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\fancyhead[CE]{\footnotesize P. T. Dang, T. T. Dao, N. M. Doan}
\fancyhead[CO]{\footnotesize The torus with asymptotically the fewest
simple closed geodesics}
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\DeclareMathOperator{\Area}{Area}
\DeclareMathOperator{\arccosh}{arccosh}
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\title{The torus with asymptotically the fewest
simple closed geodesics}
\author{Phuc Thinh Dang, Trong Toan Dao, Nhat Minh Doan}
\date{}
\begin{document}
\maketitle
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\begin{NoHyper}
\footnotetext{\textit{2020 Mathematics Subject Classification.}
Primary 30F60; Secondary 53C22, 52A10.\par
\textit{Keywords.} Hyperbolic surfaces, modular torus, simple closed
geodesics, stable norm, McShane--Rivin area conjecture, Farey polygons, Mirzakhani functions.}
\end{NoHyper}
\endgroup

\begin{abstract}
We prove that the modular torus uniquely minimizes the area of the
stable-norm unit ball among complete finite-area hyperbolic
once-punctured tori, resolving the McShane--Rivin area conjecture.
It therefore has fewer simple closed geodesics of length at most $L$
than any fixed nonmodular torus for all sufficiently large $L$.
We also bound the area increase away from the modular torus
and determine its leading growth and first correction term as
the systole tends to zero, with error bounds independent of the twist.
\end{abstract}

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\section{Introduction}

Let $S=S_{1,1}$ be a once-punctured torus. Choose an integral homology
basis to identify $H_1(S;\Z)$ with $\Z^2$ and $H_1(S;\R)$ with $\R^2$.
For a complete finite-area hyperbolic structure $X$ on $S$, let
$\|\cdot\|_X$ be its stable norm. For each nonzero primitive class $h\in H_1(S;\Z)$, $\|h\|_X$ is
the length of the unique oriented simple closed geodesic representing
$h$. The classes $h$ and $-h$ represent the same geodesic with
opposite orientations. Set
$$
\B(X)=\{w\in\R^2:\|w\|_X\leq1\},
\qquad
\A(X)=\Area(\B(X)),
$$
where $\Area$ is the usual Euclidean area. Changes of integral
homology basis have determinant $\pm1$, so $\A(X)$ is independent
of this choice. Isometries preserve the stable norm and the integral
homology lattice, so $\A$ depends only on the isometry class of $X$
and defines a function on the moduli space.

The stable norm belongs to the theory of geometric norms on homology
developed by Federer and Gromov \cite{Federer1974,Gromov1983}.
For a hyperbolic once-punctured torus $X$, McShane and Rivin showed that
its unit-ball area $\A(X)$ determines the leading term in simple-geodesic
counting \cite{McShaneRivinNorm,McShaneRivinTori}. If $N_X(L)$
counts unoriented simple closed geodesics of length at most $L$, then
\begin{equation}\label{eq:Counting}
 N_X(L)=\frac{3}{\pi^2}\A(X)L^2+O_X(L\log L).   
\end{equation}

Thus, minimizing $\A$ asks which hyperbolic torus has the fewest
simple closed geodesics in the asymptotic sense.
For the modular torus $\Xmod$, this counting problem is equivalent,
after accounting for symmetries and converting traces to lengths,
to the counting of Markoff triples studied by Zagier \cite{Zagier1982}.
McShane and Rivin conjectured that $\A$ is minimized at $\Xmod$
\cite[Conjecture~3.4]{McShaneRivinTori}.
In his lecture \cite{Bonahon2018}, Bonahon presented
numerical evidence for this conjecture, obtained in joint work
with Enriquez using shear coordinates.

Mirzakhani extended this counting result to complete finite-area
hyperbolic surfaces of genus $g$ with $n$ cusps and $3g-3+n>0$:
the growth exponent is $6g-6+2n$, and the leading coefficient
is a positive topological constant times the Mirzakhani function
\cite{MirzakhaniGrowth}.
This function is the Thurston measure of the compactly supported
measured laminations of length at most one.
With the integral multicurve lattice normalization, it equals
$\A(X)/2$ on the once-punctured torus, since each nonzero integral
multicurve corresponds to two opposite homology classes
\cite{MirzakhaniGrowth}.
Continuity and properness ensure that a minimum exists on each
moduli space, but do not identify the minimizing surfaces
\cite[Proposition~3.2 and Theorem~3.3]{MirzakhaniGrowth}.
Arana-Herrera and Athreya refined the estimates near degeneration
and proved square-integrability for Weil--Petersson measure
\cite{AranaHerreraAthreya}. The Weil--Petersson integral of this function is also related to Masur--Veech
volumes through work of Mirzakhani \cite{MirzakhaniEarthquake}
and explicit formulas of Delecroix, Goujard, Zograf, and Zorich
\cite{DelecroixGoujardZografZorich}. 

The function also appears in other counting problems.
For closed surfaces, Rafi and Souto give counting formulas for
mapping class group orbits of filling geodesic currents and for
mapping classes counted using the Thurston metric
\cite{RafiSouto}.
Aougab and Athreya obtain analogous formulas for harmonic
representatives of a fixed weighted triangulation under the
mapping class group, counted by weighted length
\cite[Theorem~1.2]{AougabAthreya}.
For two closed hyperbolic surfaces of the same genus $g\geq2$,
Lowe, Marques, and Neves showed that the leading coefficient
for counting $\pi_1$-injective minimal Lagrangian immersions
of genus $g$ into their product by area is proportional
to the product of their Mirzakhani functions
\cite[Corollary~1.5]{LoweMarquesNeves}. These connections motivate the broader question suggested in
Parlier's survey \cite{ParlierSurvey}: which surfaces minimize
the Mirzakhani function for each topological type?


\subsection*{Main results} In this paper, we prove the McShane--Rivin conjecture, establish uniqueness,
and give quantitative bounds for the increase away from
the modular torus. The symmetry of the modular torus plays an important role in our proof
and may also guide the search for minimizers in higher complexity.

\begin{theorem}
\label{thm:global-area}
\label{conj:main}
Every complete finite-area hyperbolic once-punctured torus $X$ satisfies
$$
\A(\Xmod) \le \A(X),
$$
with equality exactly when $X=\Xmod$.
Consequently, if $X \neq \Xmod$, then by \eqref{eq:Counting} $$N_{\Xmod}(L)<N_X(L)$$ for all sufficiently large $L$,
with the threshold depending on $X$.
\end{theorem}

The minimum of $\A$ has consequences for several counting problems.
In each case below, the leading coefficient is a positive constant
independent of $X$ times $\A(X)$, and hence is uniquely minimized
by the modular torus.
\begin{itemize}
\item
Closed geodesics in any fixed mapping class group orbit of a
nontrivial, nonperipheral closed curve, including curves with
self-intersections \cite{ErlandssonSouto,MirzakhaniOrbits}.
This also applies to primitive closed geodesics with at most
$k$ self-intersections, for fixed $k\geq0$, since only finitely
many orbits occur. See \cite{ErlandssonSoutoBook} for a detailed account.

\item
Essential cusp-to-cusp geodesic arcs, or nonempty finite collections
of them, in a fixed mapping class group orbit
\cite{Bell,Trin}.
Their total length is measured after truncating the ends at a
horocycle of fixed sufficiently small length.
Examples include ideal triangulations, counted by the sum of
their three truncated edge lengths.

\item
Conjugacy classes of finitely generated, noncyclic, infinite-index
subgroups of $\pi_1(S)$ in a fixed mapping class group orbit
\cite{Sasaki}.
The length assigned to a subgroup is half the total length
of the geodesic boundary of its convex core in the corresponding
covering surface.
\end{itemize}


The minimum also transfers to four-punctured spheres.
A classical correspondence pairs once-punctured tori with
four-punctured spheres, uniquely up to isometry
(see \cite{HuTan}).
Their simple closed geodesics correspond bijectively,
with lengths doubled on the sphere, so
$$
N_Y(L)=N_X(L/2).
$$
The four-punctured sphere $Y_*$ corresponding to $\Xmod$ therefore uniquely
minimizes the Mirzakhani function among four-punctured spheres.
Its leading simple-geodesic counting coefficient is one quarter
of that of $\Xmod$. Together with Mirzakhani's growth theorem \cite{MirzakhaniGrowth},
this gives the following comparison across topological types.

\begin{corollary}
\label{cor:fewest-simple-geodesics}
For every complete finite-area hyperbolic surface $Z$ without
boundary, other than $Y_*$ and the thrice-punctured sphere,
$$
N_{Y_*}(L)<N_Z(L)
$$
for all sufficiently large $L$, with the threshold depending
on $Z$.
\end{corollary}

Write $s_X=\sys(X)$ for the length of a shortest essential
simple closed geodesic on a hyperbolic surface $X$. Corollary \ref{cor:fewest-simple-geodesics} indicates that $Y_*$ asymptotically has the fewest simple closed geodesics
among complete finite-area hyperbolic surfaces without boundary,
except for the thrice-punctured sphere, which has none.
The boundary restriction is essential: one-holed tori can have
$s_X\to\infty$ \cite{Schmutz}, forcing $\A(X)\to0$ by
Minkowski's convex-body theorem
\cite[Theorem II, p.~71]{Cassels}.

 Each torus admits a reduced trace triple $(x,y,z)$,
formed from three simple closed geodesics that intersect pairwise
once. The trace of a geodesic of length $\ell$ is $2\cosh(\ell/2)$.
The triple satisfies
$$
\begin{aligned}
x^2+y^2+z^2&=xyz,\qquad 
2<x\leq y\leq z\leq\frac{xy}{2}.
\end{aligned}
$$
These coordinates are recalled in Section~\ref{sec:coordinates}
and illustrated in Figure~\ref{fig:area-comparison-domains}. For a reduced triple, set
\begin{equation}\label{eq:p_2}
p_2=\frac13\left(
\log^2\frac{x}{y}+\log^2\frac{y}{z}+\log^2\frac{z}{x}\right).
\end{equation}
This quantity measures the differences between the three trace
logarithms and vanishes exactly at the modular triple $(3,3,3)$.
Our next theorem gives matching upper and lower bounds for the
area gap throughout moduli space and a sharp asymptotic as the
systole tends to zero.

\begin{theorem}[Uniform bounds and cusp growth]
\label{thm:quantitative-area}
There are universal constants $c,C>0$ such that every $X$ satisfies
$$
c\frac{p_2e^{\sqrt{3p_2/2}}}{(1+p_2)^{3/2}}
\leq\A(X)-\A(\Xmod)
\leq C\frac{p_2e^{\sqrt{3p_2/2}}}{(1+p_2)^{3/2}}.
$$
As $s_X\to0$,
\begin{equation}\label{eq:Expansion}
\A(X)=\frac{1}{s_X\log(2/s_X)}
\left(1-\frac{\pi^2}{12\log^2(2/s_X)}
+O\bigl(\log^{-3}(2/s_X)\bigr)\right).
\end{equation}
The same expansion holds for $\A(X)-\A(\Xmod)$.
\end{theorem}

For once-punctured tori, the estimates of Arana-Herrera and
Athreya \cite[Theorem~1.5]{AranaHerreraAthreya} bound $\A(X)$
above and below by positive constant multiples of
$1/(s_X|\log s_X|)$ as $s_X\to0$.
Our expansion \eqref{eq:Expansion} determines the leading constant and the first
correction term.
\subsection*{Proof strategy} We approximate the stable-norm unit ball by a finite Farey polygon.
For each surface $X$, choose a reduced trace triple $(x,y,z)$ and
an ordered integral basis $\alpha,\beta$ realizing it, so that
$x,y,z$ are the traces of $\alpha,\beta,\alpha+\beta$. In each of the three sectors between consecutive
classes
$$
\alpha,\qquad \alpha+\beta,\qquad \beta,\qquad -\alpha,
$$
write $w=pu_0+qv_0$, where $u_0,v_0$ are the sector endpoints and $p,q\geq0$, and
define the \emph{sector height} by $h(w)=p+q$.
Let $P_6(X)$ be the centrally symmetric inscribed polygon whose vertices
are the normalized primitive classes
$
w/\|w\|_X
$
of sector height at most six, together with their negatives
(see Figure~\ref{fig:farey-p6}), and set
$$
\A_6(X)=\Area(P_6(X)).
$$
\begin{figure}[ht]
\centering
\includegraphics{figures/diagram-01.pdf}
\caption{The Farey polygon $P_6(X_{\mathrm{mod}})$.}
\label{fig:farey-p6}
\end{figure}
The polygon $P_6$ is small enough for an explicit comparison and
approximates the norm ball closely enough to control the omitted
area $$(\A-\A_6)(X):=\A(X)-\A_6(X).$$ 
Since $P_6(X)\subsetneq\B(X)$, the omitted area
$(\A-\A_6)(X)$ is positive. We use the identity
\begin{equation}
\label{eq:area-comparison-split}
\A(X)-\A(\Xmod)
=\bigl[\A_6(X)-\A_6(\Xmod)\bigr]
+\bigl[(\A-\A_6)(X)-(\A-\A_6)(\Xmod)\bigr],
\end{equation}
where the first bracket is the change in polygon area and the
second is the change in omitted area. For nonmodular surfaces
with $\sys(X)\geq7/5$, we show that the polygon area increases
by more than any possible decrease in the omitted area.
For $\sys(X)<7/5$, we estimate the full area directly.

Cyclic symmetry is central to the local comparison.
The permutation $(x,y,z)\mapsto(y,z,x)$ fixes the modular point.
For any twice continuously differentiable function invariant under
this permutation, Lemma~\ref{lem:symmetry-at-modular} gives, in
local coordinates $(x,y)$,
$$
\nabla F(3,3)=0,
\qquad
\Hess F(3,3)=\kappa
\begin{pmatrix}2&1\\1&2\end{pmatrix}.
$$
Thus there is no linear term, and the quadratic term is determined
by one number. This applies to both the polygon area $\A_6$
and the omitted area $\A-\A_6$.

For the polygon area, we compute a positive quadratic term and
show that it dominates the higher terms throughout an explicit
neighborhood. To control the uncomputed Taylor terms, we extend
the polygon formula to complex parameter values and bound the
average of its squared modulus on a circle. This controls all
remaining Taylor terms at once. Proposition~\ref{prop:uniform-p6} gives
$$
\A_6(X)-\A_6(\Xmod)>\frac7{80}p_2
$$
for nonmodular $X$ in that neighborhood.
The same average bound, combined with further Taylor coefficients,
controls the remainder farther from the modular point.

We combine these polygon estimates with bounds for the omitted
area in three regions
(Figure~\ref{fig:area-comparison-domains}).
\begin{itemize}
\item
Region $A$ consists of tori represented by reduced triples with
$\sys(X)\geq7/5$ outside $(x-3)^2+(y-3)^2\leq1/1024$.
Here the polygon increase exceeds the entire omitted area at
the modular torus. Since $p_2>1/12800$ by
\eqref{eq:outside-disk-shape-lower},
Corollary~\ref{cor:global-polygon-comparison} and
Lemma~\ref{lem:small-modular-remainder} give
$$
\A(X)-\A(\Xmod)>
\frac{19}{1000}p_2-\frac1{700000}>0.
$$

\item
Region $B$ consists of nonmodular tori represented by reduced
triples with $\sys(X)\geq7/5$ inside that disk.
The polygon increase tends to zero at the modular point, so
we must control the change in omitted area.
Its vanishing first derivatives and a lower bound for its
second derivatives limit its possible decrease to less than
$p_2/50$. Lemma~\ref{lem:lambda-local-conversion} and
Corollary~\ref{cor:local-no-grid} therefore give
$$
\A(X)-\A(\Xmod)>
\left(\frac7{80}-\frac1{50}\right)p_2
>0.
$$

\item
In region $C$, where $\sys(X)<7/5$, curves winding around a
shortest geodesic give a direct lower bound for the full area.
Theorem~\ref{thm:pure-cusp} and
Proposition~\ref{prop:modular-upper-new} yield
$$
\A(X)-\A(\Xmod)>\frac1{1000}.
$$
\end{itemize}
\begin{figure}[ht]
\centering
\begin{minipage}[b]{0.48\textwidth}
\centering
\includegraphics{figures/diagram-02.pdf}

\smallskip
\textbf{(a)} Reduced region
\end{minipage}
\hfill
\begin{minipage}[b]{0.48\textwidth}
\centering

\raisebox{0.35cm}{%
\includegraphics{figures/diagram-03.pdf}%
}

\smallskip
\textbf{(b)} Enlargement near $(3,3)$
\end{minipage}
\caption{
The three comparison regions in reduced trace coordinates.
Dashed boundaries mark $\sys(X)=7/5$ in (a) and
$(x-3)^2+(y-3)^2=1/1024$ in (b).
}
\label{fig:area-comparison-domains}
\end{figure}
These comparisons give the explicit regional bound and prove
Theorem~\ref{thm:global-area}. For Theorem~\ref{thm:quantitative-area}, the estimates in region
$B$ give the quadratic increase near the modular torus, while
the curves used in region $C$ give the expansion as $s_X\to0$.
In each limit, the ratio of the area gap to
$p_2e^{\sqrt{3p_2/2}}/(1+p_2)^{3/2}$ tends to a positive constant.
After excluding a small neighborhood of the modular point and
all sufficiently small systoles, the remaining reduced triples
form a compact set. On this set, the ratio is continuous and
positive by Theorem~\ref{thm:global-area}, so it has positive
lower and finite upper bounds. Together, these bounds prove
the global comparison.

\paragraph{Organization of the paper.}
Sections~\ref{sec:coordinates} and~\ref{sec:shared-estimates}
develop the coordinates and estimates used in the three regional
comparisons in Sections~\ref{sec:region-a}--\ref{sec:region-c}.
Section~\ref{sec:completion} completes the proofs of both main theorems.

\paragraph{Acknowledgements.}
The ideas of subdividing the reduced region for different comparisons
and estimating Taylor coefficients along two lines through the modular
point (Subsection~\ref{sec:taylor-expansion}) were developed in part from discussions with Ser Peow Tan during the last-named author's
visit to the National University of Singapore.
The last-named author thanks him for these helpful discussions and for
encouraging us to pursue the conjecture. The background and some estimates were influenced by the recent work \cite{DLN} with Xiaobin Li and Van Nguyen. The second-named author is supported by NUS President Graduate Fellowship. 

\paragraph{Declaration of AI use.}
 ChatGPT (versions 5.4, 5.5, 5.6, and 6.0) was used
extensively by the last-named author to assist with technical details, check calculations
and arguments, and prepare TikZ figures. The authors developed the proof strategy, directed this
work through questions and critical feedback, contributed ideas
along the way to improve the results, carefully checked the resulting
arguments and computations, and revised the exposition for clarity, simplicity, and accessibility to a broader readership. They take full responsibility for the content.

\section{Preliminaries}
\label{sec:coordinates}

This section introduces the notations, coordinates, area formulas,
and differential tools used in the proof.
\subsection{Trace coordinates and reduction}
Oriented essential simple closed curves on the once-punctured torus $S$ correspond to nonzero primitive classes
in $H_1(S;\Z)$.
Two primitive classes $u,v$ are Farey neighbors exactly when their
algebraic intersection number has absolute value one. Fix a complete finite-area hyperbolic structure $X$ on $S$ and let $\|\cdot\|_X$ be the stable norm on
$H_1(S;\R)$. For a primitive class $u$, define its trace by
$$
x_u=2\cosh\frac{\|u\|_X}{2}.
$$
For any Farey neighbors $u,v$, the traces
$x_u,x_v,x_{u+v}$ satisfy the Markoff equation
\begin{equation}
\label{eq:markoff}
x_u^2+x_v^2+x_{u+v}^2=x_ux_vx_{u+v},
\end{equation}
and the two traces across the corresponding Farey edge satisfy
\begin{equation}
\label{eq:vieta}
x_{u+v}+x_{u-v}=x_ux_v.
\end{equation}
See \cite{Bowditch}. Thus crossing the edge replaces the Farey triangle
$(u,v,u-v)$ by $(u,v,u+v)$ and induces the Vieta move
$$
x_{u-v}\longmapsto x_ux_v-x_{u-v}=x_{u+v}.
$$
The positive solutions of the Markoff equation give the usual trace
coordinates for marked complete hyperbolic punctured tori, see
\cite[Theorem~4.1.2 and Section~4.4]{GoldmanTrace}. We call a solution
$(x,y,z)\in(2,\infty)^3$ a \emph{positive Markoff triple}.
For $(x,y,z)=(x_u,x_v,x_{u+v})$, replacing the ordered basis
$(u,v)$ by $(v,-u-v)$ gives the three classes $v,-u-v,-u$.
Since $x_{-w}=x_w$, this change induces the cyclic permutation
\begin{equation}
\label{eq:cyclic-trace-permutation}
\rho(x,y,z)=(y,z,x).
\end{equation}
For fixed $x,y$, solving the Markoff equation for $z$ gives
$$
z_\pm(x,y)
=
\frac{xy\pm\sqrt{\mathscr D(x,y)}}2,
\qquad
\mathscr D(x,y):=x^2y^2-4x^2-4y^2.
$$
For any real Markoff triple $(x,y,z)$,
$$
\mathscr D(x,y)
=
x^2y^2-4xyz+4z^2
=
(xy-2z)^2
\geq0.
$$
Thus, on the open set
$$
\mathscr U
=
\{(x,y)\in(2,\infty)^2:\mathscr D(x,y)>0\},
$$
the two branches $z_\pm$ are distinct and real analytic. They coincide exactly on
$\mathscr D=0$, where
$$
z_+(x,y)=z_-(x,y)=\frac{xy}{2}.
$$
A positive Markoff triple $(x,y,z)$ is \emph{reduced} if
$$
2<x\leq y\leq z\leq\frac{xy}{2},
$$
see the left panel of Figure~\ref{fig:reduced-region-local}. Thus, for a
reduced triple,
$$
z=z_-(x,y).
$$
A Farey triangle is called a \emph{sink} if none of its three
adjacent Vieta moves decreases the trace it replaces.

\begin{proposition}[Markoff reduction]
\label{prop:markoff-reduction}
Every positive Markoff triple is equivalent under permutations and
Vieta moves to a reduced triple. If $(x,y,z)$ is a reduced triple
representing $X$, then
$$
x=2\cosh\frac{\sys(X)}2,
\qquad
x\leq3,
\qquad
y\leq\frac{x}{\sqrt{x-2}}.
$$
\end{proposition}

\begin{proof}
Associate $X$ with its Markoff map on the Farey tree. The reduction
argument \cite[Section~3]{Bowditch} gives a sink triangle.
Order its traces as $x\leq y\leq z$. Then $z\leq xy-z$, so the triple is reduced.

Conversely, every Vieta move from a reduced triple produces a
trace at least as large as $z$. Continue along any path without
reversing a move. At each later step
$
(U,V,W)\longmapsto(U,V,UV-W),
$
one of $U,V$ is the trace produced at the previous step.
Inductively, it is at least as large as $W$, so
$$
UV-W\geq UV-\max(U,V)>\max(U,V),
$$
since $U,V>2$. Thus $x$ is the smallest trace, so
$x=2\cosh(\sys(X)/2)$. Finally, \eqref{eq:markoff} and reducedness give
$$
x^2+y^2=z(xy-z)\geq y(xy-y)=(x-1)y^2,
$$
since $t(xy-t)$ is increasing on $(-\infty,xy/2]$ and
$
y\leq z\leq xy/2.
$
Therefore
$
y\leq x/\sqrt{x-2}.
$
Since $y\geq x$, it follows that $x\leq3$.
\end{proof}

For the local estimates near the modular torus, fix an ordered
integral basis $(\alpha,\beta)$ such that the traces of
$\alpha,\beta,\alpha+\beta$ at $\Xmod$ are all $3$.
As the hyperbolic structure varies, $x,y,z$ denote the respective
traces of these fixed classes. In particular, $x$ always denotes
the trace of $\alpha$, even when it is not the smallest trace.
These coordinates need not form a reduced triple, and each length
derivative follows a fixed curve. We use the branch $z=z_-(x,y)$ on
\begin{equation}
\label{eq:local-region}
\Rloc=
\left\{
(x,y)\in\R^2:
(x-3)^2+(y-3)^2\leq\frac1{1024},
x\leq3,\ y\geq x
\right\}.
\end{equation}
\begin{figure}[ht]
\centering

\begin{minipage}[b]{0.48\textwidth}
\centering

\includegraphics{figures/diagram-04.pdf}

\smallskip
\textbf{(a)} Reduced region
\end{minipage}
\hfill
\begin{minipage}[b]{0.48\textwidth}
\centering

% Raise only the picture; the panel labels remain bottom-aligned.
\raisebox{0.35cm}{%
\includegraphics{figures/diagram-05.pdf}%
}

\smallskip
\textbf{(b)} Enlargement at $(3,3)$
\end{minipage}

\caption{
The reduced region in the $(x,y)$-plane and the convex set
$\Rloc$ near $(3,3)$.
}
\label{fig:reduced-region-local}
\end{figure}
The set $\Rloc$ is convex, since it is the intersection of a closed
disk and two closed half-planes (see
Figure~\ref{fig:reduced-region-local}(b)).
It contains the comparison region $B$
(see Figure~\ref{fig:area-comparison-domains}(b)) and also points
whose trace triples become reduced after reordering
(Lemma~\ref{lem:local-bounds}).
Working on $\Rloc$ allows us to apply Taylor's formula along
the entire segment from $(3,3)$ to each point of region $B$.

\begin{lemma}[Local discriminant bound]
\label{lem:local-discriminant}
For every $(x,y)\in\Rloc$,
$$
\mathscr D(x,y)=x^2y^2-4x^2-4y^2>\frac{121}{16}.
$$
Consequently, $z(x,y):=z_-(x,y)$
is real analytic on a neighborhood of $\Rloc$.
\end{lemma}

\begin{proof}
The region lies in the square $[95/32,97/32]^2$. Throughout this square,
$$
0<\partial_x\mathscr D=2x(y^2-4)
\leq2\frac{97}{32}\left(\left(\frac{97}{32}\right)^2-4\right)<32,
$$
and the same bound holds for $\partial_y\mathscr D$.
Since $\mathscr D(3,3)=9$, the mean value theorem and the disk condition give
$$
\mathscr D(x,y)\geq9-32(|x-3|+|y-3|)
\geq9-\sqrt2>\frac{121}{16}.
$$
The positive lower bound persists on a neighborhood of $\Rloc$,
where the formula for $z_-$ is therefore real analytic.
\end{proof}

By Lemma~\ref{lem:local-discriminant}, the branch $z_-(x,y)$
is real analytic near $\Rloc$. On a sufficiently small open
neighborhood, $x,y,z_-(x,y)>2$, so the trace parametrization defines
a marked hyperbolic torus $X_{x,y}$ satisfying
\begin{equation}
\label{eq:marked-local-family}
x_\alpha(X_{x,y})=x,
\quad
x_\beta(X_{x,y})=y,
\quad
x_{\alpha+\beta}(X_{x,y})=z_-(x,y).
\end{equation}
Set
$$
\A(x,y):=\A(X_{x,y}).
$$
This function is real analytic on a neighborhood of $\Rloc$
by \cite[Section~3, Corollary~1]{McShaneRivinTori}.
See also Bonahon's discussion of this property in
\cite[slide~30]{Bonahon2018}.

