\section{Curve families and the RL line}\label{sec:curves} Case (ii) of Theorem~\ref{thm:S} concerns the mass vectors whose label set contains a curve set, that is, one of the nine sets $D_T$ of Table~\ref{tab:curvesets} after a relabelling. Each $D_T$ defines a curve of mass vectors up to scaling (Lemma~\ref{lem:DT}), along which a generic label set $S_C$ holds identically. The pieces enabled by $S_C$ include the pieces $\pc{d14a}$, $\pc{d14b}$ and $\pc{d15}$ of the clusters. We settle the points of these curves by excluding such pieces point by point along each curve. In this section we do so along the whole curves. We prove that every point of $T_9$ is settled (Proposition~\ref{prop:farT9}), and that every point of these curves other than the points of $T_9$ and of the special rays has finitely many central configurations (Theorem~\ref{thm:curves}). \subsection{The families}\label{ss:families} Table~\ref{tab:families} lists the families of mass vectors defined by the sets $D_T$, with their parametrizations and generic label sets. The labels follow the supplementary files. The line $\mathrm{RL}$ plays the role of a sixth family. We define the family $T_9$, with the constant $\rho_0$, in \S\ref{sec:T9} and treat it there. \begin{table}[ht] \centering \small \begin{tabular}{@{}lp{0.47\textwidth}p{0.40\textwidth}@{}} \toprule & masses $(m_1,\dots,m_5)$; parameter & generic label set $S_C$\\ \midrule $T_1$ & $(t^3,\,(1+t^2)^2,\,t^2(1+t^2)^2,\,t^2(1+t^2)^2,\,(1+t^2)^2)$; $t>0$ & $\lab{R}(23;1)$ $\lab{R}(24;1)$ $\lab{R}(35;1)$ $\lab{R}(45;1)$ $\lab{EQ}(2,5)$ $\lab{EQ}(3,4)$\\ $T_2$ & $(t^3,\,t^2(1+t^2)^2,\,(1+t^2)^2,\,(1+t^2)^2,\,(1+t^2)^2)$; $t>0$ & $\lab{R}(23;1)$ $\lab{R}(24;1)$ $\lab{R}(25;1)$ $\lab{EQ}(3,4)$ $\lab{EQ}(3,5)$ $\lab{EQ}(4,5)$\\ $T_3$ & $(a,a,a,a,1)$; $a>0$ & $\lab{EQ}(i,j)$ $(1\le i0$, $y=f^{1/2}$ & $\lab{R}(23;1)\,\lab{R}(24;1)\,\lab{R}(25;1)\,\lab{EQ}(3,4)$\\ $T_5$ & $(q^6A^2,\,(1+q^4)^2A^2,\,q^6(1+q^4)^2,\,q^4(1+q^4)^2A^2,\,q^4(1+q^4)^2A^2)$; $q\neq0$ & $\lab{E}(123)\,\lab{R}(24;1)\,\lab{R}(25;1)\,\lab{EQ}(4,5)$\\ $T_7$ & $((1-s)^2,\,s^2(1-s)^2,\,s^2,\,s^2,\,s^2(1-s)^2)$; $s\ne0,1$ & $\lab{E}(123)$ $\lab{E}(124)$ $\lab{E}(135)$ $\lab{E}(145)$ $\lab{EQ}(2,5)$ $\lab{EQ}(3,4)$\\ $T_8$ & $(q^2r^4fC^2,\,r^4g^2C^2,\,q^2r^4fg^2,\,fg^2C^2,\,r^4fg^2C^2)$; $r>0$, $q^4=f$ & $\lab{E}(123)\,\lab{R}(24;1)\,\lab{R}(25;1)$\\ $\mathrm{RL}$ & $(4,1,4,4,\mu)$; $\mu>0$ & $\lab{E}(123)$ $\lab{R}(13;2)$ $\lab{E}(124)$ $\lab{R}(14;2)$ $\lab{E}(234)$ $\lab{R}(34;2)$ $\lab{EQ}(1,3)$ $\lab{EQ}(1,4)$ $\lab{EQ}(3,4)$\\ \bottomrule \end{tabular} \caption{The families. Here $A=1+q^3+q^4$, $B=1+r+r^2$, $f=r^3+r^2+r$, $g=(1+r)(1+r^2)=1+f$ and $C=g+q^3$, with $q$ real. On $T_5$ and $T_8$ both signs of $q$ occur. Each $S_C$ also contains the labels $\lab{D6}(ab|cd)$ with $\lab{EQ}(a,b)$ and $\lab{EQ}(c,d)$ in $S_C$ (on $T_1$, $T_3$ and $T_7$), as $S(m)$ does (\S\ref{ss:pieces}).}\label{tab:families} \end{table} All masses in Table~\ref{tab:families} are positive: $A\ge\frac{229}{256}$ for real $q$, and on $T_8$, $C=1+f+q^3\ge1+f-f^{3/4}>0$. The \emph{transversal cluster} of a family is $Q_1=\{2,3,4,5\}$ for $T_1$, $T_2$, $T_4$, $T_5$, $T_7$ and $T_8$, with masses proportional to \begin{gather*} (1,c,c,1),\ c=t^2\ \text{on $T_1$};\qquad (t^2,1,1,1)\ \text{on $T_2$};\qquad (r^3,B,B,r^4B)\ \text{on $T_4$};\\ (A^2,q^6,q^4A^2,q^4A^2)\ \text{on $T_5$};\qquad (1,c,c,1),\ c=(1-s)^{-2}\ \text{on $T_7$};\\ (r^4C^2,\,q^6r^4,\,q^4C^2,\,r^4q^4C^2)\ \text{on $T_8$}. \end{gather*} On $T_3$ the clusters are $Q_k$, $k\le4$, with masses proportional to $(1,1,1,1/a)$, and $Q_5$ with masses $(1,1,1,1)$. On $\mathrm{RL}$ it is $Q_2=\{1,3,4,5\}$ with masses $(4,4,4,\mu)$. Two cluster families recur, up to relabelling and scaling: $F_A(\nu)=(1,1,1,\nu)$ and $F_B(c)=(1,c,c,1)$. \begin{lemma}\label{lem:DT} Let $T$ be one of the nine names of Table~\ref{tab:curvesets}, and let $m\in(\R_{>0})^5$ satisfy every label of $D_T$. Then a positive multiple of $m$ is a point of the family $T$, or of $T_2$ if $T=T_4$, or of $T_5$ if $T=T_8$. If $T=\mathrm{RL}$, a positive multiple of $m$ is the image of a point of $\mathrm{RL}$ under the exchange of the bodies $1$ and $2$. If $T=T_9$, it is a point of $T_9$ or such an image. \end{lemma} \begin{proof} With $x_i$ as in \S\ref{ss:conditions}, for positive masses $\lab{R}(ab;s)$ says $x_s=(x_a^2+x_b^2)/(x_ax_b)^{1/2}$, and $\lab{E}(abc)$ says that one of $x_a,x_b,x_c$ is the sum of the other two. The case $T_3$ is immediate. $T_1$ and $T_2$: we scale so that $m_2=m_5=1$ and $m_3=m_4=t^2$ with $t>0$ ($T_1$), resp.