The Geometry of Cubic Polynomials

Christopher Frayer Miyeon Kwon Christopher SchafhauserJames 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 thatthe critical points are the foci of the Steiner ellipse of 41r1r2. In this paperwe further explore the structure of these critical points. If we let Tγ be thecircle 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. Weshow that Tβ is the inversion of Tα over T1, from which many nice geometricconsequences can be drawn. For example, (1) there is a “desert” in the unitdisk, the open disk {z ∈ C : |z − 2

3| < 1

3}, in which critical points cannot

occur, and (2) a critical point of such a polynomial almost always determinesthe polynomial uniquely.

1 Introduction

In 2009, Dan Kalman won a Lester R. Ford Award for [7], published in thisMonthly. In this outstanding article, Kalman gave a new proof of Marden’stheorem,1 which he has called “the most marvelous theorem in mathematics.” [8]

Theorem 1.1 (Marden’s theorem). Given a triangle 4r1r2r3 in the complex plane,there is a unique inscribed ellipse, called the Steiner ellipse, which is tangent to thesides of the triangle at the midpoints of the sides. If

p(z) = (z − r1)(z − r2)(z − r3),

then the roots of p′(z) are the foci of the Steiner ellipse of 4r1r2r3, and the root ofp′′(z) is the centroid of 4r1r2r3.

This is indeed a marvelous theorem. What we find so attractive about it is theconnection 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 complexpolynomial was the Gauss-Lucas theorem, which guarantees that the critical points

1In accordance with Boyer’s Law [9], Marden’s theorem is originally due to J. Siebeck, accordingto [8, 11,12].

1

of any polynomial lie in the complex hull of its roots. For us, it was a specialpleasure to learn from [10] that the theorem’s namesake, Morris Marden, had helpedto found the mathematical research program at our sister school, the University ofWisconsin-Milwaukee, as well as our section of the MAA.

Kalman’s outstanding paper got us excited about cubic polynomials – partic-ularly 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 itsroots, in the case where all roots are real.

We were now curious. What happens to the critical points of a complex cubicpolynomial when its roots move? Hands-on investigation was irresistible: we drewsome figures in GeoGebra,2 set the roots in motion, and watched the critical pointsmove.

At this point we had a decision to make. On the real line, one has relatively fewoptions: 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 gotlucky. Through any three given points r1, r2, r3 in the complex plane, there is aunique circle. By changing coordinates, we can take this circle to be the unit circlewith r3 = 1, so that |rk| = 1 for each k. We set r1 and r2 in motion around theunit circle at different speeds, and traced the loci of the critical points. This articleis the result of that fortunate decision.

We were surprised to see that as the roots varied, the trajectories of the criti-cal 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 againsurprised: a critical point of such a polynomial almost always determines the poly-nomial 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 Mardenwould have enjoyed.

2 Centers

Let p(z) be a cubic polynomial with its roots on the unit circle in C. The roots ofp(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|.2GeoGebra 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 collectionof GeoGebra notebooks illustrating our main results can be found at [http://www.uwplatt.edu/∼swensonj/gocp/].

2

Before studying the zeros of p′(z), let’s take a moment to investigate the zerosof p′′(z). We’ll find some pretty geometry, and obtain Theorem 2.3, which gives usa 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 center3 of p(z) providedp′′(g) = 0.

Of course every p ∈ Γ has a unique center; recall that by Marden’s theorem, thisis the centroid of 4r1r21. Interestingly, though, we can prove a sort of converse tothis fact!

Theorem 2.3. Let g ∈ C.

• p ∈ Γ has center 13 if and only if p(z) = (z − 1)(z2 − r2) for some r with

|r| = 1.

• If 0 <∣∣g − 1

3

∣∣ ≤ 23 , then there is a unique polynomial p ∈ Γ with center g.

• If∣∣g − 1

3

∣∣ > 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 thedashed circle in Figure 1. Conversely, almost every point in that closed disk is thecenter of a unique p ∈ Γ – but the disk’s center is the center of each polynomial ina family parametrized by a circle.

