on rational radii coin representations of the wheel graph

34 Pages·2013·0.23 MB·English

Discussiones Mathematicae
General Algebra and Applications 33 (2013) 167–199
doi:10.7151/dmgaa.1200
ON RATIONAL RADII COIN REPRESENTATIONS
OF THE WHEEL GRAPH
Geir Agnarsson
Department of Mathematical Sciences
George Mason University
Fairfax, VA, USA
e-mail: [email protected]
and
Jill Bigley Dunham
Mathematics Department
Hood College
Frederick, MD, USA
e-mail: [email protected]
Abstract
A flower is a coin graph representation of the wheel graph. A petal of a
flowerisanoutercoinconnectedtothecentercoin. Theresultsofthispaper
aretwofold. Firstwederiveaparametrizationofall the rational(andhence
integer)radiicoinsofthe 3-petalflower,alsoknownasApolloniancirclesor
Soddy circles. Secondlyweconsiderageneraln-petal flowerandshowthere
is a unique irreducible polynomial Pn in n variables over the rationals Q,
the affinevarietyofwhichcontainsthe cosinusofthe internalanglesformed
by the center coin and two consecutive petals of the flower. In that process
we also derive a recursion that these irreducible polynomials satisfy.
Keywords: planar graph, coin graph, flower, polynomial ring, Galois
theory.
2010MathematicsSubjectClassification:05C10,05C25,05C31,05C35.
1. Introduction
By a coin graph we mean a graph whose vertices can be represented as closed,
non-overlapping disks in the Euclidean plane such that two vertices are adjacent
168 G. Agnarsson and J.B. Dunham
if and only if their corresponding disks intersect at their boundaries, i.e., they
touch. For n N the wheel graph W on n + 1 vertices is the simple graph
n
∈
obtained by connecting an additional center vertex to all the vertices of the cycle
C on n vertices. A coin graph representation of a wheel graph is called a flower.
n
The coins of a flower connected to the center vertex are called petals, so a coin
representation of W has n petals and is called an n-petal flower. In Figure 1 we
n
see an example of a flower on the left, and a configuration of coins that does not
form a flower on the right.
Figure 1. Examples of a flower and a non-flower.
Thestudyofflowers is centralinmany discretegeometrical settings, inparticular
in circle packings [14] and also in the study of planar graphs in general, since
every planar graph has a coin graph representation. That a coin graph is planar
is clear, but that the converse is true is a nontrivial topological result, usually
credited to Thurston [15], but is also due to both Koebe [11] and Andreev [1].
For a brief history of this result we refer to [16, p. 118]. Numerous simply stated
but extremely hard problems involving coin graphs can be found in a recent and
excellent collection ofresearchproblemsindiscretegeometry[3]. Also,Brightwell
and Scheinerman [4] explored integral representations of coin graphs, where the
radii of the coins can take arbitrary positive integer values.
In Section 2 we start by considering the case of n = 3, and we obtain all
rational (and hence integer) radii of four mutually tangent circles, sometimes
called Soddy circles after Frederick Soddy, an English chemist who rediscovered
Descartes’ Circle Theorem in 1936 [2] or Apollonian circles after Apollonius of
Perga, who first studied mutually tangent circles more than two thousand years
ago. Thiswillbeobtainedinelementaryways,involvingonlyEuclideangeometry,
high school algebra and elementary number theory. Still our parametrization is
new and differs from the one obtained by Graham et al. in [5] as we will see in
in the last subsection of Section 2.
On rational radii coin representations of the wheel graph 169
As a first step toward the study of the general case when n 3, we will in the
≥
remaining sections then study an algebraic relation the radii of general n-petal
flowersmustsatisfy,andshowthatforeveryn 3thecosinesofthecentralangles
≥
of an n-petal flower are contained in the affine variety of a uniquely determined
symmetric and irreducible polynomial P in n variables over the rationals. We
n
note that the cosines are more interesting than the sines in this investigation,
for the mere reason that cosines of the angles of an integer-sided triangle are all
rational. Hence, a necessary requirement for the radii to be integers is that the
mentioned cosines are all rational. Note that, unlike for the case n = 3, the radii
are not determined by the central angles alone for n 4, where a fixed set of
≥
central angle values can yield a wide range of radii, something we shall see in
Section 3.
