\section{The family \texorpdfstring{$(1,1,1,m_4)$}{(1,1,1,m4)}}\label{sec:family}

In this section $m=(1,1,1,m_4)$ with $m_4>0$. The point $\mu^{(4)}(m)=-(\log m_4)(1,1,1)$ lies on the axis $\rho=0$, where Theorem~\ref{thm:count} gives $c_4$ directly. As the permutations of the bodies $1$, $2$, $3$ preserve $m$, $c_1=c_2=c_3$. For body~$1$ inside, the point $\mu^{(1)}(m)=(0,0,\log m_4)$ lies on a mirror of the target of $F_1$. Theorem~\ref{thm:count} determines $c_1$ once the position of this point relative to the surface of critical values is known. Certified computations as in Section~\ref{sec:examples} decide it for one $m_4$ at a time, but for all $m_4$ at once we use the symmetry. The count $c_1$ depends on the concave CCs with body~$1$ inside that are symmetric under the exchange of the bodies $2$ and $3$. These CCs depend on one parameter, which reduces the question to functions of one variable. Theorem~\ref{thm:family} shows that they form one analytic arc, along which the mass has exactly one critical point and the Jacobian of $F_1$ has exactly two zeros. Corollary~\ref{cor:rs} gives the counts for all $m_4$, which are those of Table~\ref{tab:rs}.

\subsection{Setting}\label{sec:famsetting}
We take the outer bodies of $F_1$ in the order $2,3,4$. Thus $F_1$ is defined on an open set $\Omega_1$ of spoke vectors $a=(a_2,a_3,a_4)$, $a_j=|x_j-x_1|$. Here $F_1(a)=(\mu_2,\mu_3,\mu_4)$ with $\mu_j=\log(m_j/m_1)$. After a relabelling of the bodies, $F_1$ is the map $F$ of Section~\ref{sec:setup}, to which the results of Sections~\ref{sec:setup}--\ref{sec:thm1} apply. On $\Omega_1$ we use the coordinates $(s,x,y)$ of Section~\ref{sec:coords}, and on the target the coordinates $(A,P,Q)$. In this section $J$ is the Jacobian determinant of $F_1$, and $\Db=\{J\le0\}\subset\Omega_1$.

The transposition $\sigma$ of the bodies $2$ and $3$ acts on $\Omega_1$ by $(s,x,y)\mapsto(s,-x,y)$ and on the target by $(A,P,Q)\mapsto(A,-P,Q)$. We have $F_1\circ\sigma=\sigma\circ F_1$ and $J\circ\sigma=J$. The fixed points of $\sigma$ form the mirror
\[
\Omega_1^\sigma=\{a\in\Omega_1:\ a_2=a_3\}=\Omega_1\cap\{x=0\}.
\]
At a point of $\Omega_1^\sigma$, exchanging the labels $2$ and $3$ maps the solution of Lemma~\ref{lem:angles} to a solution with the same spokes. By uniqueness the configuration is symmetric under the reflection that exchanges the bodies $2$ and $3$. Conversely, a configuration with this symmetry has $a_2=a_3$. Thus the points of $\Omega_1^\sigma$ correspond to the symmetric concave CCs with body~$1$ inside, and $m_2=m_3$ there. As $A$ and $Q$ are even in $x$ and $P$ is odd, the derivatives $\partial_xA$, $\partial_xQ$, $\partial_sP$ and $\partial_yP$ vanish on $\Omega_1^\sigma$. The determinant of $\partial(A,P,Q)/\partial(s,x,y)$ is $J$ (proof of Lemma~\ref{lem:axis}(d)). Expanding it on $\Omega_1^\sigma$, we find
\begin{equation}\label{eq:Jsplit}
J=J_{\mathrm{sym}}\cdot\frac{\partial P}{\partial x},\qquad J_{\mathrm{sym}}=\det\frac{\partial(A,Q)}{\partial(s,y)} .
\end{equation}
We call $\partial(A,Q)/\partial(s,y)$ the \emph{symmetric block}.

In this section we do not normalize $\lambda'$. Instead we place the bodies $2$ and $3$ at $(\mp1,0)$. We write $S_{ij}=r_{ij}^{-3}-\lambda'$. Equation \eqref{eq:ccnorm} shows that a configuration is a CC with multiplier $\lambda'$ if and only if
\[
\mathsf E_j:=\sum_{i\ne j}m_iS_{ij}(x_i-x_j)=0,\qquad j=1,\dots,4 .
\]
Scaling a configuration by $t>0$ multiplies $\lambda'$ by $t^{-3}$. For $0<b<h$ let $x(b,h)$ be the configuration with the bodies $2,3$ at $(\mp1,0)$, body~$1$ at $(0,b)$ and body~$4$ at $(0,h)$. With $\hat p=\sqrt{1+b^2}$, $\hat q=\sqrt{1+h^2}$ and $d=h-b$ we have $r_{12}=r_{13}=\hat p$, $r_{23}=2$, $r_{24}=r_{34}=\hat q$ and $r_{14}=d$. Let
\begin{equation}\label{eq:famG}
\begin{gathered}
\mathsf G(b,h)=2h\hat q^{-3}+d^{-2}-\frac h4-d\hat p^{-3},\qquad
\mathsf L(b,h)=\frac{h/4+d\hat p^{-3}}{2h+d},\\
\mathsf M(b,h)=-\frac{b\,(\hat p^{-3}-\mathsf L)}{h\,(\hat q^{-3}-\mathsf L)} .
\end{gathered}
\end{equation}
For $m_4,\lambda'>0$ let $\mathsf H=(\mathsf H_{ij})$ be the symmetric $2\times2$ matrix with
\begin{equation}\label{eq:famA}
\begin{aligned}
\mathsf H_{11}&=3\lambda'm_4-2m_4\hat q^{-3}+6m_4\hat q^{-5}-m_4 d^{-3},\qquad \mathsf H_{12}=-6hm_4\hat q^{-5},\\
\mathsf H_{22}&=2\lambda'm_4+6\lambda'+6b^2\hat p^{-5}+6h^2m_4\hat q^{-5}-2m_4\hat q^{-3}-\tfrac12-2\hat p^{-3},
\end{aligned}
\end{equation}
and let $\mathsf D(b,h)=\det\mathsf H=\mathsf H_{11}\mathsf H_{22}-\mathsf H_{12}^2$ with $m_4=\mathsf M(b,h)$ and $\lambda'=\mathsf L(b,h)$. Lemma~\ref{lem:fampitch} below shows that $\mathsf H$ is the Hessian of the potential on two antisymmetric displacements of $x(b,h)$. We call $\mathsf D$ the \emph{pitchfork determinant}.

