Poncelet's Porism is a well-known and much-beloved result in plane
geometry, discovered by Jean-Victor Poncelet in 1813. Here is the
statement:
Theorem [Poncelet's Porism]
Let $C$, $D$ be non-singular conics in general position in the projective plane,
and $n \ge 3$ a natural number. If there exists an $n$-sided polygon,
all of whose vertices lie on $C$ and all of whose edges are tangent to
$D$, then there exists such a polygon with a vertex at any given point of
$C$.
If this is the first you have heard of Poncelet's Porism, pause for
thought: it is not obvious.
Poncelet's Porism has a lot going for it. It turns out to be an
archetypal result demonstrating the power of "modern" algebraic geometry,
it admits numerous fruitful generalisations, and it has connections to many
beautiful subjects including elliptic curves, modular forms, dynamical
systems, the Painlevé equations, and even instantons.
But there is another reason for its popularity: it
asserts the existence of some very appealing pictures and animations.
The animated gif at the top of this page was created using the app
embedded just below. You can use it to make your own animated gif right now.
On the left hand side, the following two conics are plotted:
$$
\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1, \qquad x^2 + y^2 = 1,
$$
together with the results of Poncelet iteration (the red lines). There are
controls for adjusting the values of $a$, $b$ and the number of edges,
as well as other elements whose functions
are hopefully self-explanatory. We defer an explanation of the curves on the
right until we discuss
Cayley's results.
For anyone interested, the source for the above is available at Microsoft's
code-hosting hub here.
Note that an instance of the porism in this real plane has two natural
topological invariants:
the vertex count, $n$,
the number of times the polygon winds around the circle, $w$.
Note that $w < n$ and that $w$, $n$ are coprime because the graph of
the polygon is connected.
These numbers will show up in different guises below.
Working over $\mathbb{C}$, let $C$, $D$ be non-singular conics in general
position in $\mathbb{P}^2$ and define the incidence space:
$$
X = \{(p, l) \in C\times D^* ~|~ p \in l\},
$$
where $D^*$ is the set of lines tangent to $D$. Because there are
two tangents to $D$ through a general point of $C$, $X$ carries a natural
involution that swaps these tangents:
$
i_D : X \to X.
$
Similarly, because there are two points of $C$ lying on a general tangent
to $D$, $X$ also carries a natural involution that swaps these points:
$
i_C : X \to X.
$
Note that $i_C$ has fixed points corresponding to the four points of
$C^*\cap D^*$, and $i_D$ has fixed
points corresponding to the four points of $C \cap D$.
Composing, we have an endomorphism:
$$
T = i_D \circ i_C : X \to X.
$$
Poncelet's Porism is the statement that $T^n$ has a fixed point iff it is
the identity map.
With this setup, we need only chain together the following basic
facts from general theory:
X is a smooth and has genus one (i.e., topologically a torus).
After choosing a point to act as identity, every smooth
curve of genus one becomes an elliptic curve, isomorphic to a quotient
of $\mathbb{C}$ by a maximal-rank lattice.
The only involutions admitting a fixed point on an elliptic curve are
of the form $u \mapsto v - u$ for some point $v$. (Alternatively, note
that for any $u \in X$, the fibre through $u$ of the natural map $X \to C$ is
$u + i_D(u)$, and that this must be constant. Similarly for $i_C$.)
Applying fact 3 to $i_C$ and $i_D$ we learn that $T$ must be a translation
and thus so also must $T^n$.
We are now done: the only tanslation with a fixed point
is the identity map.
Note that if $O\in X$ is the identity, this proof gives us a new
interpretation of $n$: it is the order of the torsion point $T(O)$. We
will depend upon this interpretation when discussing
Cayley's results below.
If you didn't know the results 1–3 already, you might not be all
that happy with this proof. The good news is, we will give another
more elementary proof, due to Jacobi,
below. In fact it's really the same proof
in different language.
Jacobi's marvellous proof of Poncelet's Porism is characteristic of its
time and benefits from judicious choice of coordinates. In preparation, we comment
on coordinate choices.
Poncelet's Porism concerns only incidence and tangency, and so
is projectively invariant. Since the group of projective transformations of
the plane is eight-dimensional, and the space of conics is five-dimensional,
the space of pairs of conics up to projective equivalence is
two-dimensional (5 + 5 - 8).