The rest of the paper is organized as follows:
In Section 2 we consider the special case of a 3-petal flower and we derive our
first main result: a free parametrization of all rational radii of the outer circles
when the inner circle has radius one, in an elementary way. This will then yield
a new free parametrization of all integer radii of four mutually tangent Soddy
circles.
InSection3weuseGaloistheorytoformallydefinethepolynomialsP (x ,...,
n 1
x ), whose affine variety contain (cosθ ,...,cosθ ) where θ ,...,θ are the in-
n 1 n 1 n
ternal angles of an n-petal flower for n 3, and prove some useful lemmas.
≥
In Section 4 we firstobtain a recursive presentation of P (x ,...,x ), and use
n 1 n
that to prove our second main result of this paper, that each symmetric P is an
n
irreducible polynomial over Q.
2. Rational radii of Soddy circles
Here we deal with the special case of n = 3, and we characterize all rational
solutions for 3-petal flowers, thereby obtaining all rational radii of four mutually
tangent Soddycircles. Thissection usesonlyelementary mathematics, butserves
as a motivation (or justification) for our discussion in the sections to follow.
In general, to find all integer-radii coins forming an n-petal flower in the
Euclidean plane, it is equivalent by scaling to find all rational radii coins where
the center coin is assumed to have radius one. Hence, assume we have a 3-petal
flower with a center radius of one and the outer radii r ,r and r . If the internal
1 2 3
angle formed by the center coin and two outer coins of radii r and r is θ ,
i i+1 i
then by the law of cosines we have
r +r r r +1
i i+1 i i+1
(1) x := cosθ = − ,
i i
r +r +r r +1
i i+1 i i+1
170 G. Agnarsson and J.B. Dunham
and the equation relating the three angles is then θ + θ + θ = 2π. Clearly,
1 2 3
if each radius r is rational, then so is each x . Isolating one angle, say θ , and
i i 3
taking the cosine of both sides we obtain
(2) x = x x (1 x2)(1 x2).
3 1 2− − 1 − 2
q
For rational x and x it is clear that x will be rational if and only if the term
1 2 3
under the radical is the square of a rational number. For i = 1,2 let x = pi for
i qi
with p ,q Z. By (2) we then obtain
i i
∈
1
x = x x (q2 p2)(q2 p2).
3 1 2− q q 1 − 1 2 − 2
1 2
q
Here (q2 p2)(q2 p2) is a square if and only if q2 p2 = s2β and s N for
1 − 1 2 − 2 i − i i i ∈
i = 1,2 where β is a square-free integer. Here we need the following result from
elementary number theory:
Theorem 2.1. Let β be a square-free integer. The integers x,y,z form a prim-
itive solution to the Diophantine equation x2+βy2 = z2 if and only if there are
positive integers m and n and a factorization β = bc where bm2 and cn2 are
relatively prime such that x = bm2−cn2, y = mn, z = bm2+cn2, where both m and
2 2
n are odd or both are even, or x =bm2 cn2, y = 2mn, z = bm2+cn2 otherwise.
−
For a proof of Theorem 2.1, see Appendix A.
Since x = pi for i =1,2 we have by Theorem 2.1 that
i qi
b m2 c n2 b m2 c n2
(3) x = 1 1− 1 1, x = 2 2− 2 2,
1 b m2+c n2 2 b m2+c n2
1 1 1 1 2 2 2 2
where β = b c = b c are two (not necessarily distinct) factorizations of the
1 1 2 2
square-free integer β, and where m ,n can be chosen from the nonnegative inte-
i i
gers. Note that either solution from Theorem 2.1 will yield the same form of x
1
and x in (3).
