DUALITY AND INSCRIBED ELLIPSES
MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE
Abstract. We give a constructive proof for the existence of inscribed
family of ellipses in convex n-gons for 3 ≤ n ≤ 5 using duality of curves.
In the case of a pentagon, we also show that there are simultaneously
two ellipses, one inscribed in the pentagon and the other inscribed in
its diagonal pentagon. The two ellipses are as intrinsically linked as are
the pentagon and its diagonal pentagon.
1. Introduction
The goal of this paper is to show how duality of curves lends itself to
investigating families of inscribed ellipses in convex n-gons. The foci of
such inscribed ellipses (or similar higher class curves) are closely related
to location of the critical points of polynomials vis-a-vis their roots. For
instance, Marden’s Theorem [3] states that the non-trivial
critical points of
Q
the product of complex power functions p(z) = nk=1 (z − zk )mk for mk ∈
R, mk 6= 0 are the foci of a curve of class n − 1 that touches the side zj zk of
the n-gon formed by z1 , z2 , · · · , zn , in the ratio mj : mk .
In this work we prove the following theorem:
Theorem 1.1 (Main Theorem). Inscribed ellipses in convex non degenerate
n-gons:
(1) In triangles there exists a unique two-parameter family of inscribed
ellipses - the parameters are the points on two sides of the triangle.
That is, by choosing two points on different sides of the triangle there
is a unique inscribed ellipse passing through those two points.
(2) In quadrilaterals there exists a unique one-parameter family of inscribed ellipses - the one parameter is the points on one side of the
quadrilateral. That is, by choosing one point on a side of the quadrilateral there is a unique inscribed ellipse that passes through that
point.
(3) In pentagons there exists precisely one inscribed ellipse - this gives a
zero-parameter family. In fact, our construction gives another ellipse
which is inscribed in the diagonal pentagon formed by the diagonals
of the original pentagon.
2010 Mathematics Subject Classification. Primary 14H52, 51E10.
Key words and phrases. Linfield’s Theorem and Ellipses.
1
2
MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE
(4) For n ≥ 6, there exist n-gons for which there are no inscribed ellipses; whenever there is an inscribed ellipse, it is unique.
Our exposition builds on the work of Linfield [2] where he provides a proof
of Marden’s theorem for n = 3. Since, we are only interested in the existence
of families of ellipses in n-gons, we generalize only the relevant aspects of
Linfield’s proof.
Besides, the already known connection between foci of ellipses and critical
points of polynomials alluded to in Marden’s Theorem, these families carry
some more interesting information. For instance, in an ongoing project the
first author [1] shows that, the loci of the foci of one of the families of ellipses
formed by keeping one point fixed on the side AB while varying the other
point on the side BC in a triangle ABC is an elliptic curve!
In the process of proving our theorem we also give an explicit construction
of the ellipses when they exist. Though the result is not new, our goal is
to provide an exposition using projective geometry (an area which has long
been neglected in the undergraduate curriculum) and some vector calculus.
Our results also follow from Brianchon’s Theorem. Since this proof is very
elementary and quick we present it at the very outset.
Theorem 1.2 (Brianchon’s Theorem). If a hexagon circumscribes an ellipse, then its three diagonals meet in a point; conversely, if the three diagonals of a hexagon meet in a point, then there exists an inscribed ellipse.
r
r
1-r
s
1-s
Figure 1. Implications of Brianchon’s Theorem for inscribed ellipses
For example, to see that there exists a two-parameter family of ellipses
inscribed in a triangle, fix 0 < r < 1 and 0 < s < 1 as in the triangle in
Figure 1. The five solid dots and one open dot form vertices of a degenerate
hexagon; these vertices satisfy the hypothesis of the converse of Brianchon’s
Theorem, guaranteeing the existence of an inscribed ellipse. For an ellipse to
be tangent to the six sides of the hexagon, it must be tangent at the points
designated by r, s, and the open dot. The fact that r and s are arbitrary
1-r
DUALITY AND INSCRIBED ELLIPSES
3
explains the existence of a two-parameter family. Similar arguments can be
made for the quadrilateral and pentagonal cases above.
