# Counting Pairs of Conics Satisfying the Poncelet Triangle Condition

Nathan Kaplan, Tianhao Wang

###### Abstract

We say that a pair of smooth plane conics satisfies the Poncelet Triangle
Condition if they intersect transversally and there is a triangle (possibly
defined over the algebraic closure instead of the original base field)
inscribed in one conic and circumscribed in the other. Chipalkatti’s paper
showed that over a finite field $\mathbb{F}_{q}$ with characteristic away from
$2,3$, a randomly chosen pairs of smooth conics over $\mathbb{F}_{q}$ with
transversal intersection has about $1/q$ chance to satisfy the PTC condition.
We will improve this result and show that the exact probability is given by
$\frac{q^{2}-5q+5}{q^{3}-5q^{2}+6q}$. We will also correct the conjecture made
in Chipalkatti’s paper, and show that the asymptotic probability for the
tetragons case is $1/q$.

###### Contents

  1. 1 Introduction
    1. 1.1 Poncelet triangle
    2. 1.2 Pencil of conics
  2. 2 Counting pairs of smooth conics satisfying the PTC in each pencil
    1. 2.1 Intersection type (1,1,1,1)
    2. 2.2 Intersection type (2,1,1)
    3. 2.3 Intersection type (2,2)
    4. 2.4 Intersection type (4)
    5. 2.5 Intersection type (3,1)
  3. 3 Generalization to tetragon case
  4. 4 Appendix
    1. 4.1 Counting the number of pencils of each intersection type
    2. 4.2 Computation data for small $q$ and conjectures for higher $n$-gons

##  1 Introduction

###  1.1 Poncelet triangle

Conics over field of characteristic $2$ is very different from conics over
other finite field. In this paper, we are always assuming that the base field
has characteristic away from $2$.

We first recall a few facts about plane conics and Poncelet Triangle.

Let $\mathcal{A}$ be a projective plane conic defined over
$\mathbb{P}^{2}(\mathbb{F}_{q})$ with defining equation given by
$f(x,y,z)=ax^{2}+bxy+cxz+dy^{2}+eyz+fz^{2}$. Then we can associate a symmetric
matrix to this conic:

$A=\begin{bmatrix}a&b/2&c/2\\\ b/2&d&e/2\\\ c/2&e/2&f\end{bmatrix}$

This matrix has the property that

  1. 1.

$\begin{bmatrix}x&y&z\end{bmatrix}A\begin{bmatrix}x\\\ y\\\
z\end{bmatrix}=f(x,y,z)$

  2. 2.

$\det A\neq 0$ if and only if $\mathcal{A}$ is smooth.

######  Definition 1.1.

Let $(\mathcal{A},\mathcal{B})$ be a pair of smooth projective plane conics
defined over $\mathbb{F}_{q}$. Then we say that it satisfies the Poncelet
Triangle Condition (PTC) if

  1. 1.

The two conics intersect transversally

  2. 2.

There is a triangle defined over $\overline{\mathbb{F}_{q}}$ that is inscribed
in one conic and circumscribed in the other

Denote $\mathcal{B}^{*}$ to be the dual conic of $\mathcal{B}$ consisting of
tangent lines to $\mathcal{B}$. Define
$E\subset\mathcal{A}\times\mathcal{B}^{*}$ consisting of $(P,\xi)$ where
$P\in\xi$. The idea of the Poncelet construction is that we start from
$(P,\xi)\in E$. Then the tangent line $\xi$ intersect $\mathcal{A}$ at another
point $P^{\prime}$, and we let $\xi^{\prime}$ to be the other tangent line to
$\mathcal{B}$ passing through $P^{\prime}$. We denote this operation by $j$

$j:(P,\xi)\mapsto(P^{\prime},\xi^{\prime})$ (1)

Then the existence of a Poncelet triangle is equivalent to there is a pair
$(P,\xi)\in E$ such that $j^{3}(P,\xi)=(P,\xi)$.

Now, we state the classical result by Cayley on the existence of Poncelet
Triangle:

Denote $A,B$ to be the symmetric matrices associated to the conics
$\mathcal{A},\mathcal{B}$. Let $\Delta(t)=\det(tA+B)$ and assuming that
$\sqrt{\Delta}$ has the following Maclaurin series expansion

$\sqrt{\Delta}=H_{0}+H_{1}t+H_{2}t^{2}+\cdots$

then we have the following criterion for a pair of conics to satisfy the PTC.

######  Theorem 1.2 (Cayley).

The pair of smooth projective plane conics $(\mathcal{A},\mathcal{B})$ with
transversal intersection satisfies the PTC if and only if $H_{2}=0$

For the proof of this theorem, we refer to a paper by Griffiths and Harris
[1]. Eventually the existence of a Poncelet Triangle will be equivalent to
certain point of the elliptic curve $y^{2}=\det(xA+B)$ being a $3$-torsion
point. Hence we expect that the characteristic $3$ case will give a different
result.

Now, we state the density result obtained in Chipalkatti’s paper [2]

######  Theorem 1.3 (Chipalkatti).

Let $\mathbb{F}_{q}$ be a finite field with characteristic away from $2,3$.
Denote $\Psi$ to be the set of pairs of smooth conics with transversal
intersection, and $\Gamma$ to be the subset of pairs satisfying PTC. Then we
have

$\frac{q-16}{q(q+1)}\leq\frac{|\Gamma|}{|\Psi|}\leq\frac{q+5}{(q-2)(q-3)}$

In particular
$\frac{|\Gamma|}{|\Psi|}=\frac{1}{q}+O\left(\frac{1}{q^{2}}\right)$. If the
ground field has characterstic $3$, then
$\frac{|\Gamma|}{|\Psi|}=\frac{2}{q}+O\left(\frac{1}{q^{2}}\right)$.

We will follow the same technique as used in Chipalkatti’s paper, and give the
exact density. We get the following results:

######  Theorem 1.4.

Assuming that the ground field has characteristic away from $2,3$.

  1. 1.

The total number of pairs of smooth conics with transversal intersection and
satisfies PTC is given by

$|\Gamma|=q^{9}-q^{8}-q^{7}+q^{5}+q^{4}-q^{3}=(q^{5}-q^{2})(q+1)q(q-1)^{2}$

We factored out $q^{5}-q^{2}$ as it is the number of smooth conics defined
over $\mathbb{F}_{q}$. When doing the counting, we can perform a change of
coordinate to fix the first conic, and count the number of second conic so
that the pair satisfies the PTC. So we do expect a factor of $q^{5}-q^{2}$
here.