2
By (1) we have
r +r r r +1 r +r r r +1 r +r r r +1
1 2 1 2 2 3 2 3 3 1 3 1
x = − , x = − , x = − .
1 2 3
r +r +r r +1 r +r +r r +1 r +r +r r +1
1 2 1 2 2 3 2 3 3 1 3 1
Rewriting each equation for x as a polynomial equation in terms of r and r
i i i+1
( where 4 1 modulo 3) and then factoring in terms of r and r we obtain
i i+1
≡
On rational radii coin representations of the wheel graph 171
x 1 x 1 2(1 x )
1 1 1
r + − r + − = − ,
1 x +1 2 x +1 (x +1)2
(cid:18) 1 (cid:19)(cid:18) 1 (cid:19) 1
x 1 x 1 2(1 x )
2 2 2
r + − r + − = − ,
2 x +1 3 x +1 (x +1)2
(cid:18) 2 (cid:19)(cid:18) 2 (cid:19) 2
x 1 x 1 2(1 x )
3 3 3
r + − r + − = − .
3 x +1 1 x +1 (x +1)2
(cid:18) 3 (cid:19)(cid:18) 3 (cid:19) 3
Now we can solve the firstand third equations for r and r respectively in terms
2 3
of r ,x ,x . Substituting these into the second equation, we can then solve that
1 1 3
for r in terms of x ,x ,x obtaining
1 1 2 3
1 x x +x +x 2(1 x )(1 x )(1 x )
1 3 3 1 1 2 3
(4) r = − − ± − − − .
1
2x x +x x 1 x
2− 1 p1 3− − 3
Putting x and x from (3) into (2) we obtain
1 2
b m2 c n2 b m2 c n2 4m m n n β
x = x x (1 x2)(1 x2)= 1 1− 1 1 2 2− 2 2 − 1 2 1 2 .
3 1 2− − 1 − 2 (b m2+c n2)(b m2+c n2)
(cid:0) 1 (cid:1)1(cid:0) 1 1 2 2(cid:1) 2 2
q
Substituting this expressions for x and those of x and x from (3) into (4), we
3 1 2
get an expression for r in terms of b ,b ,c ,c ,m ,m ,n ,n :
1 1 2 1 2 1 2 1 2
n (b c2m2n3+2βc m m n2n +βc m2n n2)
r = 1 2 1 2 1 1 1 2 1 2 2 1 1 2
1 b c c m2n2n2 b c2m2n4+c2c n4n2 2βc m m n3n +b2c m4n2
1 1 2 1 1 2− 2 1 2 1 1 2 1 2− 1 1 2 1 2 1 2 1 2
n n (b m2+c n2) c c (b c m2n2+2βm m n n +b c m2n2)
1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 2 1 2 1 .
± b c c m2n2n2 b c2m2n4+c2c n4n2 2βc m m n3n +b2c m4n2
1 1 2 1 1 2− 2 1p2 1 1 2 1 2− 1 1 2 1 2 1 2 1 2
Using the fact that β = b c = b c , the expression under the square root can be
1 1 2 2
reduced to β(c m n +c m n )2. Thus, this expression for r will only yield a
2 1 2 1 2 1 1
perfect square when β = 1. Therefore, r is rational if and only if q2 p2 = s2
1 i − i i
for i = 1,2, or in other words when 1 x2 is a rational square for i = 1,2. This
− i
means that both cosθ and sinθ are rational for i= 1,2,3.
i i
Proposition 2.2. The 3-petal flower with the center coin of radius one can have
the outer coins of rational radii r ,r ,r if and only if the internal angles θ ,θ ,θ
1 2 3 1 2 3
have both rational cosines and sines for i = 1,2,3.
172 G. Agnarsson and J.B. Dunham
Proposition 2.2 shows a property that is very special for the n-petal flower with
rational radii when n = 3. We now can write a “nice” parametrization for the
cosines x and the radii r in the case when n= 3: namely,
i i
m2 n2 m2 n2 m2 n2 m2 n2 4m m n n
x = 1− 1, x = 2− 2, x = 1− 1 2− 2 − 1 2 1 2.
