%
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We now show that the pieces $\pc{d14b}$ carry no singular place at positive masses.

\begin{theorem}\label{thm:D14}
For every $m\in(\R_{>0})^5$, no singular place of an irreducible curve $C\subseteq Y_m$ with $U\not\equiv0$ on $C$ has type $\sigma_k(j)(t)$ with $0<t<1$, for any body $j$ and any $k\in Q_j$. Hence the pieces $\pc{d14b}(j,k)$ may be removed from $\mathrm{pieces}(S(m))$ in Theorem~\ref{thm:types}, for all $m$, $j$ and $k$.
\end{theorem}

We keep the setting of Lemma~\ref{lem:d14blead}, with $Q'=Q\setminus\{k\}=\{a,b,c\}$ and positive masses. For a configuration of $Q$ with $z_k=0$, the condition $\mathrm{X3}_k$ is $\sum_{i\in Q'}m_iz_i=0$: body $k$ lies at the $z$-centre of mass of $Q$. Such a configuration \emph{has a bad shape} if $\sum_{i\in Q'}m_ix_i=0$ for some $x_i$ with $x_i^2z_i=1$. If it also satisfies $\mathrm{X3}_k$, then $\sum_{i\in Q'}m_ix_i^{-2}=0$, so that $(x_a,x_b,x_c)$ is a bad shape in the sense of \S\ref{ss:d15}. The proof of Theorem~\ref{thm:D14} has four steps. The leading cluster configuration satisfies $\mathrm{X3}_k$ and has a bad shape (Lemma~\ref{lem:d14energy}). At positive masses the only $\mathrm{K14}$ configuration with these properties is, up to symmetries, an equilateral triangle with its centre (Theorem~\ref{thm:Xhat}). This configuration is nondegenerate (Lemma~\ref{lem:Pdata}). At a nondegenerate leading cluster configuration the expansion of the place forces $\sum_{i\in Q'}m_ix_i^3=0$ unless $\sigma=(2n+1)/(n+1)$ for an integer $n\ge1$, and for these $\sigma$ it forces a coefficient to vanish that is nonzero at the equilateral configuration (Propositions~\ref{prop:D14N} and~\ref{prop:resonant}).

\paragraph{The bad shape.} We fix a square root $(-\hat w_j)^{1/2}$ and put $x_i=\hat p_{ij}(-\hat w_j)^{1/2}$ for $i\in Q'$, as in \S\ref{ss:d15}. The leading term of the row $D_{ij}$ is $\hat p_{ij}^2\hat z_i(-\hat w_j)=1$. Thus $x_i^2\hat z_i=1$. Near $\bar y$ let $w(y)$ and $x_i(y)$ be the branches of $(-y_{v_j})^{-1/2}$ and $y_{u_i}^{-1/2}$ with $w(\bar y)=(-\hat w_j)^{-1/2}$ and $x_i(\bar y)=x_i$, and put $S_r(y)=\sum_{i\in Q'}m_ix_i(y)^r$ and $\psi(y)=\sum_{i\in Q'}m_iy_{u_i}$.

\begin{lemma}\label{lem:d14exp}
At a place of type $\sigma_k(j)(t)$, with $\Pi$, $f$ and $U_Q$ as in Lemma~\ref{lem:d15expansion},
\begin{equation}\label{eq:d14F}
F(\tau,y)=\tau^{-2}U_Q(y)+\tau^{2-2\sigma}\Gamma(y)+f(\Pi)+\sum_{n\ge0}\tau^{(2n+1)(\sigma-1)}B_n(y)+R(\tau,y),
\end{equation}
where $\Gamma=-\frac{m_j}{2M}y_{v_j}\psi$, $B_n=\binom{2n}{n}4^{-n}m_jw\,y_{u_j}^nS_{2n+1}$, and $R$ is a convergent sum of terms $\tau^LR_L(y)$ with $L\ge4$. The coordinate $y_{u_j}$ occurs only in $f$, in the $B_n$ with $n\ge1$ and in $R$, and the coordinates $y_{v_i}$, $i\in Q'$, occur only in $U_Q$ and in $R$.
\end{lemma}

\begin{proof}
The cluster terms and the term $m_km_jp_{kj}$ are as in the proof of Lemma~\ref{lem:d15expansion}. For $i\in Q'$, $u_{ij}v_{ij}=-\tau^{2-2\sigma}y_{u_i}y_{v_j}(1-\tau^{2\sigma-2}y_{u_j}/y_{u_i})(1-\tau^{2\sigma+2}y_{v_i}/y_{v_j})$, so that
\[
p_{ij}=\tau^{\sigma-1}w\,x_i(y)\big(1-\tau^{2\sigma-2}y_{u_j}/y_{u_i}\big)^{-1/2}\big(1-\tau^{2\sigma+2}y_{v_i}/y_{v_j}\big)^{-1/2}
\]
with the branches of the $p_{ij}$ at $P$. The binomial series of the first factor gives the $B_n$, since $y_{u_i}^{-n}=x_i(y)^{2n}$. The second factor contributes only exponents at least $3\sigma+1>4$. Finally $I$ contributes $-\frac{m_j}My_{v_j}\psi$ at the exponent $2-2\sigma$, $\frac{m_jM_Q}M\Pi$ at $0$, and otherwise only the exponents $4$ and $2\sigma+2$.
\end{proof}

\begin{lemma}\label{lem:d14energy}
At a place of type $\sigma_k(j)(t)$:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $U_Q(y(\tau))=O(\tau^6)$;
\item $[B_0(y(\tau))]_r=0$ for $0\le r<2\sigma-2$; in particular $\sum_{i\in Q'}m_ix_i=0$, and the leading cluster configuration satisfies $\mathrm{X3}_k$ and has a bad shape;
\item $B_1(\bar y)=-\frac35\,[B_0(y(\tau))]_{2\sigma-2}$.
\end{enumerate}
\end{lemma}

\begin{proof}
As in the proof of Lemma~\ref{lem:identities}, $\nabla_yF(\tau,y(\tau))=0$ and $F(\tau,y(\tau))$ is constant. Let $Z=\sum_{i\in Q'}y_{u_i}\partial/\partial y_{u_i}$ and $W=\sum_{i\in Q'}y_{v_i}\partial/\partial y_{v_i}$. The function $U_Q$ is homogeneous of degree $-\frac12$ in the $y_{u_i}$ and in the $y_{v_i}$, $\psi$ is linear in the $y_{u_i}$, and $S_{2n+1}$ is homogeneous of degree $-n-\frac12$ in them. By Lemma~\ref{lem:d14exp}, along $y(\tau)$,
\begin{align*}
0=W\cdot\nabla_yF&=-\tfrac12\tau^{-2}U_Q+O(\tau^4),\\
0=Z\cdot\nabla_yF&=-\tfrac12\tau^{-2}U_Q+\tau^{2-2\sigma}\Gamma-\textstyle\sum_n\big(n+\frac12\big)\tau^{(2n+1)(\sigma-1)}B_n+O(\tau^4).
\end{align*}
The first identity gives (a). The $y_{u_j}$-component of $\nabla_yF=0$ reads $y_{v_j}f'(\Pi)+O(\tau^{3\sigma-3})=0$. As in the proof of Theorem~\ref{thm:order3}, $f'(\bar\Pi)=0\ne f''(\bar\Pi)$, so that $\Pi(\tau)-\bar\Pi=O(\tau^{3\sigma-3})$ and $[f(\Pi(\tau))]_q=0$ for $0<q<6\sigma-6$. Let $\sigma-1\le q\le3\sigma-3$. Then $f$, $R$ and the $B_n$ with $n\ge2$ contribute nothing at $\tau^q$. By (a) neither does $U_Q$. So the coefficients of $\tau^q$ in the second identity and in $F(\tau,y(\tau))$ are
\[
[\Gamma]_{q+2\sigma-2}-\tfrac12[B_0]_{q-\sigma+1}-\tfrac32[B_1]_{q-3\sigma+3}=0,\qquad[\Gamma]_{q+2\sigma-2}+[B_0]_{q-\sigma+1}+[B_1]_{q-3\sigma+3}=0,
\]
and their difference gives $3[B_0]_{q-\sigma+1}+5[B_1]_{q-3\sigma+3}=0$. For $q<3\sigma-3$ the second term vanishes, which is (b). For $q=3\sigma-3$ it is $5B_1(\bar y)$, which is (c). Finally $B_0(\bar y)=m_j(-\hat w_j)^{-1/2}\sum_{i\in Q'}m_ix_i$. The condition $\mathrm{X3}_k$ holds by Lemma~\ref{lem:d14blead}.
\end{proof}

