J. geom. 63 (1998) 97 - 108

0047-2468/98/020097-12 $1.50+0.20/0

[ Journal of Geometry

9 Birkh~iuser Verlag, Basel, 1998

ON PROPORTION POLYNOMIALS Martin Hintz

The proportion polynomials of maximum degree three and the proportion polynomials of class C ~ of maximum degree five are determined. As an important tool, the class of the weak proportion polynomials (which includes the class of the proportion polynomials) is introduced and examined.

1. INTRODUCTION Let D : = { ( x , y ) ~ R 2 x , y > O } a n d / : = [ 1 , oo[= {t ~ R It__ 1}, where R denotes the field of real numbers. If f : D ~ I is a function satisfying

(y2y] f(x, y) = f ~-, , f(x, y) = f(y, x), and

f(x, x) = 1 for all positive x,y, then f is called a proportion function. These functions, which were introduced by C. Alsina, E. Trillas [2] and C. Alsina [1], are important in architecture, where one likes to compare rectangles (of the real euclidean plane) from an aesthetic point of view. In this context, two rectangles are considered aesthetically equal if they are similar to each other and have at least one side length in common. (A rectangle &side lengths x,y > 0 is called similar to a rectangle of side lengths x',y' > 0 if there exists a real number c > 0 such'that (x,y) = (cx',cy') or (x,y)=(cy',cx').) Thus, if one assigns to every rectangle R a real number g(R) such that two rectangles R, S fulfil g(R) = g(S) whenever they are aesthetically equal (in the above sense), then one can define a function f : D --+ R by putting lengths

f(x,y):= g(R) for any rectangle R of side

(Y2y1

x,y; further, this function f satisfies f ( x , y ) = f --~-,

x,y. If, in addition, g(S) = 1 for every square S and

and

f(x,y)=f(y,x) for all

g(R) _>1 for every rectangle R (so that

98

Hintz

g(S) is minimum for all squares S ), then f is a proportion function. For further information on

the importance of these functions, see [2,3,4]. An example of a proportion function is the classimax{x,y} cal proportion, defined by f ( x , y ) . - min{x,y} " The proportion functions were characterized by W. Benz [5] and Z. Moszner [7]. Putting G : = { ( x , y ) E R 2 ]0
0
state the result as follows (for a proof, see [3] or [4]). THEOREM 1. Let or: G --+ I be an arbitrary function. Then there exists one and only one proportion fimction f = f(~r) such that the restriction flo of f on G is or. The following two theorems (which are proved in [3,4], too) characterize the continuous and the continuously differentiable proportion functions. Put H : = {(x, y) ~ R210 < x < 1,0 < y _
,(1,1) = 1,

(2)

q~(t, t2)=9(1, t)forall t ~ ] O , l [ = { s e R l O < s < l } ,

(3)

~o is continuous in H.

(b) Suppose that q~:H ~ I is an arbitrary function satisfying (1), (2), (3). Then the proportion function f = f(el o) is continuous and fl,, = ~"

THEOREM 3. (a) Suppose that the proportion fimction f: D --+ I has continuous partial derivatives fx,fy in D. Define ~o:= flH" Then the following properties hoM true: (1), (2), (3) (see Theorem 2) and

(4)

(p,, (py exist and are continuous in H 1 := H\{(1,1)},

(5)

the limitsof ~~ x-y

l, (ox(x,y), (a,(x,y), and x(~

x-y

yfay(x'y) exist for

(x,y) --+ (1,1) with (x,y) ~ H 1 and they are all O,

(6)

~o.(t,t =)= }q~x(l,t)+~o.(l,t)and 9.(t,t2)= -~q~x(l,t)forallte]O,l[,

(b) Suppose that q~:H-+ I is an arbitrary function satisfying properties (1), (2), (3), (4), (5), (6). Then f = f(qglo ) has continuouspartial derivatives fx,fy in D and flH = q~.

