , WN−1 = F . (See Fig. 9.)
The polygons P = {V0, . . . , Vk−1, E} and Q = {Vk, . . . , VN−1, F } are convex,
since their angle at E and F , respectively, equal (k − 1)γ = (1 − (1/ k))π . Put
q = sin α/ sin β = a/b. Then, as a and b are commensurable, q is rational. The
triangles EVi−1Vi (i = 1, . . . , k − 1) and F Vi−1Vi (i = k + 1, . . . , N ) are similar, and we
have
Vi Vi+1/Vi−1Vi = q
(i = 0, k).
(18)
This implies that P and Q are similar polygons.
Let D denote the middle point of the segment V0Vk . We show that A is centrally
symmetric with center D.
Let ei denote the line going through the vertices Vi−1 and Vi . Since each angle of
A equals α + β = (1 − (1/ k))π , it follows that the lines e0 and ek are parallel to each
other.
Let ρ denote the reflection about the point D. Then ρ(Vk) = V0 and ρ(e0) =
ek . Since VN−1 ∈ e0, it follows that ρ(VN−1) ∈ ek . Now we have VN−1V0E =
Vk−1VkF = β, and thus ρ(F ) is on the halfline starting from V0 and going
through E. Since the triangles {V0, Vk−1, E} and {Vk, VN−1, F } are similar to each
other and ρ(VN−1) is on the line ek , it follows that ρ({Vk, VN−1, F }) coincides with
ρ({V0, Vk−1, E}). Thus ρ(Q) = P and ρ(A) = A; that is, A is centrally symmetric.
By Lemma 4.7, A is a rational polygon. Then, it follows from (18) and q =
sin α/ sin β that (vi) of Theorem 2.1 holds if N = 6, and (v) of Theorem 2.1 holds
if N ≥ 8. This completes the proof of the lemma.
7 Case III: γ = 2π/3
In this case each angle of A equals π/3 or 2π/3. Thus A can be a regular triangle,
a parallelogram, a trapezoid, a pentagon or a hexagon. If A is parallelogram or a
trapezoid then two of its angles equal π/3 and the other two angles equal 2π/3; if A
is a pentagon then one of its angles equals π/3 and the other angles equal 2π/3; if A
is a hexagon, then all its angles equal 2π/3.
7.1 Subcase IIIa
First we assume that a is not a linear combination of b and c with nonnegative rational
coefficients. Then we consider the directed graph Γa .
Lemma 7.1 Suppose that a is not a linear combination of b and c with nonnegative
rational coefficients. Then
(i) the outdegree of Γa equals the indegree at each vertex;
(ii) no edge of Γa starts from the boundary of A; and
(iii) every vertex lying on the boundary of A but different from the vertices of A is
normal.
Proof It is easy to check that the equation at an exceptional point must be one of
3γ = 2π , 3α + 3β = π , 4α + 4β + γ = 2π , 6α + 6β = 2π .
In order to prove (i) it is enough to show that if the indegree of a vertex X is
positive, then so is the outdegree of X. If X is normal, then the outdegree at X is
positive by Lemma 4.2. If X is exceptional then the equation at X must be 3α +
3β = π , since X is an interior point of a side of a triangle. Let T1, . . . , T6 denote the
triangles having X as a vertex listed counterclockwise, and let hi denote the halfline
starting from X and supporting the triangles Ti and Ti+1 (i = 1, . . . , 5). Three of the
triangles Ti have angle β at X, and at least two of them are of the same orientation
in the sense that a rotation about the point X brings one of them onto the other. Let
Ti and Tj be such triangles. If j = i + 1, then an edge of Γa starts from X along the
halfline hi . If j > i + 1, then an edge of Γa starts from X along one of the halflines
hi and hj−1, depending on the location of the angle γ in Ti and Tj . This proves (i).