There are at least three natural ways to parameterise these two
dimensions:
A circle and a concentric, axis-aligned ellipse. This yields
local coordinates such as the semi-minor, semi-major radii of the
ellipse dividing by the radius of the circle.
A pair of non-concentric circles. This yields local
coordinates such as the radii of the two circles divided by the
distance between their centres.
The two cross ratios of the four intersection points of the conics:
one cross ratio on each conic.
Following Schoenberg, we will present Jacobi's argument in terms
of the parametrisation (i) even though Jacobi used (ii). Note
that (i) exists as a consequence of the well-known result
in linear algebra that for any pair of quadratic
forms, we may always find a basis in which one is a diagonal matrix of
$\pm 1$s, and the other is diagonal. We will discuss the case
over $\mathbb{R}$ with $C$ and $D$ represented by the matrices:
$$
\left[\begin{array}{ccc}
1/a^2 & 0 & 0\\
0 & 1/b^2 & 0\\
0 & 0 & -1/r^2
\end{array}\right]
\quad\mbox{and}\quad
\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{array}\right]
\quad\mbox{respectively.}\quad
$$
These matrices are of course only defined up to scale which allows us to
assume $r=1$. Note that even after this we have still not used all the
symmetry of projective invariance: we have freedom to permute the elements
of our basis. Thus the quotient by full projective invariance is naturally
(an open set of) the weighted projective space
$\mathbb{P}(1, 2, 3) \simeq \mathbb{P}^2/S_3$
where the eigenvalues $1/a^2$, $1/b^2$, $-1/r^2$ are homogeneous coordinates
on the $\mathbb{P}^2$.
Note that the same $S_3$ indeterminacy is
also visible in parameterisation (iii). A cross ratio depends on the ordering
of the four points; without an ordering we only have an orbit of the
anharmonic group (abstractly $S_3$) rather that a single value. These
cross ratios will naturally show up again below when we discuss the
natural map to the pencil of conics.
We work over $\mathbb{R}$ and take our conics
to be an ellipse and concentric circle:
$$
\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = \left(\frac{1}{r}\right)^2, \qquad x^2 + y^2 = 1.
$$
and for simplicity assume $0 < r < b \le a$. The following diagram will be useful:
We wish to study the map that sends the point $P$ to $Q$. The key is to
work in terms of the natural angular coordinate of the
circumcircle of the ellipse. More precisely, we associate points of
the ellipse with their images under the map $(x, y) \mapsto (x, ay/b)$.
This maps the points $P$, $Q$ on the ellipse to the $P'$, $Q'$ on the
circumcircle, and likewise the tangency point $M$ to $M'$. Then define
the angles:
$$
\phi_0 = \angle{AOP'},\quad \phi_1 = \angle AOQ'.
$$
We seek to understand $\phi_1$ as a function of $\phi_0$. We need another
diagram:
The ellipse in this diagram is the image of the unit circle under the same
map $(x, y) \mapsto (x, ay/b)$ used above, the line $UV$ corresponds to
the original line $P'Q'$ except corresponding to the angle $\phi_0 + \Delta\phi_0$
instead of $\phi_0$,
and the point $I$ is the intersection of $P'Q'$ and $UV$. The lengths
$\Delta\phi_0$, $\Delta\phi_1$ are the blue and green arcs shown.
Now we begin, firstly we have:
$$
\frac{d\phi_1}{d\phi_0} = \lim_{\Delta\phi_0 \to 0}\frac{\Delta\phi_1}{\Delta\phi_0}
= \lim\frac{VQ'}{UP'} = \lim\frac{IQ'}{IU} = \frac{\lim IQ'}{\lim IU} = \frac{M'Q'}{M'P'},
$$
(where we have used
$\lim\frac{\Delta\phi_0}{UP'} = \lim\frac{\Delta\phi_1}{VQ'} = 1$ as
well as the fact that the triangles $IP'U$, $IQ'V$ are similar).