\paragraph{K14 configurations with a bad shape.} By Lemmas~\ref{lem:d14blead} and~\ref{lem:d14energy}, the leading cluster configuration is a $\mathrm{K14}(m_Q)$ configuration with body $k$ at the origin that satisfies $\mathrm{X3}_k$ and has a bad shape. We determine these configurations at positive masses.

We use the following description of the configurations of $\mathrm{K14}(m)$, $m\in(\C^*)^4$. A configuration is \emph{collinear} if $w=\alpha+\beta z$ for some $\alpha,\beta\in\C$, and \emph{non-collinear} otherwise. A non-collinear configuration has $p_{il}=\lambda_0^{1/3}\eta_{il}t_it_l$ with $\lambda_0\ne0$, cube roots of unity $\eta_{il}=\eta_{li}$ and $t\in(\C^*)^4$ solving
\begin{equation}\label{eq:tsys}
\textstyle\sum_{l\ne i}m_l\eta_{il}t_l=\kappa t_i^2\quad(i=1,\dots,4),\qquad\sum_lm_lt_l^3=0
\end{equation}
for some $\kappa\in\C$. Indeed, let $A$ be the symmetric matrix with $A_{il}=m_im_lp_{il}^3$ for $i\ne l$ and zero row sums. By \eqref{eq:K14}, $A$ annihilates $\one$, $z$ and $w$, which are linearly independent for a non-collinear configuration, since no $z_{il}$ vanishes. As $A\ne0$, $A=\lambda_0 vv^{\mathsf T}$ with $\lambda_0\ne0$ and $v\cdot\one=v\cdot z=v\cdot w=0$. Since $A_{il}\ne0$, $v_i\ne0$ for all $i$. We choose $t_i$ with $t_i^3=v_i/m_i$. Then $p_{il}^3=\lambda_0 t_i^3t_l^3$, which gives the $\eta_{il}$, and $\sum_lm_lt_l^3=\sum_lv_l=0$. Finally $\sum_{l\ne i}v_lz_{il}w_{il}=\sum_lv_lz_lw_l$ does not depend on $i$. With $z_{il}w_{il}=p_{il}^{-2}=\lambda_0^{-2/3}\eta_{il}t_i^{-2}t_l^{-2}$ this is the first equation of \eqref{eq:tsys}. A gauge change $t_l\mapsto\zeta_lt_l$, $\eta_{il}\mapsto\zeta_i^{-1}\zeta_l^{-1}\eta_{il}$, with cube roots of unity $\zeta_l$, preserves \eqref{eq:tsys} and all $p_{il}$, and multiplies the three products $\eta_{12}\eta_{34}$, $\eta_{13}\eta_{24}$, $\eta_{14}\eta_{23}$ by a common factor. The gauge changes act freely on the $3^6$ patterns, since $\zeta_i\zeta_l=1$ for all $i\ne l$ forces $\zeta_1=\dots=\zeta_4=1$. Thus there are $3^6/3^4=9$ classes. The three products, up to a common factor, take $9$ values and each of them is attained. Hence two patterns are gauge equivalent if and only if their products are proportional. A pattern, and its class, is of \emph{type} I, II or III if its three products take one, two or three distinct values. There is one class of type I and there are six of type II. The two classes of type III are those of the pattern $\eta_{12}=\eta_{34}=1$, $\eta_{13}=\eta_{24}=\jj^2$, $\eta_{14}=\eta_{23}=\jj$ and of its complex conjugate.

\begin{lemma}\label{lem:typeIII}
Let $m\in(\C^*)^4$ have real ratios $m_i/m_l$. Then no non-collinear configuration of $\mathrm{K14}(m)$ has a pattern of type III.
\end{lemma}

\begin{proof}
Suppose that one has. We diagonalize the matrix of the pattern and show that $\kappa\ne0$ forces the configuration to be collinear and that $\kappa=0$ is impossible for real mass ratios. After a gauge change, $\eta$ is the pattern above or its complex conjugate. We treat the first. For the second, replace $\jj$ by $\jj^2$ throughout. Let $H=(\eta_{il})$, with zero diagonal, $u=(m_lt_l)_l$, $e_0=(1,1,1,1)$, $e_1=(1,1,-1,-1)$, $e_2=(1,-1,1,-1)$ and $e_3=(1,-1,-1,1)$. Then \eqref{eq:tsys} reads $Hu=\kappa(t_l^2)_l$ and $\sum_lu_lt_l^2=0$. We have $He_0=0$, $He_1=2e_1$, $He_2=2\jj^2e_2$, $He_3=2\jj e_3$. Hence the kernel of $H$ is $\C e_0$ and its image, spanned by $e_1,e_2,e_3$, consists of the vectors with coordinate sum $0$.

Suppose first that $\kappa\ne0$. Then $\sum_lt_l^2=0$. We write $(t_l^2)_l=\xi_1e_1+\xi_2e_2+\xi_3e_3$. We obtain $u=\frac\kappa2(\xi_1e_1+\jj\xi_2e_2+\jj^2\xi_3e_3)+\gamma e_0$ for some $\gamma\in\C$. Since $e_k\cdot e_{k'}=4\delta_{kk'}$, the last equation of \eqref{eq:tsys} becomes $2\kappa(\xi_1^2+\jj\xi_2^2+\jj^2\xi_3^2)=0$, so that $\xi_1^2+\jj\xi_2^2+\jj^2\xi_3^2=0$. For three bodies $a,b,c$ put
\[
h_{abc}=(\eta_{ab}t_c^2+\eta_{ac}t_b^2-\eta_{bc}t_a^2)^2-4\eta_{ab}\eta_{ac}t_b^2t_c^2 ,
\]
which is symmetric in $a,b,c$. For $(a,b,c)=(1,2,3)$ we have $\eta_{12}t_3^2+\eta_{13}t_2^2-\eta_{23}t_1^2=2\jj^2\xi_1+2\xi_2$ and $4\eta_{12}\eta_{13}t_2^2t_3^2=4\jj^2\big(\xi_3^2-(\xi_1-\xi_2)^2\big)$. Expanding gives
\[
h_{123}=-4\big(\xi_1^2+\jj\xi_2^2+\jj^2\xi_3^2\big)=0 .
\]
The double transpositions of the bodies fix $\eta$, change each $\xi_k$ at most in sign and permute the four sets $\{a,b,c\}$ transitively. Hence $h_{abc}=0$ for any three bodies $a,b,c$. On the other hand $d_{il}=z_{il}w_{il}=p_{il}^{-2}=\lambda_0^{-2/3}\eta_{il}t_i^{-2}t_l^{-2}$ (since $\eta_{il}^{-2}=\eta_{il}$) and $d_{ab}+d_{ac}-d_{bc}=z_{ab}w_{ac}+z_{ac}w_{ab}$, so that
\[
h_{abc}=\lambda_0^{4/3}(t_at_bt_c)^4\big((d_{ab}+d_{ac}-d_{bc})^2-4d_{ab}d_{ac}\big)=\lambda_0^{4/3}(t_at_bt_c)^4(z_{ab}w_{ac}-z_{ac}w_{ab})^2 .
\]
For a real configuration, $w=\bar z$, the factor $(z_{ab}w_{ac}-z_{ac}w_{ab})^2$ is $-16$ times the squared area of the triangle $abc$. For such a configuration this identity is Heron's formula. Since $h_{abc}=0$, $\lambda_0\ne0$ and $t\in(\C^*)^4$, we obtain $z_{ab}w_{ac}=z_{ac}w_{ab}$ for any three bodies $a,b,c$. With $a=1$ and $b=2$ this gives $w_{1c}=\beta z_{1c}$ for all $c$, where $\beta=w_{12}/z_{12}$, that is, $w=\alpha+\beta z$ with $\alpha=w_1-\beta z_1$. The configuration is collinear, a contradiction.

