NICOLAE ANGHEL
Communicated by Vasile Brnzanescu
The Erdos-Mordell inequality is analyzed from the point of view of optimization theory.
AMS 2010 Subject Classication: Primary: 51M16; Secondary: 51M04.
Key words: Erdos-Mordell inequality, strict convexity, optimization.
1. INTRODUCTION
The Erdos-Mordell inequality [6] is one of the more fundamental inequali- ties in the geometry of triangles [10]. It asserts that for every (non-degenerate) triangleT in some Euclidean plane and for every pointR inside or on its sides, when denoting by r1,r2, and r3 the distances from R to the vertices ofT, and byρ1,ρ2, andρ3 the distances fromRto its sides (cf. Figure 1, left part), then one has
(1) r1+r2+r3 ≥2(ρ1+ρ2+ρ3),
with equality precisely when the triangleT is equilateral and the pointR is its circumcenter.
The inequality (1) was conjectured by Erdos [6] in 1935, in the form of a Proposed Problem to the American Mathematical Monthly. Following the long- standing tradition of this journal, two elegant solutions by Mordell and Barrow appeared two years later [9]. The solution by Barrow actually addressed a stronger version of the inequality, where the distances fromR to the sides were replaced by the bisectors of the three angles formed by joiningRto the vertices.
Since then, some twenty new solutions and generalizations of this in- equality have appeared, another staple of an important result in mathemat- ics [1, 5, 7, 8]. Like in the proofs of the Pythagorean Theorem, they usually dier by the language used to formulate them, ranging from pure geometry, to trigonometry, to vector calculus, or by some striking and innovative proof detail. For instance, a host of proofs are geared towards showing that, in general
REV. ROUMAINE MATH. PURES APPL. 59 (2014), 4, 453464
(2) r1+r2+r3≥ a2
a3
+ a3 a2
ρ1+
a3 a1
+ a1 a3
ρ2+
a1 a2
+a2 a1
ρ3,
where a1, a2, and a3, are the lengths of the sides of the triangle, from which (1) follows immediately.
In turn, (2) follows for instance from three other inequalities, a1r1 ≥ a2ρ3+a3ρ2,a2r2 ≥a3ρ1+a1ρ3, anda3r3 ≥a1ρ2+a2ρ1. Figure1, right part, borrowed from [1], presents a visual proof of the rst one of these inequalities.
If we denote by∆the set of points inside or on the sides of a xed triangle T, the two functions
(3) ∆3R7−→r1+r2+r3 ∈(0,∞) and ∆3R7−→ρ1+ρ2+ρ3∈(0,∞), and the generalizations of them from triangles to polygons, perhaps in weighted version, play a central role in optimal location problems. There, they always seem to have been addressed separately [3, 4], while the Erdos-Mordell inequal- ity suggests that one should study them also in combination.
Fig. 1. Erdos-Mordell: Set-up (left) and Visual proof (right).
The purpose of this paper is to subject the Erdos-Mordell inequality to an optimization analysis, which will amount to the study of a certain point,Cem, and a certain non-negative constant,cem, naturally associated to the inequality.
cem, the Erdos-Mordell constant, will be the minimum value of the function (4) ∆3R7−→r1+r2+r3−2(ρ1+ρ2+ρ3)
and Cem, the Erdos-Mordell center, will be the unique point in ∆ where this minimum is attained. Obviously, they depend solely on the triangle T. In- equality (1) simply says that in general cem >0 for triangles T which are not equilateral, while for an equilateral triangle,cem = 0and Cem is its circumcen- ter.
Incidentally, an optimization `thought' can better explain how the Erdos- Mordell inequality comes to be. It is known that Erdos discovered it experi- mentally, `after having drawn many triangles' [8], while in [7] it is speculated
that he was led to it in an attempt to generalize Euler's inequality, r ≥ 2ρ, where(r, ρ) are the (circumradius, inradius) of any triangle.
Since obviously, r2 ≥ ρ1, r3 ≥ ρ1, etc., we have that, r1 +r2 + r3
≥(1/2)(ρ1+ρ2+ρ3). What is then the best constantksuch that,r1+r2+r3
≥k(ρ1+ρ2+ρ3), for any triangleT? By taking isosceles trianglesT with very small apex angles, and by taking the pointR to be the apex vertex one can see thatk≤2, and in fact by Erdos-Mordell, k= 2.