The argument above shows that if the equation at a vertex X is 3α + 3β = π ,
then an edge of Γa starts from X. Since no edge arrives at any boundary point, we
obtain (ii) from (i). Therefore, if X is a boundary point and not a vertex of A, then
the equation at X cannot be 3α + 3β = π . Since the right hand side of all other
equations at exceptional points equals 2π , it follows that X must be normal, which
proves (iii).
It is easy to check that if the equation at a vertex is X is 3γ = 2π , then the
indegree of X is zero and the outdegree of X is positive. Since this contradicts (i) of
Lemma 7.1, it follows that the equation cannot be 3γ = 2π at any vertex. Therefore,
we have ri ≤ pi for every i > N . Let P (α + β) + Rγ = (N − 2)π be the sum of
equations at the vertices of A. Then we have R ≥ P . In particular, we obtain R > 0,
and thus N = 3 is impossible.
Suppose N = 4. By (i) of Lemma 4.3, there are no adjacent vertices of A at which
the equation is γ = δ. Thus R ≤ 2, and then P ≤ R implies that P = R = 2, and that
the angles of A must be α + β, γ , α + β, γ in this order. Thus A is a parallelogram.
Since R > 0, there is a side XY of A such that the equation at X is γ = δ, and the
triangle having X as a vertex has angles γ at X and has angle β at its vertex lying on
the side XY . By (ii) of Lemma 4.3, the length of XY is k · a with a positive integer k.
Therefore, by Lemma 4.1, (i) of Theorem 2.1 holds.
Next suppose N = 5. Since there are no adjacent vertices of A at which the
equation is γ = δ, we have R ≤ 2. Then P ≥ 3, which contradicts P ≤ R.
Finally, if N = 6, then R ≤ 3, and thus there are at least three vertices of A where
the equation is 2(α + β) = δ. Thus P ≥ 6, which contradicts P ≤ R again. This
completes the proof in the subcase when a is not a linear combination of b and c with
nonnegative rational coefficients.
7.2 Subcase IIIb
Suppose that b is not a linear combination of a and c with nonnegative rational
coefficients. Since the roles of a and b are symmetric in the conditions as well as in the
statements of the theorem, this subcase can be treated similarly to Subcase IIIa.
Therefore, we may assume that a is a linear combination of b and c with
nonnegative rational coefficients, and b is a linear combination of a and c with nonnegative
rational coefficients. This implies that either a, b, c are pairwise commensurable, or
a and b are commensurable and c is not a rational multiple of a.
If a, b, c are pairwise commensurable, then so are the sides of A. We know that
A is an N gon with 3 ≤ N ≤ 6. If 4 ≤ N ≤ 6, then A has N − 2 vertices with angle
2π/3, and the other two angles of A are integer multiples of π/3. Thus, in these
cases, (ii) of Theorem 2.1 holds. If N = 3, then (iv) of Theorem 2.1 holds.
7.3 Subcase IIIc
Thus we are left with the case when a and b are commensurable and c is not a rational
multiple of a.
If A is a parallelogram, then (i) of Theorem 2.1 holds, since A is rational by
Lemma 4.7, and sin α/ sin β = a/b ∈ Q. Therefore, we may assume that A is not a
parallelogram; that is, A is a triangle, a trapezoid, a pentagon or a hexagon.
We show that α is not a rational multiple of π except when α = π/6. Indeed, we
have
and thus √3 · cot α is rational. Thus tan2 α is rational, and then so are cos2 α = 1/(1 +
tan2 α) and cos 2α. By [5, Corollary 3.12]), this implies cos 2α = 0, ±1, ±1/2. Since
α < π/3, we have 0 < 2α < 2π/3, and thus 2α = π/3 or π/2, and α = π/6 or π/4.
If α = π/4, then √3 · cot α = √3 is irrational, so the only possibility is α = π/6.
If α = π/6, then β = π/6 and, taking into consideration that A is rational by
Lemma 4.7, we find that (viii) of Theorem 2.1 holds. Therefore, we may assume that
α/π is irrational.
