HANOI PEDAGOGICAL UNIVERSITY 2
DEPARTMENT OF MATHEMATICS
———————–

PHAN THI THUY HANG

GEOMETRY OF CUBIC
POLYNOMIALS

BACHELOR THESIS
Major: Geometry

Hanoi - 2019


HANOI PEDAGOGICAL UNIVERSITY 2
DEPARTMENT OF MATHEMATICS
———————–

PHAN THI THUY HANG

GEOMETRY OF CUBIC
POLYNOMIALS

BACHELOR THESIS
Major: Geometry
Supervisor: Dr.Nguyen Tat Thang

Hanoi - 2019

Thesis Declarations
I hereby declare that the data and the results of this thesis are true and not identical
to other theses. I also assure that all the help for this thesis has been acknowledged
and that the results presented in the thesis has been identified clearly.
Ha Noi, May 5, 2019
Student

Phan Thi Thuy Hang


Thesis Acknowledgments
Firstly, I would like to take this opportunity to thank to all teachers of the
Department of Mathematics, Hanoi Pedagogical University No.2, the teachers in the
geometry group as well as the teachers involved. The lecturers have imparted valuable
knowledge and facilitated for me to complete the course and the thesis.
In particular,I would like to express my deep respect and sincere gratitude to
my supervisor Dr.Nguyen Tat Thang for the continuous support of my study as well
as related research, for his patience, motivation and immense knowledge. Without his
precious guidance in all the time of research, it would not be possible to complete this
thesis.
Due to time, capacity and conditions are limited, so the thesis cannot avoid
errors. Therefore, I look forward to receiving valuable comments and recommendations
from teachers and friends.

Ha Noi, May 5, 2019
Student

Phan Thi Thuy Hang



Contents

Preface

1

1 Preliminaries

3

1.1

Complex numbers . . . . . . . . . . . . . . . . . . . . . .

3

1.2

Barycentric coordinates . . . . . . . . . . . . . . . . . .

10

1.3

Affine transformations . . . . . . . . . . . . . . . . . . .

12

2 Geometry of cubic polynomials


15

2.1

Marden’s Projection . . . . . . . . . . . . . . . . . . . .

19

2.2

Projection of a circle is an ellipse . . . . . . . . . . . . .

21

2.3

Linear Map . . . . . . . . . . . . . . . . . . . . . . . . .

25

2.4

Siebeck - Marden Theorem . . . . . . . . . . . . . . . . .

28

3 A geometric proof of Siebeck-Marden theorem

34


3.1

Geometric properties of inellipses . . . . . . . . . . . . .

34

3.2

Inellipse and critical points of logarithmic potentials . . .

39

Conclusion

42

Bibliography

43


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PHAN THI THUY HANG

Preface
Gauss – Lucas theorem gives a relation between the roots of a polynomial with complex coefficients and the roots of its derivative which states
that the roots of the derivative polynomial are in the convex hull of the
roots of the initial polynomial. For third degree polynomials, Siebeck
– Marden theorem provides us the geometric connection between the

roots of a third degree polynomial and the roots of its derivative in a geometrical way. This geometric connection was first observed by Siebeck
(1864) and was reproved by Marden (1945). So in order to give credit
to Siebeck, who gave the intial proof, we call this result Siebeck - Marden theorem in this thesis. There has been substantial interest in this
result, and the importance as well as the effect of complex numbers in
the study of the plane geometry. From the reason above, I choose the
topic ”Geometry of cubic polynomials ” for my bachelor thesis.
Our main aim in this thesis is to give the geometrical proof of Siebeck
– Marden theorem. We use complex numbers in the study of the geometric properties of the polynomials. The content of the thesis is as
follows:
In chapter 1, we will review some mathematical knowledge needed for
the results of this thesis. Firstly, we begin with the complex numbers.
Then we define the barycentric coordinates and the proposition relating
barycentric coordinates of a point in the interior of a triangle. Finally,
we will recall some properties of the affine transformations.
In chapter 2, we will use complex numbers to prove Siebeck-Marden
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PHAN THI THUY HANG

theorem. Namely, we will recall the fundamental theorem of algebra
and represent Gauss-Lucas theorem. Then we will introduce Marden’s
project to give the relation between the Gauss-Lucas theorem and SiebeckMarden theorem. Next, we also represent the projection of a circle and
the linear map that are needed to prove Siebeck-Marden theorem.
In chapter 3, we will give a geometric proof of Seibeck-Marden theorem by representing the geometric properties of inellipses. We represent
inellipses and critical points of logarithmic potentials.