Hence $\kappa=0$. Then $u\in\ker H$, so that $t_l=\gamma/m_l$ with $\gamma\ne0$. The last equation of \eqref{eq:tsys} becomes $\gamma^3\sum_lm_l^{-2}=0$. But $\sum_lm_l^{-2}=m_1^{-2}\sum_l(m_1/m_l)^2$, where the last sum is a sum of squares of nonzero real numbers and hence positive, a contradiction.
\end{proof}

For the other types we invert the configuration at body $k$. Let $(z,w,p)$ be a non-collinear $\mathrm{K14}(m_Q)$ configuration with $m\in(\C^*)^4$ and $z_k=w_k=0$. We put $d_{il}=z_{il}w_{il}$ and, for $i\in Q'$, $z'_i=1/w_i$ and $w'_i=1/z_i$. We index the outer bodies cyclically, $a+1=b$, $b+1=c$, $c+1=a$, and put $u'_i=z'_{i+1}w'_{i+2}-z'_{i+2}w'_{i+1}$.

\begin{lemma}\label{lem:inversion}
\begin{enumerate}[label=\textup{(\alph*)}]
\item The numbers $(z'_a-z'_b)(w'_a-w'_b)$, $(z'_b-z'_c)(w'_b-w'_c)$ and $(z'_c-z'_a)(w'_c-w'_a)$ are a common nonzero factor times the products $\eta_{ab}\eta_{ck}$, $\eta_{bc}\eta_{ak}$ and $\eta_{ca}\eta_{bk}$ of the pattern.
\item There is $C\ne0$ with $m_i=C\,u'_i/p_{ik}$ for $i\in Q'$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) For $\{i,l,n\}=Q'$, $(z'_i-z'_l)(w'_i-w'_l)=d_{il}/(d_{ik}d_{lk})=d_{il}d_{nk}/(d_{ak}d_{bk}d_{ck})$, and by the derivation of \eqref{eq:tsys}, $d_{il}d_{nk}=\lambda_0^{-4/3}(t_at_bt_ct_k)^{-2}\eta_{il}\eta_{nk}$. (b) In that derivation $m_im_kp_{ik}^3=\lambda_0 v_iv_k$, and $v\perp z,w$ with $z_k=w_k=0$. The vectors $(z_a,z_b,z_c)$ and $(w_a,w_b,w_c)$ are independent, since $w=\beta z$ on $Q$ would make the configuration collinear. Hence $(v_a,v_b,v_c)$ is proportional to their cross product. Its component $i$ is $u'_i/(z'_{i+1}z'_{i+2}w'_{i+1}w'_{i+2})$. As $p_{ik}^{-2}=z_iw_i=1/(z'_iw'_i)$, $m_i=\lambda_0 v_kv_i/(m_kp_{ik}^3)$ is a common factor times $u'_i/p_{ik}$.
\end{proof}

The next theorem treats type I, in which the triangle $w'$ is equilateral (see the proof of Lemma~\ref{lem:classI}): $\mathrm{X3}_k$ and a bad shape force its centre to be the origin.

\begin{theorem}\label{thm:W}
Let $m_0,m_1,m_2>0$, $\delta\in\C$, and let $x_0,x_1,x_2\in\C^*$ satisfy $x_i^2=\delta+\jj^i$, $\sum_im_ix_i=0$ and $\sum_im_ix_i^{-2}=0$. Then $\delta=0$ and $m_0=m_1=m_2$.
\end{theorem}