Proof. By Marden’s theorem, this claim is related to the construction of a trianglegiven a vertex, the centroid, and the circumcircle.

Suppose g is the center of p(z) ∈ Γ. Then g is the centroid of triangle 41r1r2,where r1 and r2 are to be constructed.

Though we do not know where r1 and r2 are, let us denote the midpoint of r1r2by w. Of course, w lies in the closed unit disk. The segment 1w is a median of4r1r21, so g = 2

3 (w) + 13 (1). It is immediate that

∣∣g − 13

∣∣ ≤ 23 .

Consequently, to begin the construction, we solve the previous equation for wand let w = 3g−1

2 . (In geometric terms, given g, we may construct the midpoint w′

of 1g; then w is the reflection of w′ over g.)Assume first that g 6= 1

3 , so w 6= 0. We know that r1r2 is a chord of theunit circle, so its perpendicular bisector must pass through 0 – and through themidpoint w.

Hence we construct the line L through w, perpendicular to 0w. Since w liesin the closed unit disk, L intersects the unit circle in two points (counting withmultiplicity 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 r1r2, we have r2 = −r1. Conversely, in this case we can compute directly that 13

is the center of p(z) = z3 − z2 − r21z + r21.

3Calling this an inflection point might be better, but we just couldn’t do it. Perhaps we’vespent too much time telling our first-semester calculus students, “Even when f ′′(a) = 0, thatdoesn’t necessarily mean a is an inflection point of f .”

3

Figure 1: Construction of 4r1r21 given center g (Theorem 2.3)

3 Critical points

3.1 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 thecritical points determine p?

Regarding the first question, Saff and Twomey show in [13] that p has at leastone critical point in the closed disk ∆ =

{z ∈ C :

∣∣z − 12

∣∣ ≤ 12

}. Moreover, they

show that if c1 and c2 are the critical points of p, and if c1 6= 1 is on the boundaryof ∆, then c2 is the complex conjugate of c1 and hence also lies on the boundaryof ∆.

To extend the result of Saff and Twomey, we will take advantage of a delightfulgeometric symmetry. To explain this, we need some notation.

Definition 3.1. Given α > 0, we denote by Tα the circle of diameter α that passesthrough 1 and 1− α in the complex plane.4 That is,

Tα ={z ∈ C :

∣∣∣z − (1− α

2

)∣∣∣ =α

2

}.

See Figure 2 below. For example, we have seen in Theorem 2.3 that the centerof p ∈ Γ cannot lie outside T4/3.

4It may seem strange to name the circle by its diameter, rather than by its radius – but thisnotation 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

α= Re

(1

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

1− z=

1

reiθ

Re

(1

1− z

)=

cos θ

r.

The angle inscribed in Tα at z intercepts a diameter and must be a right angle.Therefore, right-triangle trigonometry gives cos θ = r

α , completing the proof.

Corollary 3.3. The complex number z 6= 1 lies in the closed unit disk if and onlyif 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 adouble 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 p′(z) = q(z) + (z− 1)q′(z).

5

Substituting z = 1, we have 0 = q(1), as we claimed. Explicitly, there is some r onthe unit circle such that

p(z) = (z − 1)2(z − r)

and

p′(z) = 3z2 − (2r + 4)z + (2r + 1)

= 3(z − 1)

(z − 2r + 1

3

).

Summarizing: p ∈ Γ has a critical point at 1 if and only if p(z) = (z− 1)2(z− r) forsome r on the unit circle, and in this case the other critical point is 1

3 + 23r ∈ T4/3.

Example 3.5. Suppose now that p ∈ Γ has a critical point at some r 6= 1 on theunit 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 adouble root at r, so:

p(z) = (z − r)2(z − 1)

and

p′(z) = 3z2 − (4r + 2)z + (r2 + 2r)

= 3(z − r)(z − r + 2

3

).

As above, we summarize: p ∈ Γ has a critical point at r 6= 1 on the unit circleif and only if p(z) = (z − r)2(z − 1), and in this case the other critical point is23 + 1

3r ∈ T2/3.