Hintz

99

We briefly fix some basic notation. By R[T] and R[X,Y], respectively, we denote the ring of polynomials in one and in two indeterminates over R , respectively. Throughout the paper, the (capital) letters

T,X,Y

denote indeterminates. We write P' for the derivative of P e R[T] and

fox,for for the partial derivatives of fo ~R[X,Y]. The degree of g

eR[T]wR[X,Y]

is denoted

by deg p' ; we put deg0 := -1. The closure and the boundary of a set A c_ R 2 with respect to the natural topology of R 2 are denoted by A and 8A, respectively.

fo(x,y)>_ l for all (x,y)~fo(x,y) defines

Assume that fo ~R[X,Y] satisfies

( x , y ) s G . Then, according to Theorem 1,

the function fo[o'G ~ R ,

a proportion function f(folG)=: f(fo), ff

k > 0 is an integer such that, for i = 0 ..... k, all partial derivatives of f(fo) of order i exist and are continuous in D, then we say that fo is a proportion polynomial of class C k ; a proportion polynomial of class C o is just called a proportion polynomial (compare [4]). This definition is fundamental for the present paper. Since H c G , Theorem 2 implies that a polynomial

fo ER[X,Y]

is a proportion polynomial if and only if

(x,y) eH,

(7)

fo(x,y)_> 1 for all

(8)

fo(1,1) = 1, and

(9)

fo(t, t2)=fo(1, t) for all

t ~10,1[.

Condition (7) is much more difficult to handle than (8) and (9). We call a polynomial fo sR[X,Y] satisfying (8) and (9) a weak proportion polynomial. In the following section 2, we determine all weak proportion polynomials of any given maximum degree N _>1. Afterwards, in the sections 3 and 4, we apply the result of section 2 to characterize the proportion polynomials of maximum degree 3 and the proportion polynomials of class C I of maximum degree 5.

2. WEAK PROPORTION POLYNOMIALS

For any integer N >- 1, let VN be the set of all polynomials ~z ~R[X,Y] that satisfy deg~z < N , ~(1,1) = 0, and

r

= ~ 0 , r ) , where r is an indeterminate. Moreover, let Vs be the set of

all ~z ~VN satisfying

fz(x,y) >_0 for all (x,y) ell.

Clearly, a given polynomial fo eR[X,Y] is a

weak proportion polynomial of maximum degree N if and only if fo - 1 e VN ; and fo is a proportion polynomial of maximum degree N if and only if fo - 1 eVs Therefore, every characterization of the set VN (V~) immediately leads to a characterization of the set of weak proportion polynomials (proportion polynomials) of maximum degree N . The following proposition can be shown without difficulty.

100

Hintz

PROPOSITION 1. For any N > 1, V~r is a subspace of the vector space R[X,Y] over R . The set VN is a convex and closed subset of VN . (Note that VN, being a real vector space of finite dimension, can be viewed as a topological space in a natural way.) If ~t,p ~V~r then ~ + p e V N. I f ~ ~V~v and 2 ~ R with 2 > 0 , then 2 ~ ~V[r I f ~r ~V[r and 2 e R ~ 0 and s < O, then Zgt ~ V[~.

with

It is our next purpose to construct, for any N _>1, a basis & t h e vector space VN . We need some notation. If (ao,...,a,,) is a finite sequence with m>_-I and ao,...,a m ~{0,1}, then we put k

a [ k ] : = ~ a i 2 k-i for k = - l , . . . , m . (Note that, if m = - l ,

then (ao,...,a_l)=() is the "empty

i=0

sequence".) Clearly, for any integer n > 1, there exists exactly one finite sequence (% ..... % ) such that m > O, a o..... a m e {0,1}, a o = 1, and a[m] = n ; we put bin(n):= (a o..... am) and call bin(n) the binary representation of n. Further, we put bin(0):=(). We call a pair (n,j) of integers admissible if n _>2 and 0 < j <_n - 2. For every admissible pair (n,j), we define a polynomial V,j e R [ X , Y ] as follows', let (z o.... ,Zm):= bin(n+ j) and (Wo,...,wv):= bin(j), and put m

