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J. Appl. Math. & Computing Vol. 17(2005), No. 1 - 2, pp. 343 - 350
TILINGS OF ORTHOGONAL POLYGONS WITH SIMILAR
RECTANGLES OR TRIANGLES
ZHANJUN SU∗ AND REN DING
Abstract. In this paper we prove two results about tilings of orthogonal
polygons. (1) Let P be an orthogonal polygon with rational vertex coordinates and let R(u) be a rectangle with side lengths u and 1. An orthogonal
polygon P can be tiled with similar copies of R(u) if and only if u is algebraic and the real part of each of its conjugates is positive; (2) Laczkovich
proved that if a triangle ∆ tiles a rectangle then either ∆ is a right triangle
or the angles of ∆ are rational multiples of π. We generalize the result of
Laczkovich to orthogonal polygons.
AMS Mathematical Subject Classification : 52C20.
Keywords and phrases : Polyomino, orthogonal polygon, tiling, field, rectangle.
1. Tilings of orthogonal polygons with similar rectangles
Let P 1 and P 2 be polygons. We say that P 1 can be tiled with similar copies
of P 2 , if P 1 can be decomposed into finitely many non-overlapping polygons
similar to P 2 . In 1990 Laczkovich considered the tilings of polygons with similar
triangles [1]. In 2001 Szegedy discussed the tilings of the square with similar
right triangles [3]. And in 1995 Laczkovish and Szekeres discussed the tilings of
the square with similar rectangles [4]. In this section we generalize the result of
[4] to orthogonal polygons.
A polyomino is the union of a finite number of standard squares (by a standard
square S in the xy-plane we mean a unit square whose corners have integer
coordinates).
An orthogonalpolygon is a polygon whose edges are horizontal or vertical.
Received September 30, 2003. Revised April 8, 2004. ∗ Corresponding author.
This research was supported by NSFH and SFHEM
c 2005 Korean Society for Computational & Applied Mathematics and Korean SIGCAM.
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Zhanjun Su and Ren Ding
Clearly, a polyomino is a special orthogonal polygon.
Let P be a polyomino and R be a rectangle. We say that a rectangle R tiles
the polyomino P if P can be decomposed into finitely many non-overlapping
rectangles similar to R. We consider, for every fixed value u, whether or not
the polyomino P can be tiled with similar copies of the rectangle R(u) with side
lengths u and 1.
Lemma 1. ([5]) Let P (z) = z n + a1 z n−1 + a2 z n−2 + · · · + an be a polynomial
with real coefficients, and let Q(z) = a1 z n−1 + a3 z n−3 + a5 z n−5 + · · · . Then all
the zeros of P (z) have negative real parts if and only if
1
Q(z)
=
P (z)
c1 z + 1 + c z+
2
,
1
1
..
. + cn1 z
where the coefficients c1 , c2 , . . . , cn are all positive.
By the definition of polyomino and Lemma 1 we have the following lemma
(see [4]).
Lemma 2. If there are positive rational numbers c1 , c2 , . . . , cn such that
c1 u +
1
c2 u +
1
= 1,
..
. + cn1 u
then every polyomino P can be tiled with similar copies of R(u).
Theorem 1. For every u > 0, a polyomino P can be tiled with similar copies
of R(u) if and only if u is algebraic and the real part of each of its conjugates is
positive.
Proof. First suppose that u is a positive algebraic number with conjugates lying
in the open half-plane Re(z) > 0. By Lemma 2 we know that the necessity of
this theorem is true.
Next suppose that P is decomposed into the non-overlapping rectangles
Ri = [ai , bi ] × [ci , di ] (i = 1, 2, . . . , n)
such that for every i either (di − ci )/(bi − ai ) = u or (di − ci )/(bi − ai ) = 1/u
holds. Let F and G be arbitrary complex-valued functions defined on the set
{ai , bi , ci , di |i = 1, 2, . . . , n}. Then
Φ([x1 , x2 ] × [y1 , y2 ]) = [F (x2 ) − F (x1 )] × [G(y2 ) − G(y1 )]
Tilings of orthogonal polygons with similar rectangles or triangles
345
defines an additive interval function.
Now let F be an arbitrary isomorphism mapping the field Q(u) into C, and
let F (u) = v. We can extend F to the field
K = Q(a1 , b1 , c1 , d1 , . . . , an , bn , cn , dn )
as an isomorphism, we also denote the extension by F . Let G = F , then G is
also an isomorphism of K, and G(u) = v. If (di − ci )/(bi − ai ) = wi , then we
have
G(di ) − G(ci ) = G(di − ci ) = G[(bi − ai )wi ] = G(bi − ai )G(wi )
= F (bi − ai )G(wi ) = F (bi ) − F (ai )G(wi ).