Admittedly, our understanding of Cem and cem is incomplete. However, in view of [2] and what follows below, it is unlikely that they can be described purely geometrically for arbitrary triangles. Nonetheless, we do present a lower bound for cem in general (Proposition2), and then proceed to identifyCem for the class of isosceles triangles (Corollary).
2. GENERALITIES ABOUT cem AND Cem
The fact that cem and Cem exist is an easy consequence of the convex properties of the continuous function
(5) EM : ∆→[0,∞), EM(R) =r1+r2+r3−2(ρ1+ρ2+ρ3).
To this end denote the vertices of T by A1, A2, and A3 and denote by H1 ∈ ←−−→
A2A3, H2 ∈ ←−−→
A3A1 and H3 ∈ ←−−→
A1A2 the feet of the perpendicular lines dropped from A1, A2 and A3 on the sides of the triangle. Then a1 = A2A3, a2 =A3A1 and a3 =A1A2. Set also h1 =H1A1,h2 =H2A2 and h3 =H3A3. Finally, for a point in ∆, say R, we denote by the same small letter with appropriate subscript indices the distances from it to the vertices of T and by the corresponding small Greek letter the distances from it to the sides of T, i.e., r1 = RA1, r2 =RA2 and r3 =RA3, and ρ1 =distance from R to ←−−→
A2A3, ρ2=distance from R to ←−−→
A3A1 and ρ3 =distance from R to ←−−→ A1A2.
Proposition 1. The continuous function EM : ∆ → [0,∞) dened by (5) is strictly convex, in the sense that if R1, R2 are two distinct points in ∆ and for a 0< t <1, R is the unique point on R1R2 such that R1R =tR1R2, then
(6) EM(R)<(1−t)EM(R1) +tEM(R2).
Consequently, the Erdos-Mordell constant cem := min{EM(R)|R ∈ ∆}
exists and it is attained at one and only one point in ∆, the Erdos-Mordell center Cem. Then the following are true about Cem and cem:
a) Cem is an interior point of ∆if and only if there is a pointM interior
to ∆such that vector-wise (7)
−−−→M A1 m1
+
−−−→M A2 m2
+
−−−→M A3 m3
+ 2
−−−→H1A1 h1
+
−−−→H2A2 h2
+
−−−→H3A3 h3
!
=~0,
in which case Cem =M and cem=m1+m2+m3−2(µ1+µ2+µ3).
b) Cem is an interior point of a side of T, say A2A3, if and only if there is a point M ∈A2A3\ {A2, A3} such that
(8)
*−−−→
M A1 m1
+ 2
−−−→H2A2 h2
+
−−−→H3A3 h3
! ,
−−−→A2A3 a1
+
= 0 and
*−−−→
M A1 m1
+ 2
−−−→H2A2 h2
+
−−−→H3A3 h3
! ,
−−−→H1A1 h1
+
≤ −2, in which case Cem =M and cem=m1+a1−2(µ2+µ3).
c) Cem is a vertex of T, say A1 if and only if when denoting by A01 the unique point such that
(9)
−−−→A1A2
a3 +
−−−→A1A3
a2 + 2
−−−→H1A1
h1 +
−−−→H2A2
h2 +
−−−→H3A3
h3
!
=−−−→
A1A01, we have that either,
c1) A01 belongs to the interior or to the sides of the ∠A2A1A3 and (10)
−−−→A1A2 a3
+
−−−→A1A3 a2
+ 2
−−−→H1A1 h1
+
−−−→H2A2 h2
+
−−−→H3A3 h3
!
≤1, or c2) A01 belongs to the exterior of the ∠A2A1A3 and
(11) 2 max
(*−−−→
H1A1 h1
+
−−−→H2A2 h2
,
−−−→A1A2 a3
+ ,
*−−−→
H1A1 h1
+
−−−→H3A3 h3
,
−−−→A1A3 a2
+)
≤ −
*−−−→
A1A2 a3
,
−−−→A1A3 a2
+ , in which case Cem =A1 and cem=a2+a3−2h1.
In the formulas above h·,·iandk·krepresent the ordinary Euclidean inner product and norm of vectors.