(10)

p

V,j(X,Y):=X"-JYJ-~X~Y~[k-q+EX~YW[k-q+(m-p-1). k=O

k=O

Note that, for given n, j , this polynomial can easily be calculated. For example, we have N2o(X,Y)=X2-X-Y+I,

~u3o(X,Y)= X 3 - X Y - X + I ,

and g 3 1 ( X , Y ) = X Z y - y 2 - y + l .

T H E O R E M 4. For any integer N >_1, the family of the polynomials ~,] with 2 < n < N and 0< j < n - 2 is a basis of VN . Proof. We use induction on N . Since it can easily be shown that V~ = {0} and since, by convention, the empty family is a basis & t h e vector space {0}, the assertion holds for N = 1. Now let N -> 2 and assume that the assertion holds for N - 1. Let (n,j) be an admissible pair and put (z 0.... ,zm) := bin(n+ j) and (wo..... wp) := bin(j). Then (10)leads to (11)

~,j(1,1) = 0,

(12)

[ifnj(T,r2)= Tn+J-ZTz'+2z[k-1] + ~P' T "~+2~[~-q+ ( m - p -

m

k=O

(13)

1), and

k=O

gt y(1,T) = T j - ~ T "[k-'] + ~ T "[k-q + ( m - p - 1 ) . k=0

Observing that z k +2z[k-1]=z[k]

k=O

for k = O , . . . , m ,

wk +2w[k-ll=w[k]

~[m] = . + jo w[p] = j , and ~[-1] = 0 = w[-1], we conclude that

for k = 0 ..... p ,

Hintz

101

(14)

,(1,rt

Moreover, since j < n - 2, it can easily be verified that z k + z[k- 1] = 89 k + z[k]) < 89 + aim])=

= 89189

for k = 0 ..... m and w , + w [ k - 1 ] < l + w [ P l = l + j < n

for

k = 0 ..... p . This implies (15)

d e g ( < , ( X , V ) - X"-JV ' ) < n and d e g < ~ = n.

From (11), (14), and (15) we conclude that g/,,j e VN holds for all admissible (n,j) with n < N . In order to prove that the V , with n < N are linearly independent, let real numbers a , with N n-2

(16)

ZZa.,v,(X.Y)

=0

n=2 j=0 N-2

be given. Subtraction of ~ c ~ . X N - J Y j from both sides of(16) leads to i=0 N-1 n-2

(17)

H-2

N-2

~ , Z a , i v , ( X , Y ) + ~ c ~ ( V ~ . ( X , Y ) - XN-sYO=-7~ c~.XN-iYJ , n=2 j=O

j=0

j=0

where, by (15), the polynomial on the left hand side has a degree less than N . Therefore, (17) implies aNj = 0 for j = 0 ..... N - 2. But now, by induction, we obtain from (16) that a , = 0 for all (n, j ) , showing that the V , with n _
Let V ~VN be given; we can write v ( X , Y ) = ~ ~ f l l , j X i Y j with fli,j ~ R . Since V ~VN, we i=0 j=o

have v(T, T 2) = V(1, T). Comparing the coefficients of T TM and T 2N-1 in the polynomials v ( T , T 2) and g/(1, T) and making use of N _>2, we obtain flo,u = fl,.N-I = 0. By (15), this iroN-2 plies that p := V - ~flN-j.iVNj is a polynomial of degree less than N . Moreover, we have j=0

/9 ~VN since p is a linear combination of elements of VN. Now degp < N impfies p ~VN_~ and, by induction, we conclude that p is a linear combination of the polynomials V , with n < N - 1. Therefore, V is a linear combination of the ~ , with n < N .

