\section{The sixteen diagrams and their types}\label{app:types}

Table~\ref{tab:diagrams} describes the diagrams of \cite[Fig.~11]{AK12} in the numbering of the figure. In diagram~3 the colours of the four sides alternate around the square, as in the figure and in the description of \cite[p.~582]{AK12} (bodies $1,3$ and $2,4$ on the diagonals, body $5$ at the centre) and of \S8.3 there. See also \S\ref{ss:akinputs}.

\begin{table}[ht]
\centering
\small
\begin{tabular}{@{}ll@{}}
\toprule
diagram & edges and circles\\
\midrule
1 & $z$-edge $12$, $z$-circles $1,2$; $w$-triangle $345$, no circles\\
2 & as 1, with $w$-circles $3,4,5$\\
3 & $z$-edges $14,23$, $w$-edges $12,34$; $1$--$4$ circled in both colours; $5$ isolated, uncircled\\
4 & $z$-edge $12$, $z$-circles $1,2$; $zw$-triangle $345$\\
5 & $1,2$ isolated, uncircled; $zw$-triangle $345$\\
6 & $z$-triangle $125$, $z$-circles $1,2$; $w$-triangle $345$, $w$-circles $3,4$\\
7 & $z$-triangle $125$, $w$-triangle $345$, no circles\\
8 & $z$-triangle $125$, no circles; $w$-triangle $345$, $w$-circles $3,4$\\
9 & $zw$-triangle $123$, all $w$-circled; $w$-edges $14,25$; $4,5$ $w$-circled\\
10 & $zw$-triangle $125$, $w$-triangle $345$, no circles\\
11 & $zw$-triangle $125$, no circles; $w$-triangle $345$, $w$-circles $3,4$\\
12 & $zw$-triangle $125$, $w$-circles $1,2,5$; $w$-triangle $345$, $w$-circle $3$\\
13 & as 12, with $w$-circles $3,4$\\
14 & $zw$-edges on the six pairs of $\{1,2,3,4\}$; $5$ isolated\\
15 & as 14, all $w$-circled, with the $w$-edge $15$\\
16 & $zw$-edges on all ten pairs\\
\bottomrule
\end{tabular}
\caption{The sixteen diagrams of \cite[Fig.~11]{AK12}.}\label{tab:diagrams}
\end{table}

We derive the entries of Table~\ref{tab:types} for diagrams $1$--$4$, $6$--$13$ and $16$, using Lemma~\ref{lem:orders} and the rules quoted after it. We use Lemma~\ref{lem:cut} only where stated, and Lemma~\ref{lem:U} and \cite[\S8.9]{AK12} only for diagram~$9$. We treat diagrams $5$, $14$ and $15$ in \S\S\ref{ss:typestable}--\ref{ss:d5}.

\emph{Diagram 1.} The ends of the $z$-stroke $12$ are $z$-circled. By Rule~1d they are not $z$-close. So $x_{12}=0$ by (O4) and $\omega_{12}=2$ by (O3). By Rule~2e the three exponents $\omega$ of the triangle $T=\{3,4,5\}$ are equal. The triangle $T$ is an isolated $w$-component without circles. By Rule~1e it has no maximal $w$-stroke. Hence $x=a>0$ on $T$. By (O4), $a<1$. On a cross pair $1t$, $\zeta_{1t}=-2$ by (O6). A $z$-stroke $1t$ would need $\omega_{1t}=2$, which shows $\mu_t:=\omega_{1t}<2$. In the triangle $12t$, (O5) gives $\omega_{2t}=\omega_{1t}$. Hence $x_{1t}=x_{2t}=\frac{\mu_t-2}2$. By (O5) in the triangles $1tt'$, where $\omega_{tt'}=3a-2$, either all $\mu_t$ equal some $\mu\le3a-2$, or two of them equal $3a-2$ and the third is larger. Lemma~\ref{lem:S1} leaves $\mu_t=3a-2$.