Let G denote the set of real numbers n · (π/3) + m · α, where n, m ∈ Z. Then G
is an additive subgroup of the reals such that π ∈ G. Recall that we denote by PG
the family of all simple, closed polygons such that, for every side XY of P , the angle
between the line going through the side XY and the xaxis belongs to G.
We may assume that the xaxis contains one of the sides of A. Then A ∈ PG and
Δi ∈ PG for every i = 1, . . . , t . Indeed, let T be any of the triangles Δi , and let e be a
line containing one of the sides of T . Then there is a sequence of triangles T0, . . . , Tk
and there is a sequence of lines e0, e1, . . . , ek such that Tk = T , e0 is the xaxis,
ek = e, and ei contains a side of both Ti−1 and Ti for every i = 1, . . . , k. Then the
angle θi of ei−1 and ei is one of α, β = (π/3) − α and 2π/3 for every i = 1, . . . , k,
and thus the angle between ek and e0 equals ik=1 θi ∈ G.
If θ = n · (π/3) + m · α where n, m ∈ Z, then we put χ (θ ) = (−1)n. Then χ is
welldefined on G. Indeed, if n · (π/3) + m · α = n · (π/3) + m · α where n, n , m, m ∈ Z,
then n = n by α/π ∈/ Q. Clearly, χ : G → {1, −1} is a multiplicative function; that
is, χ satisfies the functional equation χ (θ1 + θ2) = χ (θ1) · χ (θ2) (θ1, θ2 ∈ G). Also,
we have χ (π ) = −1, and thus χ (θ + π ) = −χ (θ ) and χ (θ + 2π ) = χ (θ ) for every
θ ∈ G. We also have χ (−θ ) = χ (θ ) for every θ ∈ G. Let Φ : PG → C be defined
by (14).
Let V1, . . . , VN = V0 be the vertices of A listed counterclockwise. We may assume
that the side X0X1 lies on the xaxis. Moreover, we shall assume that if N = 4, then
X0X1 and X2X3 are parallel sides of A, and if N = 5, then the only acute angle of A
is at the vertex X1. We put di = Xi−1Xi for every i = 1, . . . , N .
Lemma 7.2 If N = 3, then Φ(A) = 3d1. If 4 ≤ N ≤ 5, then Φ(A) = 3d4. If N = 6,
then
Φ(A) = d1 − d2 + d3 − d4 + d5 − d6.
Proof Using (15) and χ (±π/3) = −1, χ (±2π/3) = 1 we find that if N = 3, then
Φ(A) = d1 + d2 + d3 = 3d1.
If N = 4, then A is a trapezoid, and thus d2 = d4 and d1 = d3 + d4. Therefore, we
get Φ(A) = d1 + d2 − d3 + d4 = 3d4.
If N = 5, then A is a pentagon, d1 = d3 + d4 and d2 = d4 + d5. Therefore, Φ(A) =
d1 + d2 − d3 + d4 − d5 = 3d4.
It is clear that if N = 6, then (19) holds.
In Lemma 4.8 we proved that those triangles of the tiling that have their side of
length c in the interior of A come in pairs, and each pair forms a quadrilateral which
is either a parallelogram of sides a, b and of angles γ , α + β, or a kite of sides a, b
and of angles 2β, γ , 2α, γ . Let Q1, . . . , Qu be a list of these quadrilaterals. We prove
that Φ(Qi ) = 0 for every i = 1, . . . , u. This is clear if Qi is a parallelogram. Suppose
Qi is a kite. Since χ (2α) = 1, χ (2β) = χ ((2π/3) − 2β) = 1 and χ (2π/3) = 1, (15)
gives Φ(Qi ) = χ (θ1)(a − b + b − a) = 0. Therefore, if the quadrilaterals Q1, . . . , Qu
tile A, then Φ(A) = 0. By Lemma 7.2, this implies N = 6 and d1 − d2 + d3 − d4 +
d5 − d6 = 0.