[]

C O R O L L A R Y 1. For any integer N >_1, the dimension of VN is ~ N(N - 1). C O R O L L A R Y 2. The Onfinite) family of all V , with n > 2 and 0 < j < n - 2 is a basis of the

vector space ofallpolynomials V ~R[X,Y] that satisfy V(1,1) = 0 and

=

r).

102

Hintz

C O R O L L A R Y 3. For any N > 1, the weak proportion polynomials of maximum degree N w'e AT n-2

precisely all polynomials of the form 1 + ~ ~" ~z j~g,j with a,j ~ R for all (n,j).

n=2j=0

By Theorem 4, we have determined the sets VN . It is not so easy to obtain similar results for the sets V~, but the following theorem can be regarded as a first step towards this purpose. T H E O R E M 5. For any integer N >_1 and any admissible pair (n, j) with n <_N , the polyno-

mial ~,~ is an element of Vs Proof. In view of Theorem 4, it remains to show that ~,g(x,y) _>0 for all (x,y) ~ H . Choose and O
the notations as in formula (10) and let (x,y) s H . Then x > 0 from

j
that

y"['-q-J_
n+j>2j+2

and,

therefore,

Observing 2z[m-1]=n+j-Zm,

m>p+l,

z[m-1]-j>O,

and

we obtain x~'y z[~"-q = x~'yiy ~[~-~]-j <_

<_x ~ y J x 2(~[~-q-~) = x"-Jy j , that is, (18) Let

=

x"-Jy j - x ~ ' y 4"-q >_0. k s { - 1 ..... p}.

n+j-

i=k+m-p

n+j>_2j,

Since

ff

-

,=k+l

>

we 2j-

t

2"-kz[k+m-p-1]-2"-~w[k]=

have

2 "-i - j = i=k+m-v

1 - 2 p-k+1 > - 2 p-~.

J

Hence z[k + m - p - 1]- w[ k ] > - 1 , which implies (19)

z[k+m-p-1]-w[k]>_Ofor k:-l,...,p.

Now let k ~ {0..... p}. From (19) it follows that ze+m_v_l + 2(z[k + m - p - 2 ] - w[k - 1]) > w k and, therefore, x ..... "-lX2(Z[k+~'-P-q-~[k-q)_
< x ~ y w[k-q , which yields (20)

x ~ y w[k-ll-x"~*'-'-~y[k+'-p-~] __0 for k = 0,...,p.

Since m _>p + 1, the terms of (10) can be rearranged in the following way: ~nJ(X'Y):(

xn-jyj

-xZ, y z[m-q) + s

k=0

x'~y w[e-q-x~*+~-o-'y"[e+~-p-2l)

-

m~2 x'~y "[t:<]+ (m "P--

1).

k=O

Now (18) and (20), together with x ' y ~[k-q _<1 for k = 0 .... , m - p - 2 ,

imply g%j(x,y) > O.

[]

Note that it follows from Theorem 4 and Theorem 5 that, for any N _>1, no proper subspace of VN contains the set V~. Further, in view &Proposition 1, we obtain the following corollaries.

Hintz

103

C O R O L L A R Y 4. For any N > 1 and any choice o f a.j ~ R such that ct.j > 0 for all (n,j), N n-2

the polynomial 1 + ~ ~ a.j r .j is a proportion polynomial of degree at most N . n=2 j=O

COROLLARY 5. a* : {~,~,2o I'~ ~R,~_~ o}

3. P R O P O R T I O N P O L Y N O M I A L S OF M A X I M U M DEGREE TI]REE In this section, we characterize the set V3§ . By Theorem 4, the vector space V3 consists of all

polynomials of the form 0:gt20 +,8~30 +Y gt31 with a,,8,y ~ R. For the proof of the next theorem we need the following well-known facts (see, for example, [6] or any other book containing "Cardan's formula"): A cubic polynomial g(T) = T 3 + b T 2 + c T + d with b,c,d ~ R has three distinct real roots if and only if q2+p3 < 0 , where p , q are defined by 9 p = 3 c - b 2 and 54q = 2b 3 - 9 b c + 27d. Further, a polynomial g(T)= T 2 + cT + d with c,d ~ R has two distinct

real roots if and only if 4 d - c 2 < 0 .

