Tiling Polygons with Lattice Triangles
Discrete Comput Geom
Tiling Polygons with Lattice Triangles
Steve Butler 0 1
Fan Chung 0 1
Ron Graham 0 1
Miklós Laczkovich 0 1
0 M. Laczkovich Eötvös Loránd University , Budapest , Hungary
1 S. Butler University of California, Los Angeles , Los Angeles, CA , USA
Given a simple polygon with rational coordinates having one vertex at the origin and an adjacent vertex on the x-axis, we look at the problem of the location of the vertices for a tiling of the polygon using lattice triangles (i.e., triangles which are congruent to a triangle with the coordinates of the vertices being integer). We show that the coordinates of the vertices in any tiling are rationals with the possible denominators odd numbers dependent on the cotangents of the angles in the triangles.
Tiling; Lattice triangles; Rational coordinates; Stomachion
We consider the problem of tiling with triangles a simple polygon with one vertex
of the polygon at the origin and an adjacent vertex on the x-axis. We will restrict
ourselves to the case when the triangles for the tiling consist of congruent copies of
1, 2, . . . , k where each i is a lattice triangle, i.e., a triangle that can be placed
S. Butler was supported by an NSF postdoctoral fellowship.
in the plane so that the coordinates of the vertices are all integer. A simple example
of these triangles are right triangles with integer lengths ai , bi for the two legs, in this
special case we will refer to these as an (ai , bi ) triangle.
Previous papers [2–6] have looked at tiling polygons using similar triangles, i.e.,
where we start with a collection of triangles and are allowed to scale the triangles
when placing them. In this paper we do not allow for scaling but rather assume that
all triangles are congruent to one of several specified lattice triangles in our tiling. Our
prohibition of scaling is not so restrictive since it is easy to see that we can combine
j 2 copies of triangle i to make a scaled version of i where each side has been
scaled by j .
We will show in Sect. 2 that the coordinates of the vertices in our tiling are rational
numbers with the possible denominators odd numbers dependent on the cotangents
of the i . In Sect. 3 we give a construction which shows how it is possible to achieve
large terms in the denominators for some internal vertices in a packing of a large
square using right triangles (ai , bi ). Finally we give some concluding remarks.
2 Possible Vertex Locations
We first start by showing that the coordinates of the vertices in a tiling of a polygon
must be of a restricted form. We first observe that the coordinates of all the vertices
will be rational. To see this we need the following theorem.
Theorem 1 (Laczkovich ) If a simple polygon P is tiled by triangles 1, . . . , n,
then the coordinates of the vertices of each j belong to the field generated by the
coordinates of the vertices of P and the cotangents of the angles of the triangles j .
It is easy to check that a lattice triangle has all cotangents rational (in fact a triangle
has all cotangents rational if and only if it is similar to a lattice triangle). So by an
application of the above theorem if we pack our polygon with lattice triangles then
we must have only rational coordinates. But now we show that we can say even more
about the denominators of these rational numbers.
Theorem 2 Let P be a simple polygon in the plane such that the coordinates of the
vertices of P are rational, and one of the vertices of P is the origin with an adjacent
vertex on the x-axis. Suppose that P is dissected into congruent copies of lattice
triangles 1, 2, . . . , k . Let α1, α2, . . . , α3k denote the angles of the triangles i
and let cot αj = aj /bj with coprime integers aj , bj for every j = 1, 2, . . . , 3k. Then
the coordinates of each vertex in the tiling are of the form p/ j3k=1(aj2 + bj2)nj , where
p and n1, n2, . . . , n3k are integers.
The proof will make use of the following lemma.
Lemma 3 Let θ = ±t1α1 ± t2α2 ± · · · ± t3kα3k with ti nonnegative integers. Then for
some integers u, v, r, s, with r and s square-free, cos θ = u√r/ i3=k1(ai2 + bi2)ti and
sin θ = v√s/ i3=k1(ai2 + bi2)ti .
This follows since
|sin αj | =
1 + cot2 αj
aj + bj
|cos αj | =
1 + tan2 αj
aj + bj
An easy induction now shows for ti ≥ 0 that cos(ti αi ) = ui /(ai2 + bi2)ti /2 (similarly
for sin(ti αi )) and then applying the sum rule for trigonometric functions and
rationalizing gives the result.
Proof of Theorem 2 By Theorem 1 we know that all the vertices have rational
coordinates and so it remains to show that the rational coordinates are all of the correct
Looking at any vertex of the tiling that is not a vertex of P we see that there must
always be an edge of our tiling which goes from the vertex either to the left or straight
down. So starting at any internal vertex we continue taking such an edge until we get
to a vertex of the polygon (this process must stop after finitely many steps because
it is impossible to visit any vertex twice and there are only finitely many vertices).