\begin{theorem}\label{thm:family}
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2em]
\item\label{F:arc} \emph{The arc.} The set $\{\mathsf G=0,\ 0<b<h\}$ is the graph of an analytic function $b(h)$ on $(h_c,\infty)$ ($h_c=1.1394\ldots$). On it $\mathsf M>0$ exactly for $h\in(h_c,h_1)$ ($h_1=2.9373\ldots$). Every concave CC of the masses $(1,1,1,m_4)$ with body~$1$ inside that is symmetric under the reflection exchanging the bodies $2$ and $3$ is, up to similarity including reflection, the configuration $x(b,h)$ of exactly one point $(b,h)$ of the arc, and $\mathsf M(b,h)=m_4$ there. Conversely, for every point of the arc with $\mathsf M>0$ the configuration $x(b,h)$ is such a CC of the masses $(1,1,1,\mathsf M(b,h))$, with $\lambda'=\mathsf L(b,h)>0$.
\item\label{F:J} \emph{Monotonicity and degenerate points.} For $h_c<h<h_1$ write $\mathsf M(h)=\mathsf M(b(h),h)$, $\mathsf D(h)=\mathsf D(b(h),h)$, and $a(h)\in\Omega_1^\sigma$ for the spoke vector of $x(b(h),h)$ scaled to $\lambda'=1$. The function $\mathsf M(h)$ increases strictly from $0$ to $m_f$ on $(h_c,h_f]$ and decreases strictly to $0$ on $[h_f,h_1)$. The Jacobian $J$ vanishes at $a(h)$ exactly when $h\in\{h^{**},h_f\}$, where $h^{**}<\sqrt3<h_f$. In \eqref{eq:Jsplit}, $J_{\mathrm{sym}}(a(h))$ is the product of $d\mathsf M/dh$ and a continuous nonvanishing function of $h$, and $\partial P/\partial x\,(a(h))$ is the product of $\mathsf D(h)$ and a continuous nonvanishing function of $h$. These two factors vanish only at $h_f$, respectively $h^{**}$, and change sign there. At $h=\sqrt3$ the configuration is the equilateral triangle with body~$1$ at its centre and $m_4=1$, where $J<0$. Hence $J(a(h))>0$ for $h\in(h_c,h^{**})\cup(h_f,h_1)$, and $J(a(h))<0$ for $h\in(h^{**},h_f)$.
\item\label{F:val} \emph{Values.} The masses $m^{**}=\mathsf M(h^{**})$ and $m_f=\mathsf M(h_f)$ satisfy
\begin{align*}
m^{**}&=0.991842274390940915543496944364409029094559042\pm4\cdot10^{-46},\\
m_f&=1.00266054757261000068580350218611540512511700\pm5\cdot10^{-46}.
\end{align*}
In particular $m^*<m^{**}<1<m_f<(m^*)^{-3}$.
\end{enumerate}
\end{theorem}

The proofs in Sections~\ref{sec:famarc} and~\ref{sec:famdeg} are by hand. They reduce Theorem~\ref{thm:family} to statements about the zeros of two functions of $h$, namely $d\mathsf M/dh$ and $\mathsf D(h)$, and to the bounds in~\ref{F:val}. Section~\ref{sec:famcert} proves these by exact computation and ball arithmetic.

\subsection{The arc}\label{sec:famarc}

\begin{lemma}\label{lem:famcc}
Let $0<b<h$ and $m_4>0$.
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em]
\item The configuration $x(b,h)$ is a CC of the masses $(1,1,1,m_4)$ if and only if $\mathsf G(b,h)=0$ and $m_4=\mathsf M(b,h)$. The multiplier of this CC is then $\lambda'=\mathsf L(b,h)$, and $\mathsf L>0$.
\item If $\mathsf G(b,h)=0$, then
\[
\hat q^{-3}-\mathsf L=\frac{d\,(\hat q^{-3}-d^{-3})}{2h+d}<0,\qquad \mathsf M(b,h)=-\frac{2b\,(\hat p^{-3}-\frac18)}{(2h+d)(\hat q^{-3}-\mathsf L)} .
\]
In particular $\mathsf M$ has no pole on $\{\mathsf G=0\}$, and it has the sign of $\sqrt3-b$ there.
\item Every concave CC with body~$1$ inside that is symmetric under the reflection exchanging the bodies $2$ and $3$ is similar to $x(b,h)$ for exactly one pair $0<b<h$, by a similarity that may reverse the orientation.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The reflection $(u,v)\mapsto(-u,v)$ maps $x(b,h)$ to itself with the bodies $2$ and $3$ exchanged. Hence, for the masses $(1,1,1,m_4)$, the first components of $\mathsf E_1$ and $\mathsf E_4$ vanish, and $\mathsf E_3$ is the mirror image of $\mathsf E_2$. The remaining components are
\[
\mathsf E_2=\bigl(2S_{23}+S_{12}+m_4 S_{24},\ bS_{12}+m_4 hS_{24}\bigr),\quad \mathsf E_{1,2}=-2bS_{12}+m_4 dS_{14},\quad \mathsf E_{4,2}=-dS_{14}-2hS_{24},
\]
where $\mathsf E_{j,2}$ denotes the second component of $\mathsf E_j$. The sum $\sum_jm_j\mathsf E_j$ vanishes identically: its terms $m_im_jS_{ij}(x_i-x_j)$ are antisymmetric in $(i,j)$. The second component of this identity, $\mathsf E_{1,2}+2\mathsf E_{2,2}+m_4 \mathsf E_{4,2}=0$, shows that $x(b,h)$ is a CC if and only if $\mathsf E_2=0$ and $\mathsf E_{4,2}=0$. As $h\ne0$, this holds if and only if $\mathsf E_{2,2}=0$ and
\[
\text{(I)}\quad h\mathsf E_{2,1}-\mathsf E_{2,2}=2hS_{23}+dS_{12}=0,\qquad\qquad \text{(II)}\quad -\mathsf E_{4,2}=2hS_{24}+dS_{14}=0 .
\]
Both equations are linear in $\lambda'$. Equation (I) holds if and only if $\lambda'=\mathsf L(b,h)$. Equation (II) holds if and only if $\lambda'=(2h\hat q^{-3}+d^{-2})/(2h+d)$. Together they hold if and only if $\lambda'=\mathsf L$ and
\[
\mathsf G=(2h+d)\Bigl(\frac{2h\hat q^{-3}+d^{-2}}{2h+d}-\mathsf L\Bigr)=0 .
\]
At such a point $S_{24}\ne0$. Otherwise (II) gives $S_{14}=0$ and then $d^{-3}=\lambda'=\hat q^{-3}$, which is impossible for $d<h<\hat q$. Then $\mathsf E_{2,2}=0$ if and only if $m_4=-bS_{12}/(hS_{24})=\mathsf M(b,h)$. Finally $\mathsf L>0$: its numerator and its denominator are positive.

(b) On $\{\mathsf G=0\}$ the two expressions for $\lambda'$ in (a) agree. Then $\hat q^{-3}-\mathsf L=\hat q^{-3}-(2h\hat q^{-3}+d^{-2})/(2h+d)=d(\hat q^{-3}-d^{-3})/(2h+d)$. This is negative because $d<\hat q$. The formula for $\mathsf M$ follows from $\hat p^{-3}-\mathsf L=2h(\hat p^{-3}-\frac18)/(2h+d)$. The denominator $(2h+d)(\hat q^{-3}-\mathsf L)$ of this formula is negative; $\hat p^{-3}-\frac18$ has the sign of $2-\hat p$, that is, of $\sqrt3-b$.