From this one easily derives that a polynomial

g(T) = a T 3 +bT 2 + c r + d with a,b,c,d E R and 3 := degg ~{2,3} has 6 distinct real roots if

and only if di scr(g) < O, where discr(g):= 27a 2d 2 + 4b 3d - 18abcd + 4ac 3 - b 2c 2 . 6. Let ~ ~V3 and define ~z,,8,y ~ R by ~t = 0:V2o +,8~3o +Y~r Further, put A := 3a 4 + 16o~3,8+ 50a2,82 + 60o~,83 + 23,84 + 4a3},"+ 180:2,87, + 720:fl27" + 54,83y + 27,8272 .

THEOREM

Then gt belongs to the set V~ if and only if the following four conditions hold:

(21)

a+,8+r>__O,

(22)

a + fl + 2y > 0,

(23)

k>__O or max{a+,8,a+2,8}<_O,

(24)

y ;a O or ~z + ,8 ;a O or ,8 <_O.

Proof. Define P ( T ) : = 7/(T, 0) and Q(T):= r

T), where T is an indeterminate. Note that, by

definition of V3, we have Q(T) = gt(T, TZ), Q(1) = o, and P(0) = P(1)= Q(o). Using the explicit formulas for ~2o, ~t3o, ~31 given after (10), we obtain

(25)

~,(x,y) = , s x 3 + r X 2 Z + ~ , x 2 - , 8 ~ -

(26)

P(r) -- ,8r3 + 0:T~ _(~ +,8)r +(~ +,8 + r),

(27)

0(~) = -r ~ 2 - ( ~ +,8)r + (~ +,8 +r),

(28)

P(O) = P(1)= Q(O) = 0:+,8 + y ,

z z 2 - (~, + , 8 ) x - (0: + r)Z+(~ +,8 + y),

104

Hintz

(29)

O'(1) = -(ct + f l + 2 y ) ,

(30)

P'(O) = -(c~ +/3), P'(1) = a + 2,8, and discr(P) = k .

(x,y) el_if, that is, ~'(x,y)>O for all ( x , y ) e R 2 with 0 < x < l and OO for all t e[o,1]. Since Q(1) = o, we obtain from (28) and (29) that (21) and (22) hold. Further, we have P(t) > 0 for all t ~[0,11. Taking into account that 6:=degP-<3 and P(O)=P(1), we conclude that, if

For the "only if' part of the proof, assume p' eV3+ . Then p'(x,y) > 0 for all

P'(0)<0 or P'(1)>0, then 6e{2,3} and P has at most 6 - 1 distinct real roots. Hence max {-P'(0), P'(1)} < 0 or discr(P)> 0, and now (30)implies (23). Condition (24) follows from 0 <- P(~) = 89(5,8 + 6a) + y . This completes the "only if' part of the proof. For the "if' part, assume that (21), (22), (23), (24) hold. From (21), (22), (28), and (29) we obtain Q(0) _>0 and Q'(1) < o. Since Q(1) = 0 and deg Q <_2, we conclude that Q(t) >_0 for all t e [0,1]. Note that this implies (31)

~,,(1,t)>-Oand Tt(t,tz)>-O forall t ~[O,1].

We claim that (32)

P(t) > 0

for all t ~[0,1].

