$\boldsymbol{o}=(o_{1},\ldots,o_{n+1})$ will have three key properties: 1)
$\boldsymbol{o}\in P^{(p)}$, 2) $\sum_{j=1}^{n+1}o_{j}=q$, and 3)
$o_{j}=o_{j\mod k}$. Then, radius $t$ is taken as the largest value such that
$K\subset P^{(p)}$. For convenience, we will not explicit the dependence of
$K$ on integers $1\leq p<q$.

######  Lemma 40.

Let $k,n,p,q\in\mathbb{N}$ such that $1\leq p<q$ and $n+1=kp$. Recall
constants listed in Remark 38. If $\kappa_{\star}=a_{\star}/4$,
$\tau_{\star}=\max\\{\chi_{1},\ldots,\chi_{k}\\}$, $0\leq
t<t_{\star}=\kappa_{\star}q/(n+1)-\tau_{\star}$ and

$\boldsymbol{o}=(o_{1},\ldots,o_{n+1})\in\mathbb{R}^{n+1},\qquad
o_{j}=\frac{qA_{j\mod k}}{(n+1)A},$

then $o_{1}+\cdots+o_{n+1}=q$ and
$K=\\{\boldsymbol{x}\in\mathbb{R}^{n+1}:|\boldsymbol{x}-\boldsymbol{o}|_{\infty}\leq
t\\}\subset P^{(p)}$.

###### Proof.

Clearly,
$o_{1}+\cdots+o_{n+1}=\frac{q}{(n+1)A}\sum_{j=1}^{n+1}A_{j}=\frac{qp}{(n+1)A}\sum_{j=1}^{k}A_{j}=\frac{kp}{n+1}q=q$.

If $\boldsymbol{x}\in K$, then $\boldsymbol{x}=\boldsymbol{o}+t\boldsymbol{u}$
for some $\boldsymbol{u}\in\mathbb{R}^{n+1}$ such that
$|\boldsymbol{u}|_{\infty}\leq 1$. With the suitable choice of the radius $t$,
the point $\boldsymbol{x}$ satisfies the following three sets of inequalities
that define the unbounded convex polytope $P^{(p)}$ given in (16):

  * •

_First set (with $2n$ inequalities)._ Since $o_{j+1}=\alpha_{j}o_{j}$ for all
$j=1,\ldots,n$, we see that

$\alpha^{-}_{j}x_{j}+\beta^{-}_{j}<x_{j+1}<\alpha^{+}_{j}x_{j}-\beta^{+}_{j}\Leftrightarrow\left\\{\begin{array}[]{l}(\alpha^{-}_{j}u_{j}-u_{j+1})t<(\alpha_{j}-\alpha^{-}_{j})o_{j}-\beta^{-}_{j}\\\
(u_{j+1}-\alpha^{+}_{j}u_{j})t<(\alpha^{+}_{j}-\alpha_{j})o_{j}-\beta^{+}_{j}\end{array}\right.$

  * •

_Second set (with $2$ inequalities)._ Since $A_{1}=1$ and
$A_{n+1}=A_{k}=\prod_{j=1}^{k-1}\alpha_{j}=1/\alpha_{k}$, we get that
$o_{1}=\alpha_{k}o_{n+1}=\alpha_{k}o_{k}$. Besides,
$\beta^{\pm}_{n+1}=\beta^{\pm}_{k}$ and $\alpha^{\pm}_{n+1}=\alpha^{\pm}_{k}$.
Hence,

$\alpha^{-}_{n+1}x_{n+1}+\beta^{-}_{n+1}<x_{1}<\alpha^{+}_{n+1}x_{n+1}-\beta^{+}_{n+1}\Leftrightarrow\left\\{\begin{array}[]{l}(\alpha^{-}_{k}u_{n+1}-u_{1})t<(\alpha_{k}-\alpha^{-}_{k})o_{k}-\beta^{-}_{k}\\\
(u_{1}-\alpha^{+}_{k}u_{n+1})t<(\alpha^{+}_{k}-\alpha_{k})o_{k}-\beta^{+}_{k}\end{array}\right.$

  * •

_Third set (with $n+1$ inequalities)._ $x_{j}\geq\chi_{j}\Leftrightarrow-
u_{j}t\leq o_{j}-\chi_{j}$.

Let us analyze the RHS and LHS of the above $3n+3$ inequalities. Coordinates
$o_{j}$ can be as big as needed if we take $q/(n+1)\gg 1$, because quotients
$A_{j\mod k}/A$ do not depend on $p$, $q$ or $n$. Thus, using that
$\alpha^{-}_{j}<\alpha_{j}<\alpha^{+}_{j}$ for all $j=1,\ldots,k$, all RHS can
be made positive if we take $q/(n+1)\gg 1$. On the other hand, we can bound
the LHS as follows:

$(\alpha^{-}_{j}u_{j}-u_{j+1})t\leq(1+\alpha^{-}_{j})t,\qquad(u_{j+1}-\alpha^{+}_{j}u_{j})t\leq(1+\alpha^{+}_{j})t,\qquad-
u_{j}t\leq t,$

because $|u_{j}|\leq|\boldsymbol{u}|_{\infty}\leq 1$ for all $j=1,\ldots,n+1$
and $t\geq 0$. Therefore, these $3n+3$ inequalities hold when we take any
$t\in[0,t_{\star})$ with

$\displaystyle t_{\star}$
$\displaystyle=\min\left\\{\frac{(\alpha_{j}-\alpha^{-}_{j})o_{j}-\beta^{-}_{j}}{1+\alpha^{-}_{j}},\frac{(\alpha^{+}_{j}-\alpha_{j})o_{j}-\beta^{+}_{j}}{1+\alpha^{+}_{j}},o_{j}-\chi_{j}:j=1,\ldots,n+1\right\\}$
$\displaystyle=\min\left\\{\kappa^{-}_{j}q/(n+1)-\tau^{-}_{j},\kappa^{+}_{j}q/(n+1)-\tau^{+}_{j},\kappa_{j}q/(n+1)-\tau_{j}:j=1,\ldots,k\right\\},$

where

$\kappa^{\pm}_{j}=\frac{|\alpha_{j}-\alpha^{\pm}_{j}|A_{j}}{(1+\alpha^{\pm}_{j})A},\qquad\kappa_{j}=\frac{A_{j}}{A},\qquad\tau^{\pm}_{j}=\frac{\beta^{\pm}_{j}}{1+\alpha^{\pm}_{j}}=1,\qquad\tau_{j}=\chi_{j}.$

All these arguments imply that $K\subset P^{(p)}$ provided that $0\leq
t<t_{\star}:=\kappa_{\star}q/(n+1)-\tau_{\star}$, where

$\kappa_{\star}=\min\left\\{\kappa^{-}_{1},\kappa^{+}_{1},\kappa_{1},\ldots,\kappa^{-}_{k},\kappa^{+}_{k},\kappa_{k}\right\\}>0,\quad\tau_{\star}=\max\left\\{\tau^{-}_{1},\tau^{+}_{1},\tau_{1},\ldots,\tau^{-}_{k},\tau^{+}_{k},\tau_{k}\right\\}>0.$

These constants $\kappa_{\star}$ and $\tau_{\star}$ do not depend on $p$, $q$
or $n$.

We note that $(\alpha^{+}_{j}-\alpha_{j})/(1+\alpha^{+}_{j})<1$. Hence,
$\kappa^{+}_{j}<\kappa_{j}$ and we can take

$\kappa_{\star}=\min\left\\{\kappa^{-}_{1},\kappa^{+}_{1},\ldots,\kappa^{-}_{k},\kappa^{+}_{k}\right\\}=a_{\star}/4,$

see Remark 38. ∎

We look for a lower bound on $G_{q}(K)$, where $K$ is a cube of the form (21)
such that $\sum_{j=1}^{n+1}o_{j}=q$. Note that
$G_{q}\big{(}\\{0,1\\}^{n}\big{)}=\binom{n}{q}$, so
$G_{q}\big{(}\\{0,1\\}^{2q}\big{)}=\binom{2q}{q}\geq 4^{q}/(2q+1)$ grows
exponentially fast as $q\to+\infty$. We want to generalize this idea. Since
there is no standard notation for the generalized binomial coefficients that
we need —for instance, symbols $\binom{n,m}{q}$ and $\binom{n}{q}^{(m)}$ can
be found in [48, 43]—, we use our own notation. Set

$[0..m]:=\mathbb{Z}\cap[0,m]=\\{0,1,\ldots,m-1,m\\}.$

Then $G_{q}\big{(}[0..m]^{n}\big{)}$ counts the number of _weak compositions_
of $q$ into $n$ parts with no part exceeding $m$. Note that
$G_{q}\big{(}[0..m]^{n}\big{)}=0$ for any $q\not\in[0..nm]$. It is well know
[26, section I.3] that

$\sum_{q=0}^{\infty}G_{q}\big{(}[0..m]^{n}\big{)}x^{q}=(1+x+x^{2}+\cdots+x^{m})^{n}.$

Using this polynomial identity, Andrews [2] deduced that, once fixed
$m,n\in\mathbb{N}$, the sequence $G_{q}\big{(}[0..m]^{n}\big{)}$ is unimodal
on $q$ and reaches its maximum at $q=\lfloor nm/2\rfloor$.

######  Lemma 41.

$G_{\lfloor
nm/2\rfloor}\big{(}[0..m]^{n}\big{)}\geq\frac{(m+1)^{n}}{nm+1}\geq\frac{(m+1)^{n-1}}{n}$
for all $m,n\in\mathbb{N}$.

###### Proof.

It follows from $\\#[0..nm]=nm+1$,
$\sum_{q=0}^{nm}G_{q}\big{(}[0..m]^{n}\big{)}=\\#\left([0..m]^{n}\right)=(m+1)^{n}$,
and inequalities $G_{q}\big{(}[0..m]^{n}\big{)}\leq G_{\lfloor
nm/2\rfloor}\big{(}[0..m]^{n}\big{)}$ for all $q\in[0..mn]$. ∎

Now we are ready to establish the lower bound on $G_{q}(K)$ that we are
looking for.