\emph{Diagram 2.} As for diagram $1$, $x_{12}=0$. The $w$-circled bodies of $T$ are not all $w$-close (Rule~1d). By Rule~2e the triangle is maximal: $x=0$ on $T$. On the cross pairs (O6) gives $\zeta=\omega=-2$.

\emph{Diagram 3.} Since the four sides are not close by Rule~1d, they are maximal and $x=0$. On the diagonal $13$, $\zeta_{12}=2$ ($12$ is a maximal $w$-stroke) and $\zeta_{23}=-2$ ($23$ is a maximal $z$-stroke). Then (O5) gives $\zeta_{13}=-2$. Likewise $\omega_{13}=-2$. On the pairs $k5$, (O6) applies in both colours.

\emph{Diagram 4.} As for diagram $1$, with $x=1$ on $T$ and $3a-2$ replaced by $1$. Lemma~\ref{lem:S1} gives $\mu_t=1$.

\emph{Diagram 6.} The pairs $12$ and $34$ of circled bodies are not close by Rule~1d. By Rule~2e each triangle is maximal, $x=0$. On the four cross pairs (O6) applies in both colours.

\emph{Diagram 7.} By Rule~2e the exponents are equal on each of the triangles $125$ and $345$. By Rule~1e, $x=a>0$ on $125$ and $x=b>0$ on $345$. By (O4), $a,b<1$. On a cross pair, (O5) gives $\zeta_{13}=\min(\zeta_{15},\zeta_{53})=\min(3a-2,2-b)=3a-2$ and likewise $\omega_{13}=3b-2$, which gives $x_{13}=\frac32(a+b)-2$. Lemma~\ref{lem:cut} gives $a+b\le\frac43$.

\emph{Diagram 8.} On $125$, $x=a\in(0,1)$ as for diagram $7$. The edge $34$ is maximal by Rule~1d. The edges $35$, $45$ have $\omega=-2$ by (O6). Hence $x=0$ on $345$. On the cross pairs $\zeta=3a-2$ by (O5) and $\omega=-2$ by (O6).

\emph{Diagram 9.} Each body is $w$-circled. If body $4$ clustered with $w_1,w_2,w_3$ (case \S8.9.1 of \cite{AK12}), the potential would have exactly one unbounded term, $m_1m_4r_{14}^{-1}$ \cite[\S8.9.1]{AK12}, which is impossible because $U$ is constant on $C$ (Lemma~\ref{lem:U}). The same holds with $1\leftrightarrow2$, $4\leftrightarrow5$. We are therefore in case \S8.9.2, where $14$ and $25$ are maximal $w$-strokes: $\zeta=2$, $x=0$. In the equation $z_4=\sum_lm_lZ_{4l}$, $e(Z_{41})=e(Z_{52})=2$ and $Z_{42},Z_{43},Z_{51},Z_{53}$ have exponent $\frac52$, and $z_{45}=(m_4+m_5)Z_{45}+O(\varepsilon^2)$. If $e(p_{45})>0$ all terms of $z_4$ and $z_5$ would have exponent larger than $1$, and if $e(p_{45})<0$ then $e(z_{45})=e(Z_{45})<1$, both contradicting $e(z_{45})=1$, which follows from $z_{45}=z_{41}+z_{12}+z_{25}$. Thus $x_{45}=0$. Then (O5) gives the rest: the type is $\pc n(\{1,2,3\};14,25)$, with $x=1$ on $12,13,23$, $0$ on $14,25,45$ and $-\frac12$ on $15,24,34,35$.

\emph{Diagrams 10--13.} On the $zw$-triangle $125$, $x=1$. On $345$, $x=c$ by Rule~2e. On a cross pair (O5) gives $\zeta_{13}=\min(1,2-c)=1$ and $\omega_{13}=\min(1,3c-2)=3c-2$, hence $x=\frac{3c-1}2$. For diagram $10$, $0<c<1$ by Rule~1e, and Lemma~\ref{lem:cut} gives $c\le\frac13$. For diagrams $11$--$13$, $c=0$ by Rule~1d, (O6) and Rule~2e.