(c) Let $x$ be such a CC. A similarity maps the bodies $2$ and $3$ to $(\mp1,0)$. The reflection that exchanges them becomes the reflection in the vertical axis. The bodies $1$ and $4$ are fixed by it and lie on this axis, at $(0,b)$ and $(0,h)$. As body~$1$ is strictly inside the triangle of the other three, $h\ne0$. Since $c_1$ counts up to similarity including reflection, we may compose with the reflection $(u,v)\mapsto(u,-v)$ if necessary. Then $h>0$ and $0<b<h$. The only similarities that fix $(-1,0)$ and $(1,0)$ are the identity and this reflection. This makes the pair $(b,h)$ with $h>0$ unique.
\end{proof}

\begin{lemma}\label{lem:famarc}
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em]
\item For $0<b<h$, $\partial\mathsf G/\partial b=2d^{-3}+\hat p^{-3}+3bd\hat p^{-5}>0$, and $\mathsf G(b,h)\to+\infty$ as $b\to h^-$. The function $g_0(h)=\mathsf G(0,h)=2h(1+h^2)^{-3/2}+h^{-2}-\frac54h$ decreases strictly from $+\infty$ to $-\infty$ on $(0,\infty)$. Let $h_c$ be its zero. Then $h_c<\sqrt3$, and $\{\mathsf G=0,\ 0<b<h\}$ is the graph of an analytic function $b(h)$ on $(h_c,\infty)$ with $b(h)\to0$ as $h\to h_c$.
\item There is exactly one $h_1>h_c$ with $b(h_1)=\sqrt3$. Moreover $b(h)<\sqrt3$ for $h_c<h<h_1$, and $b(h)>\sqrt3$ for $h>h_1$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The formula for $\partial\mathsf G/\partial b$ follows by direct computation; its three terms are positive. As $b\to h^-$, the term $d^{-2}$ tends to $+\infty$ while the other terms stay bounded. Next,
\[
g_0'(h)=2(1-2h^2)(1+h^2)^{-5/2}-2h^{-3}-\tfrac54 .
\]
For $h\ge1/\sqrt2$ the first term is at most $0$. For $0<h<1/\sqrt2$ it is less than $2$, while $-2h^{-3}<-4\sqrt2$. In both cases $g_0'(h)<0$. Moreover $g_0\to+\infty$ as $h\to0^+$, $g_0\to-\infty$ as $h\to\infty$, and $g_0(\sqrt3)=\frac13-\sqrt3<0$. Hence $g_0$ has exactly one zero $h_c$, $h_c<\sqrt3$, and $g_0<0$ exactly on $(h_c,\infty)$. For fixed $h>0$ the function $\mathsf G(\cdot,h)$ increases strictly on $[0,h)$ from $g_0(h)$ to $+\infty$. It has a zero in $(0,h)$ if and only if $g_0(h)<0$, that is, $h>h_c$; the zero $b(h)$ is then unique. It is analytic in $h$ by the implicit function theorem. Given $\epsilon>0$, we have $\mathsf G(\epsilon,h_c)>g_0(h_c)=0$. Continuity gives $\mathsf G(\epsilon,h)>0$ for $h$ close to $h_c$, so $b(h)<\epsilon$ for these $h$.

(b) For $h>\sqrt3$ put $g_1(h)=\mathsf G(\sqrt3,h)=2h\hat q^{-3}+(h-\sqrt3)^{-2}-\frac h4-\frac18(h-\sqrt3)$. Then
\[
g_1'(h)=2(1-2h^2)(1+h^2)^{-5/2}-2(h-\sqrt3)^{-3}-\tfrac38<0,
\]
because each term is negative for $h>\sqrt3$. Since $g_1\to+\infty$ for $h\to\sqrt3^+$ and $g_1\to-\infty$ for $h\to\infty$, the function $g_1$ has exactly one zero $h_1>\sqrt3$, and $g_1>0$ on $(\sqrt3,h_1)$, $g_1<0$ on $(h_1,\infty)$. As $\mathsf G$ increases in $b$, for $h>\sqrt3$ we have $b(h)<\sqrt3$, $b(h)=\sqrt3$ or $b(h)>\sqrt3$ according as $g_1(h)>0$, $g_1(h)=0$ or $g_1(h)<0$. For $h_c<h\le\sqrt3$ we have $b(h)<h\le\sqrt3$. With $h_c<\sqrt3<h_1$, this proves (b).
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:family}\ref{F:arc}]
The first statement is Lemma~\ref{lem:famarc}(a). Lemmas~\ref{lem:famcc}(b) and~\ref{lem:famarc}(b) show that $\mathsf M>0$ on the arc exactly for $h_c<h<h_1$. Let $x$ be a symmetric concave CC of the masses $(1,1,1,m_4)$ with body~$1$ inside. Lemma~\ref{lem:famcc}(c) shows that it is similar to $x(b,h)$ for exactly one pair $0<b<h$. By Lemma~\ref{lem:famcc}(a), $\mathsf G(b,h)=0$ and $\mathsf M(b,h)=m_4$. Conversely, let $(b,h)$ be a point of the arc with $\mathsf M(b,h)>0$. Then $x(b,h)$ is a CC of the masses $(1,1,1,\mathsf M(b,h))$ with $\lambda'=\mathsf L(b,h)>0$, by Lemma~\ref{lem:famcc}(a). It is symmetric. Body~$1$ lies strictly inside the triangle of the others because $0<b<h$.
\end{proof}

Scaled by $\mathsf L^{1/3}$, the configuration $x(b(h),h)$ has $\lambda'=1$. By Lemma~\ref{lem:dziobek} the point $a(h)=\mathsf L^{1/3}(\hat p,\hat p,d)$, evaluated at $(b(h),h)$, lies in $\Omega_1^\sigma$. It depends analytically on $h\in(h_c,h_1)$, and $F_1(a(h))=(0,0,\log\mathsf M(h))$.

\subsection{The degenerate points of the arc}\label{sec:famdeg}

\begin{lemma}\label{lem:famfold}
For $h\in(h_c,h_1)$, $h\ne\sqrt3$, we have $J_{\mathrm{sym}}(a(h))=\chi_{\mathrm s}(h)\,\mathsf M'(h)$, where $\mathsf M'=d\mathsf M/dh$ and $\chi_{\mathrm s}$ is continuous and nonvanishing on $(h_c,\sqrt3)$ and on $(\sqrt3,h_1)$.
\end{lemma}

\begin{proof}
Let $0<b<h$ with $h\ne\sqrt3$, and let $\alpha,\nu>0$. The computation in the proof of Lemma~\ref{lem:famcc}(a), with the masses $(1,\alpha,\alpha,\nu)$ in place of $(1,1,1,m_4)$, shows that $x(b,h)$ is a CC of these masses with multiplier $\lambda'$ if and only if
\[
2\alpha hS_{23}+dS_{12}=0,\qquad 2\alpha hS_{24}+dS_{14}=0,\qquad bS_{12}+\nu hS_{24}=0 .
\]
The first two equations give $\lambda'=\lambda_I(\alpha):=(\alpha h/4+d\hat p^{-3})/(2\alpha h+d)$ and $\lambda'=(2\alpha h\hat q^{-3}+d^{-2})/(2\alpha h+d)$. These agree if and only if $\alpha\,e(h)=d^{-2}-d\hat p^{-3}$, where $e(h)=h/4-2h\hat q^{-3}$. As $e(h)=0$ only for $\hat q=2$, that is, $h=\sqrt3$, this means $\alpha=\tilde\alpha(b,h):=1+\mathsf G(b,h)/e(h)$. Then $\lambda'=\tilde\lambda(b,h):=\lambda_I(\tilde\alpha(b,h))$. If $\hat q^{-3}\ne\tilde\lambda$, the third equation determines $\nu=\tilde\nu(b,h):=\mathsf N(b,h,\tilde\lambda(b,h))$ with $\mathsf N(b,h,\lambda')=-b(\hat p^{-3}-\lambda')/(h(\hat q^{-3}-\lambda'))$.