2. Duality and Dual Curves
In this section we fix notations and definitions that will be used for the
rest of the paper. For more details we refer the reader to [4]. Let R be the
field of real numbers and P2 the real projective plane. To each point (x, y)
in R2 we associate a homogeneous point [x : y : 1] in P2 , and to each line
ax + by + c = 0 in R2 we associate a homogeneous line ax + by + ch = 0
in P2 . Just as λax + λby + λch = 0 for λ 6= 0 identifies the same line
ax + by + ch = 0, so it is understood that [λx : λy : λ] = [x : y : 1] for
λ 6= 0. That is, it is the ratio between components that unambiguously
distinguishes among homogeneous points.
The homogeneous point [b : −a : 0] can be seen to lie on the homogeneous
line ax + by + ch = 0, and is referred to as a point at infinity. All the points
at infinity lie on the homogeneous line at infinity, h = 0.
To each homogeneous point Q and line LP = 0 in the xyh-plane there
exists a unique line LQ = 0 and point P , respectively, in a dual αβγ-plane.
Specifically,
Q = [x : y : h] ←→ LQ := Q · (α, β, γ) = 0
LP := P · (x, y, h) = 0 ←→ P = [α : β : γ],
where LQ := Q · (α, β, γ) = xα + yβ + hγ and similarly, LP := P · (x, y, h) =
αx + βy + γh. Note that LQ is a linear homogeneous polynomial in the variables (α, β, γ), while LP is a linear homogeneous polynomial in the variables
(x, y, h). The point Q and line LQ = 0 are duals of one another; as are P
and LP = 0. Note that ∇LP = P , so that the dual of a homogeneous line
can be seen to be its gradient.
In general, to any homogenous curve ϕ = ϕ(α, β, γ) ∈ P2 of degree n −
1, n ≥ 1, we can associate a homogenous dual curve ϕ
b = ϕ(x,
b y, h) ∈ P2 ,
2
whose generic points are tangents to ϕ ∈ P . Symbolically,
ϕ
b = {[x : y : h] | P · (x, y, h) = 0, ϕ(P ) = 0}.
So the duals of the points of ϕ = 0 are tangent lines to ϕ
b = 0; analogously,
tangent lines to ϕ = 0 are duals of the points of ϕ
b = 0. To determine the
points of ϕ
b = 0 we require the set of tangent lines to ϕ = 0. To this end the
homogeneous tangent line to the curve ϕ = 0 at any point P0 on the curve
can be written as
∇ϕ(P0 ) · (P − P0 ) = 0.
According to Euler’s homogeneous function theorem, ∇ϕ(P ) · P = (n −
1)ϕ(P ), simplifying the representation of the tangent line:
∇ϕ(P0 ) · P = 0.
Thus if P0 lies on ϕ = 0, then Q0 = ∇ϕ(P0 ) lies on ϕ
b = 0.
4
MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE
Remark 2.1 (Explicit formula for a dual of a conic). If ϕ is quadratic then
(x, y, h) = ∇ϕ(α, β, γ) is a linear system. If α = α(x, y, h), β = β(x, y, h),
and γ = γ(x, y, h) is a solution to this system then the dual conic is given
by
ϕ(x,
b y, h) = ϕ(α(x, y, h), β(x, y, h), γ(x, y, h)).
This is generally not true for higher degree ϕ.
We now prove a key lemma which we need to prove our main results.
Lemma 2.2. Let P denote a convex n-gon, with vertices labeled Vk , k =
1, . . . , n and let mk , k = 1, . . . , n be positive real numbers. Then there exists
a class n − 1 curve, which is tangent to the n(n − 1)/2 lines joining the
vertices of P with the points of tangency given by:
mj
mk
(1)
Vk +
Vj .
mj + mk
mj + mk
Proof. Following Linfield [2], we associate to each vertex Vk = (xk , yk ) of
the polygon a homogeneous point and homogeneous linear polynomial
Qk = [xk : yk : 1]
←→
Lk = Qk · [α : β : γ] = xk α + yk β + γ.
The intersection point Pjk of the lines Lj = 0 and Lk = 0 is either finite or
a point at infinity:
(
[yj − yk : xk − xj : xj yk − xk yj ], Lj = 0, Lk = 0 not parallel;
Pjk =
[yk : −xk : 0],
Lj = 0, Lk = 0 parallel.
The dual of the point Pjk is the line joining Qj to Qk . Let
(2)
ϕ=
n
X
m` L1 · · · L`−1 L`+1 · · · Ln .