We shall repeatedly use the following uniform bounds on $\Rloc$.

\begin{lemma}[Local trace and length bounds]
\label{lem:local-bounds}
For $(x,y)\in\Rloc$, let $z=z_-(x,y)$. Then
$$
\frac{74}{25}<x,y,z<\frac{61}{20},
$$
and the corresponding geodesic lengths satisfy
$$
2\arccosh\frac{x}{2},
\quad
2\arccosh\frac{y}{2},
\quad
2\arccosh\frac{z}{2}
>
L_0:=\frac{236}{125}.
$$
Moreover, the fixed Farey triangle determined by
$\alpha,\beta,\alpha+\beta$ is the unique sink throughout $\Rloc$,
even when the ordered triple $(x,y,z)$ is not reduced.
Every move along a path away from this triangle produces the larger
Vieta root. The three moves adjacent to it produce traces greater
than $14279/2500$.
\end{lemma}

\begin{proof}
From the definition of $\Rloc$,
$95/32\leq x,y\leq 97/32,
$
so only $z=z_-(x,y)$ needs to be bounded. Set
$$
f_{x,y}(t)=t^2-xyt+x^2+y^2.
$$
A direct estimate on $\Rloc$ gives
$$
f_{x,y}(74/25)
\geq f_{3,97/32}(74/25)
>0,
\qquad
f_{x,y}(61/20)<0.
$$
Since both test points lie to the left of $xy/2$, the lower root satisfies
$$
\frac{74}{25}<z<\frac{61}{20}.
$$
Hence all three corresponding geodesic lengths are greater than
$$
2\arccosh\frac{37}{25}>\frac{236}{125}.
$$
The trace bounds also give
$$
2\max(x,y,z)<\frac{61}{10}
<\min(xy,yz,zx).
$$
Thus each of the three Vieta moves strictly increases the trace
it replaces, so the fixed triangle is a strict sink.
The growth argument in the proof of
Proposition~\ref{prop:markoff-reduction} shows that every subsequent
move along a path away from this triangle also increases the
replaced trace and selects the larger Vieta root. This excludes
any other sink. The three adjacent moves produce traces
$$
x_{2\alpha+\beta}=xz-y,\qquad
x_{\alpha+2\beta}=yz-x,\qquad
x_{\beta-\alpha}=xy-z.
$$
Each is greater than
$$
\left(\frac{74}{25}\right)^2-\frac{61}{20}
>\frac{61}{20}.
$$
\end{proof}

The next lemma shows that every surface with systole greater than
$381/200$ has reduced coordinates in $\Rloc$.

\begin{lemma}[High systole localization]
\label{lem:high-systole}
If $\sys(X)>381/200$, then a reduced representative of $X$ belongs to
$\Rloc$.
\end{lemma}

\begin{proof}
For a reduced triple, $x$ is the systolic trace, so
$$
x>2\cosh\frac{381}{400}>
\frac{74}{25}.
$$
Since $x\in[74/25,3]$, a direct squaring gives
$$
\frac{x}{\sqrt{x-2}}\leq6-x.
$$
Together with Proposition~\ref{prop:markoff-reduction}, this yields
$
x\leq y\leq6-x,
$
and hence
$$
(x-3)^2+(y-3)^2
\leq2(3-x)^2
<\frac1{1024}.
$$
Reducedness also gives $z=z_-(x,y)$, so $(x,y)\in\Rloc$.
\end{proof}

\subsection{The exact Farey area expansion}
\label{sec:farey-area}

To construct the Farey polygons for a surface $X$, choose a reduced
trace triple $(x,y,z)$ and an ordered integral basis $(\alpha,\beta)$
realizing it:
$$
(x_\alpha,x_\beta,x_{\alpha+\beta})=(x,y,z).
$$
The Farey triangle $(\alpha,\beta,\alpha+\beta)$ is a sink by
Proposition~\ref{prop:markoff-reduction}. We use it as the initial
triangle; this choice may depend on $X$.

When differentiating in the local family
\eqref{eq:marked-local-family}, we keep the basis already fixed
at $\Xmod$. Lemma~\ref{lem:local-bounds} ensures that its triangle
remains a sink throughout $\Rloc$, although its traces need not
be ordered. Thus the curves and sector heights remain fixed
as $(x,y)$ varies.
In either setting, a move away from the initial triangle is called
an \emph{outward Farey insertion}.

For each of the three initial sectors, let $(u_0,v_0)$ denote its
endpoints:
$$
(u_0,v_0)\in
\{(\alpha,\alpha+\beta),(\alpha+\beta,\beta),(\beta,-\alpha)\}.
$$
Every primitive class in this sector has a unique expression
$w=pu_0+qv_0$ with nonnegative integers $p,q$. Define its
\emph{sector height} by
$$
h(w)=p+q.
$$
Starting from these three intervals, recursively insert $u+v$
between the endpoints of each current interval $(u,v)$, replacing
it by $(u,u+v)$ and $(u+v,v)$. At each insertion, we call $u,v$
the \emph{parents} and $u+v$ their \emph{child}.
Let $\mathscr F$ be the set of all intervals occurring in this
subdivision, including the initial ones.
Heights are always measured relative to the fixed endpoints
$(u_0,v_0)$ of the initial sector containing the interval.
Thus, for every $(u,v)\in\mathscr F$,
$$
h(u+v)=h(u)+h(v).
$$
Every primitive class in the interior of an initial sector is
inserted exactly once.

For $K\geq1$, let $P_K(X)$ be the centrally symmetric inscribed polygon
whose vertices are the normalized primitive classes of sector height at
most $K$, together with their negatives, and set
$$
\A_K(X)=\Area(P_K(X)).
$$
For a reduced triple, this is the polygon used in the introduction
(see Figure~\ref{fig:farey-p6}). In the local family, the sectors
and heights are fixed; only the geodesic lengths used to normalize
the vertices vary.
On region $B$, the fixed traces form a reduced triple and the two
constructions agree. The local construction also defines $\A_K$
at the nonreduced triples in $\Rloc$ needed for differentiation.

For a primitive class $u$, write
$$
L_u=\|u\|_X=2\arccosh\frac{x_u}{2}.
$$
In the marked family \eqref{eq:marked-local-family}, we write
$L_w(x,y)=\|w\|_{X_{x,y}}$.
If $u+v$ is inserted between Farey neighbors $u,v$, the edge joining
$u/L_u$ to $v/L_v$ is replaced by the two edges through
$(u+v)/L_{u+v}$ (see Figure~\ref{fig:area-correction}). 
\begin{figure}[ht]
\centering
\includegraphics{figures/diagram-06.pdf}
\caption{The area added by one Farey insertion on one side of the norm ball.}
\label{fig:area-correction}
\end{figure}

Since
$$
|\det(u,v)|=|\det(u,u+v)|=|\det(u+v,v)|=1,
$$
the shaded triangle has area
$$
\frac12\left(
\frac1{L_uL_{u+v}}
+\frac1{L_{u+v}L_v}
-\frac1{L_uL_v}
\right).
$$
Including its antipodal copy, the total area gain is
\begin{equation}
\label{eq:area-gain}
S_X(u,v)
:=
\frac1{L_uL_{u+v}}
+\frac1{L_{u+v}L_v}
-\frac1{L_uL_v}
=
\frac{L_u+L_v-L_{u+v}}{L_uL_vL_{u+v}}.
\end{equation}
This is positive by the strict triangle inequality
$L_{u+v}<L_u+L_v$ \cite[Section~2]{McShaneRivinTori}. We obtained the following expansion independently.
O'Sullivan \cite[Proposition~6.5]{OSullivan2026} generalizes
Zagier's counting constant formula for Markoff triples
\cite[equation~(3)]{Zagier1982} to rising trees of
$x^2+y^2+z^2-xyz=D$, with $D\in\R$.
His $D=0$ sector formula gives our expansion after converting
half-lengths to lengths and summing over the sectors.

\begin{proposition}[Farey area expansion]
\label{prop:farey-expansion}
The polygons $P_K(X)$ are nested and
$$
\A_K(X)\nearrow\A(X).
$$
Moreover,
$$
\A_K(X)
=
\A_1(X)
+
\sum_{\substack{(u,v)\in\mathscr F\\h(u+v)\leq K}}
S_X(u,v),
$$
and hence
$$
\A(X)
=
\A_1(X)
+
\sum_{(u,v)\in\mathscr F}S_X(u,v).
$$
\end{proposition}

\begin{proof}
Increasing $K$ only adds boundary points of $\B(X)$, so the polygons are
nested. Since primitive directions are dense and the norm is continuous,
$P_K(X)$ converges to $\B(X)$ in the Hausdorff metric. Hence
$
\A_K(X)$ converges to $\A(X)
$
by \cite[Theorem~1.8.20]{Schneider}.
Summing the corrections of all insertions of height at most $K$ gives the
finite formula, and letting $K\to\infty$ gives the expansion of $\A$.
\end{proof}

We record three facts that will be used repeatedly. First,
Proposition~\ref{prop:farey-expansion} gives the tail identity
\begin{equation}
\label{eq:farey-tail}
\A(X)-\A_K(X)
=
\sum_{\substack{(u,v)\in\mathscr F\\h(u+v)>K}}
S_X(u,v).
\end{equation}

Second, at sector height $H\geq2$, each initial sector contains exactly
$\varphi(H)$ inserted classes, where
$\varphi(H)=\#\{1\leq p<H:\gcd(p,H)=1\}$ is Euler's totient function.
Indeed, these classes correspond to primitive pairs $p,q>0$ with
$p+q=H$. Hence
\begin{equation}
\label{eq:farey-height-count}
\#\{(u,v)\in\mathscr F:h(u+v)=H\}
=3\varphi(H)\leq3(H-1).
\end{equation}

Third, for an initial trace triple $(x,y,z)$, write $\A_K(x,y,z)$ for
the polygon area $\A_K(X)$ determined by that initial Farey triangle.
Permuting its three vertices only relabels the three sectors and preserves
sector height and lattice area. Hence, for every permutation $\sigma$ of the three coordinates,
\begin{equation}
\label{eq:farey-symmetry}
\A_K(\sigma(x,y,z))=\A_K(x,y,z).
\end{equation}
As for $\A$, we regard $\A_K$ as a function of the local coordinates
$(x,y)$ by writing
$$
\A_K(x,y):=\A_K(X_{x,y})
=\A_K(x,y,z_-(x,y)).
$$
In particular, $\A_K(3,3)=\A_K(\Xmod)$.

We next express the added area in terms of the two parent lengths.
For an outward insertion, the Markoff equation gives
$$
t^2-x_ux_vt+x_u^2+x_v^2=0
$$
for the two traces across the Farey edge. Whenever the inserted trace
is the larger root, substituting $x_w=2\cosh(L_w/2)$ gives
\begin{equation}
\label{eq:child-length}
\cosh\frac{L_{u+v}}{2}
=
\cosh\frac{L_u}{2}\cosh\frac{L_v}{2}
+
\sqrt{\sinh^2\frac{L_u}{2}\sinh^2\frac{L_v}{2}-1}.
\end{equation}
Lemma~\ref{lem:local-bounds} ensures this root choice for every
outward insertion in the marked family on $\Rloc$.

The formula also defines $L_{u+v}$ as a function of two independent
lengths on
\begin{equation}
\label{eq:sector-length-domain}
\mathcal E=\{(L_u,L_v)\in(0,\infty)^2:
\sinh(L_u/2)\sinh(L_v/2)\geq1\},
\end{equation}
the domain where the square root is real
(Figure~\ref{fig:sector-length-domain}). On this domain, define the
\emph{length deficit}
\begin{equation}
\label{eq:length-deficit}
\delta(L_u,L_v)=L_u+L_v-L_{u+v}
\end{equation}
and the added-area function
$$
\mathcal D(L_u,L_v)=\frac{\delta(L_u,L_v)}{L_uL_vL_{u+v}}.
$$
For an insertion using the larger root, \eqref{eq:area-gain} reads
$S_X(u,v)=\mathcal D(L_u,L_v)$. This formulation will also be used
outside $\Rloc$, after checking the root choice.

\begin{figure}[ht]
\centering
\includegraphics{figures/diagram-07.pdf}
\caption{The region $\mathcal E$ and its boundary
$L_v=2\log\coth(L_u/4)$.}
\label{fig:sector-length-domain}
\end{figure}

\subsection{Differential tools}
\label{sec:differential-tools}
We recall the gradient and Hessian and explain how cyclic symmetry
constrains them at the modular point.
\paragraph{Gradient and Hessian.}
For a twice continuously differentiable function $G(x,y)$, recall that
$$
\nabla G=
\begin{pmatrix}
G_x\\
G_y
\end{pmatrix},
\qquad
\Hess G=
\begin{pmatrix}
G_{xx}&G_{xy}\\
G_{xy}&G_{yy}
\end{pmatrix}.
$$
For a direction $v\in\R^2$, the directional derivative is
$$
D_vG=\nabla G\cdot v.
$$
Thus $\nabla G=0$ means that $G$ has zero first-order slope in every
direction. The Hessian describes the quadratic variation:
$$
G(q+v)
=
G(q)+\nabla G(q)\cdot v
+\frac12v^{\mathsf T}\Hess G(q)v
+o(|v|^2).
$$
Let
$$
\Sigma
=
\{(x,y,z)\in\R^3:x^2+y^2+z^2=xyz\}
$$
be the Markoff surface, and let
$
p=(3,3,3)
$
be the modular point (see Figure~\ref{fig:markoff-surface-tangent-plane}).
The following lemma is an elementary consequence of cyclic symmetry (see \cite{Palais1979}).

\begin{lemma}[Order-three symmetry at the modular torus]
\label{lem:symmetry-at-modular}
Suppose $\widetilde F$ is twice continuously differentiable on
$\Sigma$ near $p$ and invariant under cyclic permutation of the
coordinates. On the branch $z=z_-(x,y)$, define
$$
F(x,y)=\widetilde F(x,y,z_-(x,y)).
$$
Then
$$
\nabla F(3,3)=0,
\qquad
\Hess F(3,3)
=
\kappa
\begin{pmatrix}
2&1\\
1&2
\end{pmatrix}
$$
for some real $\kappa$.
\end{lemma}

\begin{proof}
Since $\nabla(x^2+y^2+z^2-xyz)(p)=(-3,-3,-3)$,
the tangent plane is $T_p\Sigma=\{a+b+c=0\}$.
Translating it
to $p$ gives the affine tangent plane
$
p+T_p\Sigma
$
as shown in Figure~\ref{fig:markoff-surface-tangent-plane}.


\begin{figure}[ht]
\centering

\includegraphics{figures/diagram-08.pdf}

\caption{
The Markoff surface near $p=(3,3,3)$ and its affine tangent
plane.
}
\label{fig:markoff-surface-tangent-plane}
\end{figure}

By \eqref{eq:cyclic-trace-permutation}, $\rho$ fixes $p=(3,3,3)$
and acts on $T_p\Sigma$ as a rotation through $120^\circ$.
Hence the three vectors $\xi,\rho\xi,\rho^2\xi$ sum to zero for any $\xi\in T_p\Sigma$.
Cyclic invariance makes the directional derivatives along these
vectors equal, so linearity gives
$$
3D_\xi\widetilde F(p)
=
D_{\xi+\rho\xi+\rho^2\xi}\widetilde F(p)
=
0.
$$
Thus every tangential derivative vanishes, and the chain rule yields
$\nabla F(3,3)=0$.

For the second derivatives, let $\gamma$ be a twice continuously
differentiable curve in $\Sigma$ with $\gamma(0)=p$ and
$\gamma'(0)=\xi=(a,b,-a-b)$. Since $\nabla F(3,3)=0$,
the acceleration terms vanish in the chain rule, giving
$$
\frac{d^2}{dt^2}\widetilde F(\gamma(t))\bigg|_{t=0}
=
F_{xx}(3,3)a^2+2F_{xy}(3,3)ab+F_{yy}(3,3)b^2.
$$
This depends only on $\xi$, so it defines a quadratic form
$Q$ on $T_p\Sigma$. Cyclic invariance gives $Q(\rho\xi)=Q(\xi)$. If $Q$ is not constant on the Euclidean unit circle in $T_p\Sigma$,
its maximum is attained at exactly two opposite points.
A $120^\circ$ rotation cannot preserve this pair, so cyclic
invariance forces $Q$ to be constant on the unit circle. By homogeneity,
$$
Q(\xi)=\kappa|\xi|^2
=\kappa\bigl(a^2+b^2+(a+b)^2\bigr).
$$
Comparing coefficients yields the formula for 
$
\Hess F(3,3)
$ as desired.
\end{proof}

\section{Estimates shared by the regional comparisons}
\label{sec:shared-estimates}

To apply \eqref{eq:area-comparison-split}, we bound individual
insertions, give the polygon's area formula, and estimate the modular
area and its omitted terms. We then compute the Taylor coefficients,
control the remainder, and obtain polygon bounds near the modular point.

\paragraph{Derivative notation.}
For functions of the independent lengths $L_u,L_v$, we use
$\partial_1=\partial/\partial L_u$ and
$\partial_2=\partial/\partial L_v$, and write
$f_i=\partial_i f$, $f_{ij}=\partial_i\partial_j f$
for $i,j\in\{1,2\}$. Derivatives in the trace coordinates are
indicated by $x,y$. In $\partial_x^k\partial_y^l$, the nonnegative
integers $k,l$ specify the respective derivative orders.
These conventions also apply in later sections. Indices on
sequences and coefficients are defined where they appear.


\subsection{Length and added-area estimates}
\label{sec:local}

We estimate the deficit $\delta$ and added area $\mathcal D$ defined
in Subsection~\ref{sec:farey-area}. First, their monotonicity and
convexity will control changes in the omitted area in region $B$ and
added triangles in region $C$. We then bound the deficit and its
first two length derivatives, and deduce a lower length bound in
terms of sector height. Related exponential estimates appear in
\cite{DLN}; here we need explicit constants and derivative bounds.
Derivatives with respect to $x,y$ are treated in
Subsection~\ref{sec:length-derivatives} for the comparison in region $B$.



\begin{proposition}[Properties of the deficit and added area]
\label{prop:defect-convexity}
The set $\mathcal E$ defined in \eqref{eq:sector-length-domain} is convex. On $\mathcal E$, the functions $\delta$
and $\mathcal D$ are positive, symmetric, strictly decreasing in each length,
and have convex logarithms. In particular,
$\Hess_{L_u,L_v}\mathcal D\geq0$ in the interior, and
\begin{equation}
\label{eq:sector-correction-lower}
\mathcal D(L_u,L_v)>
\frac{8e^{-L_u-L_v}}{L_uL_v(L_u+L_v)}.
\end{equation}
\end{proposition}

\begin{proof}
On $(0,\infty)^2$, the function
$\log\sinh(L_u/2)+\log\sinh(L_v/2)$ is concave. Its value is
nonnegative exactly on $\mathcal E$, so $\mathcal E$ is convex.
Work first in the interior. Set $q=e^{-L_u}$, $p=e^{-L_v}$, and put
$$
a=(1-q)(1-p),
b=2\sqrt{qp}\left(
\cosh\frac{L_u+L_v}{2}-\cosh\frac{L_{u+v}}2
\right).
$$
The addition formula and \eqref{eq:child-length} give
\begin{equation}
\label{eq:deficit-auxiliaries}
b=\frac{a-\sqrt{a^2-16qp}}2,
e^{-\delta/2}+qp e^{\delta/2}=1+qp-b.
\end{equation}
The definition of $b$ and monotonicity of $\cosh$ give $\delta>0$.
Write $\psi=1-e^{-\delta/2}$, so $0<\psi<1$ and
$\delta=-2\log(1-\psi)$. The two relations give
\begin{equation}
\label{eq:deficit-implicit-relations}
b=\frac{4qp+b^2}{a},
\psi=b+\frac{qp\psi}{1-\psi}.
\end{equation}
 The chosen roots give the positive factors
\begin{equation}
\label{eq:deficit-branch-factors}
a-2b=\sqrt{a^2-16qp}>0,
1-\frac{qp}{(1-\psi)^2}=1-e^{-L_{u+v}}>0.
\end{equation}
Since $a_i>0$ and $(qp)_i=-qp$, implicit differentiation gives,
for $i\in\{1,2\}$,
\begin{equation}
\label{eq:deficit-first-length-derivatives}
\begin{aligned}
(a-2b)b_i&=-4qp-a_i b<0,\\
\left(1-\frac{qp}{(1-\psi)^2}\right)\psi_i
&=b_i-\frac{qp\psi}{1-\psi}<0.
\end{aligned}
\end{equation}
Thus $b$, $\psi$, and $\delta=-2\log(1-\psi)$ strictly decrease
in each length.

For convexity of the logarithms, differentiate along any straight
line in the length variables. Primes denote derivatives along this line.
For positive functions $f,g$, direct differentiation gives
\begin{equation}
\label{eq:logarithmic-sum-derivative}
(\log(f+g))''
=\frac{f(\log f)''+g(\log g)''}{f+g}
+\frac{fg}{(f+g)^2}\bigl((\log f)'-(\log g)'\bigr)^2.
\end{equation}
Here $\log(qp)=-L_u-L_v$ is affine, and $-\log a$ is convex because
$(-\log(1-e^{-L}))''=e^{-L}/(1-e^{-L})^2>0$.
Applying \eqref{eq:logarithmic-sum-derivative} to the first equation
in \eqref{eq:deficit-implicit-relations} gives
$$
\left(1-\frac{2b}{a}\right)(\log b)''
\geq(-\log a)''\geq0.
$$
For the equation defining $\psi$, use
$$
\left(\log\frac{\psi}{1-\psi}\right)''
=\frac{(\log\psi)''}{1-\psi}
+\frac{\psi((\log\psi)')^2}{(1-\psi)^2}.
$$
The same identity, after collecting $(\log\psi)''$ on the left, gives
$$
\left(1-\frac{qp}{(1-\psi)^2}\right)(\log\psi)''
\geq\frac b\psi(\log b)''\geq0.
$$
The positive coefficients in \eqref{eq:deficit-branch-factors}
show that $\log b$ and $\log\psi$ are convex.
Since $\delta=-2\log(1-\psi)>2\psi$, we also have
\begin{equation}
\label{eq:deficit-log-convexity}
(\log\delta)''
=\frac{2\psi}{(1-\psi)\delta}
\left((\log\psi)''+
\frac{\delta-2\psi}{(1-\psi)\delta}((\log\psi)')^2\right)\geq0.
\end{equation}
Since $\delta$ is convex and decreasing,
$L_{u+v}=L_u+L_v-\delta$ is positive, concave, and increasing.
Consequently every term in
\begin{equation}
\label{eq:added-area-log-convexity}
\log\mathcal D
=\log\delta-\log L_u-\log L_v-\log L_{u+v}
\end{equation}
is convex. Differentiating $\mathcal D=e^{\log\mathcal D}$ twice in any
direction $\xi\in\mathbb R^2$ gives
$$
\xi^{\mathsf T}(\Hess\mathcal D)\xi
=\mathcal D\left(
\xi^{\mathsf T}(\Hess\log\mathcal D)\xi
+(\xi^{\mathsf T}\nabla\log\mathcal D)^2
\right)\geq0.
$$
Here all derivatives are in $L_u,L_v$; the first term in parentheses
is nonnegative by convexity of $\log\mathcal D$, and the second is
a square. Hence $\Hess_{L_u,L_v}\mathcal D\geq0$. The numerator of $\mathcal D$ strictly decreases
and its denominator increases in each length, so $\mathcal D$
strictly decreases as well.
Symmetry is immediate. Continuity extends convexity and
monotonicity to the boundary; strict decrease there follows
by comparing through an interior point.

Finally, $b(a-b)=4qp$ and $0<a-b<1$ give $b>4qp$.
The equation for $\psi$ gives $\psi>b$, hence
$\delta>2\psi>2b>8qp$, also on the boundary.
Together with $L_{u+v}<L_u+L_v$, this proves
\eqref{eq:sector-correction-lower}.
\end{proof}

\begin{proposition}[Bounds for the length deficit]
\label{prop:scalar-defect}
Suppose $L_u,L_v\geq L_0=236/125$ and $\max\{L_u,L_v\}\geq84/25$.
Then
$$
0<\delta<11e^{-L_u-L_v},
$$
$$
0<-\delta_1,-\delta_2<20e^{-L_u-L_v},
$$
$$
0<\delta_{11},\delta_{12},\delta_{22}<40e^{-L_u-L_v}.
$$
\end{proposition}
\begin{proof}
We bound the denominators in the implicit equations, then
use their first two derivatives. The degree-six Taylor polynomials
for the exponential give
\begin{equation}
\label{eq:basic-exponential-bounds}
e^{-L_0}<\frac{19}{125}, e^{-42/25}<\frac3{16}.
\end{equation}
By symmetry, assume $L_v\geq84/25$. Then
$q<19/125$ and $p<(3/16)^2<1/25$. Throughout this rectangle,
$a=(1-q)(1-p)>2544/3125$ and $qp<1/160$, so
$$
a^2-16qp>\frac9{16}, a-2b>\frac34.
$$
The formula $b=8qp/(a+\sqrt{a^2-16qp})$, together with
$q<19/125$ and $p<1/25$, gives $b<1/32$ and $b<(21/4)qp$.
The selected root of $\psi^2-(1-qp+b)\psi+b=0$ is the smaller
one by \eqref{eq:deficit-branch-factors}. This polynomial is
positive at zero and negative at $1/31$, since
$$
\frac1{31^2}-\frac{159}{160\cdot31}
+\frac{30}{32\cdot31}<0.
$$
Its smaller root thus satisfies
\begin{equation}
\label{eq:deficit-corner-denominators}
0<\psi<\frac1{31},
1-\frac{qp}{(1-\psi)^2}>\frac{99}{100}.
\end{equation}
The equation $\psi(1-qp/(1-\psi))=b$ now gives
$\psi<(100/99)b$. Consequently
$$
\delta<\frac{2\psi}{1-\psi}
<\frac{2\cdot31}{30}\frac{100}{99}\frac{21}{4}qp
<11qp.
$$

For $i,j\in\{1,2\}$, we have $0<a_i<1/6$ and $|a_{ij}|<1/6$.
Equation~\eqref{eq:deficit-first-length-derivatives} gives $|b_i|<7qp$.
Differentiating once more gives
$$
(a-2b)b_{ij}=2b_i b_j-a_i b_j-a_j b_i-a_{ij}b+4qp,
$$
so $0<b_{ij}<11qp$. For positivity, use $a_{ii}<0$ and
$a_{12}=qp$ with $b<1$; for the upper bound, substitute
$|b_i|<7qp$, $b<(21/4)qp$, and $qp<1/160$.
Similarly, the equation for $\psi$ gives
\begin{equation}
\label{eq:deficit-corner-derivatives}
\left(1-\frac{qp}{(1-\psi)^2}\right)\psi_{ij}
=b_{ij}+\frac{qp\psi}{1-\psi}
-\frac{qp(\psi_i+\psi_j)}{(1-\psi)^2}
+\frac{2qp\psi_i\psi_j}{(1-\psi)^3}.
\end{equation}
Using \eqref{eq:deficit-first-length-derivatives} first, and then
this formula, gives $|\psi_i|<8qp$ and $0<\psi_{ij}<12qp$.
Finally,
$$
\delta_i=\frac{2\psi_i}{1-\psi},
\delta_{ij}=\frac{2\psi_{ij}}{1-\psi}
+\frac{2\psi_i\psi_j}{(1-\psi)^2}
$$
give the stated bounds and signs.
\end{proof}

We now deduce a lower length bound in terms of sector height, using
the Farey induction of \cite[Lemma~2.1.1(2)]{Bowditch}.
Since both parent lengths are positive, \eqref{eq:child-length} gives
$L_{u+v}>\max(L_u,L_v)$ and hence $x_{u+v}>\max(x_u,x_v)$.
Thus the first-insertion bound in Lemma~\ref{lem:local-bounds}
persists under further insertions: throughout $\Rloc$, every
primitive class $w$ of sector height at least two satisfies
\begin{equation}
\label{eq:inserted-trace-bound}
x_w>\frac{14279}{2500}.
\end{equation}

\begin{lemma}[A length lower bound in terms of sector height]
\label{lem:height-lower}
On $\Rloc$, every primitive class in an initial sector satisfies
$$
L_w>\frac{42}{25}h(w)+\frac3{50}.
$$
\end{lemma}
\begin{proof}
For height one, $L_w>L_0>42/25+3/50$.
For every height at least two, \eqref{eq:inserted-trace-bound} and
$2\cosh(171/100)<14279/2500$ give $L_w>171/50$.
The latter comparison follows from \eqref{eq:cosh-remainder}
with $N=6$, and proves the claimed bound at height two.