Now consider a general p ∈ Γ. By the Gauss-Lucas theorem, both critical pointsof p will lie in the closed unit disk. We are ready to prove our symmetry result forthe critical points of p; this is our main theorem, from which many nice consequenceswill be drawn. We prove a more general statement than we need, because it is noharder 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. Thenn∑k=1

1

1− ck= 2

n∑k=1

1

1− zk

and

n∑k=1

1

αk= n. (3.2)

6

Proof. We prove (3.2) by evaluating Re(f ′′(1)f ′(1)

)in two different ways.

To begin with, f ′(z) = (n+ 1)∏nk=1 (z − ck). By logarithmic differentiation,

f ′′(z)

f ′(z)=

n∑k=1

1

z − ck

f ′′(1)

f ′(1)=

n∑k=1

1

1− ck

Re

(f ′′(1)

f ′(1)

)= Re

(n∑k=1

1

1− ck

)

Re

(f ′′(1)

f ′(1)

)=

n∑k=1

1

αk,

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 ′(z) = (z − 1)g′(z) + g(z)

f ′′(z) = (z − 1)g′′(z) + 2g′(z)

f ′′(1) = 2g′(1)

f ′′(1)

f ′(1)=

2g′(1)

g(1)

f ′′(1)

f ′(1)= 2

n∑k=1

1

1− zk,

where the last step is by logarithmic differentiation of g(z). But since zk ∈ T2, wecan apply Theorem 3.2 to obtain

Re

(f ′′(1)

f ′(1)

)= 2

n∑k=1

Re

(1

1− zk

)= 2

n∑k=1

1

2= n.

Corollary 3.7. Let p ∈ Γ, and let c1 6= 1 and c2 6= 1 be the critical points of p. Ifc1 lies on Tα and c2 lies on Tβ, then

1

α+

1

β= 2. (3.3)

Recall that we have already seen (in Example 3.4) what happens when c1 = 1or c2 = 1. Notice that (3.3) holds in Example 3.5, where {α, β} =

{2, 23}

.

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 > 1

2 . Since T1 and Tα are symmetric across the real axis and passthrough 1, the same is true of the inversion of Tα across T1. This being the case, it

5Referee’s note: Suppose first that α > 12

; we invert Tα over T1 using formulæ given in [14]. T1

has center (x0, y0) =(12, 0

)and radius k = 1

2, while Tα has center (x, y) =

(1− α

2, 0

)and radius

a = α2

. In terms of

s =k2

(x− x0)2 + (y − y0)2 − a2=

1

1− 2α,

the inversion of Tα over T1 has radius

r = |s|a =

∣∣∣∣ 1

1− 2α

∣∣∣∣ · α2 =α

2(2α− 1)

and center

(x0+s(x−x0), y0+s(y−y0)) =

(1

2+

1

1− 2α

(1

2−α

2

), 0

)=

(1 +

1

2

(1− α1− 2α

− 1

), 0

)= (1−r, 0).

That is, the inversion of Tα over T1 is Tβ , where the diameter β = 2r = α2α−1

, so that

1

α+

1

β=

1

α+

2α− 1

α= 2,

as desired.On the other hand, if α = 1

2, then the inversion of Tα is the vertical line Re(z) = 1; if α < 1

2,

then the inversion of Tα lies to the right of this line. Either way, it is impossible for a critical pointof p ∈ Γ to lie on the inversion of Tα, by the Gauss-Lucas theorem. Neither can a critical pointlie on Tβ with 1

α+ 1

β= 2, since then 1

β= 2− 1

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 14 , the square of the radius of T1. In fact, this is true, because (3.3)

gives β = α2α−1 , and then∣∣∣∣(1− α)− 1

2

∣∣∣∣ · ∣∣∣∣(1− β)− 1

2

∣∣∣∣ =

∣∣∣∣2α− 1

2· 1

2(2α− 1)

∣∣∣∣ =1

4.