`=1
The homogeneous polynomial ϕ is of degree n − 1 and interpolates Pjk since
each summand in its definition contains as a factor either Lj or Lk . Let
ϕ
b = 0 denote the dual to the curve ϕ = 0. The fact that ϕ is of degree n − 1
implies that ϕ
b is of class n − 1.
By definition, ϕ
b = 0 is tangent to the line joining Qj and Qk , since ϕ
interpolates the dual of this line, Pjk . To determine the point of tangency,
Qjk , we compute the tangent line to ϕ = 0 at Pjk . Since each of the
summands of ϕ contain at least one if not both of the factors Lj and Lk , it
can be expressed as ϕ = (mj Lk + mk Lj )Ajk + Lj Lk Bjk for appropriate Ajk
and Bjk . Thus
∇ϕ(Pjk ) = Ajk (Pjk )(mj Qk + mk Qj ).
Normalizing the homogenous component of ∇ϕ(Pjk ) we see that the dual of
the tangent line to ϕ = 0 at Pjk corresponds to
mj
mk
Qk +
Qj ,
Qjk :=
mj + mk
mj + mk
DUALITY AND INSCRIBED ELLIPSES
5
establishing (1).
Definition 2.3. The function ϕ defined in (2) plays a critical role in this
paper. In what follows we shall refer to the constants mi ’s as weights and
the function ϕ as Linfield’s function.
3. Proof of Main Theorem
In the case of a triangle (n = 3), Linfield’s function ϕ is a quadratic,
and hence its dual ϕ
b must also be a quadratic. We exhibit a simple set
of constraints on the weights mk , that depend upon two parameters, that
compel the de-homogenized quadratic to be an ellipse. In the case of a
convex quadrilateral (n = 4), Linfield’s function ϕ is a cubic. We produce
a set of weights, depending upon a single parameter, that enable ϕ to be
factored into the product of a linear and quadratic polynomial. Its dehomogenized dual, ϕ,
b in this instance is a point and an inscribed ellipse.
For a convex pentagon (n = 5), ϕ is a quartic. There exists a unique set
of constraints on the weights that permit ϕ to be factored into the product
of two quadratic polynomials. The de-homogenized duals of these factors
will be shown to be inscribed in the pentagon and in the diagonal pentagon,
respectively.
3.1. In triangles there exists a unique two-parameter family of inscribed ellipses.
Proof. Let T denote a nondegenerate triangle with homogeneous vertices
Qk , k = 1, 2, 3, labeled clockwise. Linfield’s function
ϕ = m1 L2 L3 + m2 L1 L3 + m3 L1 L2
is a quadratic in term of α, β and γ.
To establish the existence of the two-parameter family of ellipses inscribed
in T , fix two parameters 0 < r, s < 1. Geometrically, these values designate
two points Q12 = (1 − r)Q1 + rQ2 and Q23 = (1 − s)Q2 + sQ3 on adjacent
sides of T (see Figure 1). Choose positive m1 , m2 and m3 so that
r
m2
s
m1
=
, and
=
.
m2
1−r
m3
1−s
Then the dual curve ϕ
b = 0 is also quadratic, a conic upon de-homogenization.
Since the curve is tangent to the three segments, and all mi ’s are positive,
it follows that ϕ
b = 0 must be a conic and in particular an ellipse.
There are no other ellipses inscribed in the triangle other than those of
the two-parameter family. If such an ellipse did exist, then it would be
tangent to the sides Q1 Q2 , Q2 Q3 , and Q1 Q3 ; the points of contact on the
first two sides define parameters r0 , s0 . The equation of this hypothesized
ellipse satisfies five independent homogenous constraints (arising from the
3 tangency conditions and 2 points of contact on the sides Q1 Q2 , Q2 Q3 ) in
6
MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE
ΑΒΓ-plane
xyh-plane
Q1
r P12 × Hx,y,hL=0
L3 =0
P23
Q12
1-r
`
Φ=0
Φ=0
Q2
L1 =0
P13
P12
Q3
L2 =0
Q12 × HΑ,Β,ΓL=0
Figure 2. Illustration of Linfield’s theorem for n = 3
common with the de-homogenized dual ϕ
b for this choice of parameters r0 , s0 .
Hence the ellipse is among the two-parameter family.
Example 3.1. The computation of the dual of a polynomial curve is in
general a nontrivial problem. Since the coefficients of the tangent lines to
ϕ = 0 constitute homogenous points of the dual ϕ
b = 0, we must solve the
generally nonlinear system (x, y, h) = ∇ϕ(α, β, γ). In the case of a quadratic
polynomial, however, the system is linear and consequently more accessible.