  2. 2.

The overall density is given by
$\frac{|\Gamma|}{|\Psi|}=\frac{q^{2}-5q+5}{q^{3}-5q^{2}+6q}$

###  1.2 Pencil of conics

Given two plane conics with equation $F(x,y,z)\neq G(x,y,z)$, we call the
family of conics $\pi=\\{\eta F+G=0:\eta\in P^{1}(\mathbb{F}_{q})\\}$ as a
pencil of conics, and $F,G$ as generators of this pencil. We will use
$\eta\in\mathbb{F}_{q}\cup\\{\infty\\}$ where $\eta=\infty$ corresponds to the
conic $\\{F=0\\}$. We see that any two distinct members in this pencil will
have same intersections as the intersection of $F$ and $G$. If we require that
$F$, $G$ intersect transversally in $4$ points over
$\overline{\mathbb{F}_{q}}$, then these $4$ points will determine the pencil
uniquely (the pencil consists of all conics passing through the 4 points), and
there are exactly $5$ classes of such pencils up to projective automorphisms
over $\mathbb{F}_{q}$. We use the notation $(1,1,1,1)$ to denote the pencil of
conics where the two generators intersect in $4$ points, and all the $4$
points are defined over $\mathbb{F}_{q}$. Similarly, we use $(2,1,1)$ to
denote the intersection type where we have $2$ conjugate points defined over
$\mathbb{F}_{q^{2}}$, and two other $\mathbb{F}_{q}$ point.

In total, we have the following 5 classes of pencils where the, and here is
the number of pencils of each classes. I leave it unsimplified so it would be
clear where the counting comes from. The proof of this counting is included in
the Appendix.

Intersection Type | Number of pencils of this type
---|---
$(1,1,1,1)$ | $\frac{1}{4!}(q^{2}+q+1)(q^{2}+q)q^{2}(q^{2}-3(q+1)+3)$
$(2,1,1)$ | $\frac{1}{2\cdot 2}(q^{4}+q^{2}+1-q^{2}-q-1)(q^{2})(q^{2}-1)$
$(2,2)$ | $\frac{1}{2\cdot 2\cdot 2}(q^{4}-q)(q^{4}-q-(q^{2}+1-q-1))$
$(3,1)$ | $\frac{1}{3}(q^{6}+q^{3}+1-q^{2}-q-1-(q^{2}+q+1)(q^{3}-q))(q^{2}+q+1)$
$(4)$ | $\frac{1}{4}(q^{8}+q^{4}+1-q^{4}-q^{2}-1-(q^{2}+q+1)(q^{4}-q^{2}))$

The idea in Chipalkatti’s paper is that we classify pairs of smooth conics
with transversal intersection upto projective automorphisms defined over
$\mathbb{F}_{q}$, and then compute the density of pairs satisfying PTC in each
class. Chipalkatti computed the number of smooth pairs of conics satisfying
the PTC in the pencil with intersection type $(1,1,1,1)$, and did estimation
for the other pencils. We will compute that number in the pencils of the other
four intersection types.

##  2 Counting pairs of smooth conics satisfying the PTC in each pencil

From this point on, we are going to always assume that the base field has
characteristic away from $2,3$.

We will also show that pair of the same smooth conics cannot satisfy the PTC.
So when we count the number of pairs of conics satisfying the PTC, we are
always assuming that the pair consists of two distinct smooth conics.

######  Lemma 2.1.

Given a smooth conic $\mathcal{A}$ over $\mathbb{F}_{q}$ where the
characteristic is not $2,3$, then the pair $(\mathcal{A},\mathcal{A})$ does
not satisfy the PTC.

###### Proof.

Let $A$ be the symmetric matrix corresponding the smooth conic $\mathcal{A}$.
Then we have $\Delta(t)=\det(tA+A)=(t+1)^{3}\det(A)$. Then the numerator of
$H_{2}$ is given by $-a_{1}^{2}+4a_{0}a_{2}$ where $a_{i}$ is the coefficient
$t^{i}$ in $\Delta(t)$. In this case, we get $H_{2}=3\det(A)\neq 0$ as
$\mathcal{A}$ is smooth. Another consequence of this computation is that pairs
of the same singular conics will satisfy the PTC. ∎

###  2.1 Intersection type (1,1,1,1)

This is already computed in Chipalkatti’s paper. We will review the
computation again as the other cases are using exactly the same method. We
pick an explicit generators for the pencil, and assume that the first conic is
given by $C_{r}=rF+G$ and the second conic is given by $C_{s}=sF+G$ where
$r,s\in P^{1}(\mathbb{F}_{q})$. Then we compute the $H_{2}(r,s)$ (We will drop
the denominator) which will be degree $2$ in $r$ and degree $4$ in $s$. We
view this as an equation in $r$ and compute its discriminant which we denote
by ”$discH_{2}$”.

  * •

Generators of the pencil: $F=xy$, $G=z^{2}+yz+xz$

  * •

Singular members: $\eta_{1}=0$, $\eta_{2}=1$, $\eta_{3}=\infty$

  * •

$H_{2}(r,s)=r^{2}+(6s^{2}-4s^{3}-4s)r+s^{4}$

  * •

$discH_{2}=16(s^{2}-s+1)(s(s-1))^{2}$

  * •

Density of smooth pairs satisfying PTC: $\frac{q-5}{(q-2)(q-3)}$

  * •

Proof: We want to count all the $(r,s)$ with $r,s\neq 0,1$ (Ensure that
$C_{r},C_{s}$ are smooth conics) such that $H_{2}(r,s)=0$. Note that
$H_{2}(r,s)$ is always a quadratic polynomial in $r$. Hence for a fixed value
of $s$, the number of $r$ such that $H_{2}(r,s)=0$ depends on the value of
$discH_{2}(s)$.

When $s=0,1$, $discH_{2}(s)$ equal to zero, and there is one value of $r$
corresponding to each of that $s$ value, which are $r=0,1$ respectively. We
also see that when $r=0,1$, there is exactly one value of $s$ which makes
$H_{2}(r,s)=0$ which are $s=0,1$ respectively.