\ $m_3=m_4=m_5=1$ and $m_2=t^2$ ($T_2$). Then $\lab{R}(23;1)$ gives $m_1=t^3/(1+t^2)^2$. So $(1+t^2)^2m$ is the point $t$ of the family. $T_7$: scale so that $x_2=x_5=1$. Also $x_3=x_4$. If $x_1+x_3=1$ or $x_1=1+x_3$, put $s=x_1$. If $x_3=1+x_1$, put $s=-x_1$. Then $s\ne0,1$, $x_1=|s|$ and $x_3=|1-s|$. So $s^2(1-s)^2m$ is the point $s$ of $T_7$. $T_4$: we scale so that $x_5=1$, and write $x_3=x_4=r^2$, $x_2=y$ with $r,y>0$. Then $\lab{R}(23;1)$ and $\lab{R}(25;1)$ give $x_1=(y^2+r^4)/(y^{1/2}r)=(y^2+1)/y^{1/2}$, that is, $y^2(1-r)=r(1-r)(1+r+r^2)$. If $r=1$, then $m_3=m_4=m_5$ and $m$ satisfies $D_{T_2}$. Otherwise $y^2=f(r)$ and $m$ is a multiple of the point $r$ of $T_4$. $T_5$ and $T_8$: scale so that $x_5=1$ and write $x_2=q^2$, $x_4=r^2$ with $q,r>0$. On $T_5$, $\lab{EQ}(4,5)$ gives $r=1$. On $T_8$, $\lab{R}(24;1)$ and $\lab{R}(25;1)$ give $(q^4+r^4)/(qr)=(q^4+1)/q$, that is, $q^4(1-r)=r(1-r)(1+r+r^2)$. Hence $r=1$, and $m$ satisfies $D_{T_5}$, or $q^4=f(r)$. In both cases $x_1=(1+q^4)/q$ and $x_1-x_2=(1-q^3+q^4)/q>0$, since $1-u^3+u^4\ge\frac{229}{256}$ for $u>0$. Hence $\lab{E}(123)$ means $x_3=(1+q^4\pm q^3)/q$. Replacing $q$ by $-q$ in the case of the minus sign, $x_2=q^2$, $x_1=(1+q^4)/|q|$ and $x_3=(1+q^3+q^4)/|q|$ with $q\in\R\setminus\{0\}$. Hence $m$ is a multiple of the point $q$ of $T_5$, resp.\ of the point $(r,q)$ of $T_8$. $\mathrm{RL}$: since $x_2=x_3=x_4$, $\lab{E}(123)$ forces $x_1=2x_2$. Hence $m$ is a multiple of $(1,4,4,4,\mu)$; the exchange of $1$ and $2$ maps the point $\mu$ of $\mathrm{RL}$ to this vector. $T_9$: since $x_2=x_3$, $\lab{E}(123)$ forces $x_1=2x_2$. Scale so that $x_4=1$ and put $x_2=u^2$, $u>0$. Then $\lab{R}(24;1)$ reads $2u^2=(u^4+1)/u$, that is, $(u-1)(u^3-u^2-u-1)=0$. For $u=1$, $m$ is a multiple of $(1,4,4,4,\mu)$ as for $\mathrm{RL}$. Otherwise $u=\theta_0$, the real root of $u^3=u^2+u+1$, and $m=(\theta_0^{-4}/4,\theta_0^{-4},\theta_0^{-4},1,m_5)$ with $\theta_0^{-4}=\rho_0$. \end{proof} \subsection{Settling a point of a family}\label{ss:howsettled} A piece is \emph{excluded} at $m\in(\R_{>0})^5$ if no singular place $P$ of a nontrivial curve $C\subseteq Y_m$ has its type $x(P)$ in the attained set of the piece. Let $S$ be a set of labels, $E$ a set of pieces and $V$ a D5 hull. The pair $(S,E)$ is \emph{certified} relative to $V$ if there are $l_1,\dots,l_r\in\Z^{10}$ that satisfy (a) and (b) of Definition~\ref{def:cert} for the list $\cA_V(S)$ without the attained faces of the pieces of $E$ and of all pieces $\pc{d2}$ and $\pc{d9}$. A certificate for $(S,E)$ is one for every $(S',E')$ with $S'\subseteq S$ and $E'\supseteq E$ (\S\ref{ss:certs}). A \emph{minimal exclusion set} of $S$ is a set $E$ of pieces $\pc{d14a}$, $\pc{d14b}$, $\pc{d15}$, minimal under inclusion, such that $(S,E)$ is certified. In this section all certificates are relative to the hull $\Vtwo^{\mathrm w}$ or to $\VN$ (Remark~\ref{rem:hulls}). \begin{lemma}\label{lem:criterion} Let $m\in(\R_{>0})^5$, let $V$ be a D5 hull valid at $m$, let $S\supseteq S(m)$, and let $E$ be a set of pieces each of which is excluded at $m$. If $(S,E)$ is certified relative to $V$, then $m$ is settled. \end{lemma} \begin{proof} This is the proof of Proposition~\ref{prop:cert}, applied to $\cA_V(S)$ without the faces of the pieces of $E$ and of the pieces $\pc{d2}$ and $\pc{d9}$, which are excluded at every positive mass (\S\ref{ss:certs}): by Theorem~\ref{thm:types}, the validity of $V$ and the