eSource-Hadou-Complex Bézier curves

8/7/2019 eSource-Hadou-Complex Bzier curves

1/13

Computer Aided Geometric Design 27 (2010) 525537

Contents lists available at ScienceDirect

Computer Aided Geometric Design

www.elsevier.com/locate/cagd

Complex Bzier curves and the geometry of polygons

Rachid Ait-Haddou a,, Walter Herzog b, Taishin Nomura a,ca The Center for Advanced Medical Engineering and Informatics, Osaka University, Osaka, Japanb Human Performance Laboratory, Faculty of Kinesiology, The University of Calgary, Calgary, Canadac Department of Mechanical Science and Bioengineering, Graduate School of Engineering Science, Osaka University, Osaka, Japan

a r t i c l e i n f o a b s t r a c t

Article history:

Received 1 July 2009

Received in revised form 27 April 2010

Accepted 1 June 2010

Available online 4 June 2010

Keywords:

Bzier curve

Polar forms

Geometry of polygons

Discrete Fourier transform

NapoleonBarlotti Theorem

PetrDouglasNeumann Theorem

In this paper, we associate to every planar polygon a complex polynomial, in which the

blossom of the polynomial function captures the process in which linear transformations

applied to the polygon lead to regular structures. In particular, we prove, in a purely

algebraic way several well-known theorems on polygons such as the NapoleonBarlotti

Theorem, the PetrDouglasNeumann Theorem, and the Fundamental Decomposition

Theorem of polygons to regular polygons.

2010 Elsevier B.V. All rights reserved.

1. Introduction

In an anecdotal story, the Field medal laureate, Alain Connes, recalls when he first heard of the Morley Theorem (in

which one of his colleagues wrongly attributed to Napoleon) (Connes, 1998) . . . when I came back home, following one of

the advices of Littlewood, I began to look for a proof, not in books but in my head. My only motivation besides curiosity

was the obvious challenge this is one of the rare achievements of Bonaparte I should be able to compete with. Our

finding of a different proof of the Napoleon Theorem stems unexpectedly from our study of the generalization of the De

Casteljau algorithm to complex polynomials. The polar form of a complex polynomial, as well as the associated complex

De Casteljau algorithm, leads to a general framework of understanding how linear transformations applied to an arbitrary

polygon can lead to regular structures. In particular, we prove in a purely algebraic way several theorems on polygons such

as the NapoleonBarlotti Theorem (Barlotti, 1955; Fisher et al., 1985; Pech, 2001), and the PetrDouglasNeumann Theorem

(Douglas, 1940; Fisher et al., 1985; Gray, 2002).Our general framework can be understood as follows: To each polygon, we associate a complex polynomial, more pre-

cisely, the complex Bzier curve with the vertices of the polygon as its control points. For each linear transformation applied

to the polygon, there is a corresponding new polygon, and therefore, a new associated polynomial function. Such a polyno-

mial function can be readily computed using the symmetry, the multi-affinity of the polar form, as well as the geometrical

interpretation of the De Casteljau algorithm. Iterating the process leads to specific complex polynomials in which the control

polygons have regular structures.

The paper is organized as follows: In the first section, we give the technical materials related to the polar form of

complex polynomials. In the second part of the paper, we give characterizations of polynomials associated with regular

* Corresponding author.E-mail address: [email protected] (R. Ait-Haddou).

0167-8396/$ see front matter 2010 Elsevier B.V. All rights reserved.doi:10.1016/j.cagd.2010.06.002
http://www.sciencedirect.com/http://www.elsevier.com/locate/cagdmailto:[email protected]://dx.doi.org/10.1016/j.cagd.2010.06.002http://dx.doi.org/10.1016/j.cagd.2010.06.002mailto:[email protected]://www.elsevier.com/locate/cagdhttp://www.sciencedirect.com/


2/13

526 R. Ait-Haddou et al. / Computer Aided Geometric Design 27 (2010) 525537

polygons as well as affinely regular polygons. Some interesting expressions of the discrete Fourier transform of the control

polygons in terms of the polar form of the corresponding polynomials will be given. As applications, we give the proofs of

the PetrDouglasNeumann Theorem and the NapoleonBarlotti Theorem.

2. Polynomials and polygons

2.1. Polynomials associated with planar polygons

An (n + 1)-gon P in the Euclidean plane is an ordered (n + 1)-tuple of points P= (p0, p1, . . . , pn). The points pk arecalled vertices and the segments joining the points pk , pk+1 for k = 0, 1, . . . , n (with pn+1 = p0) are called sides. We canidentify each vertex pk = (p1k , p2k ) with the complex number pk = p1k + ip2k with i2 = 1. By virtue of this identification, wecan associate each polygon with a polynomial function in the following way:

Definition 1. Let P= (p0, p1, . . . , pn) be an (n + 1)-gon in the Euclidean plane. We define the polynomial P[a,b](z) of degreen associated to the polygon P over a complex interval [a, b] as

P[a,b](z) =n

i=0pi B

ni (z),

where Bni (z) is the Bernstein polynomial over the interval

[a, b

], and a and b are two distincts complex numbers, i.e.,

Bni (z) = Cni

b zb a

niz ab a

i.