In every later insertion, one parent has height at least two,
so its length exceeds $171/50>84/25$; the other length exceeds $L_0$.
Proposition~\ref{prop:scalar-defect} and
\eqref{eq:basic-exponential-bounds} therefore give
\begin{equation}
\label{eq:insertion-deficit-bound}
0<\delta<11e^{-L_u-L_v}
<11\frac{19}{125}\left(\frac3{16}\right)^2
<\frac3{50}.
\end{equation}
Thus the constant $3/50$ compensates for the loss at each insertion:
if the bound holds for both parents, then
$$
L_{u+v}=L_u+L_v-\delta
>\frac{42}{25}(h(u)+h(v))+\frac3{50}.
$$
Induction proves the result.
\end{proof}

\subsection{The finite polygon and its area formula}
\label{sec:finite}

We give an explicit area formula for the polygon $P_6$ used in
regions $A$ and $B$.
A trace $T>2$ determines the geodesic length
$$
L(T)=2\log\frac{T+\sqrt{T^2-4}}2.
$$
Write $f(T)=1/L(T)$. By Subsection~\ref{sec:farey-area}, each
consecutive vertex pair and its antipodal pair contribute $f(T)f(T')$.
Set
$$
T_n=x_{(n+1)\alpha+\beta}.
$$
These are the traces obtained by successive Dehn twists of $\beta$
about $\alpha$; we call this sequence a \emph{fan}.
Since $x=x_\alpha$, $y=x_\beta$, and $z=x_{\alpha+\beta}$,
the Vieta relation gives
\begin{equation}
\label{eq:sector-trace-recursion}
T_{-1}=y,
T_0=z,
T_{n+1}=xT_n-T_{n-1}
\text{ for }n\in\Z.
\end{equation}
For $P_6$, take $T_1,\ldots,T_5$, of sector heights $2,\ldots,6$,
and insert $5\alpha+2\beta$ between $2\alpha+\beta$ and
$3\alpha+\beta$. Its trace is $T_*=T_1T_2-x$, and its sector
height is five since
$$
5\alpha+2\beta=3\alpha+2(\alpha+\beta).
$$
The boundary order is
$$
x,T_5,T_4,T_3,T_2,T_*,T_1.
$$
The six permutations of $(x,y,z)$ give the two sides of the fans
about the three initial curves, together with these six additional
insertions (Figure~\ref{fig:p6-six-permutations}). Thus
\begin{equation}
\label{eq:height-six-polygon-area}
\A_6(X)
=\sum_{\sigma\in\mathfrak S_3}
\left(
f(x)f(T_5)+f(T_*)\bigl(f(T_1)+f(T_2)\bigr)
+\sum_{n=2}^{4}f(T_n)f(T_{n+1})
\right)\circ\sigma.
\end{equation}
At the modular point,
$$
(x,T_1,T_2,T_3,T_4,T_5,T_*)=(3,6,15,39,102,267,87).
$$
\begin{figure}[ht]
\centering
\begingroup

\begin{minipage}[b]{0.57\textwidth}
\centering
\includegraphics{figures/diagram-09.pdf}

\par\small (a) The thirty-six pairs in a half-plane
\end{minipage}
\hfill
\begin{minipage}[b]{0.40\textwidth}
\centering
\includegraphics{figures/diagram-10.pdf}

\par\small (b) The six pairs for $(x,y,z)$
\end{minipage}
\endgroup

\caption{The upper half of $P_6(\Xmod)$ in the basis
$(\alpha,\beta)$ of Figure~\ref{fig:farey-p6}.
Each numbered group contains six consecutive pairs, with
permutations $(x,y,z)$, $(z,y,x)$, $(z,x,y)$, $(y,x,z)$,
$(y,z,x)$, and $(x,z,y)$ for groups $1$--$6$, respectively.}
\label{fig:p6-six-permutations}
\end{figure}

\subsection{The modular area and its omitted terms}
\label{sec:modular-area}

The estimates below are inspired by the exponential normal-turn
bounds in \cite{DLN}: in both settings, the Farey recurrence
yields exponentially small contributions at large heights.
Here we bound added areas and compute the required constants separately. We bound the omitted area at the modular
torus by $1/700000$ for region $A$, then obtain the upper bound
for $\A(\Xmod)$ used in region $C$.

\begin{lemma}
\label{lem:small-modular-remainder}
One has $0<\A(\Xmod)-\A_6(\Xmod)<1/700000$.
\end{lemma}
\begin{proof}
We compute the insertions of heights seven and eight explicitly,
then bound the remaining contributions by a geometric series.

Let $u_0,v_0$ be the endpoints of an initial sector and write
$w=pu_0+qv_0$, so $h(w)=p+q$. The symmetry exchanging $u_0,v_0$
gives $\|pu_0+qv_0\|_{\Xmod}=\|qu_0+pv_0\|_{\Xmod}=L_w$.
The triangle inequality therefore yields
$$
(p+q)\|u_0+v_0\|_{\Xmod}
\leq \|pu_0+qv_0\|_{\Xmod}+\|qu_0+pv_0\|_{\Xmod}
=2L_w.
$$
Since $u_0+v_0$ has trace $6$, its length is
$\|u_0+v_0\|_{\Xmod}=2\operatorname{arccosh}3$. Hence
\begin{equation}
\label{eq:modular-height-bound}
L_w\geq h(w)\operatorname{arccosh}3>\frac74h(w),
\qquad
e^{-\operatorname{arccosh}3}=3-2\sqrt2<\frac{43}{250}.
\end{equation}
The Farey recurrence gives five insertion types at heights seven
and eight, each occurring six times:
\begin{equation}
\label{eq:modular-extra-insertions}
\begin{array}{c|c|c|c}
H&\text{parent traces}&\text{child trace}&10^9S_{\Xmod}(u,v)<\\
\hline
7&(3,267)&699&69\\
7&(15,39)&582&48\\
7&(6,87)&507&82\\
8&(3,699)&1830&8\\
8&(15,87)&1299&7
\end{array}
\end{equation}
Appendix~\ref{app:modular-insertion-checks} verifies the bounds of $S_{\Xmod}(u,v)$
in the table. For any parent pair $(u,v)$ with total height
$H=h(u)+h(v)\geq9$, Proposition~\ref{prop:scalar-defect}
and \eqref{eq:modular-height-bound} give
$$
S_{\Xmod}(u,v)=\frac{\delta(L_u,L_v)}{L_uL_vL_{u+v}}
<\frac{11(43/250)^H}{(7/4)^3h(u)h(v)H}\le \frac{11(43/250)^H}{(7/4)^3(H-1)H}.
$$
Since there are at most $3(H-1)$ insertions at height $H$ (see \eqref{eq:farey-height-count}), 
$$
\begin{aligned}
\A(\Xmod)-\A_8(\Xmod)
&<\frac{33}{(7/4)^3}
\sum_{H\geq9}\frac{(43/250)^H}{H}\leq\frac{33(43/250)^9}
{9(7/4)^3(1-43/250)}
<\frac{11}{10^8}.
\end{aligned}
$$
Adding the explicitly computed insertions in \eqref{eq:modular-extra-insertions} gives
$$
\A(\Xmod)-\A_6(\Xmod)
<\frac{6(69+48+82+8+7)}{10^9}+\frac{11}{10^8}
<\frac1{700000}.
$$
\end{proof}

\begin{proposition}[An elementary modular upper bound]
\label{prop:modular-upper-new}
One has $\A(\Xmod)<223/250$.
\end{proposition}
\begin{proof}
Evaluating \eqref{eq:height-six-polygon-area} at the modular torus
and bounding the logarithms by \eqref{eq:log-remainder} gives
$$
\A_6(\Xmod)<\frac{89183}{100000}.
$$
Lemma~\ref{lem:small-modular-remainder} therefore gives
$$
\A(\Xmod)<\frac{89183}{100000}+\frac1{700000}
<\frac{223}{250}.
$$
\end{proof}

\subsection{Taylor expansion of the polygon area}
\label{sec:taylor-expansion}

We now study how $\A_6$ changes near the modular point.
Lemma~\ref{lem:symmetry-at-modular} determines the form of its Hessian there.
To organize the higher-order Taylor terms
and estimate the remainder, we introduce centered logarithmic coordinates:
$$
(u,v,w):=(\log x,\log y,\log z)-\frac{\log(xyz)}3(1,1,1).
$$
Then $u+v+w=0$, and the modular triple $(3,3,3)$ corresponds to
$(u,v,w)=(0,0,0)$. These coordinates are linearly related to the shear coordinates
discussed in Bonahon's lecture \cite{Bonahon2018}. We retain $(u,v,w)$ because permutations of the traces simply
permute these coordinates. Consequently, each homogeneous Taylor
term of the polygon area $\A_6$ is symmetric in $u,v,w$, simplifying
the coefficient and remainder estimates below.
The Markoff equation gives
\begin{equation}
\label{eq:shape-trace-parametrization}
(x,y,z)=(e^{2u}+e^{2v}+e^{2w})(e^u,e^v,e^w).
\end{equation}
Conversely, this parametrization satisfies the Markoff equation
whenever $u+v+w=0$. The quantity $p_2$ introduced in \eqref{eq:p_2} becomes
$$
p_2=\frac13\bigl((u-v)^2+(v-w)^2+(w-u)^2\bigr)
=u^2+v^2+w^2.
$$
In these coordinates, we expand $\A_6$ at $u=v=w=0$.
Symmetry in $u,v,w$ makes each homogeneous Taylor term a symmetric
polynomial, hence by the fundamental theorem of symmetric polynomials
\cite[p.~21, Remark following (2.4)]{Macdonald},
each homogeneous Taylor term is a polynomial in
$u+v+w$, $uv+vw+wu$, and $uvw$.
Put $p_3=u^3+v^3+w^3$. Since $u+v+w=0$, we have
$uv+vw+wu=-p_2/2$ and $uvw=p_3/3$.
Thus, the Taylor expansion through degree eight has the form
\begin{equation}
\label{eq:degree-eight-expansion}
\begin{aligned}
\A_6(X)-\A_6(\Xmod)
={}&c_2p_2+c_3p_3+c_4p_2^2+c_5p_2p_3
+c_{6'}p_2^3+c_{6''}p_3^2\\
&+c_7p_2^2p_3+c_{8'}p_2^4+c_{8''}p_2p_3^2+R_9,
\end{aligned}
\end{equation}
where $R_9$ contains the terms of degree at least nine.
We determine the coefficients by restricting the area formula
to two lines through the modular point. On $(u,v,w)=(t,0,-t)$, we have
$p_2=2t^2$ and $p_3=0$, so the Taylor polynomial through degree eight of
\eqref{eq:degree-eight-expansion} becomes
$$
2c_2t^2+4c_4t^4+8c_{6'}t^6+16c_{8'}t^8.
$$
Comparing coefficients on the first line determines
$c_2,c_4,c_{6'},c_{8'}$.
On $(u,v,w)=(t,t,-2t)$, we have $p_2=6t^2$ and $p_3=-6t^3$,
so the Taylor polynomial through degree eight becomes
$$
\begin{aligned}
&6c_2t^2-6c_3t^3+36c_4t^4-36c_5t^5+(216c_{6'}+36c_{6''})t^6-216c_7t^7
+(1296c_{8'}+216c_{8''})t^8.
\end{aligned}
$$
The coefficients of $t^3,t^5,t^7$ determine $c_3,c_5,c_7$.
Subtracting the contributions of $c_{6'}$ and $c_{8'}$,
determined on the first line, from the coefficients of $t^6,t^8$
then determines $c_{6''},c_{8''}$.

For each substitution, we expand the traces using
\eqref{eq:sector-trace-recursion}, then expand their reciprocal
lengths and sum the products in \eqref{eq:height-six-polygon-area}.
Lemma~\ref{lem:explicit-quadratic-coefficient} in
Appendix~\ref{app:taylor-coefficients} evaluates the quadratic term
as a sum of six explicitly bounded contributions and proves
\begin{equation}
\label{eq:quadratic-area-coefficient}
c_2>\frac{23}{200}.
\end{equation}
The remaining coefficient bounds are recorded in
Lemma~\ref{lem:explicit-low-coefficients}.
The coefficient $c_2$ identifies the scalar factor in the Hessian
formula from Lemma~\ref{lem:symmetry-at-modular} and gives the
following lower bound in the trace coordinates $(x,y)$.

\begin{proposition}
\label{prop:p6}
At the modular point,
$$
\Hess\A_6(3,3)
=\frac{2c_2}{9}
\begin{pmatrix}2&1\\1&2\end{pmatrix}
>\frac{23}{900}I,
$$
where $I$ is the $2\times2$ identity matrix. The inequality means
that the second derivative in every unit direction is greater than
$23/900$.
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:symmetry-at-modular},
$$
\nabla\A_6(3,3)=0,
\qquad
\Hess\A_6(3,3)=\kappa
\begin{pmatrix}2&1\\1&2\end{pmatrix}.
$$
Let $(x(t),y(t),z(t))$ be the trace curve corresponding to
$(u,v,w)=(t,0,-t)$ under \eqref{eq:shape-trace-parametrization}.
Then $(x'(0),y'(0))=(3,0)$, and $\A_6(x(t),y(t))$ has quadratic
Taylor coefficient $2c_2$. Since the gradient vanishes, the chain rule gives
$$
4c_2
=\left.\frac{d^2}{dt^2}\A_6(x(t),y(t))\right|_{t=0}
=(3,0)\Hess\A_6(3,3)\begin{pmatrix}3\\0\end{pmatrix}
=18\kappa.
$$
Thus $\kappa=2c_2/9$. For every nonzero vector $(a,b)$,
\eqref{eq:quadratic-area-coefficient} gives
$$
\begin{aligned}
(a,b)\Hess\A_6(3,3)\begin{pmatrix}a\\b\end{pmatrix}
&=\frac{2c_2}{9}\bigl(a^2+b^2+(a+b)^2\bigr)>\frac{23}{900}(a^2+b^2).
\end{aligned}
$$
This is precisely the stated matrix inequality.
\end{proof}

\subsection{Complex extension and Taylor remainder bounds}
\label{sec:larger-neighborhood}

To control the uncomputed Taylor terms of $\A_6$, we extend
\eqref{eq:height-six-polygon-area} to complex values of $t$
along each line through the modular point in logarithmic coordinates.
We first bound the lengths away from zero, then bound the average
of $|\A_6|^2$ on the circle $|t|=3/8$.
Proposition~\ref{prop:polygon-taylor-remainder} turns this average
bound into a Taylor remainder estimate.
We apply it with $N=9$ in Proposition~\ref{prop:uniform-p6},
and with $N=20$, together with explicit bounds for degrees nine
through nineteen, in Theorem~\ref{thm:global-taylor-remainder}
for region $A$.

\begin{lemma}[Two elementary analytic estimates]
\label{lem:elementary-complex}
One has
\begin{equation}
\label{eq:complex-log-cosh-bound}
\operatorname{Re}\log\cosh z\geq
\frac{\operatorname{Re}(z^2)}2-\frac{|z|^4}{10}
 \text{ for } |z|\leq3/4,
\end{equation}
where the logarithm is zero at the origin. Also
\begin{equation}
\label{eq:linear-exponential-bound}
\operatorname{Re}e^z >\frac56-\frac{|z|}{2}
\text{ for }|z|\leq9/8.
\end{equation}
\end{lemma}
\begin{proof}
Write $z=x+iy$. For the first estimate, the identity
$
\cosh z=\cosh x\cos y+i\sinh x\sin y
$
gives
$$
|\cosh z|^2
=\cosh^2x\cos^2y+\sinh^2x\sin^2y
\geq\cosh^2x\cos^2y.
$$
Since $|y|\le |z| \le 3/4<\pi/2$, we have $\cos y>0$.
Taking square roots and logarithms therefore yields
\begin{equation}\label{eq:RelogCosh}
\operatorname{Re}\log\cosh z=\log|\cosh z|
\geq\log\cosh x+\log\cos y.  
\end{equation}
It thus suffices to bound these two real functions. For $s\geq0$, $0\leq\tanh s\leq s$ gives
$(\tanh s)'=1-\tanh^2s\geq1-s^2$.
Integrating twice from zero and using evenness gives, for every
$t\in\mathbb R$,
\begin{equation}\label{eq:logCosh}
\log\cosh t\geq\frac{t^2}{2}-\frac{t^4}{12} \geq\frac{t^2}{2}-\frac{t^4}{10}.
\end{equation}
Direct differentiation shows that
$(-\log\cos t-t^2/2)/t^4$ is increasing on $(0,\pi/2)$.
At $t=3/4$, this quotient is less than $1/10$.
Monotonicity and evenness therefore give
\begin{equation}\label{eq:logCos}\log\cos t\geq-t^2/2-t^4/10, \text{ for }|t|\leq3/4.
\end{equation}
Combining \eqref{eq:RelogCosh}, \eqref{eq:logCosh}, and
\eqref{eq:logCos} gives, for $|z|\leq3/4$,
$$
\operatorname{Re}\log\cosh z
\geq\frac{x^2-y^2}{2}-\frac{x^4+y^4}{10}
\geq\frac{\operatorname{Re}(z^2)}2-\frac{|z|^4}{10},
$$
which proves \eqref{eq:complex-log-cosh-bound}.

For the second estimate, apply $e^u\geq1+u$ with $u=x+\log 2$ gives
$$
e^x\geq\frac{1+x+\log2}{2}> \frac{x+5/3}{2}.
$$
Also,
$$
(|z|-x)(x+5/3)
\leq\frac{((|z|-x)+(x+5/3))^2}{4}<2.
$$
Multiplying by $|z|+x\geq0$ and using
$y^2=(|z|-x)(|z|+x)$ gives
$y^2(x+5/3)\le 2(|z|+x)$.
Thus, since $\cos y\geq1-y^2/2>0$,
$$
\begin{aligned}
\operatorname{Re}e^z
&=e^x\cos y>\frac{x+5/3}{2}-\frac{y^2(x+5/3)}4 \ge \frac{x+5/3}{2}-\frac{|z|+x}{2}
=\frac56-\frac{|z|}{2}.
\end{aligned}
$$
\end{proof}

\begin{proposition}[Complex extension of the polygon area]
\label{prop:complex-three-eighths}
Fix $a,b,c\in\mathbb R$ satisfy
$a+b+c=0$ and $\max\{|a|,|b|,|c|\}=1$.
For $(u,v,w)=t(a,b,c)$, define $(x,y,z)$ by
\eqref{eq:shape-trace-parametrization}. Then
$F(t)=\A_6(x(t),y(t),z(t))$, initially defined for real $t$ near zero,
extends analytically to an open disk containing $|t|\leq3/8$.
On $|t|\leq3/8$, the lengths in the polygon formula satisfy
$\operatorname{Re}L_w>l_{h(w)}$, with the bounds in
\eqref{eq:complex-length-lower-bounds}.
\end{proposition}

\begin{proof}
The area formula \eqref{eq:height-six-polygon-area} is a finite sum
of terms $1/(L_uL_v)$, so we need lower bounds for $|L_u|$ and $|L_v|$.
Throughout, assume $|t|\leq3/8$ and put
$$
p=|\operatorname{Re}t|,
\qquad
q=|\operatorname{Im}t|,
\qquad
p^2+q^2\leq9/64.
$$

Step 1: We first show $\operatorname{Re}x,\operatorname{Re}y,
\operatorname{Re}z>17/8$.
By \eqref{eq:shape-trace-parametrization} and $a+b+c=0$,
\begin{equation}
\label{eq:initial-trace-cosh-form}
x=e^{3at}+e^{(b-c)t}+e^{-(b-c)t}
=e^{3at}+2\cosh((b-c)t).
\end{equation}
Since $|b-c|+|a|=2\max\{|b|,|c|\}\leq2$,
we have $|(b-c)t|\leq3(2-|a|)/8\leq3/4$.
Using $\cosh s\geq1$ and $\cos s\geq1-s^2/2$ for real $s$ gives
\begin{equation}
\label{eq:initial-trace-cosh-bound}
\begin{aligned}
\operatorname{Re}\cosh((b-c)t)
&=\cosh((b-c)\operatorname{Re}t)
\cos((b-c)\operatorname{Im}t)\geq1-\frac9{128}(2-|a|)^2.
\end{aligned}
\end{equation}
Since $|3at|\leq9/8$, combining \eqref{eq:initial-trace-cosh-form},
\eqref{eq:initial-trace-cosh-bound}, and
\eqref{eq:linear-exponential-bound} gives
$$
\begin{aligned}
\operatorname{Re}x
&\geq\frac56-\frac{9|a|}{16}
+2-\frac9{64}(2-|a|)^2
=\frac{109}{48}-\frac9{64}|a|^2>\frac{17}{8}.
\end{aligned}
$$
The same argument applies to $y$ and $z$.

Step 2: We next show that each first child has larger modulus than both
parents. For the child $xy-z$, we use
\begin{equation}
\label{eq:first-child-quotient}
\frac{xy-z}{x}=2e^{(b-c)t}\cosh((b-a)t).
\end{equation}
It suffices to show that $\log|(xy-z)/x|>0$.
Put $d=|b-a|\leq2$.
Since $2(b-c)=(b-a)-3c$ and $|c|\leq1$,
we have $|b-c|\leq(3+d)/2$.
Combining this with $|b-c|\leq2$, using $0\leq7p/4\leq1$, gives
\begin{equation}
\label{eq:Ine}
\begin{aligned}
|b-c|
&\leq\left(1-\frac{7p}{4}\right)2
+\frac{7p}{4}\frac{3+d}{2}=2+\frac{7p}{8}(d-1)
\leq2+\frac{7p}{16}(d^2-1).
\end{aligned}
\end{equation}
Taking logarithms of the modulus in \eqref{eq:first-child-quotient}
and applying \eqref{eq:complex-log-cosh-bound} gives
$$
\log\left|\frac{xy-z}{x}\right|
\geq\log2-|b-c|p+d^2(p^2-2/25),
$$
where we used $q^2\leq9/64-p^2$, and $d^4\leq4d^2$.
Now \eqref{eq:Ine}, $d^2\leq4$, and $9p^2/16<2/25$ yield
\begin{equation}
\label{eq:complex-child-quotient-bound}
\begin{aligned}
\log\left|\frac{xy-z}{x}\right|
&\geq\log2-2p+\frac{7p^2}{16}
+d^2\left(\frac{9p^2}{16}-\frac2{25}\right)\geq\log2-\frac8{25}-2p+\frac{43p^2}{16}>0.
\end{aligned}
\end{equation}
Hence $|xy-z|>|x|$. Interchanging $x$ and $y$ gives
$|xy-z|>|y|$.

Step 3: For the later use, we will show that $|xy-z|>14/5$.
Since $|a|,|b|,|c|\leq1$,
$$
\begin{aligned}
\left|e^{2at}+e^{2bt}+e^{2ct}\right|
&\geq\cos(2q)
\left(e^{2a\operatorname{Re}t}
+e^{2b\operatorname{Re}t}
+e^{2c\operatorname{Re}t}\right)\geq3\cos(2q).
\end{aligned}
$$
Combining this with \eqref{eq:shape-trace-parametrization},
\eqref{eq:first-child-quotient}, $|a-b|\leq2-|c|$, $\operatorname{Re}t \le p$, and $\operatorname{Im}t \le q$ gives
\begin{equation}
\label{eq:complex-first-child-bound}
\begin{aligned}
|xy-z|
\geq6\cos(2q)e^{-2c\operatorname{Re}t}
|\cosh((a-b)t)|\geq6\cos(2q)e^{-2|c|p}\cos((2-|c|)q).
\end{aligned}
\end{equation}
By \eqref{eq:logCos}, $\log\cos s\geq-3s^2/5$ for $|s|\leq3/4$.
Since $(2-|c|)^2\leq4-3|c|$, it follows that
$$
\log\frac{|xy-z|}{6}
\geq-(1-|c|)\frac{24q^2}{5}
-|c|(2p+3q^2)>-\frac{19}{25},
$$
because $p^2+q^2\leq9/64$ gives
$24q^2/5\leq27/40<19/25$ and
$2p+3q^2\leq145/192-3(p-1/3)^2<19/25$.
Therefore
$$
|xy-z|>6e^{-19/25}>\frac{14}{5}.
$$
Permuting the coordinates gives the same bound for every first child.

Step 4: By Steps 1--3, the initial traces have modulus $>17/8$,
and each first child exceeds both parents and has modulus $>14/5$.
We propagate these comparisons along each outward branch.
At a later insertion $UV-W$, $W$ is a parent of the newer
trace among $U,V$. The comparison at the previous step therefore
gives $|W|<\max(|U|,|V|)$. Since $|U|,|V|>17/8$,
\begin{equation}
\label{eq:complex-trace-growth}
|UV-W|
\geq|U||V|-\max(|U|,|V|)
>\frac9{17}|U||V|
>\frac98\max(|U|,|V|).
\end{equation}
Thus the new trace again exceeds both parents, completing the induction. The height-one and height-two bounds from Steps 1 and 3,
together with the product estimate, now give
\begin{equation}
\label{eq:complex-height-trace-bounds}
|x_w|>
\begin{cases}
17/8,&h=1,\\
\displaystyle\frac{14}{5}\left(\frac98\right)^{h-2},&h\geq2,
\end{cases}
\end{equation}
for every primitive class $w$ in an initial sector, where $h=h(w)$.
Indeed, heights add. If one parent has height one, the product
estimate preserves the stated bound. If both parent heights
are at least two, substituting their stated bounds gives
an extra factor
$
\frac9{17}\frac{14}{5}\left(\frac89\right)^2
>1.
$
Any stronger bounds inherited from earlier insertions only
improve this estimate.
This follows the induction scheme of
\cite[Lemma~2.1.1(2)]{Bowditch}.