1 m2+n2 2 m2+n2 3 (m2+n2)(m2+n2)
1 1 2 2 (cid:0) (cid:1)(cid:0)1 1 (cid:1)2 2
for some m ,n ,m ,n N. Putting these into (4) and the similar equations for
1 1 2 2
∈
r and r we obtain the rational forms for r ,r ,r that contain all rational radii
2 3 1 2 3
for the outer coins of a 3-petal flower with center coin of radius one:
n (m n +m n )
1 1 2 2 1
r =
1 m n n m n2+(m2n +n2n )
− 1 1 2− 2 1 1 2 1 2
n n
1 2
r =
2
n n (m n +m n )
1 2 2 1 1 2
− ±
n (m n +m n )
2 1 2 2 1
r = .
3 m n2 m n n (n n2+m2n )
− 1 2− 2 1 2± 1 2 2 1
We will now determine a range of the parameters that will yield meaningful
solutions in the above parametrization.
Observation 2.3. If θ ,θ ,θ are the internal angles of a 3-petal flower, then
1 2 3
90◦ < θ < 180◦ for each i = 1,2,3 and these three inequalities are all sharp.
i
For a proof of Observation 2.3, see Appendix B.
By Observation 2.3 we now know that for all the angles θ , we have 90◦ <
i
θ < 180◦, and hence 1 < cosθ < 0. Hence, in the parametrization of x and
i i 1
−
x
2
m2 n2 m2 n2
x = 1− 1, x = 2− 2,
1 m2+n2 2 m2+n2
1 1 2 2
we must have n > m . In this case x in (2) must satisfy m2 n2 m2 n2
i i 3 1− 1 2− 2 −
4m m n n < 0,whichisequivalentto(m n +m n )2 > (m m n n )2. Since
1 2 1 2 1 2 2 1 1 2 1 2
(cid:0) −(cid:1)(cid:0) (cid:1)
m < n , this is equivalent to m n +m n > n n m m , or equivalently
i i 1 2 2 1 1 2 1 2
−
(5) m n +m n +m m > n n .
1 2 2 1 1 2 1 2
Looking at the expression for r ,
1
n (m n +m n )
1 1 2 2 1
r =
1 m n n m n2 (m2n +n2n )
− 1 1 2− 2 1± 1 2 1 2
we see that in order for r > 0 to hold we must have n (m2 +n2) > n (m n +
1 2 1 1 1 1 2
m n ). Re-solvingfortheradiir andr usingthepositivetermintheexpression
2 1 2 3
for r , we obtain:
1
On rational radii coin representations of the wheel graph 173
n n n (m n +m n )
1 2 2 1 2 2 1
r = , r = ,
2 m n +m n n n 3 n n2+m2n m n2 m n n
2 1 1 2− 1 2 1 2 2 1− 1 2− 2 1 2
which give ustwo additional constraints in order to ensurepositive radii: m n +
2 1
m n > n n and n (m n +m n ) > n (m2+n2). Note that the first of these
1 2 1 2 2 1 2 2 1 1 2 2
constraints is stronger than (5). Letting t = m /n for each i = 1,2,3, we can
i i i
express our first main theorem that characterizes all rational radii (and hence
integer radii by scaling) Soddy circles as follows.
Theorem 2.4. If a 3-petal flower has rational radii and the innermost coin has
radius of one, then the cosines x ,x and x of the angles at the innermost coin
1 2 3
are given by
t2 1 t2 1 (t2 1)(t2 1) 4t t
x = 1− , x = 2− , x = 1− 2− − 1 2
1 t2+1 2 t2+1 3 (t2+1)(t2 +1)
1 2 1 2
and the corresponding radii by
t +t
1 2
r = ,
1 t2 t t +1
1− 1− 2
1
r = ,
2
t +t 1
1 2
−
t +t
1 2
r = ,
3 t2 t t +1
2− 1− 2
for some t ,t Q such that 1 < t +t , t +t < t2+1, t +t < t2+1. Further,
1 2 ∈ 1 2 1 2 1 1 2 2
all rational 3-petal flowers have rational radii and internal cosines parametrized
as above in terms of such t ,t Q. This parametrization characterizes all sets
1 2
∈
of four mutually tangent Soddy circles of rational radius in the plane.
y
1
x
1
Figure 2. The region R.
