The Geometry of Cubic Polynomials
Christopher Frayer
Miyeon Kwon
Christopher Schafhauser
James A. Swenson
January 27, 2012
Abstract
Given a complex cubic polynomial p(z) = (z − 1)(z − r1 )(z − r2 ) with
|r1 | = 1 = |r2 |, where are the critical points? Marden’s Theorem tells us that
the critical points are the foci of the Steiner ellipse of 41r1 r2 . In this paper
we further explore the structure of these critical points. If we let Tγ be the
circle of diameter γ passing through 1 and 1 − γ, then there are α, β ∈ [0, 2]
such that the critical points of p lie on the circles Tα and Tβ respectively. We
show that Tβ is the inversion of Tα over T1 , from which many nice geometric
consequences can be drawn. For example, (1) there is a “desert” in the unit
disk, the open disk {z ∈ C : |z − 23 | < 31 }, in which critical points cannot
occur, and (2) a critical point of such a polynomial almost always determines
the polynomial uniquely.
1
Introduction
In 2009, Dan Kalman won a Lester R. Ford Award for [7], published in this
Monthly. In this outstanding article, Kalman gave a new proof of Marden’s
theorem,1 which he has called “the most marvelous theorem in mathematics.” [8]
Theorem 1.1 (Marden’s theorem). Given a triangle 4r1 r2 r3 in the complex plane,
there is a unique inscribed ellipse, called the Steiner ellipse, which is tangent to the
sides of the triangle at the midpoints of the sides. If
p(z) = (z − r1 )(z − r2 )(z − r3 ),
then the roots of p0 (z) are the foci of the Steiner ellipse of 4r1 r2 r3 , and the root of
p00 (z) is the centroid of 4r1 r2 r3 .
This is indeed a marvelous theorem. What we find so attractive about it is the
connection between analysis and geometry: a critical point is a focus of an ellipse!
Before learning this, all we had known about the critical points of a general complex
polynomial was the Gauss-Lucas theorem, which guarantees that the critical points
1 In accordance with Boyer’s Law [9], Marden’s theorem is originally due to J. Siebeck, according
to [8, 11, 12].
1
of any polynomial lie in the complex hull of its roots. For us, it was a special
pleasure to learn from [10] that the theorem’s namesake, Morris Marden, had helped
to found the mathematical research program at our sister school, the University of
Wisconsin-Milwaukee, as well as our section of the MAA.
Kalman’s outstanding paper got us excited about cubic polynomials – particularly those of us who were already intrigued by the “polynomial root dragging”
introduced by Bruce Anderson in [1], also in this Monthly. Anderson and others
[2–5] had investigated how the critical points of a monic polynomial depend on its
roots, in the case where all roots are real.
We were now curious. What happens to the critical points of a complex cubic
polynomial when its roots move? Hands-on investigation was irresistible: we drew
some figures in GeoGebra,2 set the roots in motion, and watched the critical points
move.
At this point we had a decision to make. On the real line, one has relatively few
options: just decide how far to drag each root to the right. In the complex plane,
a world of possibilities open up. In building our first GeoGebra notebooks, we got
lucky. Through any three given points r1 , r2 , r3 in the complex plane, there is a
unique circle. By changing coordinates, we can take this circle to be the unit circle
with r3 = 1, so that |rk | = 1 for each k. We set r1 and r2 in motion around the
unit circle at different speeds, and traced the loci of the critical points. This article
is the result of that fortunate decision.
We were surprised to see that as the roots varied, the trajectories of the critical points always avoided a certain disk. Further investigation explained this fact
(Theorem 3.8 below), and revealed a level of structure by which we were again
surprised: a critical point of such a polynomial almost always determines the polynomial uniquely (Theorem 3.13). It is our pleasure to share with you the geometry
(especially Theorem 3.6) behind these analytic facts – a result we hope Marden
would have enjoyed.
2
Centers
Let p(z) be a cubic polynomial with its roots on the unit circle in C. The roots of
p(z), and of its derivatives, are preserved when we multiply by a non-zero constant,
so we may assume that p(z) = (z − r1 )(z − r2 )(z − 1) for some ri ∈ C with |ri | = 1.
In fact, let’s name this class of polynomials.
Definition 2.1. Let Γ denote the family of cubic polynomials q : C → C such that
q(z) = (z − 1)(z − r1 )(z − r2 )
for some r1 and r2 with |r1 | = 1 = |r2 |.
2 GeoGebra
is a software package for plane geometry,
freely available at
[http://www.geogebra.org/]. We used GeoGebra to produce the figures in this paper. A collection
of GeoGebra notebooks illustrating our main results can be found at [http://www.uwplatt.edu/∼
swensonj/gocp/].