Secondly, trivially:
$$
\frac{M'Q'}{M'P'} = \frac{MQ}{MP}
$$
(both sides are equal to the corresponding ratios of $x$
coordinates.) Thirdly, by Pythagoras:
$$
\begin{align}
MP &= \sqrt{a^2/r^2 - 1}\cdot\sqrt{1 - m\sin^2\phi_0},\\
MQ &= \sqrt{a^2/r^2 - 1}\cdot\sqrt{1 - m\sin^2\phi_1}.\\
\end{align}
$$
where $m \in [0, 1)$ is given by:
$$
m = \frac{a^2 - b^2}{a^2 - r^2}.
$$
Combining these three facts, we obtain the key identity:
$$
\frac{d\phi_1}{d\phi_0} = \sqrt{\frac{1 - m\sin^2\phi_1}{1 - m\sin^2\phi_0}}.
$$
Equivalently, the function:
$$
\omega(\phi_0) = \int_{\phi_0}^{\phi_1}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}}
$$
is constant.
Now suppose
$\phi_0, \phi_1, \ldots, \phi_n$ is a sequence of angles associated with
the Poncelet construction, and that the $\phi_i$ keep track of how many
times we have wound round the circle by adding multiples of $2\pi$ if
necessary. The $\phi_i$ correspond to a Poncelet $n$-gon iff:
$$
\phi_n = \phi_0 + 2\pi w,
$$
for some natural number $w$. Since the integrand is positive, this is true
iff
$$
\int_{\phi_0}^{\phi_n}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}} =
\int_{\phi_0}^{\phi_0 + 2\pi w}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}},
$$
and this in turn is true iff:
$$
n\omega = \omega(\phi_0) + \cdots + \omega(\phi_{n-1}) = 4wK(m),
$$
where:
$$
K(m) = \int_0^{\pi/2}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}}
= \frac{1}{4}\int_0^{2\pi}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}},
$$
is the complete elliptic integral of the first kind.
We are done: the above shows that the condition for the Poncelet
construction to close is that:
$$
\omega/K(m) \in \mathbb{Q},
$$
which does not depend on $\phi_0$.
We claimed above that Griffiths and Harris's proof was really the same
as Jacobi's. The reason is that both are really just applications of the
Abel-Jacobi map.
Indeed, of the three "basic facts from general theory" to which we appealed
when giving the
Griffiths and Harris proof, number 1 is easy
and 3 is an easy consequence of 2, but 2 is a deep result. It is a
consequence of the fact that the
Abel-Jacobi map is an isomorphism for genus-one curves. (In general the
Abel-Jaocbi map is a sort-of linearisation map, but genus one curves are
already linear.)
More concretely, the point is that the key quantity:
$$
\frac{d\psi}{\sqrt{1-m\sin^2\psi}},
$$
which shows up in Jacobi's proof is the so-called "invariant differential"
(the unique, up to scale, holomorphic 1-form) on the elliptic curve,
and the map:
$$
\phi \mapsto \int_{\phi_0}^{\phi}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}},
$$
is the real component of the Abel-Jacobi map, which gives an isomorphism
between the identity component of the elliptic curve and the circle:
$$
\mathbb{R}/4K(m)\mathbb{Z}.
$$
For clarity, note that this elliptic curve has two connected components
over $\mathbb{R}$. Set-wise it is the space of pairs:
$$
X = \{(p, l) \in C\times D^* ~|~ p \in l\}.
$$
When working over $\mathbb{R}$ with an ellipse and circle as above,
the two connected components of $X$ are distinguished according to whether
$p$ is at the clockwise or anti-clockwise end of $l$ (relative to the
point of tangency with the circle).
Given two conics, how do we tell if there exists a Poncelet $n$-gon?
And if so, what is $n$?
Jacobi's proof essentially includes answers to the above questions. We
know that a necessary and sufficient condition for a pair of conics
to admit a Poncelet $n$-gon is:
$$
\frac{\omega}{4K(m)} = \frac{w}{n},
$$
where $w$ is the number of times the $n$-gon wraps round, $K$ is the
complete elliptic ingetral of the first kind, and:
$$
m = \frac{\alpha - \beta}{\alpha - 1},
$$
where for convenience we have introduced $\alpha = a^2/r^2$, $\beta = b^2/r^2$.
The only thing we don't yet know is $\omega$. However an easy
exercise in elementary geometry shows that if $\phi_0 = 0$ then:
$$
\cos\phi_1 = u,
$$
where:
$$
u = \frac{\alpha + \beta - \alpha\beta}{\alpha - \beta + \alpha\beta}.
$$
Then since $\omega$ is given in terms of the incomplete elliptic integral:
$$
\omega = F(\phi_1, m) = \int_{0}^{\phi_1}\limits\frac{d\psi}{\sqrt{1-m\sin^2\psi}}
$$
we have $\phi_1 = {\rm am}(\omega, m)$, where ${\rm am}$ is the Jacobi
amplitude function, and so:
$$
{\rm cn} (\omega, m) = u,
$$
where ${\rm cn} = \cos \circ {\rm am}$ is one of the
Jacobi elliptic functions.