Fix $h_0\in(h_c,h_1)\setminus\{\sqrt3\}$ and let $b_0=b(h_0)$. At $(b_0,h_0)$ we have $\mathsf G=0$, $\tilde\alpha=1$, $\tilde\lambda=\mathsf L>0$, $\hat q^{-3}-\tilde\lambda<0$ and $\tilde\nu=\mathsf M>0$, by Lemma~\ref{lem:famcc}. Hence on a neighbourhood $U$ of $(b_0,h_0)$ these functions are analytic and $\tilde\alpha$, $\tilde\lambda$ and $\tilde\nu$ are positive. There $x(b,h)$ is a concave CC with body~$1$ inside of the masses $(1,\tilde\alpha,\tilde\alpha,\tilde\nu)$ and with multiplier $\tilde\lambda$. By the computation above, these are the only masses of the form $(1,\alpha,\alpha,\nu)$, and $\tilde\lambda$ is the only multiplier, for which $x(b,h)$ is a CC. Scaling by $\tilde\lambda^{1/3}$ gives the point
\[
\Upsilon(b,h)=\tilde\lambda(b,h)^{1/3}\,(\hat p,\hat p,d)\in\Omega_1^\sigma,\qquad F_1(\Upsilon(b,h))=(\log\tilde\alpha,\log\tilde\alpha,\log\tilde\nu).
\]
Conversely, by Lemma~\ref{lem:famcc}(c) a point $a$ of $\Omega_1^\sigma$ is similar to $x(b,h)$ for a unique pair $(b,h)$. This pair depends analytically on $a$ through the solution of Lemma~\ref{lem:angles} and lies in $U$ if $a$ is close to $a(h_0)$. Then $x(b,h)$ is a CC for the masses of $a$. These masses have $m_2=m_3$, because $F_1\circ\sigma=\sigma\circ F_1$ and $\sigma a=a$. By the uniqueness of the masses and of the multiplier noted above, these masses are $(1,\tilde\alpha,\tilde\alpha,\tilde\nu)$, the multiplier of $x(b,h)$ is $\tilde\lambda$, and $a=\Upsilon(b,h)$. Thus $\Upsilon$ maps a neighbourhood of $(b_0,h_0)$ bijectively onto a neighbourhood of $a(h_0)$ in $\Omega_1^\sigma$; the inverse map is analytic. The Jacobian determinant $k(b,h)$ of $\Upsilon$ in the coordinates $(s,y)$ of $\Omega_1^\sigma$ does not vanish.

At $\Upsilon(b,h)$ we have $A=(2\log\tilde\alpha+\log\tilde\nu)/3$ and $Q=2(\log\tilde\alpha-\log\tilde\nu)/\sqrt6$; the linear map $(u,w)\mapsto((2u+w)/3,\,2(u-w)/\sqrt6)$ has determinant $-2/\sqrt6$. By the chain rule,
\[
J_{\mathrm{sym}}(\Upsilon(b,h))\,k(b,h)=-\frac{2}{\sqrt6}\,\det\frac{\partial(\log\tilde\alpha,\log\tilde\nu)}{\partial(b,h)} .
\]
On the arc we have $\mathsf G=0$, hence $\tilde\alpha=1$ and $d\tilde\alpha=d\mathsf G/e$. The difference $\tilde\lambda-\mathsf L=\lambda_I(1+\mathsf G/e)-\lambda_I(1)$ is $\mathsf G$ times an analytic function. So is $\tilde\nu-\mathsf M=\mathsf N(b,h,\tilde\lambda)-\mathsf N(b,h,\mathsf L)$. It follows that on the arc $\tilde\nu=\mathsf M$ and $d\tilde\nu=d\mathsf M+f\,d\mathsf G$ for an analytic function $f$, and
\[
\det\frac{\partial(\log\tilde\alpha,\log\tilde\nu)}{\partial(b,h)}=\frac{\mathsf G_b\mathsf M_h-\mathsf G_h\mathsf M_b}{e\,\mathsf M}=\frac{\mathsf G_b}{e\,\mathsf M}\,\mathsf M'(h),
\]
using $\mathsf M'(h)=\mathsf M_h+\mathsf M_b\,b'(h)=\mathsf M_h-\mathsf M_b\mathsf G_h/\mathsf G_b$. Thus $J_{\mathrm{sym}}(a(h))=\chi_{\mathrm s}(h)\mathsf M'(h)$ with $\chi_{\mathrm s}=-2\mathsf G_b/(\sqrt6\,e\,\mathsf M\,k)$, evaluated at $(b(h),h)$. Since the formula for $\Upsilon$ defines $k$ as an analytic function near each point of the arc with $h\ne\sqrt3$, the function $\chi_{\mathrm s}$ is continuous on $(h_c,\sqrt3)$ and on $(\sqrt3,h_1)$. It does not vanish because $\mathsf G_b>0$, $e\ne0$, $\mathsf M>0$ and $k\ne0$.
\end{proof}

\begin{lemma}\label{lem:fampitch}
For $h\in(h_c,h_1)$ we have $\partial P/\partial x\,(a(h))=\chi_{\mathrm a}(h)\,\mathsf D(h)$, where $\chi_{\mathrm a}$ is continuous and nonvanishing on $(h_c,h_1)$.
\end{lemma}

\begin{proof}
For masses $m$ and $\lambda'>0$ let $V_{m,\lambda'}(x)=\sum_{i<j}m_im_j\bigl(r_{ij}^{-1}+\frac12\lambda'r_{ij}^2\bigr)$ on $(\R^2)^4$. Then $\partial V_{m,\lambda'}/\partial x_j=m_j\mathsf E_j$: the CCs with multiplier $\lambda'$ are the critical points of $V_{m,\lambda'}$. For $\phi(r)=r^{-1}+\frac12\lambda'r^2$ we have $\phi''(r)-\phi'(r)/r=3r^{-3}$ and $\phi'(r)/r=-(r^{-3}-\lambda')$. These give
\begin{equation}\label{eq:famhess}
\operatorname{Hess}V_{m,\lambda'}[\xi,\xi]=\sum_{i<j}m_im_j\Bigl(3r_{ij}^{-5}\bigl((x_i-x_j)\cdot(\xi_i-\xi_j)\bigr)^2-S_{ij}\,|\xi_i-\xi_j|^2\Bigr),\qquad \xi\in(\R^2)^4 .
\end{equation}
Fix $h\in(h_c,h_1)$, and let $b=b(h)$, $m_4=\mathsf M(h)$, $\lambda'=\mathsf L(b,h)$ and $x^0=x(b,h)$, a critical point of $V_{m,\lambda'}$ for $m=(1,1,1,m_4)$. The reflection $(u,v)\mapsto(-u,v)$ combined with the exchange of the bodies $2$ and $3$ is an orthogonal map $\sigma_*$ of $(\R^2)^4$ that fixes $x^0$ and $V_{m,\lambda'}$; its $(-1)$-eigenspace, the space of antisymmetric displacements, consists of the $\xi$ with $\xi_1=(\xi_{1,1},0)$, $\xi_4=(\xi_{4,1},0)$, $\xi_2=(u,v)$ and $\xi_3=(u,-v)$. It contains the displacement $\mathbf u$, in which body~$4$ moves by $(1,0)$ and the others are fixed, the displacement $\mathbf w$, in which the bodies $2$ and $3$ move by $(0,1)$ and $(0,-1)$ and the others are fixed, the translation $\mathbf t$, in which all bodies move by $(1,0)$, and the rotation $\mathbf r$, $\mathbf r_j=(-x^0_{j,2},x^0_{j,1})$. Since $b\ne0$, these four vectors form a basis. They depend continuously on $h$. Since $V_{m,\lambda'}$ is invariant under translations and rotations and $x^0$ is a critical point, $\operatorname{Hess}V_{m,\lambda'}(x^0)\,\mathbf t=\operatorname{Hess}V_{m,\lambda'}(x^0)\,\mathbf r=0$. A direct computation with \eqref{eq:famhess}, in which only the pairs that contain body~$4$ contribute to the entries with $\mathbf u$, shows that $\operatorname{Hess}V_{m,\lambda'}(x^0)[\mathbf u,\mathbf u]=\mathsf H_{11}$, $\operatorname{Hess}V_{m,\lambda'}(x^0)[\mathbf u,\mathbf w]=\mathsf H_{12}$ and $\operatorname{Hess}V_{m,\lambda'}(x^0)[\mathbf w,\mathbf w]=\mathsf H_{22}$, with $\mathsf H$ as in \eqref{eq:famA}.