2
Before studying the zeros of p0 (z), let’s take a moment to investigate the zeros
of p00 (z). We’ll find some pretty geometry, and obtain Theorem 2.3, which gives us
a hint of the sort of thing that can be said about critical points.
Definition 2.2. Given p ∈ Γ and g ∈ C, we say g is the center 3 of p(z) provided
p00 (g) = 0.
Of course every p ∈ Γ has a unique center; recall that by Marden’s theorem, this
is the centroid of 4r1 r2 1. Interestingly, though, we can prove a sort of converse to
this fact!
Theorem 2.3. Let g ∈ C.
• p ∈ Γ has center 31 if and only if p(z) = (z − 1)(z 2 − r2 ) for some r with
|r| = 1.
• If 0 < g − 31 ≤ 23 , then there is a unique polynomial p ∈ Γ with center g.
• If g − 13 > 23 , there is no p ∈ Γ with center g.
In words: the center of any p ∈ Γ must lie in the closed disk bounded by the
dashed circle in Figure 1. Conversely, almost every point in that closed disk is the
center of a unique p ∈ Γ – but the disk’s center is the center of each polynomial in
a family parametrized by a circle.
Proof. By Marden’s theorem, this claim is related to the construction of a triangle
given a vertex, the centroid, and the circumcircle.
Suppose g is the center of p(z) ∈ Γ. Then g is the centroid of triangle 41r1 r2 ,
where r1 and r2 are to be constructed.
Though we do not know where r1 and r2 are, let us denote the midpoint of r1 r2
by w. Of course, w lies in the closed unit disk. The
segment
1w is a median of
4r1 r2 1, so g = 23 (w) + 13 (1). It is immediate that g − 13 ≤ 32 .
Consequently, to begin the construction, we solve the previous equation for w
0
and let w = 3g−1
2 . (In geometric terms, given g, we may construct the midpoint w
0
of 1g; then w is the reflection of w over g.)
Assume first that g 6= 13 , so w 6= 0. We know that r1 r2 is a chord of the
unit circle, so its perpendicular bisector must pass through 0 – and through the
midpoint w.
Hence we construct the line L through w, perpendicular to 0w. Since w lies
in the closed unit disk, L intersects the unit circle in two points (counting with
multiplicity if |w| = 1); these are r1 and r2 .
On the other hand, suppose that w = 0 and g = 13 . Since w = 0 is the midpoint
of r1 r2 , we have r2 = −r1 . Conversely, in this case we can compute directly that 13
is the center of p(z) = z 3 − z 2 − r12 z + r12 .
3 Calling this an inflection point might be better, but we just couldn’t do it. Perhaps we’ve
spent too much time telling our first-semester calculus students, “Even when f 00 (a) = 0, that
doesn’t necessarily mean a is an inflection point of f .”
3
Figure 1: Construction of 4r1 r2 1 given center g (Theorem 2.3)
3
3.1
Critical points
Inversion
Having seen Theorem 2.3, we turn to the study of the critical points of p ∈ Γ,
seeking a parallel result. Where can the critical points lie? To what extent do the
critical points determine p?
Regarding the first question, Saff and Twomey
show
in [13] that p has at least
one critical point in the closed disk ∆ = z ∈ C : z − 21 ≤ 12 . Moreover, they
show that if c1 and c2 are the critical points of p, and if c1 6= 1 is on the boundary
of ∆, then c2 is the complex conjugate of c1 and hence also lies on the boundary
of ∆.
To extend the result of Saff and Twomey, we will take advantage of a delightful
geometric symmetry. To explain this, we need some notation.
Definition 3.1. Given α > 0, we denote by Tα the circle of diameter α that passes
through 1 and 1 − α in the complex plane.4 That is,
n
α α o
Tα = z ∈ C : z − 1 −
.
=
2
2
See Figure 2 below. For example, we have seen in Theorem 2.3 that the center
of p ∈ Γ cannot lie outside T4/3 .
4 It may seem strange to name the circle by its diameter, rather than by its radius – but this
notation will simplify Theorems 3.2 and 3.6.
4
Figure 2: z lies on a unique Tα .
Theorem 3.2. Let z ∈ C with Re(z) < 1. We have z ∈ Tα if and only if
1
1
= Re
.
α
1−z
(3.1)
Proof. As shown in Figure 2, let θ denote the measure of ∠01z (taking θ > 0 ⇐⇒
Im(z) > 0), and set r = |z − 1|, so that
1 − z = re−iθ
1 iθ
1
=
e
1−z
r
1
cos θ
Re
=
.
1−z
r
The angle inscribed in Tα at z intercepts a diameter and must be a right angle.
Therefore, right-triangle trigonometry gives cos θ = αr , completing the proof.