Step 5: We now convert the trace bounds into length bounds.
For the finitely many traces in $P_6$, Step 4 gives $|x_w|>2$
on a slightly larger open disk
containing $|t|\leq3/8$. Thus we can choose an analytic square
root of $x_w^2-4$, positive at $t=0$ \cite[Theorem~6.2, p.~100]{SteinShakarchi}, and define
$$
\lambda=\frac{x_w+\sqrt{x_w^2-4}}2.
$$
Here $\lambda(x_w-\lambda)=1$, so $\lambda$ never vanishes.
Moreover, $|\lambda|=1$ would give
$x_w=\lambda+\lambda^{-1}=\lambda+\overline{\lambda}=2\operatorname{Re}\lambda\in[-2,2]$, contradicting $|x_w|>2$.
At $t=0$, the modular trace satisfies $x_w(0)>2$, so $\lambda(0)>1$.
Since $|\lambda|\neq 1$, continuity gives $|\lambda|>1$
throughout the disk.
The logarithm theorem
\cite[Theorem~6.2, p.~100]{SteinShakarchi} gives an analytic length
$L_w=2\log\lambda$ that agrees with the real length at $t=0$.
Since $\operatorname{Re}L_w=2\log|\lambda|>0$ and
$$
|x_w|=|\lambda+\lambda^{-1}|\leq|\lambda|+|\lambda|^{-1}
=2\cosh\frac{\operatorname{Re}L_w}{2},
$$
\eqref{eq:complex-height-trace-bounds} gives
$\operatorname{Re}L_w\geq2\arccosh(|x_w|/2)>l_h$, where $h=h(w)$ and
\begin{equation}
\label{eq:complex-length-lower-bounds}
\begin{array}{c|rrrrrr}
h&1&2&3&4&5&6\\
\hline
l_h&7/10&17/10&2&23/10&13/5&14/5.
\end{array}
\end{equation}
Appendix~\ref{app:complex-length-checks} verifies these comparisons.
The nonvanishing lengths make the finite sum defining $F$
analytic on the larger disk.
\end{proof}

\begin{proposition}[Taylor remainder bound for the polygon area]
\label{prop:polygon-taylor-remainder}
For every function $F$ in
Proposition~\ref{prop:complex-three-eighths}, every integer $N\geq1$,
and $0<|t|<3/8$, one has
\begin{equation}
\label{eq:mean-square-taylor-tail}
\left|F(t)-\sum_{n=0}^{N-1}\frac{F^{(n)}(0)}{n!}t^n\right|
<\sqrt{21}\frac{(8|t|/3)^N}{\sqrt{1-(8|t|/3)^2}}.
\end{equation}
\end{proposition}
\begin{proof}
The idea is to control all remaining Taylor terms through the
circle average of $|F|^2$. Averaging cancels products of different
Taylor powers, leaving the sum of their squared sizes on the
circle (see Parseval's identity
\cite[Theorem~1.3(ii), p.~80]{SteinShakarchiFourier}).

We first bound this average using the known lengths at the
modular torus.
By Proposition~\ref{prop:complex-three-eighths}, each reciprocal
length is analytic on a neighborhood of $|t|\leq3/8$.
Averaging its real part around this circle therefore gives
its value at the center
(see \cite[Corollary~7.3, p.~103]{SteinShakarchi}),
\begin{equation}
\label{eq:Mean}
\frac1{2\pi}\int_0^{2\pi}
\operatorname{Re}\frac1{L_v((3/8)e^{i\theta})}d\theta
=\operatorname{Re}\frac1{L_v(0)} = \frac1{L_v(0)}.
\end{equation}
For each of the thirty-six terms $1/(L_uL_v)$ in
\eqref{eq:height-six-polygon-area}, order $u,v$ so that
$L_v(0)\geq L_u(0)$. The bounds
$|L_u|\geq\operatorname{Re}L_u>l_{h(u)}$ and
$\operatorname{Re}L_v>l_{h(v)}$ (see \eqref{eq:complex-length-lower-bounds}) imply, for every such pair,
$$
\frac{l_{h(u)}l_{h(v)}}{|L_uL_v|^2}
=
\frac{{l_{h(u)}^2}}{|L_u|^2}
\frac{l_{h(v)}}{\operatorname{Re}L_v}
\frac{\operatorname{Re}(1/L_v)}{l_{h(u)}}
<
\frac{\operatorname{Re}(1/L_v)}{l_{h(u)}}.
$$
Thus Cauchy--Schwarz gives
\begin{equation}\label{eq:BoundF2}
|F(t)|^2
\leq
\left(\sum\frac1{l_{h(u)}l_{h(v)}}\right)
\left(\sum\frac{l_{h(u)}l_{h(v)}}{|L_uL_v|^2}\right)<
\left(\sum\frac1{l_{h(u)}l_{h(v)}}\right)
\sum\frac{\operatorname{Re}(1/L_v)}{l_{h(u)}}=\frac{94260}{10829}
\sum\frac{\operatorname{Re}(1/L_v)}{l_{h(u)}},
\end{equation}
where each sum runs over the thirty-six polygon summands.

By \eqref{eq:Mean}, using
\eqref{eq:height-six-polygon-area} at the modular point and
the bounds in \eqref{eq:complex-length-lower-bounds},
the circle average of $\sum \operatorname{Re}(1/L_v)/l_{h(u)}$ is
$$
\sum\frac1{l_{h(u)}L_v(0)}
=6\left(
\frac{165}{91L(267)}+\frac{37}{34L(87)}
+\frac1{2L(39)}+\frac{10}{23L(102)}
\right)<\frac{12}{5}.
$$
Averaging the preceding bound \eqref{eq:BoundF2} for $|F|^2$ therefore yields
$$
\frac1{2\pi}\int_0^{2\pi}
\left|F\left(\frac38e^{i\theta}\right)\right|^2d\theta
<\frac{94260}{10829}\frac{12}{5}<21.
$$
Since $F$ is analytic beyond this circle, its Taylor series
converges uniformly there. Averaging its squared modulus
cancels the mixed terms (Parseval’s identity) and gives
$$
\sum_{n=0}^{\infty}
\left|\frac{F^{(n)}(0)}{n!}\right|^2
\left(\frac38\right)^{2n}<21.
$$
Cauchy--Schwarz and the geometric series now give
$$
\begin{aligned}
\left|F(t)-\sum_{n=0}^{N-1}\frac{F^{(n)}(0)}{n!}t^n\right|
&\leq\sum_{n=N}^{\infty}
\left|\frac{F^{(n)}(0)}{n!}\right||t|^n<\sqrt{21}
\left(\sum_{n=N}^{\infty}(8|t|/3)^{2n}\right)^{1/2}=\sqrt{21}
\frac{(8|t|/3)^N}{\sqrt{1-(8|t|/3)^2}}.
\end{aligned}
$$
\end{proof}

\subsection{Polygon area increase near the modular point}
\label{sec:analytic-local-region}
To prove that the full area increases near the modular point,
we need the polygon area $\A_6$ to increase by more than the
omitted area can decrease. Here we establish the polygon bound
$\A_6(X)-\A_6(\Xmod)>7p_2/80$ and relate the logarithmic
coordinates to the trace coordinates used in region $B$.
For a real triple $(u,v,w)$, put
$$
t=\max\{|u|,|v|,|w|\}.
$$
Here $t$ is real and nonnegative. For $0<t<3/8$, write
$(u,v,w)=t(a,b,c)$ with $\max\{|a|,|b|,|c|\}=1$.
Applying \eqref{eq:mean-square-taylor-tail} with $N=9$ to the
corresponding function in Proposition~\ref{prop:complex-three-eighths}
gives
\begin{equation}
\label{eq:cauchy-remainder-bound}
|R_9|<\sqrt{21}\frac{(8t/3)^9}{\sqrt{1-(8t/3)^2}}.
\end{equation}

\begin{proposition}[A local polygon bound]
\label{prop:uniform-p6}
In the logarithmic coordinates \eqref{eq:shape-trace-parametrization},
$$
\A_6(X)-\A_6(\Xmod)>\frac7{80}p_2
\text{ whenever }0<t\leq\frac1{20}.
$$
\end{proposition}
\begin{proof}
Assume $0<t\leq1/20$. Since $u+v+w=0$, using elementary inequalities, one can show that
\begin{equation}
\label{eq:zero-sum-bounds}
\begin{aligned}
\frac32t^2&\leq p_2\leq2t^2,\qquad
|p_3|=3|uvw|\leq\frac34t^3
\leq\frac{t}{2}p_2,
\end{aligned}
\end{equation}
where $p_2$ and $p_3$ are denoted before \eqref{eq:degree-eight-expansion}. Combining \eqref{eq:cauchy-remainder-bound} with
$p_2\geq3t^2/2$ and $t\leq1/20$, we obtain
\begin{equation}\label{eq:R9bound}
    \frac{|R_9|}{p_2}
<\frac{2\sqrt{21}}{3t^2}
\frac{(8t/3)^9}{\sqrt{1-(8t/3)^2}}
<\frac1{25000}.
\end{equation}
Using the coefficient bounds \eqref{eq:degree-eight-coefficients}
in Appendix~\ref{app:taylor-coefficients}, discard the nonnegative
terms $c_{6''}p_3^2$ and $c_{8'}p_2^4$ in
\eqref{eq:degree-eight-expansion} and apply
\eqref{eq:zero-sum-bounds} to obtain
\begin{equation}
\label{eq:local-polynomial-error}
\begin{aligned}
\A_6(X)-\A_6(\Xmod)
>p_2\biggl(c_2-\frac{347}{1000}t-\frac{81}{250}t^2
-\frac{163}{200}t^3-\frac{157}{125}t^4-\frac{277}{500}t^5
-\frac{15077}{1000}t^6\biggr)+R_9.
\end{aligned}
\end{equation}
Using $0<t\leq1/20$ with \eqref{eq:quadratic-area-coefficient} and \eqref{eq:R9bound}, this gives
\begin{equation*}
\A_6(X)-\A_6(\Xmod)
>p_2\left(\frac{23}{200}-\frac{1169288557}{64000000000}\right)-\frac1{25000}p_2
>\frac7{80}p_2.
\qedhere
\end{equation*}
\end{proof}

The trace bounds in Lemma~\ref{lem:local-bounds} allow us to apply
the following lemma on the domain $\Rloc$ defined in \eqref{eq:local-region}. It places $\Rloc$ within the range
of Proposition~\ref{prop:uniform-p6} and compares the two coordinate
distances.

\begin{lemma}
\label{lem:lambda-local-conversion}
Let $(x,y,z)$ be a positive Markoff triple with
$74/25<x,y,z<61/20$, and let $(u,v,w)$ be its coordinates
in \eqref{eq:shape-trace-parametrization}. Then
$
t=\max\{|u|,|v|,|w|\}<1/60.
$
If the triple is nonmodular, then
$$
(x-3)^2+(y-3)^2+(z-3)^2<10p_2.
$$
\end{lemma}

\begin{proof}
Since $u+v+w=0$, we have
$e^{2u}+e^{2v}+e^{2w}\geq3$. From the hypothesis, $|x-3|,|y-3|,|z-3|<1/20$. The Markoff equation and
\eqref{eq:shape-trace-parametrization} give
$x^2+y^2+z^2=(e^{2u}+e^{2v}+e^{2w})^3
$ and $$
x^2+y^2+z^2=27+6\bigl((x-3)^2+(y-3)^2+(z-3)^2\bigr)-3(x+y+z-9)^2-2(x-3)(y-3)(z-3),
$$
where the right-hand side is smaller than $27+6\times 3/20^2-0+2\times 1/20^3
<(1501/500)^3.$ Consequently, $$3\le e^{2u}+e^{2v}+e^{2w}<1501/500.$$
Combining with the hypothesis $74/25<x,y,z<61/20$, we have $60/61<(74/25)/(1501/500)<e^u=x/(e^{2u}+e^{2v}+e^{2w})<(61/20)/3=61/60$. The same holds for $e^v$, and $e^w$, so $$t=\max\{|u|,|v|,|w|\}<\log\frac{61}{60}<\frac1{60}.$$
For the distance comparison, the exponential series and $u+v+w=0$ give
$$
\begin{aligned}
|e^u-1|&\leq\frac{|u|}{1-t},\qquad
0\leq e^{2u}+e^{2v}+e^{2w}-3
\leq\frac{2p_2}{1-2t},
\end{aligned}
$$ for $0\leq t<1/2$. The first bound also holds for $v,w$. By \eqref{eq:shape-trace-parametrization},
$$
(x-3,y-3,z-3)
=3(e^u-1,e^v-1,e^w-1)
+(e^{2u}+e^{2v}+e^{2w}-3)(e^u,e^v,e^w).
$$
Applying the Euclidean triangle inequality to the two terms on
the right, using the preceding bounds and
$e^u,e^v,e^w\le e^t \le 1/(1-t)$ ($0 \leq t < 1$), gives
\begin{equation}
\label{eq:trace-shape-distance}
\begin{aligned}
\sqrt{(x-3)^2+(y-3)^2+(z-3)^2}
\leq\frac{3\sqrt{p_2}}{1-t}
+\frac{2\sqrt3 p_2}{(1-t)(1-2t)}\leq\frac{3\sqrt{p_2}}{(1-t)(1-2t)},
\end{aligned}
\end{equation}
where the last inequality uses $\sqrt{p_2}\leq\sqrt3 t$.
For $t<1/60$, the coefficient is less than
$\sqrt{10}$, proving the required bound when $p_2>0$.
\end{proof}

\section{Region A: comparison outside the local disk}
\label{sec:region-a}

Region $A$ consists of tori $X$ represented by reduced triples $(x,y,z)$ with
$\sys(X)=2\arccosh(x/2)\ge7/5$ and $(x-3)^2+(y-3)^2>1/1024$.
We will show that upper bounds for $|R_9|$ and lower bounds for the polynomial
part of \eqref{eq:degree-eight-expansion} yield $\A_6(X)-\A_6(\Xmod)>
19p_2/1000$ in
Corollary~\ref{cor:global-polygon-comparison}.
Combining this with $p_2>1/12800$ in \eqref{eq:outside-disk-shape-lower}, $\A(\Xmod)-\A_6(\Xmod)<1/700000$ in
Lemma~\ref{lem:small-modular-remainder}, and \eqref{eq:area-comparison-split} implies the desired comparison for all $X$ in region $A$: $$\A(X)-\A(\Xmod)>19p_2/1000-1/700000>p_2/1400.$$

\subsection{Bounding the Taylor remainder $R_9$ on a larger neighborhood}
\label{sec:global-remainder-completion}

The direct remainder estimate \eqref{eq:R9bound} suffices for
$t\leq1/20$, but is too large at the larger radii allowed in region $A$ by
\eqref{eq:reduced-invariant-radius}. To control $R_9$ throughout
region $A$, we estimate the terms of degrees nine through
nineteen explicitly and apply \eqref{eq:mean-square-taylor-tail}
with $N=20$ to the remaining tail. The full Taylor expansion extending \eqref{eq:degree-eight-expansion} is
\begin{equation}
\label{eq:homogeneous-area-expansion}
\A_6(u,v,w)=\A_6(\Xmod)+\sum_{n=2}^{\infty}H_n(u,v,w),
\end{equation}
where $H_n$ is the homogeneous Taylor term of degree $n$.
\begin{lemma}[Four directions determine the required coefficients]
\label{lem:four-directions}
For $9\leq n\leq19$, let $\epsilon$ be zero for even $n$ and one for
odd $n$, and put $m=(n-3\epsilon)/2$. There are unique real constants
$c_{n,k}$ such that
\begin{equation}
\label{eq:invariant-homogeneous-form}
H_n=p_3^\epsilon\sum_{k=0}^{\lfloor m/3\rfloor}
c_{n,k}p_2^{m-3k}p_3^{2k}.
\end{equation}
All these constants are determined by the Taylor expansions through
degree nineteen on the four rays
$$
(u,v,w)=t(1,-s,s-1),\qquad
t\ge 0, \,\,s\in\{1/10,2/10,3/10,4/10\}.
$$
\end{lemma}

\begin{proof}
As in \eqref{eq:degree-eight-expansion}, permutation symmetry
expresses each $H_n$ as a polynomial in $p_2,p_3$.
Homogeneity gives \eqref{eq:invariant-homogeneous-form}.
We determine its coefficients by evaluating a one-variable polynomial.

Along the ray $(u,v,w)=t(1,-s,s-1)$, we have
$
p_2=t^2p_2(s),$ and $
p_3=t^3p_3(s),
$
where
$$
p_2(s):=2(1-s+s^2),
\qquad
p_3(s):=3s(1-s).
$$
Since $H_n$ is homogeneous of degree $n$, $H_n(1,-s,s-1)$ is the coefficient of $t^n$ in
$\A_6(t,-st,(s-1)t)$.
Factoring out $p_2(s)^m p_3(s)^\epsilon$, we put
$
q(s)=p_3(s)^2/p_2(s)^3
$
and obtain
\begin{equation}
\label{eq:one-variable-coefficient-polynomial}
\frac{H_n(1,-s,s-1)}{p_2(s)^m p_3(s)^\epsilon}
=\sum_{k=0}^{d}c_{n,k}q(s)^k,
\qquad d=\lfloor m/3\rfloor.
\end{equation}
Note that $q(s)>0$ on $(0,1/2)$ and its logarithmic derivative is
$$
\frac{q'(s)}{q(s)}
=(1-2s)\left(\frac2{s(1-s)}+
\frac3{1-s+s^2}\right)>0.
$$
Since $n\leq19$, one has $d\leq3$. Use $s_j=(j+1)/10$ for
$0\leq j\leq d$. The values $q(s_j)$ are distinct, and
$p_3(s_j)>0$. Note that \eqref{eq:one-variable-coefficient-polynomial} is linear with the Vandermonde determinant $\mathrm{det}(V)=\prod_{0\le i<j\le d}(q(s_j)-q(s_i))\ne 0$, hence it determines every coefficient in
\eqref{eq:invariant-homogeneous-form}. 
\end{proof}


Applying these four substitutions to the polygon formula gives
the coefficient bounds in Lemma~\ref{lem:higher-coefficients}
(Appendix~\ref{app:taylor-coefficients}). These yield the following
bounds for the remainder $R_9$.

\begin{theorem}[Two uniform Taylor remainder bounds]
\label{thm:global-taylor-remainder}
For every real zero-sum triple, the remainder beyond degree eight satisfies
$$
|R_9|<\frac3{250}p_2
\text{ whenever }0<t\leq\frac6{25},
$$
and
$$
|R_9|<\frac{11}{200}p_2
\text{ whenever }0<t\leq\frac{21}{80}.
$$
\end{theorem}

\begin{proof}
We estimate degrees nine through nineteen using
Lemma~\ref{lem:higher-coefficients}. For the remaining terms,
apply \eqref{eq:mean-square-taylor-tail} with $N=20$ along
$(a,b,c)=(u,v,w)/t$ and divide by $p_2\geq3t^2/2$ to obtain
\begin{equation}
\label{eq:tail-after-degree-nineteen}
\frac1{p_2}\left|\sum_{n=20}^{\infty}H_n\right|
<\frac{2\sqrt{21}}{3t^2}
\frac{(8t/3)^{20}}{\sqrt{1-(8t/3)^2}}.
\end{equation}

For $9\leq n\leq19$, put $\epsilon=n\bmod2$ and
$m=(n-3\epsilon)/2$. The formula
\eqref{eq:invariant-homogeneous-form} gives
\begin{equation}
\label{eq:normalized-homogeneous-bound}
\frac{|H_n|}{p_2}
\leq\frac23t^{n-2}
\sum_{k=0}^{\lfloor m/3\rfloor}
|c_{n,k}|2^{m-3k}\left(\frac34\right)^{\epsilon+2k}.
\end{equation}
This uses $p_2\leq2t^2$ and
$|p_3|\leq3t^3/4$ from \eqref{eq:zero-sum-bounds}.
Bound each $|c_{n,k}|$ by the larger absolute value of its two
endpoints in Lemma~\ref{lem:higher-coefficients}.
Both bounds increase with $t$. Substituting $6/25$ and
$21/80$ gives
\begin{equation}
\label{eq:finite-degree-group-bounds}
\begin{array}{c|cc}
t\leq
&\displaystyle\frac1{p_2}\sum_{n=9}^{19}|H_n|
&\displaystyle\frac1{p_2}\left|\sum_{n=20}^{\infty}H_n\right|\\
\hline
6/25&<1/400&<19/2000\\
21/80&<1/200&<1/20
\end{array}
\end{equation}
For completeness, let $B_n$ denote the right-hand coefficient of
$t^{n-2}$ in \eqref{eq:normalized-homogeneous-bound}, with each
$|c_{n,k}|$ replaced by the larger absolute endpoint from
Lemma~\ref{lem:higher-coefficients}. Thus
$$
B_n=\frac23\sum_{k=0}^{\lfloor m/3\rfloor}
\frac{\max(|a_{n,k}|,|a_{n,k}+1|)}{1000}
2^{m-3k}\left(\frac34\right)^{\epsilon+2k},
$$
where $a_{n,k}$ is the corresponding integer in that table.
The two finite sums are
$$
\begin{aligned}
\sum_{n=9}^{19}B_n(6/25)^{n-2}
&=\frac{335916951272113891872267}{145519152283668518066406250}
<\frac1{400},\\
\sum_{n=9}^{19}B_n(21/80)^{n-2}
&=\frac{141381022750318299872952246014482329}
{28823037615171174400000000000000000000}
<\frac1{200}.
\end{aligned}
$$
For the final column of \eqref{eq:finite-degree-group-bounds}, put
$$
V(t)^2=\frac{28}{3t^4}
\frac{(8t/3)^{40}}{1-(8t/3)^2}.
$$
This function increases for $0<t<3/8$; substituting
$t=6/25$ and $t=21/80$ gives the stated bounds.
\end{proof}

\subsection{The degree-eight comparison on the reduced region}
\label{sec:invariant-reduction}
We now prove the polygon bound needed in region $A$. It holds for
every nonmodular reduced triple with $\sys(X)\geq7/5$.
The inequalities defining a reduced triple bound $p_2$, $p_3/p_2$,
and $t$. These bounds show that the degree-eight polynomial is larger than
the possible error $|R_9|$ from Theorem~\ref{thm:global-taylor-remainder}.

\begin{lemma}[Bounds for $p_2$, $p_3/p_2$, and $t$]
\label{lem:reduced-invariants}
For a nonmodular reduced triple, one has
\begin{equation}
\label{eq:reduced-cubic-ratio}
\frac{p_3}{p_2}\leq\frac{\log2}{6}<\frac7{60}.
\end{equation}
If in addition $\sys(X)\geq7/5$, then
\begin{equation}
\label{eq:reduced-invariant-radius}
t=\max(|u|,|v|,|w|)<\frac{21}{80},
 p_2<\frac{441}{3200}.
\end{equation}
If also $t\geq6/25$, then
\begin{equation}
\label{eq:reduced-cubic-ratio-large-radius}
\frac{p_3}{p_2}<\frac1{120}.
\end{equation}
\end{lemma}
\begin{proof}
For a reduced triple, the largest of $e^{2u},e^{2v},e^{2w}$ is at
most the sum of the other two. We use this inequality to bound
$p_3/p_2$, then use the systole bound to bound $t$.
The first assertion is immediate when $p_3\leq0$. Otherwise the
coordinates are, up to permutation,
$(t,-t/2+d,-t/2-d)$ with $0\leq d<t/2$.
The preceding inequality becomes
$$
e^{3t}\leq2\cosh(2d),
3t\leq\log2+2d^2.
$$
The second inequality follows from $\log\cosh(2d)\leq2d^2$.
Since $2d\leq t$, we also have
$e^{3t}\leq2\cosh(2d)\leq2e^{t}$.
Thus $e^{2t}\leq2$ and $t<6$.
Direct algebra gives
$$
3t-6\frac{p_3}{p_2}
=\frac{24t d^2}{3t^2/2+2d^2}
\geq\frac{12d^2}{t}\ge 2d^2.
$$
Thus $6p_3/p_2\leq3t-2d^2\leq\log2<7/10$. 

To bound $t$, fix the systolic trace $x=2\cosh a$ and vary $y$
among reduced triples. Wherever $2z<xy$, the lower root $z=z_-(x,y)$
decreases with $y$. Implicit differentiation also gives
\begin{equation}
\label{eq:reduced-arc-monotonicity}
\frac{d}{dy}\log(yz)
=\frac{2(z^2-y^2)}{yz(2z-xy)}\leq0.
\end{equation}
Thus the smallest logarithmic coordinate
$(2\log x-\log y-\log z)/3$ increases with $y$, while the largest
$(2\log z-\log x-\log y)/3$ decreases. The maximum of $t$
is therefore at the smallest $y$ for which the triple is reduced. The conclusion extends
to an endpoint with $2z=xy$ by continuity.

For $x\geq\sqrt8$, that endpoint is $y=x$. Put
\begin{equation}
\label{eq:reduced-symmetric-endpoint}
q(x)=\frac{x-\sqrt{x^2-8}}2.
\end{equation}
Here $1\leq q\leq\sqrt2$ and the logarithmic coordinates are
$(-\log q/3,-\log q/3,2\log q/3)$. Hence
$t\leq(\log2)/3<21/80$.
For $2<x\leq\sqrt8$ it is
$$
y=\frac{2x}{\sqrt{x^2-4}}, z=\frac{xy}{2}.
$$
In terms of $a$ the endpoint is
$(x,y,z)=(2\cosh a,2\coth a,2\cosh a\coth a)$.
The absolute values of its smallest and largest logarithmic coordinates
are respectively
$$
\frac13\log\frac{\cosh a}{\sinh^2a},
\frac13\log\frac{\cosh^2a}{\sinh a}.
$$
Both functions decrease for $7/10\leq a\leq\operatorname{arsinh}1$. It therefore suffices to bound them at $a=7/10$.
The Taylor sums for $\cosh$ and $\sinh$, with the remainder bounds
in Appendix~\ref{app:series-estimates}, give
$$
\cosh(7/10)<\frac{157}{125},
\sinh(7/10)>\frac{379}{500}.
$$
The resulting two quotients are less than $219/100$, which is less
than $e^{63/80}$ by its Taylor sum. Thus $t<21/80$.
Finally $p_2\leq2t^2$ gives the stated bound for $p_2$.