definition of an excluded piece, at each singular place of a curve as there, the type lies in the relative interior of a face in this list. \end{proof} We exclude pieces $\pc{d14a}$, $\pc{d14b}$ and $\pc{d15}$ by the following results. We carry out the computations behind them for solutions of the rows over the field of Puiseux series with prescribed valuations. The expansions of the coordinates at a place are such solutions. \begin{enumerate}[label=\textup{(X\arabic*)}] \item By Corollary~\ref{cor:Kdrop}, if $\mathrm{K14}(m_{Q_j})$ has no solution, $\pc{d14a}(j)$ and all $\pc{d14b}(j,k)$ are excluded. At a mass point given exactly over a number field we decide this system by Theorem~\ref{thm:KA}, one run of which covers all conjugates of the point. \item By Theorem~\ref{thm:D14}, each $\pc{d14b}(j,k)$ is excluded at every $m\in(\R_{>0})^5$. \item At a place of type $\pc b_k(j)$, the leading coefficients of the cluster form a $\mathrm{K15}(m_{Q_j},k)$ solution with a bad shape (Lemma~\ref{lem:d15lead} and Theorem~\ref{thm:order3}). Thus $\pc{d15}(j,k)$ is excluded at $m$ if no $\mathrm{K15}(m_{Q_j},k)$ solution has a bad shape. If the three masses of $Q_j\setminus\{k\}$ are equal, $\pc{d15}(j,k)$ is excluded (Corollary~\ref{cor:d15equal}). \item By Corollary~\ref{cor:nod15}, each $\pc{d15}(j,k)$ is excluded at every $m\in(\R_{>0})^5$. \end{enumerate} For a family with transversal cluster $Q_1$ we write $\pi(1,k)=\{\pc{d14b}(1,k),\pc{d15}(1,k)\}$ and $\Pi(1)$ for the nine pieces $\pc{d14a}(1)$, $\pc{d14b}(1,k)$, $\pc{d15}(1,k)$, $k\in Q_1$. Similarly for other clusters. By (X2) and (X4), each pair $\pi(j,k)$ is excluded at every $m\in(\R_{>0})^5$. A \emph{special point} of a family is a point at which $\mathrm{K14}$ has a solution for its transversal cluster (on $T_3$, for one of $Q_1,\dots,Q_4$). \begin{theorem}\label{thm:SJ} Let $T$ be one of $T_1$, $T_2$, $T_3$, $T_4$, $T_5$, $T_7$, $T_8$, and read $\lab{H}$ as $\cD_3$ \textup{(}Theorem~\ref{thm:D3}\textup{)}. On each branch of $T$, $S(m)=S_C$ except at the \emph{jump points}, where a label $\lab{E}$, $\lab{R}$, $\lab{H}$ or $\lab{EQ}$ outside $S_C$ holds. At each jump point a label set containing $S(m)$ is determined. The numbers of the jump points on each family are those of Table~\ref{tab:curvecounts}. \end{theorem} \begin{table}[ht] \centering \small \begin{tabular}{@{}lrrrrrrr@{}} \toprule & $T_1$ & $T_2$ & $T_3$ & $T_4$ & $T_5$ & $T_7$ & $T_8$\\ \midrule jump points (Theorem~\ref{thm:SJ}) & $7$ & $3$ & $2$ & $8$ & $11$ & $9$ & $18$\\ \quad points of \S\ref{sec:points} & $2$ & $0$ & $1$ & $0$ & $2$ & $2$ & $0$\\ \quad points of Table~\ref{tab:noK} & $1$ & $2$ & $1$ & $0$ & $2$ & $7$ & $0$\\ \quad other jump points & $4$ & $1$ & $0$ & $8$ & $7$ & $0$ & $18$\\ special points & $16$ & $6$ & $6$ & & & $32$ & \\ sets in the list $\mathcal B$ (Theorem~\ref{thm:options}) & & & & $1992$ & $108$ & $9$ & $108$\\ \bottomrule \end{tabular} \caption{Counts along the families: the jump points, split by their treatment in Proposition~\ref{prop:exceptional}; the special points; and the numbers of sets with a witness in the proof of Theorem~\ref{thm:options}.}\label{tab:curvecounts} \end{table} \begin{proof} Each label is a disjunction of polynomial conditions (Appendix~\ref{app:labels}). We compose every polynomial exactly with the masses of the family. On $T_4$ and $T_8$ we reduce it modulo the relation $y^2=f$, resp.\ $q^4=f$, and replace it by its norm over the conjugates of $y$, resp.