From the above definition, the polygon P is the control polygon of the polynomial P(z) over the interval [a, b]. A fun-damental concept for the rest of the manuscript, in dealing with the control polygon of Bzier curves, is the notion of polar

form (or blossom) fP (Ramshaw, 1989) associated with the polynomial P[a,b](z).

Definition 2. Let P(z) be a polynomial of degree n. There exists a unique multi-affine, symmetric function in n variables

fP :Cn C such that for each z in C we have fP(z, z, . . . , z) = P(z). The function fP is called the polar form or the

blossom associated with the polynomial P.

Once we associate the polynomial P[a,b](z) with the polygon P= (p0, p1, . . . , pn), we can recover the vertices of thepolygon P using the polar form as [9]pi = fP[a,b]

a{ni}, b{i}

, i = 0, 1, . . . , n, (1)

in which the notation z{k} indicates that the complex number z is to be repeated k times.The polynomial P (z) = fP(, z, z, . . . , z) is called the polar derivative of the polynomial P(z) with respect to the pole .

It can be expressed explicitly as

P (z) = P(z) +( z)

nP(z).

We will often use the following fact within this work: if P(z) is a polynomial of degree n and P= (p0, p1, . . . , pn) itscontrol polygon over an interval [a, b], then the control polygon of the polynomial Pa(z) (resp. Pb(z)) over the interval

[a, b

]is the one obtained from P by deleting the last side of the polygon, i.e., (p0, p1, . . . , pn

1) (resp. the first side of the

polygon, i.e., (p1, . . . , pn)).

2.2. Complex De Casteljau algorithm

In this section, we extend the De Casteljau algorithm, for the geometric computation of the polar form, to the complex

plane. The notion of similarity between triangles will be important in the implementation of the algorithm. For such an

aim, we define the notion of the shape parameter of a triangle as in Lester (1996).

Definition 3. Let T = (a, b, c) be an oriented triangle in the complex plane. We define the shape T of the triangle as

T =a ba c .

It is straightforward to show that two oriented triangles T1 and T2 are similar if, and only if, they have the same shape,i.e., T1 = T2 .


3/13

R. Ait-Haddou et al. / Computer Aided Geometric Design 27 (2010) 525537 527

Lemma 1. Let P(z) be a polynomial of degree n, let fP be its polar form and u1, u2, . . . , un1 be complex numbers. Then for ev-ery complex number z1, z2 , and z3 , the two oriented triangles T1 = ( fP(u1, u2, . . . , un1, z1), fP(u1, u2, . . . , un1, z2), fP(u1, u2,. . . , un1, z3)) and T2 = (z1, z2, z3) are similar.

Proof. The shape of the triangle T1 is

T1=

fP(u1, u2, . . . , un1, z1) fP(u1, u2, . . . , un1, z2)fP(u1, u2, . . . , un1, z1) fP(u1, u2, . . . , un1, z3)

. (2)

By the multi-affinity of the polar form, we have

fP(u1, u2, . . . , un1, z1) fP(u1, u2, . . . , un1, z2) =1

n(z1 z2) fP (u1, u2, . . . , un1),

where P is the derivative of the polynomial P. Inserting the last equation into Eq. (2) leads to T1 = (z1 z2)/(z1 z3),which is the shape of the triangle T2 . 2

From the previous lemma, and given the control polygon of a polynomial over a complex interval, we can evaluate the

value of the polynomial and also the value of its polar form for arbitrary poles ui , in a similar fashion as in the case of

parametric Bzier curves, in which, instead of using the multi-affinity over a line, we use the triangle similarity over the

complex plane. The idea of the complex De Casteljau algorithm is easily illustrated using a simple example.

Example 1. Consider the cubic polynomial P(z) = z3 + 3iz 2 + 6z + 5 and consider the control polygon P of this poly-nomial over the interval [1, 1]. From the polar form we can calculate the control polygon of the polynomial P(z) as(p0, p1, p2, p3) = (2 + 3i, 4 i, 6 i, 12 + 3i) (Fig. 1). In order, for example, to calculate the value P(

3i) we proceed as

follows: Since the oriented triangle (1, 1,

3i) is equilateral, the point fP(1, 1,

3i) is such that the oriented triangle

(p0, p1, fP(1, 1,

3i)) is also equilateral. The same argument can be applied to the points fP(1, 1,

3i), fP(1, 1,

3i).

This is the level one complex De Casteljau algorithm. To proceed, we calculate the point fP(1,

3i,

3i) as the point such

that the oriented triangle ( fP(1, 1,

3i), fP(1, 1,

3i), fP(1,

3i,

3i)) is equilateral, and so on until we reach the

last level of the complex De Casteljau algorithm where we calculate fP(

3i,

3i,

3i) = P(

3i). (See Fig. 1.)