Now we assume that $s\neq 0,1$, and we know that if $H_{2}(r,s)=0$, $r\neq
0,1$ either. So we do not need to worry that smoothness of $C_{r}$ when
assuming that $s\neq 0,1$. Further, the square part of $discH_{2}(s)$ is
always a non-zero square. To use the same notation as in Chipalkatti’s paper,
we denote the non-square part of $discH_{2}(s)$ as $f(s)=s^{2}-s+1$. Then
$discH_{2}(s)$ is zero/non-zero square if and only if $f(s)$ is zero/non-zero
square.

For $f(s)=s^{2}-s+1=(s-1/2)^{2}+\frac{3}{4}$, it has two distinct roots over
$\mathbb{F}_{q}$ if and only if $-3$ is a square in $\mathbb{F}_{q}$ (we
assumed that characteristic of $\mathbb{F}_{q}$ is not $3$, so $-3\neq 0$ in
$\mathbb{F}_{q}$). So we will separate it into the corresponding two cases.

If $(s-1/2)^{2}+\frac{3}{4}=y^{2}$ for some $y\in\mathbb{F}_{q}$, then we have

$(y-s+1/2)(y+s-1/2)=\frac{3}{4}$

We call the first term by $a\in\mathbb{F}_{q}\setminus\\{0\\}$, and the second
term will be $\frac{3}{4a}$. Then by solving $s,y$ in terms of $a$, we get

$s=\frac{3+4a-4a^{2}}{8a},y=\frac{4a^{2}+3}{8a}$

Hence we parameterize all the $s$ where $f(s)$ is a square in terms of $a$

We see that $a$ and $\frac{-3}{4a}$ gives the same value for $s$. Hence each
$s$ value with $f(s)$ being a square has two distinct $a$ corresponding to it
except when $a=\frac{-3}{4a}$. In which case, $a=\pm\frac{\sqrt{-3}}{2}$, and
$s=\frac{1}{2}\pm\frac{\sqrt{-3}}{2}$ and $f(s)=0$.

In summary, if $-3$ is not a square in $\mathbb{F}_{q}$, then $f(s)$ is never
zero, and each $s$ with $f(s)$ being a non-zero square has two $a$
corresponding to it in the above parameterization. Hence there are
$\frac{q-1}{2}$ choices of $s$ value which makes $f(s)$ being a non-zero
square. We also need to exclude the value $s=0,1$, then we get $\frac{q-5}{2}$
choices of $s$ value which makes $discH_{2}(s)$ being a non-zero square. Each
$s$ value of such has two corresponding $r$ such that $H_{2}(r,s)=0$.
Therefore, we get in total $q-5$ pairs of $(r,s)$ such that the conic pair
$(C_{r},C_{s})$ satisfies the PTC.

If $-3$ is a square in $\mathbb{F}_{q}$, then there are $2$ values of $s$ that
make $f(s)=0$ and $\frac{q-3}{2}$($a\neq 0,\pm\frac{\sqrt{-3}}{2}$) choices of
$s$ value that make $f(s)$ to be a non-zero square. Then by excluding $s=0,1$
as previous, we get in total $2+2\cdot\frac{q-7}{2}=q-5$ pairs of conics in
this pencil that satisfy the PTC.

###  2.2 Intersection type (2,1,1)

  * •

Generators: $F=xy$, $G=y^{2}+yz+xz+ez^{2}$ where $T^{2}+T+e$ irreducible over
$\mathbb{F}_{q}$.

  * •

Singular members: $\eta=\infty$

  * •

$H_{2}(r,s)=r^{2}(4e-1)+r(4s^{3}e^{2}-6s^{2}e-4se+4s-2)+(-s^{4}e^{2}+6s^{2}e-4s+3)$

  * •

$discH_{2}=(16)(s^{2}e^{2}-se-3e+1)(s^{2}e-s+1)^{2}$

  * •

Density of smooth pairs satisfying PTC: $\frac{q-1}{q(q-1)}=\frac{1}{q}$

  * •

Proof: Since $T^{2}+T+e$ is irreducible over $\mathbb{F}_{q}$, we know that
$1-4e$ is not a square over $\mathbb{F}_{q}$. In particular, it is not zero.
Hence $H_{2}(r,s)$ is a quadratic polynomial in variable $r$ (The coefficient
for $r^{2}$ in $H_{2}$ is never zero).

Note that $(s^{2}e-s+1)$ (which is the square part in $discH_{2}$) has no
roots over $\mathbb{F}_{q}$ since its discriminant is $1-4e$ which cannot be a
square by assumption. Therefore, $discH_{2}$ is zero/non-zero square if and
only if $(s^{2}e^{2}-se-3e+1)$ is zero/non-zero square if and only if
$s^{2}-se^{-1}-3e^{-1}+e^{-2}$ is zero/non-zero square

Then we follow the same proof as in previous. We factor
$s^{2}-se^{-1}-3e^{-1}+e^{-2}=(s-b)^{2}+c$ where $b=\frac{e^{-1}}{2}$ and
$c=\frac{3(1-4e)}{4e^{2}}\neq 0$ (If $c=0$, then $1-4e=0$ which is a
contradiction). If $(s-b)^{2}+c=y^{2}$ for some $y\in\mathbb{F}_{q}$, then we
have $(y-s+b)(y+s-b)=c$. We call the first term as $a$ which is non-zero as
$c\neq 0$, and the second term as $c/a$. Then by solving $s$ in terms of
$\alpha$, we get

$s=\frac{c+2ba-a^{2}}{2a}\;\;\;\;\;\;\;a\in\mathbb{F}_{q}\setminus\\{0\\}$

Note that $a,-c/a$ will give the same value for $s$.

Now, we separate into two cases.

The first case is $\sqrt{-c}\in\mathbb{F}_{q}$. Then $(s-b)^{2}+c$ has two
roots over $\mathbb{F}_{q}$ which are $s=b\pm(\sqrt{-c})$. Note that happens
exactly when $a=-c/a=\pm\sqrt{-c}$. Then for the other $q-3$ choices of $a$,
two of them will give one $s$ value which makes $(s-b)^{2}+c$ to be a non-zero
square. In summary, we have $2$ $s$-values such that $\text{disc}H_{2}=0$,
which means that each of them corresponds to exactly $1$ $r$-value so that
they satisfy the PTC condition. There are also $\frac{q-3}{2}$ $s$-values such
that $\text{disc}H_{2}$ is a non-zero square, which means that each of them
corresponds to $2$ $r$-values so that they satisfy the PTC condition. Then we
have $2+\frac{q-3}{2}\cdot 2=q-1$ number of pairs of $(r,s)$ that satisfies
the PTC condition.

