\section{Diagrams with a circled cluster of each colour}\label{sec:ak50}

In this section we treat the diagrams $81$, $82$, $88$, $89$, $90$, $93$, $94$ and $98$ of Table~\ref{tab:ten} at the level of places. Each has a cluster of $z$-strokes with $z$-circled bodies and a cluster of $w$-strokes with $w$-circled bodies, and the constraint comes from the torque of one cluster \cite[\S5.0]{AK12}. Throughout, $m\in(\R_{>0})^6$, $C\subseteq Y_m$ is an irreducible curve and $P$ is a singular place of $C$. We do not assume that $U\not\equiv0$. We state each result for the labelling of Table~\ref{tab:ten}. By Lemmas~\ref{lem:slice}(b) and~\ref{lem:symm}, it then holds for all labellings, after the masses are permuted accordingly. For a body $k$ we write $\hat z_k=\lc{z_k}{a_k}$ and $\hat w_k=\lc{w_k}{b_k}$, and for a pair $\lc{p_{kl}}{}=\lc{p_{kl}}{-x_{kl}}$.

\subsection{Common facts}\label{ss:common}

\begin{lemma}\label{lem:common}
\begin{enumerate}[label=\textup{(F\arabic*)}]
\item Let $A$ be a set of two, three or four bodies whose pairs are all $z$-strokes and none a $zw$-edge, and suppose that $A$ contains a $z$-circled body and every $z$-circled body lies in $A$. Then $\zeta_{kl}=-2$, $\omega_{kl}=2$ and $x_{kl}=0$ for all pairs $kl$ in $A$.
\item In the situation of \textup{(F1)}, every $z$-circled body $k$ has $b_k\ge2$.
\item If $kl$ is a $w$-stroke with $x_{kl}=0$, then $a_k\ge\min(a_l,2)$.
\item If $a_j<a_l$ and $b_l<b_j$, then $\zeta_{jl}=a_j$, $\omega_{jl}=b_l$, $\lc{z_{jl}}{a_j}=\hat z_j$, $\lc{w_{jl}}{b_l}=-\hat w_l$, $e(p_{jl})=-\frac12(a_j+b_l)$ and $\lc{p_{jl}}{}^2\hat z_j\hat w_l=-1$. If moreover $a_l+b_j>a_j+b_l$, the term $m_jm_lp_{jl}^3(z_jw_l-z_lw_j)$ has exponent $-\frac12(a_j+b_l)$ and leading coefficient $-m_jm_l\lc{p_{jl}}{}$.
\end{enumerate}
The same holds with $z$ and $w$ exchanged.
\end{lemma}

\begin{proof}
(F1) By (E3) and the remark after Lemma~\ref{lem:rules}, the values of $\zeta$, $\omega$ and $x$ are the same on all pairs of $A$, and by (O3), (O4), $\zeta=3x-2$ and $\omega=2-x$ with $0\le x<1$. If some body of $A$ is not circled, (O6) gives $\zeta=-2$ on its pair with a circled body. If all are circled and $x>0$, then $\zeta>-2$ on $A$, so the bodies of $A$ have a common coefficient $c=\lc{z_k}{-2}\ne0$, and the coefficient of exponent $-2$ in $\sum_km_kz_k=0$ is $c\sum_{k\in A}m_k\ne0$. So $x=0$.
(F2) For $l\in A$, $e(W_{kl})=-\frac12(3\zeta_{kl}+\omega_{kl})=2$. For $l\notin A$, $l$ is not circled, so $\zeta_{kl}=-2$ by (O6) and $e(W_{kl})=3-\frac12\omega_{kl}\ge2$ by (O1). All terms of the right side of $w_k=\sum_lm_lW_{kl}$ have exponent at least $2$.
(F3) $z_k=z_l+z_{kl}$ and $\zeta_{kl}=2$ by (O3).
(F4) The first statements follow from $a_j<a_l$, $b_l<b_j$ and the distance equation at exponent $0$. The term $p_{jl}^3z_jw_l$ has exponent $-\frac12(a_j+b_l)$ and coefficient $\lc{p_{jl}}{}^3\hat z_j\hat w_l=-\lc{p_{jl}}{}$, and the term $p_{jl}^3z_lw_j$ has exponent $-\frac32(a_j+b_l)+a_l+b_j$, which is larger.
\end{proof}

\begin{lemma}\label{lem:rewrite}
Let $S$ be a set of bodies, $o\notin S$ a body, $A_z=\sum_{j\in S}m_jZ_{oj}$, $A_w=\sum_{j\in S}m_jW_{oj}$, and $B_z=z_o-A_z$, $B_w=w_o-A_w$, so that $B_z=\sum_{l\notin S\cup\{o\}}m_lZ_{ol}$ and $B_w=\sum_{l\notin S\cup\{o\}}m_lW_{ol}$ by the equations of the body $o$. Then
\[
\sum_{j\in S}m_jm_op_{jo}^3(z_jw_o-z_ow_j)=m_o(z_oA_w-w_oA_z)=m_o(w_oB_z-z_oB_w).
\]
\end{lemma}

\begin{proof}
Substitute $p_{jo}^3z_j=p_{jo}^3z_o-Z_{oj}$ and $p_{jo}^3w_j=p_{jo}^3w_o-W_{oj}$. The terms $p_{jo}^3z_ow_o$ cancel. Then substitute $A_z=z_o-B_z$ and $A_w=w_o-B_w$.
\end{proof}

This is the rewriting of \cite[\S8.6, (30)--(31)]{AK12} through the equations of the body $o$. Its point is that the individual terms of $B_w$ can have smaller exponent than $B_w$ itself. We will apply it with $o$ the body shared by the two clusters of diagrams $89$ and $94$.

\begin{lemma}\label{lem:binary}
Suppose that $5$ and $6$ are the only $z$-circled bodies, so that $m_5\hat z_5+m_6\hat z_6=0$, and let $k\notin\{5,6\}$ satisfy $a_k>-2$ and $b_k<\min(b_5,b_6)$. Then $m_5\lc{p_{5k}}{}+m_6\lc{p_{6k}}{}\ne0$.
\end{lemma}

\begin{proof}
The relation $m_5\hat z_5+m_6\hat z_6=0$ is the coefficient of exponent $-2$ in $\sum_jm_jz_j=0$. By (F4), $\lc{p_{jk}}{}^2=-1/(\hat z_j\hat w_k)$ for $j=5,6$. If $m_5\lc{p_{5k}}{}=-m_6\lc{p_{6k}}{}$, squaring gives $m_5^2/\hat z_5=m_6^2/\hat z_6$, and with $\hat z_6=-m_5\hat z_5/m_6$ this gives $m_5^3+m_6^3=0$.
\end{proof}

\subsection{Diagrams 90 and 89}
Proposition~\ref{prop:exclusion} already excludes the listed cells of diagram~$90$. We treat it first because the proof of Proposition~\ref{thm:d90} is the model for that of Theorem~\ref{thm:d89}.

\begin{proposition}\label{thm:d90}
No singular place of an irreducible curve $C\subseteq Y_m$ has diagram $90$.
\end{proposition}

\begin{proof}
Let $A=\{4,5,6\}$ and $B=\{1,2,3\}$. By (F1), $x=0$, $\zeta=-2$ and $\omega=2$ on the pairs of $A$, and $x=0$, $\omega=-2$, $\zeta=2$ on the pairs of $B$. By (O6), $\zeta=\omega=-2$ on $52$, $53$, $62$, $63$. By (F2), $b_5,b_6\ge2$ and $a_2,a_3\ge2$, and by (F3), $b_4\ge2$ and $a_1\ge2$. If $a_4\ge2$, then $\zeta_{42}\ge2$, so $\zeta_{42}=2$ and $42$ would be a $w$-stroke by (O6). Hence $-2<a_4<2$, and in the same way $-2<b_1<2$.