\begin{proof}
We show that the hypotheses force $\delta=0$ or the vanishing of an explicit polynomial in $\delta$ and $\bar\delta$, and exclude the second case with $\delta\ne0$ by writing $\delta$ in polar form and using the positivity of the masses. We may assume $\sum_im_i=1$. We put $\nu=\sum_im_i\jj^i$ and $\rho=|\delta|$. We use the basis $\one=(1,1,1)$, $\Omega=(\jj^i)_i$, $\bar\Omega=(\jj^{2i})_i$ of $\C^3$. Since $x_i\ne0$, $\delta\ne-\jj^i$, that is, $\delta^3\ne-1$.
\begin{enumerate}[label=(\arabic*)]
\item From $(\delta+\jj^i)(\delta^2-\delta\jj^i+\jj^{2i})=\delta^3+1$ we obtain $z:=(x_i^{-2})_i=(\delta^2\one-\delta\Omega+\bar\Omega)/(\delta^3+1)$. So $\sum m_ix_i^{-2}=0$ reads $\delta^2-\nu\delta+\bar\nu=0$. With its conjugate this gives $\nu(1-\rho^2)=-(\delta\rho^2+\bar\delta^2)$. If $\rho=1$, then $\delta=-\bar\delta^2$ and $\delta^3=-\delta^2\bar\delta^2=-1$, which is excluded. Hence $\rho\ne1$.
\item The plane $V=\{v:\sum_im_iv_i=0\}$ contains $z$, $\bar z$ and $x$. As the $m_i$ are real, $\bar V=V$. In the basis $(\one,\Omega,\bar\Omega)$ the coordinates of $z$ and $\bar z$ are proportional to $(\delta^2,-\delta,1)$ and $(\bar\delta^2,1,-\bar\delta)$, whose cross product $n=(\rho^2-1,\bar\delta^2+\delta^2\bar\delta,\delta^2+\delta\bar\delta^2)$ has $n_0\ne0$. So $z$ and $\bar z$ span $V$ and the coordinates $(\hat x_0,\hat x_1,\hat x_2)$ of $x$ satisfy $n_0\hat x_0+n_1\hat x_1+n_2\hat x_2=0$.
\item Since $\Omega\Omega=\bar\Omega$, $\bar\Omega\bar\Omega=\Omega$ and $\Omega\bar\Omega=\one$ componentwise, $x_i^2=\delta+\jj^i$ reads $\hat x_0^2+2\hat x_1\hat x_2=\delta$, $2\hat x_0\hat x_1+\hat x_2^2=1$ and $2\hat x_0\hat x_2+\hat x_1^2=0$. We eliminate $\hat x_0$ by (2). Then $n_0$ times the third equation and $n_0^2$ times the difference of $\delta$ times the second and the first are two binary quadratic forms in $(\hat x_1,\hat x_2)$ with a common zero $(\hat x_1,\hat x_2)\ne(0,0)$, as $x\ne0$. Their resultant, $-3\delta(\rho^2-1)^2N(\delta,\bar\delta)$ $(\dagger)$, therefore vanishes, where $N\in\Z[\delta,\bar\delta]$ has $23$ terms:
\[
\begin{aligned}
N={}&4\delta^8\bar\delta^4+4\delta^8\bar\delta+24\delta^7\bar\delta^3-3\delta^7+40\delta^6\bar\delta^5-14\delta^6\bar\delta^2+4\delta^5\bar\delta^7+19\delta^5\bar\delta^4\\
&-12\delta^5\bar\delta+54\delta^4\bar\delta^6-48\delta^4\bar\delta^3+6\delta^4+\delta^3\bar\delta^8-20\delta^3\bar\delta^5+6\delta^3\bar\delta^2+28\delta^2\bar\delta^7\\
&-18\delta^2\bar\delta^4+8\delta^2\bar\delta-12\delta\bar\delta^6+12\delta\bar\delta^3-3\delta+4\bar\delta^8-4\bar\delta^2.
\end{aligned}
\]
Hence $\delta=0$ or $N(\delta,\bar\delta)=0$.
\item Suppose that $\delta=\rho e^{i\varphi}\ne0$ and $N(\delta,\bar\delta)=0$. Since each monomial $\delta^a\bar\delta^b$ of $N$ has $a-b\equiv1\pmod3$, $N=e^{i\varphi}E$ with $E$ a polynomial in $\rho$ and $e^{\pm3i\varphi}$ with real coefficients. With $c=\cos3\varphi$, $\operatorname{Re}E$ is a polynomial in $\rho$ and $c$,
\[
\begin{aligned}
\operatorname{Re}E={}&16\rho^8c^3+(2\rho^{11}+64\rho^9-30\rho^7)c^2+(8\rho^{12}+78\rho^{10}-46\rho^8-30\rho^6+18\rho^4-4\rho^2)c\\
&+39\rho^{11}-13\rho^9-33\rho^7+6\rho^5+8\rho^3-3\rho,
\end{aligned}
\]
and
\[
\operatorname{Im}E=-2\rho^2\sin3\varphi\,K,\qquad K=8\rho^6c^2+(\rho^9+24\rho^7-9\rho^5)c+15\rho^8-5\rho^6-3\rho^4+3\rho^2-2\quad(\dagger),
\]
\[
\Res_c(\operatorname{Re}E,K)=7776\rho^{24}(\rho-1)^6(\rho+1)^6(4\rho^8+13\rho^6+27\rho^4+26\rho^2+11)\quad(\dagger).
\]
The resultant is nonzero and $K$ has the leading coefficient $8\rho^6\ne0$. Hence $\operatorname{Re}E$ and $K$ have no common zero, and $\sin3\varphi=0$. If $e^{3i\varphi}=-1$, then $\operatorname{Re}E=-\rho(\rho-1)^4S_-(\rho)$ $(\dagger)$ with $S_-=\rho^4((\rho-1)^2(8\rho+7)+4)+6\rho^3+6\rho^2+8\rho+3>0$. If $e^{3i\varphi}=1$, then $\delta=\rho\jj^l$ for some $l$, and replacing $x_i$ and $m_i$ by $\jj^lx_{i+l}$ and $m_{i+l}$ we may take $\delta=\rho$. By (1), $\nu=-\rho^2/(1-\rho)$ is real. Thus $m_1=m_2$, $3m_0=1+2\nu$ and $3m_1=1-\nu$. For $\rho>1$ this gives $3m_1(\rho-1)=-(\rho^2-\rho+1)<0$, which contradicts $m_1>0$. For $\rho<1$ it gives $3m_0(1-\rho)=(1-2\rho)(1+\rho)$, so that $\rho<\frac12$. Then $\operatorname{Re}E=\rho(\rho+1)^4S_+(\rho)$ $(\dagger)$ with $S_+=(-3+8\rho-6\rho^2)+\rho^3(6-11\rho)+\rho^5(-6+9\rho+8\rho^2)\le-\frac12+\frac{2187}{21296}+\frac1{64}<0$: on $[0,\frac12]$ the first summand is increasing, the second is at most its value $\frac{2187}{21296}$ at $\rho=\frac9{22}$, and $-6+9\rho+8\rho^2\le\frac12$. In the remaining cases $E\ne0$, a contradiction.
\item Hence $\delta=0$. Then $\bar\nu=0$ by (1), and $m_0=m_1=m_2$.
\end{enumerate}
\end{proof}

\begin{lemma}\label{lem:classI}
Let $m_a,m_b,m_c>0$ and $m_k\ne0$, and let $(z,w,p)$ be a non-collinear $\mathrm{K14}(m_Q)$ configuration of type I with $z_k=w_k=0$ that satisfies $\mathrm{X3}_k$ and has a bad shape. Then $m_a=m_b=m_c$, $(m_k/m_a)^2=\frac13$, and after a relabelling of $a,b,c$, $z=s(1,\jj,\jj^2)$ and $w=s'(1,\jj^2,\jj)$ with $s,s'\in\C^*$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:inversion}(a), $(z'_i-z'_l)(w'_i-w'_l)=\Lambda\ne0$ for the three pairs of $Q'$. Let $A=w'_a-w'_b$, $B=w'_b-w'_c$ and $C=w'_c-w'_a$. The corresponding differences of the $z'_i$ are $\Lambda/A$, $\Lambda/B$ and $\Lambda/C$. As their sum is $0$, $AB+BC+CA=0$. With $A+B+C=0$, the numbers $A$, $B$, $C$ are the three roots of $X^3=ABC$. So $w'$ is an equilateral triangle and $z'$ is one of the opposite orientation. Scaling $z$ and $w$, which preserves $\mathrm{K14}$, $\mathrm{X3}_k$ and the bad shape, and labelling the outer bodies $0,1,2$, we obtain $w'_i=\delta+\jj^i$ and $z'_i=\gamma+\jj^{-i}$ with $\gamma,\delta\in\C$. Then $x_i^2=1/z_i=\delta+\jj^i$. The condition $\mathrm{X3}_k$ is $\sum m_ix_i^{-2}=0$. Theorem~\ref{thm:W} gives $\delta=0$ and $m_0=m_1=m_2$. Now $u'_i=(\jj^2-\jj)(\gamma\jj^i-1)$ and $p_{ik}^2=z'_iw'_i=\gamma\jj^i+1$. By Lemma~\ref{lem:inversion}(b), the three numbers $(\gamma\jj^i-1)^2/(\gamma\jj^i+1)$ equal one number $\varkappa$. If $\gamma\ne0$, the three distinct numbers $\gamma\jj^i$ are roots of the quadratic $(X-1)^2-\varkappa(X+1)$, which is impossible. Hence $\gamma=0$, $z=(1,\jj^2,\jj)$ and $w=(1,\jj,\jj^2)$. The exchange of $b$ and $c$ gives the stated form. Finally, $m_am_bp_{ab}^3=\lambda_0 v_av_b$ and $m_bm_kp_{bk}^3=\lambda_0 v_bv_k$ give $(m_k/m_a)^2=(v_k/v_a)^2(d_{bk}/d_{ab})^3$. Here $v_a=v_b=v_c=-v_k/3$ by the cross product in the proof of Lemma~\ref{lem:inversion}, $d_{bk}=1$ and $d_{ab}=3$, so that $(m_k/m_a)^2=\frac13$.
\end{proof}

\begin{theorem}\label{thm:typeII}
Let $m_a,m_b,m_c,m_k\in\C^*$ with $m_b/m_a,m_c/m_a\in\R$. No non-collinear $\mathrm{K14}(m_Q)$ configuration of type II with $z_k=w_k=0$ satisfies $\mathrm{X3}_k$ and has a bad shape.
\end{theorem}