######  Lemma 42.

Let $n,q\in\mathbb{N}$ and $t>0$. If $K$ is a cube of the form (21) such that
$\sum_{j=1}^{n+1}o_{j}=q$ and $t\geq 3/2$, then

$G_{q}(K)\geq\frac{(2t-3)^{n}}{n+1}.$ (22)

###### Proof.

There exists an integer point $\boldsymbol{o}^{\prime}\in\mathbb{Z}^{n+1}$
such that $|\boldsymbol{o}-\boldsymbol{o}^{\prime}|_{\infty}\leq 1$ and
$\sum_{j=1}^{n+1}o^{\prime}_{j}=q$. If $\boldsymbol{o}\in\mathbb{Z}^{n+1}$, we
take $\boldsymbol{o}^{\prime}=\boldsymbol{o}$. If
$\boldsymbol{o}\not\in\mathbb{Z}^{n+1}$, we can take, for instance,

$o^{\prime}_{j}=\begin{cases}\lfloor o_{j}\rfloor+1,&\text{for }j\leq i,\\\
\lfloor o_{j}\rfloor,&\text{otherwise},\end{cases}$

where $i=q-\sum_{j=1}^{n+1}\lfloor o_{j}\rfloor\in[1..n]$, so that
$\sum_{j=1}^{n+1}o^{\prime}_{j}=i+\sum_{j=1}^{n+1}\lfloor o_{j}\rfloor=q$.

Set $m=\lfloor t\rfloor-1\in\mathbb{N}\cup\\{0\\}$ and
$v_{j}=o^{\prime}_{j}-m$. Clearly, $[v_{j},v_{j}+2m]\subset[o_{j}-t,o_{j}+t]$.
Hence, given any $\boldsymbol{y}\in[0..2m]^{n+1}$ such that
$\sum_{j=1}^{n+1}y_{j}=(n+1)m$, the sum of the components of the vector
$\boldsymbol{x}=\boldsymbol{y}+\boldsymbol{v}\in K\cap\mathbb{Z}^{n+1}$ is
equal to

$\sum_{j=1}^{n+1}x_{j}=(n+1)m+\left(\sum_{j=1}^{n+1}o^{\prime}_{j}\right)-(n+1)m=q.$

Besides, the correspondence
$[0..2m]^{n+1}\ni\boldsymbol{y}\mapsto\boldsymbol{x}=\boldsymbol{y}+\boldsymbol{v}\in
K\cap\mathbb{Z}^{n+1}$ is injective, which implies that

$G_{q}(K)\geq
G_{(n+1)m}\big{(}[0..2m]^{n+1}\big{)}=G_{\lfloor(n+1)2m/2\rfloor}\big{(}[0..2m]^{n+1}\big{)}\geq\frac{(2m+1)^{n}}{n+1}\geq\frac{(2t-3)^{n}}{n+1}.$

We have used Lemma 41 and $m=\lfloor t\rfloor-1\geq t-2$ in the last two
inequalities. ∎

To end, we prove the first lower bound stated in Theorem 37.

_Proof of the polynomial general lower bound._ This bound follows from bound
(15), inclusion $K\subset P^{(p)}$, bound (22), condition $0\leq
t<t_{\star}:=\kappa_{\star}q/(n+1)-\tau_{\star}$ required in Lemma 40, and
identities $a_{\star}=4\kappa_{\star}$, $b_{\star}=4\tau_{\star}+6$ and
$n+1=kp$. Namely,

$\displaystyle\\#\Pi(p,q)$ $\displaystyle\geq
2^{n+1}G_{q}(P^{(p)})\geq\max_{t\in[3/2,t_{\star})}\big{\\{}2^{n+1}G_{q}(K)\big{\\}}\geq\max_{t\in[3/2,t_{\star})}\frac{2(4t-6)^{n}}{n+1}=\frac{2(4t_{\star}-6)^{n}}{n+1}$
$\displaystyle=\frac{2}{n+1}\left(\frac{4\kappa_{\star}q}{n+1}-4\tau_{\star}-6\right)^{n}=\frac{2}{n+1}\left(\frac{a_{\star}q}{n+1}-b_{\star}\right)^{n}=\frac{2}{kp}\left(\frac{a_{\star}q}{kp}-b_{\star}\right)^{kp-1}.$

Note that $[3/2,t_{\star})\neq\emptyset$ since $q>b_{\star}kp/a_{\star}$
implies that $t_{\star}=\kappa_{\star}q/(n+1)-\tau_{\star}>3/2$. ∎

###  8.3 Proof of the two asymptotic lower bounds

We describe the exponentially fast growth of $\\#\Pi(p,q)$ when $p=\lfloor
xq/k\rfloor$ and $q\to+\infty$ for some fixed limit ratio $x>0$. We shall also
determine the limit ratio $x_{\star}>0$ that gives the largest exponent
$h_{\star}$ in the exponential bound.

######  Lemma 43.

Let $0<a<b$ and $k\in\mathbb{N}$. If

  * •

$M(x)=2(a/x-b)^{-k-1}/x$ for $0<x<a/b$;

  * •

$h(x)=x\log(a/x-b)$ for $0<x<a/b$; and

  * •

$G(p,q)=2(aq/kp-b)^{kp-1}/kp$ for $p,q\in\mathbb{N}$ such that $0<kp/q<a/b$,

then

$G(\lfloor xq/k\rfloor,q)\geq M(x){\rm e}^{h(x)q}/q,\qquad\forall
q\geq(1+b)kp/a,\ \forall x\in\big{(}0,a/(b+1)\big{]}.$

The exponent $h:(0,a/b)\to\mathbb{R}$ reaches its maximum value
$h_{\star}=h(x_{\star})=aW_{0}(b/{\rm e})/b>0$ at the point
$x_{\star}=aW_{0}(b/{\rm e})/((1+W_{0}(b/{\rm e}))b)$.

###### Proof.

If $q\in\mathbb{N}$, $x\in(0,a/(b+1)]$ and $p=\lfloor xq/k\rfloor$, then
$aq/kp-b\geq 1$, $xq-k<kp\leq xq$ and

$G(p,q)=\frac{2}{kp}\left(\frac{aq}{kp}-b\right)^{kp-1}\geq\frac{2}{kp}\left(\frac{aq}{kp}-b\right)^{xq-k-1}\geq\frac{2}{xq}\left(\frac{a}{x}-b\right)^{xq-k-1}=\frac{1}{q}M(x){\rm
e}^{h(x)q}.$

Next, we look for the global maximum of $h(x)$. After the changes of variable

$\displaystyle(0,+\infty)\ni\hat{x}$ $\displaystyle\leftrightarrow
x=\frac{a}{b+{\rm e}^{\hat{x}}}\in\big{(}0,a/(b+1)\big{)},$
$\displaystyle(0,+\infty)\ni\hat{h}$ $\displaystyle\leftrightarrow
h=a\hat{h}\in(0,+\infty),$

we get that $h(x)=x\log(a/x-b)=x\hat{x}=a\hat{x}/(b+{\rm e}^{\hat{x}})$, so
$\hat{h}(\hat{x})=\hat{x}/(b+{\rm e}^{\hat{x}})$. We have reduced the search
of the global maximum point $x_{\star}\in(0,a/(b+1))$ of $h(x)$ to the search
of the global maximum point $\hat{x}_{\star}>0$ of $\hat{h}(\hat{x})$. Since

$\frac{\,{\rm d}\hat{h}}{\,{\rm d}\hat{x}}(\hat{x})=\frac{b+(1-\hat{x}){\rm
e}^{\hat{x}}}{(b+{\rm e}^{\hat{x}})^{2}}=0\Leftrightarrow(\hat{x}-1){\rm
e}^{\hat{x}-1}=b/{\rm e}\Leftrightarrow\hat{x}=\hat{x}_{\star}:=1+W_{0}(b/{\rm
e}),$

we deduce that $\hat{h}(\hat{x})$ reaches its maximum value

$\hat{h}_{\star}=\hat{h}(\hat{x}_{\star})=\frac{\hat{x}_{\star}}{b+{\rm
e}^{\hat{x}_{\star}}}=\frac{1}{{\rm e}^{\hat{x}_{\star}}}=\frac{1}{{\rm
e}}\frac{1}{{\rm e}^{W_{0}(b/{\rm e})}}=\frac{W_{0}(b/{\rm e})}{b}$

at the point $\hat{x}=\hat{x}_{\star}$. In order to compute $\hat{x}_{\star}$,
we have used that $W_{0}(b/{\rm e}){\rm e}^{W_{0}(b/{\rm e})}=b/{\rm e}$.
Expressions for $x_{\star}$ and $h_{\star}$ are obtained by undoing both
changes of variable. ∎

We can prove now the second and third lower bounds stated in Theorem 37.

_Proof of both asymptotic lower bounds._ The second bound of Theorem 37
follows from the first one by applying Lemma 43 with $a=a_{\star}$ and
$b=b_{\star}$. Analogously, the third bound of Theorem 37 follows from the
first one by taking $c_{\star}(p)=2(a_{\star})^{kp-1}/(kp)^{kp}$. ∎

##  9 The length spectrum of circular polygons

The purpose of this section is to prove Theorem E, which shows an unusual
feature of the length spectrum of billiards in circular polygons. In
particular, it shows that the well-known results of Marvizi-Melrose [45] fail
to hold for circular polygons. This was expected because there are so many
periodic billiard trajectories inside circular polygons —as we have seen in
the previous section— that we can construct sequences of them whose lengths
have rather different asymptotic behaviors.