Corollary 3.3. The complex number z 6= 1 lies in the closed unit disk if and only
if there is a unique α ∈ (0, 2] for which z lies on Tα .
Let’s walk through a couple of examples.
Example 3.4. Suppose p ∈ Γ has a critical point at 1. Then we know that p has a
double root at 1. This is a familiar fact, but just in case: Since p(z) = (z − 1)q(z)
for some quadratic polynomial q, the product rule gives p0 (z) = q(z) + (z − 1)q 0 (z).
5
Substituting z = 1, we have 0 = q(1), as we claimed. Explicitly, there is some r on
the unit circle such that
p(z)
=
(z − 1)2 (z − r)
p0 (z)
=
3z 2 − (2r + 4)z + (2r + 1)
2r + 1
.
3(z − 1) z −
3
and
=
Summarizing: p ∈ Γ has a critical point at 1 if and only if p(z) = (z − 1)2 (z − r) for
some r on the unit circle, and in this case the other critical point is 31 + 23 r ∈ T4/3 .
Example 3.5. Suppose now that p ∈ Γ has a critical point at some r 6= 1 on the
unit circle. By the Gauss-Lucas theorem, r is in the convex hull of the roots of p,
and so r must be a root of p. It follows, as in example 3.4, that in fact p has a
double root at r, so:
p(z)
=
(z − r)2 (z − 1)
p0 (z)
=
3z 2 − (4r + 2)z + (r2 + 2r)
r+2
3(z − r) z −
.
3
and
=
As above, we summarize: p ∈ Γ has a critical point at r 6= 1 on the unit circle
if and only if p(z) = (z − r)2 (z − 1), and in this case the other critical point is
2
1
3 + 3 r ∈ T2/3 .
Now consider a general p ∈ Γ. By the Gauss-Lucas theorem, both critical points
of p will lie in the closed unit disk. We are ready to prove our symmetry result for
the critical points of p; this is our main theorem, from which many nice consequences
will be drawn. We prove a more general statement than we need, because it is no
harder to do so.
Theorem 3.6. Let f (z) = (z − 1)(z − z1 ) · · · (z − zn ), where zk = eiθk for each k.
Let c1 , . . . , cn denote the critical numbers of f (z), and suppose that 1 6= ck ∈ Tαk
for each k. Then
n
n
X
X
1
1
= 2
1 − ck
1 − zk
k=1
k=1
and
n
X
1
αk
= n.
k=1
6
(3.2)
00 (1)
Proof. We prove (3.2) by evaluating Re ff 0 (1)
in two different ways.
Q
n
0
To begin with, f (z) = (n + 1) k=1 (z − ck ). By logarithmic differentiation,
f 00 (z)
f 0 (z)
n
X
=
k=1
n
X
00
1
z − ck
1
f (1)
=
f 0 (1)
1 − ck
k=1
!
00 n
X
f (1)
1
Re
= Re
f 0 (1)
1 − ck
k=1
00 n
X
f (1)
1
Re
=
,
f 0 (1)
αk
k=1
where the last step follows from Theorem 3.2.
On the other hand, if we write f (z) = (z − 1)g(z), then the product rule yields
f 0 (z)
=
(z − 1)g 0 (z) + g(z)
f 00 (z)
=
(z − 1)g 00 (z) + 2g 0 (z)
00
f (1) =
f 00 (1)
=
f 0 (1)
f 00 (1)
f 0 (1)
=
2g 0 (1)
2g 0 (1)
g(1)
n
X
1
2
,
1 − zk
k=1
where the last step is by logarithmic differentiation of g(z). But since zk ∈ T2 , we
can apply Theorem 3.2 to obtain
Re
f 00 (1)
f 0 (1)
=2
n
X
Re
k=1
1
1 − zk
=2
n
X
1
k=1
2
= n.
Corollary 3.7. Let p ∈ Γ, and let c1 6= 1 and c2 6= 1 be the critical points of p. If
c1 lies on Tα and c2 lies on Tβ , then
1
1
+ = 2.
α β
(3.3)
Recall that we have already seen (in Example 3.4) what happens
c1 = 1
when
or c2 = 1. Notice that (3.3) holds in Example 3.5, where {α, β} = 2, 32 .
7
Figure 3: Tβ is the inversion of Tα across T1 .
Geometrically, (3.3) means that Tβ is the inversion of the circle Tα across T1 .
(See Figure 3.)5 To see this, suppose that (3.3) holds, and that c1 ∈ Tα and c2 ∈ Tβ .