Proof. As in ([4], p. 240), the continuous function∆3R7−→r1+r2+r3∈ (0,∞) is strictly convex. Using triangle similarity it is also obvious that the function∆3R7−→ρ1+ρ2+ρ3 ∈(0,∞) is linear, with linearity being dened as in (6), but replacing the inequality sign<with equality. Combining the two functions as in the denition of EM proves then the strict convexity of EM, and therefore the existence of cem and CEM.
Actually, by considering signed distances to the sides of T, EM extends naturally to a continuous strictly convex function on the whole plane containing T, which is dierentiable everywhere, except at the vertices of T. It is easily seen that the gradient (of this extension) of EM is given by the formula (12)
∇EM(R) =−
−−→RA1
r1
+
−−→RA2
r2
+
−−→RA3
r3
+2
−−−→H1A1
h1
+
−−−→H2A2
h2
+
−−−→H3A3
h3
!!
, R6=A1, A2,andA3. A fundamental property of continuous strictly convex functions dened on compact convex sets [11] says that EM attains its minimum value at the point M ∈ ∆ where the directional derivatives (which always exist, even at non-dierentiability points) have the property that
(13) D~vEM(M)≥0, for every~v=
−−→M R
M R, R∈∆, R6=M.
If the minimum point M ∈∆is a dierentiability point forEM, i.e.,M is an interior point of ∆or an interior point of a side ofT, then from equation (12) we get
(14)
D~vEM(M) =−
* −−−→
M A1 m1
+
−−−→M A2 m2
+
−−−→M A3 m3
+2
−−−→H1A1 h1
+
−−−→H2A2 h2
+
−−−→H3A3 h3
!!
, ~v +
.
In case a) when M is an interior point to ∆, equation (14) yields the familiar vanishing (7) of the gradient ∇EM(M) .
In caseb), whenM is a point interior to the sideA2A3,∇EM(M)reduces to
∇EM(M) =−
−−−→M A1
m1 + 2
−−−→H1A1
h1 +
−−−→H2A2
h2 +
−−−→H3A3
h3
!!
.
A little thought shows that in this case equation (14) reduces to the pair of equations (8).
The most interesting case is case c), whenM =A1. AlthoughEM is not dierentiable at A1, a direct calculation shows that
D~vEM(A1) = 1−
*−−−→
A1A2 a3
+
−−−→A1A3 a2
+ 2
−−−→H1A1 h1
+
−−−→H2A2 h2
+
−−−→H3A3 h3
! , ~v
+ .
Therefore,D~vEM(A1)≥0, for every vector~v=−−→
A1R/r1, R∈∆,R 6=A1 if and only if
max (*−−−→
A1A2 a3 +
−−−→A1A3 a2 +2
−−−→H1A1 h1 +
−−−→H2A2 h2 +
−−−→H3A3 h3
! ,~v
+
~ v=
−−→A1R
r1 , R∈∆, R6=A1 )
≤1.
At which vector ~v the above maximum is attained depends on the loca- tion of the vector −−−→
A1A2/a3+−−−→
A1A3/a2+ 2(−−−→
H1A1/h1+−−−→
H2A2/h2+−−−→
H3A3/h3), translated to originate at A1, with respect to the ∠A2A1A3.
In sub-case c1) obviously ~v must be taken to be the unit vector in the direction of−−−→
A1A2/a3+−−−→
A1A3/a2+ 2
−−−→
H1A1/h1+−−−→
H2A2/h2+−−−→
H3A3/h3
, if the latter vector is non-zero.
In sub-case c2)~v must be taken to be either −−−→
A1A2/a3 or −−−→
A1A3/a2. The content of equations c1) and c2), then follows.
We conclude this section by giving a lower bound forcem. Since it follows from the proof of the Erdos-Mordell inequality alluded to in the Introduction, and not from the Proposition above, it is never optimal, except for equilateral triangles.
Proposition 2. The notations being as above, we have
(15) cem ≥ 2S
a1a2a3min
(a1−a2)2,(a2−a3)2,(a3−a1)2 , where S is the area of the triangle T.
Proof. According to equation (2) for every R∈∆we have (16) EM(R)≥
a2
a3 +a3
a2 −2
ρ1+ a3
a1 +a1
a3 −2
ρ2+ a1
a2 + a2
a1 −2
ρ3. A standard linear programming approach (the simplex method) or [3]
shows that the RHS of (16) attains, as function of R ∈ ∆, its minimum at one of the vertices of T. (15) then follows. Notice that this Proposition is more relevant than the Erdos-Mordell inequality only for triangles which are not isosceles or equilateral.