2



Chapter 1
Preliminaries
1.1

Complex numbers

Definition 1.1. Consider the complex number z = x + iy(x, y ∈ R.)
In the plane, let Ox, Oy be two coordinate axes such that Ox is
perpendicular to Oy. A point Z(x;y) is called the representative point,
or imagine of the complex number z.
Conversely, for each real point Z1 (x1 ; y1 ) in the coordinate plane,
there is unique a complex number z1 = x1 + iy1 , z1 is called the complex
coordinate of the point Z1 .
The plane whose each real point is viewed as a representative point
of a complex number is called the complex plane.
Definition 1.2. 1. The axis Ox is the locus of imagines of the real
numbers. The axis Oy is the locus of imagines of the imaginary numbers.
So, Ox is also called the real axis and Oy is also called the imaginary
axis.
2. The number -z is the complex coordinate of the symmetric point of
the point Z passing through the origin O.
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Definition 1.3. The conjugate complex number of z = x + iy is defined
and denoted by z = x − iy. The representative point Z of the complex
number z is the symmetric point of Z passing through Ox.
Definition 1.4. Let z=x+iy be a complex number. Then the trigono√
metric form of z is z = r(cos θ + i sin θ), where r = x2 + y 2 ∈ [0; ∞)
is called a modulus of z and θ ∈ [0; 2π) is the value of the angle be−→
tween vector OZ and Ox according to the positive direction is called an
argument of z.
So, if z ̸= 0, the modulus and argument of z are uniquely defined.
Consider z = r(cos θ + i sin θ) and let t = θ + k2π, k ∈ Z. Then we
have
z = r[cos(t − k2π) + i sin(t − k2π)] = r(cos t + i sin t),
that is the arbitrary complex number z is represented of the form r(cos t+
i sin t), where r ≥ 0, t ∈ R.
Put Argz = {t : θ + 2kπ, k ∈ Z} is called the extended argument of the
complex number z.
Definition 1.5. Let z1 , z2 ̸= 0 be two complex numbers of the trigonometric forms
z1 = r1 (cos t1 + i sin t1 ); z2 = r2 (cos t2 + i sin t2 ).

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PHAN THI THUY HANG

Then z1 , z2 is said to be equal if and only if

r =r
1

2
 t − t = k2π, k ∈ Z
1

2

Definition 1.6. Using the Euler’s formula cos θ + i sin θ = eiθ , the complex number z=x + iy can be written of the form z = reiθ . This form is
called the exponential form of z.
Remark 1.7. 1. If z=0, we choose r=0 and θ be arbitrary.
2. We have z = re−iθ .
Definition 1.8. The representative point Z of the complex number z is
−→
defined when the vector OZ is defined. Then we say that the complex
−→
number z is expressed by OZ. The complex number and its representative vector have the same modulus and we can say that the argument of
the complex number is also the argument of its representative vector.
Definition 1.9. Let zk = xk + iyk (k = 1, 2, ..., n) be n complex numbers and Zk be its n representative points of them. Then the sum of
z1 , z2 , ..., zn is

z = z1 + z2 + ... + zn (1)
with its representative point Z is defined by
−→ −−→ −−→
−−→
OZ = OZ1 + OZ2 + ... + OZn (2).
Suppose that Z(x;y). We find the coordinates of Z depending on
the equation (2) as follows
x + iy =

∑


(xi + iyi ) =
5

∑

xi + i

∑

yi .


Bachelor thesis

PHAN THI THUY HANG

This is the equation (1).
Definition 1.10. Let z1 , z2 be the complex numbers expressed by the
−−→ −−→
−→
vectors OZ1 , OZ2 . Then the difference z = z1 − z2 is expressed by OZ =
−−→ −−→
OZ1 − OZ2 of the associated vectors.