Then β > 0, so α > 21 . Since T1 and Tα are symmetric across the real axis and pass
through 1, the same is true of the inversion of Tα across T1 . This being the case, it
note: Suppose first that α > 12 ; we invert Tα over T1 using formulæ given in [14]. T1
, 0 and radius
has center (x0 , y0 ) = 21 , 0 and radius k = 21 , while Tα has center (x, y) = 1 − α
2
a= α
. In terms of
2
1
k2
=
,
s=
2
(x − x0 ) + (y − y0 )2 − a2
1 − 2α
the inversion of Tα over T1 has radius
1 α
α
·
r = |s|a = =
1 − 2α 2
2(2α − 1)
5 Referee’s
and center
(x0 +s(x−x0 ), y0 +s(y−y0 )) =
1
1
+
2
1 − 2α
1
α
−
2
2
1
1−α
,0 = 1 +
− 1 , 0 = (1−r, 0).
2 1 − 2α
That is, the inversion of Tα over T1 is Tβ , where the diameter β = 2r =
α
,
2α−1
so that
1
1
2α − 1
1
+ =
+
= 2,
α
β
α
α
as desired.
On the other hand, if α = 21 , then the inversion of Tα is the vertical line Re(z) = 1; if α < 21 ,
then the inversion of Tα lies to the right of this line. Either way, it is impossible for a critical point
of p ∈ Γ to lie on the inversion of Tα , by the Gauss-Lucas theorem. Neither can a critical point
1
1
lie on Tβ with α
+ β1 = 2, since then β1 = 2 − α
< 0, contrary to Corollary 3.3.
8
Figure 4: No p ∈ Γ has a critical point in the desert (Theorem 3.8).
is enough to show that the product of the distances D1 and D2 to 12 from 1 − α and
1 − β equals 41 , the square of the radius of T1 . In fact, this is true, because (3.3)
α
, and then
gives β = 2α−1
(1 − α) −
1 · (1 − β) −
2 1
1 2α − 1
1
= .
=
·
2
2
2(2α − 1) 4
Expressed another way, (3.3) says that the radius of the unit circle is the harmonic mean of the diameters of Tα and Tβ . In this form, the statement generalizes
to an arbitrary coordinate system: given 4ABC in the complex plane, let Tα and
Tβ be the circles tangent to the circumcircle at A that pass through the foci of the
Steiner ellipse. The radius of the circumcircle then equals the harmonic mean of
the diameters of Tα and Tβ .
We think Corollary 3.7 is intrinsically attractive, but even better, it is useful!
First, it allows us to prove our
observation:
there is a “desert” in the unit
original
disk, the open disk z ∈ C : z − 32 < 13 , in which critical points cannot occur.
Theorem 3.8. No polynomial p ∈ Γ has a critical point strictly inside T2/3 .
Proof. If c is strictly inside T2/3 , then c lies on Tα for some α ∈ 0, 23 . Suppose for
contradiction that c is a critical point of some polynomial p ∈ Γ. Then the other
critical point of p lies on Tβ , where (by Corollary 3.7) β1 = 2 − α1 < 2 − 23 = 12 – but
then β > 2, which is impossible by Corollary 3.3.
Recall that Saff and Twomey had shown that every p ∈ Γ has at least one critical
point on or inside T1 . Corollary 3.7 lets us say more.
9
Theorem 3.9 (cf. [13]). Let c1 6= 1 and c2 6= 1 be the critical points of p ∈ Γ. If
c1 lies on T1 , then c2 also lies on T1 . Otherwise, c1 and c2 are on opposite sides
of T1 .
Proof. Let c1 ∈ Tα and c2 ∈ Tβ . Then
α < 1 if and only if β > 1.
3.2
+ β1 = 2, so α = 1 if and only if β = 1 and
1
α
A critical point determines p ∈ Γ (almost always)
Once again, suppose p ∈ Γ, with roots r1 , r2 , and 1, and let c be a critical point
of p. Then
p(z)
=
(z − r1 )(z − r2 )(z − 1)
= z 3 − (r1 + r2 + 1)z 2 + (r1 + r2 + r1 r2 )z − r1 r2
so that
p0 (z)
=
3z 2 − 2(r1 + r2 + 1)z + (r1 + r2 + r1 r2 )
0
=
3c2 − 2c(r1 + r2 + 1) + (r1 + r2 + r1 r2 ).
and
Assuming that r1 6= 2c − 1, we get
r2 =
(2c − 1)r1 + (2c − 3c2 )
,
r1 + (1 − 2c)
(3.4)
which motivates the following definition.
Definition 3.10. Given c ∈ C, we define the Möbius transformation
fc (z) =
(2c − 1)z + (2c − 3c2 )
.
z + (1 − 2c)
We let Sc denote the image of the unit circle under fc .
A Möbius transformation is a function of the form f (z) = az+b
cz+d . When ad − bc 6=
dz−b
0, one has f −1 (z) = −cz+a
. In the case of fc , we have ad − bc = −(c − 1)2 , so when
z−1
c 6= 1, we have (fc )−1 = fc . [In the exceptional case, we have f1 (z) = z−1
= 1
−1
when z 6= 1, and of course (f1 ) does not exist.]