3. A TRIGONOMETRIC ANALYSIS
The content of Proposition 1becomes more transparent if we translate it into trigonometric language. This is only tting sinceCemandcem depend only on the similarity class of a triangle, in the sense that the scaling of a triangle preserves the relative position of Cem and scales cem accordingly. Although they can be expressed in terms of two of the angles of T, for aesthetic reasons our results will also include the third angle.
Proposition 3. The following is true about the triangle A1A2A3 whose respective angles have measures α1, α2 and α3.
a) Cem is a vertex of T if and only if there is a relabeling of the vertices of T, making α2≤α3, such that either
(17) sinα1−2 cosα1+ 2≥2 cosα2 and
4(sinα2+ sinα3) + 8(cosα2+ cosα3)≥13 + 8 sinα1−6 cosα1, or
(18) sinα1−2 cosα1+ 2<2 cosα2 and 2 sinα1+ cosα1≤2 sinα2. In this case, Cem =A1.
b) Cem is a point interior to a side of T if and only if there is a relabeling of the vertices of T, making α2≤α3, such that
(19) 2(sinα3−sinα2)<cosα2 and
8(cosα1+ 1)<8(cosα1+ cosα2+ cosα3)≤11.
In this case, Cem∈A2A3\ {A2, A3} and necessarily the angles α2 andα3 are acute angles.
c) Cem is an interior point of ∆ if and only if neither a) nor b) above holds, when applied to any vertex, respectively side of T. In such a case, for any relabeling of the vertices ofT makingα2 ≤α3, there are angle measuresβ2
and β3, 0< β2 < α2,0< β3< α3 such that
(20) 4(sinα3−sinα2)(cosβ2−cosβ3)+ 4(cosα2+ cosα3−1)(sinβ2+ sinβ3)
−2 cos(β2+β3) = 8(cosα1+ cosα2+ cosα3)−13.
There are three distinct equalities of type (20), so in principle Cem can be realized as the intersection point of three complicated curves.
Cem is geometrically constructible in caseb), however unlikely to be so in case c), in general.
Proof. a) Relabeling the vertices of T if necessary we can assume that Cem=A1 andα2≤α3. Choose a coordinate system in the plane ofT centered at A1, with thex-axis in the direction of the bisector line of the∠A2A1A3 and the y-axis such that A3 belongs to the half-plane y > 0. In this coordinate system,
(21)
−−−→A1A3 a2 =
cosα1
2 ,sinα1 2
,
−−−→A1A2 a3 =
cosα1
2 ,−sinα1 2
,
−−−→H3A3
h3 = sinα1
2 ,cosα1
2
,
−−−→H2A2
h2 = sinα1
2 ,−cosα1
2
,
−−−→H1A1
h1 =
−sinα1
2 +α3
,cosα1
2 +α3
.
Following Proposition 1c), then (22) −−−→
A1A01 =
2 cosα1
2 + 4 sinα1
2 −2 sinα1
2 +α3
,2 cosα1
2 +α3 . Under the hypothesis α2 ≤α3,A01 belongs to the interior or to the sides of the ∠A2A1A3 if and only if sinα1−2 cosα1+ 2≥2 cosα2. Also,
(23)
−−−→
A1A01
2
= 14+8 sinα1−6 cosα1−4(sinα2+sinα3)−8(cosα2+cosα3), and equation (11) is equivalent to 2 sinα1+ cosα1≤2 sinα2. Proposition3a) follows now from Proposition1c).
b) Again, subject to relabeling we can assume that Cem =M ∈ A2A3\ {A2, A3} and α2 ≤ α3. Choose now a coordinate system with origin in M, x-axis directed along −−−→
A2A3, and y-axis such that A1 belongs to the half-plane y >0. Denoting bytthe measure of the∠A1M A3we have thatα2 < t < π−α3, and in this coordinate system,
(24)
−−−→M A1 m1
= (cost, sint),
−−−→H1A1 h1
= (0,1),
−−−→H2A2 h2
= (−sinα3,−cosα3),
−−−→H3A3
h3 = (sinα2,−cosα2),
−−−→A2A3
a1 = (1,0).