It is easy to check, using the fact that each angle of A equals 2π/3, that d1 + d2 =
d4 + d5 and d2 + d3 = d5 + d6. Thus d1 − d2 + d3 − d4 + d5 − d6 = 0 implies d1 =
d4, d2 = d5 and d3 = d6. Therefore, A is centrally symmetric. Since A is rational by
Lemma 4.7 and sin α/ sin β = a/b ∈ Q, it follows that (vi) of Theorem 2.1 holds.
Next we suppose that Q1, . . . , Qu do not tile A. Then, by Lemma 4.8, there is a
triangle of the tiling that has its side of length c on the boundary of A. Let T1, . . . , Tv
be a list of all these triangles. Then we have
Φ(A) =
u
i=1
j=1
j=1
Φ(Qi ) +
Φ(Tj ) =
Φ(Tj ).
Since the graph Γc is empty by Lemma 4.6, it follows from Lemma 4.5 that there is a
side XY of A which is covered by the sides of length c of the triangles Tj supported
by the side XY . Since A is rational by Lemma 4.7, we find that each side of A has
this property; that is, the boundary of A is covered by the sides of length c of the
triangles T1, . . . , Tv . Thus di = ki · c (i = 1, . . . , N ), where k1, . . . , kN are positive
integers.
It is easy to check, using χ (α) = 1, χ (β) = −1 and χ (γ ) = 1, that if a triangle
Tj is supported by X0X1, then Φ(Tj ) = c + a − b. Then we find by checking each
Since a = b by α/π ∈/ Q, it follows from (20) and Lemma 7.2 that these cases are
impossible. Thus we have N = 6, when
Φ(Tj ) = 3k1(c + a − b) = 3d1 + 3k1(a − b),
Φ(Tj ) = 3k4(c + a − b) = 3d4 + 3k4(a − b).
case that if N = 3, then
and if 4 ≤ N ≤ 5, then
v
v
Φ(Tj ) = (k1 − k2 + k3 − k4 + k5 − k6) · (c + a − b)
= (d1 − d2 + d3 − d4 + d5 − d6) + k · (a − b),
where k = k1 − k2 + k3 − k4 + k5 − k6. By (20) and Lemma 7.2 we have k = 0, and
thus d1 − d2 + d3 − d4 + d5 − d6 = k · c = 0. As we proved above, this implies that
A is centrally symmetric, and that (vi) of Theorem 2.1 holds.
8 Case IV: γ = π/3
If γ = π/3 then, similarly to the previous case, each angle of A equals π/3 or
2π/3. Thus A can be a regular triangle, a parallelogram, a trapezoid, a pentagon
or a hexagon. If A is parallelogram or a trapezoid then two of its angles equals π/3
and the other two angles equal 2π/3; if A is a pentagon then one of its angles equals
π/3 and the other angles equal 2π/3; if A is a hexagon, then all its angles equal
2π/3.
8.1 Subcase IVa
First we assume that a is not a linear combination of b and c with nonnegative
rational coefficients. Then we consider the directed graph Γa . Unfortunately, it can happen
that the indegree is different from the outdegree at certain points (unlike in the
previous case). Therefore, we define another directed graph as well. We shall denote by
Γk the set of those directed segments X−→Y for which the segment XY is the common
side of two triangles having angles γ at X and β at Y . We shall consider the set
Γa ∪ Γk of all directed edges belonging to either Γa or Γk .
We show that the indegree of Γa ∪ Γk is zero or one at each vertex. Indeed, if
the indegree of Γa is positive at a vertex V , then V is an inner point of a side of a
triangle, and then either V is normal or the equation at V is 3γ = π . The indegree
of Γk at the vertex V is zero in both cases. On the other hand, it is easy to see that the
indegree of Γk is at most one at every vertex.