Remark 1. The polynomial P(z) expressed in the Bernstein basis have geometrical properties similar to the usual parametric

Bzier curves, in terms of De Casteljau algorithm, subdivision. For this reason, we can view the polynomial P(z), expressed

in the Bernstein basis, as a complex Bzier curve, even though the polynomial P(z) assign a complex output point to each

complex value of the input parameter.

In the following, we will show that the location of the roots of the polynomial P[a,b](z) associated with a polygonP= (p0, p1, . . . , pn) over an interval [a, b] already convey some geometrical informations about the shape of the polygon.In order to do so, we recall the notion of circular region and the well-known Walsh Coincidence Theorem (Marden, 1966).

Definition 4. A circular region of the complex plane is defined as the closed interior or exterior of a circle or a closed

half-plane.

Theorem 1 (Walsh Coincidence Theorem). Let f be a symmetric multi-affine function of n complex variables. Let u1, u2, . . . , un be n

complex numbers which lie in a circular region C. Then there exists a in C such that

f(u1, u2, . . . , un)

=f( , , . . . , ) .

Let P= (p0, p1, . . . , pn) be a planar polygon and denote by P[a,b](z) the associated complex polynomial over an interval[a, b], then we have

Theorem 2. If the polynomial P[a,b](z) has no complex roots in the disk of diameter [a, b] then the polygon P= (p0, p1, . . . , pn)satisfies pi , pi+1 = Re(pi pi+1) > 0, for i = 0, . . . , n 1, i.e., the unsigned angle at the origin of the triangle (pi , O , pi+1) is strictlysmaller than /2.

Proof. Let us assume that for a certain index i (i n 1) we have pi , pi+1 0. Then the disk of diameter [pi , pi+1]contains the origin. Therefore, there exists a complex number in the disk of diameter [a, b] such that

fPa{ni1}, b{i},

= 0. (3)

In fact, the complex number in (3) is the unique point such that the oriented triangles (O , pi, pi+1) and (, a, b) aresimilar. Therefore, since O belongs to the disc of diameter [pi , pi+1], the point belongs to the disc of diameter [a, b].


4/13


Fig. 1. The complex De Casteljau algorithm applied to the polynomial P(z) of Example 1. Refer to the example for further explanations.

Fig. 2. For a polynomial with no critical points in the unit disk, the control polygon over the interval [1, 1] satisfies the property that all the angles i arebigger than /2.

By the Walsh Coincidence Theorem, relation (3) indicates that the polynomial P[a,b](z) has a root in the disk of diameter[a, b], which contradicts our initial assumption. 2

If we apply Theorem 2 to the derivative of the polynomial P

[a,b

](z), we obtain that if the polynomial P

[a,b

](z) has no

critical point in the disk of diameter [a, b], the unsigned interior angle at pi , i = 1, 2, . . . , n1, of the triangle (pi1, pi , pi+1)is strictly bigger than /2 (Fig. 2).


5/13


There are several geometrical characteristics of the polygon P that can be deduced from the location of the roots of the

corresponding polynomial P[a,b](z). However, since this is not the main goal we set in our introduction, such a topic will bethe subject of another manuscript.

For our subsequent discussions, we need the following two propositions, which are also of interest on their own.

Proposition 1. Let P(z) be a polynomial of degree n. If, for a complex number , we have fP(, z, z, . . . , z) 0, then the polynomialP(z) is of the form P(z)

=(z

)n for a complex constant.

Proof. Let fP(u1, u2, . . . , un) be the polar form of the polynomial P(z), viewed as a multivariate polynomial in the vari-

ables u1, u2, . . . , un . If for a complex number , we have fP(, z, z, . . . , z) = 0 for all z, then we would have fP(, u2, u3,. . . , un) = 0, for all tuples (u2, u3, . . . , un). It then follows that fP(u1, u2, . . . , un) must be divisible, as a polynomial, by(u1 ). In a similar way, fP(u1,u2, . . . , un) must also be divisible by the polynomials (u2 ),(u3 ) , . . . , (un ). Sincemultivariate polynomials form a UFD, we deduce that fP(u1,u2, . . . , un) = (u1 )(u2 ) (un ), for some complexnumber , from which it follows that P(z) = (z )n. 2

We can generalize the last proposition as follows:

Corollary 1. Let P(z) be a polynomial of degree n. Let us assume that there exist k (k n) distinct complex numbers 1, 2, . . . , ksuch that fP(1, 2, . . . , k, z, z, . . . , z) = 0, for every complex number z. Then the polynomial P(z) can be written as

P(z) = 1(z 1)n

+ 2(z 2)n

+ + k(z k)n

,

where i , i = 1, . . . , k are complex numbers.