If we are not at (0, 0) then we can move along the edges of P to get to (0, 0). In
particular, we can get from the origin to any vertex in the tiling by moving along
edges of triangles in the tiling.
For a vertex (x, y) in the tiling fix a path that goes from the origin (0, 0) to (x, y)
that moves along edges in the tiling. We show by induction that at each vertex in the
path we have coordinates of the desired form and so in particular the terminal vertex
(x, y) is also of the desired form. We trivially have that (0, 0) is of the desired form.
Now suppose that (xi , yi ) is a point on the path with rational coordinates and that
(xi+1, yi+1) is the next point on the path and that the distance between them is √a
for some integer a (which holds since all triangles are congruent to lattice triangles).
We first note that the line segment joining them will form an angle with the x-axis of
θ = s1α1 + s2α2 + · · · + s3kα3k
for some appropriate choice of si . This follows since the initial angle as well as the
angle between any two consecutive line segments is composed of some combination
of angles of triangles in the tiling. So applying Lemma 3 we have
xi+1 = xi +
√a cos θ
yi+1 = yi +
√a sin θ
= xi +
= yi +
i3=k1(ai2 + bi2)si
i3=k1(ai2 + bi2)si
Since both xi and xi+1 are rational we must have that √ar is an integer, similarly
since yi and yi+1 are rational we must have √as is an integer. By induction since
(xi , yi ) is of the desired form we also have that (xi+1, yi+1) is of the desired form,
concluding the proof.
This shows that the denominators of the coordinates are dependent on the
cotangents. Since no prime of the form 4k + 3 can divide a2 + b2 for a, b coprime integers
we see that we will never have a prime of that form in the denominator. We can
actually show a little more.
Theorem 4 Let P be a simple polygon in the plane such that the coordinates of the
vertices of P are rational and one of the vertices of P is the origin. Suppose that
P is dissected into lattice triangles. Then the denominators of the coordinates of the
vertices are products of primes of the form 4k + 1.
Proof By Theorem 1 we know that the coordinates of the vertices are rational and so
we only need to show that the terms in the denominator are of the correct form.
As in Theorem 2 we can find a path from (0, 0) to any vertex moving along edges
of the triangles in the dissection. We now proceed by induction along the vertices in
the path. Clearly it holds for (0, 0). Suppose that it now holds for point (xi , yi ) on the
path and consider the point (xi+1, yi+1). Since the distance between these two is the
side of a lattice triangle they are distance √a apart with a an integer.
It suffices to show that xi+1 − xi and yi+1 − yi both have denominators which are
products of primes of the form 4k + 1. Let p/q = xi+1 − xi with p and q = 0 coprime
integers and r = yi+1 − yi rational. Then (p/q)2 + r2 = a. Since p2 + (rq)2 = aq2
it follows that s = rq is an integer. Let p = p1d and s = s1d where d is the greatest
common divisor of p and s. Then d and q are coprime and thus if t is a prime divisor
of q, then t divides p12 + s12. Since p1 and s1 are coprime, this implies that t is a prime
of the form 4k + 1 or t = 2. But the latter is impossible. Indeed, if t = 2 then q is
even, then p is odd, then s is odd, then p2 + s2 is of the form 4k + 2, while aq2 is
divisible by 4, a contradiction.
3 Constructing Vertices with Large Denominators
We have seen that in our tiling with lattice triangles when the coordinates of vertices
are rational the denominators must be of a restricted form. In this section we consider
the special case when the triangles i = (ai , bi ) are right triangles (recall that ai , bi
are integers which are the length of the legs), and show how one could build a large
denominator using congruent copies of these triangles. We will let βi be the smallest
angle in the triangle i .
Looking at the proof in Theorem 2 we see that the difference between xi and xi+1
is of the form √a cos θ . So if we want to get a large denominator for a vertex then we
should get cos θ to have a large denominator. Again looking at Lemma 3 we see that
to achieve this we want to form an edge which has an angle of the form
θ = s1α1 + s2α2 + · · · + s3kα3k
with the si large. We show how to form a packing of some [0, N ]×[0, N ] square
(N to be chosen later) which has an edge in the tiling with slope
θ = 2s1β1 + 2s2β2 + · · · + 2skβk
and the si ≥ 0 arbitrarily large (we also assume βi = π/4).