Let $|\Gamma|$ be the length of $\Gamma$. Let $\kappa(s)$ be the curvature of
$\Gamma$ as a function of an arc-length parameter $s\in[0,|\Gamma|)$. If
$z,z^{\prime}\in\Gamma$ are any two consecutive impact points of a billiard
trajectory $g$, then the segment $[z,z^{\prime}]\subset\mathbb{R}^{2}$ is a
_link_ of $g$ and $\int_{z}^{z^{\prime}}\,{\rm d}s$ is the distance from $z$
to $z^{\prime}$ along $\Gamma$. Note that
$|z^{\prime}-z|<\int_{z}^{z^{\prime}}\,{\rm d}s$ by convexity. If
$g=\\{z_{0},\ldots,z_{q-1}\\}\subset\Gamma$ is a $q$-periodic billiard
trajectory, let $L(g)=|z_{1}-z_{0}|+\cdots+|z_{q-1}-z_{0}|$ be its _length_.
Let

$\underline{L}_{q}=\inf\\{L(g):g\in\Pi(1,q)\\},\qquad\overline{L}_{q}=\sup\\{L(g):g\in\Pi(1,q)\\}.$

To begin with, let us recall the Marvizi-Melrose results for smooth ovals. A
_smooth oval_ is a regular, simple, closed, oriented $C^{\infty}$ curve with
positive curvature everywhere.

######  Theorem 44 (Marvizi & Melrose [45]).

Let $\Gamma$ be any smooth oval.

  1. (a)

$\lim_{q\to+\infty}q^{i}\big{(}\overline{L}_{q}-\underline{L}_{q}\big{)}=0$
for all $i\in\mathbb{N}$.

  2. (b)

There are asymptotic coefficients $c_{i}\in\mathbb{R}$ such that if
$g_{q}\in\Pi(1,q)$, then

$L(g_{q})\asymp|\Gamma|+\sum_{i=1}^{\infty}\frac{c_{i}}{q^{2i}},\quad\mbox{as
$q\to+\infty$.}$

  3. (c)

$c_{1}=-\frac{1}{24}\left[\int_{\Gamma}\kappa^{2/3}(s)\,{\rm
d}s\right]^{3}<0$.

  4. (d)

If $[z,z^{\prime}]$ is a link of $g_{q}\in\Pi(1,q)$, then

$\int_{z}^{z^{\prime}}\,{\rm d}s\asymp\frac{1}{q}|\Gamma|,\quad\mbox{uniformly
as $q\to+\infty$.}$

The symbol $\asymp$ means that the RHS is asymptotic to the LHS. The first
property implies that the Melrose-Marvizi asymptotic coefficients $c_{i}$ do
not depend on the choice of the sequence of periodic trajectories
$(g_{p})_{q}$. All of them can be explicitly written in terms of the
curvature. For instance, the formulas for $c_{1}$, $c_{2}$, $c_{3}$, and
$c_{4}$ can be found in [56]. Property (d) means that not only are the lengths
of $g_{q}$ asymptotically well-behaved, but as $q\to+\infty$, the distribution
of the points in $g_{q}$ is asymptotically well-behaved with respect to any
one point. Hence, property (d) is like a weak local version of property (b).
There is also a strong local version in [45, Theorem 5.9].

We will check that none of properties (a)–(d) of Theorem 44 hold for circular
polygons.

From now on, we assume that $\Gamma$ is a circular polygon with arcs
$\Gamma_{j}$, radii $r_{j}$, central angles $\delta_{j}$ and polar
parametrisation $z(\varphi)$ introduced in Definition 3. As usual, factors
$\alpha^{\pm}_{j}$, addends $\beta_{j}^{\pm}=\alpha_{j}^{\pm}+1$ and integers
$\chi_{j}\geq 2$ were introduced in Lemma 17 and Corollary 19, although we are
now only interested in $(1,q)$-periodic trajectories, so $1\leq j\leq k$ and
we no longer need to consider $j$ modulo $k$. Recall that $p=1$ along this
section.

######  Remark 45.

Corollary 33 implies that there are at least $2^{k}$ generic sliding periodic
billiard trajectories $g_{q}\in\Pi(1,q)$ with _exactly_ $x_{j}\in\mathbb{N}$
impacts in the arc $\Gamma_{j}$, $j=1,\ldots,k$, for any integer point
$\boldsymbol{x}=(x_{1},\ldots,x_{k})\in
P^{(1)}\cap\mathbb{Z}^{k}=P^{(1)}\cap\mathbb{N}^{k}$ such that
$x_{1}+\cdots+x_{k}=q$. Here, $P^{(1)}$ is the unbounded convex polytope of
$\mathbb{R}^{k}$ defined in (16) for $p=1$.

We need a couple of technical results before tackling the proof of Theorem E.

First, we compute the lengths of generic sliding $(1,q)$-periodic billiard
trajectories. By definition, they impact all arcs but no nodes. Angles similar
to $\varphi_{j}^{\pm}$ and $\psi_{j}$ below were considered in the proof of
part (c) in Lemma 11.

######  Lemma 46.

If $g_{q}\in\Pi(1,q)$ is a generic sliding periodic billiard trajectory inside
$\Gamma$, then

$L(g_{q})=\sum_{j=1}^{k}\big{(}\ell^{-}_{j}+(x_{j}-1)\ell_{j}+\ell^{+}_{j}\big{)},\qquad\ell_{j}=2r_{j}\sin\psi_{j},\qquad\ell^{\pm}_{j}=\frac{r_{j}\sin\varphi^{\pm}_{j}}{\cos(\psi_{j}-\varphi_{j}^{\pm})},$
(23)

where

  * (i)

$x_{j}\in\mathbb{N}$ is the number of impact points in $\Gamma_{j}$;

  * (ii)

$\psi_{j}>0$ is the constant angle of reflection along the $x_{j}$ impacts in
$\Gamma_{j}$; and

  * (iii)

$\varphi^{\pm}_{j}\in(0,2\psi_{j})$ are the impact angles such that
$[z(b_{j}-\varphi^{+}_{j}),z(a_{j+1}+\varphi^{-}_{j+1})]$ is the transition
link connecting $\Gamma_{j}$ and $\Gamma_{j+1}$.

Besides, $\varphi^{-}_{j}+2(x_{j}-1)\psi_{j}+\varphi^{+}_{j}=\delta_{j}$ for
all $j=1,\ldots,k$.

Figure 6: Isosceles triangle $\Delta OPQ$ and a point
$R\in[P,Q]\setminus\\{P,Q\\}$ such that $\angle QPO=\angle OQP=\beta$,
$|\vec{OP}|=|\vec{OQ}|=r$, $\angle POQ=2\psi=\varphi^{+}+\varphi^{-}=\angle
POR+\angle ROQ$, and
$|\vec{PQ}|=\ell=\ell^{+}+\ell^{-}=|\vec{PR}|+|\vec{RQ}|$.

###### Proof.

If we apply the law of sinus to the three triangles $\Delta POQ$, $\Delta POR$
and $\Delta ROQ$ displayed in Figure 6 (see its caption for the definition of
each quantity), we get that

$\frac{\ell}{2\sin\psi\cos\psi}=\frac{\ell}{\sin
2\psi}=\frac{r}{\sin\beta}=\frac{r}{\cos\psi},\qquad\frac{\ell^{\pm}}{\sin\varphi^{\pm}}=\frac{r}{\sin(\pi-\beta-\varphi^{\pm})}=\frac{r}{\cos(\psi-\varphi^{\pm})},$

since $\beta=\pi/2-\psi$, $\angle ORP=\pi-\beta-\varphi^{+}$ and $\angle
QRO=\pi-\beta-\varphi^{-}$. Therefore,

$\ell=2r\sin\psi,\qquad\ell^{\pm}=\frac{r\sin\varphi^{\pm}}{\cos(\psi-\varphi^{\pm})}.$

We deduce (23) from those three formulas.

If $g_{q}$ has $x_{j}$ impacts in $\Gamma_{j}$ with constant angle of
reflection $\psi_{j}$, then it has $x_{j}-1$ circular links with a certain
constant length $\ell_{j}$. Each one of these circular links $[P,Q]$ is the
base of an isosceles triangle $\Delta OPQ$ like the one displayed in Figure 6,
with $O=O_{j}$, $r=r_{j}$ and $\psi=\psi_{j}$. Hence
$\ell_{j}=2r_{j}\sin\psi_{j}$.

Let us consider the transition link
$[z(b_{j}-\varphi^{+}_{j}),z(a_{j+1}+\varphi^{-}_{j+1})]$ connecting
$\Gamma_{j}$ and $\Gamma_{j+1}$ and the isosceles triangle $\Delta OPQ$ with

$O=O_{j},\qquad P=z(b_{j}-\varphi^{+}_{j})\in\Gamma_{j},\qquad
Q=O_{j}+r_{j}{\rm e}^{\mskip 2.0mu{\rm i}\mskip
1.0mu(b_{j}-\varphi^{+}_{j}+2\psi_{j})}.$

We stress that $Q$ is an auxiliary point: $Q\not\in\Gamma$. Let
$R=[P,Q]\cap[O,z(b_{j})]$. Then $r=r_{j}$, $\varphi^{+}=\varphi^{+}_{j}$ and
$\ell^{+}=\ell^{+}_{j}$. Therefore,
$\ell^{+}_{j}=r_{j}\sin\varphi^{+}_{j}/\cos(\psi_{j}-\varphi^{+}_{j})$. The
formula for $\ell^{-}_{j+1}$ is deduced in a similar way, but taking

$O=O_{j+1},\qquad P=O_{j+1}+r_{j+1}{\rm e}^{\mskip 2.0mu{\rm i}\mskip
1.0mu(a_{j+1}-\varphi^{-}_{j+1}-2\psi_{j+1})},\qquad
Q=z(a_{j+1}+\varphi^{-}_{j+1})\in\Gamma_{j+1},$

$R=[P,Q]\cap[O,z(a_{j+1})]$, $r=r_{j+1}$, $\varphi^{-}=\varphi^{-}_{j+1}$ and
$\ell^{-}=\ell^{-}_{j+1}$. (In this case, the auxiliary point is $P$:
$P\not\in\Gamma$). By construction, the transition link
$[z(b_{j}-\varphi^{+}_{j}),z(a_{j+1}+\varphi^{-}_{j+1})]$ has length
$\ell^{+}_{j}+\ell^{+}_{j+1}$. This proves (23).