Next we differentiate along $\partial/\partial x$. For small $\tau$ let $X^\tau$ be the configuration of the point $a(h)+\tau E_1$ of $\Omega_1$, placed with body~$1$ at the origin, body~$4$ on the positive vertical half-axis and body~$2$ to its left, then scaled by $\mathsf L^{-1/3}$ and translated so that $X^0=x^0$. It is a CC with multiplier $\mathsf L$ for the masses $m(\tau)=(1,e^{F_1(a(h)+\tau E_1)})$: $\nabla V_{m(\tau),\lambda'}(X^\tau)=0$. Differentiating at $\tau=0$ gives
\[
\operatorname{Hess}V_{m,\lambda'}(x^0)\,\dot X+\sum_k\dot m_k\,\frac{\partial\nabla V_{m,\lambda'}}{\partial m_k}(x^0)=0 .
\]
Here $\dot m_1=0$, and $\dot m_4=m_4\,\partial_x\mu_4=0$ because $\mu_4$ is $\sigma$-invariant. Since $m_2=m_3=1$ and $\partial_x\mu_3=-\partial_x\mu_2$ on $\Omega_1^\sigma$, we have $\dot m_2=-\dot m_3=\partial_x\mu_2=\frac1{\sqrt2}\,\partial P/\partial x$. The point $a(h)-\tau E_1$ is $\sigma(a(h)+\tau E_1)$. Thus $X^{-\tau}=\sigma_*X^\tau$ and $\dot X$ is antisymmetric:
\[
\dot X=z_1\mathbf u+z_2\mathbf w+z_3\mathbf t+z_4\mathbf r .
\]
We have $(z_1,z_2)\ne0$. Otherwise $\dot X$ would be an infinitesimal isometry, and the spokes $\mathsf L^{-1/3}(a(h)+\tau E_1)$ of $X^\tau$ would be stationary at $\tau=0$. Taking the inner product with $\mathbf u$ and $\mathbf w$ gives
\[
\mathsf H\begin{pmatrix}z_1\\ z_2\end{pmatrix}+\dot m_2\begin{pmatrix}\beta_1\\ \beta_2\end{pmatrix}=0,\quad \beta_1=\Bigl\langle\mathbf u,\Bigl(\frac{\partial}{\partial m_2}-\frac{\partial}{\partial m_3}\Bigr)\nabla V_{m,\lambda'}\Bigr\rangle,\quad \beta_2=\Bigl\langle\mathbf w,\Bigl(\frac{\partial}{\partial m_2}-\frac{\partial}{\partial m_3}\Bigr)\nabla V_{m,\lambda'}\Bigr\rangle .
\]
At $x^0$ and for arbitrary $m_2$, $m_3$, the formulas for $\mathsf E_2$, $\mathsf E_3$ and $\mathsf E_4$ give $\langle\mathbf u,\nabla V_{m,\lambda'}\rangle=m_4\mathsf E_{4,1}=m_4\,(m_3-m_2)S_{24}$ and $\langle\mathbf w,\nabla V_{m,\lambda'}\rangle=m_2\mathsf E_{2,2}-m_3\mathsf E_{3,2}=(m_2-m_3)(bS_{12}+m_4 hS_{24})$. Hence $\beta_1=-2m_4 S_{24}$ and $\beta_2=2(bS_{12}+m_4 hS_{24})=2\mathsf E_{2,2}=0$. By Lemma~\ref{lem:famcc}, $\beta_1>0$. Moreover $\mathsf H_{12}=-6hm_4\hat q^{-5}<0$. The second row of the system, $\mathsf H_{12}z_1+\mathsf H_{22}z_2=0$, shows that $z_2\ne0$. Multiplying the first row, $\mathsf H_{11}z_1+\mathsf H_{12}z_2=-\beta_1\dot m_2$, by $\mathsf H_{12}$ and substituting $\mathsf H_{12}z_1=-\mathsf H_{22}z_2$ gives $(\mathsf H_{11}\mathsf H_{22}-\mathsf H_{12}^2)\,z_2=\mathsf H_{12}\beta_1\dot m_2$, that is,
\[
\frac{\partial P}{\partial x}\,(a(h))=\sqrt2\,\dot m_2=\frac{\sqrt2\,z_2}{\mathsf H_{12}\,\beta_1}\,\mathsf D(h).
\]
Adding an infinitesimal isometry to $\dot X$ does not change $(z_1,z_2)$, and both $\dot X$ and the basis depend continuously on $h$. Thus $z_2$ is a continuous nonvanishing function of $h$, so $\chi_{\mathrm a}=\sqrt2\,z_2/(\mathsf H_{12}\beta_1)$ is continuous and nonvanishing.
\end{proof}

\subsection{The certified computations}\label{sec:famcert}

\begin{proposition}\label{prop:famcert}
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em]
\item $h_c=1.139428224956\ldots$ and $h_1=2.937347078968\ldots$
\item $\mathsf M'$ has exactly one zero $h_f$ in $(h_c,h_1)$, and $h_f=1.774177389967\ldots$
\item $\mathsf D$ has exactly one zero $h^{**}$ in $(h_c,h_1)$, and $h^{**}=1.690172438713\ldots$ Moreover $\mathsf D(1.60)>0>\mathsf D(1.75)$.
\item $m^{**}$ and $m_f$ satisfy the bounds of Theorem~\ref{thm:family}\ref{F:val}.
\end{enumerate}
\end{proposition}

\begin{proof}
The existence of the zero of $\mathsf M'$ is elementary. Lemma~\ref{lem:famcc}(b) shows that $(2h+d)(\hat q^{-3}-\mathsf L)=d(\hat q^{-3}-d^{-3})$ on the arc. As $h\to h_c$, this tends to $h_c(\hat q(h_c)^{-3}-h_c^{-3})\ne0$, while $b(h)\to0$. The formula for $\mathsf M$ in Lemma~\ref{lem:famcc}(b) then implies $\mathsf M(h)\to0$ as $h\to h_c$. Also $\mathsf M(h_1)=0$ and $\mathsf M>0$ on $(h_c,h_1)$. Rolle's theorem then yields a zero of $\mathsf M'$ in $(h_c,h_1)$. The zero of $\mathsf D$ exists by the signs in (c). The rest is computed. All ball arithmetic uses Arb \cite{Joh17} in FLINT \cite{FLI25} through python-flint and starts from exact rational constants.