Lemma 8.1 Suppose that a is not a linear combination of b and c with nonnegative
rational coefficients. Then
(i) the outdegree of Γa ∪ Γk equals the indegree at each vertex;
(ii) no edge of Γa ∪ Γk starts from the boundary of A; and
(iii) every vertex lying on the boundary of A but different from the vertices of A is
normal.
Proof It is easy to check that the equation at an exceptional point must be one of
3γ = π , 6γ = 2π , 3α + 3β = 2π , α + β + 4γ = 2π .
In order to prove (i) it is enough to show that if the indegree of a vertex X is
positive, then so is the outdegree of X.
First assume that X is the endpoint of an edge of Γa . If X is normal, then the
outdegree at X is positive by Lemma 4.2. If X is exceptional then the equation at X
must be 3γ = π , since X is an interior point of a side of a triangle. It is easy to see
that in this case either an edge of Γa or an edge of Γk starts from X.
Next assume that X is the endpoint of an edge Z−→X of Γk . Then the equation at X
is 2α + 2β + 2γ = 2π or 3α + 3β = 2π . Let T1, . . . , T6 denote the triangles having X
as a vertex listed counterclockwise, and such that the common side of T1 and T6 is the
segment ZX. Then the angles of T1 and T6 at Z and X equal γ and β, respectively.
Let hi denote the halfline starting from X and supporting the triangles Ti and Ti+1
(i = 0, . . . , 5), where we put T0 = T6.
Suppose that the equation at X is 3α + 3β = 2π . Then the angle between hi
and hj is different from π for every i = j . Indeed, otherwise pα + qβ = π would
hold for some integers 0 ≤ p, q ≤ 3. Since α + β = 2π/3, this implies either α =
β = π/3 or {α, β} = {π/6, π/2}. However, in each case we would have a/b ∈ Q,
which contradicts the assumption that a is not a linear combination of b and c with
nonnegative rational coefficients. Therefore, no two of the halflines h0, . . . , h5 can
form a line.
Since three of the triangles Ti have angle β at X, at least two of them are of the
same orientation in the sense that a rotation about the point X brings one of them
onto the other. Let Ti and Tj be such triangles. If j = i + 1, then an edge of Γa starts
from X along the halfline hi . If j > i + 1, then an edge of Γa starts from X along one
of the halflines hi and hj−1, depending on the location of the angle γ in Ti and Tj .
Next suppose that the equation at X is 2α + 2β + 2γ = 2π . One can check, by
considering the possible cases, that if the angle between hi and hj is different from
π for every i = j then an edge of Γa ∪ Γk must start from X. If there are i = j such
that the angle between hi and hj equals π , then pα + qβ + rγ = π holds for some
integers 0 ≤ p, q, r ≤ 2. It is easy to see that the only possibility is p = q = r = 1.
In this case the triangles T1 and T3 are supported by a line which is the union of the
halflines h0 and h3. It is easy to check that in this case either an edge of Γa starts from
X, or the common side of T3 and T4 is an edge of Γk starting from X. This proves (i).
Since no edge arrives at any boundary point, we obtain (ii) from (i). Therefore,
if X is a boundary point and not a vertex of A, then the equation at X cannot be
3γ = π . Since the right hand side of all other equations at exceptional points equals
2π , it follows that X must be normal, which proves (iii).
The equations at the vertices of A must be of the form α + β = δ, γ = δ or 2γ = δ.
Suppose that the angle of A at the vertex V1 equals π/3. Then the equation at V1
is γ = δ, and there is a triangle T1 of the tiling such that V1 is a vertex of T1 and
the angle of T1 at V1 is γ . We may assume that T1 has angle β at its vertex lying on
the side V1V2. By (ii) of Lemma 4.3, every triangle supported by V1V2 is a translated
copy of T , and thus the length of V1V2 is an integer multiple of a.