Finally, relation
$\varphi^{-}_{j}+2(x_{j}-1)\psi_{j}+\varphi^{+}_{j}=\delta_{j}$ is
geometrically evident. ∎

Next, we need a technical result about the extreme values of the
differentiable strict concave function (25) over a bounded convex polytope
$P^{(1)}_{\infty}$ related to the unbounded convex polytope $P^{(1)}$ of
$\mathbb{R}^{k}$ defined in (16) for $p=1$. Recall that
$\Delta_{k-1}=\\{\boldsymbol{x}\in\mathbb{R}^{k}:\boldsymbol{x}>0,\
x_{1}+\cdots+x_{k}=1\\}$ is the open $(k-1)$-simplex and
$H_{k}=\\{\boldsymbol{x}\in\mathbb{R}^{k}:x_{1}+\cdots+x_{k}=1\\}$.

######  Lemma 47.

The bounded convex polytope
$P^{(1)}_{\infty}=\lim_{q\to\infty}\left(\big{\\{}\boldsymbol{x}/q:\boldsymbol{x}\in
P^{(1)}\big{\\}}\cap H_{k}\right)$ is given by

$P^{(1)}_{\infty}=\left\\{\boldsymbol{x}\in\mathbb{R}^{k}:\begin{array}[]{l}\alpha^{-}_{j}x_{j}<x_{j+1}<\alpha^{+}_{j}x_{j},\quad\forall
j=1,\ldots,k-1\\\ \alpha^{-}_{k}x_{k}<x_{1}<\alpha^{+}_{k}x_{k}\\\
x_{j}>0,\quad\forall j=1,\ldots,k\\\ x_{1}+\cdots+x_{k}=1\end{array}\right\\}$
(24)

and its compact closure $K^{(1)}_{\infty}$ is contained in the open simplex
$\Delta_{k-1}$. Let

$h:\Delta_{k-1}\to(-\infty,0),\qquad
h(\boldsymbol{y})=-\frac{1}{24}\sum_{j=1}^{k}\frac{\delta_{j}^{3}r_{j}}{y^{2}_{j}}.$
(25)

Set $I_{1}=h\big{(}P^{(1)}_{\infty}\big{)}$, $c^{-}_{1}=\inf I_{1}$ and
$c^{+}_{1}=\sup I_{1}$. Then $c_{1}^{+}\in I_{1}$ and

$-\infty<c_{1}^{-}\leq-\pi^{2}|\Gamma|/6<c_{1}^{+}=\frac{1}{24}\left[\int_{\Gamma}\kappa^{2/3}(s)\,{\rm
d}s\right]^{3}<0.$ (26)

###### Proof.

Expression (24) is trivial. We check that
$K^{(1)}_{\infty}\subset\Delta_{k-1}$ by a reductio ad absurdum argument. Let
us assume that $\boldsymbol{x}=(x_{1},\ldots,x_{k})\in K^{(1)}_{\infty}$ and
$x_{i}=0$ for some $i$. Then inequalities $\alpha^{-}_{j}x_{j}\leq
x_{j+1}\leq\alpha^{+}_{j}x_{j}$ for $j=1,\ldots,k-1$ and
$\alpha^{-}_{k}x_{k}\leq x_{1}\leq\alpha^{+}_{k}x_{k}$ imply that

$x_{i+1}=\cdots=x_{k}=x_{1}=\cdots=x_{i-1}=0,$

so identity $x_{1}+\cdots+x_{k}=1$ fails. Contradiction.

The image of a compact convex set by a continuous function that only takes
negative values is a compact interval of $(-\infty,0)$, so
$\overline{I_{1}}=h\big{(}K_{\infty}^{(1)}\big{)}=[c^{-}_{1},c^{+}_{1}]$ for
some numbers $-\infty<c^{-}_{1}\leq c^{+}_{1}<0$. Let us estimate the minimum
value $c^{-}_{1}$, compute exactly the maximum value $c^{+}_{1}$, prove that
$c^{-}_{1}<c^{+}_{1}$, and check that $c^{+}_{1}\in I_{1}$.

We claim that function (25) attains its maximum value only at
$\boldsymbol{y}=\boldsymbol{w}(1/3)$, where

$\boldsymbol{w}(\xi)=\frac{1}{S(\xi)}\big{(}s_{1}(\xi),\ldots,s_{k}(\xi)\big{)}\in\Delta_{k-1},\qquad
s_{j}(\xi)=\delta_{j}r^{\xi}_{j},\qquad
S(\xi)=\sum_{j=1}^{k}\delta_{j}r^{\xi}_{j}$

for all $\xi\in\mathbb{R}$. On the one hand, the gradient of
$\mathbb{R}_{+}^{k}\ni\boldsymbol{y}\mapsto\sum_{j=1}^{k}\delta^{3}_{j}r_{j}/y^{2}_{j}$
is the vector with components $-2\delta^{3}_{j}r_{j}/y^{3}_{j}$, so
$\boldsymbol{y}\in\Delta_{k-1}$ is a critical point of (25) if and only if

$\big{(}s_{i}(1/3)/y_{i}\big{)}^{3}=\delta^{3}_{i}r_{i}/y^{3}_{i}=\delta^{3}_{j}r_{j}/y^{3}_{j}=\big{(}s_{j}(1/3)/y_{j}\big{)}^{3},\qquad\forall
i\neq j.$

This means that $\boldsymbol{w}(1/3)$ is the only critical point of (25). On
the other hand, $\sum_{j=1}^{k}\delta^{3}_{j}r_{j}/y^{2}_{j}$ is a nonnegative
weighted sum of convex terms $1/y^{2}_{j}$, so
$-\frac{1}{24}\sum_{j=1}^{k}\delta^{3}_{j}r_{j}/y^{2}_{j}$ is strict concave
function on $\mathbb{R}_{+}^{k}$ and (25) is a differentiable strict concave
function. Hence, the local maximum $\boldsymbol{w}(1/3)$ is a strict global
maximum. This proves the claim. Besides,

$h(\boldsymbol{w}(\xi))=-\frac{1}{24}\sum_{j=1}^{k}\frac{\delta^{3}_{j}r_{j}}{\big{(}s_{j}(\xi)/S(\xi)\big{)}^{2}}=-\frac{1}{24}S(\xi)^{2}\sum_{j=1}^{k}\delta_{j}r^{1-2\xi}_{j}=-\frac{1}{24}S(\xi)^{2}S(1-2\xi).$

In particular, $h(\boldsymbol{w}(0))<h(\boldsymbol{w}(1/3))$ and

$\displaystyle h(\boldsymbol{w}(0))$
$\displaystyle=-S(0)^{2}S(1)/24=-(2\pi)^{2}|\Gamma|/24=-\pi^{2}|\Gamma|/6,$
$\displaystyle h(\boldsymbol{w}(1/3))$
$\displaystyle=-S(1/3)^{3}/24=-\frac{1}{24}\left[\sum_{j=1}^{k}\delta_{j}r^{1/3}_{j}\right]^{3}=-\frac{1}{24}\left[\int_{\Gamma}\kappa^{2/3}(s)\,{\rm
d}s\right]^{3}.$

Here we have used that $|\Gamma_{j}|=\delta_{j}r_{j}$ and
$\int_{\Gamma_{j}}\kappa^{2/3}(s)\,{\rm
d}s=|\Gamma_{j}|/r_{j}^{2/3}=\delta_{j}r_{j}^{1/3}$ since $\Gamma_{j}$ is a
circular arc of radius $r_{j}$ and central angle $\delta_{j}$.

Hence, property $c_{1}^{+}\in I_{1}$ and inequalities (26) hold provided
$\boldsymbol{w}(0)\in K^{(1)}_{\infty}$ and $\boldsymbol{w}(1/3)\in
P_{\infty}^{(1)}$. It turns out that $\boldsymbol{w}(\xi)$ satisfies the
$3k+1$ conditions listed in (24), so that $\boldsymbol{w}(\xi)\in
P^{(1)}_{\infty}$, for all $\xi\in(0,1/2]$. For instance,
$\boldsymbol{w}(\xi)$ satisfies the first $2k-2$ inequalities:

$\displaystyle\xi\in(0,1/2]$ $\displaystyle\Rightarrow
r^{\xi}_{j}\min\left\\{1,\sqrt{r_{j+1}/r_{j}}\right\\}<r^{\xi}_{j+1}<r^{\xi}_{j}\max\left\\{1,\sqrt{r_{j+1}/r_{j}}\right\\}$
$\displaystyle\Rightarrow\alpha^{-}_{j}w_{j}(\xi)<w_{j+1}(\xi)<\alpha^{+}_{j}w_{j}(\xi)\mbox{
for $j=1,\ldots,k-1$}.$

Inequalities $\alpha^{-}_{k}w_{k}(\xi)<w_{1}(\xi)<\alpha^{+}_{k}w_{k}(\xi)$
are proved in a similar way. Inequalities $w_{j}(\xi)>0$ and identity
$w_{1}(\xi)+\cdots+w_{k}(\xi)=1$ are trivial. Finally,
$\boldsymbol{w}(0)=\lim_{\xi\to 0^{+}}\boldsymbol{w}(\xi)\in K^{(1)}_{\infty}$
∎

The relation between the left endpoint $c_{1}^{-}$ of interval $I_{1}$ and
$-\pi^{2}|\Gamma|/6$ is an open problem. For instance, both quantities
coincide for squared pseudo-ellipses. To be precise, let
$\Gamma=E_{\pi/2,r,R}$ be a squared pseudo-ellipse of radii $r$ and $R>r$, see
Section 2. That is, $\Gamma=E_{\pi/2,r,R}$ is the circular $4$-gon with radii
$r_{1}=r_{3}=r$ and $r_{2}=r_{4}=R$, and central angles
$\delta_{1}=\delta_{2}=\delta_{3}=\delta_{4}=\pi/2$. A tedious computation
that we omit for the sake of brevity shows that

$\displaystyle c^{-}_{1}$
$\displaystyle=-\pi^{2}|E_{\pi/2,r,R}|/6=-\pi^{3}(R+r)/6,$ $\displaystyle
c^{+}_{1}$
$\displaystyle=-\frac{1}{24}\left[\int_{E_{\pi/2,r,R}}\kappa^{2/3}(s)\,{\rm
d}s\right]^{3}=-\frac{1}{24}\left[\sum_{j=1}^{k}\delta_{j}\sqrt[3]{r_{j}}\right]^{3}=-\frac{\pi^{3}}{24}\left[\sqrt[3]{R}+\sqrt[3]{r})\right]^{3}.$

These two expressions above coincide when $R=r$. In general,
$c^{+}_{1}-c^{-}_{1}$ tends to zero when $\Gamma$ tends to a circle of finite
radius.