\emph{(a)} Interval Newton steps \cite{Moo66,Neu90} for $g_0$ and for $g_1$ of Lemma~\ref{lem:famarc} give $h_c\in1.139428224956200914514853\pm2\cdot10^{-25}$ and $h_1\in2.9373470789684912304\pm2\cdot10^{-20}$.

\emph{Polynomial form.} The substitution $b=(t^2-1)/(2t)$, $\hat p=(t^2+1)/(2t)$, $h=(v^2-1)/(2v)$, $\hat q=(v^2+1)/(2v)$ maps $\{1<t<v\}$ onto $\{0<b<h\}$, and $b<\sqrt3$ if and only if $t<2+\sqrt3$. It turns $\mathsf G$, $\mathsf L$, $\mathsf M$ and $\mathsf D$ into rational functions of $(t,v)$ with rational coefficients, which we compute exactly with SymPy. We write $\mathsf M=M_n/M_d$ in lowest terms. Then $(2h+d)(\hat q^{-3}-\mathsf L)=-M_d/\bigl(8tv(t^2+1)^3(v^2+1)^3\bigr)$. Lemma~\ref{lem:famcc}(b) shows that $M_d$ does not vanish on the arc. Let $\mathcal G$ be the numerator of $\mathsf G$, a polynomial of bidegree $(10,12)$ in $(t,v)$. The denominator $8v(t-v)^2(t^2+1)^3(v^2+1)^3(tv+1)^2$ of $\mathsf G$ does not vanish for $1<t<v$: the arc is $\{\mathcal G=0,\ 1<t<v\}$, and $\partial_t\mathcal G\ne0$ there by Lemma~\ref{lem:famarc}(a). By the chain rule, at a point of the arc $\mathsf M'(h)=0$ if and only if
\[
(\partial_tM_n\,M_d-M_n\,\partial_tM_d)\,\partial_v\mathcal G-(\partial_vM_n\,M_d-M_n\,\partial_vM_d)\,\partial_t\mathcal G=0 .
\]
For $1<t<v$ the zeros of this polynomial are those of a factor $\mathcal F$ of bidegree $(25,22)$: its other irreducible factors are among $t$, $v$, $t\pm1$, $v\pm1$, $t^2+1$, $v^2+1$, $tv+1$ and $v-t$, none of which vanishes there. The numerator of $\mathsf D$ is $-12v(t-1)^2(t+1)^2(t^2-4t+1)(t^4+4t^3+18t^2+4t+1)\,\mathcal P$ for a polynomial $\mathcal P$ of bidegree $(29,25)$. The denominator of $\mathsf D$ is a product of powers of $t-v$, $tv+1$, $t^2+1$, $v^2+1$, $M_d$ and $t^2v-3tv^2+3t-v=-2tv(3h-b)$. On the arc with $1<t<2+\sqrt3$, that is, $h_c<h<h_1$, all these factors except $\mathcal P$ are nonzero: there $\mathsf D(h)=0$ if and only if $\mathcal P=0$.

\emph{Resultants and root isolation.} We prove the uniqueness of these zeros in two independent ways. The first, below, uses resultants of degree up to $550$, computed with Singular. The second, the certificate without resultants at the end of this proof, works along the arc in the variables $(b,h)$, independently of the resultants. If $(t_0,v_0)$ is a common zero of $\mathcal G$ and of $\mathcal X\in\{\mathcal F,\mathcal P\}$, then $v_0$ is a root of the resultant $\operatorname{Res}_t(\mathcal G,\mathcal X)\in\mathbb Z[v]$ and $t_0$ is a root of $\operatorname{Res}_v(\mathcal G,\mathcal X)\in\mathbb Z[t]$: each resultant lies in the ideal generated by $\mathcal G$ and $\mathcal X$. We compute the four resultants exactly with Singular. Their degrees are $474$ and $484$ for $\mathcal F$, and $502$ and $550$ for $\mathcal P$. We factor them over $\mathbb Z$. The routine \texttt{arb\_fmpz\_poly\_complex\_roots} of FLINT, called through \texttt{python-flint} with target precision $800$ bits, encloses the complex roots of each irreducible factor in disjoint balls, each certified to be real (imaginary part exactly $0$) or nonreal (imaginary part excluding $0$). For $\mathcal F$ there are $6$ real roots $v>1$ and $6$ real roots $t>1$, and for $\mathcal P$ there are $3$ and $2$. The real roots equal to $1$ are exact roots of the factors $t-1$ or $v-1$, that is, $b=0$ or $h=0$, outside the open domain. For each pair of a $t$-ball and a $v$-ball we evaluate $\mathcal G$ and $\mathcal X$ on their product in ball arithmetic at $800$ bits. We discard the pair if one of the two enclosures excludes $0$ or if $t>v$ on the product. The only pair whose order is not decided is $t=v=2+\sqrt3$ for $\mathcal F$, the same root of the factor $t^2-4t+1$ of both resultants, which lies on the boundary $b=h$. Exactly one pair with $1<t<v$ survives for each $\mathcal X$. Hence $\mathsf M'$ has at most one zero on the whole arc $h>h_c$, and $\mathsf D$ has at most one zero on $(h_c,h_1)$. These zeros lie in the surviving boxes, at $(b,h)=(0.61349\ldots,1.77417\ldots)$ for $\mathcal F$ and $(0.54129\ldots,1.69017\ldots)$ for $\mathcal P$. With the existence shown above, this proves (b) and the first part of (c).

\emph{The values.} The coordinates of the surviving pairs are simple roots of irreducible factors of the resultants, of degree $240$ for $\mathcal F$ and $126$ for $\mathcal P$. These are the degrees of the coordinates $t$ and $v$, not of $m_f$ and $m^{**}$. An interval Newton step on these factors at $400$ bits encloses the coordinates in balls of radius less than $10^{-100}$. Evaluating $M_n/M_d$ on these balls proves (d).

\emph{The signs.} Bisection with certified signs of $\mathcal G(\cdot,v)$ encloses the points of the arc at $h=1.60$ and $h=1.75$. On these enclosures $\mathcal P$ has opposite strict signs. As the other factors of $\mathsf D$ do not vanish on the arc, $\mathsf D$ changes sign between $h=1.60$ and $h=1.75$. The certificate without resultants below certifies the signs themselves, $\mathsf D(1.60)>0>\mathsf D(1.75)$, which is the second part of (c).

\emph{A certificate without resultants.} The program \texttt{arc\_cert.py} encloses the arc by interval Newton steps for $\mathsf G(\cdot,h)$ on subintervals of $[h_c-10^{-4},h_1+10^{-4}]$ and, by certified signs and monotonicity along the arc, verifies the uniqueness of the zeros in (b) and (c) a second time, in the variables $(b,h)$ (README of the supplement). It also shows that $\widehat{\mathsf D}=\mathsf D/(b\,\mathsf M)$, which has the sign of $\mathsf D$ on $(h_c,h_1)$, satisfies $\widehat{\mathsf D}(1.60)>0>\widehat{\mathsf D}(1.75)$.