We have z = z1 + (−z2 ) = z1 + z ′2 . The point Z ′2 is symmetric with
−→ −−→ −−→ −−→ −−→
Z2 passing through O. So, we get OZ = OZ1 + OZ ′2 = OZ1 − OZ ′3 (3).
−→ −−→
Corollary 1.11. The equation (3) can be written as OZ = Z2 Z1 . So,
−−→

Z2 Z1 also expresses the difference z1 − z2 .
Definition 1.12. Let z1 , z2 be the complex numbers expressed by the
−−→ −−→
vectors OZ1 , OZ2 . Then the product z = z1 z2 is expressed by the vector
−→
−→
OZ. The vector OZ is found as follows:
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PHAN THI THUY HANG

−−→
1. Rotate the vector OZ1 around O with the angle equal to the argument
−−→
of the vector OZ2 .
2. Multiply the obtained vector above with the modulus of the vector
−−→
OZ2 .
If r1 , r2 and θ1 , θ2 are the modulus and arguments of z1 , z2 , respectively, then we get z1 = r1 eiθ1 , z2 = r2 eiθ2 . Thus, z = z1 z2 = r1 r2 ei(θ1 +θ2 ) .
Then arg(z) = θ1 + θ2 and the modulus of z is r1 r2 = OZ1 .OZ2 .
We take the point U ∈ Ox having the horizontal coordinate 1. Then the
point Z is the third vertex of the triangle △OZ1 Z congruent with the
−→
OZ
1
triangle △OU Z2 , with (Ox; OZ) = θ1 + θ2 and OZ
= OZ

OU .
2

Definition 1.13. Let z1 , z2 be the complex numbers expressed by the
−−→ −−→
−→
vectors OZ1 , OZ2 . Then the ratio z = zz21 is expressed by the vector OZ.
−→
The vector OZ is found as follows:
−−→
−−→
1. Rotate the vector OZ1 around O with the angle equal to −arg(OZ2 ).
−−→
2. Divide the obtained vector above by the modulus of the vector OZ2 .
By the previous definition, we get z =

r1 i(θ1 −θ2 )
r2 e

and the point Z

is the third vertex of the triangle △OZ1 Z congruent with the triangle
△OU Z2 .
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PHAN THI THUY HANG


Definition 1.14. Consider the positive integer n ≥ 2 and a complex
number z0 ̸= 0 and the equation Z n − z0 = 0 (1). Then we say that The
root Z of the equation (1) is a the nth root of the complex number z0 .
Proposition 1.15. Let z0 = r(cos θ + i sin θ) be a complex number with
r > 0 and θ ∈ [0; 2π). The nth root of z0 contains n distinct roots of the
following formula:
Zk =

√
n

r(cos

θ + 2kπ
θ + 2kπ
+ i sin
), k = 0, 1, ..., n − 1.
n
n

Definition 1.16. Each root of the equation Z n − 1 = 0 is called a nth
root of the unit.
Since 1 = cos 0+i sin 0, by the formula of the nth root of the complex
number, we obtain the nth roots of the unit is
εk = cos

2kπ
2kπ
+ i sin
, k ∈ {0, 1, 2, ..., n − 1} .

n
n

Then we have
ε0 = cos 0 + i sin 0; ε1 = cos

8

2π
2π
+ i sin
= ε;
n
n


Bachelor thesis

PHAN THI THUY HANG

4π
2(n − 1)π
2(n − 1)π
4π
+i sin
= ε2 ; εn−1 = cos
+i sin
= εn−1 .
n
n

n
n
{
}
We denote Un = 1, ε, ε2 , ..., εn−1 .

ε2 = cos

The geometric expressions of the nth root of the unit are the vertices
of an equilateral polygon with n sides inscribed the unit circle with the
center O and the radius 1. We consider some special values of n.
1. For n=2, the square roots of 1 are (the roots of the equation Z 2 − 1 =
0) -1 and 2. For n=3, the cube roots of the unit are (the roots of the
equation Z 3 − 1 = 0):
εk = cos

2kπ
2kπ
+ i sin
, k ∈ {0, 1, 2} .
3
3

2π
1
So, ε0 = 1, ε1 = cos 2π
3 + i sin 3 = − 2 + i

− 21


−i

√

3
2

√

3
2

4π
= ε, ε1 = cos 4π
3 + i sin 3 =

= ε2 .