The second case is $\sqrt{-c}\notin\mathbb{F}_{q}$. Then in this case,
$(s-b)^{2}+c$ has no roots over $\mathbb{F}_{q}$, and there are
$\frac{q-1}{2}$ choice of $s$-values which makes it a non-zero square. In
total, there are $\frac{q-1}{2}\cdot 2=q-1$ smooth pairs of conics that
satisfy the PTC in this pencil.

###  2.3 Intersection type (2,2)

  * •

Generators: $F=xy$, $G=e_{1}x^{2}+e_{2}y^{2}+xz+yz+z^{2}$ where
$T^{2}+T+e_{1}$, $T^{2}+T+e_{2}$ are irreducible over $\mathbb{F}_{q}$.

  * •

Singular members: $\eta_{3}=\infty$,
$\eta_{1}=\frac{1+\sqrt{(1-4e_{1})(1-4e_{2})}}{2}$,
$\eta_{2}=\frac{1-\sqrt{(1-4e_{1})(1-4e_{2})}}{2}$

  * •

$H_{2}(r,s)=r^{2}(-16e_{1}e_{2}+4e_{1}+4e_{2}-1)+r(16se_{1}e_{2}+4s^{3}-6s^{2}-4se_{1}-4se_{2}+8e_{1}e_{2}+4s-2e_{1}-2e_{2})+(-s^{4}-24s^{2}e_{1}e_{2}+48e_{1}^{2}e_{2}^{2}+6s^{2}e_{1}+6s^{2}e_{2}+16se_{1}e_{2}-24e_{1}^{2}e_{2}-24e_{1}e_{2}^{2}-4se_{1}-4se_{2}+3e_{1}^{2}+3e_{2}^{2}+6e_{1}e_{2})$

  * •

$discH_{2}=(16)(s^{2}+12e_{1}e_{2}-s-3e_{1}-3e_{2}+1)(-s^{2}+4e_{1}e_{2}+s-e_{1}-e_{2})^{2}$

  * •

Density of smooth pairs satisfying PTC: $\frac{q-5}{(q-3)(q-2)}$

  * •

Proof: We know that $1-4e_{1},1-4e_{2}$ are not square over $\mathbb{F}_{q}$
by the assumption. In particular, neither of them can be zero. Then the
coefficient for $r^{2}$ in $H_{2}$ is
$(-16e_{1}e_{2}+4e_{1}+4e_{2}-1)=-(4e_{1}-1)(4e_{2}-1)\neq 0$. Hence $H_{2}$
is always a quadratic polynomial in $r$.

The square term in $discH_{2}$ is $(-s^{2}+4e_{1}e_{2}+s-e_{1}-e_{2})$. It has
two roots which are exactly $\eta_{1},\eta_{2}$. Hence for
$s\neq\eta_{1},\eta_{2}$, we know that $discH_{2}$ is zero/non-zero square if
and only if $s^{2}+12e_{1}e_{2}-s-3e_{1}-3e_{2}+1$ is zero/non-zero square.

When $s=\eta_{1},\eta_{2}$, we know that $discH_{2}=0$. Hence there are one
$r$-value each corresponding to the $s$-value above. We see that
$H_{2}(\eta_{1},\eta_{1})=0$ and $H_{2}(\eta_{2},\eta_{2})=0$. Further,
$H_{2}(\eta_{1},s)=(s-\eta_{1})^{4}$ and $H_{2}(\eta_{2},s)=(s-\eta_{2})^{4}$.
Hence for $s\neq\eta_{1},\eta_{2}$, if $H_{2}(r,s)$ has solutions for $r$,
then $r\neq\eta_{1},\eta_{2}$ either.

We follow the same strategy as above, We write
$s^{2}+12e_{1}e_{2}-s-3e_{1}-3e_{2}+1=(s-b)^{2}+c$ where $b=\frac{1}{2}$,
$c=\frac{3}{4}(4e_{1}-1)(4e_{2}-1)\neq 0$. We still separate it into two
cases, but we need to add one more condition that $s\neq\eta_{1},\eta_{2}$.
This will remove $4$ choices for $\alpha$ (Not two choices since
$b\pm\sqrt{-c}\neq\eta_{1},\eta_{2}$). Hence there are either $\frac{q-5}{2}$
choices of $s\in\mathbb{F}_{q}\setminus\\{\eta_{1},\eta_{2}\\}$ which makes
$discH_{2}$ to be non-zero square ($\sqrt{-c}\notin\mathbb{F}_{q}$), or
$\frac{q-7}{2}$ choices of
$s\in\mathbb{F}_{q}\setminus\\{\eta_{1},\eta_{2}\\}$ makes $discH_{2}$ into
non-zero square, and $2$ choices of $s$ makes $discH_{2}$ into zero
($\sqrt{-c}\in\mathbb{F}_{q}$).

In summary, we have $q-5$ smooth pairs of conics that satisfy the PTC in this
pencil.

###  2.4 Intersection type (4)

  * •

Generators: $F=x^{2}-ay^{2}$, $G=z^{2}-by^{2}+2cxy$ where
$\sqrt{a}\notin\mathbb{F}_{q}$, $\sqrt{b^{2}-4ac^{2}}\notin\mathbb{F}_{q}$.

  * •

Singular members: $\eta=\infty$

  * •

$H_{2}(r,s)=r^{2}(4ac^{2}-b^{2})+r(4s^{3}a^{2}+6s^{2}ab-4sac^{2}+4sb^{2}+2bc^{2})+(-s^{4}a^{2}+6s^{2}ac^{2}+4sbc^{2}+3c^{4})$

  * •

$discH_{2}=(16)(s^{2}a^{2}+sab-3ac^{2}+b^{2})(s^{2}a+sb+c^{2})^{2}$

  * •

Density of smooth pairs satisfying PTC: $\frac{q-1}{q(q-1)}=\frac{1}{q}$

  * •

Proof: This is very similar to the $(2,1,1)$ case. So we sketch the proof
instead of writing it in details.

    1. 1.