\emph{Checks.} msolve solves the systems $\mathcal G=\mathcal F=0$ and $\mathcal G=\mathcal P=0$ without resultants. Each has $21$ real solutions, isolated in boxes with rational endpoints. Comparing these endpoints exactly with the domain shows that exactly one box of each system lies in $\{1<t<v\}$, at the point found above, and that the others lie outside, except for one box of the second system that meets the line $t=1$. This box contains the end $b=0$, $h=h_c$ of the arc, which is a common zero of $\mathcal G$ and $\mathcal P$. A floating-point computation at $70$ digits with the mass map of Section~\ref{sec:setup}, which uses neither $\mathsf G$ nor $\mathsf M$, locates the zeros of $J_{\mathrm{sym}}$ and of $\partial P/\partial x$ along the arc and reproduces $m_f$ and $m^{**}$ to within $5\cdot10^{-46}$.
\end{proof}

\subsection{End of the proof}
\emph{Proof of Theorem~\ref{thm:family}\ref{F:J} and~\ref{F:val}.} By the proof of Proposition~\ref{prop:famcert}, $\mathsf M(h)\to0$ as $h\to h_c$, $\mathsf M(h_1)=0$ and $\mathsf M>0$ on $(h_c,h_1)$. Proposition~\ref{prop:famcert}(b) adds that $\mathsf M'$ vanishes only at $h_f$. Hence $\mathsf M'>0$ on $(h_c,h_f)$ and $\mathsf M'<0$ on $(h_f,h_1)$. This is the monotonicity; the maximum is $m_f=\mathsf M(h_f)$.

The point $(b,h)=(1/\sqrt3,\sqrt3)$ has $\hat p=d=2/\sqrt3$, $\hat q=2$, $\mathsf G=0$, $\mathsf L=(3+3\sqrt3)/32$ and $\mathsf M=1$. Then $b(\sqrt3)=1/\sqrt3$ and $\mathsf M(\sqrt3)=1$. The configuration $x(1/\sqrt3,\sqrt3)$ is the equilateral triangle of side $2$ with body~$1$ at its centre; $a(\sqrt3)$ is the point of the axis of $\Omega_1$ for four equal masses. As $1\ne m^*$, it is not the umbilic, so $J(a(\sqrt3))<0$ by Proposition~\ref{prop:ax}. Equation \eqref{eq:Jsplit} then gives $J_{\mathrm{sym}}(a(\sqrt3))\ne0$. Since $\sqrt3\in(h_c,h_f)$ (Lemma~\ref{lem:famarc}(a) and Proposition~\ref{prop:famcert}(b)), also $\mathsf M'(\sqrt3)\ne0$. Hence the quotient $J_{\mathrm{sym}}(a(h))/\mathsf M'(h)$ is continuous and nonvanishing near $\sqrt3$. It agrees with $\chi_{\mathrm s}$ of Lemma~\ref{lem:famfold} for $h\ne\sqrt3$ and extends it to a continuous nonvanishing function on $(h_c,h_1)$ with $J_{\mathrm{sym}}(a(h))=\chi_{\mathrm s}(h)\mathsf M'(h)$.

Equation \eqref{eq:Jsplit} and Lemmas~\ref{lem:famfold} and~\ref{lem:fampitch} show that $J(a(h))=\chi_{\mathrm s}(h)\chi_{\mathrm a}(h)\,\mathsf M'(h)\,\mathsf D(h)$ on $(h_c,h_1)$. Here $\chi_{\mathrm s}\chi_{\mathrm a}$ is continuous and nonvanishing. The factor $\mathsf M'$ vanishes only at $h_f$ and changes sign there. The factor $\mathsf D$ vanishes only at $h^{**}$ and changes sign there, by Proposition~\ref{prop:famcert}(c) and $1.60<h^{**}<1.75$. Proposition~\ref{prop:famcert}(b),(c) shows that $h^{**}<\sqrt3<h_f$. It follows that $J(a(h))$ vanishes exactly at $h^{**}$ and $h_f$ and changes sign at both. As $J(a(\sqrt3))<0$, we get $J<0$ on $(h^{**},h_f)$ and $J>0$ on $(h_c,h^{**})$ and on $(h_f,h_1)$. This proves~\ref{F:J}.

The bounds of~\ref{F:val} are Proposition~\ref{prop:famcert}(d). Finally $m^*=(81+64\sqrt3)/249=0.7704\ldots$ and $(m^*)^{-3}=2.1862\ldots$\qed

\subsection{The counts}\label{sec:famcount}

\begin{corollary}\label{cor:rs}
Let $m_4>0$.
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em]
\item $c_4(1,1,1,m_4)=4$ for $m_4\ne m^*$, and $c_4(1,1,1,m^*)=1$.
\item $c_1=c_2=c_3$, and this number is $2$ for $m_4\in(0,m^{**}]$, $4$ for $m_4\in(m^{**},m_f)$, $3$ for $m_4=m_f$, and $2$ for $m_4\in(m_f,\infty)$.
\item $N(1,1,1,m_4)$ is $38$ on $(0,m^*)$, $32$ at $m^*$, $38$ on $(m^*,m^{**}]$, $50$ on $(m^{**},m_f)$, $44$ at $m_f$, and $38$ on $(m_f,\infty)$. These are the values of Table~\ref{tab:rs}.
\item $(1,1,1,m_4)\in Z_4$ if and only if $m_4=m^*$, and for $i\in\{1,2,3\}$, $(1,1,1,m_4)\in Z_i$ if and only if $m_4\in\{m^{**},m_f\}$.
\end{enumerate}
\end{corollary}

\begin{proof}
As the permutations of the bodies $1$, $2$, $3$ preserve the masses, $c_1=c_2=c_3$, and $(1,1,1,m_4)$ lies either in all of $Z_1$, $Z_2$, $Z_3$ or in none of them. We consider body~$1$ and the map $F_1$ of Section~\ref{sec:famsetting}. Then $\mu':=\mu^{(1)}(m)=(0,0,\log m_4)$ has the coordinates $P=0$, $A=\frac13\log m_4$, $Q=-\frac2{\sqrt6}\log m_4$ and $\rho=|Q|$, with $\Theta=90^\circ$ for $m_4<1$ and $\Theta=270^\circ$ for $m_4>1$.

\emph{Step 1: the fibre.} Recall that $F_1$, like $F$, is injective on $\Db$ (Theorem~\ref{thm:inj}). As $\sigma$ fixes $\mu'$, it maps $F_1^{-1}(\mu')$ to itself. Let $p\in F_1^{-1}(\mu')$ with $\sigma p\ne p$. Then $J(\sigma p)=J(p)$, and $J(p)>0$: otherwise $p$ and $\sigma p$ would be two points of $\Db$ with the same image. The points of the fibre off $\Omega_1^\sigma$ form pairs $\{p,\sigma p\}$ with $J>0$. The points of the fibre on $\Omega_1^\sigma$ are the symmetric concave CCs of the masses $(1,1,1,m_4)$ with body~$1$ inside, that is, by Theorem~\ref{thm:family}\ref{F:arc}, the points $a(h)$ with $\mathsf M(h)=m_4$. By Theorem~\ref{thm:family}\ref{F:J} there are two of them for $0<m_4<m_f$ (one with $h<h_f$ and one with $h>h_f$), one for $m_4=m_f$ and none for $m_4>m_f$.