Proof. We will proceed by induction on the integer k. For k = 1, we have the statement already proved in Proposition 1.Let us assume that the corollary is true for any k complex numbers, and consider (k + 1) complex numbers 1, 2, . . . , k+1 ,such that fP(1, 2, . . . , k+1, z, z, . . . , z) 0. Let Q(z) and R(z) be the polynomials of degree n 1 defined as the polarderivative of the polynomial P(z) with respect to the poles k and k+1 respectively, i.e., Q(z) = Pk (z) and R(z) = P

k+1 (z).

Then, for every complex number z, we have

fQ(1, . . . , k1, k+1, z, z, . . . , z) = 0 and fR (1, 2, . . . , k, z, z, . . . , z) = 0.By the induction hypothesis, there exist complex numbers i , i such that

Q(z) =k+1

i=1,i =ki (z i)n1 and R(z) =

ki=1

i(z i )n1. (4)

Let h be an integer such that 1 h k 1. We have

fQ

1, . . . , h1, h+1, . . . , k1, k+1, z{nk}= h(z h)nk k+1

i=1,i =h,k(i h) (5)

and

fR

1, . . . , h1, h+1, . . . , k1, k, z{nk}= h(z h)nk k

i=1,i =h(i h). (6)

As the two expressions in (5) and (6) are equal due to the definition of the polynomials Q and R , we found that forh = 1, . . . , k 1h(k+1 h) = h(k h). (7)

Moreover, by the multi-affinity of the polar form, we have

P(z) = z kk+1 k

R(z) + k+1 zk+1 k

Q(z). (8)

Inserting Eqs. (4) into Eq. (8), leads to the fact that the polynomial P(z) can be written as

P(z) = k(z k)n + k+1(z k+1)n + H(z), (9)where H(z) is the polynomial of degree n given by

H(z) = 1k+1 k

k

1

i=1

(i i)z + (k+1i i k)(z i )n1.


6/13


Noticing from (7) that (i i)i + (k+1i i k) = 0 for i = 1, . . . , k 1 and inserting H(z) into (9) lead to the desiredresult. 2

In order to prove our next needed proposition, we introduce a bracket operator on the space of polynomials. Given two

polynomials P(z), Q(z) of degree n, we define (Goldman, 1994)

P(z), Q(z)n =

n

k=0

(1)n

n!P(k)() Q(nk)(),

where R( j)(z) stand for the jem derivative of a polynomial R(z). The operator [P(z), Q(z)]n is a bilinear form in the spaceof polynomials of degree n ([P(z), Q(z)]n is independent of ). Moreover, we have (Goldman, 1994)

P(z), (u1 z)(u2 z) (un z)

n= fP(u1, u2, . . . , un), (10)

where fP is the polar form of the polynomial P(z).

Proposition 2. Let P(z) be a polynomial of degree n and fP its blossom. Let (uj1, u

j2, . . . , u

jn) be a set of n-tuple complex numbers,

j = 1, . . . , l. Then for every set of complex number s 1, s2, . . . , sl such that S = s1 + s2 + +sl = 0, we havel

j=1

s j fPu j1, u j2, . . . , u jn=S fP(1, . . . , n),

where i are the roots of the polynomial

H(z) =l

j=1s j

ni=1

z u j

i

.

Proof. From (10), we have

lj=1

s j fP

uj1, u

j2, . . . , u

jn

= lj=1

s j

P(z),

uj1 z

u

j2 z

u jn zn.By the bilinearity of the bracket operator, we get

lj=1

s j fP

uj1, u

j2, . . . , u

jn

=

P(z),

lj=1

s j

uj1 z

u

j2 z

u jn z

n

,

Using again Eq. (10) lead to the proof of the proposition. 2

Remark 2. Proposition 2 can be recast as expressing the fact that affine combination of polar values of a univariate poly-

nomial is itself a polar value, with polar arguments given as the roots of another polynomial, say, H(z). Note that even if

the poles uji

in Proposition 2 are reals it does not guarantee that the roots of the polynomial H(z) are reals. Therefore, the

statement of Proposition 2 is true only in the complex plane, as it is algebraically closed.

3. Polynomials associated with regular polygons

In this section, we will first characterize the polynomials associated with regular and affinely regular polygons. We will

show how such a characterization leads to a purely algebraic proof of the PetrDouglasNeumann Theorem, as well as the

NapoleonBarlotti Theorem.

3.1. Polynomials associated with k-regular polygons

We define a (n + 1)-gon P= (p0, p1, . . . , pn) to be a k-regular polygon if there exists a point, A, such that the rotationthrough the angle 2k /(n + 1) about A moves p0 to p1 , p1 to p2, . . . , pn to p0 . We denote by Rk the special k-regular(n + 1)-gon with vertices (1, k, 2k, . . . , nk) and is the complex number = e2i /n+1 .

Theorem 3. The polynomial O k(z) (1 k n) associated with the k-regular polygon Rk over a complex interval [a, b] is given by

O k(z) = z ka kn

where k =ak

b

k 1 . (11)


7/13


Fig. 3. Polynomial associated with a 2-regular 7-gon. The level one De Casteljau algorithm with respect to the pole 2 leads to the zero polynomial. (Seeproof of Theorem 3 for more explanations.)