If P(0)= P(1)> 0, this can easily be derived from (23) and (30) together with the fact that degP _<3. Now assume P(0) = P(1) _0 or P'(1)=0 and P'(0)<0. In the case 0=P'(0)=-(~z+,8) and 0 < P'(1) = a + 2,8 we obtain ,8 > 0, and further y = 0 since a +,8 + y = 0. This contradicts (24). In the case 0=P'(1) and 0 > e ' ( 0 ) = - ( . + , 8 ) we conclude from that ?" = - ( a +,8) and, therefore, a +,8 + 2r = - ( a +,8) < 0, contradicting (22). Hence we have proved (32). From (31) and (32) we obtain (33)

V(x,y) > 0 for all

Let (.g,.9) ~ r

(x,y)eSH.

be a point where the continuous function g has its global minimum on the com-

pact set H . It suffices to prove that V(g,.v) _>0. Suppose and (25), the polynomial t s[0,,2], and therefore sume (34)

y < 0.

g(~,y)

< 0. If y > 0, then, by (33)

h(T):=W(2,T ) satisfies h(0)_> 0, h(22)> 0, h"(t)=-2y <0 for all h(t)>O for all t s[0,,~2], contradicting h(y)< 0. Hence we may as-

Hintz

105

By (33), we have 0 < ~ < 1, 0 < 7 < 22, and hence 0 = ~ r ( g , y ) = - 2 r Y + 7 ~ 2 - P ~ - ( a + r ) . Defining R(T):= ~(T 2 - p y < T - ( a y - l + (35)

1)) and S(T) := ~/(T,R(T)), we see that ~ = R(~) and

S has a local minimum at ~.

From (22) and (34) one can derive that R(1) _>1; find since R(~) < ~2, we conclude that there exists a z s l t

suchthat R ( t ) < t 2 for Y~<_t
(36)

g
and

(37)

R(z)=z 2 .

Since R(T) - T 2 is a polynomial of degree 2 with a negative leading coefficient, it follows that

R(t) < t 2 for all t ~ R with t
a+y <0.

Note that 2y R(T) = 7"T2 - f i T - (a + y), which in view of (25) implies (39)

vr(T,R(T))=O.

By applying the chain rule to the definition of S and using (39) and (25), we obtain

(40)

3+

pr

(41)

S"(T) = 37T 2 + 3 f l T + ( a -Z" +17pz02.-I~],and

(42)

S'"(T)=67T+3/~.

Differentiation of ~,(r,r:)=~,(a,T) yields (40), (37), (39), and (25), we have

~,~(r,r~)+2r~,~(r,r2)=~,,(1,r).

Hence, by

s'(4--~,x(~,R(z))=~,.(z,~)=~,,(1,~)-2~@,~:)=

=~,~(L~)-2~,~(z,R(z))=~,,(1,z)=-2rz-(a+p).

From (34) and (36) we obtain

s'(~) _<-at. 1- (= +/~), and now (22) leads to (43)

S'(z) < O.

From (37) and the definition of R it follows that 7z 2 = - / 3 z - (co + 7). Therefore, (41) implies (44)

S"(z) = -2~z - 4y + 89

-1 .

Note that (40) and (34) show that S' is a polynomial of degree 3 with a negative leading coefficient. Because of(35), it follows that S' must have three distinct real roots. By means of a simple discussion of the relative positions of the roots of S',S", S " on the real line, we conclude from (36) and (43) that S"(z) < 0 and S'"(1) < 0. Hence (44) and (42) lead to

106

I-Iintz

(45)

-2a-47+ 89

(46)

67 + 3,l? < 0.

-~ <0 and

Formulas (22), (34), and (38) imply f l > - - ( a + 7 ) - ? ' > 0 .

Hence we obtain from (46) that

0 < fl < - 2 7 and, therefore, ]3z < 47 ~ . This, together with (45), (34), and (38), leads to the contradiction 0 > -2~z - 47 + 89

> - 2 a - 47 + 27 = -2 (a + y) > 0, which completes the proof. []

4. P R O P O R T I O N P O L Y N O M I A L S OF CLASS C 1 OF M A X I M U M DEGREE FIVE In the following, we use the explicit formulas for the polynomials ~zo, ~3o, ~/3~ mentioned after

(t0); further, we need the following polynomials, which can easily be calculated:

p',o(X,Y)= X ' - X y 2 - X - Y + 2 , r

p',I(X,Y): X 4 y - Y s - X Y + I ,

(X,F) : X3Y 2 - X Y 3 - X Y + Y , and N,3(X,Y) =XaY s _y4 +Xy_y2 - F + l .