Consequently, (25)
−−−→M A1 m1
+ 2
−−−→H2A2 h2
+
−−−→H3A3 h3
!
= (cost−2 sinα3+ 2 sinα2,sint−2 cosα3−2 cosα2), and so the equations (8) of Proposition 1b) become
(26) cost= 2(sinα3−sinα2) and sint≤2(cosα2+ cosα3−1).
However, α2 < t < π−α3 is equivalent tocosα2 >cost > −cosα3, or by (26),
(27) −cosα3 <2(sinα3−sinα2)<cosα2. Also, sincesint >0, the second part of (26) implies that (28) cosα2+ cosα3 >1,
and in fact is equivalent to
(29) sin2t= 1−cos2t≤4(cosα2+ cosα3−1)2.
(28) can happen only if α2 and α3 are acute angles, and since α2 ≤ α3 impliessinα2≤sinα3, the rst inequality in (27) is redundant.
Putting together (26), (27), (28), and (29), yields then (19). All the above arguments are reversible and this completes the proof of the b) part of Proposition 3, via Proposition1b).
c) Obviously, Cem must be an interior point to∆ if it is neither a vertex of T nor an interior point to some side of T. For a labeling of the vertices of T withα2 ≤α3,Cem =M is the intersection point of a ray originating at A2 and making some angle of measureβ2,0< β2< α2, with −−−→
A2A3, with a similar ray, originating atA3and making some angle of measureβ3,0< β3< α3, with
−−−→A3A2.
Choose now a coordinate system originating at the intersection point of
←−→M A1 with A2A3, having x-axis in the direction of −−−→
A2A3 and y-axis such that A1 belongs to the half-plane y > 0. Then, denoting by t the measure of the angle formed by −−−→
M A1 with−−−→
A2A3 we have
(30)
−−−→M A1
m1
= (cost,sint),
−−−→M A2
m2
= (−cosβ2,−sinβ2),
−−−→M A3 m3
= (cosβ3,−sinβ3),
−−−→H1A1 h1
= (0,1),
−−−→H2A2 h2
= (−sinα3,−cosα3),
−−−→H3A3 h3
= (sinα2,−cosα2).
Equation (7) is then equivalent to
(31) cost= cosβ2−cosβ3+ 2(sinα3−sinα2) sint= sinβ2+ sinβ3+ 2(cosα2+ cosα3−1)
(20) follows now by eliminating t between the two equations (31), via sin2t+ cos2t= 1.
The constructibility of Cem in caseb) follows from the constructibility of an anglet such that (cf. (26))
cost= 2(sinα3−sinα2) = 2H1A1(A1A2−A1A3) A1A2·A1A3
.
However, the constructibility of Cem in casec) seems unlikely, given the complexity of equation (20).
Corolarry. 1)For an isosceles triangleA1A2A3withα2 =α3 = π 2 −α1
2 , Cem always belongs toH1A1, as follows:
a) Cem=H1 if and only if 3
4 ≤sinα1 2 <1. b) Cem =A1 if and only if 0<sinα1
2 ≤ 6−√ 11 10 .
c) Cem ∈ H1A1\ {H1, A1} if and only if 6−√ 11
10 <sinα1 2 < 3
4. More precisely,Cem is the point M interior to H1A1 such that if β is the measure of
∠M A2A3, then sinβ = 3−4 sinα21
2 . Cem can be constructed in the following way: Construct rst the point P on the ray originating at A2 in the direction of −−−→
A2A3 such that A2P = 3A1A2−4H1A2. On the same ray, construct then the point Q such that A2Q=A1A2. The circles of center A2 and radiusA2P, respectively of center Q and radius A1A2, intersect at a point S on the same side of ←−−→
A2A3 as A1. The parallel line to ←−−→
A2A3 through the point S intersects the circle of centerQ and radiusA1A2 a second time at point T. Then the line
←−→A2T intersects H1A1 at the point Cem.
2) For a 30−60−90 triangle A1A2A3 with α1 = π
2, α2= π
6, α3= π 3, Cem ∈ A2A3\ {A2, A3}. More precisely, Cem is the point M interior to the hypotenuseA2A3 such that iftis the measure of∠A1M A3, thencost=√
3−1.