174 G. Agnarsson and J.B. Dunham
We note thattheconditions therational parameters t andt satisfy intheabove
1 2
Theorem 2.4 is simply (t ,t ) R Q2 where
1 2
∈ ∩
R = (x,y) R2 :1 < x+y, y < x2 x+1, x < y2 y+1 ,
{ ∈ − − }
is an open region symmetric about the line x = y, which looks like a “hyperbolic
triangle”withvertices (0,1),(1,0) and(1,1), andwheretheinternalangleateach
of the vertices is 0 (see Figure 2.) By the 2-dimensionality of the region R, we
note that the parametrization of the radii and the internal cosines in the above
Theorem 2.4 in terms of t ,t Q is free in the sense that they don’t satisfy any
1 2
∈
equation.
Example. Considert =2/3andt = 1/2. Weseethat(t ,t ) = (2/3,1/2) R,
1 2 1 2
∈
and are hence legitimate rational parameters by Theorem 2.4. Here we have
x = 5 , x = 3, x = 33, and the corresponding radii r = 21, r = 6,
1 −13 2 −5 3 −65 1 5 2
r = 14. By scaling by the factor of 5 we obtain an integral flower with center
3
radius of r = 5 and the outer radii r = 21, r = 30 and r = 70. See Figure 3.
1 2 3
Remark. Considera 3-flower with the center coin of radius one. Assumefurther
thecentercoiniscenteredattheoriginofthecomplexplaneC. Theinversionz
7→
1/z of the center circle will map all the outer disks to corresponding disks inside
the center disk. This configuration has three non-overlapping disks completely
inside the center disk, and does therefore not represent a 3-flower. However, this
inversionwillkeepalltheanglesθ ,θ andθ andhencetheircosinesx ,x andx
1 2 3 1 2 3
intact. The radii r ,r and r will however change to r′,r′ and r′ respectively,
1 2 3 1 2 3
where (1 + r )(1 r′) = 1 or r′ = ri for each i. By Theorem 2.4 this will
i − i i 1+ri
automatically yieldafreeparametrization ofthreerationalradii,non-overlapping
mutually touching coins, all three touching the unit circle on its inside where
(t ,t ) R Q2 as before.
1 2
∈ ∩
2.1. Descartes’ circle theorem and another parametrization
A nice relation connecting the radii of four mutually tangent Soddy circles in the
Euclidean plane is given by Descartes’ circle theorem [2].
Theorem 2.5 (Descartes). A collection of four mutually tangent circles in the
plane, where b = 1/r denotes the curvatures of the circles, satisfies the relation
i i
1
b2+b2+b2+b2 = (b +b +b +b )2.
1 2 3 4 2 1 2 3 4
Remark. Theorem 2.5 has been generalized to higher dimensions.
Itisstraightforward tocheck thatourrationalparametrization fromTheorem2.4
satisfies Descartes’ circle theorem. Another elegant parametrization of integer
Soddy circles are given by Graham et al. in [5] in the following theorem.
On rational radii coin representations of the wheel graph 175
Theorem 2.6 (Graham, Lagarias, Mallows, Wilks, Yan). The following para-
metrization characterizes the integral curvatures of a set of Soddy circles:
b = x, b = d x, b = d x, b = 2m+d +d x,
1 2 1 3 2 4 1 2
− − − −
where x2+m2 = d d and 0 2m d d .