It is well-known that any (invertible) Möbius transformation maps circles (and
lines) to circles (and lines), so Sc is a circle – or a line when there is some z ∈ T2
for which the denominator z + (1 − 2c) = 0. (Recall that T2 is the unit circle; this
is an ugly notation, but it will do.) In other words, Sc is a line when
1
1
⇐⇒ c ∈ T1 .
|1 − 2c| = 1 ⇐⇒ − c =
2
2
Let’s pause to study an important example.
10
Example 3.11. Suppose that c 6= ±1 and c ∈ T2 . We already know that Sc is a
circle. Direct calculations yield
fc (c) = c,
fc (1) =
3c2 − 1
3c − 1
and fc (−1) =
.
2
2c
Therefore for z ∈ {1, −1, c}
fc (z) − 3 c = 1 ,
2 2
so Sc is a circle of radius 12 centered at 32 c. A similar argument shows that S(2+c)/3
is a circle of radius 12 centered at 2c .
Summarizing: If c 6= ±1 and c ∈ T2 , then Sc is a circle of radius 12 that is
externally tangent to T2 at c and S(2+c)/3 is a circle of radius 12 that is internally
tangent to T2 at c.
Returning to our study of p, let’s record what we have seen in (3.4).6
Theorem 3.12. Suppose p(z) = (z − r1 )(z − r2 )(z − 1) ∈ Γ and 1 6= c ∈ C. Then
p has a critical point at c if and only if fc (r1 ) = r2 .
Now, if c 6= 1, then fc maps T2 onto Sc , and since (fc )−1 = fc , fc also maps Sc
onto T2 . Hence fc restricts to a permutation of Sc ∩ T2 , and if c is a critical point
of p, we have {r1 , r2 } ⊆ Sc ∩ T2 . We can use this fact, given any c 6= 1 in the closed
unit disk, to classify the polynomials p ∈ Γ having a critical point at c.
Because Sc and T2 are circles, there are four cases to consider:
1. Sc and T2 are disjoint;
2. Sc and T2 are tangent;
3. Sc and T2 intersect in two distinct points;
4. Sc = T2 .
In the first case, there can be no p ∈ Γ with a critical point at c, because no
point in C is eligible to be r1 (or r2 ).
In the second case, if Sc ∩ T2 = {r}, then it is necessary that r1 = r = r2 and
p(z) = (z − 1)(z − r)2 . Conversely, when p is of this type, we have seen in Examples
3.5 and 3.11 that Sc ∩ T2 = {r}.
In the third case, we assume Sc ∩ T2 = {a, b} for some a 6= b. Suppose for the
sake of contradiction that fc (a) = a. Since fc is a permutation of Sc ∩ T2 , we have
fc (b) = b. By Theorem 3.12, c is a critical point of pa (z) = (z − a)2 (z − 1) and
of pb (z) = (z − b)2 (z − 1). From Example 3.11, c ∈ {a, (2 + a)/3} ∩ {b, (2 + b)/3}.
However, (2 + a)/3 is on the segment 1a, and similarly (2 + b)/3 ∈ 1b; since a 6= b,
{a, (2 + a)/3} ∩ {b, (2 + b)/3} = ∅ – a contradiction. It follows that fc (a) = b and
6 Referee’s note: (3.4) does not apply when r = 2c − 1. In this exceptional case, substitution
1
of c = (1 + r1 )/2 into the equation p0 (c) = 0 yields 0 = −(r1 − 1)2 /4. Thus r1 = 1, which implies
c = 1.
11
fc (b) = a, and so p(z) = (z − 1)(z − a)(z − b) is the only polynomial with a critical
point at c.
Thus, in each of the first three cases, there is at most one polynomial p ∈ Γ with
a critical point at c.
To handle the last case, suppose Sc = T2 . In this case, whenever |r| = 1, we
have |fc (r)| = 1. In particular, |fc (1)| = 1 = |fc (−1)|. Now,
fc (1)
=
=
=
−3c2 + 4c − 1
2 − 2c
(3c − 1)(c − 1)
2(c − 1)
3c − 1
,
2
where the last equality follows since c 6= 1 by assumption. Set c = x + iy and
simplify |fc1 (1)| = 1 to obtain
y2 =
2
1
4
− x−
.
9
3
(3.5)
2
Likewise, fc (−1) = 3c2c−1 . Again, we write c = x + iy in the known equation
|fc1 (−1)| = 1, which yields7
9(x2 + y 2 )2 − 6(x2 − y 2 ) + 1 = 4(x2 + y 2 ).