Proof. Let O k(z) be the polynomial associated with the k-regular polygon Rk over the interval [a, b]. The shape of theoriented triangles Ti = (ik, (i+1)k, 0) are Ti = 1 k , for i = 0, . . . , n. The shape of the oriented triangle S = (a, b, k)is S = 1 k . Thus all the triangles Ti are similar and also similar to the triangle S . Therefore, the level one complexDe Casteljau algorithm associated with the polygon Rk and the pole u1 = k leads to the fact that the control pointsfO k (k, a

{ni1}, b{i}) = 0 for i = 0, . . . , n 1 (Fig. 3). Consequently, from the multi-affine property of the polar form, wehave fO k (k, z, z, . . . , z) 0. Therefore, by Proposition 1, the polynomial O k(z) can be expressed as O k(z) = (z k)n.Since O k(a) =1 (being the first control point), we have = 1/(a k)n . 2

Remark 3. Note that Theorem 3 can be proved in a more straightforward way by computing, using the polar form, the

control polygon of the polynomial O k(z) over the interval [a, b] and comparing the control points with the vertices of thek-regular (n + 1)-gon Rk . The main difference in the two proofs, is that in our given proof, we deduce the polynomialO k(z) from a purely geometric point of view, while in the suggested proof of this remark, we have to guess the polynomial

O k(z) before analytically showing that it is the right polynomial. Within this work, we always try to opt for the geometrical

method, as it give more insights into the problem.

Theorem 4. The set of polynomials B = (1, O 1(z) , . . . , O n(z)) form a basis of the linear space of polynomials of degree n.

Proof. We just need to show that the polynomials 1, O 1(z) , . . . , O n(z) are linearly independent. Let us assume that we are

given (n + 1) complex numbers 0, 1, . . . , n such that the polynomial Q(z) = 0 + 1 O 1(z) + +n O n(z) 0. The polarforms of the polynomials O k are given by

fO k

(u1

, u2

, . . . , un

)=

ni=1ui

k

a k .Therefore, we have

fQ(1, 2, . . . , n) = 0 = 0.Moreover, we have

fQ(1, 2, . . . , i1, a, i+1, . . . , n) = 0 + i

k=1,...,n, k =i

k ia i

= 0,

which implies that 0 = 1 = = n = 0. 2

Given an (n

+1)-gon P, we can write its associated polynomial in the basis B. This leads to a decomposition of the

initial polygon into linear combinations of regular (n + 1)-gons. Therefore, we have the following fundamental (Schoenberg,1950):


8/13


Corollary 2 (Fundamental Theorem of Decomposition of n-polygons). Every n-gon can be written as a linear combination of regular

n-gons.

Let P be a (n + 1)-gon (p0, p1, . . . , pn) with associated polynomial P(z) over an interval [a, b]. By the regular decompo-sition, we have

P(z) = 0 + 1 O 1(z) + + n O n(z).By considering the polar form on both sides of the last equation, we have

fP(1, 2, . . . , n) = 0.Moreover,

fP(1, . . . , i1, a, i+1, . . . , n) = 0 + i

k=1,...,n, k =i

k ia i

.

Since

k ia i

= k i

k 1 ,

we obtain

i = hi

fP(1, . . . , i1, a, i+1, . . . , n) 0

where

hi =n

k=1,k=i

k 1k i =

k=1,n,k =i(

k 1)(n1)i

k=1,n;k=i (ki 1)

. (12)

Noticing that we have

n

k=1, k=iki 1

=

i

k=1ki 1

n

k=i+1ki 1

=

n

j=2(ni)+ j 1

ni

j=1 j 1

,

and inserting the last equation into (12), leads to

hi =(i 1)

(n1)i(i 1) = i .

Therefore,

i =1

n

(a b)ii 1 fP (1, . . . , i1, i+1, . . . , n).

Moreover, let us calculate the control points pi as a function of the complex numbers i . From Eq. (1), we have

pk = fPa{nk}, b{k},

but since we have

fO j

a{nk}, b{k}= b j

a j

k= kj ,

we obtain

pk =n

j=0 j

jk =n

j=0 je

2 ijkn+1 .

Therefore, the vector (n + 1)(0, 1, . . . , n) is the discrete Fourier transform of the vector (p0, p1, . . . , pn).

Proposition 3. Let P(z) be a polynomial of degree n, with control polygon P

=(p0, p1, . . . , pn) over an interval

[a, b

]and letY

=(0, 1, . . . , n) be the coefficients of the decomposition of the polynomial P in the basis B. Then (n + 1)Y is the discrete Fouriertransform ofP.