1 2 1 2
≤ ≤ ≤
Weconcludethissection bybrieflycomparingourrationalparametrization tothe
one given by Theorem 2.6. Suppose we have a 3-petal flower, the coins of which
have integer radii. Further, assume the center coin is the coin with curvature b .
1
By conveniently permuting indices, the remaining outer coins have radii r ,r ,r
1 2 3
given by
b b b
1 1 1
r = , r = , r = .
1 2 3
b b b
2 4 3
Using the rational parametrization of r ,r and r given by Theorem 2.4 and
1 2 3
replacing each b with the integer parametrization from Theorem 2.6, we can
i
solve for d /x,d /x and m/x in terms of t and t and obtain
1 2 1 2
m 1 t t d t2+1 d t2+1
= − 1 2, 1 = 1 , 2 = 2 .
x t +t x t +t x t +t
1 2 1 2 1 2
From this it is clear that 1 + (m/x)2 = (d /x)(d /x), so the quadratic equa-
1 2
tion relating the parameters in Theorem 2.6 is automatically satisfied. Secondly,
the assumption 0 2m d d in Theorem 2.6 translates to the natural
1 2
≤ ≤ ≤
assumption that the radii are ordered by r r r .
2 3 1
≤ ≤
Figure 3. An example Soddy circle obtained from the parametrization.
176 G. Agnarsson and J.B. Dunham
3. The inner cosines of the general n-petal flower
In Section 2 we established a parametrization of all the outer radii of a 3-petal
flower if the inner coin was assumed to have radius one. In doing so we first
established all rational solutions of the inner cosines x ,x and x from the de-
1 2 3
termining relation (2). In this special case of n = 3, (where the inner radius is
assumed to be one) the radii r ,r and r are uniquely determined by the x ,
1 2 3 i
so next we expressed each r in terms of these rational x ,x and x , thereby
i 1 2 3
obtaining free parametrization of the radii r ,r and r . Hence, for n 4 a nec-
1 2 3
≥
essary first step in determining all the rational radii of an n-petal flower with the
inner coin of radius one is to determine all the rational inner cosines given by (1)
where now θ ,...,θ form the inner angles of the n-petal flower in a clockwise
1 n
order. When n 4, however, the radii r are not uniquely determined by the
i
≥
inner cosines x = cosθ , i 1,...,n .
i i
∈ { }
Example. Consider the case of a 4-petal flower with innerradius one and θ =
1
θ = θ = θ = π/2. In this case we have by Pythagoras’s Theorem (r +r )2 =
2 3 4 i i+1
(r +1)2 +(r +1)2 or r = (r +1)/(r 1) for each i = 1,2,3,4, and so a
i i+1 i+1 i i
−
complete rational parametrization of the outer radii are given by
r+1
r =r = r Q, r = r = .
1 3 2 4
∈ r 1
−
Whether a similar game can be played for any given inner angles θ ,...,θ of
1 n
an n-petal flower with rational cosines x = cosθ Q is not what we will
i i
∈
attempt to answer, but rather determining a “minimal” relation the rational
inner cosines must satisfy in general. This is the topic for the remainder of this
paper. We will show that for n 3, the x lie in an affine variety of a symmetric
i
≥
polynomial P (x ,...,x ) (see [10, p. 252] for more information and general
n 1 n
algebraic properties of symmetric polynomials.) The polynomial P in x ,...,x
n 1 n
will be the minimal polynomial of cosθ ,...,cosθ over Q where
1 n
n
(6) θ = 2π.
i
i=1
X
Note that not all sets of n angles θ ,...,θ satisfying (6) can be internal angles
1 n
of an n-petal flower, not even when n = 3, as we saw in Observation 2.3. We
will however see, by the way we defineP here below, that P is symmetric when
n n
n 3. This will be the main task of this section. We then turn to our second
≥
main contribution of the article: to prove that P is irreducible over Q. That we
n
will do in the following Section 4.
Remark. As we remarked after Theorem 2.4, we can assume a setup where the
center coin of radius one is centered at the origin of the complex plane C. The