\ $q$, a nonzero polynomial in $r$ because the relation is irreducible over $\Q(r)$. We factor the result over $\Q$. We isolate the real roots of the factors in ball arithmetic at $400$ bits or more and show that each root ball lies on one side of every boundary point of a branch, unless the boundary point is itself a root. On $T_4$ and $T_8$ a root $r_0$ is a jump point only if the label polynomial vanishes at the conjugate of the branch. We decide this by balls at the conjugates or, where they do not separate, by an exact gcd over $\Q(r_0)$. We decide an $\lab{H}$ label at a root of its polynomial by its two conditions $\mathrm{Tri}$ in ball arithmetic, and the $\lab{H}$ labels whose polynomial vanishes identically on a family through the exact gcd of their two $\mathrm{Tri}$ conditions. No label outside $S_C$ holds identically on a branch: we show each of them false in ball arithmetic at two sample points of every branch. The identifications of jump points with the points of \S\ref{sec:points} and with the special rays are exact: the mass ratios have the same minimal polynomials, with a common isolating ball. \end{proof} \subsection{The exceptional points}\label{ss:exceptional} Table~\ref{tab:noK} lists the jump points that we treat individually. We will count three of them directly (\S\ref{sec:rays}). \begin{table}[ht] \centering \small \begin{tabular}{@{}llp{0.25\textwidth}l@{}} \toprule point & masses, up to scaling and relabelling & on the families & finiteness\\ \midrule $\bone$ & $(1,4,4,4,4)$ & $T_1$ $t=1$; $T_2$ $t=1$; $T_3$ $a=4$; $T_5$ $q=-1$; $T_7$ $s=2$; $\mathrm{RL}$ $\mu=4$ & Theorem~\ref{thm:count1}\\ $\btwo$ & $(1,1,4,4,4)$ & $T_7$ $s=\frac12,-1$; $\mathrm{RL}$ $\mu=1$ & Theorem~\ref{thm:count2}\\ $P_1$ & $(9,36,36,36,4)$ & $T_5$ $q=1$; $\mathrm{RL}$ $\mu=\frac49$ & Theorem~\ref{thm:countP1}\\ $P_2$ & $(9,36,36,4,9)$ & $T_7$ $s=\frac32,3$ & settled\\ $\hone$ & $((1+t_*)^{-2},t_*^{-2},t_*^{-2},1,1)$ & $T_7$ $s\approx1.5548,\ 2.8025$ & settled\\ $\htwo$ & $(\frac{25}8,\frac{25}2,\frac{25}2,\frac{25}2,1)$ & $T_2$ $t=\frac12$; $\mathrm{RL}$ $\mu=\frac8{25}$ & settled\\ \bottomrule \end{tabular} \caption{The jump points treated individually. Here $t_*\approx1.8024986256$ is the root greater than $1$ of $2t^8+t^6-8t^5-6t^4-8t^3+t^2+2$.}\label{tab:noK} \end{table} At $P_2$ the label set contains $\lab{H}(12|35)$ and $\lab{H}(13|25)$, whose conditions $\mathrm{Tri}$ fail. So Theorem~\ref{thm:D3} removes the two pieces $\pc{d3}$. It also enables $\pc{d6}(145|234)$, which is removed by Proposition~\ref{prop:d6P2}. $\mathrm{K14}$ of the cluster $Q_4$, with masses $(9,36,36,9)$, has no solution, the pieces $\pc{d15}(4,2)$ and $\pc{d15}(4,3)$ are excluded by (X4), and the remaining pair is certified relative to $\VN$. At $\hone$ the label set contains two labels $\lab{H}$ whose conditions $\cD_3$ hold. So Theorem~\ref{thm:D3} does not remove the two pieces $\pc{d3}$. Proposition~\ref{prop:d3tstar} does. $\mathrm{K14}$ of the cluster of the four heavier bodies has no solution, the four pieces $\pc{d15}$ of this cluster are excluded by (X4), and the remaining pair is certified. $\htwo$ is the point $\mu=8/25$ of $\mathrm{RL}$ after the exchange of the bodies $1$ and $2$. It is settled by Theorem~\ref{thm:RL}. \begin{proposition}\label{prop:exceptional} On $T_1$, $T_2$, $T_3$, $T_4$, $T_5$, $T_7$ and $T_8$, every jump point of Theorem~\ref{thm:SJ} has finitely many central configurations, and every such point is settled except the points $\bone$, $\btwo$ and $P_1$ of Table~\ref{tab:noK}. \end{proposition} \begin{proof} Table~\ref{tab:curvecounts} splits the jump points into three kinds. The points of \S\ref{sec:points} are K-certified and settled there. On $T_3$ this is the point of equal masses. We have treated the points of Table~\ref{tab:noK} above. Every other jump point carries a point certificate: in an exact description over a number field, $\mathrm{K14}$ of the transversal cluster has no solution (Theorem~\ref{thm:KA}), the pieces $\pc{d15}$ needed of this cluster are excluded by (X4), and the pair formed by the label set of the point and the pieces so excluded is certified. On $T_8$ the cluster is $Q_1$ and the pair is $(S,\Pi(1))$, with $S$ the label set of the point. Thus Lemma~\ref{lem:criterion} applies. Each certificate in this proof holds relative to $\Vtwo^{\mathrm w}$ or to $\VN$ (\S\ref{ss:checks}). \end{proof} \subsection{Four-body configurations at the masses $F_A$ and $F_B$}\label{ss:FAFB} \begin{lemma}\label{lem:K14FAFB} $\mathrm{K14}(1,1,1,\nu)$, $\nu>0$, has a solution for exactly six values of $\nu$, and $\mathrm{K14}(1,c,c,1)$, $c>0$, for exactly sixteen values of $c$. They are listed in Table~\ref{tab:K14values}. \end{lemma} \begin{proof} We enumerate the configurations exactly: the components of the K14 system in Dziobek form are eliminated by resultants and norms over their coefficient fields, all zeros are enclosed by Krawczyk boxes, each rejected candidate is shown to be degenerate or to give a value of $\nu$, resp.