\emph{Diagram 16.} Since every pair is a $zw$-edge, $x=\pc 1$, which is excluded by Lemma~\ref{lem:cut}.

\input{sec/a_d5hulls}

\section{Proof of Theorem~\ref{thm:D3}}\label{app:D3}

By relabelling we take $s=5$, $D_1=\{1,3\}$, $D_2=\{2,4\}$. The sides are $12,14,23,34$. Let $P$ have type $\pc h(5;13|24)$.

\emph{Orders and leading solutions.} By Table~\ref{tab:types} and (O3)--(O6), after the rotation the coordinates relative to body $5$ have $e(u_k)=e(v_k)=-2$ for $k\le4$, $e(p)=0$ on the sides and $e(p)=2$ on the diagonals and on the pairs $k5$. We write the coordinates as $\tau$-powers times $y$ and divide each row by its lowest power. This gives a weighted system $G(\tau,y)=G_0(y)+\tau^2G_2(y)+\tau^3G_3(y)$, to be solved with $y(0)$ a \emph{leading solution}, $G_0(y(0))=0$ with all coordinates nonzero. We normalize $y_{u_1}=y_{v_1}=1$. The leading solutions then exist only when $m_1m_3=m_2m_4$. They are explicit: in case A, $y_u=(1,U,U,1)$, $y_v=(1,1,V,V)$ with $U=-m_1/m_2$, $V=-m_2/m_3$, $k_e^3=1/(m_k+m_l)$ on each side $e=kl$, and $q_{13}^2=m_3/M_Q$, $q_{24}^2=-m_3/M_Q$, $q_{15}^2=1$, $q_{25}^2=-m_2/m_1$, $q_{35}^2=m_3/m_1$, $q_{45}^2=-m_4/m_1$, where $k_e$, $q_e$ are the leading coefficients of $p_e$ on the sides and on the other pairs. Case B exchanges $u$ and $v$.

\emph{Rank and obstruction.} At each leading solution the Jacobian $J_0$ of $G_0$ has rank $16$: one checks by a direct elimination that on the side rows and the rows $Z_k$, $W_k$ the reduced Jacobian in the variables $y_{u_2},y_{u_3},y_{u_4},y_{v_2},y_{v_3},y_{v_4}$ and the four side values is injective. The six other rows have the invertible diagonal block $\operatorname{diag}(2/q_e)$. The left kernel of $J_0$ is spanned by $\lambda_Z=-Z_1-\frac{m_2}{m_1}Z_2+\frac{m_2}{m_1}Z_3+Z_4$ and $\lambda_W=\frac{m_2}{m_3}W_1-\frac{m_2}{m_3}W_2-W_3+W_4$ (in case A). Since $\lambda$ vanishes on the side rows, $\lambda\cdot G_2=0$. The first obstruction is
\begin{gather*}
\lambda_Z\cdot G_3(y_0)=\frac{O}{m_1m_2},\\
O=m_2\big(m_1m_3q_{13}+m_2m_4q_{24}+m_1m_5q_{15}+m_2m_5q_{25}+m_3m_5q_{35}+m_4m_5q_{45}\big),
\end{gather*}
and $\lambda_W\cdot G_3(y_0)=-O/(m_1m_3)$. If $O\ne0$, we compare coefficients in $G(\tau,y(\tau))\equiv0$ through the order $\tau^3$, using that $J_0$ restricted to the rows other than $Z_4$, $W_2$ is invertible, and obtain a contradiction. Hence $O=0$.