Now we prove $\eqref{eq:reduced-cubic-ratio-large-radius}$, suppose $t\ge 6/25$ and $p_3>0$. Using the representation
$(t,-t/2+d,-t/2-d),$ since $p_2=3t^2/2+2d^2\leq2t^2$ and
$2d^2\geq3t-\log2$, we have
$$
\frac{p_3}{p_2}
=\frac{t}{2}-\frac{4t d^2}{p_2}
\leq\frac{t}{2}-3+\frac{\log2}{t}.
$$
The last expression decreases for $6/25\leq t\leq21/80$.
The Taylor sum of degree nine for the exponential gives
$e^{1733/2500}>2$, so $\log2<1733/2500$. Consequently
$$
\frac{p_3}{p_2}
<\frac3{25}-3+\frac{1733}{600}
=\frac1{120}.
$$
\end{proof}

Let $T_8$ be the sum of the Taylor terms of degrees two through
eight in \eqref{eq:degree-eight-expansion}. Thus
\begin{equation}
\label{eq:polynomial-remainder-split}
\A_6-\A_6(\Xmod)=T_8+R_9.
\end{equation}

\begin{proposition}[A lower bound for the degree-eight polynomial]
\label{prop:global-polynomial}
Suppose $(x,y,z)$ is a reduced triple for $X$ and $\sys(X)\geq7/5$.
If $0<t\leq6/25$, then
$$
T_8(u,v,w)>\frac{31}{1000}p_2.
$$
If instead $t\geq6/25$, then
$$
T_8(u,v,w)>\frac2{25}p_2.
$$
\end{proposition}
\begin{proof}
We reduce the comparison to one variable: first show that $T_8/p_2$
decreases with $p_3/p_2$, then replace that ratio by its upper bound.
The resulting polynomial in $p_2$ can be bounded at its endpoints.
Regard $T_8/p_2$ as a quadratic polynomial in $p_3/p_2$, with
$p_2$ fixed. Its quadratic coefficient is
$c_{6''}p_2+c_{8''}p_2^2>0$ for $0<p_2\leq441/3200$.
On the range $p_3/p_2\leq7/60$ from
\eqref{eq:reduced-cubic-ratio}, its derivative with respect to this ratio
is at most
$$
-\frac{692}{1000}+\frac{815}{1000}p_2
+\frac{277}{1000}p_2^2
+2\frac{6367}{1000}p_2\frac7{60}<0.
$$
The last expression increases with $p_2$; at $441/3200$ it equals
$-3785555003/10240000000<0$.
Replacing $p_3/p_2$ by $7/60$ therefore gives a lower bound.
Apply \eqref{eq:quadratic-area-coefficient} and the lower bounds in
\eqref{eq:degree-eight-coefficients}. With $a=7/60$, the constant,
linear, quadratic, and cubic coefficients are, respectively,
$$
\begin{aligned}
\frac{23}{200}-\frac{694}{1000}a&=\frac{1021}{30000},\\
-\frac{162}{1000}+\frac{812}{1000}a+\frac{6354}{1000}a^2
&=\frac{11531}{600000},\\
-\frac{314}{1000}+\frac{270}{1000}a-\frac{15077}{1000}a^2
&=-\frac{1755773}{3600000},\\
\frac{1118}{1000}&=\frac{559}{500}.
\end{aligned}
$$
Consequently,
\begin{equation}
\label{eq:reduced-polynomial-lower-bound}
\frac{T_8}{p_2}>
\frac{1021}{30000}
+\frac{11531}{600000}p_2
-\frac{1755773}{3600000}p_2^2
+\frac{559}{500}p_2^3.
\end{equation}
For $t\leq6/25$, we have $p_2\leq72/625$.
The cubic on the right is concave on this interval: its second
derivative is increasing and is negative at $72/625$.
Its endpoint values exceed $31/1000$. At the nonzero endpoint its value is
$$
\frac{46119182759}{1464843750000}>\frac{31}{1000}.
$$
Concavity proves the first bound throughout the interval.

If $t\geq6/25$, use
\eqref{eq:reduced-cubic-ratio-large-radius} instead. Since $T_8/p_2$ decreases as $p_3/p_2$ increases,
substituting $p_3/p_2=1/120$ again gives a lower bound. At this value the terms with
coefficients $c_5,c_7,c_{8'}$ are positive, as is the combined quadratic
term $(c_{6''}p_2+c_{8''}p_2^2)(p_3/p_2)^2$. Discarding them gives
$$
\frac{T_8}{p_2}
>\frac{23}{200}-\frac{347}{60000}
-\frac{81}{500}p_2-\frac{157}{500}p_2^2
>\frac2{25}.
$$
The last expression decreases with $p_2$. At $p_2=441/3200$ it equals
$$
\frac{1243045849}{15360000000}>\frac2{25},
$$
which proves the final comparison.
\end{proof}

\begin{corollary}[Uniform polygon comparison]
\label{cor:global-polygon-comparison}
For every nonmodular reduced triple with $\sys(X)\geq7/5$,
$$
\A_6(X)-\A_6(\Xmod)>\frac{19}{1000}p_2.
$$
\end{corollary}
\begin{proof}
Use \eqref{eq:polynomial-remainder-split},
Proposition~\ref{prop:global-polynomial}, and
Theorem~\ref{thm:global-taylor-remainder}.
For $0<t\leq6/25$, they give
$$
\A_6(X)-\A_6(\Xmod)
>\left(\frac{31}{1000}-\frac3{250}\right)p_2
=\frac{19}{1000}p_2.
$$
For $6/25<t<21/80$, they give
$$
\A_6(X)-\A_6(\Xmod)
>\left(\frac2{25}-\frac{11}{200}\right)p_2
=\frac1{40}p_2>\frac{19}{1000}p_2.
$$
The radius bound \eqref{eq:reduced-invariant-radius} covers the whole
reduced region under consideration.
\end{proof}

We now compare the full areas on region $A$. If $t\leq1/20$,
\eqref{eq:trace-shape-distance}, $e^{t}<20/19$, and
$e^{2t}<10/9$ give
$$
\frac1{1024}<(x-3)^2+(y-3)^2
<\left(\frac{200}{57}\right)^2p_2<\frac{25}{2}p_2.
$$
If $t>1/20$, then \eqref{eq:zero-sum-bounds} gives
$p_2\geq3t^2/2>3/800$. Thus throughout region $A$,
\begin{equation}
\label{eq:outside-disk-shape-lower}
p_2>\frac1{12800}.
\end{equation}
Corollary~\ref{cor:global-polygon-comparison},
Lemma~\ref{lem:small-modular-remainder}, and
\eqref{eq:area-comparison-split} therefore imply
\begin{equation}
\label{eq:region-a-comparison}
\A(X)-\A(\Xmod)>
\frac{19}{1000}p_2-\frac1{700000}
>\left(\frac{19}{1000}-\frac{12800}{700000}\right)p_2
=\frac1{1400}p_2>0.
\end{equation}
Section~\ref{sec:completion} uses the sharper polygon bound near
the disk to obtain the coefficient $1/80$.

\section{Region B: comparison inside the local disk}
\label{sec:region-b}

Region $B$ consists of nonmodular tori represented by reduced triples with
$\sys(X)\geq7/5$ and $(x-3)^2+(y-3)^2\leq1/1024$. Reducedness gives $x\leq3$ and
$y\geq x$, so these points lie in $\Rloc$.
Here we compare the change in omitted area with the polygon gain.
Its zero gradient and a lower Hessian bound give
$$
(\A-\A_6)(X)-(\A-\A_6)(\Xmod)
>-\frac{(x-3)^2+(y-3)^2}{500}
$$
away from the modular point. We prove this by bounding length
derivatives, differentiating the area series, and summing the
possible negative contributions. Combining it with the polygon
bound will yield $\A(X)-\A(\Xmod)>27p_2/400$ throughout $\Rloc$,
which is stronger than needed on region $B$.

\subsection{Derivatives of all simple lengths}
\label{sec:length-derivatives}

To control the change of the omitted area in region $B$, we need
bounds for length derivatives in $x,y$. Differentiating the Farey
recursion expresses them as the parent derivatives plus exponentially
small corrections, bounded in Subsection~\ref{sec:local}.

\begin{proposition}[Length derivatives bounded by sector height]
\label{prop:length-derivatives}
On $\Rloc$, every simple curve $w$ in an initial Farey sector satisfies
$$
\max\{|\partial_xL_w|,|\partial_yL_w|\}<\frac32h(w),
$$
$$
\max\{|\partial_{xx}L_w|,|\partial_{xy}L_w|,
|\partial_{yy}L_w|\}<12h(w).
$$
\end{proposition}
\begin{proof}
We first check heights one and two, then sum the corrections
introduced by later insertions.

By Lemmas~\ref{lem:local-discriminant} and~\ref{lem:local-bounds},
$\sqrt{\mathscr D}>11/4$, $95/32\leq x,y\leq97/32$, and
$74/25<x,y,z<61/20$. Implicit differentiation gives
$$
z_x=\frac{2x-yz}{\sqrt{\mathscr D}},
z_y=\frac{2y-xz}{\sqrt{\mathscr D}},
$$
so $-5/4<z_x,z_y<0$. Differentiating once more gives
$$
z_{xx}=\frac{2-2yz_x+2z_x^2}{\sqrt{\mathscr D}},
z_{xy}=\frac{-z-yz_y-xz_x+2z_xz_y}{\sqrt{\mathscr D}},
$$
and the analogous formula for $z_{yy}$. The preceding bounds imply
$|z_{xx}|,|z_{xy}|,|z_{yy}|<6$.
For $L(t)=2\arccosh(t/2)$,
$$
L'(t)=\frac2{\sqrt{t^2-4}},
L''(t)=-\frac{2t}{(t^2-4)^{3/2}}.
$$
On $[74/25,61/20]$, these satisfy $0<L'<1$ and $|L''|<3/5$.
Thus the three initial lengths have first derivatives bounded by
$5/4$ and second derivatives bounded by
$6+(3/5)(5/4)^2<7$.

The first insertion traces $xy-z,yz-x,zx-y$ exceed $14279/2500$.
Their first derivatives have absolute value less than $5$ and
second derivatives less than $21$, by the bounds for $z$ above.
For the first derivatives use $z_x,z_y<0$; for example,
$|z+yz_y|\leq\max\{z,-yz_y\}$.
On $t\geq14279/2500$, $L'(t)<3/8$ and $|L''(t)|<2/25$.
The chain rule therefore bounds the first derivatives by
$15/8<5/2$ and the second by $(3/8)21+(2/25)5^2<14$.
We have proved, at heights one and two,
\begin{equation}
\label{eq:initial-length-derivatives}
\max\{|\partial_xL_w|,|\partial_yL_w|\}<\frac54h(w),
\max\{|\partial_{xx}L_w|,|\partial_{xy}L_w|,|\partial_{yy}L_w|\}<7h(w).
\end{equation}
All comparisons for $L'$ and $L''$ follow by squaring positive quantities.

For a later insertion $w=u+v$, put $H=h(u)+h(v)\geq3$.
For example,
$$
\partial_xL_w=(1-\delta_1)\partial_xL_u
+(1-\delta_2)\partial_xL_v,
$$
with the same formula for $y$. Proposition~\ref{prop:scalar-defect}
and \eqref{eq:basic-exponential-bounds} give
\begin{equation}
\label{eq:length-recursion-coefficients}
1<1-\delta_1,1-\delta_2<1+20\left(\frac3{16}\right)^H,
0<\delta_{11},\delta_{12},\delta_{22}
<40\left(\frac3{16}\right)^H.
\end{equation}
After division by $H$, the parent bounds are averaged with weights
$h(u)/H$ and $h(v)/H$. Since both parent heights are smaller than $H$,
induction bounds each first derivative divided by $h(w)$ by
$5/4$ times the following product:
\begin{equation}
\label{eq:length-derivative-product}
\prod_{H=3}^{\infty}\left(1+20\left(\frac3{16}\right)^H\right)
<e^{135/832}<\frac65.
\end{equation}
Here $\log(1+t)<t$ bounds the product, and
$\log(6/5)>1/5-(1/5)^2/2>135/832$.
This proves the first-derivative bound $3h(w)/2$.

A second differentiation has the same inherited parent terms,
plus products of two first derivatives. For example,
$$
\partial_{xx}L_w=(1-\delta_1)\partial_{xx}L_u
+(1-\delta_2)\partial_{xx}L_v
-\delta_{11}(\partial_xL_u)^2
-2\delta_{12}\partial_xL_u\partial_xL_v
-\delta_{22}(\partial_xL_v)^2.
$$
For each of $\partial_{xx},\partial_{xy},\partial_{yy}$,
the extra terms have total absolute value less than
$40(3/2)^2H^2(3/16)^H$.
Divide by $H$, iterate from the base bound $7$, and use
\eqref{eq:length-derivative-product}. Each second derivative
in absolute value, divided by $h(w)$, is less than
$$
\frac65\left(7+90\sum_{H=3}^{\infty}
H\left(\frac3{16}\right)^H\right)
=\frac65\left(7+90\frac{567}{21632}\right)<12.
$$
\end{proof}

\subsection{Differentiating the area and bounding the omitted terms}
\label{sec:tail}

Proposition~\ref{prop:farey-expansion} gives
$$
\A-\A_6=
\sum_{\substack{(u,v)\in\mathscr F\\h(u+v)>6}}S_X(u,v).
$$
We show that this sum can be differentiated twice term by term,
then bound its Hessian to control the possible decrease in omitted area.
\begin{proposition}[Differentiating the area formula twice]
\label{prop:c2}
For all nonnegative integers $k,l$ with $k+l\leq2$, the series
\begin{equation}
\label{eq:sixseries}
\sum_{(u,v)\in\mathscr F}
\partial_x^k\partial_y^l S_X(u,v)
\end{equation}
converges uniformly on $\Rloc$ and locally uniformly on an open
neighborhood of $\Rloc$. Thus the area series and its first and
second derivative series converge, and the area formula may be
differentiated twice term by term. In particular, on $\Rloc$,
\begin{equation}
\label{eq:area-hessian-tail}
\Hess(\A-\A_K)=
\sum_{\substack{(u,v)\in\mathscr F\\h(u+v)>K}}
\Hess S_X(u,v).
\end{equation}
\end{proposition}
\begin{proof}
We group the insertions by height $H=h(u)+h(v)$.
For $H\geq3$, Proposition~\ref{prop:scalar-defect} bounds $\delta$
and its first two derivatives in $L_u,L_v$ by constants
times $e^{-L_u-L_v}$. Since $L_{u+v}=L_u+L_v-\delta$, its first derivatives are
$1-\delta_1,1-\delta_2$, and its second derivatives are
$-\delta_{11},-\delta_{12},-\delta_{22}$, all uniformly bounded.
Together with $L_u,L_v,L_{u+v}\geq L_0$, this implies that
$(L_uL_vL_{u+v})^{-1}$ and its first two derivatives are
uniformly bounded. The product rule therefore bounds $\mathcal D=\delta/(L_uL_vL_{u+v})$
and its first and second partial derivatives in $L_u,L_v$ by
uniform constants times $e^{-L_u-L_v}$.

Proposition~\ref{prop:length-derivatives} bounds every first and second
partial derivative of $L_u,L_v$ in $x,y$ by a constant times $H$.
For the second derivative in $x$, the chain rule gives
$$
\partial_{xx}S_X(u,v)
=\mathcal D_{11}(\partial_xL_u)^2
+2\mathcal D_{12}(\partial_xL_u)(\partial_xL_v)
+\mathcal D_{22}(\partial_xL_v)^2
+\mathcal D_1\partial_{xx}L_u
+\mathcal D_2\partial_{xx}L_v.
$$
The first three terms are bounded by a constant times
$H^2e^{-L_u-L_v}$, and the last two by a constant times
$He^{-L_u-L_v}$. The same argument applies to $\partial_{xy}S_X(u,v)$ and $\partial_{yy}S_X(u,v)$, so all second derivatives are bounded by a constant times $H^2e^{-L_u-L_v}$. First derivatives involve only one
parent-length derivative and are bounded by a constant times
$He^{-L_u-L_v}$.

Together with the bound for $S_X(u,v)$ obtained above and
$L_u+L_v>(42/25)H$ from Lemma~\ref{lem:height-lower}, this gives
a constant $C$, independent of the insertion and the point in
$\Rloc$, such that, for $H\geq3$,
\begin{equation}
\label{eq:area-correction-derivatives}
|\partial_x^k\partial_y^l S_X(u,v)|
\leq C H^{k+l}e^{-42H/25},
\qquad k,l\geq0,\quad k+l\leq2.
\end{equation}
The finitely many height-two terms and their derivatives are
continuous on the compact set $\Rloc$, hence bounded.
Increasing $C$ if necessary extends this estimate to every $H\geq2$. There are at most $3H$ insertions of height $H$
by \eqref{eq:farey-height-count}. Since
$\sum_{H\geq2}H^3e^{-42H/25}<\infty$, the Weierstrass test
gives uniform convergence of all six series in \eqref{eq:sixseries}.

To differentiate the sum at boundary points of $\Rloc$, we also
need convergence on an open neighborhood. The bounds used to start
the height and derivative inductions concern only the finitely many
curves of heights one and two. These bounds, including the
inequalities selecting the larger child root, are strict and
depend continuously on $(x,y)$. By compactness, they remain valid
on a common open neighborhood of $\Rloc$. Repeating the inductions
there gives \eqref{eq:area-correction-derivatives}, after increasing
$C$ if necessary, and hence uniform convergence of all six series.

Applying the termwise differentiation theorem successively in $x$
and $y$ \cite[Theorem~7.17]{Rudin}, we may differentiate the area
formula twice term by term. Applied to \eqref{eq:farey-tail},
this gives \eqref{eq:area-hessian-tail} for every integer $K\geq1$.
\end{proof}

\begin{proposition}[A lower Hessian bound for the omitted area]
\label{prop:height-six-lower-tail}
Throughout $\Rloc$,
$$
\Hess(\A-\A_6)>-\frac1{250}I,
$$
where $I$ is the $2\times2$ identity matrix.
\end{proposition}
\begin{proof}
We bound each omitted insertion, then sum over their heights.
Fix $m=h(u)$, $n=h(v)$, and $H=m+n\geq7$.
Since $S_X(u,v)=\mathcal D(L_u,L_v)$, the chain rule gives
\begin{equation}
\label{eq:area-hessian-chain-rule}
\Hess S_X(u,v)
=J^{\mathsf T}(\Hess_{L_u,L_v}\mathcal D)J
+\mathcal D_1\Hess L_u+\mathcal D_2\Hess L_v,
\end{equation}
where $J$ is the derivative matrix of $(x,y)\mapsto(L_u,L_v)$.
The first term is nonnegative in every direction by
Proposition~\ref{prop:defect-convexity}. Proposition~\ref{prop:length-derivatives} bounds each entry of
$\Hess L_u$ by $12m$ in absolute value. Thus
$$
|\xi^{\mathsf T}(\Hess L_u)\xi|
\leq12m(|\xi_1|+|\xi_2|)^2\leq24m|\xi|^2
$$
for every $\xi\in\mathbb R^2$, and similarly for $L_v$.
Multiplying by $\mathcal D_1,\mathcal D_2$ and using absolute
values to bound each possible negative contribution gives
$$
\xi^{\mathsf T}(\Hess S_X(u,v))\xi
\geq-24\bigl(m|\mathcal D_1|+n|\mathcal D_2|\bigr)|\xi|^2.
$$
Since this holds in every direction,
$$
\Hess S_X(u,v)\geq
-24\bigl(m|\mathcal D_1|+n|\mathcal D_2|\bigr)I.
$$
We now bound these two derivatives. Proposition~\ref{prop:scalar-defect},
Lemma~\ref{lem:height-lower}, and \eqref{eq:basic-exponential-bounds} give
$\delta<11e^{-L_u-L_v}$, $|\delta_1|,|\delta_2|<20e^{-L_u-L_v}$,
and
$$
0<1-\delta_1,1-\delta_2<4/3.
$$
Differentiating $\mathcal D=\delta/(L_uL_vL_{u+v})$, with
$\partial_1L_{u+v}=1-\delta_1$, gives
$$
|\mathcal D_1|
<\frac{e^{-L_u-L_v}}{L_uL_vL_{u+v}}
\left(20+\frac{11}{L_u}+\frac{44}{3L_{u+v}}\right).
$$
The bound for $|\mathcal D_2|$ has $L_v$ in place of $L_u$.
Multiplying by $m,n$ and adding, we obtain
$$
m|\mathcal D_1|+n|\mathcal D_2|
<\frac{e^{-L_u-L_v}}{L_uL_vL_{u+v}}
\left(20H+11\left(\frac m{L_u}+\frac n{L_v}
+\frac{4H}{3L_{u+v}}\right)\right).
$$
The height bounds $L_u>(42/25)m$, $L_v>(42/25)n$, and
$L_{u+v}>(42/25)H$ bound the second term in parentheses by
$11(25/42)(2+4/3)<22$. Therefore
\begin{equation}
\label{eq:weighted-correction-gradient}
m|\mathcal D_1|+n|\mathcal D_2|
<\left(\frac{25}{42}\right)^3
\frac{20H+22}{mnH}e^{-L_u-L_v}.
\end{equation}
Since $mn\geq H-1$ and $H\geq7$,
$$
24\left(\frac{25}{42}\right)^3\frac{20H+22}{mnH}
<20.
$$
Thus each omitted insertion satisfies
$$
\Hess S_X(u,v)>-20e^{-L_u-L_v}I
>-20\left(\frac3{16}\right)^H I.
$$
There are at most $3(H-1)$ insertions of height $H$ by
\eqref{eq:farey-height-count}. Thus \eqref{eq:area-hessian-tail} gives
$$
\Hess(\A-\A_6)
\geq-60\sum_{H=7}^{\infty}(H-1)\left(\frac3{16}\right)^H I
>-\frac1{250}I.
$$
Here the sum is evaluated by differentiating
$\sum_{H=7}^{\infty}t^{H-1}=t^6/(1-t)$
and multiplying by $t^2$, at $t=3/16$.
\end{proof}

\subsection{Completing the comparison in region B}
\label{sec:region-b-comparison}

Combining the polygon gain with the omitted-area estimate gives
the following inequality throughout $\Rloc$, and hence on region $B$.

\begin{corollary}[Uniform local area inequality]
\label{cor:local-no-grid}
For $(x,y)\in\Rloc\setminus\{(3,3)\}$,
$$
\A(x,y)-\A(3,3)>\frac7{80}p_2
-\frac1{500}\bigl((x-3)^2+(y-3)^2\bigr).
$$
\end{corollary}

\begin{proof}
By Proposition~\ref{prop:c2}, $\A-\A_6$ is $C^2$ near
$\Rloc$. Its permutation symmetry and
Lemma~\ref{lem:symmetry-at-modular} give
$\nabla(\A-\A_6)(3,3)=0$, while
Proposition~\ref{prop:height-six-lower-tail} gives
$\Hess(\A-\A_6)>-I/250$ on $\Rloc$.
For $(x,y)\neq(3,3)$, put $d=(x-3,y-3)$.
Taylor's formula along the segment in $\Rloc$ gives
$$
(\A-\A_6)(x,y)-(\A-\A_6)(3,3)>-\frac1{500}|d|^2.
$$
By Lemma~\ref{lem:lambda-local-conversion}, $t<1/60$ and
$|d|^2<10p_2$. Adding the polygon bound from
Proposition~\ref{prop:uniform-p6} in \eqref{eq:area-comparison-split}
gives
$$
\A(x,y)-\A(3,3)>\frac7{80}p_2-\frac1{500}|d|^2.
$$
\end{proof}

\subsubsection{Local convexity and the area inequality}

\begin{theorem}[Local estimate near the modular torus]
\label{thm:pure-local}
Every $X$ with $\sys(X)>381/200$ admits reduced coordinates in $\Rloc$.
Moreover,
$$
\A(x,y)-\A(3,3)\geq
\frac{27}{4000}\bigl((x-3)^2+(y-3)^2\bigr),
  (x,y)\in\Rloc,
$$
and
$$
\nabla\A(3,3)=0,
\Hess\A(3,3)>\frac{97}{4500}I.
$$
In the centered logarithmic coordinates, as $p_2\to0$,
$$
\A(X)-\A(\Xmod)=\kappa p_2+o(p_2),
\kappa>\frac{97}{1000}.
$$
In particular, $\A$ is strictly convex on a sufficiently small disk
centered at $(3,3)$.
\end{theorem}

\begin{proof}
Corollary~\ref{cor:local-no-grid} and Lemma~\ref{lem:lambda-local-conversion} give the area inequality and its
equality case. Since $\A$ is symmetric in the three traces,
Lemma~\ref{lem:symmetry-at-modular} gives
$$
\nabla\A(3,3)=0.
$$
Propositions~\ref{prop:p6} and~\ref{prop:height-six-lower-tail} give
$$
\Hess\A(3,3)
>
\left(
\frac{23}{900}-\frac1{250}
\right)I
=
\frac{97}{4500}I.
$$
By Proposition~\ref{prop:c2}, the Hessian is continuous on an open
neighborhood of $(3,3)$. Hence there exists $\varepsilon>0$ such that
$$
\Hess\A(x,y)>\frac1{100}I
\quad\text{for }(x-3)^2+(y-3)^2<\varepsilon^2.
$$
This proves strict convexity on that disk.

In centered logarithmic coordinates, cyclic symmetry makes the
quadratic term a multiple $\kappa p_2$. At the modular point,
$d(u,v,w)=(dx,dy,-dx-dy)/3$, so this term has Hessian
$\frac{2\kappa}{9}\begin{pmatrix}2&1\\1&2\end{pmatrix}$ in $(x,y)$.
Its smaller eigenvalue is $2\kappa/9$; the preceding Hessian bound
therefore gives $\kappa>97/1000$.
Finally, Lemma~\ref{lem:high-systole} places every torus with
$\sys(X)>381/200$ in $\Rloc$.
\end{proof}

\begin{remark}
Figure~\ref{fig:area-local-mesh} numerically illustrates
the area above its modular value on the disk of radius $1/32$ centered at
$(3,3)$, with vertical scale enlarged by $10^5$. The graph suggests strict convexity throughout this disk. Our analytic proof establishes it on a sufficiently small
neighborhood of the modular point.
\end{remark}
\begin{figure}[ht]
\centering
\includegraphics{figures/diagram-11.pdf}
\caption{Area above its modular value.}
\label{fig:area-local-mesh}
\end{figure}

\section{Region C: comparison when the systole is short}
\label{sec:region-c}

We base the fan construction from Subsection~\ref{sec:finite}
on a systole. After establishing its area bounds and cusp growth,
we prove the comparison for region $C$ successively for systole
at most $2/3$, $6/5$, and $7/5$.