The main result of this section is nothing more than a reformulation of
Theorem E.

######  Theorem 48.

Let $P^{(1)}_{\infty}\subset\Delta_{k-1}$,
$I_{1}=h\big{(}P^{(1)}_{\infty}\big{)}\subset(-\infty,0)$, $c_{1}^{-}=\inf
I_{1}$ and $c_{1}^{+}=\max I_{1}$ be the open bounded convex polytope of
$\mathbb{R}^{k}$, the image interval, and the extreme values introduced in
Lemma 47, respectively. Extreme values $c_{1}^{\pm}$ satisfy inequalities
(26). For any fixed $c\in[c_{1}^{-},c_{1}^{+}]$ there exist a period
$q_{0}\in\mathbb{N}$ and a sequence $(g_{q})_{q\geq q_{0}}$ of generic sliding
periodic billiard trajectories $g_{q}\in\Pi(1,q)$ such that

$L(g_{q})=|\Gamma|+c/q^{2}+\mathop{\rm O}\nolimits(1/q^{3}),\quad\mbox{as
$q\to+\infty$.}$

Consequently, there exist a sequence $(h_{q})_{q}$, with $h_{q}\in\Pi(1,q)$,
such that

$c_{-}=\liminf_{q\to+\infty}\big{(}(L(h_{q})-|\Gamma|)q^{2}\big{)}<\limsup_{q\to+\infty}\big{(}(L(h_{q})-|\Gamma|)q^{2}\big{)}=c_{+},\quad\mbox{as
$q\to+\infty$.}$

###### Proof.

If $c\in(c_{1}^{-},c_{1}^{+}]$, then $c=h(\boldsymbol{y})$ for some
$\boldsymbol{y}\in P^{(1)}_{\infty}$. If $q\in\mathbb{N}$ is big enough, then
there exists a point $\boldsymbol{x}=(x_{1},\ldots,x_{k})\in\mathbb{N}^{k}$
such that $|q\boldsymbol{y}-\boldsymbol{x}|_{\infty}\leq 1$ and
$\boldsymbol{x}\in P^{(1)}\cap qH_{k}$, where $P^{(1)}$ is the unbounded
convex polytope defined in (16) for $p=1$. Let us prove this claim.

First, we observe that $y_{j}>0$, so $qy_{j}\geq 1$ when $q\gg 1$. If
$q\boldsymbol{y}\in\mathbb{N}^{k}$, then we take
$\boldsymbol{x}=q\boldsymbol{y}$. If $q\boldsymbol{y}\not\in\mathbb{N}^{k}$,
then we can take, for instance,

$x_{j}=\begin{cases}\lfloor qy_{j}\rfloor+1,&\text{for }j\leq i,\\\ \lfloor
qy_{j}\rfloor,&\text{otherwise},\end{cases}$

where $i=q-\sum_{j=1}^{k}\lfloor qy_{j}\rfloor\in\\{1,\ldots,k-1\\}$, so that
$\sum_{j=1}^{k}x_{j}=i+\sum_{j=1}^{k}\lfloor qy_{j}\rfloor=q$. This means that
$\boldsymbol{x}\in qH_{k}$. To end the proof of the claim, we deduce that
$\boldsymbol{x}\in P^{(1)}$ from limits
$\lim_{q\to+\infty}\boldsymbol{x}/q=\boldsymbol{y}\in P^{(1)}_{\infty}$ and
$P^{(1)}_{\infty}=\lim_{q\to\infty}\left(\big{\\{}\boldsymbol{x}/q:\boldsymbol{x}\in
P^{(1)}\big{\\}}\cap H_{k}\right)$. Recall that $P_{\infty}^{(1)}$ is an open
set in $H_{k}$.

As we have explained before, see Remark 45, if $q\gg 1$ then there are at
least $2^{k}$ generic sliding periodic billiard trajectories
$g_{q}\in\Pi(1,q)$ with exactly $x_{j}\in\mathbb{N}$ impacts on the arc
$\Gamma_{j}$ and length (23). The numbers $x_{j}\in\mathbb{N}$, the constant
angles of reflection $\psi_{j}>0$ and the impact angles
$\varphi^{\pm}_{j}\in(0,2\psi_{j})$ described in Lemma 46 satisfy identity
$\varphi^{-}_{j}+2(x_{j}-1)\psi_{j}+\varphi^{+}_{j}=\delta_{j}$ and uniform
estimates $x_{j}=qy_{j}+\mathop{\rm O}\nolimits(1)$,
$\varphi^{\pm}_{j}=\mathop{\rm O}\nolimits(1/q)$ and
$\psi_{j}=\delta_{j}/2(x_{j}-1)+\mathop{\rm O}\nolimits(1/q^{2})=\mathop{\rm
O}\nolimits(1/q)$ as $q\to+\infty$. Therefore,

$\ell^{\pm}_{j}=\frac{r_{j}\sin\varphi^{\pm}_{j}}{\cos(\psi_{j}-\varphi_{j}^{\pm})}=r_{j}\varphi^{\pm}_{j}+\mathop{\rm
O}\nolimits\Big{(}(\varphi^{\pm}_{j})^{3},\varphi^{\pm}_{j}|\psi_{j}-\varphi^{\pm}_{j}|^{2}\Big{)}=r_{j}\varphi^{\pm}_{j}+\mathop{\rm
O}\nolimits(1/q^{3})$

and

$\displaystyle(x_{j}-1)\ell_{j}$
$\displaystyle=2r_{j}(x_{j}-1)\sin\psi_{j}=2r_{j}(x_{j}-1)\Big{(}\psi_{j}-\psi_{j}^{3}/6+\mathop{\rm
O}\nolimits\big{(}\psi^{5}_{j}\big{)}\Big{)}$
$\displaystyle=2r_{j}(x_{j}-1)\psi_{j}-r_{j}(x_{j}-1)\psi_{j}^{3}/3+\mathop{\rm
O}\nolimits(1/q^{4})$
$\displaystyle=2r_{j}(x_{j}-1)\psi_{j}-\frac{\delta^{3}_{j}r_{j}}{24(x_{j}-1)^{2}}+\mathop{\rm
O}\nolimits(1/q^{3})$
$\displaystyle=2r_{j}(x_{j}-1)\psi_{j}-\frac{\delta^{3}_{j}r_{j}}{24y^{2}_{j}}\frac{1}{q^{2}}+\mathop{\rm
O}\nolimits(1/q^{3}).$

Finally, we estimate the total length (23) as follows:

$\displaystyle L(g_{q})$
$\displaystyle=\sum_{j=1}^{k}\left(\ell^{-}_{j}+(x_{j}-1)\ell_{j}+\ell^{+}_{j}\right)$
$\displaystyle=\sum_{j=1}^{k}r_{j}(\varphi^{-}_{j}+2(x_{j}-1)\psi_{j}+\varphi^{+}_{j})+h(\boldsymbol{y})/q^{2}+\mathop{\rm
O}\nolimits(1/q^{3})$ $\displaystyle=|\Gamma|+c/q^{2}+\mathop{\rm
O}\nolimits(1/q^{3}).$

We have used that
$\varphi^{-}_{j}+2(x_{j}-1)\psi_{j}+\varphi^{+}_{j}=\delta_{j}$,
$|\Gamma|=\sum_{j=1}^{k}\delta_{j}r_{j}$ and $c=h(\boldsymbol{y})$ in the last
line. Function $h(\boldsymbol{y})$ was defined in (25). This ends the proof of
the case $c\in(c_{1}^{-},c_{1}^{+}]$.

The case $c=c_{1}^{-}$ can be obtained from the case
$c\in(c_{1}^{-},c_{1}^{+}]$ by using a classical diagonalization argument
about sequences of sequences of lengths.

Finally, sequence $(h_{q})_{q}$ is constructed by interleaving two sequences
of generic sliding $(1,q)$-periodic billiard trajectories associated with the
asymptotic coefficients $c_{1}^{-}$ and $c_{1}^{+}$ respectively. ∎

We observe that generic sliding $(1,q)$-periodic billiard trajectories inside
circular polygons are _asymptotically shorter_ than the ones inside smooth
ovals, since $c_{1}^{+}$ in (26) has the same formula that constant $c_{1}$ in
part (c) of Theorem 44.

The generic sliding periodic billiard trajectories analyzed in the proof of
Theorem 48 do not impact the set of nodes. Next we consider other sliding
periodic billiard trajectories with the opposite property. They impact _all
nodes_ of the circular polygon in such a way that the angle of reflection
remains _constant_ along the whole trajectory. After thinking about it a bit,
the reader will realize that these _nodal_ sliding periodic trajectories can
only take place for certain circular polygons, that we call _rational_.