Taking into consideration that for every z ∈ N , F (z) = z and G(z) = z, we
obtain
n
n
X
X
[F (bi ) − F (ai )] × [G(di ) − G(ci )] =
|F (bi − ai )|2 G(wi ) = A,
i=1
i=1
where A is the area of P.
For every i we have either wi = u or wi = 1/u. In the first case G(wi ) = v
and Re[G(wi )] = Re(v), while in the second case
G(wi ) = 1/v = v/|v|2 and Re[G(wi )] = Re(v)/|v|2 .
Therefore the real part of each form on the left-hand of the above equation
is of the form cRe(v), where c ≥ 0. Since the value of the sum is A, this
gives Re(v) > 0. It is well known that the set of numbers F (u), when F runs
through the isomorphism of Q(u), equals the set of all transcendental numbers
if u is transcendental, and equals the set of conjugates of u if u is algebraic.
Since there are transcendental numbers with negative real part, this implies
that u is algebraic and each of its conjugates has positive real part. The proof
is complete.
According to the technique of Theorem 1 we know that if every vertex of orthogonal polygon P has rational vertex coordinates, then for orthogonal polygon
P we have the following theorem.
Theorem 2. For every u > 0, the orthogonal polygon P with rational vertex
coordinates can be tiled with similar copies of R(u) if and only if u is algebraic
and the real part of each of its conjugates is positive.
For rectangle R(a) (a is not rational), we have√not the same result as Theorem
1 and√Theorem 2. For example rectangle R( 2) can be tiled by itself, but
Re(− 2) < 0 (since x2 − 2 = 0).
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Zhanjun Su and Ren Ding
2. Tilings of orthogonal polygons with similar triangles
We will use the notation of [1], and we say that a triangle ∆ tiles the polygon
P if P can be decomposed into finitely many non-overlapping triangles similar
to ∆.
In 1990 M. Laczkovich [1] proved that if a triangle ∆ tiles a rectangle then
either ∆ is a right triangle or the angles of ∆ are rational multiples of π. We
prove that the above result about tilings of rectangles with similar triangles is
also true for orthogonal polygons.
The following polygon P is an orthogonal polygon.
Lemma 3. For every orthogonal polygon P with n vertices, it holds that n =
2m + 4, where m is the number of the angle 32 π [2].
Theorem 3. If a triangle ∆ with angles α, β, γ (α ≤ β ≤ γ) tiles an orthogonal
polygon P, then either ∆ is a right triangle or the angles of ∆ are rational
multiples of π.
Proof. Let P be an orthogonal polygon and ∆ be a triangle with angles α, β, γ
(α ≤ β ≤ γ).
Let P be decomposed into the triangles ∆1 , ∆2 , . . . , ∆N such that each ∆i is
similar to ∆, and let V be a vertex of one of the triangles ∆i (i = 1, 2, . . . , N ).
If V is a vertex of P with angle π2 , then we have
π
(1)
pα + qβ + rγ = .
2
If V is a vertex of P with angle
3π
2 ,
then we obtain
3π
.
2
If V is not a vertex of P, then we have either
pα + qβ + rγ =
(2)
pα + qβ + rγ = π.
(3)
pα + qβ + rγ = 2π,
(4)
or
where p, q, and r are all integers.
Let
L∆ = {r1 α + r2 β + r3 γ|ri ∈ Q, i = 1, 2, 3}.
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We denote by dimL∆ the dimension of L∆ as a linear space over Q.
If the angles of ∆ are not rational multiples of π, then we have the following
cases:
Case I : dimL∆ = 3.
In this case, α, β, γ are linearly independent over Q. Hence if pα+qβ +rγ = π
holds, where p, q, r are integers, then p = q = r = 1. Consequently, π = α+β+γ.
So we obtain
π
1
1
1
= α + β + γ.
2
2
2
2
Since α, β, γ are linearly independent over Q and π2 is uniquely representable
in the above form, the numbers p, q, r in the equations (1) can not be integers,
which is a contradiction. Hence when dimL∆ = 3, the triangle ∆ can not tile
the orthogonal polygon P.
Case II : dimL∆ = 2.
Since 3γ ≥ π and α ≤ β ≤ γ, each of the equations (1) must take the form of
one of the following equations:
γ=
π
.
2
In this case, ∆ is a right triangle.
or
π
, max{p1 , q1 } ≥ 2.
2
Each of the equations (2) is of the form
p1 α + q1 β =
3π
, 0 ≤ r2 ≤ 4.