Cemcan be constructed in the following way: Construct rst the pointP ∈A1A2
such that A2P = 1
2A2A3. Construct then the point Q on ←−−→
A1A3 such that A1 is between Q and A3, and P Q= 1
2A2A3. Denote by S the intersection point of ←→
P Q and the bisector line of ∠A1A2A3. Next, let T be the symmetrical of the point S with respect to the line ←−−→
A2A3. Then Cem is the intersection point of A2A3 and the line through A1 parallel to ←−→
T M1, where M1 is the mid-point of A2A3.
Proof. 1) First, it is easy to see why Cem ∈ H1A1. If not, Cem and its symmetrical with respect to H1A1 are two distinct points in ∆ where EM attains its minimum value, a contradiction.
a) The equations (19) in Proposition3b) are equivalent to sinα1
2 ≥ 3 4. b) Only (17) apply to this case. However, it is easier to see that since α1
2 +α3=π 2,−−−→
A1A01in equation (22) becomes−−−→
A1A01 =
2 cosα1
2 +4 sinα1
2 −2,0 and so
−−−→A1A01
= 2 cosα1
2 + 4 sinα1
2 −2. Then
−−−→A1A01
≤1 is equivalent to sinα1
2 ≤ 6−√ 11 10 .
c) In equations (31), β2 = β3 implies cost = 0 and sint = 2 sinβ2 +2
2 sinα1
2 −1
, and so t= π
2 andsinβ2 = 3−4 sinα21
2 . The proposed con- struction matches then the trigonometric expression of sinβ, β = β2, since
sinα1
2 = A2A3
2A1A2.
2)The three inequalities (19) are trivially satised whenα1 = π
2, α2 = π 6, α3= π
3. The angle measure t in the proof of Proposition 3b) satises cost=
√3−1, by the rst of the two equations (26), and the proposed construction mirrors just this, since the measure of∠A1P Qist.
4. FINAL REMARKS
1)All the techniques employed above apply verbatim to the more general optimization function
(32) EMλ,µ: ∆→R,
EMλ,µ(R) =λ(r1+r2+r3) +µ(ρ1+ρ2+ρ3), λ, µ∈R, λ≥0.
Notice that the study of EMλ,µ addresses the classical Fermat-Torricelli problems [3, 4], when λ= 1, µ= 0, or λ= 0, µ= 1.
2) Since in any triangleT,a1ρ1+a2ρ2+a3ρ3 = 2S, whereS is the area of T, the level curves of the function (of R ∈ ∆) ρ1+ρ2+ρ3 are segments.
Asρ1+ρ2+ρ3 is constant on these level curves it is tempting to rst restrict EM to them and study apparently simpler Fermat-Torricelli-type problems constrained to lines, in an attempt to get more insight whenCem is an interior point to∆. It is known however [2] that such problems are equally intractable from a constructive viewpoint, in general.
Acknowledgments. The paper beneted from the insightful remarks of the referee.
REFERENCES
[1] C. Alsina and R.B. Nelson, A Visual Proof of the Erdos-Mordell Inequality. Forum Geom. 7(2007), 99102.
[2] C. Bajaj, The Algebraic Degree of Geometric Optimization Problems . Discrete Comput.
Geom. 3(1988), 177191.
[3] R. Barbara, The Fermat-Torricelli Points of n Lines. Math. Gazette 84(2000), 499, 2429.
[4] V. Boltyanski, H. Martini and V. Soltan, Geometric Methods and Optimization Problems.
Kluwer, Dordrecht, 1999.
[5] S. Dar and S. Gueron, A Weighted Erdos-Mordell Inequality. Amer. Math. Monthly 108(2001), 165168.
[6] P. Erdos, Problem 3740. Amer. Math. Monthly42(1935), 396.
[7] N.D. Kazarino, Geometric Inequalities. MAA, Washington, 1961.
[8] V. Komornik, A Short Proof of the Erdos-Mordell Theorem. Amer. Math. Monthly104 (1997), 5760.
[9] L.J. Mordell and D.F. Barrow, Solution 3740. Amer. Math. Monthly 44 (1937), 252254.
[10] V. Pambuccian, The Erdos-Mordell Inequality Is Equivalent to Non-Positive Curvature . J. Geom. 88(2008), 134139.
[11] R.T. Rockafellar, Convex Analysis. Princeton Univ. Press, Princeton, N. J., 1970.
Received 31 January 2014 University of North Texas, Department of Mathematics,
Denton, TX 76203 [email protected]