For $j\in A$ and $l\in B$ we have $a_j<2\le a_l$, $b_l<2\le b_j$ and $a_l+b_j\ge4>a_j+b_l$, so (F4) applies to the nine pairs between $A$ and $B$. In the torque $T_A$ the term of the pair $jl$ has exponent $E_{jl}=-\frac12(a_j+b_l)$ and a nonzero leading coefficient: $E_{41}=-\frac12(a_4+b_1)$, $E_{4l}=1-\frac12a_4$ for $l=2,3$, $E_{j1}=1-\frac12b_1$ for $j=5,6$, and $E_{jl}=2$ for $j=5,6$, $l=2,3$. Since $E_{4l}-E_{41}=1+\frac12b_1$, $E_{j1}-E_{41}=1+\frac12a_4$ and $2-E_{41}=2+\frac12(a_4+b_1)$ are positive, $E_{41}$ is attained by a single term, contradicting Lemma~\ref{lem:initial}.
\end{proof}

\begin{theorem}\label{thm:d89}
No singular place of an irreducible curve $C\subseteq Y_m$ has diagram $89$.
\end{theorem}

\begin{proof}
Here $A=\{4,5,6\}$ carries the $z$-strokes with circles at $5$, $6$, and $W=\{2,3,4\}$ the $w$-strokes with circles at $2$, $3$. Body $1$ is \emph{free}: it has no stroke and no circle.

\emph{Exponents.} By (F1), $x=0$ on the six strokes, with $\zeta=-2$, $\omega=2$ on $45,46,56$ and $\omega=-2$, $\zeta=2$ on $23,24,34$. By (O6), $\zeta=\omega=-2$ on $25,26,35,36$, $\zeta_{15}=\zeta_{16}=-2$ and $\omega_{12}=\omega_{13}=-2$. By (F2) and (F3), $b_5,b_6,b_4\ge2$ and $a_2,a_3,a_4\ge2$. If $a_1\ge2$, then $\zeta_{12}=2$ and $12$ would be a $w$-stroke. If $b_1\ge2$, then $\omega_{15}=2$ and $15$ would be a $z$-stroke. So $a_1,b_1\in(-2,2)$, and $\zeta_{1l}=a_1$ for $l=2,3,4$, $\omega_{1j}=b_1$ for $j=4,5,6$.

\emph{$a_1+b_1=0$.} By Lemma~\ref{lem:U}(b), $U$ is constant on $C$. In $U=\sum m_km_lp_{kl}$ the exponents $-x_{kl}$ are $0$ on the strokes, $2$ on $25,26,35,36$, $1-\frac12a_1>0$ on $12$, $13$, $1-\frac12b_1>0$ on $15$, $16$, and $-\frac12(a_1+b_1)$ on $14$. If $a_1+b_1>0$, the term of $14$ is the only term of negative exponent. In $\sum_km_kz_kw_k=U$ the exponents $a_k+b_k$ are at least $0$ for $k\ne1$, so if $a_1+b_1<0$, the term of $k=1$ is the only one of negative exponent. Both contradict Lemma~\ref{lem:initial}.

\emph{The torque of $S=\{5,6\}$.} The pairs $(j,l)$ with $j\in S$, $l=2,3$ give terms of exponent at least $2$. The pairs $(j,1)$ satisfy (F4), since $a_j=-2<a_1$, $b_1<2\le b_j$ and $a_1+b_j>-2+b_1$. Their terms have exponent $1-\frac12b_1=1+\frac12a_1$ and coefficients $-m_jm_1\lc{p_{j1}}{}$. By Lemma~\ref{lem:rewrite} with $o=4$, the pairs $(j,4)$ contribute $m_4(w_4B_z-z_4B_w)$ with $B_z=m_1Z_{41}+m_2Z_{42}+m_3Z_{43}$ and $B_w=w_4-A_w$. Here $e(Z_{42})=e(Z_{43})=2$ and $e(Z_{41})=-\frac12(a_1+3b_1)=a_1$, so $e(B_z)\ge a_1$. Also $e(A_w)\ge2$, so $e(B_w)\ge2$, although $W_{42}$ and $W_{43}$ have exponent $-2$. Since $a_4,b_4\ge2$, this contribution has exponent at least $\min(2+a_1,4)$, which exceeds $1+\frac12a_1$ for $-2<a_1<2$, and so does $2$. The coefficient of exponent $1+\frac12a_1$ in $T_S$ is therefore $-m_1(m_5\lc{p_{51}}{}+m_6\lc{p_{61}}{})$, which is not zero by Lemma~\ref{lem:binary}. This contradicts $T_S=0$.
\end{proof}

\begin{remark}\label{rem:d89lead}
The terms of the pairs $(j,4)$ in $T_{\{5,6\}}$ have exponent $b_4-2$, which can be smaller than $1+\frac12a_1$. Their initial form $m_4\hat w_4\sum_jm_j\lc{p_{j4}}{}^3\hat z_j$ is a multiple of the initial form of the $z$-equation of body $4$ at exponent $-2$, which reads $\sum_jm_j\lc{p_{4j}}{}^3(-\hat z_j)=0$. So when $b_4-2<1+\frac12a_1$, the initial form of the torque is implied by that of the equation of body $4$. Lemma~\ref{lem:rewrite} removes the cancellation before the exponents are compared. In five bodies the corresponding diagram is diagram~$6$ of \cite[Fig.~11]{AK12}, without the free body, and the same torque gives the conditions $m_1=m_2$, $m_3=m_4$ of \cite[(32)]{AK12}.
\end{remark}

\subsection{Diagrams 82 and 81}
Proposition~\ref{prop:exclusion} also excludes all listed cells of these two diagrams. We include them because the arguments are short and hold at every positive mass.

\begin{proposition}\label{prop:d82}
No singular place of an irreducible curve $C\subseteq Y_m$ has diagram $82$.
\end{proposition}

\begin{proof}
\emph{Exponents.} By (F1), $x=0$ on $56$ and on $23,24,34$, where $\zeta=2$, $\omega=-2$. By (O6), $\zeta=\omega=-2$ on $\{5,6\}\times\{3,4\}$, $\zeta_{j1}=\zeta_{j2}=-2$ for $j=5,6$, and $\omega_{l1}=-2$ for $l=3,4$. By (F2) and (F3), $b_5,b_6\ge2$ and $a_3,a_4,a_2\ge2$. As in Theorem~\ref{thm:d89}, $a_1,b_1,b_2\in(-2,2)$, since otherwise $13$ would be a $w$-stroke or $15$, $25$ a $z$-stroke. Hence $\omega_{jk}=b_k$ for $j=5,6$, $k=1,2$, and $\zeta_{1l}=a_1$ for $l=2,3,4$.

\emph{$b_1=b_2$.} In $T_{\{5,6\}}$ the pairs $(j,l)$, $l=3,4$, give terms of exponent at least $2$, and the pairs $(j,k)$, $k=1,2$, satisfy (F4) and give terms of exponent $1-\frac12b_k<2$ and coefficients $-m_km_j\lc{p_{jk}}{}$. If $b_1\ne b_2$, the body $k$ with the larger $b_k$ gives the coefficient $-m_k(m_5\lc{p_{5k}}{}+m_6\lc{p_{6k}}{})$ of least exponent, which is not zero by Lemma~\ref{lem:binary}. So $b_1=b_2=:b$.

\emph{$a_1=-b$.} All pairs other than $12$ have $x\le0$. Since $U$ is constant, the term of $12$ cannot be the only one of negative exponent, so $2x_{12}=a_1+\omega_{12}\le0$, and $\omega_{12}\ge b$ gives $a_1+b\le0$. In $\sum_km_kz_kw_k=U$ the exponents $a_k+b_k$ are at least $0$ for $k\ne1$, so $a_1+b\ge0$. Hence $a_1=-b$, $\omega_{12}=b$, and $e(Z_{21})=-\frac12(\zeta_{12}+3\omega_{12})=-b$.