Since ${\rm cn}$ has period $4K(m)$, this determines $\omega$ as:
$$
\omega = cn^{-1}\left(\frac{\alpha + \beta - \alpha\beta}{\alpha - \beta + \alpha\beta}, m\right)
$$
As an example, suppose we wish to determine whether a pair of conics
admit an inscribed quadrilateral, winding around once. By the above, this
occurs iff $\omega = K(m)$, or equivalently $\phi_1 = \pi/2$ which means
$u = 0$, i.e.,
$$
\alpha + \beta = \alpha\beta,
$$
which is the same as:
$$
1/a^2 + 1/b^2 = 1/r^2.
$$
Although it was easy to determine this condition for the existence of a
quadrilateral using the expression for $\omega$ above, in general it is
difficult to determine the equations for existence of Poncelet $n$-gons
this way. Fortunately there is another much easier method, due to
Cayley.
In preparation for Cayley's result, and in danger of being
embarrassingly explicit, we spell out a relevant elementary result.
Proposition
Let $C$, $D$ be a general pair of conics in the projective plane,
together with a distinguished point $P \in C\cap D$. Let $K$
be the pencil of conics containing $C$, $D$ and let $X$ be the
associated incidence curve
(introduced above)
together with its natural endomorphism $T$.
Then we have a natural double cover:
$$
\pi_P : X \to K,
$$
whose ramification locus in $X$ naturally corresponds with $C\cap D$, and
whose branch locus in $K$ is the four points that are three singular
conics in $K$, together with $C \in K$. Furthermore, if $O \in X$ is the
ramification point over the branch point $C \in K$, then:
$$
\pi_P(T(O)) = D.
$$
Proof (sketch) Firstly, given a non-singular conic $C$ with distinguished point
$P \in C$, let $L_P$ be the set of lines through $P$, and let:
$$
\begin{align}
\rho_P : C &\to L_P,\\
Q &\mapsto \mbox{line $PQ$ if $P \ne Q$, otherwise the tangent at $P$},
\end{align}
$$
be the natural isomorphism.
Secondly, given a
general pair of conics $C$, $D$, together with a distinguished point
$P \in C\cap D$, let $K$ denote the pencil of conics containing $C$, $D$.
Because of generality, $C_1^*\cap C_2^*$ is not incident with
$C_1\cap C_2$ for any $C_1, C_2 \in K$ and so we have a well-defined
natural isomorphism:
$$
\begin{align}
\tau_P : L_P &\to K,\\
l &\mapsto \mbox{unique conic in $K$ with tangent $l$ at $P$.}
\end{align}
$$
As zero set of the determinant, the space of
singular conics is a cubic hypersurface in the space of all conics and so
meets $K$ in three points. Under $\tau_P$, these three singular conics in
$K$ naturally correspond to the lines joining $P$ to the three points of
$C\cap D \setminus \{P\}$ and of course $C$ itself, as point in $K$,
correponds to the tangent to $C$ at $P$.
Thirdly, recall the incidence curve $X$ defined above,
together with its natural involutions $i_C$, $i_D$ and their composition
$t = i_D\circ i_C$. Let:
$$
\gamma : X \to C,
$$
be the natural double cover (i.e., the quotient by $i_D$) and note that its
branch locus is $C\cap D$. The map we seek is:
$$
\pi_P = \tau_P \circ \rho_P \circ \gamma
$$
and is trivially verified to have the stated properties. ∎
As an aside, note that the above is related to the
"cross ratio coordinates" mentioned above
as parametrisation (iii). Indeed if we choose an ordering of the three
singular conics in $K$ then we have a unique isomorphism
$K \simeq \mathbb{P}^1$ under which the singular conics correspond to
$0, 1, \infty$. This allows us to regard $C, D \in K$ as elements of
$\mathbb{P}^1 \setminus \{0, 1, \infty\}$, i.e., as two numbers (differing
from 0, 1). This connects with the natural coordinates (iii) and some
related systems.
In 1853, Cayley proved the following beautiful result:
Theorem
Let $C$, $D$ be $3\times 3$ matrices defining quadratic
forms which cut out a general pair of conics in the projective plane.