\subsection{Area bounds from curves around a systole}
\label{sec:global}

Mirzakhani's cusp estimate implies $\A(X)\to\infty$ as
$\sys(X)\to0$ \cite[Proposition~3.6]{MirzakhaniGrowth}.
To join this region to our polygon comparison, we need an explicit
cutoff. Twisting a curve about a systole gives vertices of an inscribed
polygon and hence a lower area bound. We refine this bound in
Subsections~\ref{sec:stronger-pure-cusp} and~\ref{sec:new-cusp}
to prove the following theorem.

\begin{theorem}[Short-systole estimate]
\label{thm:pure-cusp}
If $\sys(X)\leq7/5$, then
$$
\A(X)>\frac{893}{1000}>\A(\Xmod),
$$
where $\A(\Xmod)<223/250$.
\end{theorem}

The recurrence from Subsection~\ref{sec:finite} can be solved
explicitly, expressing every fan trace in terms of the base length
and one twist parameter.

\begin{lemma}[Trace formula for a fan]
\label{lem:balanced-fan}
Let $\gamma$ be a simple class of length $2a$ and let $\gamma'$ be
a Farey neighbor.
After replacing $\gamma'$ by $\gamma'+k\gamma$, there is
$|\theta|\leq a/2$ such that, for every $n\in\Z$,
$$
x_{\gamma'+n\gamma}
=x_n(a,\theta)
:=2\coth a\cosh\left(\left(n+\frac12\right)a+\theta\right).
$$
\end{lemma}
\begin{proof}
The recurrence \eqref{eq:sector-trace-recursion}, with
$x=2\cosh a$, has general solution $x_n=Ae^{na}+Be^{-na}$.
Positivity for all $n\in\Z$ gives $A,B\geq0$, and the Markoff
identity for $(2\cosh a,x_n,x_{n+1})$ gives $AB=x^2/(x^2-4)=\coth^2a$.
Thus $x_n=2\coth a\cosh(na+a/2+\theta)$ for some real $\theta$.
Replacing $\gamma'$ by $\gamma'+k\gamma$ changes $\theta$ by $ka$,
so we may arrange $|\theta|\leq a/2$.
\end{proof}

From now on, take $\gamma$ to be a systole. Consecutive fan classes,
and each fan class paired with $\gamma$, have absolute determinant
one. Including the opposite vertices and closing at $\pm\gamma/(2a)$
gives a centrally symmetric inscribed polygon. Its area is a sum
of reciprocal length products, as in
\eqref{eq:height-six-polygon-area}. Upper bounds for its lengths
give a lower area bound; further Farey insertions increase the area
by \eqref{eq:area-gain}.

\begin{proposition}[A lower bound independent of the twist]
\label{prop:twist-free-fan}
For $n\ge0$, put
\begin{equation}
\label{eq:fan-length-function}
l_a(t)=2\operatorname{arccosh}(\coth a\cosh t),
c_n=l_a(na),
m_n=\frac{c_n+c_{n+1}}2.
\end{equation}
For every nonnegative integer $M$,
\begin{equation}
\label{eq:twist-free-fan-bound}
\A(X)\ge
\frac1{m_0^2}
+2\sum_{n=0}^{M-1}\frac1{m_nm_{n+1}}
+\frac1{a m_M}.
\end{equation}
The sum is empty when $M=0$. These lower bounds increase strictly with
$M$ and converge to
$$
\frac1{m_0^2}+2\sum_{n\ge0}\frac1{m_nm_{n+1}}.
$$
\end{proposition}

\begin{proof}
Convexity bounds each length above by the line joining two
consecutive endpoint values. After this substitution, we pair the
area terms with twist parameters $\theta$ and $-\theta$. Each pair
is smallest at $\theta=0$, giving a bound independent of the twist.
Set $k=\coth a$. Direct differentiation gives
\begin{equation}
\label{eq:fan-length-curvature}
l_a''(t)=
\frac{2k(k^2-1)\cosh t}{(k^2\cosh^2t-1)^{3/2}}>0.
\end{equation}
Write $s=1/2+\theta/a\in[0,1]$, and put
$$
p_n(s)=(1-s)c_n+s c_{n+1}.
$$
Convexity gives
$$
l_a((n+s)a)\le p_n(s).
$$
Consider the finite fan with endpoint parameters $-(M+1-s)a$ and
$(M+s)a$, closed by the two points $\pm\gamma/(2a)$. Its area is
$$
\begin{aligned}
&\frac1{l_a(sa)l_a((1-s)a)}\\
&+\sum_{n=0}^{M-1}\left(
\frac1{l_a((n+s)a)l_a((n+1+s)a)}
+\frac1{l_a((n+1-s)a)l_a((n+2-s)a)}
\right)\\
&+\frac1{2a}\left(
\frac1{l_a((M+s)a)}+\frac1{l_a((M+1-s)a)}
\right).
\end{aligned}
$$
Replace $l_a((n+s)a)$ by its upper bound $p_n(s)$, and likewise for
$1-s$. This gives a lower bound for the area. Since
$p_0(s)+p_0(1-s)=2m_0$, its central term is at least $m_0^{-2}$.
For positive affine functions $u(s),v(s)$, direct differentiation gives
\begin{equation}
\label{eq:reciprocal-affine-convexity}
\frac{d^2}{ds^2}\frac1{uv}
=\frac2{uv}\left(
\left(\frac{u'}{u}\right)^2+
\left(\frac{v'}{v}\right)^2+
\frac{u'}{u}\frac{v'}{v}
\right)\ge0.
\end{equation}
By \eqref{eq:reciprocal-affine-convexity}, the middle terms at $s$
and $1-s$ have sum at least twice their value at $s=1/2$, namely
$2/(m_nm_{n+1})$. Convexity of the reciprocal gives
at least $1/(a m_M)$ for the two terms from edges ending at
$\pm\gamma/(2a)$. This proves the bound.

For $t\ge0$ we have $0\le l_a'(t)<2$, so
$0<m_{M+1}-m_M<2a$. The increment of the lower bound is exactly
$$
\frac{2a-(m_{M+1}-m_M)}{a m_Mm_{M+1}}>0.
$$
Finally $m_n=2na+O_a(1)$, which proves convergence.
\end{proof}

\subsubsection{Two-sided bounds and growth at the cusp}

The same twisting curves give an upper bound as well: their limiting
directions determine a diamond containing the norm ball. The full fan
then approximates its area closely enough to determine the first
correction in Theorem~\ref{thm:quantitative-area}.

\begin{proposition}
\label{prop:cusp-growth}
Every $X$ satisfies
$$
0\leq1-s_X\log\coth(s_X/2)\A(X)
\leq\frac{\pi^2}{12\log^2\coth(s_X/2)}
+\frac{s_X^2\cosh(s_X/2)}{32\log\coth(s_X/2)}.
$$
The cusp expansion in Theorem~\ref{thm:quantitative-area}
holds uniformly in the twist.
\end{proposition}

\begin{proof}
Put $a=s_X/2$ and $k=\coth a$, and use the classes and parameter
$\theta$ from Lemma~\ref{lem:balanced-fan}. Their lengths are
$L_n=l_a((n+1/2)a+\theta)$. As $n\to\pm\infty$,
$$
L_n-2a|n|\longrightarrow2\log k\pm(a+2\theta).
$$
For $q>0$ and $n\in\Z$, the triangle inequality gives
$\|p\gamma+q\gamma'\|_X\geq qL_n-2a|qn-p|$.
Taking the two limits, and using central symmetry for $q<0$, gives
$$
\|p\gamma+q\gamma'\|_X
\geq|2ap+(a+2\theta)q|+2\log k|q|.
$$
The diamond defined by the right-hand side has area
$1/(2a\log k)$, which proves the upper bound for $\A(X)$.

Let $\A_{\mathrm{fan}}$ denote the area of the full inscribed fan,
and let $\widetilde l_a$ interpolate $l_a$ affinely on the grid
$a\Z+a/2+\theta$. The area formula and integration on each grid
interval give
$$
\A_{\mathrm{fan}}=\sum_{n\in\Z}\frac1{L_nL_{n+1}}
=\frac1a\int_{\R}\frac{dt}{\widetilde l_a(t)^2}
\leq\A(X).
$$
The closing edges at $\pm\gamma/(2a)$ tend to zero as the finite
fans exhaust this polygon. Differentiating
\eqref{eq:fan-length-function} gives $0<l_a''(t)\leq2\cosh a$,
and therefore
$$
0\leq\widetilde l_a(t)-l_a(t)\leq\frac{a^2\cosh a}{4}.
$$
For $g(t)=\log(1+e^{-2t})$, $t\geq0$, elementary bounds for
$\arccosh$ give
$$
2(\log k+|t|)\leq l_a(t)
\leq2(\log k+|t|+g(|t|)).
$$
Using $(1+z)^{-2}\geq1-2z$ for $z\geq0$ in the integral above,
together with
$$
\int_0^\infty g(t)dt
=\frac12\sum_{j\geq1}\frac{(-1)^{j+1}}{j^2}
=\frac{\pi^2}{24},
$$
gives
$$
\A_{\mathrm{fan}}\geq\frac1{2a\log k}
\left(1-\frac{\pi^2}{12\log^2k}
-\frac{a^2\cosh a}{8\log k}\right).
$$
This proves the stated two-sided bound.

For the correction term we must also control the area missed by
the fan. For sufficiently small $a$, all fan lengths are at least
$l_a(0)>4$. The opposite curve is the short systole,
so every outward insertion uses the larger Markoff root; its trace
exceeds both parent traces, and this persists under insertion.
By Proposition~\ref{prop:scalar-defect}, every length deficit is
at most $11e^{-2l_a(0)}$. Farey induction shows that subtracting
this amount from each endpoint length gives a lower linear bound
for the norm in that sector: at a class $pu+qv$, the accumulated
loss is at most $(p+q-1)11e^{-2l_a(0)}$.
Homogeneity and continuity extend this bound to the whole sector.
Comparing sector areas therefore gives
$$
\A_{\mathrm{fan}}\leq\A(X)
\leq\left(1-\frac{11e^{-2l_a(0)}}{l_a(0)}\right)^{-2}
\A_{\mathrm{fan}}.
$$
Since $e^{-2l_a(0)}=O(a^4)$, the omitted area is
$O(a^3/\log^2k)$. The interpolation estimate also gives
$$
0\leq\frac1a\int_{\R}\frac{dt}{l_a(t)^2}
-\A_{\mathrm{fan}}
\leq\frac{a\cosh a}{16\log^2k}.
$$
Finally,
$2\log(2k\cosh t)-l_a(t)=O(a^2\operatorname{sech}^2t)$,
uniformly in $t$. Hence
$$
\A(X)=\frac1{2a}\int_0^\infty
\frac{dt}{(\log k+t+g(t))^2}+O(a/\log^2k).
$$
Expanding in $g(t)$ first, whose first moment and square are
integrable, yields
$$
\begin{aligned}
\int_0^\infty\frac{dt}{(\log k+t+g(t))^2}
&=\frac1{\log k}-2\int_0^\infty
\frac{g(t)}{(\log k+t)^3}dt+O(\log^{-4}k)\\
&=\frac1{\log k}-\frac{\pi^2}{12\log^3k}
+O(\log^{-4}k).
\end{aligned}
$$
Since $\log k=\log(1/a)+O(a^2)$, this is the claimed expansion.
Subtracting the fixed value $\A(\Xmod)$ does not change it.
All error bounds are independent of $\theta$.
\end{proof}

\subsubsection{Area bound for systole at most $2/3$}

We first bound the lengths of all fan curves when the fan is based
on a systole.
This will also let us apply Proposition~\ref{prop:defect-convexity}
to the area added by an insertion between two fan classes.

\begin{lemma}[A lower bound for fan lengths]
\label{lem:systolic-fan-domain}
Suppose the fan is based on a systole of length $2a$. Every length of a
curve in this fan, written as
$$
l_a(t)=2\operatorname{arccosh}(\coth a\cosh t)
$$
is at least $2\operatorname{arsinh}1$.
\end{lemma}

\begin{proof}
If $a\leq\operatorname{arsinh}1$, then $\coth a\geq\sqrt2$, so
$l_a(t)\geq2\operatorname{arccosh}\sqrt2=2\operatorname{arsinh}1$. If
$a\geq\operatorname{arsinh}1$, the systolic assumption gives
$l_a(t)\geq2a\geq 2\operatorname{arsinh}1$ for every curve in the fan.
\end{proof}

\begin{proposition}[Area bound for systole at most $2/3$]
\label{prop:pure-cusp-exclusion}
If $\sys(X)\le2/3$, then
$$
\A(X)>
\frac{893}{1000}.
$$
\end{proposition}

\begin{proof}
Here the smallest fan polygon already suffices. We bound its mean
length $m_0$ above, then show that the resulting area bound is smallest
at $a=1/3$.
Put $a=\sys(X)/2\le1/3$. Taking $M=0$ in
Proposition~\ref{prop:twist-free-fan} gives
$$
\A(X)\ge\frac1{a m_0}+\frac1{m_0^2}.
$$
To bound $m_0=(c_0+c_1)/2$ above, we use
$$
\coth x\le\frac1x+\frac x3,
\log\cosh x\le\frac{x^2}2
(x>0).
$$
For the first, the function
$(1+x^2/3)\sinh x-x\cosh x$ vanishes at zero and has derivative
$x(x\cosh x-\sinh x)/3\ge0$. The second follows by integrating
$\tanh x\le x$. We also use
$$
2\operatorname{arccosh}z
\le2\log(2z)-\frac1{2z^2}
(z>1).
$$
To see this, apply
$\sqrt{1-u}\le1-u/2$ and $\log(1-u/4)\le-u/4$ with $u=z^{-2}$.

Since $c_0=2\log\coth(a/2)$, these estimates imply
$$
c_0\le2\log(2/a)+\frac{a^2}6.
$$
For $z=\coth a\cosh a$ we have
$$
z\le a^{-1}\exp(5a^2/6).
$$
Therefore, using $e^{-x}\ge1-x$,
$$ c_1 \le2\log(2/a)+\frac{5a^2}3-\frac{a^2}2e^{-5a^2/3} \le2\log(2/a)+\frac{7a^2}6+\frac{5a^4}6.$$
Consequently
$$
m_0\le b(a):=2\log(2/a)+\frac{2a^2}3+\frac{5a^4}{12}.
$$
The function
$$
f(a)=\frac1{a b(a)}+\frac1{b(a)^2}
$$
is decreasing on $(0,1/3]$. Indeed $b>3$, $ab'\ge-2$, and
$2a^2b'\ge-4a$, while
$$
f'(a)=-\frac{b(b+ab')+2a^2b'}{a^2b^3}.
$$
Its numerator satisfies $b(b+ab')+2a^2b'\ge b(b-2)-4a>3-\frac43>0,$ where $b>3$ follows from $b(a)>2\log(2/a)\ge2\log(6)>3$, thus $\A(X)\ge f(1/3)>\frac{893}{1000}$. 
\end{proof}

\subsection{Area bounds for systole at most 6/5}
\label{sec:stronger-pure-cusp}

We extend the short-systole estimate from $2/3$ to $6/5$ in one step.
Convexity in $a$ bounds the fan lengths above by the lines joining
their values at $a=1/3$ and $a=3/5$. Inserting one vertex between
the two central fan classes adds enough area to complete the comparison.

\begin{lemma}[Convexity in the systole and twist parameters]
\label{lem:joint-fan-convexity}
For every real $k$, the function
$$
(a,\theta)\longmapsto l_a(ka+\theta)
=2\operatorname{arccosh}(\coth a\cosh(ka+\theta))
$$
is convex on $(0,\infty)\times\R$. In particular, for fixed $k$,
$$
\frac{d^2}{da^2}l_a(ka)\geq\frac2{\sinh^2a}.
$$
For $k\geq1$, the function $a\mapsto l_a''(ka)$ is decreasing, where
the primes in $l_a''$ refer to the second argument.
\end{lemma}

\begin{proof}
Write $b(a)=\log\coth(a/2)$ and
$h(b,t)=\operatorname{arccosh}(\cosh b\cosh t)$, so
$l_a(t)=2h(b(a),t)$. We prove that $h$ is convex and increasing in
$b$; convexity of $b(a)$ will then give the required convexity by
the chain rule. If $J=\sinh h$, direct differentiation gives
$$
\Hess_{b,t}h=\frac1{J^3}
\begin{pmatrix}
\cosh b\cosh t\sinh^2t&-\sinh b\sinh t\\
-\sinh b\sinh t&\cosh b\cosh t\sinh^2b
\end{pmatrix}.
$$
Its diagonal entries are nonnegative and its determinant is
$\sinh^2b\sinh^2t/J^4\geq0$. Also $h_b>0$,
$b'=-1/\sinh a$ and $b''=\cosh a/\sinh^2a>0$.
The second-order chain rule therefore proves convexity. Since
$$
h_b=\frac{\sinh b\cosh t}{J}\geq\tanh b=\frac1{\cosh a},
$$
the second derivative of $l_a(ka)=2h(b(a),ka)$ is at least
$2h_b b''\geq2/\sinh^2a$.

For the last assertion, write
\begin{equation}
\label{eq:fan-curvature-function}
g_a(t)=l_a''(t)
=\frac{2\cosh a\cosh t}
{(1+\cosh^2a\sinh^2t)^{3/2}}.
\end{equation}
For $u=\tanh a$, $v=\tanh(ka)$ and
$d=1-u^2+u^2v^2\leq1$, logarithmic differentiation gives
$$
\frac{d}{da}\log g_a(ka)
=\frac{u(d-3v^2)+kv(d-3)}d
\leq\frac{u-2kv}d<0,
$$
because $k\geq1$ and $v\geq u>0$.
\end{proof}

One central insertion improves the finite fan-polygon bound.
Its two parent lengths $L_\pm=l_a(a/2\pm\theta)$ have mean at most
$m_0(a)$, by convexity of $l_a$. The opposite trace is systolic,
so the inserted trace is the larger root in \eqref{eq:child-length}.
Lemma~\ref{lem:systolic-fan-domain} puts these lengths, and the
pair obtained by replacing both by their mean, in $\mathcal E$.
By Proposition~\ref{prop:defect-convexity}, symmetry and convexity
show that this replacement cannot increase $\mathcal D$.
Increasing the common length to $m_0$ cannot increase it either. Thus
\eqref{eq:sector-correction-lower} gives
\begin{equation}
\label{eq:central-fan-insertion-bound}
\mathcal D(L_-,L_+)
\geq\mathcal D(m_0,m_0)>
\frac{4e^{-2m_0}}{m_0^3}.
\end{equation}
This holds for any fan based on a systole and will also be used in
Subsection~\ref{sec:new-cusp}.


\begin{proposition}[Area bound for systole at most $6/5$]
\label{prop:cusp-six-fifths}
If $\sys(X)\leq6/5$, then
$$
\A(X)>\frac{893}{1000}.
$$
\end{proposition}

\begin{proof}
Proposition~\ref{prop:pure-cusp-exclusion} covers systole at most $2/3$.
For the remaining range, put $L=1/3$, $U=3/5$ and
$a=\sys(X)/2\in[L,U]$.
Let $b_n(a)$ be affine with the following endpoint values:
\begin{equation}
\label{eq:cusp-six-fifths-affine-bounds}
\begin{array}{c|cc}
n&b_n(L)&b_n(U)\\\hline
0&73189/20000&13313/5000\\
1&38763/10000&66299/20000\\
2&8537/2000&86537/20000\\
3&95591/20000&109373/20000\\
4&53619/10000&6651/1000
\end{array}
\end{equation}
These values bound $m_n$ at the endpoints, as verified in
Appendix~\ref{app:cusp-six-fifths-checks}. Convexity of $m_n$
(Lemma~\ref{lem:joint-fan-convexity}) therefore gives
$m_n(a)<b_n(a)$ throughout $[L,U]$.

Since $m_0<b_0$, \eqref{eq:central-fan-insertion-bound} bounds
the added area below by $4e^{-2b_0}/b_0^3$.
Adding this to the finite fan-polygon bound in the proof of
Proposition~\ref{prop:twist-free-fan}, with $M=4$, gives
$$
\A(X)>Q(a):=
\frac1{b_0(a)^2}
+2\sum_{n=0}^{3}\frac1{b_n(a)b_{n+1}(a)}
+\frac1{a b_4(a)}
+\frac{4e^{-2b_0(a)}}{b_0(a)^3}.
$$

We need only one tangent line to bound $Q$ on the whole interval.
Each term involving reciprocal lengths is convex by
\eqref{eq:reciprocal-affine-convexity}; its last term has convex
logarithm $\log4-2b_0-3\log b_0$, hence is convex as well.
At $a_*=279/500$, exact rational bounds give
\begin{equation}
\label{eq:cusp-six-fifths-tangent-values}
Q(a_*)>\frac{8931}{10000},
-\frac1{1000}<Q'(a_*)<0.
\end{equation}
The evaluation is given in Appendix~\ref{app:cusp-six-fifths-checks}.
Since a convex function lies above its tangent, for every
$a\in[L,U]$ we have
$$
Q(a)\geq Q(a_*)+Q'(a_*)(a-a_*)
>\frac{8931}{10000}-\frac1{1000}\left(\frac35-\frac{279}{500}\right)>\frac{893}{1000}.
$$
This proves the assertion.
\end{proof}

\subsection{Area bounds for systole at most 7/5}
\label{sec:new-cusp}
To reach systole $7/5$, we improve the polygon bound by considering
pairs of lengths with twist parameters $\theta$ and $-\theta$.
We estimate both their mean and their difference, then add one
Farey vertex between the two central fan classes.

For consecutive endpoint lengths, write
$$
c_n(a)=l_a(na),
m_n(a)=\frac{c_n(a)+c_{n+1}(a)}2,
\Delta_n(a)=\frac{c_{n+1}(a)-c_n(a)}2.
$$
Thus $m_n$ is their mean and $\Delta_n$ is half their difference.
We compare these values with the lengths at twist parameters
$\theta$ and $-\theta$.

\begin{lemma}[Lengths at opposite twist parameters]
\label{lem:reflected-fan-lengths}
Let $0\leq\theta\leq a/2$, set $\tau=2\theta/a$, and put
$$
\mu_n=\frac{l_a((n+\tfrac12)a+\theta)+l_a((n+\tfrac12)a-\theta)}2,
$$
$$
d_n=\frac{l_a((n+\tfrac12)a+\theta)-l_a((n+\tfrac12)a-\theta)}2.
$$
Then $d_n\geq \Delta_n(a)\tau$. Moreover $\Delta_n(a)/a$ is increasing in $a$.
If $\kappa_n\leq g_a(t)$ on $na\leq t\leq(n+1)a$, then
\begin{equation}
\label{eq:reflected-mean-bound}
\mu_n\leq m_n(a)-\frac{\kappa_na^2}{8}(1-\tau^2).
\end{equation}
The same mean bound holds with
$$
\kappa_0=\frac{33}{20},
\kappa_n=g_a((n+\tfrac12)a)\text{ for }n\geq1
$$
whenever $2/3\leq a\leq7/10$.
\end{lemma}

\begin{proof}
Dividing the half-difference by $\theta$ gives the average of
$l_a'$ on the interval between the two arguments. We first bound
this average to estimate $d_n$, then use $l_a''$ to estimate how
far $\mu_n$ lies below the line joining the endpoint lengths.
The function $l_a'$ is increasing and concave on $[0,\infty)$, since
$g_a>0$ and $g_a'(t)<0$ for $t>0$. For a concave function, the average on an interval is at least the
average of its endpoint values. Differentiating the interval average
with respect to its half-length therefore shows that the average
decreases as the interval expands about its center. Apply this to
$l_a'$ on the intervals centered at $(n+1/2)a$, with half-lengths
$\theta$ and $a/2$. It gives $d_n/\theta\geq2\Delta_n/a$, including the limit at $\theta=0$.
Furthermore,
$$
\frac{\Delta_n(a)}a
=\int_n^{n+1}
\frac{\sinh(av)}{\sqrt{\sinh^2(av)+\operatorname{sech}^2a}}dv
$$
is increasing, because the integrand is increasing in $a$ for $v\geq0$.

If $l_a''\geq \kappa_n$, subtract the quadratic $\kappa_nt^2/2$ and use the
fact that the resulting convex function lies below the line joining
its endpoint values. Averaging at the two arguments equally spaced
about $(n+1/2)a$ proves the bound for $\mu_n$.

For the stronger mean bounds, we improve the lower bound for
$g_a=l_a''$ before integrating twice. Put $w=\sinh^2a$ and
$\xi=(1+\cosh^2a\sinh^2t)^{-1}$. Direct differentiation gives
\begin{equation}
\label{eq:fan-curvature-second-derivative}
\frac{g_a''(t)}{g_a(t)}
=4+(12w-6)\xi-15w\xi^2.
\end{equation}

For $n\geq1$, we use the tangent to $g_a$ at the midpoint of
$[na,(n+1)a]$. It lies below $g_a$ because $g_a$ is convex for $t\geq a$.
Indeed $w>1/2$, and $0<\xi\leq(1+w+w^2)^{-1}$ on this range.
The quadratic in \eqref{eq:fan-curvature-second-derivative} is concave
in $\xi$, so it suffices to check its two endpoint values. At zero
its value is $4$; at the other endpoint, its sign is that of
$$
4w^4+20w^3+18w^2-w-2>18w^2-w-2\geq2.
$$
Integrating the tangent twice, with the endpoint lengths fixed,
gives a cubic upper bound for $l_a$. Averaging its values at
$(n+1/2)a\pm\theta$ gives \eqref{eq:reflected-mean-bound} with
$\kappa_n=g_a((n+1/2)a)$.