\emph{Body 2.} In $z_2=m_1Z_{21}+m_3Z_{23}+m_4Z_{24}+m_5Z_{25}+m_6Z_{26}$ the exponents are $-b$, $2$, $2$, $1-\frac32b$, $1-\frac32b$ on the right and at least $2$ on the left. For $-2<b<2$ the exponent $-b$ is attained by a single term, contradicting Lemma~\ref{lem:initial}.
\end{proof}

\begin{proposition}\label{prop:d81}
If a singular place of an irreducible curve $C\subseteq Y_m$ has diagram $81$, then $m_1=m_2$, $m_3=m_4$ and $m_5=m_6$.
\end{proposition}

\begin{proof}
\emph{Exponents.} By (F1), $x_{56}=x_{34}=0$, and by (O6), $\zeta=\omega=-2$ on $\{5,6\}\times\{3,4\}$. By (F2), $b_5,b_6,a_3,a_4\ge2$, and as before $a_k,b_k\in(-2,2)$ for $k=1,2$. So $\zeta_{jk}=-2$, $\omega_{jk}=b_k$ for $j\in\{5,6\}$, and $\omega_{lk}=-2$, $\zeta_{lk}=a_k$ for $l\in\{3,4\}$.

\emph{Equal exponents.} As in Proposition~\ref{prop:d82}, the torque of $\{5,6\}$ gives $b_1=b_2=:b$ and the torque of $\{3,4\}$ gives $a_1=a_2=:a$. Since $U$ is constant and all pairs other than $12$ have $x\le0$, $a+b\le\zeta_{12}+\omega_{12}=2x_{12}\le0$.

\emph{The pair $12$.} The equation of body $1$ in $z$ has the terms $z_1$ (exponent $a$), $m_2Z_{12}$ (exponent $e_Z=-\frac12(\zeta_{12}+3\omega_{12})$), $m_5Z_{15}+m_6Z_{16}$ (exponent $1-\frac32b$ exactly, since its leading coefficient is $(m_5\lc{p_{15}}{}+m_6\lc{p_{16}}{})/\hat w_1\ne0$ by (F4) and Lemma~\ref{lem:binary}), and $m_3Z_{13}+m_4Z_{14}$ (exponent at least $3-\frac12a>a$). The least exponent is attained twice among the first three. If $a=1-\frac32b\le e_Z$, then $a+b\le0$ gives $b\ge2$. If $e_Z=1-\frac32b<a$, then $e_Z\le-\frac12(a+3b)$ gives $a\le-2$. Both are impossible, so $a=e_Z$, and in the same way $b=-\frac12(3\zeta_{12}+\omega_{12})$. Adding, $a+b=-2(\zeta_{12}+\omega_{12})=-4x_{12}\ge0$. Hence $a+b=0$, $x_{12}=0$, $\zeta_{12}=a$ and $\omega_{12}=-a$.

\emph{The binary $\{1,2\}$.} At exponent $a$ in $z$ and $-a$ in $w$, the equations of bodies $1$ and $2$ contain only $z_k$, $w_k$ and their mutual terms, since $1+\frac32a>a$ and $3-\frac12a>a$ for $-2<a<2$. Hence $m_1\hat z_1+m_2\hat z_2=0$ and $m_1\hat w_1+m_2\hat w_2=0$.

\emph{The torques.} In $T_{\{5,6\}}$ the two bodies $k=1,2$ tie at exponent $1-\frac12b$, below the terms of $3$, $4$. Write $\hat z_5=cm_6$, $\hat z_6=-cm_5$ with $c\ne0$. By (F4), $m_5\lc{p_{5k}}{}+m_6\lc{p_{6k}}{}=(c\hat w_k)^{-1/2}(m_5m_6)^{-1/2}u_k$ with $u_k\in\{\pm m_6^{3/2}\pm im_5^{3/2}\}$ for fixed branches of the square roots. The coefficient of least exponent gives $m_1\hat w_1^{-1/2}u_1+m_2\hat w_2^{-1/2}u_2=0$. Since $|u_1|=|u_2|$ and $\hat w_1/\hat w_2=-m_2/m_1$, this gives $(u_1/u_2)^2=-m_2^3/m_1^3$, so $m_1=m_2$ and $u_1/u_2=\pm i$, which forces $m_5^{3/2}=m_6^{3/2}$. The torque of $\{3,4\}$ gives $m_3=m_4$ in the same way.
\end{proof}

\subsection{Complex central configurations of three bodies}\label{ss:three}
For $\mu=(\mu_1,\mu_2,\mu_3)$ with $\mu_k\ne0$ and $M_\mu=\mu_1+\mu_2+\mu_3\ne0$, let $N(\mu)$ be the system
\[
z_k=\sum_{l\ne k}\mu_lp_{kl}^3(z_k-z_l),\quad w_k=\sum_{l\ne k}\mu_lp_{kl}^3(w_k-w_l),\quad p_{kl}^2(z_k-z_l)(w_k-w_l)=1
\]
in $z,w\in\C^3$ and $p_{12},p_{13},p_{23}$. It implies $\sum\mu_kz_k=\sum\mu_kw_k=0$, and the rotations $(z,w)\mapsto(sz,s^{-1}w)$ act on its solutions. With the slice $z_1-z_2=1$ and the \emph{shape} $\tau=z_1-z_3$, the positions are $z_k=\ell_k(\tau)/M_\mu$, where $\ell_1(t)=\mu_2+\mu_3t$, $\ell_2=\ell_1-M_\mu$ and $\ell_3=\ell_1-M_\mu t$. Put $\Lambda_k=\prod_{k'\ne k}\ell_{k'}$,
\[
N_{01}=t^4-5t^3+9t^2-5t+1,\quad N_{10}=t^4+t^3-2t+1,\quad N_{11}=t^4-2t^3+t+1,
\]
and, for the four classes $\sigma=(\sigma_{12},\sigma_{13},\sigma_{23})\in\{(1,1,1),(1,1,-1),(1,-1,-1),(1,-1,1)\}$ of signs modulo $\pm1$, let $Q_\sigma(x)=x^2(1+x)^2\big(x(F_1-F_2)-(F_2-F_3)\big)$, where $F_k=\sum_{l\ne k}\mu_l\sigma_{kl}z_{kl}^{-2}$ with $z_{12}=1$, $z_{23}=x$, $z_{13}=1+x$ and $\sigma_{lk}=-\sigma_{kl}$. Let
\[
S_\mu(t)=(t^2-t+1)\,N_{01}(t)\,N_{10}(t)\,N_{11}(t)\prod_\sigma Q_\sigma(t-1),
\]
a polynomial of degree $34$ with leading coefficient $(\mu_1+\mu_2)^4$.

\begin{lemma}\label{lem:three}
Every solution of $N(\mu)$ is of \emph{Lagrange type}, $p_{12}^3=p_{13}^3=p_{23}^3=1/M_\mu$, or of \emph{collinear type}, $w=\lambda z$ with $\lambda\ne0$. The shape of each solution is a root of $S_\mu$.
\end{lemma}

\begin{proof}
Let $L$ be the matrix with entries $L_{kl}=\mu_lp_{kl}^3$ for $k\ne l$ and zero row sums. The equations say $(L+I)z=(L+I)w=0$, and $\mu^{\mathsf T}L=0$, so $z$ and $w$ lie in the plane $H=\{v:\sum\mu_kv_k=0\}$. If $z$ and $w$ are independent, $L+I$ vanishes on $H$, so $L+I=u\mu^{\mathsf T}$. Applied to $(1,1,1)$, this gives $u=M_\mu^{-1}(1,1,1)$, and the off-diagonal entries give $p_{kl}^3=1/M_\mu$. Otherwise $w=\lambda z$, since $z\ne0$.