Let:
$$
\sqrt{\det(tC + D)} = a_0 + a_1 t + a_2 t^2 + \cdots
$$
be the taylor series of a branch of the square root of the cubic
$\det(tC + D)$. Then there exists a Poncelet polygon,
inscribed in $C$ and circumscribed about $D$, with vertex count
dividing $n$ iff the determinant of an associated $\left[\frac{n-1}{2}\right]$-dimensional
Hankel matrix
vanishes, specifically iff:
$$
\left|\begin{array}{cccc}
a_2 & \cdots & a_{p+1}\\
\cdot & & \cdot\\
\cdot & & \cdot\\
a_{p+1} & \cdots & a_{2p}
\end{array}\right| = 0,~\mbox{if $n = 2p+1$,}
\qquad \mbox{or:}\qquad
\left|\begin{array}{cccc}
a_3 & \cdots & a_{p+1}\\
\cdot & & \cdot\\
\cdot & & \cdot\\
a_{p+1} & \cdots & a_{2p-1}
\end{array}\right| = 0,~\mbox{if $n = 2p$}.
$$
The choice of matrices $C$, $D$ for our conics (which by abuse of notation
we continue to denote $C$, $D$) determines a natural affine coordinate
$t$ on $K$, the pencil of conics containing $C$, $D$:
$$
t \mapsto \mbox{conic cut out by $tC + D$},
$$
and so provides an isomorphism $K \simeq \mathbb{P}^1$, under which $C$,
$D$ correspond to $t = \infty, 0$ respectively, and under which the
singular conics correspond to the roots of $\det(tC + D)$.
If we pick a point $P \in C\cap D$ and, in the notation of the
proposition above, take $O \in X$ as
identity, rendering $X$ an elliptic curve, then by
Griffiths and Harris's proof, we know that
$C$, $D$ admit a Poncelet $n$-gon iff $T(O)$ has order $n$. Cayley's result
is thus equivalent to the following:
Theorem [restatement of Cayley's result]
Let $X \to \mathbb{P}^1$ be a double covering by an elliptic curve $X$,
under which the identity is sent to the branch point $\infty$, and the
remaining branch points are the values
$t_1, t_2, t_3 \in \mathbb{P}^1 - \{0,\infty\}$.
Then the points of $X$ over $t=0$
have order dividing $n$ iff the Cayley-type Hankel matrix,
associated to the taylor series:
$$
\sqrt{(t - t_1)(t - t_2)(t - t_3)} = a_0 + a_1 t + a_2 t^2 + \cdots
$$
has vanishing determinant.
By the above, when the identity is taken to be a point of $C\cap D$
(together with the tangent to $D$ at that point) our
incidence curve $X$ is naturally isomorphic to the elliptic curve with
Weierstrass equation:
$$
s^2 = \det(tC + D).
$$
Since the invariant differential of an elliptic curve in this form
is $dt/s$, the invariant
differential of $X$ (with this choice of identity) is:
$$
\frac{dt}{\sqrt{\det(tC + D)}}.
$$
We should be able to use this together with Jacobi's closure condition
given above to
obtain a proof of Cayley's result but we prefer a slick proof due to
Atiyah, presented by Hitchin in [14].
Proof (sketch) Recall that a point $x$ on an elliptic curve has
order dividing $n$
iff there exists a meromorphic function with a zero of order $n$ at $x$,
a pole of order $n$ at the identity and no other zeros or poles. We will
show that the vanishing of the Cayley-type Hankel determinant is sufficient
for the existence of such a function and leave necessity as an exercise.
To avoid case analysis, assume $n = 2p + 1$ is odd;
the argument for even $n$ requires only trivial modifications.
Represent $X$ by the plane curve with equation:
$$
s^2 = (t-t_1)(t-t_2)(t-t_3),
$$
represent finite points on $X$ using coordinates $(t, s)$, let $O$ be the
identity, and let $a_0 = \sqrt{-t_1 t_2 t_3}$ be a choice of square root.
Consider the two natural meromorphic functions $s$, $t$ on $X$ and note
that:
$s$ has a pole of order 3 at $O$, and simple zeros at the three
points $(t_i, 0)$,
$t$ has a pole of order 2 at $O$, and simple zeros at the two
points $(0, \pm a_0)$,
and that neither function has any other zeros or poles.