For the central interval $[0,a]$, a quadratic lower bound for $g_a$
is enough. On $0\leq\xi\leq1$, the same concave quadratic is at least
$-2-3w$. Since $0<g_a(t)\leq2\cosh a$, we obtain
$$
g_a''(t)\geq-2\cosh a(3\cosh^2a-1).
$$
Using $g_a'(0)=0$ and integrating twice gives
$$
g_a(t)\geq2\cosh a-\cosh a(3\cosh^2a-1)t^2.
$$
Integrate this quadratic twice, again matching the endpoint lengths.
The resulting quartic lies above $l_a$. Averaging at
$a/2-\theta$ and $a/2+\theta$ gives
$$
\mu_0\leq m_0-
\frac{a^2(1-\tau^2)}8
\left(2\cosh a-
\frac{a^2(7+\tau^2)}{24}\cosh a(3\cosh^2a-1)\right).
$$
Here $0\leq\tau\leq1$, so the expression in parentheses is at least
$$
2\cosh a-\frac{a^2}3\cosh a(3\cosh^2a-1)
>\frac{246}{100}
-\frac{49}{300}\frac{63}{50}
\left(3\left(\frac{63}{50}\right)^2-1\right)
>\frac{33}{20}.
$$
We used $123/100<\cosh a<63/50$ on $[2/3,7/10]$;
the endpoint comparisons are recorded in
\eqref{eq:fan-curvature-endpoint-checks}. This proves the bound for $\mu_0$.
\end{proof}

Substituting these bounds for the means and differences into the
polygon area gives a lower bound that is quadratic in the twist
parameter $\theta$. The next lemma records this bound.

\begin{lemma}[An area bound using pairs of fan lengths]
\label{lem:paired-fan-lower}
Suppose the mean bounds in Lemma~\ref{lem:reflected-fan-lengths} hold
with nonnegative constants $\kappa_0,\ldots,\kappa_M$. Set
$$
\widetilde m_n=m_n-\frac{\kappa_na^2}8,
$$
and assume $\widetilde m_n>0$. Define
\begin{equation}
\label{eq:paired-fan-base-area}
F(a)=\frac1{\widetilde m_0^2}+2\sum_{n=0}^{M-1}\frac1{\widetilde m_n\widetilde m_{n+1}}
+\frac1{a\widetilde m_M},
\end{equation}
\begin{equation}
\label{eq:paired-fan-quadratic-coefficients}
\begin{aligned}
N(a)={}&\frac{\kappa_0}{4\widetilde m_0^3}
+\frac14\sum_{n=0}^{M-1}
\frac{\kappa_n/\widetilde m_n+\kappa_{n+1}/\widetilde m_{n+1}}{\widetilde m_n\widetilde m_{n+1}}
+\frac{\kappa_M}{8a\widetilde m_M^2},\\
P(a)={}&\frac{(\Delta_0/a)^2}{m_0^4}\\
&+2\sum_{n=0}^{M-1}\left(
\frac{(\Delta_n/a)^2}{m_n^3m_{n+1}}
+\frac{(\Delta_n/a)(\Delta_{n+1}/a)}{m_n^2m_{n+1}^2}
+\frac{(\Delta_{n+1}/a)^2}{m_nm_{n+1}^3}\right)\\
&+\frac{(\Delta_M/a)^2}{a m_M^3}.
\end{aligned}
\end{equation}
Then the finite fan polygon, before further insertions and with end
vertices joined to $\pm\gamma/(2a)$, has area at least
$$
F(a)+a^2\tau^2 \bigl(P(a)-N(a)\bigr).
$$
If additionally $m_0<8/3$, one insertion in the central sector gives
$$
\A(X)>F(a)+a^2\tau^2 \bigl(P(a)-N(a)\bigr)+\frac1{1000}.
$$
\end{lemma}

\begin{proof}
For the terms paired at $\theta$ and $-\theta$, increasing the
mean length lowers the area, while increasing the half-difference
raises it. The possible loss gives $N(a)$; the guaranteed increase
gives $P(a)$.
By \eqref{eq:reflected-mean-bound}, the means satisfy
$\mu_n\leq \overline m_n:=\widetilde m_n+\kappa_na^2\tau^2/8$. For the central term, this bound and
$d_0\geq \Delta_0\tau$ give
$$
\frac1{\mu_0^2-d_0^2}
\geq\frac1{\overline m_0^2-\Delta_0^2\tau^2}
\geq\frac1{\widetilde m_0^2}-\frac{\kappa_0a^2}{4\widetilde m_0^3}\tau^2
+\frac{\Delta_0^2}{m_0^4}\tau^2.
$$
The last inequality uses the tangent bound for
$(\widetilde m_0+\kappa_0a^2\tau^2/8)^{-2}$ and the elementary
inequality $(1-t)^{-1}\geq1+t$ for $0\leq t<1$.

For the two side terms corresponding to $\theta$ and $-\theta$,
their sum decreases with either mean and increases with either
nonnegative half-difference. These facts follow
by differentiating the two reciprocal products. Replace the means by
$\overline m_n,\overline m_{n+1}$ and the differences by $\Delta_n\tau,\Delta_{n+1}\tau$.
The substituted lengths remain positive: in each pair, the smaller
one is at least the smaller actual length. For $x,y\in[0,1)$,
$$
\frac12\left(\frac1{(1-x)(1-y)}+
\frac1{(1+x)(1+y)}\right)\geq1+x^2+xy+y^2.
$$
Combining the fractions gives
$(1+xy)/((1-x^2)(1-y^2))$; apply $(1-t)^{-1}\geq1+t$ to both
reciprocal factors. Since $\overline m_n\leq m_n$, this gives the
three positive terms in $P$. The tangent inequality
$$
\frac1{\overline m_n\overline m_{n+1}}
\geq\frac1{\widetilde m_n\widetilde m_{n+1}}
-\frac{a^2}{8}\frac{\kappa_n/\widetilde m_n+\kappa_{n+1}/\widetilde m_{n+1}}{\widetilde m_n\widetilde m_{n+1}}\tau^2
$$
gives the corresponding negative terms. For the two edges ending at $\pm\gamma/(2a)$, use
$$
\frac{\overline m_M}{a(\overline m_M^2-\Delta_M^2\tau^2)}
\geq\frac1{a\widetilde m_M}-\frac{\kappa_Ma}{8\widetilde m_M^2}\tau^2
+\frac{\Delta_M^2}{a m_M^3}\tau^2.
$$
Adding these inequalities gives \eqref{eq:paired-fan-base-area}
and the coefficients in \eqref{eq:paired-fan-quadratic-coefficients},
proving the first assertion.

For the central insertion, \eqref{eq:central-fan-insertion-bound}
and $m_0<8/3$ give an added area greater than
$$
\frac{27}{128}e^{-16/3}>\frac1{1000}.
$$
The last elementary inequality follows from
$e^{2/3}<39/20$ and $(39/20)^8<3375/16$.
\end{proof}

\subsubsection{Applying the bound on two intervals}
It remains to apply Lemma~\ref{lem:paired-fan-lower} for
$3/5\leq a\leq7/10$. We use $M=4$ and divide this range at $a=2/3$,
where the stronger mean bound in Lemma~\ref{lem:reflected-fan-lengths} applies.
We will prove $P(a)>N(a)$, so the twist term is nonnegative, and
$F(a)+1/1000>893/1000$. We first choose the constants $\kappa_n$
for the bounds on the means, then prove these two inequalities
on each interval.
The following rows give $[L,U]$ and the constants $1000\kappa_n$:
\begin{equation}
\label{eq:fan-interval-curvature-constants}
\begin{array}{c|cc|rrrrr}
\text{row}&L&U&0&1&2&3&4\\\hline
1&3/5&2/3&1276&365&96&25&6\\
2&2/3&7/10&1650&621&153&37&9
\end{array}
\end{equation}
For row 1, $\kappa_n<g_U((n+1)U)$, so the first mean bound applies
throughout the interval by monotonicity of $a\mapsto g_a(ka)$.
For row 2 and $n\geq1$, use
$\kappa_n<g_U((n+1/2)U)$ and the stronger mean bound.
Its central bound is $\kappa_0=33/20$ from
Lemma~\ref{lem:reflected-fan-lengths}.

To bound $N$ above, we bound its length denominators below.
To bound $P$ below, we bound its lengths above and $\Delta_n/a$
below. The endpoint calculations in Appendix~\ref{app:fan-interval-checks}
give
\begin{equation}
\label{eq:paired-fan-coefficient-bounds}
\begin{array}{c|cc}
\text{row}&N(a)<&P(a)>\\\hline
1&9/200&59/1000\\
2&131/2000&667/10000
\end{array}
\end{equation}
Thus $P(a)>N(a)$ on both intervals. The same length bounds
give $m_0<8/3$, so the central insertion adds more than $1/1000$.

We now bound the remaining term $F(a)$ using affine upper bounds
for its lengths.
By Lemma~\ref{lem:joint-fan-convexity}, every
$\widetilde m_n(a)=m_n(a)-\kappa_na^2/8$ is convex on these intervals, since
$$
\widetilde m_n''(a)\geq\frac2{\sinh^2a}-\frac{\kappa_n}4>0.
$$
Let $\widehat m_n$ be the
affine function joining the rational upper bounds for
$\widetilde m_n(L)$ and $\widetilde m_n(U)$ specified in
Appendix~\ref{app:fan-interval-checks}.
Convexity of $\widetilde m_n$ gives
$\widetilde m_n\leq\widehat m_n$ throughout $[L,U]$. Hence
\begin{equation}
\label{eq:paired-fan-affine-area}
F(a)\geq Q(a):=
\frac1{\widehat m_0(a)^2}
+2\sum_{n=0}^3\frac1{\widehat m_n(a)\widehat m_{n+1}(a)}
+\frac1{a\widehat m_4(a)}.
\end{equation}
The function $Q$ is convex by
\eqref{eq:reciprocal-affine-convexity}. Its endpoint value and derivative satisfy
\begin{equation}
\label{eq:paired-fan-tangent-values}
\begin{array}{c|cc}
\text{row}&Q(U)>&Q'(U)<\\\hline
1&2231/2500&0\\
2&8929/10000&0
\end{array}
\end{equation}
The rational evaluations are given in Appendix~\ref{app:fan-interval-checks}.
Convexity gives $Q(a)\geq Q(U)+Q'(U)(a-U)\geq Q(U)$ for $a\in[L,U]$,
since both $Q'(U)$ and $a-U$ are nonpositive.

\begin{proof}[Proof of Theorem~\ref{thm:pure-cusp}]
For $a\leq3/5$, Proposition~\ref{prop:cusp-six-fifths} gives
$\A(X)>893/1000$.
For $3/5\leq a\leq2/3$, \eqref{eq:paired-fan-coefficient-bounds}
gives $P-N>0$, and we have $m_0<8/3$ as shown above.
Combining Lemma~\ref{lem:paired-fan-lower} with the first row of
\eqref{eq:paired-fan-tangent-values} gives
$$
\A(X)>\frac{2231}{2500}+\frac1{1000}
>\frac{893}{1000}.
$$
For $2/3\leq a\leq7/10$, the second row of
\eqref{eq:paired-fan-coefficient-bounds} also gives $P-N>0$.
The second row of \eqref{eq:paired-fan-tangent-values} and
Lemma~\ref{lem:paired-fan-lower} therefore give
$$
\A(X)>\frac{8929}{10000}+\frac1{1000}
>\frac{893}{1000}.
$$
This proves the claimed strict inequality for every $\sys(X)=2a\leq7/5$.
The modular upper bound is Proposition~\ref{prop:modular-upper-new}.
\end{proof}

\section{Proofs of the main theorems}
\label{sec:completion}

We first combine the regional comparisons into an explicit lower
bound, which proves the minimum theorem. We then establish the
global comparison using the local and cusp estimates.

\begin{proposition}[Explicit regional estimate]
\label{prop:regional-area}
Every nonmodular torus $X$ satisfies
$$
\A(X)-\A(\Xmod)>
\begin{cases}
p_2/80,&\text{if }s_X\geq7/5,\\
1/1000,&\text{if }s_X<7/5.
\end{cases}
$$
\end{proposition}

\begin{proof}
In region $C$, Theorem~\ref{thm:pure-cusp} and
Proposition~\ref{prop:modular-upper-new} give
$$
\A(X)-\A(\Xmod)>
\frac{893}{1000}-\frac{223}{250}
=\frac1{1000}.
$$
In region $B$, Lemma~\ref{lem:lambda-local-conversion} and Corollary~\ref{cor:local-no-grid} give
$\A(X)-\A(\Xmod)>27p_2/400>p_2/80$.

In region $A$, \eqref{eq:outside-disk-shape-lower} gives
$p_2>1/12800$. If $t\leq1/20$, the local polygon bound
in Proposition~\ref{prop:uniform-p6} gives
$$
\A(X)-\A(\Xmod)>
\frac7{80}p_2-\frac1{700000}
>\left(\frac7{80}-\frac{12800}{700000}\right)p_2
>\frac1{80}p_2.
$$
If $t>1/20$, then $p_2>3/800$, and
\eqref{eq:region-a-comparison} gives
$$
\A(X)-\A(\Xmod)>
\frac{19}{1000}p_2-\frac1{700000}
>\left(\frac{19}{1000}-\frac{800}{3\cdot700000}\right)p_2
>\frac1{80}p_2.
$$
These cases cover every nonmodular torus.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:global-area}]
For a nonmodular torus, $p_2>0$, so
Proposition~\ref{prop:regional-area} gives
$\A(X)>\A(\Xmod)$. Equality holds at the modular torus.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:quantitative-area}]
Proposition~\ref{prop:cusp-growth} proves the cusp expansion.
Near the modular point, Theorem~\ref{thm:pure-local} gives
$\A(X)-\A(\Xmod)=\kappa p_2+o(p_2)$ with $\kappa>0$,
so the required quotient tends to $\kappa$.
At the other end, the discriminant and
Proposition~\ref{prop:markoff-reduction} give
$$
\frac{2x}{\sqrt{x^2-4}}\leq y\leq\frac{x}{\sqrt{x-2}},
1\leq\frac zy\leq\frac x2.
$$
With $x=2\cosh(s_X/2)$, these imply
$y/x=(2/s_X)(1+O(s_X^2))$ and $z/y=1+O(s_X^2)$.
Since
$$
\log(y/x)\leq\sqrt{3p_2/2}\leq\log(z/x),
$$
we obtain $\sqrt{3p_2/2}=\log(2/s_X)+O(s_X^2)$.
The cusp expansion consequently gives
$$
\frac{\A(X)-\A(\Xmod)}
{p_2e^{\sqrt{3p_2/2}}/(1+p_2)^{3/2}}
\longrightarrow\frac1{\sqrt6}
\text{ as }s_X\to0.
$$
Finally, the displayed trace bounds make the reduced region with
$s_X\geq\varepsilon$ compact for each $\varepsilon>0$.
Outside a fixed small neighborhood of the modular point, continuity and
Theorem~\ref{thm:global-area} bound this quotient above and below
by positive constants. Together with the two limits, this proves
the uniform bounds.
\end{proof}
\appendix
\section*{Appendix}
\addcontentsline{toc}{section}{Appendix}
% Number appendix parts independently of Regions A, B, and C.
\renewcommand{\thesubsection}{A\arabic{subsection}}
\renewcommand{\theHsubsection}{appendix.\arabic{subsection}}
\renewcommand{\theHsubsubsection}{\theHsubsection.\arabic{subsubsection}}
\numberwithin{theorem}{subsection}
\renewcommand{\theHtheorem}{\theHsubsection.\arabic{theorem}}
\subsection{Elementary bounds for series remainders}
\label{app:series-estimates}

The numerical comparisons use rational arithmetic and explicit
bounds for the remaining series terms. We collect those bounds here.
Appendix~\ref{app:taylor-coefficients} applies them to lengths and
Taylor coefficients; Appendix~\ref{app:numerical-checks} records the
remaining evaluations used in the main text.

For a nonnegative integer $N$ and $0\leq t<N+2$,
\begin{equation}
\label{eq:exp-remainder}
0\leq e^t-\sum_{j=0}^{N}\frac{t^j}{j!}
\leq\frac{t^{N+1}}{(N+1)!}\frac1{1-t/(N+2)}.
\end{equation}
Indeed, after the first omitted term, the ratio of successive terms
is at most $t/(N+2)$, so their sum is bounded by a geometric series.

The same argument applied to the even terms gives, for real $t$ with
$t^2<(2N+3)(2N+4)$,
\begin{equation}
\label{eq:cosh-remainder}
0\leq\cosh t-\sum_{j=0}^{N}\frac{t^{2j}}{(2j)!}
\leq\frac{t^{2N+2}}{(2N+2)!}
\frac1{1-t^2/((2N+3)(2N+4))}.
\end{equation}
For example, $N=8$ gives $\cosh(118/125)<37/25$, the inequality
used to bound $L_0$ in Lemma~\ref{lem:local-bounds}.

For the fan endpoint calculations, it is convenient to use fixed
polynomials. Put
$$
\begin{aligned}
C(x)&=\sum_{j=0}^{12}\frac{x^{2j}}{(2j)!},&
S(x)&=\sum_{j=0}^{12}\frac{x^{2j+1}}{(2j+1)!},\\
C^+(x)&=C(x)+\frac{2x^{26}}{26!},&
S^+(x)&=S(x)+\frac{2x^{27}}{27!}.
\end{aligned}
$$
Then, for $0\leq x\leq5$,
\begin{equation}
\label{eq:fan-series-bounds}
C(x)\leq\cosh x\leq C^+(x),
S(x)\leq\sinh x\leq S^+(x).
\end{equation}
The ratios after the first omitted terms are at most
$25/(27\cdot28)$ and $25/(28\cdot29)$, respectively. Both are
less than $1/2$, so twice the first omitted term bounds each remainder.

For logarithms, let $v\geq1$ and put $u=(v-1)/(v+1)$. Then
$0\leq u<1$, and
\begin{equation}
\label{eq:log-remainder}
0\leq\log v-2\sum_{j=0}^{N}\frac{u^{2j+1}}{2j+1}
\leq\frac{2u^{2N+3}}{(2N+3)(1-u^2)}.
\end{equation}
This follows from the positive series
$\log v=2\sum_{j\geq0}u^{2j+1}/(2j+1)$ by bounding the omitted
denominators below by $2N+3$ and summing the remaining geometric
series. For an arbitrary positive argument, first write it as
$2^k v$ with $1\leq v<2$ and use the same bound at $v$ and at $2$.
All finite sums and remainder bounds are rational when their
arguments are rational.

\subsection{Length bounds and Taylor coefficients of the polygon area}
\label{app:taylor-coefficients}

The polygon calculations use $f(T)=1/L(T)$, where
$L(T)=2\arccosh(T/2)$. We give the modular length bounds and their
refinements, an explicit evaluation of the quadratic term, the
recursions and two-line calculation through degree eight, and the
coefficient table used in the degree-nineteen remainder estimate.

\paragraph{The modular length bounds.}
For $t\in\{3,6,15,39,87,102,267\}$, enclose $\sqrt{t^2-4}$
between consecutive multiples of $10^{-6}$ by squaring.
Substitute these bounds into $L(t)=2\log((t+\sqrt{t^2-4})/2)$
and apply \eqref{eq:log-remainder} with $N=11$, after reducing
its arguments to $[1,2]$ by powers of two. The resulting rational
bounds give the numerical comparisons in
Proposition~\ref{prop:modular-upper-new} and
Proposition~\ref{prop:polygon-taylor-remainder}.

\paragraph{Rational input bounds.}
Put $S_T=\sqrt{T^2-4}$. The following integers satisfy
$$
\frac{R_T}{10^{20}}<S_T<\frac{R_T+1}{10^{20}},
\quad
\frac{J_T}{10^{20}}<L(T)<\frac{J_T+1}{10^{20}}.
$$
\begin{center}
\small
\begin{tabular}{r|r|r}
$T$&$R_T$&$J_T$\\
\hline
3&223606797749978969640&192484730023841378999\\
6&565685424949238019520&352549434807817210093\\
15&1486606874731850552261&540715166186280463466\\
39&3894868418830089310004&732580706919993661986\\
87&8697700845625814003528&893155194923016687644\\
102&10198039027185569660056&924975336509101048101\\
267&26699250925821870330699&1117446926139105944726\\
\end{tabular}
\end{center}
The square-root bounds follow by checking
$R_T^2<10^{40}(T^2-4)<(R_T+1)^2$.
For the length bounds, refine the square-root intervals as in
\eqref{eq:coefficient-root-brackets} below and apply
\eqref{eq:log-remainder} with $N=99$ to $(T+S_T)/2$, first dividing
by a power of two to reduce the argument to $[1,2]$.
Bound $\log2$ by the same formula.

For the Taylor calculations through degree nineteen, use the same
procedure with $D=10^{100}$ and
$$
K_T=\max\{k\in\Z_{\geq0}:k^2\leq D^2(T^2-4)\}.
$$
Then
\begin{equation}
\label{eq:coefficient-root-brackets}
K_T/D\leq S_T\leq(K_T+1)/D.
\end{equation}
Apply \eqref{eq:log-remainder}, again with $N=99$, to these endpoints
to bound $L(T)$. Every subsequent operation uses interval arithmetic:
addition adds endpoints, multiplication takes the minimum and maximum
of the four endpoint products, and
$[a,b]^{-1}=[1/b,1/a]$ when $0\notin[a,b]$.
After an operation, an interval $[a,b]$ may be replaced by
$$
\left[\frac{\lfloor Da\rfloor}{D},
\frac{\lceil Db\rceil}{D}\right].
$$
Thus all endpoints are rational and every rounding is outward.

\paragraph{The quadratic coefficient as six contributions.}
\begin{lemma}
\label{lem:explicit-quadratic-coefficient}
The coefficient in \eqref{eq:degree-eight-expansion} satisfies
$$
c_2>\frac{23}{200}.
$$
\end{lemma}
\begin{proof}
Along $(u,v,w)=(t,0,-t)$ the initial traces are
$$
x(t)=e^{3t}+e^t+e^{-t},\quad
y(t)=e^{2t}+1+e^{-2t},\quad
z(t)=e^t+e^{-t}+e^{-3t}.
$$
Their values, first derivatives, and second derivatives at zero are
$(3,3,3)$, $(3,0,-3)$, and $(11,8,11)$.
For each of the six permutations, differentiate the trace recurrence:
$$
\begin{aligned}
T_{n+1}'&=x'T_n+xT_n'-T_{n-1}',\\
T_{n+1}''&=x''T_n+2x'T_n'+xT_n''-T_{n-1}'',\\
T_*'&=T_1'T_2+T_1T_2'-x',\\
T_*''&=T_1''T_2+2T_1'T_2'+T_1T_2''-x''.
\end{aligned}
$$
All values in these formulas are taken at $t=0$.
For a modular trace $T$, write
$$
A_T=\frac12\sum_{\sigma\in\mathfrak S_3}T_\sigma'(0)^2,
\quad
B_T=\frac12\sum_{\sigma\in\mathfrak S_3}T_\sigma''(0),
$$
where $T_\sigma$ denotes the corresponding trace after permutation.
For a consecutive pair put
$D_{T,U}=\sum_{\sigma\in\mathfrak S_3}T_\sigma'(0)U_\sigma'(0)$.
The derivative recurrences give
$$
\begin{array}{c|rrrrrrr}
T&3&6&15&39&87&102&267\\
\hline
A_T&18&288&4446&57258&409014&648288&6705198\\
B_T&30&132&618&2622&7746&10236&37554
\end{array}
$$
and the six values of $D_{T,U}$ displayed below.
Differentiating $f(T)=1/L(T)$ gives
$$
f'(T)=-\frac{2}{S_TL(T)^2},
\quad
f''(T)=\frac{2T}{S_T^3L(T)^2}
+\frac{8}{S_T^2L(T)^3}.
$$
The contribution of the pair $(T,U)$, summed over all permutations,
to the coefficient of $t^2$ is therefore
$$
\begin{aligned}
Q_{T,U}={}&A_Tf''(T)f(U)+A_Uf(T)f''(U)
+D_{T,U}f'(T)f'(U)\\
&+B_Tf'(T)f(U)+B_Uf(T)f'(U).
\end{aligned}
$$
Substituting the rational input intervals above gives the following
strict bounds. The final two columns bound $10^9Q_{T,U}$.
$$
\begin{array}{c|r|r|r}
(T,U)&D_{T,U}&\text{lower}&\text{upper}\\
\hline
(3,267)&21078&144338437&144338438\\
(6,87)&21096&31985371&31985372\\
(15,39)&31338&16852085&16852086\\
(15,87)&84222&12863676&12863677\\
(39,102)&383112&12782608&12782609\\
(102,267)&4160232&11679178&11679179
\end{array}
$$
If $r_2$ is the coefficient of $t^2$ in
$\A_6(t,0,-t)-\A_6(0,0,0)$, then $r_2=2c_2$ and
$$
r_2=\sum_{(T,U)}Q_{T,U}
>\frac{230501355}{10^9}
>\frac{23}{100}.
$$
Dividing by two proves the claim.
\end{proof}

\paragraph{Taylor coefficient recursions.}
Suppose the trace series is real analytic near $t=0$ and has the form
$$
x(t)=\sum_{j\geq0}x_jt^j,
x_0>2.
$$
Set $s(x)=\sqrt{x^2-4}$. We compute the square root, its reciprocal,
the length, and finally the reciprocal length in that order.

First expand the square root and its reciprocal:
$$
s(x(t))=\sum_{j\geq0}s_jt^j,
\frac1{s(x(t))}=\sum_{j\geq0}v_jt^j.
$$
Their constant terms are $s_0=\sqrt{x_0^2-4}$ and $v_0=1/s_0$.
Squaring the first series and multiplying the two series give, for $j\geq1$,
\begin{equation}
\label{eq:square-root-coefficients}
s_j=
\frac{\displaystyle\sum_{i=0}^jx_ix_{j-i}
-\displaystyle\sum_{i=1}^{j-1}s_is_{j-i}}{2s_0},
v_j=-\frac1{s_0}\sum_{i=1}^js_iv_{j-i}.
\end{equation}
Each right-hand side uses only coefficients already computed.

Next write
$$
L(x(t))=\sum_{j\geq0}l_jt^j,
f(x(t))=\sum_{j\geq0}f_jt^j,
$$
with $l_0=L(x_0)$ and $f_0=1/l_0$.
The chain rule
$$
\frac{d}{dt}L(x(t))=\frac{2x'(t)}{s(x(t))}
$$
and the identity $L(x(t))f(x(t))=1$ give
\begin{equation}
\label{eq:reciprocal-length-coefficients}
l_j=\frac2j\sum_{i=1}^ji x_iv_{j-i},
f_j=-\frac1{l_0}\sum_{i=1}^jl_if_{j-i},
j\geq1.
\end{equation}
Thus the trace coefficients determine the reciprocal-length coefficients
successively, using only finite sums and division by the positive
constants $s_0,l_0$.