7 Referee’s
(3.6)
note:
3(x + iy)2 − 1 (x − iy) 2(x + iy)(x − iy)
3(x + iy)(x2 + y 2 ) − (x − iy)
x[3(x2 + y 2 ) − 1] + iy[3(x2 + y 2 ) + 1]
=
1
=
2(x2 + y 2 )
=
2(x2 + y 2 )
x2 [9(x2 + y 2 )2 − 6(x2 + y 2 ) + 1] + y 2 [9(x2 + y 2 )2 + 6(x2 + y 2 ) + 1]
=
4(x2 + y 2 )2
9(x2 + y 2 )3 − 6(x2 + y 2 )(x2 − y 2 ) + (x2 + y 2 )
=
4(x2 + y 2 )2
=
4(x2 + y 2 )
2
2 2
2
2
9(x + y ) − 6(x − y ) + 1
To confirm that c 6= 0 in this case, note that 0 ∈ T1 , so S0 is a line and hence S0 6= T2 .
12
(x2 + y 2 6= 0)
We use (3.5) to simplify this equation; eventually,8 we get
0 = 3x2 − 2x − 1 = (3x + 1)(x − 1).
Since c 6= 1 by assumption, x = − 31 and (by (3.5)) y = 0. That is, if Sc = T2 , then
c = − 13 .
Conversely, we claim that S−1/3 = T2 . Clearing fractions, f−1/3 (z) = −5z−3
3z+5 ,
and a routine calculation shows that
f−1/3 (i) = −
8
15
− i
17 17
which has modulus 1. Now we know that S−1/3 is a circle containing f−1/3 (1),
f−1/3 (−1), and f−1/3 (i), each of which has modulus 1. This implies the claim.
It follows that if c = −1
3 is a critical point of p, then there is some r 6= 1 on the
unit circle for which p(z) = (z − 1)(z − r) z − f−1/3 (r) . On the other hand, it is
easy to check9 that every such polynomial has a critical point at c = −1
3 .
Assembling our results, we have the following theorem.
Theorem 3.13. Let c ∈ C.
8 Referee’s
note: from (3.5),
9y 2
=
4 − (3x − 1)2
9y
2
=
4 − 9x2 + 6x − 1
9x + 9y
2
=
2
2
=
2
2
=
6x + 3
2x + 1
and
3
2
2
2x − (x + y 2 )
x2 − y 2
=
(2x + 1)2 − 2(6x2 − 2x − 1) + 1
=
2
x +y
x −y
3(4x + 4x + 1) − 6(6x − 2x − 1) + 3
=
6x2 − 2x − 1
,
3
4
(2x + 1)
3
4(2x + 1)
−24x2 + 16x + 8
=
0
2
9 Referee’s
2
so from (3.6),
note:
p(z)
=
p0 (z)
=
p0 (−1/3)
=
p0 (−1/3)
=
p0 (−1/3)
=
p0 (−1/3)
=
5r + 3
(z 2 − z − rz + r) z +
3r + 5
5r + 3
(2z − 1 − r) z +
+ (z 2 − z − rz + r)
3r + 5
−5
−1
5r + 3
1
1
−r
+
+
+ (1 + r) + r
3
3
3r + 5
9
3
−1
−1
5r + 3
4
4
(5 + 3r)
+
+
+ r
3
3
3r + 5
9
3
5
1
1
4
4
+ r − (5r + 3) +
+ r
9
3
3
9
3
0
13
Figure 5: If 1 6= c1 ∈ T1 , then c2 = c1 (Theorem 3.14).
• If c 6∈ {1, − 13 }, there is at most one p ∈ Γ with a critical point at c.
• If c lies strictly inside T2/3 , or strictly outside T2 , then there is no p ∈ Γ with
a critical point at c.
• p ∈ Γ has a critical point at 1 if and only if p(z) = (z − 1)2 (z − r) for some r
on the unit circle.
5r+3
• p ∈ Γ has a critical point at −1
if
and
only
if
p(z)
=
(z
−
1)(z
−
r)
z
+
3
3r+5
for some r on the unit circle.
As an application of Theorem 3.13, we can give an independent proof of Saff
and Twomey’s theorem that c2 = c1 when c1 , c2 are critical points of p ∈ Γ and
1 6= c1 ∈ T1 : see Figure 5.
Theorem 3.14 ([13]). Let c1 and c2 be the critical points of p ∈ Γ. If 1 6= c1 ∈ T1 ,
then c2 = c1 .
Proof. Let c1 = x + iy ∈ T1 .
We pull a rabbit out of a hat: let r = eiθ , where cos θ = 32 x − 21 ∈ −1
2 ,1 .