There is a triangle Tk such that Tk is supported by V1V2, V2 is a vertex of Tk , and
Tk is a translated copy of T1. Then the angle of Tk at V2 is β, and thus the equation
at V2 is α + β = δ2. Then there is a triangle T1 supported by V2V3 such that V2 is
a vertex of T1 and the angle of T1 at V2 equals α. Let V2 = U0, . . . , U = V3 be a
division of V2V3 such that each Ui−1Ui is the side of a triangle Ti of the tiling. Then
every triangle Ti has angle α at the vertex Ui−1 (see Fig. 3). In particular, T has
angle α at U −1, and then the angle of T at V3 is different from α. If it is β, then the
equation at V3 is α + β = δ2. However, in this case an edge of Γa would start from
V3 which is impossible. Thus the angle of T at V3 is γ . Then the equation at V3 is
γ = δ or 2γ = δ. The latter is impossible, because in that case an edge of Γa ∪ Γk
would start from V3. Thus the angle of A at V3 equals π/3. Since the angle of A at V1
is also π/3, it follows that A is a parallelogram such that the length of its side V1V2
is an integer multiple of a. Then, by Lemma 4.1, (i) of Theorem 2.1 holds.
Next suppose that no angle of A equals π/3. Then each angle of A equals 2π/3,
and the equations at the vertices of A are 2γ = δ or α + β = δ.
Suppose that there is a vertex of A where the equation is 2γ = δ. Let V1 be such a
vertex. Since no edge of Γk starts from V1, one of the triangles having V1 as a vertex
must have angle β at its vertex lying on the boundary of A. Let T1 be such a triangle;
we may assume that T1 is supported by V1V2. By (ii) of Lemma 4.3, each triangle
supported by V1V2 is a translated copy of T1. Then the equation at V2 is α + β = δ2.
Then there is a triangle T1 supported by V2V3 such that V2 is a vertex of T1 and the
angle of T1 at V2 equals α. Following the argument above we can see that in this case
the angle of A at V3 is π/3 which is impossible.
Finally, suppose that the equation at every vertex of A is α + β = δ. Then, since
no edge of Γa starts from V1, one of the triangles having V1 as a vertex must have
angle β at V1 and angle γ at its vertex lying on the boundary of A. Let T1 be such a
triangle; we may assume that T1 is supported by V1V2. By (ii) of Lemma 4.3, each
triangle supported by V1V2 is a translated copy of T1. Then the equation at V2 is
2γ = δ2 which is impossible. This completes the proof in the subcase when a is not
a linear combination of b and c with nonnegative rational coefficients.
8.2 Subcase IVb
Suppose that b is not a linear combination of a and c with nonnegative rational
coefficients. Since the roles of a and b are symmetric in the conditions as well as in the
statements of the theorem, this subcase can be treated similarly to Subcase IVa.
Therefore, we may assume that a is a linear combination of b and c with
nonnegative rational coefficients, and b is a linear combination of a and c with nonnegative
rational coefficients. This implies that either a, b, c are pairwise commensurable, or
a and b are commensurable and c is not a rational multiple of a.
If a, b, c are pairwise commensurable, then so are the sides of A. We know that
A is an N gon with 3 ≤ N ≤ 6. If 4 ≤ N ≤ 6, then A has N − 2 vertices with angle
2π/3, and the other two angles of A are integer multiples of π/3. Thus, in these
cases, (iii) of Theorem 2.1 holds. If N = 3, then (iv) of Theorem 2.1 holds.
8.3 Subcase IVc
Thus we are left with the case when a and b are commensurable and c is not a rational
multiple of a.
If A is a parallelogram, then (i) of Theorem 2.1 holds, since A is rational by
Lemma 4.7, and sin α/ sin β = a/b ∈ Q. Therefore, we may assume that A is not a
parallelogram.