$H_{2}$ is always a degree $2$ polynomial in $r$ by the assumption.

    2. 2.

The square term of $discH_{2}$ has no roots over $\mathbb{F}_{q}$. $discH_{2}$
is zero/non-zero square if and only if $s^{2}a^{2}+sab-3ac^{2}+b^{2}$ is
zero/non-zero square.

    3. 3.

There is only one singular member $\eta=\infty$. We follow the same proof as
in the pencil with intersection type $(2,1,1)$, and we get $q-1$ pairs of
smooth conics in this pencil satisfying the PTC.

###  2.5 Intersection type (3,1)

  * •

Generators: $F=y^{2}-xz$, $G=x^{2}+by^{2}+cxy+yz$ where $T^{3}+bT^{2}+cT+1$ is
irreducible over $\mathbb{F}_{q}$.

  * •

Singular members: None

  * •

$H_{2}(r,s)=r^{2}(3s^{4}+4s^{3}b+6s^{2}c-c^{2}+12s+4b)+r(2s^{4}b+4s^{3}b^{2}-4s^{3}c+6s^{2}bc+4sc^{2}-18s^{2}-4sb+2c)+(-s^{4}b^{2}+4s^{4}c+12s^{3}+6s^{2}b+4sc+3)$

  * •

$H_{2}(\infty,s)=3s^{4}+4s^{3}b+6s^{2}c-c^{2}+12s+4b$ (This is exactly the
coefficient of $r^{2}$ term in $H_{2}(r,s)$)

  * •

$H_{2}(r,\infty)=3r^{2}+2rb-b^{2}+4c$ with discriminant $16(b^{2}-3c)$

  * •

$discH_{2}=(16)(s^{2}(b^{2}-3c)+s(bc-9)+c^{2}-3b)(s^{3}+s^{2}b+sc+1)^{2}$

  * •

Density of smooth pairs satisfying PTC: $\frac{q+1}{q(q+1)}=\frac{1}{q}$

  * •

Proof: The key difference with the previous case is that $H_{2}$ could be non-
quadratic in terms of $r$, and $discH_{2}(s)$ could also be either linear or
quadratic. Also, this pencil has no singular members, so we also need to
consider the case when $r,s=\infty$. Also, by Lemma 2.1, we know that
$H_{2}(\infty,\infty)\neq 0$. Hence we do not need to worry about the case
when both $r,s=\infty$.

We will first see that we always treat $H_{2}$ as a quadratic polynomial in
$r$ by ignoring the case when $r=\infty$.

The coefficient for $r^{2}$ in $H_{2}(r,s)$ is exactly $H_{2}(\infty,s)$.
Hence if $H_{2}(\infty,s)$ has a root $s_{0}\in\mathbb{F}_{q}$, then we get
$(\infty,s_{0})$ when counting the the pairs of conics satisfying PTC. For
that value of $s_{0}$, we see that $discH_{2}$ is the square of the
coefficient of the linear term in $H_{2}$.

If this number is not zero, then $discH_{2}$ is a non-zero square, and we
would get two values of $r$ such that $H_{2}(r,s_{0})=0$ if we are assuming
that $H_{2}$ is a quadratic polynomial. What really happens here is that
$H_{2}(r,s_{0})$ is a linear polynomial, and there is one value
$r_{0}\in\mathbb{F}_{q}$ such that $H_{2}(r_{0},s_{0})=0$. Note that we also
have $H_{2}(\infty,s_{0})=0$, and hence we can view $r_{0}$ and $\infty$ as
the two solutions to $H_{2}(r,s_{0})=0$.

If the linear term of $H_{2}$ is zero, then $discH_{2}=0$, and we are
expecting one value of $r$ such that $H_{2}(r,s_{0})=0$ if we are assuming
that $H_{2}$ is quadratic. In this case, we claim that $H_{2}(r,s_{0})$ equal
to a non-zero number. If this were zero, then any values of
$r\in\mathbb{F}_{q}$ will make $H_{2}(r,s_{0})=0$. In particular,
$H_{2}(s_{0},s_{0})=0$, but this is a contradiction to Lemma 2.1, which states
that the pair of the the same smooth conics cannot satisfy the PTC. Therefore,
in this case, $H_{2}(r,s_{0})$ is a fixed non-zero number, and
$H_{2}(r,s_{0})=0$ has no solutions for $r\in\mathbb{F}_{q}$. The one
hypothetical solution is exactly given by $r=\infty$.

In summary, we can always assume that $H_{2}(r,s)$ is a quadratic polynomial,
and ignore the case where $r=\infty$.

Then we follow the same method as previous. The first observation is that the
square factor in $discH_{2}$ is never zero by assumption. Hence $discH_{2}$ is
zero/non-zero square if and only if $s^{2}(b^{2}-3c)+s(bc-9)+c^{2}-3b$ is
zero/non-zero square.

As pointed out in Chipalkatti’s paper ([2], Page 11), $b^{2}-3c$ and $bc-9$
cannot be both zero. Hence $discH_{2}$ is either quadratic or linear, but
never a constant polynomial. We separate it into three cases.

In the first case, if $b^{2}-3c$ is zero, then $discH_{2}$ is a linear
function, and there are $\frac{q-1}{2}$ choice of $s$ which makes $discH_{2}$
to be non-zero square, and $1$ choice of $s$ makes it zero. This in total
gives $q$ pairs of $(r,s)$. Note that the discriminant of $H_{2}(r,\infty)$ is
also zero in this case. Hence there is one more pair $(r_{0},\infty)$ coming
from $s=\infty$. In total, we get $q+1$ pairs of smooth conics in this pencil
satisfying the PTC.

In the second case, we assume that $b^{2}-3c$ is a non-zero square. Then for
$s=\infty$, there are two pairs $(r_{0},\infty)$, $(r_{1},\infty)$ that
satisfies the PTC. For $s\neq\infty$, we follow the same method as in the
$(2,1,1)$ case, and get $q-1$ pairs of $(r,s)$ with $H_{2}(r,s)=0$. This in
total gives $q+1$ pairs again.

