Poring over Poncelet

Poncelet's Porism

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.

Create your own animated gif

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:

  1. the vertex count, $n$,
  2. 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.

Griffiths and Harris proof

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:

  1. X is a smooth and has genus one (i.e., topologically a torus).
  2. 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.
  3. 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.

Projective invariance

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:

  1. 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.
  2. 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.
  3. 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.

Bertrand-Jacobi-Schoenberg proof

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 $\phi_0$.

Jacobi vs Griffiths and Harris

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).

Jacobi closure condition

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.

Branching over the pencil

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.

Cayley closure conditions

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:

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.)

  1. [pdf] Avksentyev, E.A. "A universal measure for a pencil of conics and the Great Poncelet Theorem" Sbornik: Math., 205:5, 613–632 (2014).
  2. [pdf] Avksentyev, E.A. and Protasov, V.Yu. "Universal measure for Poncelet-type theorems" Proc. Amer. Math. Soc. (to appear) (2018).
  3. [pdf] Barth, W., and Bauer Th. "Poncelet theorems", Expo. Math., 14, 125–144 (1996).
  4. [pdf] Barth, W., and Michel, J. "Modular curves and Poncelet polygons", Math. Ann., 295, 25–49 (1993).
  5. [pdf] Bos, H.J.M., Kers, C., Oort, F., and Raven, D.W. "Poncelet's closure theorem", Expo. Math., 5, 289–364 (1987).
  6. [pdf] Del Centina, A. "Poncelet's Porism: a long story of renewed discoveries, I" Arch. Hist. Exact Sci., 70(1) (2014).
  7. [pdf] Del Centina, A. "Poncelet's Porism: a long story of renewed discoveries, II" Arch. Hist. Exact Sci., 70(2) (2014).
  8. [pdf] Cayley, A. "Note on the Porism of the in-and-circumscribed Polygon", Philos. Mag., 6, 99–102, (1853).
  9. [pdf] Cayley, A. "On the Porism of the in-and-circumscribed Polygon", Phil. Trans. Roy. Soc. London, 151, 225–239, (1861).
  10. [pdf] Griffiths, P., and Harris, J. "On Cayley's explicit solution to Poncelet's Porism", Enseign. Math., 24, 31–40 (1978).
  11. [pdf] Griffiths, P., and Harris, J. "A Poncelet theorem in space", Commentarii Mathematici Helvetici, 52 (1977).
  12. [pdf] Halbeisen, L., Hungerbühler N. "A Simple Proof of Poncelet’s Theorem (on the occasion of its bicentennial)", Amer. Math. Monthly, 121, 1–14 (2014).
  13. [pdf] Hartshorne, R. "Stable vector bundles and Instantons", Commun. Math. Phys., 59, 1–15 (1978).
  14. [pdf] Hitchin, N.J. "Poncelet Polygons and the Painlevé Equations", Geometry and Analysis (Bombay, 1992) Tata Inst. Fund. Res., Bombay, 151–185 (1995).
  15. [pdf] Jacobi, C.G.J. "Ueber die Anwendung der elliptischen Transcendenten auf ein bekanntes Problem der Elementargeometrie", J. Reine Angew. Math., 3, 376–387 (1828). [I am very grateful to D. Reznik for providing me with an English translation [pdf] of this paper due to Prof. H. Stachel.]
  16. [pdf] Kerawala, S.M. "Poncelet Porism in Two Circles", Bull. Calcutta Math. Soc. 39, 85–105 (1947). [I am very grateful to Michael Borcherds for answering an appeal and supplying me with this difficult-to-obtain reference.]
  17. [pdf] King, J.L. "Three problems in Search of a Measure" Amer. Math. Monthly, 101, 609–628 (1994).
  18. [pdf] Poncelet, J.V. "Reflexions sur l'usage de l'analise algebrique dans la geometrie", Annales de Mathematique, 8, 143–155 (1817).
  19. [pdf] Poncelet, J.V. "Traité des propriétés projectives des figures" Paris: Gauthier-Villars (1865, first ed. 1822).
  20. [pdf] Richelot, F.J. "Anwendung der der elliptischen Transcendenten auf die sphärischen Polygone, welche zugleich einem kleinen Kreise der Kugel eingeschrieben und einem andern umgeschrieben sind", J. Reine Angew. Math., 5, 250–267, (1830).
  21. [pdf] Schoenberg, I.J. "On Jacobi-Bertrand's proof of a Theorem of Poncelet", in: Erdős P., Alpár L., Halász G., Sárközy A. (eds) Studies in Pure Mathematics. Birkhäuser, Basel (1983).
  22. [pdf] Vallès, J. "A vector bundle proof of Poncelet's closure theorem", Expo. Math., 30, 399–405 (2012).