(In general we do not need to have all of the coefficients of the angles be even,
for instance for (
) triangles where the hypotenuse is integer we can have any
coefficient. It is also possible to combine the angles on two different lattice triangles
(see Fig. 5 for an example). For our purpose in constructing a large denominator what
we have is sufficient.)
The basic construction is shown in Fig. 1. Namely we build a packing of a square
one “layer” at a time. Given a square that we want to place in the middle, the idea
is to “blow up” triangle (a, b) to triangle (ka, kb) where k is large enough so that
we can place the square inside the middle region. Note that the dimensions of the
square in the middle is k(b − a)×k(b − a) and in the worst case scenario we can
choose k to be the side length of Ti and then pack (b − a)2 copies in the middle.
After packing Ti inside of Ti+1 we see that the angle of every edge in Ti now has an
angle of 2α = 2 arctan(a/b) added to it.
So to construct an angle of θ we find a packing of some square, call it T0. We
then apply this process s1 times using the (a1, b1) triangles, then s2 times using the
(a2, b2) triangles, . . . , and finally sk times using the (ak, bk) triangles.
An example of this construction is shown in Fig. 2 where we apply the procedure
3 times using (
) triangles. The initial packing of [0, 2]×[0, 2] has all vertices at
integer coordinates. The second generation is a packing of [0, 6]×[0, 6] and has all
vertices with integer/5 coordinates including a vertex located at (16/5, 8/5). The
third generation is a packing of [0, 18] × [0, 18] and has all vertices with integer/25
coordinates including a vertex located at (256/25, 208/25). The fourth generation is a
packing of [0, 54]×[0, 54] and has all vertices with integer/125 coordinates including
a vertex located at (3536/125, 3448/125).
4 Concluding Remarks
Looking at the proof of Theorem 2 we see that the coordinates of the vertices can be
written in the form ( cos θj ej , sin θj ej ) where cos θj , sin θj ∈ Z[cos αi , sin αi :
i = 1, 2, . . . , 3k] and ej comes from the set of edge lengths. In other words the
coordinates lie in the R-module where our ring is generated by the sines and cosines
of the angles of the triangles and the abelian group is generated by the edge lengths
of the triangles. This is more restrictive than Theorem 1 where the vertices are in the
field generated by the cotangents of the angles.
We can also consider tilings where instead of using congruent lattice triangles we
use congruent polygons. For example, Archimedes’ Stomachion is a subdivision of
the square into fourteen polygonal pieces (see Fig. 3). It is known  that there are
268 ways to arrange the pieces of the Stomachion to tile a square. Further if we put
any of these arrangements of the Stomachion on [0, 12]×[0, 12] each vertex will be
located at integer coordinates.
We can consider tilings of polygons P where we tile using congruent copies of
the fourteen Stomachion pieces. Since each piece in the Stomachion can be divided
into lattice triangles with the only divisors of aj2 + bj2 being 2, 5, 13, 17, we get the
following by using Theorems 2 and 4.
Corollary 5 Let P be a simple polygon in the plane such that the coordinates of
the vertices of P are rational, and one of the vertices of P is the origin with an
adjacent vertex on the x-axis. Suppose that P is dissected into congruent copies of
the fourteen Stomachion pieces. Then the coordinates of the vertices are of the form
p/(5q 13r 17s ).
Since the Stomachion pieces include a (
) and a (
) triangle we see that
the powers of 5 and 13 can be arbitrarily large in the denominator. However, it is
unknown whether the power 17 can be arbitrarily large (or even 1).
If we examine the packing in Fig. 2 we see that all the triangles could be glued
back to back on their hypotenuses. It is a simple exercise to show using Theorem 1
that for (
) packings we will always be able to glue triangles on their hypotenuses,
so in this case we are actually packing with “boxes” and “kites” (see Fig. 4).
This more generally will occur whenever the triangles i = (ai , bi ) have irrational
hypotenuses which are linearly independent (over Q). However when we have two
hypotenuses which are rational multiples of each other, or a rational hypotenuse, more
interesting behavior can occur. As an example consider the tiling in Fig. 5 which uses
) and (1, 7) triangles (with hypotenuses √2 and 5√2 respectively). In this case
our tiling does not consist only of boxes and kites.
We can use lattice triangles to tile polygons that do not have all rational
coordinates. For example we can use two (
) triangles and four (
) triangles to
tile [0, √5]×[0, √5]. More generally, the proofs that we have given can be used
to show in the case when our polygon does not have rational coordinates that the
coordinates of the vertices also have some restricted forms. As an example, it can
be shown in the tiling of [0, √r ]×[0, √r ] (r not a square) with right triangles
k = (ak , bk ) the coordinates of the vertices are of the form
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