\begin{proof}
We parametrize the inverted configuration by three unknowns, derive two polynomial equations in them from the bad shape and $\mathrm{X3}_k$, solve the resulting system exactly over $\Q$, and show in ball arithmetic that at each of its solutions one of the two squared ratios of the outer masses is not positive. We label the outer bodies $0,1,2$. By Lemma~\ref{lem:inversion}(a), $(z'_i-z'_l)(w'_i-w'_l)=\theta_{il}\Lambda$, where $\Lambda\ne0$ and $\theta_{01},\theta_{12},\theta_{20}$ are cube roots of unity, exactly two of them equal. After a relabelling, which preserves $\mathrm{K14}$, $\mathrm{X3}_k$ and the bad shape, and a change of $\Lambda$, $(\theta_{01},\theta_{12},\theta_{20})=(1,1,\varpi)$ with $\varpi\in\{\jj,\jj^2\}$. Let $A=w'_0-w'_1$ and $B=w'_1-w'_2$. Then $z'_0-z'_1=\Lambda/A$, $z'_1-z'_2=\Lambda/B$ and $z'_2-z'_0=-\varpi\Lambda/(A+B)$. As their sum is $0$, $A^2+(2-\varpi)AB+B^2=0$ $(\dagger)$. So $t=A/B$ satisfies $t^2+(2-\varpi)t+1=0$, that is, $\varpi=(t+1)^2/t$. The condition $\varpi^2+\varpi+1=0$ becomes
\[
f(t)=t^4+5t^3+9t^2+5t+1=0\quad(\dagger).
\]
Scaling $z$ and $w$ we take $B=\Lambda=1$. Then $w'=(\delta,\delta-t,\delta-t-1)$ and $z'=(\gamma,\gamma-1/t,\gamma-1/t-1)$ with $\gamma,\delta\in\C$. The $u'_i$ are affine in $(\gamma,\delta)$ $(\dagger)$. With $\xi_i=p_{ik}/x_i$ we have $\xi_i^2=z'_i$, as $x_i^2=w'_i$. Lemma~\ref{lem:inversion}(b) gives $m_i=C\,u'_i/(x_i\xi_i)$. So the bad shape and $\mathrm{X3}_k$ read
\[
\textstyle\sum_iu'_i/\xi_i=0,\qquad\sum_iu'_i/(x_i^3\xi_i)=0.
\]
For $r\in\C^3$, some choice of signs gives $r_0\pm r_1\pm r_2=0$ if and only if $H(r_0^2,r_1^2,r_2^2)=0$, where $H(h)=\sum_ih_i^2-2\sum_{i<l}h_ih_l$, because $H(r_0^2,r_1^2,r_2^2)$ is the product of the four sums $r_0\pm r_1\pm r_2$ $(\dagger)$. Hence $H(h)=H(h')=0$ for $h_i=u_i'^2/z'_i$ and $h'_i=u_i'^2/(w_i'^3z'_i)$. Clearing denominators, we obtain the polynomials $H_1=t^8(z'_0z'_1z'_2)^2H(h)$ and $H_2=t^8(w_0'^3w_1'^3w_2'^3z'_0z'_1z'_2)^2H(h')$ in $\Q[t,\gamma,\delta]$ $(\dagger)$. Let $S=t\prod_iw'_i\prod_i(tz'_i)\prod_i(tu'_i)$. It does not vanish at a configuration: $t\ne0$ as $f(0)\ne0$, the $z'_i$ and $w'_i$ are finite and nonzero, and $u'_i\ne0$ as $m_i\ne0$. If the outer masses have real ratios, then
\[
R_i=(m_i/m_0)^2=\frac{u_i'^2w'_0z'_0}{u_0'^2w'_iz'_i}\in(0,\infty)\qquad(i=1,2).
\]
It therefore suffices to show that no solution $(t,\gamma,\delta,s)$ of
\begin{equation}\label{eq:heron}
f(t)=H_1=H_2=sS-1=0
\end{equation}
has $R_1,R_2\in(0,\infty)$.

We solve \eqref{eq:heron} exactly over $\Q$. A rational parametrization by the coordinate $\delta$, with a squarefree eliminant of degree $280$ and a denominator that is invertible modulo the eliminant, satisfies the four equations exactly modulo the eliminant. So \eqref{eq:heron} has at least $280$ solutions, with distinct values of $\delta$. To show that it has no others we use a second system $E$, in the unknowns $t,\gamma,\delta$ and $e_i=x_{i+1}x_{i+2}$, $f_i=t\xi_{i+1}\xi_{i+2}$ (indices modulo $3$). It consists of $f(t)$, the relations $e_i^2=w'_{i+1}w'_{i+2}$, $e_ie_l=w'_ne_n$, $f_i^2=(tz'_{i+1})(tz'_{i+2})$ and $f_if_l=(tz'_n)f_n$ for $\{i,l,n\}=\{0,1,2\}$, and the two conditions in the form $\sum_i(tu'_i)f_i=0$ and $\sum_i(tu'_i)e_i^3f_i=0$. Each solution of \eqref{eq:heron} gives a point of $V(E)$ with the same $(t,\gamma,\delta)$, at which $S\ne0$: we choose the signs of the $\xi_i$ so that the first sum vanishes, and then those of the $x_i$ for the second. A standard basis of $E$ over $\Q$ shows that $E$ has dimension $0$ and that the quotient ring of the polynomial ring by $E$ has dimension $472$ over $\Q$. Multiplication by $S$ on this quotient ring has rank $280$ over $\Q$. Over $\overline{\Q}$, the quotient ring is the product of the local rings at the points of $V(E)$. On the local ring at a point where $S\ne0$ multiplication by $S$ is invertible. So the rank is at least the total multiplicity of the points of $V(E)$ at which $S\ne0$, there are at most $280$ such points, and the $280$ solutions are all. In ball arithmetic at $800$ bits, with disjoint balls around the roots of the eliminant, at each solution $|\operatorname{Im}R_i|\ge0.0469$ or $\operatorname{Re}R_i\le-0.0469$ for $i=1$ or $i=2$. The programs and their logs are in the supplementary material.
\end{proof}

\begin{theorem}\label{thm:col}
Let $m_a,m_b,m_c,m_k>0$. No collinear $\mathrm{K14}(m_Q)$ configuration with $z_k=0$ satisfies $\mathrm{X3}_k$ and has a bad shape.
\end{theorem}

\begin{proof}
Let $w=\alpha+\beta z$. Then $\beta\ne0$, as no $w_{il}$ vanishes, $w_{il}=\beta z_{il}$, and $p_{il}=s_{il}\beta^{-1/2}/z_{il}$ with $s_{il}=\pm1$ and $s_{li}=-s_{il}$. So $F^w_i=\beta F^z_i$ and $F^z_i=\beta^{-3/2}\sum_{l\ne i}m_ls_{il}/z_{il}^2$. Thus $\mathrm{K14}$ holds if and only if
\[
\textstyle(C_i)\qquad\sum_{l\ne i}m_ls_{il}/z_{il}^2=0\qquad(i\in Q).
\]
We write $z_i=x_i^{-2}$ for $i\in Q'$, with the $x_i$ of the bad shape, $\chi_i=x_i^3$ and $\epsilon_i=s_{ki}$, and $m$, $x$, $x^{-2}$ for the vectors $(m_i)$, $(x_i)$, $(x_i^{-2})$, $i\in Q'$. Then $\mathrm{X3}_k$ and the bad shape say $m\cdot x^{-2}=m\cdot x=0$. The row $(C_k)$ says $\sum_{i\in Q'}m_i\epsilon_ix_i^4=0$.