######  Definition 49.

We say that a circular polygon $\Gamma$ is _rational_ when all its central
angles are rational multiples of $\pi$, so

$\delta_{j}=m_{j}\delta,$

for some $\delta=\gcd(\delta_{1},\ldots,\delta_{k})$ and
$m_{j}=\delta_{j}/\delta\in\mathbb{N}$. Set $M=\sum_{j=1}^{k}m_{j}$. Then
$M\delta=2\pi$. A billiard trajectory inside a rational circular polygon is
_nodal_ when it impacts all nodes (interspersed with possibly many other non-
nodal impacts) in the counter-clockwise ordering.

Squared pseudo-ellipses and Moss’s eggs are rational circular polygons, see
Section 2. Any nodal orbit in a rational circular polygon has constant angle
of reflection, is sliding, and is periodic with a rotation number of the form
$1/q$, $q$ being the period.

It is easy to compute the length of a nodal trajectory. Recall that
$\delta_{j}=b_{j}-a_{j}$, $b_{j}=a_{j+1}$ and $f:\mathcal{M}\to\mathcal{M}$ is
the billiard map.

######  Proposition 50.

Let $\Gamma$ be a rational circular $k$-gon with arcs $\Gamma_{j}$, radii
$r_{j}$ and central angles $\delta_{j}$. Set
$\delta=\gcd(\delta_{1},\ldots,\delta_{k})$. Fix some $\psi=\delta/2i$ with
$i\in\mathbb{N}$. Let $g_{q}$ be the billiard trajectory generated by

$(\varphi_{n},\theta_{n})=f^{n}(a_{1},\psi),\qquad\forall n\in\mathbb{Z}.$

  1. (a)

The billiard trajectory $g_{q}$ is nodal and $g_{q}\in\Pi(1,q)$ with period
$q=Mi$.

  2. (b)

$L(g_{q})=|\Gamma|-\pi^{2}|\Gamma|/6q^{2}+\mathop{\rm O}\nolimits(1/q^{4})$ as
$q=Mi\to+\infty$.

  3. (c)

If $[z,z^{\prime}]$ is a circular link of $g_{q}$ associated to the arc
$\Gamma_{j}$, then

$\int_{z}^{z^{\prime}}\,{\rm d}s=\frac{1}{q}2\pi
r_{j}\neq\frac{1}{q}|\Gamma|.$

###### Proof.

  1. (a)

Once fixed the index $i\in\mathbb{N}$, we deduce that
$\varphi_{im_{1}}=a_{1}+2im_{1}\psi=a_{1}+\delta_{1}=b_{1}=a_{2}$. That is,
the first $im_{1}$ links of the billiard trajectory connect both nodes of the
first arc. In particular, the angle of reflection does no change when we enter
the second arc, so the next $im_{2}$ links connect its both nodes, and so on.
This means that the orbit is nodal and periodic with rotation number $1/q$ and
period $q=Mi$.

  2. (b)

The length of each link in the $j$-th arc is equal to
$\ell_{j}=2r_{j}\sin\psi$, so

$\displaystyle L(g_{q})$
$\displaystyle=\sum_{j=1}^{k}im_{j}\ell_{j}=\sum_{j=1}^{k}2im_{j}r_{j}\left[\psi-\frac{1}{6}\psi^{3}+\mathop{\rm
O}\nolimits\big{(}\psi^{5}\big{)}\right]$
$\displaystyle=\sum_{j=1}^{k}m_{j}\delta
r_{j}-\frac{\delta^{2}}{24}\left[\sum_{j=1}^{k}m_{j}\delta
r_{j}\right]\frac{1}{i^{2}}+\mathop{\rm
O}\nolimits\big{(}i^{-4}\big{)}=|\Gamma|-\frac{\pi^{2}|\Gamma|}{6q^{2}}+\mathop{\rm
O}\nolimits(1/q^{4}),$

where we have used that $\delta_{j}=m_{j}\delta$,
$|\Gamma|=\sum_{j=1}^{k}\delta_{j}r_{j}$, $q=Mi$, and $M\delta=2\pi$.

  3. (c)

If two consecutive impact points $z$ and $z^{\prime}$ belong to the arc
$\Gamma_{j}$, then

$\int_{z}^{z^{\prime}}\,{\rm d}s=2\psi r_{j}=\frac{1}{i}\delta
r_{j}=\frac{1}{q}M\delta r_{j}=\frac{1}{q}2\pi
r_{j}\neq\frac{1}{q}|\Gamma|.\qed$

Nodal billiard trajectories give the simplest samples of sequences of sliding
periodic billiard trajectories in circular polygons where properties (c) and
(d) of Theorem 44 fail. They can be obtained without the heavy machinery
developed in this paper, but they only take place on rational circular
polygons. If property $c_{1}^{-}=-\pi^{2}|\Gamma|/6$ were true, nodal billiard
trajectories would be the _asymptotically shortest_ sliding $(1,q)$-periodic
billiard trajectories as $q\to+\infty$. This is the reason to take them into
account.

## Acknowledgments

A. C. has received funding for this project from the European Research Council
(ERC) under the European Union’s Horizon 2020 research and innovation
programme (Grant Agreement No 757802). R. R.-R. was supported in part by the
grant PID-2021-122954NB-100 which was funded by MCIN/AEI/10.13039/501100011033
and “ERDF: A way of making Europe”. Thanks to Aida Chaikh and Pau Martín for
useful and stimulating conversations.

##  Appendix A Proof of Lemma 24

Along this proof we shall freely use the natural convention that objects
$\Xi^{i}_{k+1},\zeta^{i}_{k+1},\xi^{i}_{k+1}$ should be identified with
$\Xi^{i+1}_{1},\zeta^{i+1}_{1},\xi^{i+1}_{1}$, respectively.

  1. (a)

The key observation is that functions $\zeta^{i}_{j}(n)$ and $\xi^{i}_{j}(n)$
can be recursively bounded. To be precise, since $\zeta^{i}_{j+1}(n)$ is the
smallest integer such that
$\zeta^{i}_{j+1}(n)>\alpha_{j}^{-}\zeta^{i}_{j}(n)+\beta_{j}^{-}$,
$\xi^{i}_{j+1}(n)$ is the largest integer such that
$\xi^{i}_{j+1}(n)<\alpha_{j}^{+}\xi^{i}_{j}(n)-\beta_{j}^{+}$ and
$\beta_{j}^{+}=\alpha_{j}^{+}+1>1$, we deduce that

$\left.\begin{array}[]{rl}\zeta^{0}_{j+1}(n)\leq\alpha_{j}^{-}\zeta^{0}_{j}(n)+\beta_{j}^{-}+1,&\quad\forall
j=1.\ldots,k\\\
\alpha_{j}^{+}\xi^{i}_{j}(n)-\beta_{j}^{+}-1\leq\xi^{i}_{j+1}(n)\leq\alpha^{+}_{j}\xi^{i}_{j}(n),&\quad\forall
j=1,\ldots,k\ \forall i\geq 0.\end{array}\right\\}$ (27)

(A comment is in order. The careful reader may notice that, by definition of
alphabet $\boldsymbol{Q}$, $\zeta^{0}_{j}(n)\geq\chi_{j}$. Thus, it looks like
we should have written bound

$\zeta^{0}_{j+1}(n)\leq\max\big{\\{}\chi_{j+1},\alpha_{j}^{-}\zeta^{0}_{j}(n)+\beta_{j}^{-}+1\big{\\}},\qquad\forall
j=1,\ldots,k$

instead of the first bound in (27). However, we do not need it when
$\chi_{1}\gg\chi_{2},\ldots,\chi_{k}$. Under this assumption, which is the
second part of hypothesis (X), we know that the first $k-1$ minima
$\zeta^{0}_{2}(n),\ldots,\zeta^{0}_{k}(n)$ are not affected by restrictions
$\xi_{j}^{0}(n)\geq\chi_{j}$, and, if necessary, we replace
$\zeta^{1}_{1}(n)=\zeta_{k+1}^{0}(n)$ —which is the last minimum that we need
to take care of— by $\chi_{1}$.)

If we apply recursively $k$ times bounds (27), we get the cyclic bounds

$\left.\begin{array}[]{l}\zeta^{1}_{1}(n)\leq\max\\{\chi_{1},\zeta^{0}_{1}(n)/\alpha+\gamma^{-}_{1}\\}\\\
\alpha\xi^{i}_{j}(n)-\gamma^{+}_{j}\leq\xi^{i+1}_{j}(n)\leq\alpha\xi^{i}_{j}(n),\quad\forall
j=1,\ldots,k,\forall i\geq 0\end{array}\right\\}$ (28)

where $\alpha=\prod_{j=1}^{k}\alpha_{j}^{+}$,
$1/\alpha=\prod_{j=1}^{k}\alpha_{j}^{-}$ and

$\gamma^{\pm}_{j}=\sum_{m=1}^{k}\left(\prod_{l=1}^{m-1}\alpha_{j-l}^{\pm}\right)\big{(}\beta_{j-m}^{\pm}+1\big{)},\qquad\forall
j=1,\ldots,k.$

    1. i)

If we apply recursively $j-1$ times the bounds for the maxima in the second
line of equation (27), we get

$\lambda_{j}n-\gamma_{j}=\lambda_{j}\xi^{0}_{1}(n)-\gamma_{j}\leq\xi^{0}_{j}(n)\leq\lambda_{j}\xi^{0}_{1}(n)=\lambda_{j}n,\qquad\forall
j=1,\ldots,k,$ (29)

where $\lambda_{1}=1$, $\gamma_{1}=0$, and

$\lambda_{j}=\prod_{l=1}^{j-1}\alpha_{l}^{+},\qquad\gamma_{j}=\sum_{m=1}^{j-1}\left(\prod_{l=1}^{m-1}\alpha_{j-l}^{\pm}\right)\big{(}\beta_{j-m}^{\pm}+1\big{)},\qquad\forall
j=2,\ldots,k.$