\ $c$, that is not positive real, and each value found is identified as a root of an exact polynomial. \end{proof} \begin{table}[ht] \centering \small \begin{tabular}{@{}lp{0.30\textwidth}p{0.52\textwidth}@{}} \toprule & values (to $4$ decimals) & polynomial\\ \midrule $F_A$, $\nu$ & $0.2462$, $12.1840$ (collinear) & $x^8-116x^6-4810x^4-1044x^2+81$\\ & $0.2540$, $19.6837$ (collinear) & $x^8-32x^7+300x^6-864x^5-5098x^4-4320x^3+7500x^2-4000x+625$\\ & $1/\sqrt3$ & $3x^2-1$\\ & $6.0193$ & $3x^6-106x^4-97x^2-25$\\ \midrule $F_B$, $c$ & $0.1125$, $3.2140$ (collinear) & $x^8+16x^6-40x^5-2x^4-512x^3+192x^2-24x+1$\\ & $0.1387$, $7.2098$ & $x^4-52x^2+1$\\ & $0.1801$, $5.5533$ (collinear) & degree $16$\\ & $0.2490$; $4.0154$; $0.6012$; $1.6633$ (collinear) & $x^4+4x-1$; $x^4-4x^3-1$; $x^4+4x^3-1$; $x^4-4x-1$\\ & $0.2510$, $3.9843$ (collinear) & $x^{16}-16x^{14}-4x^{12}+528x^{10}-6906x^8+528x^6-4x^4-16x^2+1$\\ & $0.3111$, $8.8913$ (collinear) & $x^8-24x^7+192x^6-512x^5-2x^4-40x^3+16x^2+1$\\ & $0.4429$; $2.2579$ & $x^4+10x^2-2$; $2x^4-10x^2-1$\\ \bottomrule \end{tabular} \caption{The positive masses $\nu$ and $c$ at which $\mathrm{K14}(1,1,1,\nu)$, resp.\ $\mathrm{K14}(1,c,c,1)$, has a solution, with their minimal polynomials. The entry ``collinear'' marks the values at which the configurations are collinear.}\label{tab:K14values} \end{table} The special points of $T_1$, $T_2$, $T_3$ and $T_7$ are therefore the points $t=\sqrt c$, $t=\sqrt\nu$, $a=1/\nu$ and $s=1\pm c^{-1/2}$, respectively, with $\nu$ and $c$ from Table~\ref{tab:K14values}. Table~\ref{tab:curvecounts} gives their numbers. \begin{proposition}\label{prop:d14a} At each of the $32$ special points of $T_7$ the piece $\pc{d14a}(1)$ is excluded, and at each of the $6$ special points of $T_3$ every $\pc{d14a}(k)$, $k\le4$, is excluded. \end{proposition} \begin{proof} Since the masses of $T_3$ are symmetric in the bodies $1,\dots,4$, it suffices to treat $\pc{d14a}(1)$. At a place of type $\pc u_0(1)$, after a rotation and possibly an exchange of the colours, $\val z_{il}=\val w_{il}=1$ inside $Q=Q_1$ and $\val z_{1l}=\alpha$, $\val w_{1l}=-\alpha$ for one $\alpha\in[0,1]$. The leading coefficients of the cluster form a K14 configuration in a gauge, together with explicit data for body $1$ with signs $s_l=\pm1$ of the leading $p_{1l}$. The configurations at the special values are known completely (Lemma~\ref{lem:K14FAFB}). In a weighted form of the rows, the Jacobian $J$ of the cluster rows has rank $10$ at every configuration (a nonzero $10\times10$ minor in ball arithmetic). The first-order forcing must lie in the image of $J$. This image is described by two linear conditions, the pairings with the two cluster scalings, which lie in the kernel of the Hessian of the cluster potential (Remark~\ref{rem:d14a}). For $\alpha<1$ this forces, according to the sign pattern and to whether $\alpha=0$, the vanishing of a signed dipole $\sum_is_im_i\tilde\zeta_i$, the quadrupole $\sum_im_i\tilde\zeta_i^2$ or $A^2Q_{zz}Q_{ww}-B^2I^2$, where $\tilde\zeta$, $\tilde\omega$ are the centred leading positions, $Q_{zz}=\sum m_i\tilde\zeta_i^2$, $Q_{ww}=\sum m_i\tilde\omega_i^2$, $I=\sum m_i\tilde\zeta_i\tilde\omega_i$, $A=3m_1$ and $B=3m_1+2M_Q$. In $400$-bit ball arithmetic each of these quantities is nonzero at all configurations of all special values, except the dipole of the sign class $(+,-,-,+)$ at sixteen centrally symmetric configurations of $T_7$, where it vanishes by the symmetry (proved by a Krawczyk uniqueness argument), and the quadrupole