In the third case, we assume that $b^{2}-3c$ is not a square over
$\mathbb{F}_{q}$. Then there is no $r$ such that $(r,\infty)$ satisfies the
PTC. Further by dividing out by $b^{2}-3c$, we see that $discH_{2}$ is
zero/non-zero square if and only if
$s^{2}+s\frac{bc-9}{b^{2}-3c}+\frac{c^{2}-3b}{b^{2}-3c}$ is zero/non-square.
We still use the same method as before. In one case, there are $2$ values of
$s$ makes this polynomial zero, and $q-(\frac{q-3}{2}+2)=\frac{q-1}{2}$ values
of $s$ makes it into non-square. In total, there are $2+\frac{q+1}{2}\cdot
2=q+1$ pairs of $(r,s)$ that satisfies the PTC. The other case is that
$s^{2}+s\frac{bc-9}{b^{2}-3c}+\frac{c^{2}-3b}{b^{2}-3c}$ has no roots in
$\mathbb{F}_{q}$, and there are $q-(\frac{q-1}{2})=\frac{q+1}{2}$ number of
$s$ makes it non-square. Hence there are in total $q+1$ pairs of $(r,s)$ that
satisfies the PTC.

In all cases, there are in total $q+1$ pairs of smooth conics in this pencil
satisfying the PTC.

By combing the number of smooth pairs of conics satisfying the PTC in each
pencil class with the number of pencils of each class, we get the Theorem 1.4
stated in the section 1.1.

##  3 Generalization to tetragon case

The construction of the Poncelet Triangle using the $j$ map also extends to
arbitrary $n$-gons (See equation (1)). There is also a corresponding Cayley’s
Criterion on the existence of Poncelet $n$-gons involving the Hankel
determinant with entries in $H_{2},H_{3},\ldots$. For example, the existence
of a Poncelet tetragon is equivalent to $H_{3}=0$. Chipalkatti conjectured
based on computational data that the density of pairs of smooth conics having
an Poncelet tetragon is asymptotically111We fix a prime $p$ away from finitely
many bad characteristics, then let $q=p^{k}$. The way we are taking this limit
is by taking $k\rightarrow\infty$ $3/q$. However, we found that the
computation data is only valid for pairs of conics in the pencil of
intersection type $(1,1,1,1)$. We will show that the true asymptotic density
is actually $1/q$.

######  Theorem 3.1.

Given a random pair of smooth conics with transversal intersection, then the
probability that it has a Poncelet tetragon is asymptotically $1/q$. Further,
the asymptotic probabilities in each pencil class are given by

Pencil intersection type | Density of pencils in this type | Probability of having a Poncelet tetragon
---|---|---
$(1,1,1,1)$ | $1/24$ | $3/q$
$(2,1,1)$ | $1/4$ | $1/q$
$(2,2)$ | $1/8$ | $3/q$
$(3,1)$ | $1/3$ | $0$
$(4)$ | $1/4$ | $1/q$

###### Proof.

A pair of smooth conics having a Poncelet tetragon is equivalent to $H_{3}=0$.
What we will do is to factor $H_{3}(r,s)$ in each pencil class and use the
Hasse-Weil bound to estimate the number of rational points on $H_{3}(r,s)$.
The integer before $1/q$ is actually the number of $\mathbb{F}_{q}$-components
of $H_{3}(r,s)$. We also dropped the smoothness requirement as smoothness is a
generic condition, and it will not affect the asymptotic probability
calculation.

In the pencil with intersection type $(1,1,1,1)$ with generators same as
previous, we compute that

$H_{3}(r,s)=(-2rs+s^{2}+r)(s^{2}-r)(s^{2}+r-2s)$

We omit some details here, but the 3 components here is indicating that
$H_{3}(r,s)$ in other pencils will also have $3$ components defined over
$\overline{\mathbb{F}_{q}}$. This is proved by using change of coordinates
over $\overline{\mathbb{F}_{q}}$ to change from one pencil to another and
change of generators inside the pencil, and keeping track of the changes in
the variables $r,s$.

Note that each component is a geometrically irreducible quadratic. Hence
$H_{3}(r,s)$ has $3$ geometrically irreducible components and each of them are
defined over $\mathbb{F}_{q}$. By the Hasse-Weil estimate, we know that
$H_{3}(r,s)$ has $3q+O(\sqrt{q})$ number of rational points, and hence the
density of pairs of conics having a Poncelet tetragon is
$\frac{3}{q}+O(q^{-3/2})$ in this pencil.

Similarly, in the pencil with intersection type $(2,1,1)$, we get

$H_{3}(r,s)=(-2rse+s^{2}e+r-1)(s^{4}e^{2}-2s^{3}e+4r^{2}e-8rse+6s^{2}e-r^{2}+2rs-2s+1)$

Over $\mathbb{F}_{q^{2}}$, the second term factor as

$e^{2}(s^{2}+Ar+Bs+e^{-1})(s^{2}-Ar+(-2e^{-1}-B)s+e^{-1})$

where $A=e^{-1}\sqrt{1-4e}\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$ by the
assumption that $T^{2}+T+e$ is irreducible over $\mathbb{F}_{q}$, and
$B=-e^{-1}-e^{-1}(\sqrt{1-4e})$

Therefore, we know that $H_{3}$ has $3$ components over
$\overline{\mathbb{F}_{q}}$. One of the component is defined over
$\mathbb{F}_{q}$, contributing $q+O(\sqrt{q})$ number of rational points. The
rational points of the second $\mathbb{F}_{q}$ component is a subset of the
intersection of its two $\overline{\mathbb{F}_{q}}$ components, which has at
most $4$ points. Therefore, $H_{3}(r,s)$ has $q+O(\sqrt{q})$ number of
rational points, and the density of pairs conics with a Poncelet tetragon is
given by $\frac{1}{q}+O(q^{-3/2})$

The density results for the other pencils follows from the same computation.
We will just list the factorization of the corresponding $H_{3}(r,s)$ here:

For the pencil class with intersection type $(2,2)$, we have

$\displaystyle H_{3}(r,s)$
$\displaystyle=(-2rs+s^{2}+r+4e_{1}e_{2}-e_{1}-e_{2})(s^{2}+Ar+Bs+C)(s^{2}-Ar+(-2-B)s+C)$

where $A=\sqrt{(1-4e_{1})(1-4e_{2})}$, $B=-1-\sqrt{(1-4e_{1})(1-4e_{2})}$,
$C=e_{1}+e_{2}-4e_{1}e_{2}$. Note that $A,B\in\mathbb{F}_{q}$ by the
assumption that $T^{2}+T+e_{1},T^{2}+T+e_{2}$ are irreducible over
$\mathbb{F}_{q}$ ($1-4e_{1}$, $1-4e_{2}$ are non-square in $\mathbb{F}_{q}$,
and hence their product is a square in $\mathbb{F}_{q}$).