In the Lagrange case, put $R_{kl}=p_{kl}^{-2}=M_\mu^{2/3}\mathrm{j}^{a_{kl}}$ with $\mathrm{j}=e^{2\pi i/3}$. With $z_{12}=1$, $z_{13}=\tau$, $w_{12}=R_{12}$ and $w_{13}=R_{13}/\tau$, the distance equation of $23$ becomes $R_{12}\tau^2-(R_{12}+R_{13}-R_{23})\tau+R_{13}=0$, that is $t^2-(1+\mathrm{j}^b-\mathrm{j}^c)t+\mathrm{j}^b=0$ for some $(b,c)\in(\Z/3)^2$. The classes $(0,0)$, $(1,2)$, $(2,1)$ give $t^2-t+1$, and the three pairs of conjugate classes give the products $N_{01}$, $N_{10}$, $N_{11}$ of conjugate quadratics. In the collinear case $p_{kl}=\sigma_{kl}\kappa/z_{kl}$ with $\kappa^2=1/\lambda$ and $\sigma$ antisymmetric, so $z_k=\kappa^3F_k$, and $z_{12}=1$, $z_{23}=x$ give $x(F_1-F_2)=F_2-F_3$, the vanishing of $Q_\sigma(x)$ with $x=\tau-1$. The patterns $\sigma$ and $-\sigma$ give the same solutions.
\end{proof}

The real solutions are the Lagrange and Euler configurations. The dichotomy is the one of \cite[Lemma~E.2]{Tom26}, and the factors of $S_\mu$ are those of the complex three-body shapes listed in the appendix of \cite{AK12} ($S_1$--$S_8$ there, written in the $w$-coordinates of the bodies): $t^2-t+1$ is the equilateral factor, the $N$'s are the three other Lagrange factors, three $Q_\sigma$ are the Euler quintics, and the fourth is divisible by $x^2+x+1$. For $\mu=(1,2,7)$, $(1,2,8)$ and $(2,3,11)$, an elimination in Singular of all unknowns except $\tau$ gives an eliminant whose square-free part equals that of $S_\mu$, of degree $32$, the number of distinct shapes, and the system with the slice $z_1-z_2=1$ has $102$ solutions, $114$ with multiplicity.\cert

\subsection{Diagram 98}\label{ss:d98}
Let $A=\{4,5,6\}$ be the $z$-circled triangle and $B=\{1,2,3\}$ the $w$-circled one. In a general labelling $A$ is any set of three bodies and $B$ its complement, and the exchange of the colours exchanges the roles of $A$ and $B$.

\begin{lemma}\label{lem:d98type}
At a singular place with diagram $98$:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\zeta=-2$, $\omega=2$, $x=0$ on the pairs of $A$, and $\omega=-2$, $\zeta=2$, $x=0$ on the pairs of $B$;
\item $\zeta=\omega=-2$ and $x=-2$ on the nine pairs $jl$ with $j\in A$, $l\in B$;
\item $a_j=-2$ and $b_j\ge2$ for $j\in A$, and $b_l=-2$ and $a_l\ge2$ for $l\in B$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) is (F1), (b) is (O6) in both colours, and (c) is (F2).
\end{proof}

For $j,j'\in A$ and $l,l'\in B$ put $\hat z_j=\lc{z_j}{-2}$, $\tilde w_j=\lc{w_j}{2}$, $\hat p_{jj'}=\lc{p_{jj'}}{0}$, $\hat w_l=\lc{w_l}{-2}$, $\tilde z_l=\lc{z_l}{2}$, $\hat p_{ll'}=\lc{p_{ll'}}{0}$ and $\hat p_{jl}=\lc{p_{jl}}{2}$.

\begin{lemma}\label{lem:d98lead}
At a singular place with diagram $98$:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $(\hat z_A,\tilde w_A,\hat p_A)$ is a solution of $N(m_A)$, and $\hat z_j\ne0$ for $j\in A$;
\item $(\hat w_B,\tilde z_B,\hat p_B)$ is a solution of $N(m_B)$, and $\hat w_l\ne0$ for $l\in B$;
\item $\hat p_{jl}^2\hat z_j\hat w_l=-1$ for $j\in A$, $l\in B$;
\item $\sum_{j\in A,\,l\in B}m_jm_l\hat p_{jl}=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) In the $z$-equation of $j\in A$ the terms $Z_{jj'}$, $j'\in A$, have exponent $-2$ and the terms $Z_{jl}$, $l\in B$, exponent $4$. At exponent $-2$ it gives $\hat z_j=\sum_{j'}m_{j'}\hat p_{jj'}^3(\hat z_j-\hat z_{j'})$, where $\hat z_j-\hat z_{j'}=\lc{z_{jj'}}{-2}\ne0$. The $w$-equation of $j$ has terms of exponent $2$ ($W_{jj'}$) and $4$ ($W_{jl}$), and $b_j\ge2$. At exponent $2$ it gives the $w$-equation of $N(m_A)$. The distance equation of $jj'$ at exponent $0$ gives $\hat p_{jj'}^2(\hat z_j-\hat z_{j'})(\tilde w_j-\tilde w_{j'})=1$. (b) is (a) with the colours exchanged. (c) is (F4). (d) In $T_A$ all terms have exponent at least $2$ by (F4), and the coefficient of exponent $2$ is $-\sum m_jm_l\hat p_{jl}$.
\end{proof}

\begin{theorem}\label{thm:d98}
There is a nonzero polynomial $P\in\Z[\mu_1,\mu_2,\mu_3,\mu_1',\mu_2',\mu_3']$ such that for every singular place with diagram $98$, $P(m_A,m_B)=0$, where $m_A$ and $m_B$ list the masses of $A$ and $B$ in increasing order of the labels. Hence diagram $98$ occurs at no singular place for masses with $P_{98}(m)=\prod_{|A|=3}P(m_A,m_{A^c})\ne0$.
\end{theorem}

\begin{proof}
Let $\Phi(Y)=\prod_\varepsilon\sum_{j,l=1}^3\varepsilon_{jl}\sqrt{Y_{jl}}$, the product over the $256$ sign matrices $\varepsilon\in\{\pm1\}^{3\times3}$ with $\varepsilon_{11}=1$. A sign change of one $\sqrt{Y_{jl}}$ permutes the factors up to sign, and $256$ is even, so $\Phi$ is a polynomial in the $Y_{jl}$ with integer coefficients, homogeneous of degree $128$. Put
\begin{align*}
\Psi(t,t';\mu,\mu')&=\Phi\Big(\big(\mu_j^2\mu_l'^2\Lambda_j(t;\mu)\Lambda_l(t';\mu')\big)_{j,l}\Big),\\
R(t)&=\Res_{t'}\big(S_{\mu'}(t'),\Psi\big),\qquad P=\Res_t\big(S_\mu(t),R(t)\big).
\end{align*}
Resultants of polynomials with integer coefficients have integer coefficients.

Take a place with diagram $98$. By Lemmas~\ref{lem:d98lead}(a) and~\ref{lem:three}, a rotation of the $A$-data has shape $\tau$, a root of $S_{m_A}$, with $\hat z_j=s\,\ell_j(\tau)/M_A$ for some $s\ne0$. Likewise the $w$-shape $\tau'$ of the $B$-data is a root of $S_{m_B}$, with $\hat w_l=s'\ell_l(\tau';m_B)/M_B$. By Lemma~\ref{lem:d98lead}(c), $c_{jl}=m_jm_l\hat p_{jl}$ satisfies $c_{jl}^2=\kappa\,m_j^2m_l^2\Lambda_j(\tau)\Lambda_l(\tau')$ with the common factor $\kappa=-M_AM_B/(ss'\prod_j\ell_j(\tau)\prod_l\ell_l(\tau'))\ne0$. By Lemma~\ref{lem:d98lead}(d), $\sum c_{jl}=0$, so one factor of $\Phi(c^2)$ vanishes. Since $\Phi$ is homogeneous, $\Psi(\tau,\tau')=\kappa^{-128}\Phi(c^2)=0$. Hence $R(\tau)=0$ and then $P(m_A,m_B)=0$, since a resultant lies in the ideal of its two arguments. The $20$ choices of $A$ cover all labellings and the exchange of the colours.