9/13


From the inverse Fourier transform, we know that 0 is the centroid of the polygon (p0, p1, . . . , pn) i.e.; 0 = (p0 + p1 + + pn)/(n + 1). Let us give a geometric interpretation of the expression

fP(1, 2, . . . , n) = 0. (13)The oriented triangles (a, k, b) are isosceles since a k = (b a)/(k 1) and b k = (b a)k/(k 1). Moreover,we have (a k)/(b k) = k , thus the angle of the triangle (a, k, b) at the vertex k is 2k /(n + 1). Therefore, thegeometric interpretation of the complex De Casteljau algorithm for the expression (13), and the symmetry of the polar form,

leads naturally to the following part of the PetrDouglasNeumann Theorem:

Theorem 5 (A part of PetrDouglasNeumann Theorem). On the sides of (n + 1)- polygon P construct isosceles triangles with vertexangle 2 j1/(n + 1) where j1 in {1, , . . . , n}. The resulting vertices form a new n-polygon Pj1 . On the sides of this new polygon,construct isosceles triangles with vertex angle 2 j2/(n + 1), where j2 in {1, . . . , n}/{ j1}. We get a (n 1)-polygonPj1, j2 . Continuethis construction for all values 1, . . . , n. The final pointPj1, j2,..., jn is the centroid of the initial polygon.

Remark 4. Proposition 2 gives a more direct proof of the relation (13) and therefore of Theorem 5. In fact, we have

0 =1

n + 1n

i=0pi =

1

n + 1n

i=0fP

a{ni}, b{i}

.

Therefore, from Proposition 2, 0 = fP(1, 2, . . . , n), where i are the roots of the polynomial H(z) with

H(z) =n

i=0(z a)ni (z b)i .

Since H(a) = (a b)n = 0, the function R(z) = H(z)/(z a)n has the same zeros as H(z). The function R(z) can be rewrittenas

R(z) =n

i=0

z bz a

n,

in which the zeros are given by the complex numbers satisfying ( zbza )

n+1 = 1, i.e., the complex numbers i as given inEq. (11). This lead to the expression (13).

In the following we will show that the centroid of a closed polygon is preserved under any level one De Casteljau

subdivision.

Lemma 2. On the side of an (n + 1)-gon P = (p0, p1, . . . , pn) construct points qi , i = 0, . . . , n such that all the triangles(pi, qi , pi+1), i = 0, . . . , n are similar, where pn+1 = p0 . Then the centroid of the polygon Q = (q0, q1, . . . , qn) is the same as thecentroid of the polygon P.

Proof. Let P(z) (resp. H(z)) be the polynomial associated with the polygon P= (p0, p1, . . . , pn) (resp. H = (p0, p1, . . . ,pn, p0)). The polar forms of the polynomials P(z) and H(z) are related by

fP(u1, u2, . . . , un) = fH (a, u1, u2, . . . , un).

Each point qi can be expressed as fH (, a{n

i}, b{

i}) where is the unique complex number such that the oriented triangle

(a, , b) is similar to the oriented triangle (pi , qi , pi+1). From Eq. (13), the centroid of the polygon Q is given by

fH (, 1, 2, . . . , n),

where the i are defined in (11). The polygon T1 (resp. T2) obtained by taking the first vertex (resp. the last vertex) fromthe polygon H has the same centroid as the polygon P. In term of the polar form this can be expressed as

fH (a, 1, . . . , n) = fP(1, . . . , n) and fH (b, 1, . . . , n) = fP(1, . . . , n).By the multi-affinity of the polar form, the last equations mean that for every complex number we have

fH (, 1, 2, . . . , n) = fP(1, 2, . . . , n).In particular, this is true for the value of . 2

We are in a position now to prove the complete PetrDouglasNewmann Theorem (Fig. 4):


10/13


Fig. 4. The PetrDouglasNeumann Theorem for 4-gon. The red polygon is a square and the star point is the common centroid of the red, green and blue

polygons. (See proof of Theorem 6 for more detailed explanation.) (For interpretation of the references to color in this figure legend, the reader is referred

to the web version of this paper.)

Theorem 6 (PetrDouglasNeumann Theorem). On the sides of a closed polygon P construct isosceles triangles with vertex angle

2 j1/(n + 1) where j1 in {1, 2, . . . , n}. The resulting vertices form a new polygon Pj1 . On the sides of this new polygon constructisosceles triangles with vertex angle 2 j2/(n + 1), where j2 {1, . . . , n}/{ j1}. We get a polygon Pj1, j2 . Continue this constructionfor all values 1, . . . , n. Then

i) The final polygon Pj1, j2,..., jn is a point.

ii) The pointPj1, j2,..., jn is the common centroid of all polygons P, Pj1 , Pj1, j2 , . . . ,Pj1, j2,..., jn

1.

iii) The polygon Pj1, j2,..., jn1 is jn-regular.

Proof. Let us denote the (n + 1)-polygon P by (p0, p1, . . . , pn) and we denote by Q the 2(n + 1)-polygon (p0, p1, . . . , pn,p0, p1, . . . , pn). Let Q(z) be the polynomial associated with the polygon Q over an interval [a, b]. The polar forms of thepolynomials P(z) and Q(z) are related by

fP(u1, u2, . . . , un) = fQ

a{n+1}, u1, u2, . . . , un

.