With the setup in place, we're ready for Atiyah's trick. Consider the
following $p$ functions:
$$
f_i = t^{p-i}(s - a_0 - a_1t - a_2t^2 - \cdots - a_i t^i)
$$
for $i = 1, \ldots, p$. Note that $f_1$ has a pole of order
$2p+1$ at $O$ and that $f_i$ has a pole of order $2p$ at $O$ for
$i = 2, \ldots, p$.
Now, in a neighbourhood of $(0, a_0)$ we have:
$$
s = a_0 + a_1 t + a_2 t^2 + \cdots,
$$
and so:
$$
\begin{array}{cccccccc}
f_1 & = & a_{2}t^{p+1} & + & a_{3}t^{p+2} & + ~\cdots~ + & a_{p+1}t^{2p} & + ~\cdots,\\
f_2 & = & a_{3}t^{p+1} & + & a_{4}t^{p+2} & + ~\cdots~ + & a_{p+2}t^{2p} & + ~\cdots,\\
&\cdots\\
f_p & = & a_{p+1}t^{p+1} & + & a_{p+2}t^{p+2} & + ~\cdots~ + & a_{2p}t^{2p} & + ~\cdots.\\
\end{array}
$$
Thus, if the relevant determinant vanishes, we can find scalars $\lambda_i$,
$i=1, \ldots, p$ not all zero such that:
$$
f = \lambda_1 f_1 + \cdots + \lambda_p f_p
$$
has a local expansion with no terms of degree less than $t^{2p+1}$, i.e.,
a zero of order at least $2p+1$ at $(0, a_0)$. Since $f$ has a unique
pole of order at most $2p+1$, it must in fact have a unique zero, and
pole, both of order exactly $2p+1$, as required. ∎
We can finally explain the curves on the right hand side of the
app above. These are the curves
corresponding Poncelet configurations for triangles, quadrilaterals,
pentagons etc. plotted in the weighted projective space
$\mathbb{P}(1, 2, 3)$ of Poncelet configurations with local coordinates:
$$
\begin{align}
x &= e_2 / e_1^2,\\
y &= e_3 / e_1^3,
\end{align}
$$
where $e_i$ are the elementary symmetric functions in
$1/a^2$, $1/b^2$, $1/r^2$, i.e.,:
$$
-\det(tC + D) = e_3 t^3 + e_2 t^2 + e_1 t + 1.
$$
Thus, in these coordinates, the series for Cayley's result is:
$$
\sqrt{yt^3 + xt^2 + t + 1} = 1 + \frac{1}{2}t + \frac{1}{8}(-1 + 4x)t^2
+ \frac{1}{16}(1 - 4x + 8y)t^3
+ \frac{1}{128}(-5 + 24x - 16x^2 - 32y)t^4 + \cdots
$$
and the curves are:
$$
\begin{array}{ll}
\mbox{triangles} & 1 - 4x = 0,\\
\mbox{quadrilaterals} & 1 - 4x + 8y = 0,\\
\mbox{pentagons} & 1 - 12 x + 48 x^2 - 64 x^3 - 32 y + 128 x y - 256 y^2 = 0,\\
\mbox{hexagons} & 3 - 20 x + 16 x^2 + 64 x^3 + 96 y - 384 x y + 512 y^2 = 0,\\
\end{array}
$$
etc.
I have collected relevant references below together with unpaywalled
links to their pdfs.
References to Bertrand's work are conspicuously absent. This is because,
though I could locate the works of Bertrand that Schoenberg mentions, I
could not find any mention of Poncelet's porism within them! In particular
I could not find anything relevant at the page locations provided by
Schoenberg. ¯\_(ツ)_/¯
Note that there are numerous beautiful expository
articles on Poncelet's Porism, its history, and its generalisations. (I have
yet to read most of these myself.)
[pdf] Avksentyev, E.A. "A universal measure for a pencil of conics and the Great Poncelet Theorem" Sbornik: Math., 205:5, 613–632 (2014).
[pdf] Avksentyev, E.A. and Protasov, V.Yu. "Universal measure for Poncelet-type theorems" Proc. Amer. Math. Soc. (to appear) (2018).
[pdf] Barth, W., and Bauer Th. "Poncelet theorems", Expo. Math., 14, 125–144 (1996).
[pdf] Barth, W., and Michel, J. "Modular curves and Poncelet polygons", Math. Ann., 295, 25–49 (1993).
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