\paragraph{The coefficients through degree eight.}
\begin{lemma}
\label{lem:explicit-low-coefficients}
The quadratic coefficient satisfies
$$
\frac{115250679305}{10^{12}}<c_2
<\frac{115250679306}{10^{12}}.
$$
The remaining coefficients through degree eight satisfy
\begin{equation}
\label{eq:degree-eight-coefficients}
\begin{array}{c|rrrrrrrr}
j&3&4&5&6'&6''&7&8'&8''\\
\hline
1000c_j>&-694&-162&812&-314&6354&270&1118&-15077\\
1000c_j<&-692&-161&815&-312&6367&277&1128&-14964.
\end{array}
\end{equation}
\end{lemma}
\begin{proof}
Let $r_j$ and $q_j$ denote the coefficients of $t^j$ in
$$
\A_6(t,0,-t)-\A_6(0,0,0),
\quad
\A_6(t,t,-2t)-\A_6(0,0,0),
$$
respectively. For the first line, use the initial trace formulas in
Lemma~\ref{lem:explicit-quadratic-coefficient}. For the second, they are
$$
x(t)=y(t)=2e^{3t}+e^{-3t},
\quad
z(t)=2+e^{-6t}.
$$
Their Taylor coefficients are rational: replace each $e^{at}$ by
$\sum_{j=0}^8a^jt^j/j!$, discarding terms of degree greater than eight
after each multiplication. For every permutation, apply the trace
recurrence, then \eqref{eq:square-root-coefficients} and
\eqref{eq:reciprocal-length-coefficients}, and sum the six products
in \eqref{eq:height-six-polygon-area}.
This gives the following explicit intervals. Each entry is a strict
lower bound for $10^{12}r_j$ or $10^{12}q_j$; adding one gives a
strict upper bound. The odd $r_j$ vanish because the first restriction
is even.
\begin{center}
\small
\begin{tabular}{r|r|r}
$j$&$10^{12}r_j$&$10^{12}q_j$\\
\hline
2&230501358611&691504075834\\
3&--&4158193745227\\
4&-646178134908&-5815603214170\\
5&--&-29291413002270\\
6&-2503246530490&161369799006231\\
7&--&-59139571788055\\
8&17964933816381&-1789267475604971\\
\end{tabular}
\end{center}
The conversion to the invariant coefficients is
$$
\begin{aligned}
c_2&=r_2/2,&c_3&=-q_3/6,&c_4&=r_4/4,\\
c_5&=-q_5/36,&c_{6'}&=r_6/8,
&c_{6''}&=(q_6-27r_6)/36,\\
c_7&=-q_7/216,&c_{8'}&=r_8/16,
&c_{8''}&=(q_8-81r_8)/216.
\end{aligned}
$$
In particular, the first row gives
$$
\frac{230501358611}{2\cdot10^{12}}<c_2
<\frac{230501358612}{2\cdot10^{12}}.
$$
For the two coefficients obtained by subtraction, the same table gives
$$
\begin{aligned}
\frac{228957455329434}{36\cdot10^{12}}
&<c_{6''}<\frac{228957455329462}{36\cdot10^{12}},\\
-\frac{3244427114731913}{216\cdot10^{12}}
&<c_{8''}<-\frac{3244427114731831}{216\cdot10^{12}}.
\end{aligned}
$$
These bounds and the remaining divisions give every entry of
\eqref{eq:degree-eight-coefficients}.
\end{proof}

\paragraph{Coefficients of degrees nine through nineteen.}
The following table supplies the bounds used in
Theorem~\ref{thm:global-taylor-remainder}. Its computation uses the
recursions above and the four substitutions of Lemma~\ref{lem:four-directions}.

\begin{lemma}[Explicit higher-coefficient bounds]
\label{lem:higher-coefficients}
For $9\leq n\leq19$, each table entry is a strict lower bound for
$1000c_{n,k}$; adding one gives a strict upper bound.
A dash means that the coefficient does not occur.
\begin{center}
\footnotesize\normalfont
\setlength{\tabcolsep}{4pt}
\begin{tabular}{r|rrrr}
$n$&$k=0$&$k=1$&$k=2$&$k=3$\\
\hline
9&2407&-37499&--&--\\
10&-1373&14063&--&--\\
11&-13233&135121&--&--\\
12&985&-39829&286516&--\\
13&22984&-201453&--&--\\
14&-1948&185062&-1374518&--\\
15&-20400&405076&-1909443&--\\
16&6289&-450043&3030012&--\\
17&31287&-1786116&11542319&--\\
18&-12548&721576&-6517347&14598450\\
19&-111754&5301317&-32106332&--\\
\end{tabular}
\end{center}
\end{lemma}
\begin{proof}
Use the finer square-root bounds \eqref{eq:coefficient-root-brackets}
and the length bounds derived from them. For each ordered permutation of
the initial traces, generate $T_0,\ldots,T_5$ by
\eqref{eq:sector-trace-recursion}, set $T_*=T_1T_2-x$, and sum the
six contributions in
\eqref{eq:height-six-polygon-area}.

On a ray $t(1,-s,s-1)$ the initial traces are
$$
\begin{aligned}
x(t)&=e^{3t}+e^{(1-2s)t}+e^{(2s-1)t},\\
y(t)&=e^{(2-s)t}+e^{-3st}+e^{(s-2)t},\\
z(t)&=e^{(1+s)t}+e^{-(1+s)t}+e^{3(s-1)t}.
\end{aligned}
$$
At the four rational values of $s$ in Lemma~\ref{lem:four-directions},
their Taylor coefficients are rational. Products are evaluated by
$$
(ab)_j=\sum_{i=0}^j a_i b_{j-i},
0\leq j\leq19.
$$
The trace recurrence, followed by the reciprocal-length recursions
\eqref{eq:square-root-coefficients}--\eqref{eq:reciprocal-length-coefficients},
therefore bounds each $h_n(s)$ above and below.
With $m,\epsilon,d$ as in Lemma~\ref{lem:four-directions} and
$s_j=(j+1)/10$, Lagrange interpolation gives $c_{n,k}$ as the
coefficient of $q^k$ in
\begin{equation}
\label{eq:invariant-coefficient-interpolation}
\sum_{j=0}^{d}
\frac{h_n(s_j)}{p_2(s_j)^m p_3(s_j)^\epsilon}
\prod_{\substack{0\leq l\leq d\\l\ne j}}
\frac{q-q(s_l)}{q(s_j)-q(s_l)}.
\end{equation}
Substituting the bounds for $h_n(s_j)$ in this formula gives the table.
All divisions are by the positive constant terms of the square-root
and length series or by the nonzero rational interpolation differences.
Thus the table follows from the stated rational bounds by keeping
lower and upper bounds at every arithmetic step.
\end{proof}

\subsection{Verification of the numerical bounds}
\label{app:numerical-checks}

We collect the endpoint calculations used in the modular-area,
complex-length, and short-systole estimates. Each calculation uses
the rational series bounds in Appendix~\ref{app:series-estimates}.

\subsubsection{The omitted area at the modular torus}
\label{app:modular-insertion-checks}

In \eqref{eq:modular-extra-insertions}, the area added by parent
traces $a,b$ and child trace $c$ is
$$
\frac1{L(a)L(c)}+\frac1{L(b)L(c)}-\frac1{L(a)L(b)}.
$$
To check the stated upper bounds, bracket each square root in $L(T)$
between consecutive multiples of $10^{-20}$ and apply
\eqref{eq:log-remainder} with $N=24$, after rescaling its argument
into $[1,2]$. In the added-area formula, use lower length bounds
in the two positive terms and upper length bounds in the
subtracted term. These are rational comparisons.

\subsubsection{The complex length bounds}
\label{app:complex-length-checks}

To check \eqref{eq:complex-length-lower-bounds}, compare the corresponding trace bound in
\eqref{eq:complex-height-trace-bounds} with $2\cosh(l_h/2)$.
Each comparison follows from \eqref{eq:cosh-remainder} with $N=8$:
$$
\cosh s<\sum_{j=0}^{8}\frac{s^{2j}}{(2j)!}
+\frac{s^{18}}{18!}\frac1{1-s^2/(19\cdot20)}(0<s\leq3/2).
$$

\subsubsection{The fan bound for systole at most 6/5}
\label{app:cusp-six-fifths-checks}

We verify the endpoint bounds and tangent-line values used in
Proposition~\ref{prop:cusp-six-fifths}. Here $L=1/3$, $U=3/5$, and
$c_n(a)=2\operatorname{arccosh}(\coth a\cosh(na))$.

The following integer entries are strict upper bounds for
$10000c_n(a)$ at the two endpoints:
$$
\begin{array}{c|rrrrrr}
a&0&1&2&3&4&5\\\hline
1/3&36020&37169&40357&45013&50578&56660\\
3/5&24668&28584&37715&48822&60551&72469
\end{array}
$$
To check an entry $r$ in column $n$, use the polynomial bounds
\eqref{eq:fan-series-bounds}. In every case, rational arithmetic gives
$$
S(a)C(r/20000)-C^+(a)C^+(na)>\frac1{100000}>0.
$$
Thus $\sinh a\cosh(r/20000)>\cosh a\cosh(na)$, which is equivalent
to the stated upper bound for $c_n(a)$.

Averaging consecutive entries and dividing by $10000$ gives
the endpoint bounds for $m_n=(c_n+c_{n+1})/2$ in
\eqref{eq:cusp-six-fifths-affine-bounds}.

For the function $Q$ in that proof, take $a_*=279/500$.
All $b_n(a_*)$ and $b_n'$ are rational, and
$2b_0(a_*)=22556831/4000000$. Use \eqref{eq:exp-remainder} with
this argument and $N=20$, then take reciprocals to bound
$e^{-2b_0(a_*)}$ above and below. Substitution in $Q$ and its derivative proves
\eqref{eq:cusp-six-fifths-tangent-values}.

\subsubsection{The fan bounds on the two remaining intervals}
\label{app:fan-interval-checks}

As in \eqref{eq:fan-interval-curvature-constants}, rows 1 and 2
refer to $[L,U]=[3/5,2/3]$ and $[2/3,7/10]$, respectively.
We verify the curvature constants in
\eqref{eq:fan-interval-curvature-constants}, the length and difference
bounds in \eqref{eq:fan-interval-arrays}, and the tangent-line values
in \eqref{eq:paired-fan-tangent-values}.

To bound $N$ above we need lower bounds for its length denominators.
To bound $P$ below we need upper bounds for the lengths and lower
bounds for $\Delta_n/a$. The following table gives all three
bounds throughout each interval:
$$
\widetilde m_n>\frac{\lambda_n}{1000},
m_n<\frac{\eta_n}{1000},
\frac{\Delta_n}a>\frac{\pi_n}{1000}.
$$
\begin{equation}
\label{eq:fan-interval-arrays}
\begin{array}{c|c|rrrrr}
\text{row}&\text{array}&0&1&2&3&4\\\hline
1&\lambda&2440&3270&4300&5440&6630\\
1&\eta&2663&3315&4472&5762&7084\\
1&\pi&326&760&925&977&993\\\hline
2&\lambda&2347&3259&4458&5756&7080\\
2&\eta&2514&3303&4556&5920&7311\\
2&\pi&366&810&949&986&996
\end{array}
\end{equation}
We verify these bounds and the curvature constants using the polynomials
$C,S,C^+,S^+$ from \eqref{eq:fan-series-bounds} on $[0,5]$.
The integers in this table mean $v_n/10000<c_n(a)<(v_n+1)/10000$:
\begin{equation}
\label{eq:fan-endpoint-length-table}
\begin{array}{c|rrrrrr}
a&v_0&v_1&v_2&v_3&v_4&v_5\\\hline
3/5&24667&28583&37714&48821&60550&72468\\
2/3&22694&27581&38382&51040&64195&77482\\
7/10&21790&27204&38845&52261&66118&80083
\end{array}
\end{equation}
For every entry the two required rational inequalities are
$$
C^+(v_n/20000)S^+(a)<C(a)C(na),
$$
$$
C((v_n+1)/20000)S(a)>C^+(a)C^+(na).
$$
Since $\cosh$ is increasing on $[0,\infty)$, these inequalities
prove the lower and upper bounds for $c_n(a)$.
At argument zero use its exact values $\cosh0=1$, $\sinh0=0$.
The bounds for $g_a=l_a''$ follow by substituting $C,S,C^+,S^+$
in \eqref{eq:fan-curvature-function} and squaring positive quantities.
The same polynomials give the remaining comparisons used above:
\begin{equation}
\label{eq:fan-curvature-endpoint-checks}
\begin{gathered}
S(2/3)^2>\frac12,
C(2/3)>\frac{123}{100},
C^+(7/10)<\frac{63}{50},\\
C^+(2/3)+S^+(2/3)<\frac{39}{20}.
\end{gathered}
\end{equation}

Since $\widetilde m_n$ is convex, its tangent at $U$ gives a lower bound on $[L,U]$.
Explicit upper bounds for $\widetilde m_n'(U)$ are
$$
\begin{array}{c|rrrrr}
\text{row}&0&1&2&3&4\\\hline
1&0&0&12/5&47/10&67/10\\
2&0&14/100&261/100&481/100&688/100
\end{array}
$$
They follow from
$$
c_n'(a)=
\frac{2(-\cosh(na)/\sinh a+n\cosh a\sinh(na))}
{\sqrt{1+\cosh^2a\sinh^2(na)}},
\widetilde m_n'=\frac{c_n'+c_{n+1}'}2-\frac{\kappa_na}4.
$$
If the corresponding slope bound is $s_n\geq0$, the length bounds give
$$
\widetilde m_n(a)>
\frac{v_n(U)+v_{n+1}(U)}{20000}
-\frac{\kappa_nU^2}8-s_n(U-L)
>\frac{\lambda_n}{1000}.
$$
Convexity of $m_n$ bounds it above by the larger endpoint value, giving
$\eta_n$. The monotonicity of $\Delta_n/a$ gives $\pi_n$ from the two
strict length bounds at $L$. For example, in row 2,
$$
\frac{\Delta_1(2/3)}{2/3}
>\frac{(38382-27582)/20000}{2/3}
=\frac{81}{100}.
$$

Substitute \eqref{eq:fan-interval-curvature-constants} and
\eqref{eq:fan-interval-arrays} into
\eqref{eq:paired-fan-quadratic-coefficients}.
Specifically use $\lambda_n/1000$ and $a\geq L$ to bound $N$ above,
and $\eta_n/1000$, $\pi_n/1000$ and $a\leq U$ to bound $P$ below.
Evaluating these rational expressions gives
\eqref{eq:paired-fan-coefficient-bounds}. The bounds
$m_0<2663/1000$ in row 1 and $m_0<2514/1000$ in row 2 also
give $m_0<8/3$ on both intervals.

Finally, we specify the affine upper bounds used in
\eqref{eq:paired-fan-affine-area}. Using
\eqref{eq:fan-endpoint-length-table}, define $\widehat m_n$ by
$$
\widehat m_n(L)=\frac{v_n(L)+v_{n+1}(L)+2}{20000}-\frac{\kappa_nL^2}8,
$$
$$
\widehat m_n(U)=\frac{v_n(U)+v_{n+1}(U)+2}{20000}-\frac{\kappa_nU^2}8.
$$
These bound $\widetilde m_n$ at both endpoints. The resulting
values at $U$ and slopes are
$$
\begin{array}{c|r|r|r|r}
&\multicolumn{2}{c|}{\text{row 1}}&\multicolumn{2}{c}{\text{row 2}}\\
n&\widehat m_n(U)&\widehat m_n'&\widehat m_n(U)&\widehat m_n'\\\hline
0&439733/180000&-145997/60000&187899/80000&-17627/8000\\
1&118007/36000&-7399/24000&2611611/800000&1833/80000\\
2&16747/3750&43001/20000&3636823/800000&199989/80000\\
3&1036883/180000&105457/24000&4733427/800000&1130323/240000\\
4&425017/60000&64933/10000&5847679/800000&542757/80000
\end{array}
$$
Substituting these rational numbers in \eqref{eq:paired-fan-affine-area}
and its derivative gives both rows of \eqref{eq:paired-fan-tangent-values}.

\small
\begin{thebibliography}{MR95b}

\bibitem[ABC+21]{AndersenEtAlKontsevich}
J. E. Andersen, G. Borot, S. Charbonnier, A. Giacchetto,
D. Lewa\'nski, and C. Wheeler,
On the Kontsevich geometry of the combinatorial Teichm\"uller space,
arXiv:2010.11806, version 3, 2021,
\url{https://arxiv.org/abs/2010.11806v3}.

\bibitem[AA26]{AougabAthreya}
T. Aougab and J. S. Athreya,
Currents with corners and counting weighted triangulations,
\emph{J. Lond. Math. Soc.} \textbf{114} (2026), e70675,
\url{https://doi.org/10.1112/jlms.70675}.

\bibitem[AHA20]{AranaHerreraAthreya}
F. Arana-Herrera and J. S. Athreya,
Square-integrability of the Mirzakhani function and statistics of simple
closed geodesics on hyperbolic surfaces,
\emph{Forum Math. Sigma} \textbf{8} (2020), e9, 33 pp.,
\url{https://doi.org/10.1017/fms.2019.49}.

\bibitem[Bel23]{Bell}
N. Bell,
Counting arcs on hyperbolic surfaces,
\emph{Groups Geom. Dyn.} \textbf{17} (2023), no. 2, 459--478,
\url{https://doi.org/10.4171/GGD/705}.

\bibitem[Bon18]{Bonahon2018}
F. Bonahon,
\emph{The Mirzakhani function on the moduli space of a surface},
recorded lecture at the 2018 Fields Medal Symposium: Maryam Mirzakhani,
Fields Institute, Toronto, November 7, 2018,
reporting joint work with Sabrina Enriquez,
\url{https://www.fields.utoronto.ca/talks/Mirzakhani-function-moduli-space-surface}.

% \bibitem[BCD+23]{BorotEtAlUnitBall}
% G. Borot, S. Charbonnier, V. Delecroix, A. Giacchetto, and C. Wheeler,
% Around the combinatorial unit ball of measured foliations on bordered surfaces,
% \emph{Int. Math. Res. Not. IMRN} \textbf{2023} (2023), no. 17,
% 14464--14514,
% \url{https://doi.org/10.1093/imrn/rnac231}.

\bibitem[Bow98]{Bowditch}
B. H. Bowditch,
Markoff triples and quasifuchsian groups,
\emph{Proc. Lond. Math. Soc.} (3) \textbf{77} (1998), no. 3, 697--736,
\url{https://doi.org/10.1112/S0024611598000604}.


\bibitem[Cas59]{Cassels}
J. W. S. Cassels,
\emph{An Introduction to the Geometry of Numbers},
Grundlehren der mathematischen Wissenschaften, vol. 99,
Springer-Verlag, Berlin, 1959.

\bibitem[DGZZ21]{DelecroixGoujardZografZorich}
V. Delecroix, \'E. Goujard, P. Zograf, and A. Zorich,
Masur--Veech volumes, frequencies of simple closed geodesics, and
intersection numbers of moduli spaces of curves,
\emph{Duke Math. J.} \textbf{170} (2021), no. 12, 2633--2718,
\url{https://doi.org/10.1215/00127094-2021-0054}.

\bibitem[DLN26]{DLN}
N. M. Doan, X. Li, and V. Nguyen,
McShane--Rivin norm balls and simple-length multiplicities,
arXiv:2605.14574, version 2, 2026,
\url{https://arxiv.org/abs/2605.14574v2}.

\bibitem[ES16]{ErlandssonSouto}
V. Erlandsson and J. Souto,
Counting curves in hyperbolic surfaces,
\emph{Geom. Funct. Anal.} \textbf{26} (2016), no. 3, 729--777,
\url{https://doi.org/10.1007/s00039-016-0374-7}.

\bibitem[ES22]{ErlandssonSoutoBook}
V. Erlandsson and J. Souto,
\emph{Mirzakhani's Curve Counting and Geodesic Currents},
Progress in Mathematics \textbf{345}, Birkh\"auser, Cham, 2022,
\url{https://doi.org/10.1007/978-3-031-08705-9}.

\bibitem[Fed74]{Federer1974}
H. Federer,
Real flat chains, cochains and variational problems,
\emph{Indiana Univ. Math. J.} \textbf{24} (1974/75), 351--407.

\bibitem[FM12]{FarbMargalit}
B. Farb and D. Margalit,
\emph{A Primer on Mapping Class Groups},
Princeton Mathematical Series \textbf{49}, Princeton University Press,
Princeton, 2012.

\bibitem[HT14]{HuTan}
H. Hu and S. P. Tan,
\emph{New identities for small hyperbolic surfaces},
Bull. Lond. Math. Soc. \textbf{46} (2014), no.~5, 1021--1031.

\bibitem[Gol03]{GoldmanOneHoled}
W. M. Goldman,
The modular group action on real $\mathrm{SL}(2)$-characters of a one-holed torus,
\emph{Geom. Topol.} \textbf{7} (2003), 443--486,
\url{https://doi.org/10.2140/gt.2003.7.443}.

\bibitem[Gol09]{GoldmanTrace}
W. M. Goldman,
Trace coordinates on Fricke spaces of some simple hyperbolic surfaces,
in \emph{Handbook of Teichm\"uller Theory}, Vol. II,
IRMA Lect. Math. Theor. Phys. \textbf{13}, European Mathematical Society,
Z\"urich, 2009, 611--684,
\url{https://doi.org/10.4171/055-1/16}.

\bibitem[Gro83]{Gromov1983}
M. Gromov,
Filling Riemannian manifolds,
\emph{J. Differential Geom.} \textbf{18} (1983), no. 1, 1--147,
\url{https://doi.org/10.4310/jdg/1214509283}.

\bibitem[LMN26]{LoweMarquesNeves}
B. Lowe, F. C. Marques, and A. Neves,
Counting minimal Lagrangians via Mirzakhani functions,
arXiv:2605.04614, version 3, 2026,
\url{https://arxiv.org/abs/2605.04614v3}.

\bibitem[Ma09]{MaHigherChain}
T.-W. Ma,
Higher chain formula proved by combinatorics,
\emph{Electron. J. Combin.} \textbf{16} (2009), no. 1, N21, 7 pp.,
\url{https://doi.org/10.37236/259}.

\bibitem[Mac95]{Macdonald}
I. G. Macdonald,
\emph{Symmetric Functions and Hall Polynomials}, second edition,
Oxford Mathematical Monographs, Clarendon Press, Oxford, 1995.

\bibitem[Mir07]{MirzakhaniRandom}
M. Mirzakhani,
Random hyperbolic surfaces and measured laminations,
in \emph{In the Tradition of Ahlfors--Bers, IV},
Contemp. Math. \textbf{432}, American Mathematical Society,
Providence, RI, 2007, 179--198.

\bibitem[Mir08a]{MirzakhaniGrowth}
M. Mirzakhani,
Growth of the number of simple closed geodesics on hyperbolic surfaces,
\emph{Ann. of Math.} (2) \textbf{168} (2008), no. 1, 97--125,
\url{https://doi.org/10.4007/annals.2008.168.97}.

\bibitem[Mir08b]{MirzakhaniEarthquake}
M. Mirzakhani,
Ergodic theory of the earthquake flow,
\emph{Int. Math. Res. Not. IMRN} \textbf{2008} (2008),
Art. ID rnm116, 39 pp.,
\url{https://doi.org/10.1093/imrn/rnm116}.


\bibitem[Mir16]{MirzakhaniOrbits}
M. Mirzakhani,
Counting mapping class group orbits on hyperbolic surfaces,
arXiv:1601.03342, 2016,
\url{https://arxiv.org/abs/1601.03342}.

\bibitem[MR95a]{McShaneRivinNorm}
G. McShane and I. Rivin,
A norm on homology of surfaces and counting simple geodesics,
\emph{Internat. Math. Res. Notices} \textbf{1995} (1995), no. 2, 61--69,
\url{https://doi.org/10.1155/S1073792895000055}.

\bibitem[MR95b]{McShaneRivinTori}
G. McShane and I. Rivin,
Simple curves on hyperbolic tori,
\emph{C. R. Acad. Sci. Paris S\'er. I Math.} \textbf{320} (1995), no. 12,
1523--1528; electronic text corresponding to the published article:
arXiv:math/0005220.

\bibitem[O'S26]{OSullivan2026}
C. O'Sullivan,
Patterns in the Markov numbers and their generalizations,
arXiv:2609.14149, 2026,
\url{https://arxiv.org/abs/2609.14149}.

\bibitem[Pal79]{Palais1979}
R. S. Palais,
The principle of symmetric criticality,
\emph{Communications in Mathematical Physics}
\textbf{69} (1979), no. 1, 19--30.

\bibitem[Par14]{ParlierSurvey}
H. Parlier,
Simple closed geodesics and the study of Teichm\"uller spaces,
in \emph{Handbook of Teichm\"uller Theory}, Vol. IV,
IRMA Lect. Math. Theor. Phys. \textbf{19}, European Mathematical Society,
Z\"urich, 2014, 113--134,
\url{https://doi.org/10.4171/117-1/3}.

\bibitem[RS19]{RafiSouto}
K. Rafi and J. Souto,
Geodesic currents and counting problems,
\emph{Geom. Funct. Anal.} \textbf{29} (2019), 871--889,
\url{https://doi.org/10.1007/s00039-019-00502-7}.

\bibitem[Roc70]{Rockafellar}
R. T. Rockafellar,
\emph{Convex Analysis},
Princeton Mathematical Series \textbf{28}, Princeton University Press,
Princeton, 1970.

\bibitem[Rud76]{Rudin}
W. Rudin,
\emph{Principles of Mathematical Analysis}, third edition,
McGraw--Hill, New York, 1976.

\bibitem[Sas25]{Sasaki}
D. Sasaki,
Counting subgroups via Mirzakhani's curve counting,
arXiv:2409.08109, version 3, 2025,
\url{https://arxiv.org/abs/2409.08109v3}.

\bibitem[Sch93]{Schmutz}
P. Schmutz,
\emph{Riemann surfaces with shortest geodesic of maximal length},
Geometric and Functional Analysis \textbf{3} (1993), 564--631.

\bibitem[Sch14]{Schneider}
R. Schneider,
\emph{Convex Bodies: The Brunn--Minkowski Theory}, second expanded edition,
Encyclopedia of Mathematics and its Applications \textbf{151},
Cambridge University Press, Cambridge, 2014.

\bibitem[Ser85]{Series}
C. Series,
The geometry of Markoff numbers,
\emph{Math. Intelligencer} \textbf{7} (1985), no. 3, 20--29.

\bibitem[SS03a]{SteinShakarchiFourier}
E. M. Stein and R. Shakarchi,
\emph{Fourier Analysis: An Introduction},
Princeton Lectures in Analysis, vol. I,
Princeton University Press, Princeton, 2003.

\bibitem[SS03b]{SteinShakarchi}
E. M. Stein and R. Shakarchi,
\emph{Complex Analysis},
Princeton Lectures in Analysis, vol. II,
Princeton University Press, Princeton, 2003.

\bibitem[Thu88]{Thurston1988}
W. P. Thurston,
On the geometry and dynamics of diffeomorphisms of surfaces,
\emph{Bull. Amer. Math. Soc.} (N.S.) \textbf{19} (1988), no. 2,
417--431.

\bibitem[Tri24]{Trin}
M. Trin,
Counting arcs of the same type,
\emph{Int. Math. Res. Not. IMRN} \textbf{2024} (2024),
no. 16, 11749--11765,
\url{https://doi.org/10.1093/imrn/rnae143}.

\bibitem[Zag82]{Zagier1982}
D. Zagier,
On the number of Markoff numbers below a given bound,
\emph{Math. Comp.} \textbf{39} (1982), no. 160, 709--723,
\url{https://doi.org/10.1090/S0025-5718-1982-0669663-7}.

\end{thebibliography}

\bigskip
\noindent
\begin{minipage}{\textwidth}
\small
\textsc{Phuc Thinh Dang}\par
Vietnam National University, Ho Chi Minh City, Vietnam\par
\textit{Email address:} 
\href{mailto:dpthinhmaths@gmail.com}{\nolinkurl{dpthinhmaths@gmail.com}}
\end{minipage}

\medskip
\noindent
\begin{minipage}{\textwidth}
\small
\textsc{Trong Toan Dao}\par
National University of Singapore, Singapore \par
\textit{Email address:}
\href{mailto:daotrongtoan.dtt4@gmail.com}{\nolinkurl{daotrongtoan.dtt4@gmail.com}} and \href{mailto:daotrongtoan@u.nus.edu}{\nolinkurl{daotrongtoan@u.nus.edu}}
\end{minipage}

\medskip
\noindent
\begin{minipage}{\textwidth}
\small
\textsc{Nhat Minh Doan}\par
Institute of Mathematics, Vietnam Academy of Science and Technology,
Hanoi, Vietnam\par
\textit{Email address:}
\href{mailto:dnminh@math.ac.vn}{\nolinkurl{dnminh@math.ac.vn}}\par
\textit{URL:}
\href{https://sites.google.com/view/dnminh}{\nolinkurl{https://sites.google.com/view/dnminh}}
\end{minipage}

\end{document}