Therefore, $H_{3}(r,s)$ has $3$ components over $\mathbb{F}_{q}$ (and also
over $\overline{\mathbb{F}_{q}}$). It follows that the density in this pencil
is $3/q+O(q^{-3/2})$

For the pencil class with intersection type $(4)$, we have

$H_{3}(r,s)=a^{2}(2rsa-s^{2}a+rb+c^{2})(s^{2}+Ar+Bs+a^{-1}c^{2})(s^{2}-Ar+(2a^{-1}b-B)s+a^{-1}c^{2})$

where $A=a^{-1}\sqrt{b^{2}-4ac^{2}}$, $B=a^{-1}b-a^{-1}\sqrt{b^{2}-4ac^{2}}$.
Note that $A,B\notin\mathbb{F}_{q}$ by the assumption. Hence $H_{3}$ has $3$
components over $\overline{\mathbb{F}_{q}}$, and only one of them is defined
over $\mathbb{F}_{q}$, and the other two are conjugate $\mathbb{F}_{q^{2}}$
components. Similar to the $(2,1,1)$ case, we get the density
$1/q+O(q^{-3/2})$ in this pencil.

For the pencil class with intersection type $(3,1)$, we get

$\displaystyle H_{3}(r,s)=r^{3}s^{6}$
$\displaystyle+2r^{3}s^{5}b+r^{2}s^{6}b+4r^{2}s^{5}b^{2}-rs^{6}b^{2}+2rs^{5}b^{3}-s^{6}b^{3}+5r^{3}s^{4}c$
$\displaystyle-6r^{2}s^{5}c+4rs^{6}c+10r^{2}s^{4}bc-8rs^{5}bc+4s^{6}bc+5rs^{4}b^{2}c-2s^{5}b^{2}c-5r^{3}s^{2}c^{2}$
$\displaystyle+20r^{2}s^{3}c^{2}-20rs^{4}c^{2}+8s^{5}c^{2}-2r^{3}sbc^{2}+5r^{2}s^{2}bc^{2}+20r^{3}s^{3}-45r^{2}s^{4}+36rs^{5}-8s^{6}$
$\displaystyle+20r^{3}s^{2}b-40r^{2}s^{3}b+30rs^{4}b-4s^{5}b+8r^{3}sb^{2}-20r^{2}s^{2}b^{2}+20rs^{3}b^{2}-5s^{4}b^{2}-r^{3}c^{3}$
$\displaystyle+2r^{2}sc^{3}-4r^{3}sc+30r^{2}s^{2}c-40rs^{3}c+20s^{4}c+4r^{3}bc-8r^{2}sbc+10rs^{2}bc-r^{2}c^{2}+4rsc^{2}$
$\displaystyle-8r^{3}+36r^{2}s-45rs^{2}+20s^{3}+4r^{2}b-6rsb+5s^{2}b+rc+2sc+1$

Over $\mathbb{F}_{q^{3}}$, it factor as

$\displaystyle H_{3}(r,s)=$
$\displaystyle(rs^{2}+(-2\alpha)rs+(2\alpha+b)s^{2}+(-2\alpha^{2}-2b\alpha-c)r+(2\alpha^{2}+2b\alpha+2c)s+1)$
$\displaystyle\cdot(rs^{2}+(-2\alpha^{\prime})rs+(2\alpha^{\prime}+b)s^{2}+(-2\alpha^{\prime
2}-2b\alpha^{\prime}-c)r+(2\alpha^{\prime 2}+2b\alpha^{\prime}+2c)s+1)$
$\displaystyle\cdot(rs^{2}+(-2\alpha^{\prime\prime})rs+(2\alpha^{\prime\prime}+b)s^{2}+(-2\alpha^{\prime\prime
2}-2b\alpha^{\prime\prime}-c)r+(2\alpha^{\prime\prime
2}+2b\alpha^{\prime\prime}+2c)s+1)$

where $\alpha\in\mathbb{F}_{q^{3}}$ is a root to $T^{3}+bT^{2}+cT+1$, and
$\alpha^{\prime},\alpha^{\prime\prime}$ are its conjugates.

The rational points of $H_{3}(r,s)$ will be a subset of the intersection of
these $3$ conjugate cubic curves defined over $\mathbb{F}_{q^{3}}$, which is a
finite number less than or equal to $9$. Hence the density in this pencil is
$0+O(q^{-2})$.

∎

##  4 Appendix

###  4.1 Counting the number of pencils of each intersection type

  1. 1.

$(1,1,1,1)$: There is no restriction on first point. The second point cannot
be the same as the first point. The third point not on the line passing
through the first two. The last point is not on any of the $3$ lines passing
through two previous points. We divide by $4!$ as the order of the $4$ points
does not matter.

$\frac{1}{4!}(q^{2}+q+1)(q^{2}+q)q^{2}(q^{2}-3(q+1)+3)$

  2. 2.

$(2,1,1)$: We pick first point in $\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$,
and the second point is the conjugate to the first point. We require that the
third point not on the line passing though first two points. Note that the
line passing through the first two points is a $\mathbb{F}_{q}$ line, and
hence we remove $q+1$ choices for the third point.

Note that the line connecting the first/second point with the third point is
not defined over $\mathbb{F}_{q}$ (those two lines are conjugate to each
other), and it has at most $1$ $\mathbb{F}_{q}$ point (a line with two
$\mathbb{F}_{q}$ points are defined over $\mathbb{F}_{q}$) which is the third
point. Hence the last point is not on the line passing through the first two,
and not equal to the third point.

$\frac{1}{2\cdot 2}(q^{4}+q^{2}+1-q^{2}-q-1)(q^{2})(q^{2}-1)$

  3. 3.

$(2,2)$: We pick first point in $\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$,
and the second point is conjugate to first point. The third point is also a
$\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$ point, that is not on the line
passing through the first two points. The last point is the conjugate to the
third points. Note that the last point is not on the line passing through any
two points of the previous. This is because that the line connecting the first
two points, and the last two points are $\mathbb{F}_{q}$ lines.