If $\lambda=\max\\{\lambda_{1},\ldots,\lambda_{k}\\}$ and we choose any $\nu$
such that $0<\nu<\min\\{\lambda_{1},\ldots,\lambda_{k}\\}$, then (29) implies
that $\nu n\leq\xi^{0}_{j}(n)\leq\lambda n$ for all $j=1,\ldots,k$ provided
that $\chi_{1}$ is large enough, as it is assumed in hypothesis (X). To be
precise, if we assume that $n\geq\chi_{1}\gg 1$, then $\nu
n\leq\lambda_{j}n-\gamma_{j}$ for all $j=1,\ldots,k$. It suffices to take

$\chi_{1}\geq\max\big{\\{}\gamma_{j}/(\lambda_{j}-\nu):j=1,\ldots,k\big{\\}}.$

    2. ii)

We assume that $i\geq 0$. The upper bound
$\xi^{i}_{j}(n)\leq\alpha^{i}\xi^{0}_{j}(n)$ follows directly from (28). The
lower bound $\xi^{i}_{j}(n)\geq\tau\alpha^{i}\xi^{0}_{j}(n)$ for some
$\tau\in(0,1)$ is more tricky. First, we realize that if we choose any
$\kappa\in(1,\alpha)$, then (28) implies the weaker lower bound
$\xi^{i}_{j}(n)\geq\kappa^{i}\xi^{0}_{j}(n)$ provided
$n\geq\chi_{1}\geq\max\big{\\{}\gamma_{1}^{+},\ldots,\gamma_{k}^{+}\big{\\}}/(\alpha-\kappa)$.
This means that $\xi^{i}_{j}(n)$ grows geometrically as $i\to+\infty$. Second,
we know that

$\xi^{i}_{j}(n)\geq\alpha\xi^{i-1}_{j}(n)-\gamma_{j}^{+}=\left(1-\frac{\gamma_{j}^{+}}{\alpha\xi^{i-1}_{j}(n)}\right)\alpha\xi^{i-1}_{j}(n)\geq\cdots\geq\tau_{i,j}\alpha^{i}\xi^{0}_{j}(n),$

where

$0<\prod_{l=0}^{+\infty}\left(1-\frac{\gamma_{j}^{+}}{\alpha\xi^{l}_{j}(n)}\right)=:\tau_{j}<\tau_{i,j}=\prod_{l=0}^{i-1}\left(1-\frac{\gamma_{j}^{+}}{\alpha\xi^{l}_{j}(n)}\right)<1,\qquad\forall
i\geq 0.$

The above infinite product converges to a non-zero value $\tau_{j}$ because

$\sum_{l=0}^{+\infty}\frac{\gamma_{j}^{+}}{\alpha\xi^{l}_{j}(n)}\leq\frac{\gamma_{j}^{+}}{\alpha\xi^{0}_{j}(n)}\sum_{l=0}^{+\infty}\kappa^{-l}<+\infty.$

If we set $\tau=\min\\{\tau_{1},\ldots,\tau_{k}\\}$, then
$\xi^{i}_{j}(n)\geq\tau\alpha^{i}\xi^{i}_{0}(n)$. This ends the proof for the
forward case $i\geq 0$. The backward case $i<0$ is proved in a similar way.

    3. iii)

Inequality $\rho^{i}(n)\leq\rho^{i+1}(n)$ is trivial. Using the already proved
parts (ai) and (aii) of this lemma and the formula for geometric sums, we get

$\displaystyle\frac{\rho^{i+1}(n)}{\xi^{i}_{j}(n)}$
$\displaystyle\leq\frac{\sum_{j=1}^{k}\sum_{m=0}^{i}\alpha^{m}\xi^{0}_{j}(n)}{\tau\alpha^{i}\xi^{0}_{j}(n)}\leq\frac{\sum_{j=1}^{k}\sum_{m=0}^{i}\alpha^{m}\lambda
n}{\tau\alpha^{i}\nu
n}=\frac{k\lambda(\alpha^{i+1}-1)}{\tau\nu(\alpha-1)\alpha^{i}}$
$\displaystyle\leq\frac{k\lambda\alpha}{\tau\nu(\alpha-1)}=:\lambda^{\prime},$
$\displaystyle\frac{\rho^{i}(n)}{\xi^{i}_{j}(n)}$
$\displaystyle\geq\frac{\sum_{j=1}^{k}\sum_{m=0}^{i-1}\tau\alpha^{m}\xi^{0}_{j}(n)}{\alpha^{i}\xi^{0}_{j}(n)}\geq\frac{\sum_{j=1}^{k}\sum_{m=0}^{i-1}\tau\alpha^{m}\nu
n}{\alpha^{i}\lambda
n}=\frac{k\tau\nu(\alpha^{i}-1)}{\lambda(\alpha-1)\alpha^{i}}$
$\displaystyle\geq\frac{k\tau\nu}{\lambda\alpha}=:\nu^{\prime}.$

    4. iv)

Inequalities $n/\alpha+\gamma^{-}\leq n-1<n+1\leq\alpha n-\gamma^{+}$ for all
$n\geq\chi_{1}$ follow from hypotheses (B) and (X). It suffices to take

$\chi_{1}\geq\max\\{(1+\gamma^{+})/(\alpha-1),(1+\gamma^{-})/(1-1/\alpha)\\}.$

Set $\gamma^{\pm}=\gamma^{\pm}_{1}$. Inequalities
$\zeta^{1}_{1}(n)\leq\max\\{\chi_{1},n/\alpha+\gamma^{-}\\}$ and $\alpha
n-\gamma^{+}\leq\xi^{1}_{1}(n)$ follow directly by taking $i=0$ in (28),
because $\zeta^{0}_{1}(n)=n=\xi^{0}_{1}(n)$ by definition.

    5. v)

If we take
$n\geq\max\\{(\chi_{1}-\gamma^{-})\alpha,(N+\gamma^{+})/(\alpha-1),(N+\gamma^{-})/(1-1/\alpha)\\}$,
then $\chi_{1}\leq n/\alpha+\gamma^{-}\leq n-N<n+N\leq\alpha n-\gamma^{+}$.

  2. (b)

Let us check that sets $\Xi^{i}_{j}(n)$ have no gaps in $\mathbb{N}$. That is,
we want to check that $[n^{-},n^{+}]\cap\mathbb{N}\subset\Xi^{i}_{j}(n)$ for
all $n^{\pm}\in\Xi^{i}_{j}(n)$ such that $n^{-}\leq n^{+}$.

First, we consider the forward case $i\geq 0$. We prove it by induction in the
ordering

$\Xi^{0}_{1},\ldots,\Xi^{0}_{k},\Xi^{1}_{1}=\Xi^{0}_{k+1},\ldots,\Xi^{1}_{k},\Xi^{2}_{1}=\Xi^{1}_{k+1},\ldots,\Xi^{2}_{k},\ldots,\Xi^{i}_{1}=\Xi^{i-1}_{k+1},\ldots,\Xi^{i}_{j},\Xi^{i}_{j+1},\ldots.$

The base case is trivial: $\Xi^{0}_{1}(n)=\\{n\\}$. Let us perform now the
inductive step. We assume that $\Xi^{i}_{j}(n)$ has no holes in $\mathbb{N}$
for some $i\geq 0$ and $1\leq j\leq k$. The next set is

$\Xi^{i}_{j+1}(n)=\left\\{n^{\prime\prime}\in\mathbb{N}:n^{\prime\prime}\geq\chi_{j+1},\
\exists n^{\prime}\in\Xi^{i}_{j}(n)\mbox{ s. t.
$\alpha_{j}^{-}n^{\prime}+\beta_{j}^{-}<n^{\prime\prime}<\alpha_{j}^{+}n^{\prime}-\beta_{j}^{+}$}\right\\}.$

If $\Xi^{i}_{j+1}(n)$ has a hole in $\mathbb{N}$, there is
$n^{\prime}\geq\chi_{j}$ such that
$\alpha_{j}^{+}n^{\prime}-\beta_{j}^{+}\leq\alpha_{j}^{-}(n^{\prime}+1)+\beta_{j}^{-}$,
which is impossible by hypotheses (A) and (X). It suffices to take

$\chi_{j}>(\alpha_{j}^{-}+\beta_{j}^{-}+\beta_{j}^{+})/(\alpha_{j}^{+}-\alpha_{j}^{-}).$

Property
$\big{[}\max\\{\chi_{1},n-|i|\\},n+|i|\big{]}\cap\mathbb{N}\subset\Xi^{i}_{1}(n)$
for all $i\in\mathbb{Z}$ and $n\geq\chi_{1}$ follows by induction from part
(aiv) of this lemma and the fact that $\Xi^{i}_{1}(n)$ has no gaps in
$\mathbb{N}$.

This ends the proof for the forward case $i\geq 0$. The backward case $i<0$ is
similar. ∎

##  Appendix B Proof of Proposition 39

Fix any $p\in\mathbb{N}$. We look for the optimal value of $c_{\star}(p)>0$
such that

$\\#\Pi(p,q)\geq 2^{kp}G_{q}\big{(}P^{(p)}\big{)}\geq
c_{\star}(p)q^{n}+\mathop{\rm O}\nolimits(q^{n-1})\quad\mbox{ as
$q\to+\infty$}.$

Therefore, we want to count as many integer points in
$P^{(p)}\subset\mathbb{R}^{n+1}$ whose coordinates sum $q\in\mathbb{N}$ as
possible. We shall put these points in a 1-to-1 correspondence with the
integer points of a $q$-dilated bounded convex polytope of $\mathbb{R}^{n}$ by
means of a projection.

We shall use a lower bound established by Wills [58]. Let us briefly describe
it. If $t>0$ and $P\subset\mathbb{R}^{n}$, then
$tP=\\{y\boldsymbol{x}:\boldsymbol{x}\in P\\}$ and
$P/t=\\{\boldsymbol{x}/t:\boldsymbol{x}\in P\\}$ are the _$t$ -dilation_ and
_$t$ -contraction_ of $P$. The _inradius_ $\varrho(K)$ of a proper compact
convex set $K\subset\mathbb{R}^{n}$ is the biggest number $\varrho>0$ such
that $K$ contains a ball of radius $\varrho$. Note that $0<\varrho(K)<\infty$
for any proper compact $K$.