Expressed another way, (3.3) says that the radius of the unit circle is the har-monic mean of the diameters of Tα and Tβ . In this form, the statement generalizesto an arbitrary coordinate system: given 4ABC in the complex plane, let Tα andTβ be the circles tangent to the circumcircle at A that pass through the foci of theSteiner ellipse. The radius of the circumcircle then equals the harmonic mean ofthe 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 original observation: there is a “desert” in the unitdisk, the open disk

{z ∈ C :

∣∣z − 23

∣∣ < 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 othercritical point of p lies on Tβ , where (by Corollary 3.7) 1

β = 2− 1α < 2− 3

2 = 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 criticalpoint 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. Ifc1 lies on T1, then c2 also lies on T1. Otherwise, c1 and c2 are on opposite sidesof T1.

Proof. Let c1 ∈ Tα and c2 ∈ Tβ . Then 1α + 1

β = 2, so α = 1 if and only if β = 1 andα < 1 if and only if β > 1.

3.2 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 pointof p. Then

p(z) = (z − r1)(z − r2)(z − 1)

= z3 − (r1 + r2 + 1)z2 + (r1 + r2 + r1r2)z − r1r2so that

p′(z) = 3z2 − 2(r1 + r2 + 1)z + (r1 + r2 + r1r2)

and

0 = 3c2 − 2c(r1 + r2 + 1) + (r1 + r2 + r1r2).

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 Mobius transformation

fc(z) =(2c− 1)z + (2c− 3c2)

z + (1− 2c).

We let Sc denote the image of the unit circle under fc.

A Mobius transformation is a function of the form f(z) = az+bcz+d . When ad−bc 6=

0, one has f−1(z) = dz−b−cz+a . In the case of fc, we have ad− bc = −(c− 1)2, so when

c 6= 1, we have (fc)−1 = fc. [In the exceptional case, we have f1(z) = z−1

z−1 = 1

when z 6= 1, and of course (f1)−1 does not exist.]It is well-known that any (invertible) Mobius transformation maps circles (and

lines) to circles (and lines), so Sc is a circle – or a line when there is some z ∈ T2for which the denominator z + (1− 2c) = 0. (Recall that T2 is the unit circle; thisis an ugly notation, but it will do.) In other words, Sc is a line when

|1− 2c| = 1 ⇐⇒∣∣∣∣12 − c

∣∣∣∣ =1

2⇐⇒ c ∈ T1.

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 acircle. Direct calculations yield

fc(c) = c, fc(1) =3c− 1

2and fc(−1) =

3c2 − 1

2c.

Therefore for z ∈ {1,−1, c} ∣∣∣∣fc(z)− 3

2c

∣∣∣∣ =1

2,

so Sc is a circle of radius 12 centered at 3

2c. A similar argument shows that S(2+c)/3

is a circle of radius 12 centered at c

2 .Summarizing: If c 6= ±1 and c ∈ T2, then Sc is a circle of radius 1

2 that isexternally tangent to T2 at c and S(2+c)/3 is a circle of radius 1

2 that is internallytangent 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. Thenp 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 pointof p, we have {r1, r2} ⊆ Sc ∩T2. We can use this fact, given any c 6= 1 in the closedunit 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 nopoint 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 andp(z) = (z−1)(z−r)2. Conversely, when p is of this type, we have seen in Examples3.5 and 3.11 that Sc ∩ T2 = {r}.

In the third case, we assume Sc ∩ T2 = {a, b} for some a 6= b. Suppose for thesake of contradiction that fc(a) = a. Since fc is a permutation of Sc ∩ T2, we havefc(b) = b. By Theorem 3.12, c is a critical point of pa(z) = (z − a)2(z − 1) andof 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

6Referee’s note: (3.4) does not apply when r1 = 2c − 1. In this exceptional case, substitutionof c = (1 + r1)/2 into the equation p′(c) = 0 yields 0 = −(r1 − 1)2/4. Thus r1 = 1, which impliesc = 1.