We express the equilateral triangle inscribed the circle (O;1) as the below
figure.

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1.2

PHAN THI THUY HANG


Barycentric coordinates

Lemma 1.17. Let A1 , A2 , ..., An be n points of E2 and k1 , k2 , ..., kn ∈ R
n
∑
ki = k ̸= 0. Then there exists a unique point G satisfying
such that
i=1

−−→ −
−−→
−−→
→
k1 GA1 + k2 GA2 + ... + kn GAn = 0 .
Proof. Indeed, taking an arbitrary point O, we have
−−→ −
−−→
−−→
→
k1 GA1 + k2 GA2 + ... + kn GAn = 0
−−→ −→
−−→ −→
−−→ −→
−
→
⇔ k1 (OA1 − OG) + k2 (OA2 − OG) + ... + kn (OAn − OG) = 0
−→
−−→
−−→
−−→

⇔ OG = k1 (k1 OA1 + k2 OA2 + ... + kn OAn ).
Thus, G is defined and unique.

Definition 1.18. In the previous lemma, the point G is called the
n
∑
barycenter of A1 , A2 , ..., An with the coefficients k1 , k2 , ..., kn . If
ki = 1,
i=1

we call (k1 , k2 , ..., kn ) the barycentric coordinates of the barycenter G.
Proposition 1.19. Let A, B, C be distinct points in E2 and P be an
arbitrary point in the interior of the triangle ABC. Then the barycentric
coordinates of P with respect to A, B, C is
(

S∆P BC S∆P AC S∆P BA
;
;
S∆ABC S∆ABC S∆ABC

)
.

Proof. Let S∆P BC = S1 , S∆P AC = S2 , S∆P AB = S3 . So we need to prove
that

−→
−−→
−→ −

→
S1 .P A + S2 .P B + S3 .P C = 0.

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PHAN THI THUY HANG

Assume that A’ is the intersection point of AP and BC. Then we have
−−→
−→
P A = −P A′ .

S3
S∆P BA′

−−→
= −P A′ .

S2
S∆P CA′

.

This implies that
−−→
−−→
−→

−→
−→
S1 .P A = S∆P BA′ .P A + S∆P CA′ .P A = −P A′ .S3 − P A′ .S2 .
Now, we need to show that
−−→
−−→
−−→
−→
−P A′ .S3 − P A′ .S2 = −P B.S2 − P C.S3 .
Indeed,

−−→
−−→
−−→
−→
−P A′ .S3 − P A′ .S2 = −P B.S2 − P C.S3
−−→ −−→
−−→ −→
−
→
⇔ S2 .(P A′ − P B) + S3 .(P A′ − P C) = 0
−−→
−−→ −
→
⇔ S2 .BA′ + S3 .CA′ = 0 .

This is true. Because we have
that

S∆BAA′

S∆CAA′

=

A′ B
A′ C

and

S∆BM A′
S∆CM A′

S∆BAA′ − S∆BM A′
S3
A′ B
=
=
.
A′ C
S∆CAA′ − S∆CM A′
S2

11

=

A′ B
A′ C ,

it follows



Bachelor thesis

1.3

PHAN THI THUY HANG

Affine transformations

Definition 1.20. Let A, A′ be the affine spaces over a field K and f:A→A′
−
→
−
→
be a mapping. If there exists a linear mapping φ : A → A′ satisfying
−−−−−−−→
−−→
f (M )f (N ) = φ(M N ) for all M, N ∈ A, then f is called the affine mapping
associated with φ. The map φ is called the linear mapping associated
−
→
with f and denoted by f .
Example 1.21. Identity mapping IdA : A → A, f(M)=M, ∀M ∈ A of
the affine space A is an affine mapping. The linear mapping associated
−
→
→ of A .
with IdA is Id−
A