We illustrate the construction, by using Linfield’s function corresponding to
triangle with vertices (0, 0), (1, 0), and (0, 1). In this instance Q1 = [0 : 0 : 1],
Q2 = [1 : 0 : 1], Q3 = [0 : 1 : 1], and Linfield’s function is
ϕ(α, β, γ) = m3 γ(α + γ) + m2 γ(β + γ) + m1 (α + γ)(β + γ).
In this example, the system (x, y, h) = ∇ϕ(α, β, γ) is
x = m3 γ + m1 (β + γ),
y = m2 γ + m1 (α + γ),
h = m1 (α + β + 2γ) + m2 (β + 2γ) + m3 (α + 2γ).
and it always has a solution provided m1 m2 m3 6= 0. Here,
α = hm1 (m1 + m2 ) − (m1 + m2 )2 x − (m21 − m2 m3 + m1 (m2 + m3 ))y,
β = hm1 (m1 + m3 ) − (m21 − m2 m3 + m1 (m2 + m3 ))x − (m1 + m3 )2 y,
γ = m1 (−hm1 + (m1 + m2 )x + (m1 + m3 )y).
Substituting these values for α, β, γ into ϕ(α, β, γ) yields the dual of ϕ as a
function of x, y, h:
ϕ(x,
b y, h) = −m21 (−1+x+y)2 −(m2 x−m3 y)2 −2m1 (−1+x+y)(m2 x+m3 y).
DUALITY AND INSCRIBED ELLIPSES
7
3.2. In convex quadrilaterals there exists a one-parameter family
of inscribed ellipses.
Proof. Let Q denote a convex nondegenerate quadrilateral with homogeneous vertices Qk , k = 0, . . . , 3, labeled clockwise. In this instance Linfield’sfunction
(3)
ϕ = m0 L1 L2 L3 + m1 L0 L2 L2 + m2 L0 L1 L3 + m3 L0 L1 L2
is cubic. Without loss of generalization we assume that the intersection of
the diagonals of Q lies at the origin and that the vertices Q1 and Q3 lie on
the vertical axis. Thus there exist constants 0 < θ, φ < 1 for which
(4)(1 − θ)Q1 + θQ3 = [0 : 0 : 1]
and
(1 − φ)Q2 + φQ0 = [0 : 0 : 1].
For a parameter 0 < r < 1. Choose positive real numbers m0 , m1 , m2 and
m3 such that
(5)
m1
r
,
=
m2
1−r
m1
θ
,
=
m3
1−θ
and
m2
φ
.
=
m0
1−φ
Conditions (4) imply (1 − θ)L1 + θL3 = γ = (1 − φ)L2 + φL0 . So, upon
writing (3) as
ϕ = m2 L0 + m0 L2 L1 L3 + m1 L3 + m3 L1 L2 L0 ,
and taking into account the constraints (5), it can be verified that Linfield’sfunction can be factored into a linear term times a quadratic one:
(6)
γ
(1 − r)θL1 L3 + rφL2 L0 .
(1 − r)θφ
The dual of the linear factor γ/((1 − r)θφ) is the origin; the dual of the
quadratic is a conic. Consequently, the dual of ϕ is a conic, tangent to the
interior of the four segments defining Q, and hence must be an ellipse.
As in the case of the triangle, any ellipse inscribed in the quadrilateral
would necessarily share five independent conditions with ϕ
b for a particular
choice of r. Hence this family is unique.
Remark 3.2. The linear factor in the function ϕ in (6) corresponds to the
line at infinity because we constrained the intersection of the diagonals of
Q to lie at the origin. If this were not the case, then the linear factor of ϕ
would correspond to the dual of that point of intersection.
Remark 3.3. Since we can translate and rotate a quadrilateral Q so that
it fulfills the conditions in the above proof, we can apply the rotation and
translation to the computed two-parameter family in reverse to obtain the
desired family inscribed in Q.