and the dipole of $(+,+,+,-)$ at $a=\sqrt3$, where the configuration is an equilateral triangle with its centre. In these cases, and for $\alpha=1$ at all points, we compute the obstruction series of a Lyapunov--Schmidt reduction to higher order in ball arithmetic at all leading points, with the vanishing of the lower orders proved exactly. A nonzero coefficient excludes each case. \end{proof} \subsection{The family $T_3$}\label{ss:T3} \begin{theorem}\label{thm:T3} Every point of $T_3$ has finitely many central configurations, and every point other than $a=4$ is settled. \end{theorem} \begin{proof} For each $k\in\{1,2,3,4\}$, the pair $(S_C,\{\pc{d14a}(k)\}\cup\Pi(5))$ is certified. The cluster $Q_5$ has the masses $(1,1,1,1)$, at which K14 has no solution (\S\ref{sec:points}). The pieces $\pc{d15}(5,k)$ are excluded by (X4). Hence $\Pi(5)$ is excluded at every point. Off the six special points $\mathrm{K14}(m_{Q_k})$ has no solution. At the special points $\pc{d14a}(k)$ is excluded by Proposition~\ref{prop:d14a}. By Theorem~\ref{thm:SJ} and Lemma~\ref{lem:criterion} each point with label set $S_C$ is settled. Proposition~\ref{prop:exceptional} covers the other points. \end{proof} \subsection{The families $T_1$ and $T_7$}\label{ss:T1T7} On $T_1$ and $T_7$ the transversal cluster is $F_B(c)$, a relabelling of $(1,1,c,c)$. By (X4) all pieces $\pc{d15}(1,k)$ are removed on these two families. \begin{theorem}\label{thm:T1T7} Every point of $T_1$ and of $T_7$ has finitely many central configurations. Every point of $T_1$ other than $t=1$, and every point of $T_7$ other than $s\in\{2,\frac12,-1\}$, is settled. \end{theorem} \begin{proof} On $T_7$, off the sixteen special values of $c$, $\mathrm{K14}(m_{Q_1})$ has no solution. Thus (X1) excludes $\pc{d14a}(1)$. At the special values $\pc{d14a}(1)$ is excluded by Proposition~\ref{prop:d14a}. Thus, by \S\ref{ss:howsettled}, $\Pi(1)$ is excluded at all points of $T_7$, and $\pi(1,2)\cup\pi(1,5)$ at all points of $T_1$. The pairs $(S_C,\Pi(1))$ on $T_7$ and $(S_C,\pi(1,2)\cup\pi(1,5))$ on $T_1$ are certified, and $\Pi(1)$ is the only minimal exclusion set of $S_C$ on $T_7$ (Theorem~\ref{thm:options}). We conclude as in the proof of Theorem~\ref{thm:T3}. \end{proof} \subsection{The family $T_2$}\label{ss:T2} \begin{theorem}\label{thm:T2} Every point of $T_2$ has finitely many central configurations, and every point other than $t=1$ is settled. Every piece $\pc{d15}(1,k)$, $k\in Q_1$, is excluded at every point of $T_2$. \end{theorem} \begin{proof} The pair $(S_C,\pi(1,3)\cup\pi(1,4))$ is certified. Each pair $\pi(1,k)$, $k\in Q_1$, is excluded at every point of $T_2$ (\S\ref{ss:howsettled}). We conclude as in the proof of Theorem~\ref{thm:T3}. \end{proof} \subsection{The families $T_4$, $T_5$ and $T_8$}\label{ss:T458} The following two results describe, for the families $T_4$, $T_5$, $T_7$ and $T_8$, all ways of settling a point with label set $S_C$ by Lemma~\ref{lem:criterion} with $V=\Vtwo^{\mathrm w}$ and pieces $\pc{d14a}$, $\pc{d14b}$, $\pc{d15}$. A \emph{witness} for $(S,E)$ is a nonempty set $W$ of faces in $\cA_V(S)$ outside the pieces of $E$ and outside the pieces $\pc{d2}[T]$ and $\pc{d9}(T;ax,by)$, together with rationals $c_v>0$ on the vertices $v$ of the faces in $W$ such that $\sum_vc_vv=0$. Any functional that is nonnegative on these vertices vanishes on all of them. So, by induction on $i$, every functional $l_i$ of a certificate for $(S,E)$ vanishes on them, against (b) of Definition~\ref{def:cert} for the faces of $W$. Hence $(S,E)$ is not certified relative to $V$, and neither is $(S,E')$ for any $E'\subseteq E$. \begin{theorem}\label{thm:options} For each of the families $T_4$, $T_5$, $T_7$ and $T_8$, the minimal exclusion sets of $S_C$ relative to $\Vtwo^{\mathrm w}$, among the $45$ pieces $\pc{d14a}(j)$, $\pc{d14b}(j,k)$, $\pc{d15}(j,k)$, are the following. \begin{enumerate}[label=\textup{(\roman*)}] \item $T_4$ \textup{(}$19$ sets\textup{)}: $\pi(1,A)\cup\pi(1,B)$ for $A0$ and $m=(4,1,4,4,\mu)$. The planar five-body problem with masses $m$ has finitely many central configurations up to similarity, and $m$ is settled unless $\mu\in\{1,4,\frac49\}$. \end{theorem} The proof uses three results valid for every $\mu>0$: Proposition~\ref{prop:nod2} excludes the piece $\pc{d2}[134]$, and (X2) and (X4) exclude every $\pc{d14b}(2,k)$ and every $\pc{d15}(2,k)$, $k\in Q_2$, respectively. \begin{proof}[Proof of Theorem~\ref{thm:RL}] Each label polynomial restricted to the line is a polynomial in $\mu$. The label polynomials that vanish identically give $S_C$. The other nonzero restrictions have $6$ irreducible factors over $\Q$ with a positive root (Table~\ref{tab:RLfactors}). We isolate these roots in ball arithmetic. Off the positive roots of these factors $S(m)\subseteq S_C$. At a root of a factor $g$, $S(m)\subseteq S(g)$, the union of $S_C$ and the labels with a polynomial divisible by $g$. Let $U=\pi(2,1)\cup\pi(2,3)\cup\pi(2,4)\cup\pi(2,5)\cup\{\pc{d2}[134]\}$, excluded at all $\mu>0$ by the results above. We treat the factors as follows. \begin{enumerate}[label=\textup{(\roman*)}] \item The pair $(S_C,U)$ is certified, and so is $(S(g),U)$ for the factors $g$ of the first row of Table~\ref{tab:RLfactors}. \item At the root $\mu=\frac8{25}$ of the factor in the second row, K14 has no solution for any of the five clusters (Theorem~\ref{thm:KA}), the pieces $\pc{d15}$ are removed by (X4), and the pair formed by $S(g)$ and all pieces $\pc{d14a}$, $\pc{d14b}$, $\pc{d15}$ together with $\pc{d2}[134]$ is certified. \item The roots of the factors of the third row are $1$, $4$ and $\frac49$. At $\mu=1$ and $\mu=4$, $m$ is a permutation of $(1,1,4,4,4)$, resp.\ of $(1,4,4,4,4)$. At $\mu=\frac49$, $9m$ is a permutation of $P_1$. Theorems~\ref{thm:count2}, \ref{thm:count1} and~\ref{thm:countP1} apply. \end{enumerate} By Lemma~\ref{lem:criterion}, $m$ is settled at every $\mu>0$ other than $1$, $4$ and $\frac49$. \begin{table}[ht] \centering \small \begin{tabular}{@{}lrl@{}} \toprule treatment & factors & degrees\\ \midrule (i) $(S(g),U)$ certified & $2$ & $1$ (root $\mu=16$), $3$\\ (ii) all K14 systems fail & $1$ & $1$ (root $\mu=\frac8{25}$)\\ (iii) counted & $3$ & $1$, $1$, $1$ ($\mu-1$, $\mu-4$, $9\mu-4$)\\ \bottomrule \end{tabular} \caption{The factors with a positive root of the label polynomials restricted to $\mathrm{RL}$, by their treatment in the proof of Theorem~\ref{thm:RL}.}\label{tab:RLfactors} \end{table} \end{proof} \subsection{The family $T_9$}\label{sec:T9}\label{ss:farstrata} Theorem~\ref{thm:curves} treats the mass vectors whose label set contains a curve set, except the points of the special rays and of the family $T_9$ below. In this subsection we settle the points of $T_9$ (Proposition~\ref{prop:farT9}). For $T_9$ we need Theorem~\ref{thm:KApar}. Table~\ref{tab:farclose} collects the numbers of the computation. \paragraph{The curve $T_9$.} Let $\rho_0=0.0873780\ldots$ be the real root of $p^3+5p^2+11p-1$, and let $T_9$ be the curve $m(s)=(\rho_0/4,\rho_0,\rho_0,1,s)$, $s>0$. The cluster $Q_1$ has masses $(\rho_0,\rho_0,1,s)$, with body $b$ in position $b-1$. The cluster $Q_5$ has the constant masses $(\rho_0/4,\rho_0,\rho_0,1)$. \begin{proposition}\label{prop:farT9} Every point of $T_9$ is settled. \end{proposition} \begin{proof} \emph{Labels.} We reduce each label polynomial composed with $m(s)$ modulo the minimal polynomial of $\rho_0$, isolate the positive roots of its norm in ball arithmetic, and decide at each of them, by a ball or by an exact gcd over $\Q(\rho_0)$, whether the polynomial vanishes there. As in the proof of Theorem~\ref{thm:SJ}, a label $\lab H$ holds only where its two conditions $\mathrm{Tri}$ hold as well, which we decide in the same way. At each $s>0$ the label set is $S_C=\{\lab E(123),\lab R(23;1),\lab R(24;1),\lab R(34;1),\lab{EQ}(2,3)\}$, except at $13$ jump points, all in $[0.0030920,2091.64]$. They fall into $10$ classes by their label sets. \emph{Pieces removed at every $s$.