2
If r2 = 3, then we consider the following equations:
p1 α + q1 β = π2 , max{p1 , q1 } ≥ 2,
p2 α + q2 β + 3γ = 3π
2 .
p2 α + q2 β + r2 γ =
Since γ = π − α − β, we have
p1 α + q1 β = π2 ,
(3 − p2 )α + (3 − q2 )β =
(5)
(6)
3π
2 .
Because the two equations above are not independent, we have
p2 + 3p1 = 3, q2 + 3q1 = 3.
By max{p1 , q1 } ≥ 2, and p1 , p2 , q1 , q2 are integers, we know that the equations
above can not hold.
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Zhanjun Su and Ren Ding
If r2 = 4, then we consider the following equations:
p1 α + q1 β = π2 , max{p1 , q1 } ≥ 2,
p2 α + q2 β + 4γ = 3π
2 .
Since γ = π − α − β, we have
p1 α + q1 β = π2 ,
(4 − p2 )α + (4 − q2 )β =
5π
2 .
Because the two equations above are not independent, we have
p2 + 5p1 = 4, q2 + 5q1 = 4.
By max{p1 , q1 } ≥ 2, and p1 , p2 , q1 , q2 are integers, we know that the equations
above also can not hold.
Then in the equations (6), we need consider the case of 0 ≤ r2 ≤ 2. So we
consider the following equations:
p1 α + q1 β = π2 , max{p1 , q1 } ≥ 2,
p2 α + q2 β + r2 γ = 3π
2 , 0 ≤ r2 ≤ 2.
Applying Lemma 3, we know that in the orthogonal polygon P the number
n−4
m of angle 3π
and the number t of angle π2 is equal to n+4
2 is equal to
2
2 .
n+4
n−4
Since 2 > 2 , and max{p1 , q1 } ≥ 2, then in the equations (1) and (2) the
total number of α0 s or β 0 s is strictly greater than the number of γ 0 s.
Let
π
(7)
kα + mβ =
2
be one of these equations in (5). Then either k ≥ max{m, 2} or m ≥ max{k, 2}.
Suppose the former (the latter can be considered similarly). Then the total
number of α0 s is strictly greater than the number of γ 0 s in the equations (1) and
(2) and hence there must be an equation among the equations (3) and (4) such
that p < r. If p, q, r are all positive, then we can subtract α + β + γ = π from
(3) or (4) so that we may assume min{p, q} = 0. Subtracting rα + rβ + rγ = rπ
from the following equation
pα + qβ + rγ = vπ (v = 1 or v = 2),
we obtain that
(p − r)α + (q − r)β = (v − r)π.
(8)
Consider the equations (7) and (8). Since the equations (7) and (8) are not
independent, there is a number c such that
ck = p − r, cm = q − r, c/2 = v − r.
Tilings of orthogonal polygons with similar rectangles or triangles
349
Since p < r, we have c < 0, and by m ≤ k, we obtain p ≤ q and v < r. This
gives, by min{p, q} = 0, p = 0. We also have
v−r =
1
−r
c=
, 2k(r − v) = r.
2
3k
If v = 1, then
2k(r − 1) = r, k =
r
.
2(r − 1)
Since k ≥ 2,
r
4
≥ 2, r ≥ 4r − 4, r ≤ , r = 0, 1.
2(r − 1)
3
Because v = 1 and v < r, this is impossible. If v = 2, then
r
r
2k(r − 2) = r, k =
,
≥ 2, r ≥ 4r − 8, 3r ≤ 8, r = 0, 1, 2.
2(r − 2) 2r − 4
Since v = 2 and v < r, this is also impossible. So in this case the triangle ∆ can
not tile the orthogonal polygon P. The proof is complete.
Acknowledgement
The authors wish to thank the anonymous referees for their careful reading
and helpful suggestions which we accepted gratefully.
References
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3. B. Szegedy, Tilings of the Square with Similar Right Triangles, Combinatorica 21 (2001)
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4. M. Laczkovich and G. Szekeres, Tilings of the Square with Similar Rectangles, Discrete
Comput. Geom. 13 (1995), 569-572.
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Zhanjun Su and Ren Ding
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Zhanjun Su received his BSc and MSc from the Hebei Normal University. Since 2001
he has been a graduate student for Ph. D. degree at Hebei Normal University under the
direction of Prof. Ren Ding. His research interests focus on discrete geometry, convex
geometry and combinatorial geometry.
Department of Mathematics, Hebei Normal University, Shijiazhuang 050016, P. R. China
e-mail: [email protected]
Ren Ding is a professor of mathematics, supervising Ph. D. programs at Hebei Normal
University. His research interests focus on discrete geometry, convex geometry and combinatorial geometry.
Department of Mathematics, Hebei Normal University, Shijiazhuang 050016, P. R. China
e-mail: [email protected]