$P\ne0$: at points with $(\mu_1+\mu_2)(\mu_1'+\mu_2')\ne0$, specialization commutes with the resultants and
\[
P(\mu,\mu')=(\mu_1+\mu_2)^{4D_1}(\mu_1'+\mu_2')^{4\cdot34\cdot D_2}\prod_{S_\mu(t)=0}\ \prod_{S_{\mu'}(t')=0}\Psi(t,t';\mu,\mu')
\]
over the roots with multiplicity, where $D_1$ and $D_2$ are the formal degrees of $R$ in $t$ and of $\Psi$ in $t'$. By Lemma~\ref{lem:dominant} below, $P((1,2,40),(1,2,40))\ne0$.
\end{proof}

For a root $t$ of $S_\mu$ put $A_j(t)=\mu_j|\Lambda_j(t)|^{1/2}$ $(j=1,2,3)$ and, if $A_3(t)\ne0$, $\rho(t)=A_3/(A_1+A_2+A_3)$.

\begin{lemma}\label{lem:dominant}
For $\mu=(1,2,40)$ every root $t$ of $S_\mu$ satisfies $|t|>\frac1{10}$ and $|t-1|>\frac1{10}$, and $\rho(t)>\frac9{10}$. Consequently $\Psi(t,t')\ne0$ for all roots $t$, $t'$ of $S_\mu$, and $P((1,2,40),(1,2,40))\ne0$.
\end{lemma}

\begin{proof}
The four Lagrange factors $t^2-t+1$, $N_{01}$, $N_{10}$, $N_{11}$ of $S_\mu$ take the value $1$ at $t=0$ and at $t=1$. If $p$ is one of them, $c\in\{0,1\}$ and $|h|\le\frac1{10}$, then $|p(c+h)-1|\le\sum_{k\ge1}|a_k|10^{-k}\le0.5951$, where the $a_k$ are the coefficients of $p(c+h)$ as a polynomial in $h$. So $p$ has no root in the two disks. For the collinear factors put $t=1+x$, so that $x=0$ and $x=-1$ correspond to $t=1$ and $t=0$. At $\mu=(1,2,40)$,
\begin{gather*}
Q_\sigma=40\,C_\sigma+D_\sigma,\qquad C_\sigma=\sigma_{13}x^3-\sigma_{23}(1+x)^3,\\
D_\sigma=3\sigma_{12}x^3(1+x)^2+\sigma_{12}x^2(1+x)^2-\sigma_{13}x^2-2\sigma_{23}(1+x)^2.
\end{gather*}
For $c\in\{0,-1\}$ we have $|C_\sigma(c)|=1$, and on the disk $|x-c|\le\frac1{10}$ the triangle inequality, applied to the coefficients of the expansions in powers of $x-c$, gives $|C_\sigma-C_\sigma(c)|\le0.332$ and $|D_\sigma|\le2.446$. So $40|C_\sigma|>|D_\sigma|$ there, and $Q_\sigma$ has no root in the disk. This proves the first statement.

Now $\ell_1=2+40t$, $\ell_2=40(t-1)-1$ and $\ell_3=2-3t$. Hence $|\ell_1|\ge20|t|$, $|\ell_2|\ge20|t-1|$, $|\ell_3|\le2+3|t|$ and $|\ell_3|\le1+3|t-1|$. In particular $\Lambda_3=\ell_1\ell_2\ne0$, and
\[
\frac{A_1}{A_3}=\frac1{40}\Big|\frac{\ell_3}{\ell_1}\Big|^{1/2}\le\frac1{40}\Big(\frac{2+3|t|}{20|t|}\Big)^{1/2}<0.03,\qquad \frac{A_2}{A_3}=\frac1{20}\Big|\frac{\ell_3}{\ell_2}\Big|^{1/2}\le\frac1{20}\Big(\frac{1+3|t-1|}{20|t-1|}\Big)^{1/2}<0.05.
\]
So $\rho(t)>1/1.08>\frac9{10}$.

For roots $t$, $t'$ the moduli of the nine numbers $\sqrt{Y_{jl}}$ in $\Psi(t,t')$ are $A_j(t)A_l(t')$. Each signed sum therefore satisfies
\[
\Big|\sum\varepsilon_{jl}\sqrt{Y_{jl}}\Big|\ge|\sqrt{Y_{33}}|-\sum_{(j,l)\ne(3,3)}|\sqrt{Y_{jl}}|=(2\rho(t)\rho(t')-1)\sum|\sqrt{Y_{jl}}|>0.
\]
So $\Psi(t,t')\ne0$, and $P((1,2,40),(1,2,40))\ne0$ by the formula in the proof of Theorem~\ref{thm:d98}.
\end{proof}

The bound says that the heaviest body of each triangle has the dominant cross coefficient in each shape. It is a two-triangle form of the T-test of \cite[\S E.9]{Tom26}. We take a heavy third body because it makes the proof short: $\ell_3$ does not depend on $\mu_3$, while $\ell_1$ and $\ell_2$ grow like $\mu_3$, and for large $\mu_3$ the shapes stay away from the collisions $t=0$ and $t=1$ of the third body with the other two. At $\mu=(1,2,7)$ we have checked the same conclusion by computation. At this mass point $S_\mu$ has $34$ roots counted with multiplicity, $32$ distinct, all with $\ell_1\ell_2\ell_3\ne0$, and the minimum of $\rho$ over them is $0.7641079385\ldots>1/\sqrt2$, attained at a root of $N_{11}$, $t\approx-0.47356+0.44477i$. Two independent programs enclose the roots and bound $\rho$, one in ball arithmetic in Arb with certified root isolation and one in exact rational arithmetic with Rouch\'e disks of radius $2^{-120}$. A direct evaluation of all $1024\times256$ signed sums at $\mu=\mu'=(1,2,7)$ gives $|\sum\varepsilon c|/\sum|c|\ge0.4537$, and the bound $\rho\rho'>\frac12$ also holds at $\mu=(1,2,7)$, $\mu'=(1,2,8)$.\cert

\begin{remark}\label{rem:P98set}
A sign matrix $\varepsilon_{jl}=\delta_j\eta_l$ splits the signed sum, for suitable branches, into
\[
\Big(\sum_j\delta_j\mu_j\sqrt{\Lambda_j(t)}\Big)\Big(\sum_l\eta_l\mu'_l\sqrt{\Lambda_l(t')}\Big).
\] Hence, for positive $\mu$ and $\mu'$, $P(\mu,\mu')=0$ whenever $P_{93}(\mu)=0$ or $P_{93}(\mu')=0$, with $P_{93}$ as in Theorem~\ref{thm:d93}. Since $P_{93}$ vanishes at equal masses, $P_{98}(m)=0$ for every $m$ in which three masses are equal.
\end{remark}

\subsection{Diagram 93}\label{ss:d93}
Here $B=\{1,2,3\}$ is the fully $w$-circled triangle and $A=\{4,5,6\}$ the $z$-triangle, in which $4$ is not $z$-circled. Let $\Phi_3=Y_1^2+Y_2^2+Y_3^2-2(Y_1Y_2+Y_2Y_3+Y_3Y_1)$, the product of the four sums $\sqrt{Y_1}\pm\sqrt{Y_2}\pm\sqrt{Y_3}$, and put $P_{93}(\mu)=\Res_t\big(S_\mu(t),\Phi_3(t;\mu)\big)$ with $Y_l=\mu_l^2\Lambda_l(t;\mu)$.

\begin{theorem}\cert\label{thm:d93}
For every singular place with diagram $93$, $P_{93}(m_B)=0$. Up to a constant factor and a power of $\mu_1+\mu_2+\mu_3$, $P_{93}$ is the product $L_1L_2L_3L_4(L_5L_5')^2L_6L_7L_8$ of the polynomials of \cite[\S8.2.1]{AK12}, and $P_{93}(1,2,40)\ne0$.
\end{theorem}

\begin{proof}
By (F1), $x=0$ on the pairs of $A$ and $B$. By (O6), $\zeta=\omega=-2$ on the pairs $(j,l)$ with $j\in\{5,6\}$, $l\in B$, and $\omega_{4l}=-2$. By (F2) and (F3), $b_5,b_6,b_4\ge2$ and $a_l\ge2$ for $l\in B$, and $-2<a_4<2$, since otherwise $4l$ would be a $w$-stroke. So $\zeta_{4l}=a_4$. In the torque $T_A$, the pairs $(4,l)$ satisfy (F4) with exponent $1-\frac12a_4<2$, and the pairs $(j,l)$ with $j=5,6$ give exponent $2$. Hence $\sum_{l\in B}m_l\lc{p_{4l}}{}=0$, and with $\lc{p_{4l}}{}^2=-1/(\hat z_4\hat w_l)$,
\[
\sum_{l\in B}\varepsilon_lm_l\hat w_l^{-1/2}=0
\]
for some signs, the relation \cite[(24)]{AK12}. The leading system of $B$ in both colours is complete, in the sense that the cross terms in the $w$-equations of $B$ at exponent $-2$ and in the $z$-equations at exponent $2$ have exponents at least $1-\frac32a_4>-2$ and $3-\frac12a_4>2$. So $(\hat w_B,\lc{z_B}{2},\hat p_B)$ solves $N(m_B)$, its shape is a root of $S_{m_B}$ by Lemma~\ref{lem:three}, and the displayed relation makes $\Phi_3$ vanish there. As in Theorem~\ref{thm:d98}, $P_{93}(m_B)=0$.