\emph{The condition $O=0$.} The values $q_{13}$, $q_{15}$, $q_{35}$ are real and $q_{24}$, $q_{25}$, $q_{45}$ are imaginary, with all signs occurring among the leading solutions. Therefore $O=0$ for some leading solution if and only if both $m_1m_3q_{13}\pm m_1m_5q_{15}\pm m_3m_5q_{35}$ and $m_2m_4q_{24}\pm m_2m_5q_{25}\pm m_4m_5q_{45}$ vanish for some signs. We divide the first triple of absolute values $m_1m_3^{3/2}M_Q^{-1/2}$, $m_1m_5$, $m_3^{3/2}m_5m_1^{-1/2}$ by $m_1m_3^{3/2}m_5$ and obtain $x_5^2/\sqrt{M_Q}$, $x_3^3$, $x_1^3$. A signed sum of three positive numbers vanishes for some signs if and only if one is the sum of the other two. For the first triple this is $\mathrm{Tri}(13;5)$. The second triple gives $\mathrm{Tri}(24;5)$ in the same way. \qed

\section{The labels}\label{app:labels}

The polynomials are those of \cite[\S8]{AK12} (see \S\ref{ss:akinputs}). The files of the supplementary material contain them with checksums. With $x_i$ as in \S\ref{ss:conditions}, they read as follows.
\begin{itemize}
\item $\lab{E}(\{a,b,c\})$: $m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2-2m_am_bm_c(m_a+m_b+m_c)=0$, the product over the signs of $x_a\pm x_b\pm x_c$ times $(m_am_bm_c)^2$.
\item $\lab{R}(ab;s)$: $m_s^2(m_a+m_b)^4-m_a^3m_b^3=0$.
\item $\lab{H}(ij|kl)$: $m_im_j-m_km_l=0$; and $\mathrm{Tri}(ij;s)$: for positive masses equivalent to
\[
\Big(\frac1{m_s^2M_{Q_s}}-\frac1{m_i^3}-\frac1{m_j^3}\Big)^2-\frac4{m_i^3m_j^3}=0 .
\]
\end{itemize}

\input{sec/app_d2}

\section{The polynomial $P_{11}$}\label{app:P11}
In terms of $e_1=m_a+m_b+m_c$, $e_2=m_am_b+m_bm_c+m_cm_a$ and $e_3=m_am_bm_c$,
\begin{multline*}
P_{11}=e_1^{11}-13e_1^9e_2+17e_1^8e_3+63e_1^7e_2^2-162e_1^6e_2e_3-135e_1^5e_2^3+105e_1^5e_3^2+513e_1^4e_2^2e_3\\
+108e_1^3e_2^4-651e_1^3e_2e_3^2-540e_1^2e_2^3e_3+250e_1^2e_3^3+1000e_1e_2^2e_3^2-675e_2e_3^3 .
\end{multline*}
Expanded in the masses it has $78$ terms, and $P_{11}(1,1,1)=9$. The substitution $m_a=c+p+q$, $m_b=c+p$, $m_c=c$ gives a polynomial with $72$ terms, all coefficients positive and the coefficient of $c^{11}$ equal to $9$. With the symmetry of $P_{11}$, this gives $P_{11}(m)\ge9\min(m)^{11}>0$ on the open positive orthant.

\section{Diagrams 6 and 3 at two exceptional mass points}\label{app:exceptional}

At the mass point $P_2$ and at the mass point $\hone$ of Table~\ref{tab:noK} a piece of diagram $6$, respectively of diagram $3$, is enabled by the mass conditions. We exclude these diagrams there by computing the leading solutions of a place and an obstruction at a higher order.

At a place with a fixed diagram and type, the exponents of the coordinates (relative to body $5$) are determined, up to the colour exchange. We write the coordinates as $\tau^{2e(q)}y_q$ as in \S\ref{ss:expansion}, divide each row by its lowest power of $\tau$, and normalize two coordinates as there. This gives a weighted system $G(\tau,y)=\sum_{n\ge0}\tau^nG_n(y)$. The point $y(0)$ is a \emph{leading solution}: $G_0(y(0))=0$, with all coordinates nonzero. Let $R$ be a set of rows such that the Jacobian $J_0[R]$ of $G_0$ restricted to these rows is invertible at $y_0$. Then the rows of $R$ determine a unique formal solution $Y(\tau)$ of $G_R(\tau,Y)=0$ with $Y(0)=y_0$, by the implicit function theorem for formal series. Then every solution $y(\tau)$ of $G=0$ with leading solution $y_0$ equals $Y$ after the normalization. Let $\eta_{R,n}(y_0)$ be the coefficient of $\tau^n$ in the remaining rows evaluated at $Y(\tau)$. If $\eta_{R,n}(y_0)\ne0$ for some $n$, no place of the given type has leading solution $y_0$.