Let q(z) = (z − 1)(z − r) (z − r) ∈ Γ. Then q(z) = z 3 − Az 2 + Az − 1, where
A = 1 + 2 cos θ = 3x. Differentiating, q 0 (z) = 3 z 2 − 2xz + x .
2
Observe now that since c1 ∈ T1 , we have x − 21 + y 2 = 14 , and so c1 c1 =
x2 + y 2 = x, while of course c1 + c1 = 2x. Consequently, q 0 (z) = 3(z − c1 )(z − c1 ).
Now q ∈ Γ has critical points c1 and c2 = c1 . Then, by uniqueness (Theorem
3.13), q = p, completing the proof.
14
The calculation works for a geometric reason, as we see in Figure 5: in an
isosceles triangle, the axes of the Steiner ellipse are parallel and perpendicular to
the base. This, together with the uniqueness from Theorem 3.13, is the idea of the
proof.
1
lies
We can still improve on
Theorem
3.13:
it
remains
to
show
that
if
c
∈
6
1,
−
3
2
on Tα for some α ∈ 3 , 2 , then |Sc ∩ T2 | = 2, so that there exists some (necessarily
unique) p ∈ Γ with a critical point at c. Suppose to the contrary that no such p ∈ Γ
exists. Then Sc and T2 are disjoint. So Sc lies either entirely inside or entirely
outside the circle T2 . Without loss of generality we will assume that Sc lies inside
T2 .
We will study how Sc changes as we ‘drag’ c 6= 1 through X := {z : z ∈ Tα , α ∈
( 23 , 2]}. Define a path γ : [0, 1] → X with γ(0) = c, γ(1) ∈ T2 and γ(t) ∈ Int(X)
for t < 1. Since Sγ(1) is externally tangent to T2 (see Example 3.11) and Sγ(0)
lies inside T2 , by continuity, there must be a t0 ∈ (0, 1) where Sγ(t0 ) is internally
tangent to T2 . But then γ(t0 ) ∈ T2/3 , a contradiction.
We can now strengthen the first part of Theorem 3.13.
Theorem 3.15. If c 6∈ {1, − 13 } lies on Tα for some α ∈ [ 32 , 2], then there is a
unique p ∈ Γ with a critical point at c.
4
Conclusions
We began in Section 1 by thinking of a cubic polynomial as a triangle, inscribing
it in a circle, and choosing coordinates in such a way that the polynomial belonged
to Γ. Our results can be pulled back to this general context. Let’s emphasize the
geometry behind these results in a pair of summary theorems.
Theorem 4.1. The triangle 4ABC can be constructed given a vertex and any
two of the following five points: the other vertices, the centroid, and the foci of the
Steiner ellipse.
Theorem 4.2. The triangle 4ABC can be constructed given the circumcircle, a
vertex, and either the centroid or a focus of the Steiner ellipse.
We have shown in previous sections how various sets of data might determine
a polynomial, but we have not insisted on constructibility, so for completeness, let
us indicate the ideas on which these constructions are founded. Recall that it is
possible to construct sums, differences, products, quotients, and square roots of
known points in the complex plane.
The next four constructions prove Theorem 4.1.
Construction 4.3. Construct 4ABC, given vertex A and the foci F1 and F2 of
the Steiner ellipse.
We work backwards using Marden’s theorem and the quadratic formula. Treating A, F1 , and F2 as points in the complex plane, construct σ1 = A − 23 (F1 + F2 )
√
−σ1 ± σ12 −4σ2
and σ2 = r1 σ1 + 3F1 F2 . Then construct B and C at
.
X
2
15
Construction 4.4. Construct 4ABC given vertex A, vertex B, and centroid G.
Construct C = 3G − A − B.
X
Construction 4.5. Construct 4ABC given vertex A, vertex B, and a focus F of
the Steiner ellipse.
2
(A+B)+AB
.
X
As in (3.4), construct C = 3F −2F
2F −A−B
Construction 4.6. Construct 4ABC given vertex A, centroid G, and a focus F1
of the Steiner ellipse.
Let F2 be the reflection of F1 over G. Apply construction 4.3 with the points
F1 and F2 to find B and C.
X
This completes the proof of Theorem 4.1; with the next two constructions, we
prove Theorem 4.2.
Construction 4.7. Construct 4ABC given vertex A, circumcircle S, and centroid G.
We proceed as in Theorem 2.3. Construct the point O at the center of S. Draw
line L through A and G, and let M be the midpoint of AG. With center G, draw a
circle S 0 of radius AM . Let S 0 intersect L at the point X 6= M . Draw line XO and
construct the line L0 perpendicular to XO through X. Then L0 intersects S at the
points B and C.