The factorization of $P_{93}$ is computed in $\Q[\mu_1,\mu_2,\mu_3]$ by Singular, one factor of $S_\mu$ at a time.\cert{} The factor $t^2-t+1$ of $S_\mu$ gives $L_5L_5'$ and a power of $M_\mu$; each of the three Lagrange quartics gives a factor of degree $28$ with $433$ terms ($L_6$, $L_7$, $L_8$); each Euler quintic gives a factor of degree $36$ with $565$ terms ($L_1$, $L_2$, $L_3$); and the fourth quintic gives $L_5L_5'$ and a factor of degree $22$ with $264$ terms ($L_4$). The two terms of each factor printed in \cite[p.~574]{AK12} agree with ours. By the proof of Lemma~\ref{lem:dominant}, at $\mu=(1,2,40)$ every root $t$ of $S_\mu$ has $A_3(t)>A_1(t)+A_2(t)$. Since $|\sqrt{Y_l}|=A_l(t)$, no sum $\sqrt{Y_1}\pm\sqrt{Y_2}\pm\sqrt{Y_3}$ vanishes, and $P_{93}(1,2,40)$, a nonzero multiple of the product of $\Phi_3(t;\mu)$ over the roots $t$ of $S_\mu$, is not zero. An exact evaluation shows that $P_{93}(1,2,7)\ne0$ as well.
\end{proof}

The factor $L_5=\frac12\sum(\mu_i-\mu_j)^2$ vanishes at equal masses, where the equilateral shape satisfies the displayed relation. Numerical minimization, which the proof does not use, finds further positive zeros of $P_{93}$ on twelve rays in $\mu$-space besides $(1,1,1)$, but we do not know whether there are others.

\subsection{Diagrams 88 and 94}\label{ss:d88}
In diagram $88$ the binary $\{5,6\}$ is $z$-circled and the four bodies of $K=\{1,2,3,4\}$ are $w$-circled and pairwise joined by $w$-strokes. In diagram $94$ the triangle $\{4,5,6\}$ carries $z$-strokes with circles at $5$, $6$, and $Q=\{1,2,3,4\}$ carries $w$-strokes on all pairs with circles at $1$, $2$, $3$.

For masses $\mu=(\mu_1,\dots,\mu_4)$, let $N_4(\mu)$ be the system of \S\ref{ss:three} for four bodies, in unknowns $X,Y\in\C^4$ and $p_{kl}$:
\[
X_k=\sum_{l\ne k}\mu_lp_{kl}^3(X_k-X_l),\qquad Y_k=\sum_{l\ne k}\mu_lp_{kl}^3(Y_k-Y_l),\qquad p_{kl}^2(X_k-X_l)(Y_k-Y_l)=1.
\]
The \emph{centred system} $\mathrm{KC}(\mu)$ consists of $X_4=0$, the $X$-equations of $N_4(\mu)$ for $k=1,2,3$, the equation $0=\sum_{l\le3}\mu_lp_{4l}^3(X_4-X_l)$, the $Y$-equations for $k=1,2,3$, and the six distance equations. It has no $Y$-equation for body $4$.

\begin{lemma}\label{lem:d88lead}
\begin{enumerate}[label=\textup{(\alph*)}]
\item At a singular place with diagram $88$, put $X_l=\hat w_l$, $Y_l=\lc{z_l}{2}$, $\hat p_{ll'}=\lc{p_{ll'}}{0}$ for $l,l'\in K$ and $\hat p_{jl}=\lc{p_{jl}}{2}$ for $j\in\{5,6\}$, $l\in K$. Then $m_5\hat z_5+m_6\hat z_6=0$, $(X,Y,\hat p_K)$ solves $N_4(m_K)$ with all $X_l\ne0$, $\hat p_{jl}^2\hat z_jX_l=-1$, and
\begin{equation}\label{eq:t88}
\sum_{j\in\{5,6\},\,l\in K}m_jm_l\hat p_{jl}=0.
\end{equation}
\item At a singular place with diagram $94$, put $X_l=\lc{w_l}{-2}$, $Y_l=\lc{z_l}{2}$ for $l\in Q$ and $\hat p$ as in \textup{(a)}. Then $m_5\hat z_5+m_6\hat z_6=0$, $(X,Y,\hat p_Q)$ solves $\mathrm{KC}(m_1,\dots,m_4)$ with $X_1X_2X_3\ne0$, $\hat p_{jl}^2\hat z_jX_l=-1$ for $l\le3$, and
\begin{equation}\label{eq:t94}
\sum_{j\in\{5,6\},\,l\le3}m_jm_l\hat p_{jl}=0.
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
(a) By (F1), $x=0$ on $56$ and on the pairs of $K$, with $\omega=-2$, $\zeta=2$ there. By (O6), $\zeta=\omega=-2$ on the eight pairs between $\{5,6\}$ and $K$, and by (F2), $b_5,b_6\ge2$ and $a_l\ge2$ for $l\in K$. The equations of the bodies of $K$ at exponent $-2$ in $w$ and $2$ in $z$, and their distance equations at exponent $0$, form $N_4(m_K)$, since the cross terms have exponent $4$. The relation $m_5\hat z_5+m_6\hat z_6=0$ is the coefficient of exponent $-2$ in $\sum_km_kz_k=0$, the relation for $\hat p_{jl}$ is (F4), and the sum over $j$ and $l$ is the coefficient of exponent $2$ in $T_{\{5,6\}}$.
(b) By (F1) and (O6), $x=0$ on all strokes and $\zeta=\omega=-2$ on $\{5,6\}\times\{1,2,3\}$. By (F2) and (F3), $b_5,b_6\ge2$, $a_1,a_2,a_3\ge2$, $a_4\ge2$ ($z_4=z_1-z_{14}$) and $b_4\ge2$ ($w_4=w_5-w_{54}$). So $X_4=0$. The $w$-equations of $1,2,3$ at exponent $-2$, their $z$-equations at exponent $2$ (with $\lc{z_{l4}}{2}=Y_l-Y_4$) and the distance equations form $\mathrm{KC}$. The $w$-equation of body $4$ at exponent $-2$ has the terms $W_{4l}$, $l\le3$, of exponent $-2$ and the terms $W_{4j}$, $j=5,6$, of exponent $2$, and gives the equation of the centre. The $z$-equation of body $4$ at exponent $2$ contains subleading terms of the triangle and is not used. In $T_{\{5,6\}}$ the pairs $(j,l)$, $l\le3$, give exponent $2$ by (F4). By Lemma~\ref{lem:rewrite} with $o=4$, the pairs $(j,4)$ contribute $m_4(w_4B_z-z_4B_w)$ with $B_z=\sum_{l\le3}m_lZ_{4l}$ and $B_w=w_4-A_w$ of exponent at least $2$, although each $W_{4l}$ has exponent $-2$. So this contribution has exponent at least $4$.
\end{proof}

\begin{lemma}\label{lem:Ttest}
\begin{enumerate}[label=\textup{(\alph*)}]
\item In \textup{(a)} of Lemma~\ref{lem:d88lead}, if $m_k|X_k|^{-1/2}>\sum_{l\in K,\,l\ne k}m_l|X_l|^{-1/2}$ for some $k\in K$, then \eqref{eq:t88} fails for every choice of the signs of the $\hat p_{jl}$ and all $m_5,m_6>0$.
\item In \textup{(b)}, with $\beta_l=m_l/\sqrt{X_l}$ for fixed branches and $\rho=(m_6/m_5)^{3/2}$, \eqref{eq:t94} holds for some signs of the $\hat p_{jl}$ if and only if $\sum_{l\le3}\varepsilon_l\beta_l(1+is_l\rho)=0$ for some $\varepsilon,s\in\{\pm1\}^3$.
\end{enumerate}
\end{lemma}