\emph{Step 2: the critical values.} The point $\mu'$ is a critical value of $F_1$ if and only if $m_4\in\{m^{**},m_f\}$. Indeed, $a(h^{**})$ and $a(h_f)$ are preimages of $\mu^{(1)}(1,1,1,m^{**})$ and $\mu^{(1)}(1,1,1,m_f)$ with $J=0$, by Theorem~\ref{thm:family}\ref{F:J}. Conversely, let $q\in F_1^{-1}(\mu')$ with $J(q)=0$. Then $q$ and $\sigma q$ lie in $\Db$ and have the same image, forcing $q=\sigma q$ by injectivity. By Step~1, $q=a(h)$ with $\mathsf M(h)=m_4$. Then, by Theorem~\ref{thm:family}\ref{F:J}, $J(a(h))=0$ implies $h\in\{h^{**},h_f\}$. This proves (d) for $i=1$.

\emph{Step 3: the regular values.} Let $m_4\notin\{m^{**},m_f\}$. Step~2 shows that $\mu'$ is a regular value. Recall from Theorem~\ref{thm:degtwo} that at a regular value $c_1=2+2n_-$, where $n_-$ is the number of preimages with $J<0$. Step~1 identifies $n_-$ with the number of points $a(h)$ with $\mathsf M(h)=m_4$ and $J(a(h))<0$. For $0<m_4<m^{**}$, Theorem~\ref{thm:family}\ref{F:J} shows that the two such points have $h<h^{**}$ and $h>h_f$, so $J>0$ at both and $c_1=2$. For $m^{**}<m_4<m_f$ they have $h^{**}<h<h_f$ and $h>h_f$. Then $n_-=1$ and $c_1=4$. For $m_4>m_f$ there are none, so $c_1=2$.

\emph{Step 4: the critical values.} At $m_4=m^{**}$ we have $A=\frac13\log m^{**}<0<A^*$ and $\Theta=90^\circ$. As $\mu'$ is a critical value, $\rho=R(A,\Theta)$ by Theorem~\ref{thm:count}. This is the cuspidal edge, where $c_1=2$ by Theorem~\ref{thm:count}. At $m_4=m_f$ we have $0<A=\frac13\log m_f<-\log m^*=A^*$ by Theorem~\ref{thm:family}\ref{F:val}, $\Theta=270^\circ$, and again $\rho=R(A,\Theta)$. This is the fold sheet, where Theorem~\ref{thm:count} gives $c_1=3$.

\emph{Step 5: body $4$, and $N$.} The point $\mu^{(4)}(m)=-(\log m_4)(1,1,1)$ has $\rho=0$ and $A=-\log m_4$. For $m_4\ne m^*$ we have $A\ne A^*$ and $\rho=0<R(A,\Theta)$ for all $\Theta$. Theorem~\ref{thm:count} then gives $c_4=4$ and shows that $\mu^{(4)}$ is a regular value. For $m_4=m^*$ we have $(A,\rho)=(A^*,0)$, so $c_4=1$ by Theorem~\ref{thm:count} (Proposition~\ref{prop:umbcount}). This value is critical: it is the image of the umbilic $p^*$, where $J=0$. This proves (a) and (d). Finally $N=12+6+2(c_1+c_2+c_3+c_4)$ by Theorem~\ref{thm:N}, and $m^*<m^{**}$ by Theorem~\ref{thm:family}\ref{F:val}. Substituting the counts, we obtain (c): $18+2(6+4)=38$, $18+2(6+1)=32$, $18+2(12+4)=50$ and $18+2(9+4)=44$.
\end{proof}

Theorem~\ref{thm:family}\ref{F:val} rests on the resultants of Section~\ref{sec:famcert}. The certificate without resultants does not give these values, but the decimals of $m^{**}$ and $m_f$ printed in Theorem~\ref{thm:N}(e) also follow without resultants. The program \texttt{family.py} (README of the supplement) certifies, by the procedure of Section~\ref{sec:decide}, opposite statuses of body~$1$ at the two ends of each of two intervals of width $5.4\cdot10^{-25}$. Steps~1--3 of the proof above do not use Theorem~\ref{thm:family}\ref{F:val}. These steps show that $c_1$ changes only at $m^{**}$ and $m_f$, so one interval contains $m^{**}$ and the other $m_f$.

As a check, floating-point counts of all concave CCs with body~$1$ inside, by Newton's method from many starting points, both in the variables of the mass map and in Cartesian coordinates, agree with Corollary~\ref{cor:rs}(b) at all sampled values of $m_4$, including the signs of $J$.

\subsection{Relation to earlier work}\label{sec:famlit}
Bernat, Llibre and P\'erez-Chavela \cite{BLP09} computed numerically the number of kite CCs of the masses $(1,1,1,m_4)$ and the values of $m_4$ at which it changes. Rusu and Santoprete counted all CCs of these masses numerically and obtained the values of $N$ in Table~\ref{tab:rs} \cite[\S4.2]{RS16}. They note \cite[\S4.3]{RS16} that their value of the fold bifurcation differs slightly from that of \cite{BLP09}. In \cite[\S\S4.3--4.4]{RS16} they certified the two degenerate CCs by a Krawczyk computation \cite{Kra69} on the full system of Dziobek equations together with the determinant of its Jacobian matrix. Their enclosures, printed as $1.00266054757261000068580350?$ and $0.99184227439094091554349?$, contain the values of Theorem~\ref{thm:family}\ref{F:val}. They also proved that the bifurcations at $m_f$ and at $m^{**}$ are a fold and a supercritical pitchfork. Liu, Poudel, Raut and Xie \cite{LPRX26} classify the concave kite CCs of three equal masses by a computer-assisted proof with interval arithmetic and the Krawczyk operator. For a unit body inside, which is their outer case, they show that the symmetric CCs form a one-parameter family \cite[Lemmas~3.1--3.2]{LPRX26}, by the argument of Lemma~\ref{lem:famarc}(a) in other coordinates. They show that there are two such CCs for $m_4<m_f$, one for $m_4=m_f$ and none for $m_4>m_f$ \cite[Theorem~1.2]{LPRX26}. They also show that the mass has a unique maximum $m_f$ along the family (their $m_O$) and that this maximum is a fold in the reduced symmetric space \cite[\S\S3.2--3.3]{LPRX26}. For the body of mass $m_4$ inside, which is their inner case, the kite CCs with a given axis are the equilateral one and one isosceles family. Shi and Xie \cite[Theorem~1.4]{SX10} obtained this description of the inner case, as \cite[Remark~1.3]{LPRX26} points out. The computer-assisted proof of \cite{LPRX26} rederives it in other variables. It finds two CCs for $m_4\ne m^*$, one for $m_4=m^*$ and a transcritical bifurcation at $m^*$ in the reduced space \cite[Theorem~1.2 and \S4.3]{LPRX26}. The asymmetric CCs and the pitchfork at $m^{**}$ are treated numerically in \cite[\S5]{LPRX26}. In \cite[\S3.2]{LPRX26} the value of $m_f$ is printed as $1.00266054757261000068580350216?$ and stated to lie in $[1.\cdots21,\,1.\cdots22]$. In SageMath's convention the printed value denotes $[1.\cdots350215,\,1.\cdots350217]$, which excludes $m_f$ (its $29$th decimal is $8$); read at $28$ decimals, the stated interval contains $m_f$. Relative to \cite{RS16}, the uniqueness of the fold point and of the pitchfork point on the whole arc is new here, and relative to \cite{LPRX26} the uniqueness of the pitchfork point. The counts of Corollary~\ref{cor:rs}, which include the asymmetric CCs, are new as theorems.