Suppose that N ≤ 4; that is, A is a triangle or a trapezoid. Then the angle of A
equals π/3 at two consecutive vertices, say, X and Y , and the corresponding
equations at X and Y must be γ = δ. Since Γc is empty, it follows from (i) of Lemma 4.4
that there is an exceptional vertex in the interior of XY . Let X = U0, . . . , Uk = Y be a
division of XY such that each Ui−1Ui is the side of a triangle Ti of the tiling. Let i be
the smallest positive index such that Ui is exceptional. Since T1 has angle γ at U0, it
follows that Tj has angle γ at Uj−1 for every j = 1, . . . , i (see Fig. 4). In particular,
Ti has angle γ at Ui−1, and thus the angle of Ti at Ui is different from γ . However,
as Ui is exceptional, the equation at Ui must be 3γ = π , which is a contradiction.
Therefore, we have N ≥ 5, and thus A is a pentagon or a hexagon.
We prove that A must be a centrally symmetric hexagon, and thus (vi) of
Theorem 2.1 holds.
First we assume that α/π is irrational. Then we consider the group G, the
multiplicative function χ : G → {−1, 1} and the additive function Φ : PG → R as in
the Subcase IIIc. Repeating the argument of IIIc we find that A must be a centrally
symmetric hexagon.
Finally, we consider the case when α/π is rational. We prove that this happens
only if α = π/6 or α = π/2. Indeed, we have
b sin β
a = sin α =
sin((2π/3) − α)
sin α
and thus √3 · cot α is rational. Thus tan2 α is rational, and then so are cos2 α =
1/(1 + tan2 α) and cos 2α. By [5, Corollary 3.12]), this implies cos 2α = 0, ±1, ±1/2.
Since α < 2π/3, we have 0 < 2α < 4π/3, and thus 2α ∈ {π/3, π/2, 2π/3, π } and
α ∈ {π/6, π/4, π/3, π/2}. If α = π/4 then √3 · cot α = √3 is irrational, which is
impossible. If α = π/3, then α = β = γ and a = b = c, which contradicts the
condition that c/a is irrational. Therefore, we have either α = π/6 or is α = π/2. If
α = π/6 then β = π/2. Since the roles of α and β are symmetric, we may assume
that α = π/2 and β = π/6. Then we have a = 2b.
Now we prove that A has to be a hexagon.
By Lemma 4.8, the triangles having their side of length c in the interior of A come
in pairs. The triangles of each pair have a common side of length c, and thus they
form a quadrilateral which is either a parallelogram of sides a, b and of angles γ =
π/3, α + β = 2π/3, or a kite of sides a, b and of angles 2β = π/3, γ = π/3, 2α =
π, γ = π/3. Note that the kites are, in fact, regular triangles of side 2b. For every kite
Q we denote by V (Q) the vertex of Q with angle labeled with 2β, and by W (Q)
the middle point of the side opposite to V (Q). Since the tiling is regular, it has the
property that for every kite Q, we have V (Q) = W (Q ) and W (Q) = V (Q ) for
some kites Q and Q .
We divide each parallelogram into two rhombuses of side b and of angles
π/3, 2π/3. Thus A is tiled with regular triangles of side 2b and rhombuses of side
b and of angles π/3, 2π/3. In addition, each triangle Q of the tiling has a selected
vertex V (Q) and a point W (Q) opposite to V (Q) with the property described above.
We show that the existence of such a tiling implies that A is either a parallelogram or
a hexagon.
Suppose this is not true, and let A be a counterexample with a minimal number of
pieces of the tiling. Since A is not a hexagon, it has a vertex X where the angle of A
is π/3. Then X cannot be a vertex of a triangle Q, because in that case either V (Q)
or W (Q) would be in the boundary of A, which is impossible. Thus X is the vertex of
a rhombus R1. Let Y be a vertex of A adjacent to X. Then R1 has a vertex U1 on the
side XY . If U1 = Y , then U1 is also the vertex of a triangle Q1 or a rhombus R2. But
U1 cannot be the vertex of a triangle Q1, because in that case either V (Q1) or W (Q1)
would be a point of XY , which is impossible. Thus U1 is the vertex of a rhombus R2.