Suppose first that $x$ and $x^{-2}$ are dependent. Then $\chi_a=\chi_b=\chi_c$, that is, $x_i=\zeta_ix_a$ with $\zeta_i^3=1$. As $m_i>0$, the relation $\sum m_i\zeta_i=0$ forces $\{\zeta_a,\zeta_b,\zeta_c\}=\{1,\jj,\jj^2\}$ and $m_a=m_b=m_c=:\mu$. Now $(C_k)$ reads $\mu x_a^4\sum\epsilon_i\zeta_i=0$, which forces $\epsilon_a=\epsilon_b=\epsilon_c$. After a scaling of $z$ and a relabelling of $b$ and $c$, $z=(1,\jj,\jj^2)$. As $(1-\jj)^2=-3\jj$ and $(1-\jj^2)^2=-3\jj^2$, the row $(C_a)$ reads $m_ks_{ak}=\frac\mu3(s_{ab}\jj^2+s_{ac}\jj)$. Since $m_k$ is real, $s_{ab}=s_{ac}$ and $m_k=-s_{ak}s_{ab}\mu/3$. In the same way $(C_b)$ and $(C_c)$ give $s_{ba}=s_{bc}$, $m_k=-s_{bk}s_{bc}\mu/3$, and $s_{cb}=s_{ca}$, $m_k=-s_{ck}s_{ca}\mu/3$. As $s_{ik}=-\epsilon_a$ for all $i\in Q'$, $m_k>0$ forces $s_{ab}=s_{bc}=s_{ca}=\epsilon_a$. Then $s_{ac}=s_{ab}=\epsilon_a$ contradicts $s_{ca}=\epsilon_a$.

Hence $x$ and $x^{-2}$ are independent and $m$ is proportional to their cross product, that is, to $(x_a^2(\chi_b-\chi_c),x_b^2(\chi_c-\chi_a),x_c^2(\chi_a-\chi_b))$. In particular the $\chi_i$ are distinct. Now $(C_k)$ says that the rows $(x_i,x_i^{-2},\epsilon_ix_i^4)$, $i\in Q'$, have determinant $0$, that is, after multiplying the rows by $x_i^2$, $\det(\chi_i,1,\epsilon_i\chi_i^2)_{i\in Q'}=0$. If $\epsilon_a=\epsilon_b=\epsilon_c$, this is $\pm$ the Vandermonde determinant of the $\chi_i$ $(\dagger)$, which is nonzero. Otherwise, after the reversal $s\mapsto-s$, which preserves the rows $(C_i)$, and a relabelling, $\epsilon=(1,1,-1)$. The determinant is then $(\chi_b-\chi_a)(2\chi_c^2-(\chi_a-\chi_c)(\chi_b-\chi_c))$ $(\dagger)$. So $(\chi_a-\chi_c)(\chi_b-\chi_c)=2\chi_c^2$. We divide $x$ by $x_c$, which preserves all conditions and the mass ratios, and put $A=\chi_a-1$ and $B=\chi_b-1$. Then $AB=2$, $A\notin\{0,-1,-2,\pm\sqrt2\}$ since the $\chi_i$ are nonzero and distinct, and
\[
m_a/m_c=2x_a^2/(A^2-2),\qquad m_b/m_c=-x_b^2A^2/(A^2-2).
\]
As both are positive, $(x_bA/x_a)^2<0$, $x_b=i\gamma x_a/A$ with $\gamma\in\R\setminus\{0\}$, and $x_a^2=\rho(A^2-2)$ with $\rho>0$. Taking cubes, with $\chi_a=1+A$ and $\chi_b=(A+2)/A$, we obtain
\[
\textup{(E1)}\quad\gamma^3=iA^2(A+2)/(A+1),\qquad\textup{(E2)}\quad(A+1)^2/(A^2-2)^3=\rho^3>0.
\]
Let $A=u+iv$ with $u,v$ real. Then (E1) gives $P_1=0$ and (E2) gives $P_2=0$, where
\[
P_1=\operatorname{Re}\big(A^2(A+2)(\bar A+1)\big)=u^4+3u^3+2u^2-uv^2-v^4-2v^2\quad(\dagger),\qquad P_2=\operatorname{Im}\big((A+1)^2(\bar A^2-2)^3\big).
\]
Their resultant is $\Res_v(P_1,P_2)=4096u^8(u+1)^3(u+2)h(u)^2$ $(\dagger)$ with $h=2u^3+4u^2-u-4$. The polynomial $P_1$ has the leading coefficient $-1$ in $v$. So $u\in\{0,-1,-2\}$ or $h(u)=0$. For $u=0,-1,-2$, $P_1=-v^2(v^2+2),-v^2(v^2+1),-v^4$, so $v=0$ and $A$ is excluded. The cubic $h$ has discriminant $-104<0$ and hence a single real root $u_0$, and $h(0.917)<0<h(0.918)$, so that $u_0\in(0.917,0.918)$. At $u=u_0$, $P_1=0$ is a quadratic equation in $v^2$ whose roots have the product $-u_0^2(u_0+1)(u_0+2)<0$. Its positive root is $v_0^2=2-u_0^2$, because $P_1=2h(u)$ on the circle $u^2+v^2=2$ $(\dagger)$. Thus $A$ is $A_0=u_0+iv_0$ or $\bar A_0$. If $A=A_0$, then (E1) and (E2) determine $\gamma\in\R$ and $\rho>0$, hence $x_a$ up to sign, $x_b=i\gamma x_a/A_0$, the positions $z_i=x_i^{-2}$ and the outer mass ratios. Since $m_k$ has the coefficient $s_{ak}/z_a^2\ne0$ in the row $(C_a)$, this row determines $m_k/m_a$ for each choice of $s_{ab},s_{ac}$. Certified in ball arithmetic, the four values are $\pm(0.0059+0.4768i)$ and $\pm(0.1808+0.6951i)$, none of them real. At $\bar A_0$ they are the conjugates. This contradicts $m_k>0$.
\end{proof}

\begin{theorem}\label{thm:Xhat}
Let $m_a,m_b,m_c,m_k>0$, and let $(z,w,p)$ be a $\mathrm{K14}(m_Q)$ configuration with $z_k=w_k=0$ that satisfies $\mathrm{X3}_k$ and has a bad shape. Then $(m_a,m_b,m_c,m_k)$ is a positive multiple of $(\sqrt3,\sqrt3,\sqrt3,1)$, and after a relabelling of $a,b,c$, $(z,w,p)=(sz_{\mathcal E},s'w_{\mathcal E},\zeta p_{\mathcal E})$ with $s,s'\in\C^*$ and $\zeta^2ss'=1$. Here $z_{\mathcal E}=(1,\jj,\jj^2)$ and $w_{\mathcal E}=(1,\jj^2,\jj)$ on $(a,b,c)$, and $p_{\mathcal E}$ is $1/\sqrt3$ on the pairs of $Q'$ and $-1$ on the pairs $ik$.
\end{theorem}

\begin{proof}
Theorem~\ref{thm:col} excludes collinear configurations. A non-collinear one has a pattern as in \eqref{eq:tsys}. Lemma~\ref{lem:typeIII} and Theorem~\ref{thm:typeII} exclude the types III and II. For type I, Lemma~\ref{lem:classI} gives the masses and, after a relabelling and a scaling of $z$ and $w$, $z=z_{\mathcal E}$ and $w=w_{\mathcal E}$. We scale the masses to $(\sqrt3,\sqrt3,\sqrt3,1)$. Then $d_{il}=3$ on the pairs of $Q'$ and $d_{ik}=1$, so that $p_{il}=e_{il}/\sqrt3$ and $p_{ik}=f_i$ with $e_{il},f_i\in\{\pm1\}$. The row $F^z_a=0$ reads $(e_{ab}(1-\jj)+e_{ac}(1-\jj^2))/3+f_a=0$. Its imaginary part gives $e_{ab}=e_{ac}$, and its real part $f_a=-e_{ab}$. The rows of $b$ and $c$, divided by $\jj$ and $\jj^2$, give the same relations, so that $p=\pm p_{\mathcal E}$. Conversely, $\pm p_{\mathcal E}$ solve the rows $F^z_i=0$ for $i\in Q'$ and therefore also $F^z_k=0$. The rows $F^w=0$ are their complex conjugates.
\end{proof}