\begin{proposition}\label{prop:d6P2}
Let $m=(9,36,36,4,9)$. No singular place of an irreducible curve $C\subseteq Y_m$ has diagram $6$ with centre $4$ and wings $\{1,4,5\}$, $\{2,3,4\}$.
\end{proposition}

\begin{proof}
We renumber the bodies so that $m=(9,9,36,36,4)$ and the diagram has centre $5$ and wings $\{1,2,5\}$, $\{3,4,5\}$. At such a place the type is $\pc v(5;125|345)$. The exponents of the coordinates are fixed by it (after exchanging the colours if necessary). The leading solutions of the weighted system are given by closed formulas in $10$ roots (cube and square roots of explicit functions of the masses), which satisfy a square system of $10$ polynomial relations. We check:
\begin{enumerate}[label=\textup{(\roman*)}]
\item in exact arithmetic, that each point of the relations, substituted into the closed formulas, is a leading solution;
\item by a standard basis over $\Q$, that the leading system has $5184$ solutions counted with multiplicity;
\item in ball arithmetic, with Krawczyk's test, that the relations have $5184$ distinct simple zeros, so that the closed-form points are all the leading solutions and each is simple, and $J_0$ has rank $16$ at each of them;
\item in ball arithmetic, that at each leading solution, for a set $R$ of $16$ rows with $J_0[R]$ invertible, some $\eta_{R,n}$ with $n\le6$ is nonzero.
\end{enumerate}
The first nonzero obstruction has order $3$ at $3888$ leading solutions, order $4$ at $576$, order $5$ at $576$ and order $6$ at $144$. By the remark above, no place has this diagram.
\end{proof}

Chen and Tong state the stronger result that no curve has a place of diagram~$6$ at any positive mass \cite[Thm.~9.10]{CT26}.

Let $t_*\approx1.8024986256$ be the root greater than $1$ of $p(t)=2t^8+t^6-8t^5-6t^4-8t^3+t^2+2$. The other positive root of $p$ is $1/t_*$.

\begin{proposition}\label{prop:d3tstar}
Let $t\in\{t_*,1/t_*\}$ and $m=(1,1,t^{-2},t^{-2},(1+t)^{-2})$, so that $(x_1,x_3,x_5)=(1,t,1+t)$ with $x_i=m_i^{-1/2}$. No singular place of an irreducible curve $C\subseteq Y_m$ has as its type one of the $15$ points $\pc h(s;D_1|D_2)$ of diagram $3$.
\end{proposition}

\begin{proof}
Since $\lab{E}(\{1,3,5\})$ holds at $m$, Theorem~\ref{thm:D3} leaves only the two points $\pc h(5;13|24)$ and $\pc h(5;14|23)$, at which $\cD_3$ holds. The relabelling $(34)$ fixes $m$ and exchanges them. At $\pc h(5;13|24)$ the leading solutions are the $10368$ points of Appendix~\ref{app:D3}. The first obstruction vanishes exactly at the $648$ of them with $O=0$. We check:
\begin{enumerate}[label=\textup{(\roman*)}]
\item an enclosure of $t$ of width $2^{-440}$, by exact rational bisection on a sign change of $p$;
\item enclosures of the $10368$ leading solutions in balls, from their closed forms;
\item in ball arithmetic at $400$ bits, that at each leading solution, for a set $R$ of $16$ rows with $J_0[R]$ invertible, some $\eta_{R,n}$ with $n\le5$ is nonzero.
\end{enumerate}
The first nonzero obstruction has order $3$ at $9720$ leading solutions, order $4$ at $432$ and order $5$ at $216$.
\end{proof}

The mass points of Proposition~\ref{prop:d3tstar} are, up to relabelling, the point $\hone$ of Table~\ref{tab:noK}.