We define (47)

~ := 3 7e2o- P'3o - 10p'31 +5p'4o +8N41 +2N42 - 4 ~50 - N51,

which more explicitly means (48)

~ X , ~ = - 4 X s - X 4 y + 5 x 4 +8X3Y+2X2Y2 - X 3- 10X2Y-4XY: - 2 ,3 +3X 2 +5Y 2 - 3 X + l .

T I t E O R E M 7. The proportion polynomials of class C ~ of maximum degree 5 are precisely the

polynomials of the form 1+ a ~ with a ~ R, a >-O. Proof. Let q7 be an arbitrary proportion polynomial of class C t of maximum degree 5. By CorNlary 3, there exist a2o,..., ~zs3 s R such that 5

(49)

n-2

(o(X,Y)=I+ZZc%.~,j(X,Y). n=2 J=0

Since cp is continuous in H , it follows from Theorem 3 (a) that c,o satisfies (5) and (6). By applying the second parts of (5) and (6) and taking into account that cp is a polynomial, we obtain cPx(1,1) = 0 and T2~or(T, T2)+Ox(1,T) = 0, where T is an indeterminate. By (49), this yields (50)

ZZ~zj n=2 j = 0

gtj x(1,1)=O and Z Z ~ z . , n=2 j = 0

T gt.j r T, T + ~/.j x (,

=0.

Hintz

107

Using the explicit formulas for the polynomials V,~ given above, we can find the values (V,i)x(1,1) appearing in the first part of (50) and, moreover, we can calculate the coefficients of

T~ T 1..... T s in the second part of (50), thus transforming (50) into the following ten equations for the a ,s : ~2o +~3o +20:31 +3a4o +20:41 -I- ~42 +30:50 + 30:5~ -~"0:52 + 36t53 = O, 0:20 + 20:30 + 3~4o + 40:50 = 0, -a3o + 2a3~ + 3~41 -0~42 + 3~51 -0:52 + 0:53 = 0, --0:20 -- ~31 -- 0:40 -- 20:41 + 30:42 -- 20:50 + 40:52 -- 0~53= 0, --~30 -- O:42 -- 0:51 -- 2~52 + 30:53 = 0, --0:3~ -- 2~Z4O-- 20:53 = 0, --0:4I -- 20:50 = 0, --0:42 -- 2C~51 = 0, --0:52 = 0, --0:53 = 0. By solving this system of linear equations and comparing with (49) and (47) we obtain (p = 1 + m,~, where 0: := -0:51. Moreover, we have (p(0, 0) _ 1 since (p is a proportion polynomial; this, together with (48), implies 0: _>0. Thus we have shown that every proportion polynomial of class C ~ of maximum degree 5 is of the form 1 + c@ with ~z _>0. For the proof of the converse, let 0: _>0 be given and put (p : = 1 + ~z~. From (48) one obtains

~/(x,y)=(x2-y)2(5-4x-y)+(1-x)3>_O

for x , y < l

and, therefore, (p(x,y)_>l for all

(x,y) ~H. Hence (Pin maps H into I . By Theorem 3 (b), it sumces to show that (P satisfies the conditions (1)-(6). Conditions (1), (2) immediately follow from Corollary 3, while (3), (4) are trivial. For the proof of(5) and (6), note first (51)

~'x(X,Y)=4X(X2-Y)(5-4X-Y)-4(X2-Y)2-3(1-X)

(52)

~r(X,Y)=-2(X2-Y)(S-4X-Y)-(X2-Y)

2 and

2.