We write $\mathcal E$ for the configuration $(z_{\mathcal E},w_{\mathcal E},p_{\mathcal E})$ with the masses $(\sqrt3,\sqrt3,\sqrt3,1)$. It is the equilateral triangle $1,\jj,\jj^2$ with body $k$ at its centre.

\paragraph{Nondegeneracy and the slice expansion.} Let $(z,w,p)$ be a $\mathrm{K14}(m_Q)$ configuration with body $k$ at the origin and $m\in(\C^*)^4$. Since $\partial U_Q/\partial u_i=-\frac12m_iF^w_i$ and $\partial U_Q/\partial v_i=-\frac12m_iF^z_i$ (\S\ref{ss:order5}), it is a critical point of $U_Q$ in the six coordinates $(u_i,v_i)_{i\in Q'}$. Let $H$ be the Hessian of $U_Q$ there, $Z=(z,0)$ and $W=(0,w)$. As $U_Q$ is homogeneous of degree $-\frac12$ in $u$ and in $v$ separately, $HZ=HW=0$. We call a $\mathrm{K14}$ configuration \emph{nondegenerate} if $\ker H$ is spanned by $Z$ and $W$. For $a\in Q'$ let $\mathcal S_a=\{\xi\in\C^6:\xi_{u_a}=\xi_{v_a}=0\}$. The components of a vector along $\mathcal S_a$ are its four components other than those of $u_a$ and $v_a$. The matrix $H_a$ is the block of $H$ on $\mathcal S_a$. Let $g=\nabla\psi=(m_a,m_b,m_c,0,0,0)$.

\begin{lemma}\label{lem:d14slice}
Let the configuration be nondegenerate and satisfy $\mathrm{X3}_k$, and let $a\in Q'$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $H_a$ is invertible, and there is a unique analytic map $\lambda\mapsto Y(\lambda)\in(z,w)+\mathcal S_a$, for small $\lambda\in\C$, with $Y(0)=(z,w)$ and $\partial_l(U_Q-\lambda\psi)(Y(\lambda))=0$ for the four coordinates $l\ne u_a,v_a$.
\item Let $U_Q(Y(\lambda))=\sum_h\alpha_h\lambda^h$ and $\psi(Y(\lambda))=\sum_h\psi_h\lambda^h$. Then $\alpha_0=\psi_0=0$ and $h\alpha_h=(h-1)\psi_{h-1}$ for $h\ge1$. If $\xi\in\mathcal S_a$ and the components of $H\xi-g$ along $\mathcal S_a$ vanish, then $\psi_1=g\cdot\xi$ and $\psi_2=-\frac12D^3U_Q[\xi,\xi,\xi]$.
\item Replace $(z,w,p)$ by $(sz,s'w,\zeta p)$ with $\zeta^2ss'=1$ and $m$ by $cm$ with $c\ne0$, and relabel the outer bodies together with $a$. The new configuration is nondegenerate, and $\psi_h$ is multiplied by $c^{1-h}s(s/\zeta)^h$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) Let $\xi\in\mathcal S_a$ with $H_a\xi=0$. Then $H\xi=\beta e_{u_a}+\beta'e_{v_a}$. Pairing with $Z$ and $W$ gives $\beta z_a=\beta'w_a=0$. So $H\xi=0$ and $\xi\in\C Z+\C W$ with $\xi_{u_a}=\xi_{v_a}=0$. Since $z_a,w_a\ne0$, $\xi=0$. The implicit function theorem gives $Y$.
(b) $\alpha_0=U_Q(z,w)=0$ (\S\ref{ss:Kgating}). By $\mathrm{X3}_k$, $\psi_0=0$. Let $\hat F(\lambda)=(U_Q-\lambda\psi)(Y(\lambda))=\sum_hf_h\lambda^h$. The gradient of $U_Q-\lambda\psi$ at $Y(\lambda)$ has only the components $u_a$ and $v_a$, and $Y'(\lambda)\in\mathcal S_a$. So $\hat F'(\lambda)=-\psi(Y(\lambda))$ and $U_Q(Y(\lambda))=\hat F-\lambda\hat F'$. Hence $\alpha_h=(1-h)f_h$ and $\psi_{h-1}=-hf_h$. Next, $Y(\lambda)=(z,w)+\lambda\xi+\lambda^2\eta+O(\lambda^3)$, where $\eta\in\mathcal S_a$ and the components of $H\eta+\frac12D^3U_Q[\xi,\xi]$ along $\mathcal S_a$ vanish. As $\psi$ is linear and $H$ is symmetric, $\psi_1=g\cdot\xi$ and $\psi_2=g\cdot\eta=\xi\cdot H\eta=-\frac12D^3U_Q[\xi,\xi,\xi]$.
(c) Near the two configurations, $U_Q(sz',s'w')=\zeta\,U_Q(z',w')$ for the branches of the $p_{il}$, and $\psi(sz')=s\psi(z')$. So the equations of (a) for the new configuration at $\lambda$ are those for the old one at $(s/\zeta)\lambda$, after the substitution $(z',w')\mapsto(sz',s'w')$. This multiplies $\psi_h$ by $s(s/\zeta)^h$. Under $m\mapsto cm$, $U_Q$ and $\psi$ are multiplied by $c^2$ and $c$. Thus $Y(\lambda)$ becomes $Y(\lambda/c)$ and $\psi_h$ is multiplied by $c^{1-h}$. The Hessian becomes $\zeta D^{-1}HD^{-1}$ with $D=\mathrm{diag}(s,s,s,s',s',s')$, resp.\ $c^2H$, and a relabelling permutes the coordinates, so that nondegeneracy is preserved.
\end{proof}

\begin{lemma}\label{lem:Pdata}
The configuration $\mathcal E$ is nondegenerate. Its Hessian $H$ has rank $4$ and $\det H_a=81/256$, and for the slice of $a$, $\psi_1=0$ and $\psi_2=40/\sqrt3$.
\end{lemma}

\begin{proof}
By the end of the proof of Theorem~\ref{thm:Xhat}, $\mathcal E$ is a $\mathrm{K14}$ configuration, and it satisfies $\mathrm{X3}_k$ as $1+\jj+\jj^2=0$. Hence Lemma~\ref{lem:d14slice} applies. As $HZ=HW=0$, rank $4$ means that $\ker H$ is spanned by $Z$ and $W$. The values are exact computations in $\Q(\sqrt3,i)$. We take $\nabla U_Q$, $H$ and $D^3U_Q$ at $\mathcal E$ from the exact Taylor expansion of each $p_{il}$ in the form $p_{il}(\mathcal E)(1+\epsilon)^{-1/2}$, with $\epsilon$ a polynomial in the displacement. We check $\nabla U_Q=0$ and evaluate the formulas of Lemma~\ref{lem:d14slice}(b). The two programs that make these computations also test $\det H_a\ne0$. They are in the supplementary material.
\end{proof}