} Theorem~\ref{thm:D14} removes the pieces $\pc{d14b}(1,k)$. $\mathrm{K14}$ fails at the masses of $Q_5$ (Theorem~\ref{thm:KA}) and Corollary~\ref{cor:nod15} removes the pieces $\pc{d15}(5,k)$. Hence no piece of the cluster $Q_5$ is attained. \emph{The generic set.} Two certificates for $S_C$ relative to $\Vtwo^{\mathrm w}$ remove $\pc{d14b}(1,k)$, $k=2,\dots,5$, and in addition $\pc{d15}(1,5)$, respectively $\pc{d15}(1,2)$, $\pc{d15}(1,3)$, $\pc{d15}(1,4)$. By (X2) and (X4) these pieces are removed at all $s$, so that both certificates apply. Hence $m(s)$ is settled by Proposition~\ref{prop:cert} with these pieces removed, as in the proof of Proposition~\ref{prop:Kcert}. \emph{The jump points.} Each class has a certificate whose removed pieces are removed at its points: $\pc{d14b}(1,k)$ and $\pc{d15}(1,k)$ by (X2) and (X4), the pieces of $Q_5$ as above, and, in the classes whose certificate removes it, $\pc{d14a}(1)$ because $\mathrm{K14}(m_{Q_1})$ has no solution at the roots of their jump factors, by Theorem~\ref{thm:KApar}(b). We check the hypothesis of that theorem by a unit standard basis with the factor adjoined, or through a polynomial in $s$ alone in the ideal of the full system that is prime to the factor. At $s=\rho_0/4$, where $m_1=m_5$, the certificate also removes the piece $\pc{d6}(145|234)$. After relabelling to the masses $(1,1,4,4,4/\rho_0)$, we carry out the test for diagram $6$ used at $P_2$ in \S\ref{ss:exceptional} with $\Q(4/\rho_0)$ in place of $\Q$, with the standard basis and the closed forms exact over $\Q(4/\rho_0)$: the leading system is zero-dimensional, and each of its solutions has a nonzero obstruction coefficient; the leading solutions and their obstructions are certified in ball arithmetic. Hence no place of this diagram exists. The certificates at $s=\rho_0/4$ and at the two jump points $s\approx0.0520549$ and $s\approx0.176101$, where $\lab E(245)$ and $\lab E(345)$ hold, are relative to $\VN$, the others relative to $\Vtwo^{\mathrm w}$. \end{proof} \begin{table}[ht] \centering \small \begin{tabular}{@{}lp{0.62\textwidth}r@{}} \toprule & quantity & number\\ \midrule $T_9$ & jump points; classes & $13$; $10$\\ & certificates for $T_9$ & $12$\\ & solutions of the leading system of diagram $6$ at $s=\rho_0/4$ & $5184$\\ \bottomrule \end{tabular} \caption{The computation along $T_9$ in numbers.}\label{tab:farclose} \end{table} \subsection{The curve sets}\label{ss:HC} \begin{theorem}\label{thm:curves}\label{prop:HC} Let $m\in(\R_{>0})^5$ satisfy every label of $\sigma(D_T)$ for one of the nine names $T$ of Table~\ref{tab:curvesets} and a relabelling $\sigma$. Suppose that $m$ is not a positive multiple of a permutation of $(1,4,4,4,4)$ or of $(1,1,4,4,4)$, and, if $T=T_9$, that $m$ is not a point of the family $T_9$ up to scaling and relabelling. Then the planar five-body problem with masses $m$ has finitely many central configurations up to similarity, and $m$ is settled unless it is a positive multiple of a permutation of $P_1$ \textup{(}\S\ref{ss:twopoints}\textup{)}. \end{theorem} \begin{proof} By Lemma~\ref{lem:DT}, applied to $\sigma^{-1}m$, a positive multiple of $m$ is obtained by a relabelling from a point on one of the following families: \begin{enumerate}[label=\textup{(\roman*)}] \item $T$ itself, if $T$ is $T_1$, $T_2$, $T_3$, $T_5$ or $T_7$; \item $T_4$ or $T_2$, if $T=T_4$, and $T_8$ or $T_5$, if $T=T_8$; \item $\mathrm{RL}$, if $T$ is $\mathrm{RL}$ or $T_9$. For $T_9$ the other case of Lemma~\ref{lem:DT}, a point of $T_9$, is excluded by hypothesis. \end{enumerate} As in the proof of Corollary~\ref{cor:S}, positive scaling and relabelling of the masses preserve finiteness and settledness. Theorems~\ref{thm:T3}, \ref{thm:T1T7}, \ref{thm:T2} and~\ref{thm:RL} and Proposition~\ref{prop:T458}, which hold on the whole families, give the claim. The points of these families that are only counted are the points of the special rays, excluded here, and the positive multiples of permutations of $P_1$, which are counted points. \end{proof}