X
In order to complete the proof of Theorem 4.2 we need to show that we can
construct 4ABC given a vertex, circumcircle and a focus of the Steiner ellipse. In
order to understand the main idea of the construction, let’s make the connection to
section 3.2. With this in mind, assume that A = 1, S = T2 is the circumcircle and
F ∈
/ {1, 31 } is a focus of the Steiner ellipse. If X is any point on S other than A = 1,
use Construction 4.5 to construct a point X 0 such that 4AXX 0 has a Steiner ellipse
with focus at F . Construction 4.5 does not guarantee that X 0 lies on S, so this
may not be the desired triangle. However, as we let X trace out the circle S, X 0
traces out the circle SF . Therefore, since F ∈
/ {1, 31 } is the focus of a Steiner ellipse,
S ∩ SF = {B, C}.
Construction 4.8. Construct 4ABC given vertex A, circumcircle S, and a focus
F of the Steiner ellipse.
Let distinct points X, Y , and Z, different from A, be given on the circle S.
Following Construction 4.5, construct points X 0 , Y 0 , and Z 0 such that 4AXX 0 ,
4AY Y 0 and 4AZZ 0 each have a Steiner ellipse with a focus at F . Construct S 0 to
be the circumcircle of 4X 0 Y 0 Z 0 . Then S 0 and S intersect at the points B and C.
X
This completes the last of our proofs, but, as at the end of every paper, some
questions remain unanswered. It would be especially nice to learn more about
polynomials of higher degree. Preliminary results suggest that some subset of the
polynomials of the form p(z) = (z − 1)j (z − r1 )k (z − r2 )` , with {r1 , r2 } ⊆ T2 and
{j, k, `} ⊆ N, should be amenable to the same type of analysis. For example, if
p(z) = (z − 1)(z − r1 )k (z − r2 )` , we have the following critical points:
16
• c1 = r1 with multiplicity k − 1;
• c2 = r2 with multiplicity ` − 1;
• two non-trivial critical points, c3 ∈ Tα and c4 ∈ Tβ .
The analogue of (3.3) in this case is
1
k+`
1
+ =1+
.
α β
2
We are sure that much more is waiting to be discovered.
Acknowledgment: We are grateful to Ryan Knuesel for his work on the software
that first helped us visualize the trajectories of the critical points of polynomials
p ∈ Γ.
References
[1]
Bruce Anderson, Polynomial root dragging, Amer. Math. Monthly 100 (1993), no. 9, 864–866.
MR1247536 (94i:26006)
[2]
Matthew Boelkins, Justin From, and Samuel Kolins, Polynomial root squeezing, Math. Mag.
81 (2008), no. 1, 39–44. MR2380056
[3]
Christopher Frayer, More Polynomial Root Squeezing, Mathematics Magazine 83 (2010),
no. 3, 218–222.
[4]
, Squeezing polynomial roots a nonuniform distance, Missouri J. Math. Sci. 22 (2010),
no. 2, 124–129. MR2675407 (2011e:26016)
[5]
Christopher Frayer and James A. Swenson, Polynomial root motion, Amer. Math. Monthly
117 (2010), no. 7, 641–646, DOI 10.4169/000298910X496778. MR2681525
[6]
Pamela Gorkin and Elizabeth Skubak, Polynomials, ellipses, and matrices: two
questions, one answer, Amer. Math. Monthly 118 (2011), no. 6, 522–533, DOI
10.4169/amer.math.monthly.118.06.522. MR2812283
[7]
Dan Kalman, An elementary proof of Marden’s theorem, Amer. Math. Monthly 115 (2008),
no. 4, 330–338. MR2398412 (2009c:30018)
[8]
, The Most Marvelous Theorem in Mathematics, Journal of Online Mathematics and
its Applications 8 (April 2008).
[9]
H. C. Kennedy, Classroom Notes: Who Discovered Boyer’s Law?, Amer. Math. Monthly 79
(1972), no. 1, 66–67, DOI 10.2307/2978134. MR1536593
[10] Albert Marden and Philip Marden, Morris Marden (1905-1991), Complex Variables 26
(1994), 183–186. http://www4.uwm.edu/letsci/math/alumni/history/marden.cfm
[11] Morris Marden, Geometry of polynomials, Second edition. Mathematical Surveys, No. 3,
American Mathematical Society, Providence, R.I., 1966. MR0225972 (37 #1562)
[12] D. Minda and S. Phelps, Triangles, ellipses, and cubic polynomials, Amer. Math. Monthly
115 (2008), no. 8, 679–689. MR2456092 (2009h:12003)
[13] E. B. Saff and J. B. Twomey, A note on the location of critical points of polynomials, Proc.
Amer. Math. Soc. 27 (1971), 303–308. MR0271312 (42 #6195)
[14] Eric W. Weisstein, Inversion, MathWorld – A Wolfram Web Resource (January 4, 2012).
http://mathworld.wolfram.com/Inversion.html
17