By the Fundamental Theorem of the regular decomposition of the polynomial Q(z), we have

Q(z) = 0 +2n

+1

k=1k

z ka k

2n+1, where k = ak bk 1 and = e2i /2(n+1).


11/13


Since (0, 1, . . . , 2n+1) is the discrete Fourier transform of the 2-period sequence (p0, p1, . . . , pn, p0, p1, . . . , pn), we have1 = 3 = = 2k+1 = = 2n+1 = 0. Furthermore, direct inspection shows that 2k = k , where k is the complexnumber defined in (11). Therefore, the regular decomposition of the polynomial Q(z) can be rewritten as

Q(z) = 0 +n

k=12k

z ka k

2n+1. (14)

From Eq. (14), the polynomial associated with the polygon Pj1, j2,..., jn over the interval [a, b] is given byL j1, j2,..., jn (z) = fQ(1, 2, . . . , n, a, z, z, . . . , z) = 0.

Therefore, Pj1, j2,..., jn is a point, which is also the centroid of the polygon P (since the two polygons P and Q have the

same centroid). That completes the proof of part i) of the Theorem.

ii) The second part of the Theorem is just an iteration of Lemma 2.

Finally, to prove the third statement of the Theorem, we can just note that the polynomial associated with the polygon

Pj1, j2,..., jn1 is given by

L j1, j2,..., jn1 (z) = fQ( j1 , . . . , jn1 , a, a, z, . . . , z) = 0 + K

z jna jn

n,

which is just a scale, rotation and translation of the polygon R jn and therefore jn-regular. 2

3.2. Polynomials associated to k-affinely regular n-gons

Definition 5. A (n + 1)-gon is said to be k-affinely regular if it is the affine image of a k-regular (n + 1)-gon.

As we have done for regular polygons, in the following we will characterize the form of the polynomials associated with

affinely regular polygons. Let P= (p0, p1, . . . , pn) be an (n + 1)-gon, normalized by a translation such that its centroid islocated at the origin. It is a simple geometric fact (Fisher et al., 1985) that the polygon P is k-affinely regular if and only if

the vertices ofP satisfies the following relation

pi+2 + pi2

= cos

2k

n + 1

pi+1, i = 0, . . . , n, with pn+1 = p0 and pn+2 = p1. (15)

In fact, the last relation can be easily checked for k-regular polygons, and being an affine relationship, it then extend tok-affinely regular polygons. The following shows that this last relation allows us to characterize the set of polynomials

associated with affinely regular polygons:

Theorem 7. The polynomial Ak(z) associated to a k-affinely regular(n + 1)-gon Pwith centroid at the origin, over an interval [a, b]is of the form

Ak(z) = 1(z k)n + 2(z n+1k)n (16)where k are defined in (11).

Proof. Let P(z) be the polynomial of degree n associated with the k-affinely regular polygon P= (p0, p1, . . . , pn) anddenote by Q(z) the polynomial of degree (n + 2) associated with the control polygon (q0, . . . , qn+2) = (p0, . . . , pn, p0, p1).Since P is k-affinely regular, we have

q j+2 2cos

2k

n + 1

q j+1 + q j = 0, for all j = 0, . . . , n. (17)

In term of the polar form, relation (17) can be rewritten as

fQ

a{n j}, b{ j+2} 2cos 2k

n + 1

fQ

a{n j+1}, b{ j+1}+ fQa{n j+2}, b{ j}= 0.

Since

(z b)2 2cos

2k

n + 1

(z a)(z b) + (z a)2 =

2 2cos

2k

n + 1

(z k)(z n+1k),

by Proposition 2, we have for every i in

{1, . . . , n

}fQ

a{ni}, b{i}, k, n+1k= 0. (18)


12/13


Fig. 5. The NapoleonBarlotti Theorem for triangles. Note that every triangle is an affinely regular polygon. Therefore, the black triangle is equilateral. (See

proof of Theorem 8 for more detailed explanation.) (For interpretation of the references to color in this figure legend, the reader is referred to the web

version of this paper.)

From the multi-affinity of the polar form, relation (18) implies that for every complex number z, we have

fQ(z, z, . . . , z, a, a, k, n+1k) = fP(z, z, . . . , z, k, n+1k) 0.Therefore from Corollary 1, there exist complex numbers 1, 2 such that

P(z) = 1(z k)n + 2(z n+1k)n.Moreover, the control polygon of a polynomial of the form (16) satisfy relation (15) as the control polygon is the sum of

two regular polygons, each of them satisfy (15). 2

Theorem 8 (NapoleonBarlotti). For a given (n + 1)-gon P= (p0, p1, . . . , pn), letB1 (resp. B2) be the (n + 1)-gon whose verticesare the centroids of k-regular (n + 1)-gons erected externally (resp. internally) on the sides ofP. Then B1 is k-regular (resp. B2 is(n k)-regular) if and only ifP is k-affinely regular(Fig. 5).