######  Lemma 51.

If $K$ is a proper compact convex subset of $\mathbb{R}^{n}$, then

$\\#(tK\cap\mathbb{Z}^{n})\geq\mathop{\rm
V}\nolimits(K)\big{(}t-\sqrt{n}/2\varrho(K)\big{)}^{n},\qquad\forall
t\geq\sqrt{n}/2\varrho(K).$

###### Proof.

The case $t=1$ is proved in [58], assuming that $\varrho(K)\geq\sqrt{n}/2$.
The general case follows directly from this case since $tK$ is a proper
compact convex subset of $\mathbb{R}^{n}$, $\mathop{\rm
V}\nolimits(tK)=t^{n}\mathop{\rm V}\nolimits(K)$, and
$\varrho(tK)=t\varrho(K)\geq 2/\sqrt{n}$ if $t\geq\sqrt{n}/2\varrho(K)$. ∎

The convex polytope (16) is not closed, so the convex polytopes
$\tilde{P}^{(p)}_{q}$ defined in Proposition 39 are not closed either.
However, they are the projection of some convex polytopes contained in the
open simplex
$\Delta_{n}=\\{\boldsymbol{x}\in\mathbb{R}^{n+1}:\boldsymbol{x}>0,\
x_{1}+\cdots+x_{n}=1\\}$, which implies that they are bounded. Hence we need
to extend Lemma 51 to proper bounded convex subsets of $\mathbb{R}^{n}$.

######  Corollary 52.

If $P$ is a proper bounded convex subset of $\mathbb{R}^{n}$ and $K=\bar{P}$,
then

$\\#(tP\cap\mathbb{Z}^{n})\geq\mathop{\rm
V}\nolimits(K)\big{(}s-\sqrt{n}/2\varrho(K)\big{)}^{n},\qquad\forall
t>s\geq\sqrt{n}/2\varrho(K).$

###### Proof.

The closure $K=\bar{P}$ is compact. Let $\bar{B}$ be a closed ball of radius
$\varrho(K)>0$ contained in $K$. Let $B=\mathop{\rm Int}\nolimits\bar{B}$.
Given any point $-\boldsymbol{x}\in B$, we have that
$s(\boldsymbol{x}+K)\subset t(\boldsymbol{x}+P)$ for all $t>s>0$. If
$t>\sqrt{n}/2\varrho(K)$, then there is a point $-\boldsymbol{x}_{t}\in B$
such that $t\boldsymbol{x}_{t}\in\mathbb{Z}^{n}$. Then

$\displaystyle\\#(tP\cap\mathbb{Z}^{n})$
$\displaystyle=\\#\big{(}(t\boldsymbol{x}_{t}+tP)\cap\mathbb{Z}^{n}\big{)}=\\#\big{(}t(\boldsymbol{x}_{t}+P)\cap\mathbb{Z}^{n}\big{)}\geq\\#\big{(}s(\boldsymbol{x}_{t}+K)\cap\mathbb{Z}^{n}\big{)}$
$\displaystyle\geq\mathop{\rm
V}\nolimits(\boldsymbol{x}_{t}+K)\big{(}s-\sqrt{n}/2\varrho(\boldsymbol{x}_{t}+K)\big{)}^{n}=\mathop{\rm
V}\nolimits(K)\big{(}s-\sqrt{n}/2\varrho(K)\big{)}^{n},$

for all $t>s\geq\sqrt{n}/2\varrho(K)$. ∎

###### Proof of Proposition 39..

  1. (a)

Let
$H_{n+1}=\left\\{\boldsymbol{x}\in\mathbb{R}^{n+1}:x_{1}+\cdots+x_{n+1}=1\right\\}$.
The cardinality of a finite set is invariant under $q$-dilations,
$q$-contractions, and 1-to-1 projections. Thus,

$\displaystyle\\#\Pi(p,q)$ $\displaystyle\geq
2^{kp}G_{q}\big{(}P^{(p)}\big{)}$
$\displaystyle=2^{kp}\\#\left\\{\boldsymbol{x}=(x_{1},\ldots,x_{n+1})\in
P^{(p)}\cap\mathbb{Z}^{n+1}:x_{1}+\cdots+x_{n+1}=q\right\\}$
$\displaystyle=2^{kp}\\#\big{(}P^{(p)}\cap\mathbb{Z}^{n+1}\cap
qH_{n+1}\big{)}$
$\displaystyle=2^{kp}\\#\big{(}(P^{(p)}/q)\cap(\mathbb{Z}^{n+1}/q)\cap
H_{n+1}\big{)}$
$\displaystyle=2^{kp}\\#\big{(}P^{(p)}_{q}\cap(\mathbb{Z}^{n+1}/q)\big{)}$
$\displaystyle=2^{kp}\\#\big{(}qP^{(p)}_{q}\cap\mathbb{Z}^{n+1}\big{)}$
$\displaystyle=2^{kp}\\#\big{(}q\tilde{P}^{(p)}_{q}\cap\mathbb{Z}^{n}\big{)}$
$\displaystyle\geq 2^{kp}\mathop{\rm
V}\nolimits\big{(}\tilde{K}^{(p)}_{q}\big{)}\Big{(}q-1-\sqrt{n}/2\varrho\big{(}\tilde{K}^{(p)}_{q}\big{)}\Big{)}^{n}$
$\displaystyle\geq
2^{kp}V\big{(}\tilde{K}^{(p)}_{\infty}\big{)}q^{kp-1}+\mathop{\rm
O}\nolimits(q^{kp-2})\quad\mbox{ as $q\to+\infty$},$

where $\tilde{K}^{(p)}_{q}$ is the closure of $\tilde{P}^{(p)}_{q}$. We have
used Corollary 52 with $t=q$ and $s=q-1$ in the second to last inequality. In
the last inequality, we have used estimates

$V\big{(}\tilde{K}^{(p)}_{q}\big{)}=V\big{(}\tilde{K}^{(p)}_{\infty}\big{)}+\mathop{\rm
O}\nolimits(1/q),\qquad\varrho\big{(}\tilde{K}^{(p)}_{q}\big{)}=\varrho\big{(}\tilde{K}^{(p)}_{\infty}\big{)}+\mathop{\rm
O}\nolimits(1/q).$

These estimates follow from the fact that each facet of the limit compact
polytope $\tilde{K}^{(p)}_{\infty}$ is at an $\mathop{\rm
O}\nolimits(1/q)$-distance of the corresponding facet of the polytope
$\tilde{K}^{(p)}_{q}$, which can be easily seen by comparing the half-space
representation (20) of $\tilde{K}^{(p)}_{\infty}$ with the half-space
representation

$\tilde{K}^{(p)}_{q}=\left\\{\tilde{\boldsymbol{x}}\in\mathbb{R}^{n}:\begin{array}[]{l}\alpha^{-}_{j}x_{j}+\beta^{-}_{j}/q\leq
x_{j+1}\leq\alpha^{+}_{j}x_{j}-\beta^{+}_{j}/q,\quad\forall j=1,\ldots,n-1\\\
\alpha^{-}_{n}x_{n}+\beta^{-}_{n}/q\leq
1-\varsigma(\tilde{\boldsymbol{x}})\leq\alpha^{+}_{n}x_{n}-\beta^{+}_{n}/q\\\
\alpha^{-}_{n+1}(1-\varsigma(\tilde{\boldsymbol{x}}))+\beta^{-}_{n+1}/q\leq
x_{1}\leq\alpha^{+}_{n+1}(1-\varsigma(\tilde{\boldsymbol{x}}))-\beta^{+}_{n+1}/q\\\
x_{j}\geq\chi_{j}/q,\quad\forall j=1,\ldots,n\\\
\varsigma(\tilde{\boldsymbol{x}})\leq 1-\chi_{n+1}/q\end{array}\right\\}.$
(30)

  2. (b)

All convex bodies are Jordan measurable and

$\mathop{\rm
V}\nolimits(J)=\lim_{q\to+\infty}q^{-n}\\#\left(J\cap(\mathbb{Z}^{n}/q)\right)=\lim_{q\to+\infty}q^{-n}\\#\left(qJ\cap\mathbb{Z}^{n}\right)$

for any Jordan measurable set $J\subset\mathbb{R}^{n}$, see [31, section 7.2].
Therefore,

$\displaystyle\lim_{q\to+\infty}q^{-n}G_{q}\big{(}P^{(p)}\big{)}$
$\displaystyle\leq\lim_{q\to+\infty}q^{-n}\\#\big{(}q\tilde{K}^{(p)}_{q}\cap\mathbb{Z}^{n}\big{)}$
$\displaystyle\leq\lim_{q\to+\infty}q^{-n}\\#\big{(}q\tilde{K}^{(p)}_{\infty}\cap\mathbb{Z}^{n}\big{)}=\mathop{\rm
V}\nolimits\big{(}\tilde{K}^{(p)}_{\infty}\big{)},$
$\displaystyle\lim_{q\to+\infty}q^{-n}G_{q}\big{(}P^{(p)}\big{)}$
$\displaystyle\geq\lim_{q\to+\infty}\Big{(}V\big{(}\tilde{K}^{(p)}_{\infty}\big{)}+\mathop{\rm
O}\nolimits(1/q)\Big{)}=\mathop{\rm
V}\nolimits\big{(}\tilde{K}^{(p)}_{\infty}\big{)}.$

We have used that $\tilde{K}^{(p)}_{q}\subset\tilde{K}^{(p)}_{\infty}$
—compare half-space representations (30) and (20)— in the first line and the
lower bound obtained at the beginning of this proof in the second one.

  3. (c)

It is a simple computation using that
$x_{n+1}=1-\varsigma(\tilde{\boldsymbol{x}})$ when
$\boldsymbol{x}=(\tilde{\boldsymbol{x}},x_{n+1})\in H_{n+1}$. ∎

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