8
MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE
3.3. In pentagons there exists a unique inscribed ellipse.
Proof. In this case Linfield’s function takes the form
ϕ = m1 L2 L3 L4 L5 + m2 L1 L3 L4 L5 + m3 L1 L2 L4 L5 +
m4 L1 L2 L3 L5 + m5 L1 L2 L3 L4 .
and we shall show that ϕ can be expressed as the product of two quadratic
polynomials for appropriate choices of the weights m1 , . . . , m5 . To see this
let P denote a convex nondegenerate pentagon. Extend two of the sides of P
to form a convex quadrilateral Q from which the pentagon can be thought to
have been cut (see Figure 3). Label the vertices of Q in a clockwise fashion
Qk , for k = 0, 1, 2, 3, beginning with the vertex not among those of P. Label
the remaining vertices of P consecutively Q4 and Q5 .
Without loss of generalization we assume that, as in the n = 4 case, the
diagonals of Q intersect at the origin, and additionally that the vertices
Q1 and Q3 lie on the vertical axis (see Figure 3). With this structure in
place there exist constants 0 < θ, φ < 1 for which (1 − θ)Q1 + θQ3 = 0 and
(1 − φ)Q0 + φQ2 = 0.
The first factor of ϕ can be modeled on the quadratic factor in (6),
ϕP = (1 − r)θL1 L3 + rφL0 L2 .
(7)
and for the second factor of ϕ, we let QA , QB , and QC denote the clockwise
labeling of the homogeneous vertices of the diagonal pentagon that do not
lie on the vertical axis, and define
ϕDP = (1 − s)LA LC + sLB L2 .
Q1
PC24
ΦDP =0
Q5
QA
ΦP =0P
P23
Q0
QB
0
`
ΦDP =0
QC
Q4
@0:0:1D
Q2
`
ΦP =0
034
P12
PBC35
P45
P015
PAB14
PA25
Q3
Figure 3. Illustration of Linfield’s theorem for n = 5, here
P23 denotes the intersection of lines L2 = 0, L3 = 0.
DUALITY AND INSCRIBED ELLIPSES
9
We start with ϕP . Since each Lk vanishes at any point with k among
its subscript, ϕ∞ = 0 can be seen to interpolate the four points P12 , P23 ,
P034 , and P015 for any choice of the free parameter r. We now show r in (7)
can be chosen so that ϕP also interpolates a fifth point, P45 . Once we have
shown that ϕP = 0 interpolates the duals of the five sides of P, it follows
immediately that ϕ
bP = 0 must be tangent to the sides of P.
Because Q0 and Q1 are multiples of Q2 and Q3 , respectively, it follows
that L0 = 0 and L2 = 0 are parallel, as are L1 = 0 and L3 = 0. Furthermore,
since Q1 and Q3 are assumed to lie on the vertical axis, L1 = 0 and L3 = 0
must be parallel to the α-axis, and intersect at P13 = [1 : 0 : 0]. Because the
dual image of a strictly convex polygon is also strictly convex, the point P45
must lie between L1 = 0 and L3 = 0, and to one side of L0 = 0 and L2 = 0
(see Figure 3). Since each of the homogeneous linear polynomials L0 , L1 ,
L2 , and L3 is positive at the origin [0 : 0 : 1], it follows that the product
L1 L3 at P45 is positive, and the product L0 L2 at P45 is negative. Thus there
exists a unique parameter value 0 < r0 < 1 for which (1 − r0 )θL1 L3 (P45 ) +
r0 φL0 L2 (P45 ) = 0, hence ϕP (P45 ) = 0.
Likewise, for the second factor of ϕ, we let QA , QB , and QC denote the
clockwise labeling of the homogeneous vertices of the diagonal pentagon that
do not lie on the vertical axis, and define
ϕDP = (1 − s)LA LC + sLB L2 .
By analogy with ϕP above, ϕDP interpolates the points PA25 , PAB14 , PBC35 ,
and PC24 , duals of four of the sides of the diagonal pentagon DP. Let xA ,
xB , xC denote the first components of QA , QB , and QC respectively and
recall that P13 = [1 : 0 : 0] is the dual of the side of DP containing the
origin. Since LA LC (P13 ) = xA xC > 0 and LB L2 (P13 ) = xB x2 < 0, there
exists a unique parameter 0 < s0 < 1 for which ϕDP (P13 ) = 0. With this
parameter choice, ϕDP interpolates the duals of all five sides of the diagonal
pentagon of P.