\begin{proof}
Write $\hat z_5=cm_6$, $\hat z_6=-cm_5$ with $c\ne0$. Then $\hat p_{6l}/\hat p_{5l}=\pm(\hat z_5/\hat z_6)^{1/2}=\pm i(m_6/m_5)^{1/2}$, so $m_l(m_5\hat p_{5l}+m_6\hat p_{6l})=m_lm_5\hat p_{5l}(1\pm i\rho)$, with $|\hat p_{5l}|=|cm_6X_l|^{-1/2}$. (a) The $l$-th group has modulus $K'm_l|X_l|^{-1/2}$ with $K'=|c|^{-1/2}(m_5m_6)^{-1/2}(m_5^3+m_6^3)^{1/2}$, independent of $l$ and of the signs, and the dominant group cannot be cancelled. (b) The sum is $m_5(-cm_6)^{-1/2}\sum_l\varepsilon_l\beta_l(1+is_l\rho)$.
\end{proof}

The inequality in (a) is the T-test of \cite[\S E.9]{Tom26} for a cluster of four bodies, and (b) is the reduction of \cite[\S8.2]{AK12}. The test in (a) does not involve the binary masses and is invariant under rotation. For the computations we use the following systems. For $N_4$: the slice $X_1-X_2=1$, the unknowns $v=(X_2,X_3,Y_1,Y_2,Y_3,p_{12},\dots,p_{34})$, and $X_4$, $Y_4$ eliminated by the centre of mass. The system $S_{12}$ consists of the $X$- and $Y$-equations of bodies $1,2,3$ and the six distance equations, and $S_{11}$ omits the $Y$-equation of body $3$. Let $\mathrm{Xeq}_k=X_k-\sum_{l\ne k}\mu_lp_{kl}^3(X_k-X_l)$ and $\mathrm{Yeq}_k$ likewise. Since $\sum_k\mu_k(Y_k\,\mathrm{Xeq}_k-X_k\,\mathrm{Yeq}_k)=0$ identically, $\mu_3(X_4-X_3)\mathrm{Yeq}_3$ lies in the ideal of $S_{11}$, and $X_4-X_3$ is invertible modulo the distance equation of $34$. So $S_{11}$ and $S_{12}$ generate the same ideal when $\mu_3$ and $\mu_4$ are units. For $\mathrm{KC}$: the slice $X_1=1$, the twelve unknowns $X_2,X_3,Y_1,\dots,Y_4,p_{12},\dots,p_{34}$, and the system $\mathrm{KC}_{12}$ of the four $X$-equations, the $Y$-equations of bodies $1,2$ and the six distances, which generates the same ideal as $\mathrm{KC}$ when $\mu_3$ is a unit, by the same torque identity. In the coordinates $U_k=Y_k-Y_4$ $(k\le3)$, with $U_4=0$, the $Y$-equation of body $1$ reads $Y_4=-U_1+\sum_{l\ne1}\mu_lp_{1l}^3(U_1-U_l)$. Substituting this into the $Y$-equations of bodies $2$ and $3$ gives a system $\mathrm{KC}'$ of twelve equations in the eleven unknowns $X_2,X_3,U_1,U_2,U_3,p_{12},\dots,p_{34}$: the four $X$-equations, these two $Y$-equations and the six distances. Its coordinate ring is isomorphic to that of $\mathrm{KC}$.

\begin{lemma}\label{lem:goodred}
Let $p$ be a prime, $\mu_0\in\Z^4$, $\cO=\Z[\mu]_{\mathfrak m}$ with $\mathfrak m=(p,\mu-\mu_0)$, and let $I\subseteq\cO[v]$ be an ideal. Let $\xi_1,\dots,\xi_N$ be polynomials in $v$ over $\cO$ that are units modulo $I$, such that each $v_i$ is congruent modulo $I$ to a polynomial in $\xi$ over $\cO$, and let $G\subseteq I$ be a finite set of polynomials in $\xi$. Suppose that all coefficients of the polynomials in $G$ are units of $\cO$. Let $\Sigma$ be a fan with support the tropical prevariety of $G$ that contains all faces of its cones and on the relative interior of each cone of which the initial form of each $g\in G$ is constant. Suppose that for each cone $\Gamma$ of $\Sigma$ that meets $\{w:\min_iw_i<0\}$, the initial forms of $G$ at a relative interior point of $\Gamma$, reduced modulo $\mathfrak m$, that is, specialized at $\mu=\mu_0$ and reduced modulo $p$, have no common zero in $(\bar\F_p^*)^N$. Then $\cO[v]/I$ is a finite $\cO$-module.
\end{lemma}

\begin{proof}
It suffices that each $v_i$ is integral over $\cO$ modulo each minimal prime $\mathfrak q$ of $I$, since a product of monic relations modulo the minimal primes is nilpotent modulo $I$. Let $\cO'$ be the image of $\cO$ in $R'=\cO[v]/\mathfrak q$, a local domain with maximal ideal $\mathfrak m'$, and suppose $y=v_i$ is not integral over $\cO'$. Then $\mathfrak m'\cO'[1/y]+(1/y)$ is a proper ideal: a relation $1=a_0+a_1y^{-1}+\dots+a_ry^{-r}$ with $a_0\in\mathfrak m'$ would make $y$ integral, since $1-a_0$ is a unit. A maximal ideal $\mathfrak n$ containing it gives a Noetherian local domain $\cO'[1/y]_{\mathfrak n}$ that dominates $\cO'$, and a discrete valuation $\nu$ of the fraction field of $R'$ dominating it \cite[II, 7.1.7]{EGA2}. Then $\nu(y)<0$, $\nu\ge0$ on $\cO'$, the residue field $\kappa$ of $\nu$ is an extension of $\F_p$, and the masses have residues $\mu_0$ modulo $p$. The point $\xi$ of the generic point of $R'$ has $w=\nu(\xi)\in\Z^N$ with some $w_i<0$, because $\nu(y)<0$ and $y$ is congruent modulo $I\subseteq\mathfrak q$ to a polynomial in $\xi$ over $\cO$. Each $g\in G$ vanishes at $\xi$ and has unit coefficients, so its terms of least weight $w$ cancel. Hence $w$ lies in the prevariety, and the initial forms of $G$ at $w$, reduced modulo $\mathfrak m$, vanish at the residues in $\kappa$ of the $\xi_i\pi^{-w_i}$, where $\pi$ is a uniformizer of $\nu$. These residues are nonzero. So the ideal of $\F_p[\xi,t]$ generated by the reduced initial forms and $1-t\prod_i\xi_i$ has a zero over $\kappa$, hence is proper, and by the Nullstellensatz it has a zero over $\bar\F_p$: the reduced initial forms have a common zero in $(\bar\F_p^*)^N$. The point $w$ lies in the relative interior of a cone of $\Sigma$ that meets $\{\min w_i<0\}$, and the initial forms are constant on that relative interior. This contradicts the hypothesis.
\end{proof}