Now condition (6) can easily be verified; further, it is clear that ~x(X,y)--+ q~x(1,1)= 0 and (Pr (x,y) --+ (Pr(l,l) = 0 for (x,y)--~ (1,1). It remains to show that (53)

( P ( x ' y ) - I - - + 0 and X(~176 x-y x-y

where H I : H \ { ( 1 , 1 ) } .

~0 for (x, y) ---~(1,1) with (x,y)~H1,

It is easy to verify that

~(X,Y):(X-Y)Q(X,Y)+R(X)

Q(X,Y):= X 4 - 1 0 X 3 - 2 X 2 Y + 1 5 X 2 + 5 X Y + Y 2 - 5 X - 5 Y For all (x,y)~H 1 with

x
we have

y<_x z < x

= (5x 2 + ~ ( 1 - x)2. Since R(1)=0, we thus have

and

with

R(X):=(5X2+I)(1-X) 3.

and therefore

R(x) < R(x) = x-y - x-x 2

R(x) <_- ( 5 x 2 + l-) ( 1 - x ) 2 =: S(x) for all x-y x

108

Hintz

(x,y) ~H~. Now let ( x , y ) - + (1,1) be a sequence with (x,y)~H1, where the indices of the quence are omitted. Since S(x)-+ S(1) = 0, we obtain

R(x)

se-

-+ 0. Further, it is clear that

x-f

Q(x,y)-+Q(1,1) = 0. Hence ~ p ( x , y ) - l _ a Q ( x , y ) + a R(x) @ 0, which shows the first part of x -y x -y

(53).

The proof of the second part is very similar: observe first that X~Yx(X,Y ) +Y~'r(X,Y)

= - 2 0 X s - 5 X 4 y + 2 0 X "4 +32X3Y+8XaY 2 - 3 X 3 - 3 0 X Z Y - 12XY 2 - 3 2 ,3 + 6 X 2 +10Y 2 - 3 X . It can be verified that X ~ x ( X , Y ) + Y ~ r ( X , Y ) = ( X - Y ) Q I ( X , Y ) + R z ( X ) with Q,(X,Y):=

and Like above, we see

Rl(x) x-y

_
(:sx3-35x 2+l

X-#x-x2)

for all (x,y) sHI.

Let again

(x,y) -+ (1,1) be a sequence with ( x , y ) ~ H 1. Since Sl(x)-+O and Q~(x,y)--~o, we obtain

Xf~

Y{~ X-y

=~ZQl(X,y)+ocR'(x) ---~0. x-y

[]

C O R O L L A R Y 6. There exists no plvportion polynomial of class C ~ of degree at most 4 ex-

cept the constant polynomial 1.

REFERENCES [1]

ALSrNA,C.: ProblemP272 (Problemsand Remarks, No. 27), Aequat. Math. 29 (1985), p. 100.

[2]

ALSINA,C. and TRILLAS,E.: Lecciones de Algebray Geometria (Cursopard estudiantes de Arquitectura), Ed. Gustavo Gili, S.A., Barcelona, 1984.

[3]

Bt~NZ,W.: Afimctional equationsproblem in architecture, Arch. Marl1. 47 (1986), pp. 165-181.

[4]

BENZ,W.: Real Geometries,BI Wissenschaftsverlag,Mannheim, Leipzig, Wien, Zfirich, 1994.

[5]

BENZ,W.: RemarkP272S1 (Problemsand Remarks, No. 29), Aeqnat. Math. 29 (1985), pp. 101-102.

[6]

BRONSHTEIN,I.N. and SEMENDYAYEV,K.A.: A Guide Book to Mathematics, Verlag Harri Deutsch, Ztnich, Frankfurt, and Springer-Verlag, New York, 1973.

[7]

MOSZNER,Z.: RemarkP272S2 (Problemsand Remarks, No. 30), Aequat. Math. 29 (1985), pp. 102-104.

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