11

fc(b) = a, and so p(z) = (z − 1)(z − a)(z − b) is the only polynomial with a criticalpoint at c.

Thus, in each of the first three cases, there is at most one polynomial p ∈ Γ witha critical point at c.

To handle the last case, suppose Sc = T2. In this case, whenever |r| = 1, wehave |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 andsimplify |fc1(1)| = 1 to obtain

y2 =4

9−(x− 1

3

)2

. (3.5)

Likewise, fc(−1) = 3c2−12c . Again, we write c = x + iy in the known equation

|fc1(−1)| = 1, which yields7

9(x2 + y2)2 − 6(x2 − y2) + 1 = 4(x2 + y2). (3.6)

7Referee’s note: ∣∣∣∣∣(3(x+ iy)2 − 1

)(x− iy)

2(x+ iy)(x− iy)

∣∣∣∣∣ = 1∣∣3(x+ iy)(x2 + y2)− (x− iy)∣∣ = 2(x2 + y2)∣∣x[3(x2 + y2)− 1] + iy[3(x2 + y2) + 1]∣∣ = 2(x2 + y2)

x2[9(x2 + y2)2 − 6(x2 + y2) + 1] + y2[9(x2 + y2)2 + 6(x2 + y2) + 1] = 4(x2 + y2)2

9(x2 + y2)3 − 6(x2 + y2)(x2 − y2) + (x2 + y2) = 4(x2 + y2)2

9(x2 + y2)2 − 6(x2 − y2) + 1 = 4(x2 + y2) (x2 + y2 6= 0)

To confirm that c 6= 0 in this case, note that 0 ∈ T1, so S0 is a line and hence S0 6= T2.

12

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 = − 13 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−33z+5 ,

and a routine calculation shows that

f−1/3(i) = −15

17− 8

17i

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 = −13 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 = −13 .

Assembling our results, we have the following theorem.

Theorem 3.13. Let c ∈ C.

8Referee’s note: from (3.5),

9y2 = 4− (3x− 1)2

9y2 = 4− 9x2 + 6x− 1

9x2 + 9y2 = 6x+ 3

x2 + y2 =2x+ 1

3and

x2 − y2 = 2x2 − (x2 + y2)

x2 − y2 =6x2 − 2x− 1

3, so from (3.6),

(2x+ 1)2 − 2(6x2 − 2x− 1) + 1 =4

3(2x+ 1)

3(4x2 + 4x+ 1)− 6(6x2 − 2x− 1) + 3 = 4(2x+ 1)

−24x2 + 16x+ 8 = 0

9Referee’s note:

p(z) = (z2 − z − rz + r)

(z +

5r + 3

3r + 5

)p′(z) = (2z − 1− r)

(z +

5r + 3

3r + 5

)+ (z2 − z − rz + r)

p′(−1/3) =

(−5

3− r

)(−1

3+

5r + 3

3r + 5

)+

(1

9+

1

3(1 + r) + r

)p′(−1/3) =

−1

3(5 + 3r)

(−1

3+

5r + 3

3r + 5

)+

(4

9+

4

3r

)p′(−1/3) =

5

9+

1

3r −

1

3(5r + 3) +

(4

9+

4

3r

)p′(−1/3) = 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 witha 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 ron the unit circle.

• p ∈ Γ has a critical point at −13 if and only if p(z) = (z−1)(z− r)(z + 5r+3

3r+5

)for some r on the unit circle.

As an application of Theorem 3.13, we can give an independent proof of Saffand Twomey’s theorem that c2 = c1 when c1, c2 are critical points of p ∈ Γ and1 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 =

(32x−

12

)∈[−1

2 , 1].

Let q(z) = (z − 1)(z − r) (z − r) ∈ Γ. Then q(z) = z3 − Az2 + Az − 1, whereA = 1 + 2 cos θ = 3x. Differentiating, q′(z) = 3

(z2 − 2xz + x

).

Observe now that since c1 ∈ T1, we have(x− 1

2

)2+ y2 = 1

4 , and so c1c1 =x2 + y2 = x, while of course c1 + c1 = 2x. Consequently, q′(z) = 3(z − c1)(z − c1).