Let U2 be a vertex of R2 lying on XY and different from U1. If U2 = Y then, by
repeating the argument we find that U2 is the vertex of a rhombus R3 etc. In this way
we obtain a division X = U0, U1, . . . , Uk = Y of the segment XY and a sequence of
rhombuses R1, . . . , Rk such that Ui−1Ui is a side of Ri for every i = 1, . . . , k.
Then P = R1 ∪ · · · ∪ Rk is a parallelogram. Since, by assumption, A is not a
parallelogram, we have P = A. Then A \ P is a convex polygon tiled with triangles
and rhombuses satisfying the condition described above. Now, the number of pieces
in the tiling of A \ P is smaller than that of the tiling of A, and thus A \ P is either
a parallelogram or a hexagon. However, A \ P has an angle π/3, and thus it cannot
be a hexagon. Therefore, A \ P is a parallelogram, and then so is A, which is a
contradiction. This proves that A is a hexagon, and thus (ix) of Theorem 2.1 holds.
9 Case V: γ = π/2
9.1 Subcase Va
Under this assumption each angle of A equals π/2, and thus A is a rectangle. Our
aim is to prove that in this case (i) of Theorem 2.1 holds.
First we assume that a is not a linear combination of b and c with nonnegative rational
coefficients. Then we consider the directed graph Γa .
Lemma 9.1 Suppose that γ = π/2 and a is not a linear combination of b and c with
nonnegative rational coefficients. Then
(i) the outdegree of Γa equals the indegree at each vertex;
(ii) no edge of Γa starts from the boundary of A; and
(iii) the length of one of the sides of A is an integer multiple of a.
Proof In order to prove (i) it is enough to show that if the indegree of a vertex
Y is positive, then so is the outdegree of Y . If Y is normal, then the outdegree
at Y is positive by Lemma 4.2. If Y is exceptional then the equation at Y must be
2α + 2β = π or 2γ = π , since Y is an interior point of a side of a triangle. An
inspection of all possible cases shows that at least one edge of Γa starts from Y .
Since no edge arrives at any boundary point, (ii) follows from (i).
Let X be a vertex of A. In order to prove (iii), first we show that there is a triangle
T such that X is a vertex of T and the side of T of length a is on the boundary
of A. This is clear if the equation at X is γ = δ. Therefore, we may assume that the
equation at X is α + β = δ. Let T be the triangle having X as a vertex and having
angle β at X. Since, by (ii), no edge of Γa starts from X, it follows that the vertex of
T with angle α is not on the boundary of A. Then, the side of T of length a is on the
boundary of A.
We proved that there is a side XY of A and there is a division X = U0, . . . , Uk = Y
of XY such that each Ui−1Ui is the side of a triangle Ti of the tiling, and U0U1 = a.
We prove that Ui−1Ui = a for every i. Suppose this is not true. Then there is an
1 ≤ i < k such that Ui−1Ui = a and Ui Ui+1 = a. Then an edge of Γa must start
from Ui ; this can be shown by considering the same cases as in the proof of (i). This,
however, contradicts (ii). Thus Ui−1Ui = a for every i, which proves (iii).
Now Lemma 4.1 and (iii) of Lemma 9.1 imply that (i) of Theorem 2.1 holds,
which completes the proof in the subcase when a is not a linear combination of b
and c with nonnegative rational coefficients. The same argument applies if b is not a
linear combination of a and c with nonnegative rational coefficients.
Therefore, we may assume that a is a linear combination of b and c with
nonnegative rational coefficients, and b is a linear combination of a and c with nonnegative
rational coefficients. This implies that either a, b, c are pairwise commensurable, or
a and b are commensurable and c is not a rational multiple of a.
If a, b, c are pairwise commensurable, then so are the sides of A, and we find that
(i) of Theorem 2.1 holds again.
Finally, if a and b are commensurable and c is not a rational multiple of a, then
the sides of A are commensurable by Lemma 4.7, and we have the same conclusion.
This completes the proof of Theorem 2.1.
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