$\frac{1}{2\cdot 2\cdot 2}(q^{4}-q)(q^{4}-q-(q^{2}+1-q-1))$

  4. 4.

$(3,1)$: We pick first point in $\mathbb{F}_{q^{3}}\setminus\mathbb{F}_{q}$.
All its conjugate lie on the same line if and only if they lie on a
$\mathbb{F}_{q}$ line. Passing through each
$\mathbb{F}_{q^{3}}\setminus\mathbb{F}_{q}$ point, there is at most one
$\mathbb{F}_{q}$ line (two points determine a line, $P,\sigma(P)$ must all on
that line). There are in total $q^{2}+q+1$ number of $\mathbb{F}_{q}$ lines,
and each of them has $q^{3}-q$ number of
$\mathbb{F}_{q^{3}}\setminus\mathbb{F}_{q}$ point, and there is no repetition
among them. Therefore, there are in total
$q^{6}+q^{3}+1-q^{2}-q-1-(q^{2}+q+1)(q^{3}-q)$ number of choice for the first
point.

If the last $\mathbb{F}_{q}$ point is on one of the line passing through two
points of the first three, then it is on all those 3 lines since those 3 lines
are conjugate. However, this is impossible as those 3 lines does not intersect
at $1$ point. So there is no restriction to the last $\mathbb{F}_{q}$ point.

$\frac{1}{3}(q^{6}+q^{3}+1-q^{2}-q-1-(q^{2}+q+1)(q^{3}-q))(q^{2}+q+1)$

  5. 5.

$(4)$: Same as previous, the first
$\mathbb{F}_{q^{4}}\setminus(\mathbb{F}_{q}\cup\mathbb{F}_{q^{2}})$ point
cannot lies on a $\mathbb{F}_{q}$ line. If 3 of them lies on the same line,
the all of them must lies on the same line, and that line is a
$\mathbb{F}_{q}$ line.

$\frac{1}{4}(q^{8}+q^{4}+1-q^{4}-q^{2}-1-(q^{2}+q+1)(q^{4}-q^{2}))$

###  4.2 Computation data for small $q$ and conjectures for higher $n$-gons

All the below computations are done by SageMath.

We will always fix the conic $\mathcal{B}$ to be to $x^{2}+y^{2}+z^{2}$ (which
is equivalent to fixing the second matrix to be $I$), and count the number of
first conic $\mathcal{A}$ such that there is a Poncelet n-gon in the pair
$(\mathcal{A},\mathcal{B})$. We will forget the requirement of transversal
intersection and smoothness. Both of the conditions are generic, and hence
they will not influence the main term when we are computing the asymptotic
density.

The computation is done by running over all possible symmetric matrices
$3\times 3$ matrices $A$ over $\mathbb{F}_{q}$, and count the numbers of $A$’s
such that the pair $(A,I)$ satisfying the Poncelet n-gon condition. When
stating the density, we divide the previous number by $q^{5}$, so that the
approximate density would that number times $1/q$.

  1. 1.

$n=4$ (tetragon case):

$q$ | Density ($\times 1/q$)
---|---
3 | 29/27
5 | 129/125
7 | 349/343
9 | 737/729
11 | 1341/1331
13 | 2209/2197
17 | 4929/4913

  2. 2.

$n=5$

$q$ | Density ($\times 1/q$)
---|---
3 | 65/81
5 | 549/625
7 | 2113/2401
9 | 6065/6561
11 | 13661/14641
13 | 26545/28561

  3. 3.

$n=6$

$q$ | Density ($\times 1/q$)
---|---
3 | 125/81
5 | 1049/625
7 | 4213/2401
9 | 11825/6561
11 | 26861/14641

We see that the density for the tetragon case is approximately $1/q$, which is
similar to what we obtained in the previous section.

We denote $f_{n}(r,s)$ to be the polynomial that determine if a pair of conics
$(C_{r}=rF+G,C_{s}=sF+G)$ in a pencil with generator $F,G$ satisfies the
Poncelet $n$-gon condition or not. For example, $f_{3}(r,s)=H_{2}(r,s)$,
$f_{4}(r,s)=H_{3}(r,s)$, and $f_{5}(r,s)=H_{2}H_{4}-H_{3}^{2}$. Then the
integer sequence $\mu_{n}$ in Chipalkatti’s paper ([2], Page 13, Conjecture
4.1) is exactly the number of $\mathbb{F}_{q}$ components of the corresponding
$f_{n}$ in the pencil with intersection type $(1,1,1,1)$.

We already see that in section $3$ that $f_{4}=H_{3}$ in the pencil class with
intersection type $(1,1,1,1)$ has $3$ geometrically irreducible
$\mathbb{F}_{q}$ components. This explains $\mu_{4}=3$ in Chipalkatti’s
conjecture. Similarly, the $\mu_{6}=4$ in the conjecture is coming from the
$4$ $\mathbb{F}_{q}$ components of $f_{6}(r,s)$ in the pencil class with
intersection type $(1,1,1,1)$:

$\displaystyle f_{6}(r,s)=(s^{4}-4r^{2}s+$ $\displaystyle
6rs^{2}-4s^{3}+r^{2})(s^{4}+4r^{2}s-6rs^{2}-3r^{2}+4rs)$
$\displaystyle\cdot(-4rs^{3}+s^{4}+6rs^{2}+r^{2}-4rs)(-4rs^{3}+3s^{4}+6rs^{2}-4s^{3}-r^{2})$

We correct Chipalkatti’s conjecture regarding the proportion of conic pairs in
$\mathbb{P}^{2}(\mathbb{F}_{q})$ that satisfies the $n$-gon condition as
following:

######  Conjecture 4.1.

The proportion of conic pairs in $\mathbb{P}^{2}(\mathbb{F}_{q})$ that
satisfies the $n$-gon condition is asymptotically equal to $\mu_{n}/q$ where
$\mu_{n}$ are given by

$\mu_{5}=1,\mu_{6}=2,\ldots$

Note that we arelady proved that $\mu_{3}=1$ and $\mu_{4}=1$.

## References

  * [1] P. Griffiths, J. Harris, _On Cayley’s explicit solution to Poncelet’s porism_ , Enseign. Math. (2) 24(1–2) (1978) 31–40.
  * [2] J. Chipalkatti, _On the Poncelet triangle condition over finite fields_ , Finite Fields and Their Applications. 45 (2017) 59-72.