\paragraph{The exponent.} At a nondegenerate leading cluster configuration the place follows the map $Y$ of Lemma~\ref{lem:d14slice}, with $\lambda$ replaced by a series of order $4-2\sigma$.

\begin{lemma}\label{lem:d14follow}
Let the leading cluster configuration $\bar y_Q$ at a place of type $\sigma_k(j)(t)$ be nondegenerate, let $a\in Q'$, and let $Y$ be as in Lemma~\ref{lem:d14slice}. After a reparametrization and a rotation as in \S\ref{ss:expansion}, $y_{u_a}(\tau)\equiv\hat z_a$ and $y_{v_a}(\tau)\equiv\hat w_a$. Then, with $y_Q=(y_{u_i},y_{v_i})_{i\in Q'}$ and $\lambda(\tau)=\frac{m_j}{2M}y_{v_j}(\tau)\,\tau^{4-2\sigma}$,
\[
y_Q(\tau)-Y(\lambda(\tau))=O(\tau^{1+\sigma}),\qquad U_Q(y_Q(\tau))-U_Q(Y(\lambda(\tau)))=O(\tau^{5-\sigma}).
\]
\end{lemma}

\begin{proof}
The conditions are $\mu\varphi^{-2}y_{u_a}=\hat z_a$ and $\mu^{-1}\varphi^{-2}y_{v_a}=\hat w_a$, which determine $\varphi^4$ and then $\mu$. By Lemma~\ref{lem:d14exp}, the cluster components of $\tau^2\nabla_yF(\tau,y(\tau))=0$ read $\nabla(U_Q-\lambda(\tau)\psi)(y_Q(\tau))=O(\tau^{1+\sigma})$. Both $y_Q(\tau)$ and $Y(\lambda(\tau))$ lie in $\bar y_Q+\mathcal S_a$. The components of $\nabla(U_Q-\lambda\psi)$ along $\mathcal S_a$ vanish at $(Y(\lambda),\lambda)$, they are $O(\tau^{1+\sigma})$ at $(y_Q(\tau),\lambda(\tau))$, and their derivative at $(\bar y_Q,0)$ is the invertible $H_a$. This gives the first estimate. For the second, let $\delta=y_Q(\tau)-Y(\lambda(\tau))\in\mathcal S_a$. Along $\mathcal S_a$ the components of $\nabla U_Q(Y(\lambda))$ and $\lambda g$ agree. Hence $U_Q(y_Q)-U_Q(Y)=\lambda(\tau)\,g\cdot\delta+O(|\delta|^2)=O(\tau^{5-\sigma})$, since $2+2\sigma>5-\sigma$.
\end{proof}

In both propositions below we use $\lambda(\tau)=\bar\lambda\tau^{4-2\sigma}(1+o(1))$ with $\bar\lambda=m_j\hat w_j/(2M)\ne0$. For a convergent power series $\sum_hc_h\lambda^h$, the first $h$ with $c_h\ne0$ contributes $c_h\bar\lambda^h$ at the exponent $h(4-2\sigma)$, and all other contributions have larger exponents.

\begin{proposition}\label{prop:D14N}
Let $\sigma\ne(2n+1)/(n+1)$ for all integers $n\ge1$. At positive masses, the leading cluster configuration at a place of type $\sigma_k(j)(t)$, $\sigma=1+t$, is degenerate.
\end{proposition}

\begin{proof}
Suppose that it is nondegenerate, and normalize the place as in Lemma~\ref{lem:d14follow}. Lemma~\ref{lem:d14energy} holds in this parametrization, since its proof uses only $\nabla_yF=0$ and the constancy of $F$ along the place. Let $S_1(Y(\lambda))=\sum_hs_h\lambda^h$, with $s_0=0$ by the bad shape. Then $S_1(y_Q(\tau))=\sum_hs_h\lambda(\tau)^h+O(\tau^{1+\sigma})$, where $1+\sigma>2\sigma-2$. Since $B_0=m_jwS_1$ and $w(y(\tau))\to(-\hat w_j)^{-1/2}\ne0$, Lemma~\ref{lem:d14energy}(b) gives $[S_1(y_Q(\tau))]_r=0$ for $r<2\sigma-2$ and $[B_0]_{2\sigma-2}=m_j(-\hat w_j)^{-1/2}[S_1]_{2\sigma-2}$. So the first $h$ with $s_h\ne0$, if there is one, has $h(4-2\sigma)\ge2\sigma-2$, with equality only for $\sigma=(2h+1)/(h+1)$. Hence $h(4-2\sigma)>2\sigma-2$ and $[S_1(y_Q(\tau))]_{2\sigma-2}=0$. By Lemma~\ref{lem:d14energy}(c), $B_1(\bar y)=\frac12m_j(-\hat w_j)^{-1/2}\hat z_j\sum_{i\in Q'}m_ix_i^3=0$. Thus $\sum_{i\in Q'}m_ix_i^3=0$. With $\sum_{i\in Q'}m_ix_i=\sum_{i\in Q'}m_ix_i^{-2}=0$, this contradicts Proposition~\ref{prop:P11}.
\end{proof}

\begin{proposition}\label{prop:resonant}
Let $\sigma=(2n+1)/(n+1)$ for an integer $n\ge1$. If the leading cluster configuration at a place of type $\sigma_k(j)(t)$, $\sigma=1+t$, is nondegenerate, then $\psi_1=\psi_2=0$ for each $a\in Q'$.
\end{proposition}

\begin{proof}
We normalize the place for $a$ as in Lemma~\ref{lem:d14follow}. By Lemmas~\ref{lem:d14energy}(a) and~\ref{lem:d14follow}, $\sum_h\alpha_h\lambda(\tau)^h=U_Q(Y(\lambda(\tau)))=O(\tau^{5-\sigma})$. So $\alpha_h=0$ whenever $h(4-2\sigma)<5-\sigma$. As $\sigma\ge\frac32>\frac75$, this holds for $h=2$ and $h=3$. Then Lemma~\ref{lem:d14slice}(b) gives $\psi_1=2\alpha_2=0$ and $\psi_2=\frac32\alpha_3=0$.
\end{proof}

\begin{proof}[Proof of Theorem~\textup{\ref{thm:D14}}]
Let a singular place $P$ have type $\sigma_k(j)(t)$ with $0<t<1$, and let $\sigma=1+t$. By Lemmas~\ref{lem:d14blead} and~\ref{lem:d14energy}(b), the leading cluster configuration is a $\mathrm{K14}(m_Q)$ configuration that satisfies $\mathrm{X3}_k$ and has a bad shape. By Theorem~\ref{thm:Xhat}, $m_Q=c(\sqrt3,\sqrt3,\sqrt3,1)$ with $c>0$, and after a relabelling of $a,b,c$ the leading cluster configuration is $(sz_{\mathcal E},s'w_{\mathcal E},\zeta p_{\mathcal E})$ with $\zeta^2ss'=1$. By Lemmas~\ref{lem:Pdata} and~\ref{lem:d14slice}(c) it is nondegenerate. If $\sigma$ is not of the form $(2n+1)/(n+1)$, Proposition~\ref{prop:D14N} gives a contradiction. Otherwise we take the slice of the outer body that corresponds to $a$ in $\mathcal E$. By Lemmas~\ref{lem:Pdata} and~\ref{lem:d14slice}(c), $\psi_2=c^{-1}s(s/\zeta)^2\cdot40/\sqrt3\ne0$, which contradicts Proposition~\ref{prop:resonant}.
\end{proof}