Definition 1.22. If the affine mapping f is injective, we say that f is
affine monomorphism. If the affine mapping f is surjective, we say that
f is affine epimorphism . If the affine mapping f is bijective, we say that
f is affine isomorphism .
In particular, when the map f : A→A, we say that f is called affine
self-homomorphism of A.
If there is an affine isomorphism from A to A′ , we say that A and A′
are isomorphism together and denoted by A ∼
= A′ . If f : A→A is affine
isomorphism, we say that f is an affine self-isomorphism of A.
Proposition 1.23. The affine mapping preserves the barycenter of every
system of finite points.
Proof. Suppose that f:A →A′ is an affine mapping, G is the barycenter
of points P1 , P2 , ..., Pm with coefficients λ1 , λ2 , ..., λm . Then we have
m
∑

−−→ −
→
λi GPi = 0

i=1

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PHAN THI THUY HANG


It follows that
m
∑

m
∑
−
→−−→
−−−−−−−→ −
→
λi f (GPi ) =
λi f (G)f (Pi ) = 0 .

i=1

i=1

Thus, f(G) is the barycenter of points f (P1 ), f (P2 ), ..., f (Pm ) with coefficients λ1 , λ2 , ..., λm .
Definition 1.24. Every self-isomorphism of the affine space A is called
an affine transformation of A.
Proposition 1.25. The affine transformation maps the simplex of dimension m to the simplex of dimension m.
Proof. Suppose that f:A→A is the affine transformation of A. Consider
a simplex of dimension m
{
C=

−−→ ∑ −−→
M ∈ A : OM =
ti OAi , ti ≥ 0, t0 + t1 + ... + tm = 1
m


}

i=0

We will show that f(C) is also a simplex of dimension m. Indeed, we put
I = f (O), Bi = f (Ai ), i = 1, 2, ..., m.
We have
m
m
−−−−→ −
→ (−−→) ∑ −
→ (−−→) ∑ −→
If (M ) = f OM =
ti f OAi =
ti IBi .
i=0

i=0

This implies that
{
f (C) =

}
m
∑
−−→
−→
M ′ ∈ A : IM ′ =

ti IBi , ti ≥ 0, t0 + t1 + ... + tm = 1 .
i=0

In other words, f(C) is the simplex of dimension m with the vertices
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PHAN THI THUY HANG

B0 , B1 , ..., Bm .

14


Chapter 2
Geometry of cubic polynomials
Imagine a sphere floating in 3-space. By inscribing one of its great
circles within an equilateral triangle, we can use the linear projection
in map to the complex plane to project the vertices of the equilateral
triangle onto the roots of a given cubic polynomial p(z). This discovery
helps us to find a proof of the Siebeck-Marden theorem, the surprising
result that the roots of the derivative p’(z) are the foci of the ellipse
inscribed in and tangent to the midpoints of the triangle in complex
plane determined by the roots of the polynomial p(z).

Figure 2.1: An equilateral triangle and its inscribed sphere.

In order to prove Siebeck-Marden theorem, we need to find some

different aspects of the geometry of the floating sphere in figure 2.1.
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Bachelor thesis

PHAN THI THUY HANG

Siebeck-Marden theorem states this figure projects to the roots of the
polynomial, viewed as points in the complex plane. To do this, firstly
we will recall the Fundamental Theorem of Algebra.
Theorem 2.1. (Fundamental Theorem of Algebra) Any polynomial of
degree n has exactly n roots. The polynomial is assumed to have complex
coefficients and the roots are complex as well.
By the Fundamental Theorem of Algebra, a cubic polynomial with
real or complex coefficients has precisely three (possibly complex) roots.
So, suppose that r, s, t are the roots of the cubic polynomial p(x). Then
p(x) can be written in the form
p(x) = a(x − r)(x − s)(x − t)
where a is a constant.
We can see that the roots r, s, t form a triangle in the complex plane.

In this case, Carl Friedrich Gauss, a German, and Felix Lucas, a
Frenchman give us a geometrical relation between the roots of a polynomial P and the roots of its derivative P’.

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PHAN THI THUY HANG

Theorem 2.2. (Gauss-Lucas Theorem) If P is a (non-constant) polynomial with complex coefficients, all zeros of P’ belong to the convex hull
of the set of zeros of P.
We can see Gauss-Lucas Theorem implies that the two roots of the
derivative of the cubic polynomial must lie inside the triangle determined
by the roots of this cubic polynomial in the complex plane. So, we will
prove the theorem for the cubic case by using proof by contradiction.
Proof.