By Lemma 2.2 and the arguments above, ϕ and the product ϕP · ϕDP
both interpolate the duals of the ten segments joining the vertices Qj to Qk ,
for distinct j, k = 1, 2, . . . , 5. With the constraints
(8)
(9)
r0
m1
=
m2
1 − r0
m1 Q5 + m5 Q1 = ∇ϕP (P015 )
m1
θ
=
m3
1−θ
m3 Q4 + m4 Q3 = ∇ϕP (P034 )
we shall show that ϕ and the product ϕP · ϕDP additionally have common
gradients at four specific points. It is this observation that forces the two
quartic polynomials ϕ and ϕP · ϕDP to be proportional to one another.
The gradient of Linfield’s function ϕ at the points P12 , P23 , P015 , and P034
corresponds to points of interpolation for ϕ
b = 0. Each of these interpolated
points subdivides a line segment into fixed proportions according the the
values of the weights. With the weights defined by (8) and (9), it follows
10
MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE
that
∇ϕ(P12 ) ∝ m1 Q2 + m2 Q1 ,
∇ϕ(P015 ) ∝ m1 Q5 + m5 Q1 ,
∇ϕ(P23 ) ∝ m2 Q3 + m3 Q2 ,
∇ϕ(P034 ) ∝ m3 Q4 + m4 Q3
We next compute the gradient of the product ϕP · ϕDP . In general
∇(ϕP · ϕDP ) = ∇ϕP · ϕDP + ϕP · ∇ϕDP ,
but, since ϕP is zero at the points P12 , P23 , P015 , and P034 , satisfies
∇(ϕP · ϕDP )(P ) ∝ ∇ϕP (P )
at those points. The proof of the theorem is complete if we show that
∇ϕ ∝ ∇ϕP at the four points P12 , P23 , P015 , and P034 . The proportionality
is obvious at the latter two points from the constraints on m1 , m3 , m4 , and
m5 in (9).
To see that ϕ and ϕP are proportional at P12 we note that ∇ϕ(P12 )
and ∇ϕP (P12 ) each correspond to a point between Q1 and Q2 . That is,
comparison of the properties of the weights in (8) with those in (5), with
r = r0 , imply that ∇ϕ(P12 ) ∝ ∇ϕP (P12 ). A similar argument holds at the
point P23 .
Thus Linfield’s function ϕ is proportional to the product of ϕP ·ϕDP . The
dual of each factor is an ellipse: that of ϕP is inscribed in the pentagon,
and that of ϕDP is inscribed in its diagonal pentagon. The uniqueness of
each inscribed ellipse follows because another ellipse would need to share
five independent tangent conditions, forcing them to be the same.
3.4. If there exists an inscribed ellipse in a convex n-gon for n ≥ 6,
then it is unique.
Proof. If a convex hexagon contains an inscribed ellipse, extend two suitable
sides of the hexagon to form a pentagon from which the hexagon can be
thought to have been cut out. The ellipse that is inscribed in the hexagon,
is also inscribed in the pentagon. Since any pentagon can only inscribe
a unique ellipse, this ellipse must be unique. A similar argument can be
applied to n-gons for n > 6.
3.5. For n ≥ 6 there exist convex n-gons with no inscribed ellipses.
Proof. By Brianchon’s theorem, a convex hexagon contains an inscribed
ellipse if and only if the diagonals intersect at a point. It is unlikely, that a
generic hexagon has an inscribed ellipse. This follows from the observation
that three generic lines (these can be extended to form the diagonals of a
hexagon) are unlikely to be concurrent (see Figure 1).
DUALITY AND INSCRIBED ELLIPSES
11
References
[1] M. Agarwal and N. Natarajan. Elliptic curves from ellipses. In preparation.
[2] Ben-Zion Linfield. On the relation of the roots and poles of a rational function to the
roots of its derivative. Bull. Amer. Math. Soc., 27(1):17–21, 1920.
[3] Morris Marden. Geometry of polynomials. Second edition. Mathematical Surveys, No.
3. American Mathematical Society, Providence, R.I., 1966.
[4] Joseph H. Silverman and John Tate. Rational points on elliptic curves. Undergraduate
Texts in Mathematics. Springer-Verlag, New York, 1992.
Department of Mathematics and Statistics, University of Michigan, Dearborn, Dearborn, MI 48128
E-mail address: [email protected]
Department of Mathematics and Statistics, University of Michigan-Dearborn,
Dearborn, MI 48128
E-mail address: [email protected]
Department of Mathematics and Statistics, University of Michigan-Dearborn,
Dearborn, MI 48128
E-mail address: [email protected]