Now q ∈ Γ has critical points c1 and c2 = c1. Then, by uniqueness (Theorem3.13), q = p, completing the proof.

14

The calculation works for a geometric reason, as we see in Figure 5: in anisosceles triangle, the axes of the Steiner ellipse are parallel and perpendicular tothe base. This, together with the uniqueness from Theorem 3.13, is the idea of theproof.

We can still improve on Theorem 3.13: it remains to show that if c 6∈{

1,− 13

}lies

on Tα for some α ∈(23 , 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 ∈ Γexists. Then Sc and T2 are disjoint. So Sc lies either entirely inside or entirelyoutside the circle T2. Without loss of generality we will assume that Sc lies insideT2.

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 internallytangent 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 ∈ [ 23 , 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, inscribingit in a circle, and choosing coordinates in such a way that the polynomial belongedto Γ. Our results can be pulled back to this general context. Let’s emphasize thegeometry behind these results in a pair of summary theorems.

Theorem 4.1. The triangle 4ABC can be constructed given a vertex and anytwo of the following five points: the other vertices, the centroid, and the foci of theSteiner ellipse.

Theorem 4.2. The triangle 4ABC can be constructed given the circumcircle, avertex, and either the centroid or a focus of the Steiner ellipse.

We have shown in previous sections how various sets of data might determinea polynomial, but we have not insisted on constructibility, so for completeness, letus indicate the ideas on which these constructions are founded. Recall that it ispossible to construct sums, differences, products, quotients, and square roots ofknown 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 ofthe Steiner ellipse.

We work backwards using Marden’s theorem and the quadratic formula. Treat-ing A, F1, and F2 as points in the complex plane, construct σ1 = A − 3

2 (F1 + F2)

and σ2 = r1σ1 + 3F1F2. Then construct B and C at−σ1±

√σ21−4σ2

2 . X

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 ofthe Steiner ellipse.

As in (3.4), construct C = 3F 2−2F (A+B)+AB2F−A−B . X

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, weprove Theorem 4.2.

Construction 4.7. Construct 4ABC given vertex A, circumcircle S, and cen-troid G.

We proceed as in Theorem 2.3. Construct the point O at the center of S. Drawline L through A and G, and let M be the midpoint of AG. With center G, draw acircle S′ of radius AM . Let S′ intersect L at the point X 6= M . Draw line XO andconstruct the line L′ perpendicular to XO through X. Then L′ intersects S at thepoints B and C. X

In order to complete the proof of Theorem 4.2 we need to show that we canconstruct 4ABC given a vertex, circumcircle and a focus of the Steiner ellipse. Inorder to understand the main idea of the construction, let’s make the connection tosection 3.2. With this in mind, assume that A = 1, S = T2 is the circumcircle andF /∈ {1, 13} 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 ′ such that4AXX ′ has a Steiner ellipsewith focus at F . Construction 4.5 does not guarantee that X ′ lies on S, so thismay not be the desired triangle. However, as we let X trace out the circle S, X ′

traces out the circle SF . Therefore, since F /∈ {1, 13} is the focus of a Steiner ellipse,S ∩ SF = {B,C}.

Construction 4.8. Construct 4ABC given vertex A, circumcircle S, and a focusF 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 ′, Y ′, and Z ′ such that 4AXX ′,4AY Y ′ and 4AZZ ′ each have a Steiner ellipse with a focus at F . Construct S′ tobe the circumcircle of 4X ′Y ′Z ′. Then S′ 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, somequestions remain unanswered. It would be especially nice to learn more aboutpolynomials of higher degree. Preliminary results suggest that some subset of thepolynomials 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, ifp(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

α+

1

β= 1 +

k + `

2.

We are sure that much more is waiting to be discovered.

Acknowledgment: We are grateful to Ryan Knuesel for his work on the softwarethat first helped us visualize the trajectories of the critical points of polynomialsp ∈ Γ.

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