Let r, s, t be the complex roots of the cubic polynomial p(x).

Then p(x) can be written in the form p(x) = a(x − r)(x − s)(x − t),
where a is a constant. It implies that
p′ (x) = a (x − s) (x − t) + a (x − r) (x − t) + a (x − r) (x − s) .
Assume p’(u)=0 where u is not in the complex hull of r, s, t. So, we can
see that u lies outside the triangle formed by the roots of the polynomial
p(x). Since r, s, t have the same role, without loss of generality, we may
suppose u is divided from vertex s and by side rt of the triangle.

This complex plane can be rotated by some angle such that the
dividing side rt is vertical, with u on the right side and the third vertex
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PHAN THI THUY HANG

s is on the left. We call this angle θ. Note that the vectors, drawn green

in the diagram, from the roots of the polynomial to our roots of the
derivative are all pointing towards the right, meaning they have positive
real part.

We have
p′ (u) a [(u − s)(u − t)] + a [(u − r)(u − t)] + a [(u − r)(u − s)]
=
p(u)
a(u − r)(u − s)(u − t)
1
1
1
+
+
.
=
u−r u−s u−t
If we take the conjugate of both side of the equation, we get the
sum of the reciprocal of the conjugate:
p′ (u)
1
1
1
1
1
1
=
+
+
=

+
+
.
p(u)
u−r u−s u−t u−r u−s u−t
Recall that a value multiplied by its conjugate equals the magnitude
squared: e.g. x ∗ x = |x|2 . So we multiply each summand by 1, in the
form of their respective u-root and we obtain
[

] [
] [
]
1 u−r
u−r
u−s
u−t
1 u−s
1 u−t
.
.
+
.
+
=
+
+
.
u−r u−r
u−s u−s

u−t u−t
|u − r|2 |u − s|2 |u − t|2
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PHAN THI THUY HANG

If we multiply each summand by eiθ , we obtain the vectors that
are illustrated in the diagram above, point to the right. These vectors
have positive real part. The contradiction comes when p′ (u) = 0. So
our result must equal zero. Moreover, we stated that each summand has
positive real part, and so must their sum. This contradicts the fact the
result must be zero.
0=e

iθ p

′ (u)

p(u)

= eiθ

u−r
iθ u − s
iθ u − t
+
e

+
e
.
|u − r|2
|u − s|2
|u − t|2

The proof can be extended for polynomials of higher orders.

2.1

Marden’s Projection

The sphere with the inscribed triangle is floating above a copy of the
complex plane. The vertices of the triangle project onto the roots of the
cubic polynomial on the (x,y)-plane, viewed as a copy of the complex
numbers. The triangle, which is the image of the projection, represents
the convex hull of the three roots of p(x). Gauss-Lucas theorem states
the two roots of the derivative of the cubic lies within that comvex hull.
From this, we can begin to make connections for Siebeck-Marden theorem. Siebeck-Marden theorem states if we have any three real numbers,
not all equal, then they are the projections of the vertices of some equilateral triangle in the complex plane.
For a cubic polynomial p(x) with three roots (not all equal), the
inscribed circle of the equilateral triangle that projects onto those roots
itself projects to an interval with endpoints equal to the roots of p’(x).
The theorem says that the roots of p’(x), where z is a root of the
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PHAN THI THUY HANG

derivative of the cubic, are the foci of the ellipse inscribed in that triangle
tangent to the midpoints of the sides.

Figure 2.2: Roots of p’ and foci of ”midpoint” ellipse.

Figure 2.2 illustrates Siebeck-Marden theorem. As depicted in the
image, the roots of p’ are the foci of the midpoint of two roots of the
quadratic polynomial p(x).

Consider an equilateral triangle in a copy of the complex plane. Let
the vertices of the triangle be 1, ω and ω. As in figure 2.1, we have an
inscribed sphere within the equilateral triangle. If we take the projection of the triangle at the equator of the sphere to the triangle on the
complex plane. The resulting projection of that circle (that represented
20