\begin{lemma}\cert\label{lem:d88comp}
Let $\mu_0=(1,2,3,30)$ and $p=65521$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item The hypotheses of Lemma~\ref{lem:goodred} hold for the ideal of $S_{12}$ and for the ideal of $\mathrm{KC}_{12}$, with $17$ coordinates $\xi$ (the differences of the positions in each colour and the $p_{kl}$), with $G$ of $30$ and of $101$ polynomials, respectively, and with prevarieties of $2149$ and $224$ cones, of which $2148$ and $223$, respectively, meet $\{\min w_i<0\}$.
\item $\dim_{\F_p}\F_p[v]/\langle S_{12}(\mu_0)\rangle=20{,}316$ and $\dim_{\F_p}\F_p[v']/\langle\mathrm{KC}'(\mu_0)\rangle=4788$, where $v'$ are the unknowns of $\mathrm{KC}'$.
\item There are $20{,}316$ pairwise disjoint boxes in $\C^{11}$, each containing exactly one zero of $S_{11}(\mu_0)$, at which the Jacobian is invertible and $X_3-X_4\ne0$, and on each of which the test of Lemma~\ref{lem:Ttest}\textup{(a)} holds for some $k$ with $m_k|X_k|^{-1/2}-\sum_{l\ne k}m_l|X_l|^{-1/2}\ge0.1756\sum_lm_l|X_l|^{-1/2}$.
\item There are $4788$ pairwise disjoint boxes in $\C^{12}$, each containing exactly one zero of $\mathrm{KC}_{12}(\mu_0)$, at which the Jacobian is invertible and $X_3\ne0$, and on each of which, for $(m_5,m_6)=(1,2)$, all $32$ sums of Lemma~\ref{lem:Ttest}\textup{(b)} have modulus at least $1.47\cdot10^{-6}(1+\rho^2)^{1/2}\sum_l|\beta_l|$.
\end{enumerate}
\end{lemma}

\begin{proof}[Description of the computations]
(a) The polynomials of $G$ are the linear relations among the differences, the equations of the bodies, the distance equations, torques $T_J$, subset sums and the identity of Lemma~\ref{lem:U}(a), written in the coordinates $\xi$, and for $\mathrm{KC}$ also the eliminations of $Y_4$ through the $Y$-equation of each of the bodies $1,2,3$. Each lies in the ideal by an explicit identity over $\Z[\mu,1/M,1/\mu_4]$. A symbolic check shows that no coefficient vanishes at $\mu_0$ modulo $p$ ($354$ and $1133$ terms). The prevarieties are computed by gfan. The listed cones are all faces of the common refinement, so the initial forms are constant on their relative interiors. Since the distance equations force $2w_{p_{kl}}+w_{X_{kl}}+w_{Y_{kl}}=0$, the prevariety meets the nonnegative orthant only at $0$. For each cone that meets $\{\min w_i<0\}$, the initial system at a relative interior point is saturated by the product of the coordinates and found to be the unit ideal over $\F_p$ in Singular. (b) Gr\"obner bases modulo $p$ in msolve, and the count of standard monomials of the leading ideal. (c), (d) Krawczyk's test \cite{Kra69} in ball arithmetic (Arb), at $256$ and $384$ bits, on boxes of radius below $6\cdot10^{-47}$ and $5\cdot10^{-54}$, respectively, around approximate zeros. The inequalities are lower bounds in ball arithmetic. The Arb programs do not test that the boxes are pairwise disjoint, but the two checks of Table~\ref{tab:verif2} do.
\end{proof}

Section~\ref{sec:verification} lists the checks of these computations. By (a) and Lemma~\ref{lem:goodred}, the coordinate rings of $S_{12}(\mu_0)$ and $\mathrm{KC}_{12}(\mu_0)$ over $\Z_{(p)}$, obtained by the base change $\mu\mapsto\mu_0$ from $\cO$, are finite $\Z_{(p)}$-modules. Their fibres over $\F_p$ have the dimensions in (b), since $\mu_3$ is a unit modulo $p$, so that $\mathrm{KC}_{12}$, $\mathrm{KC}$ and $\mathrm{KC}'$ have isomorphic coordinate rings. By Nakayama's lemma the two systems have at most $20{,}316$ and $4788$ complex zeros at $\mu_0$, respectively, so the boxes of (c) and (d) contain all of them.

\begin{theorem}\label{thm:d88}
There are nonzero polynomials $P_{88},P_{94}\in\Z[\mu_1,\dots,\mu_4;\mu_5,\mu_6]$ with the following properties. If a singular place has diagram $88$ with $z$-binary $\{a,b\}$ and $w$-cluster $K=\{1,\dots,6\}\setminus\{a,b\}$, then $P_{88}(m_K;m_a,m_b)=0$. If a singular place has diagram $94$ with the labelling of Table~\ref{tab:ten}, then $P_{94}(m_1,m_2,m_3,m_4;m_5,m_6)=0$.
\end{theorem}

\begin{proof}
Let $V\subseteq\{\mu\in\C^4:\mu_4\ne0\}\times\C^{11}$ be the zero set of $S_{12}$, whose coefficients involve $1/\mu_4$, over the space of the masses $\mu$. By Lemmas~\ref{lem:goodred} and~\ref{lem:d88comp}(a), $R=\cO[v]/\langle S_{12}\rangle$ is a finite $\cO$-module, so by Nakayama's lemma $\dim_{\Q(\mu)}R\otimes\Q(\mu)\le\dim_{\F_p}R/\mathfrak mR=20{,}316$ (Lemma~\ref{lem:d88comp}(b)). Hence there are $d\le20{,}316$ and a dense Zariski-open set $W\subseteq\{\mu_4\ne0\}$ over which all fibres of $V$ have exactly $d$ points. By Lemma~\ref{lem:d88comp}(c) and the implicit function theorem for $S_{11}$, there are a neighbourhood $N$ of $\mu_0$ and $20{,}316$ disjoint analytic branches of zeros of $S_{12}$ over $N$. For $\mu\in N\cap W$ the fibre contains these branches and has $d\le20{,}316$ points, so $d=20{,}316$ and the fibre consists of the branches. The test of Lemma~\ref{lem:Ttest}(a) holds on each branch at $\mu_0$ with a positive margin, hence on a smaller neighbourhood $N'$. By Lemma~\ref{lem:Ttest}(a), for $\mu\in N'\cap W\cap\R^4$ and all $m_5,m_6>0$ the leading data of Lemma~\ref{lem:d88lead}(a) do not exist.

Let $\mathrm{Bad}\subseteq\C^6$ be the set of $(\mu,m_5,m_6)$ for which the leading data of Lemma~\ref{lem:d88lead}(a) exist, with any signs. It is the projection of a constructible set defined over $\Q$, hence constructible. It misses $(N'\cap W\cap\R^4)\times\R_{>0}^2$, which contains a nonempty open subset of $\R^6$, since the real points of $\C^4\setminus W$ are nowhere dense in $\R^4$. A constructible set that is Zariski dense in $\C^6$ contains a nonempty Zariski-open set, whose real points are dense in $\R^6$. So the Zariski closure of $\mathrm{Bad}$ is a proper subvariety defined over $\Q$. We take for $P_{88}$ a nonzero integer polynomial vanishing on it. Other labellings follow by permuting the masses, and the exchange of the colours gives the same leading system with $z$ and $w$ exchanged.

For diagram $94$ the same argument applies to $\mathrm{KC}_{12}$ with the count $4788$ and Lemma~\ref{lem:d88comp}(d): the $32$ sums of Lemma~\ref{lem:Ttest}(b) are continuous along the branches and bounded away from zero at $(\mu_0,1,2)$, so they do not vanish on a neighbourhood of $(\mu_0,1,2)$ in $\C^4\times\R_{>0}^2$, intersected with $W\times\R_{>0}^2$.
\end{proof}

Note that for $\mu_0$ the test of Lemma~\ref{lem:Ttest}(a) holds at every solution of $N_4$, whereas the same test over the bodies $1,2,3$ fails at $1092$ of the $4788$ solutions of $\mathrm{KC}$.\cert{} For this reason we test the sums of Lemma~\ref{lem:d88comp}(d) at a single ratio of the binary masses. At $36$ solutions of $\mathrm{KC}$ the smallest of the $32$ sums has relative margin below $10^{-4}$ in the sense of Lemma~\ref{lem:d88comp}(d). These solutions have $|Y_4|>5\cdot10^3$, and there $|Y_l/Y_4|<3\cdot10^{-5}$ for $l\le3$.\cert{} At these solutions the equation of the centre, which reads $\sum_{l\le3}\varepsilon_lm_lX_l^{-1/2}(1-Y_l/Y_4)^{-3/2}=0$ for suitable signs, is close to $\sum_{l\le3}\varepsilon_lm_lX_l^{-1/2}=0$, which says that a sum of Lemma~\ref{lem:Ttest}(b) with $s_1=s_2=s_3$ vanishes.