Proof. Let P(z) (resp. Q(z)) be the polynomial associated with the (n

+1)-gon P

=(p0, p1, . . . , pn) (resp. the (n

+2)-gon

Q= (p0, p1, . . . , pn, p0) over the interval [a, b] = [k, 1]. Note that in this case n+1k = 0 and k = 1 + k . The centroidof the k-regular (n + 1)-gon erected externally on the side [pi , pi+1] has coordinates ri = fQ(a{ni}, b{i}, 0). Therefore thepolynomial associated with the polygon B1 = (r0, r1, . . . , rn) over the same interval [a, b] is

R(z) = fQ(z, z, . . . , z, 0).If the polygon P is k-affinely regular (without loss of generality, we can assume that the centroid of the polygon P is at

the origin), then by Theorem 7, the polynomial Pd(z) associated with the control polygon (p0, p1, . . . , pn, p0, p1, . . . , pn) is

given by Pd(z) = 1(z k)2n+1 + 2z2n+1 . As

Q(z) = fPd

k{n}

, z{n+1}

,

we have Q(z) = 1(1)n(z k)n+1 + 2kn zn+1. Consequently,R(z)

=fQ(z, z, . . . , z, 0)

= k(

1)n1(z

k)

n,

which shows that the polygon B1 is k-regular.


13/13


Conversely, ifB1 is k-regular, then there exist complex numbers and C such that R(z) = (z k)n + C. Consider thepolynomial

H(z) = Q(z) + k

(z k)n+1 C. (19)

We have

fH (z, z, . . . , z, 0) = fQ(z, z, . . . , z, 0) (z k)n

C 0.Therefore, by Proposition 1, the polynomial H(z) can be written as H(z) = zn+1 , which shows from Eq. (19) that thepolynomial fQ (a, z, z, . . . , z) = P(z) can be written as

P(z) = k

(a k)(z k)n + azn + C.

Therefore, the polygon P= (p0, p1, . . . , pn) is k-affinely regular.The proof for the case of the centroids of the internally erected sides polygon B2 proceed with the same lines as in

the previous case of externally erected sides, in which this time we take the interval [a, b] to be [n+1k, 1] instead of[k, 1]. 2

4. Conclusions

In this paper, we showed how the polar form of a complex polynomial can be instrumental in revealing the process

in which linear transformations on polygons can lead to regular structures. Some interesting expressions for the discrete

Fourier transform of the control polygon in terms of the polar form have been obtained. One of the recurrent themes in

this work is the similitude between triangles for the implementation of the complex De Casteljau algorithm. For the case

of rational Bzier curves, the triangle similitude is replaced by the invariance of the cross-ratio. Therefore, we suspect that

generalizing the concepts of the paper to rational Bzier curves can lead to far-reaching generalizations of the PetrDouglas

Neumann Theorem as well as the NapoleonBarlotti Theorem.

Acknowledgements

The authors are sincerely grateful to the anonymous referees whose important suggestions allowed us to substantially

improve the quality of this work. The authors want also to thank the anonymous referees for finding a shorter proof to

Proposition 1 than the one initially suggested.

Grant: This work was supported in part by the MEXT Global COE Program at Osaka University, Japan.

References

Barlotti, A., 1955. Una proprieta degli n-agoni che si ottengono transformando in una anita un n-agono regolare. Boll. Un. Mat. Ital. 10, 9698.

Connes, A., 1998. A new proof of Morleys Theorem. In: Les relations entre les mathmatiques et la physique thorique. IHES Publications, pp. 4346.

Douglas, J., 1940. Geometry of polygons in the complex plane. J. Math. Phys. 19, 93130.

Fisher, J.C., Ruoff, D., Shiletto, J., 1985. Perpendicular polygons. Amer. Math. Monthly 92 (1), 2337.

Goldman, R.N., 1994. Dual polynomial bases. J. Approx. Theory 79, 311346.

Gray, S.B., 2002. Generalizing the PetrDouglasNeumann Theorem on n-gons. Amer. Math. Monthly 110, 210226.

Lester, J.A., 1996. Triangles I: Shapes. Aequation Math. 52, 3054.

Marden, M., 1966. Geometry of Polynomials. Amer. Math. Soc., Providence.

Pech, P., 2001. The harmonic analysis of polygons and Napoleons theorem. J. Geom. Graph. 5 (1), 1322.

Ramshaw, L., 1989. Blossoms are polar forms. Comp. Aided Geom. Design 6 (4), 323358.

Schoenberg, I.J., 1950. The finite Fourier series and elementary geometry. Amer. Math. Monthly 57, 390404.

Date post:	08-Apr-2018
Category:	Documents
Upload:	jeremy-brown
View:	237 times
Download:	0 times

eSource-Hadou-Complex Bézier curves

Documents