\section{Diameter extensions of eleven-point cores} \label{sec:eleven-extensions} We next give a finite extension argument for the configurations in Lemma~\ref{lem:eleven-classification}. Throughout this section a palette consists of \emph{squared} distances. This convention does not change the number of distinct distances. \begin{lemma}\label{lem:eleven-extension} Let $A\subset X\subset\mathbb R^2$, where $|A|=11$, $A$ determines at most six distances, and $X$ determines at most seven distances. If $\operatorname{diam}(A)<\operatorname{diam}(X)$, then $|X|\le16$. \end{lemma} \begin{proof} Apply the classification in Lemma~\ref{lem:eleven-classification}. We treat its fifty-two lattice configurations and ten circular pattern families below. The lattice calculations are unchanged by similarity, so use the coordinates of Tables~\ref{tab:lattice-cores} and~\ref{tab:eleven-deletions}. Write $D=\operatorname{diam}(X)$ and let $S$ be the squared-distance palette of $A$. If $X$ has at most six distances, then the published bound $g_2(6)=13$ suffices. We may therefore assume that $X$ has exactly seven distances. \subsection*{Nonextreme anchors} Let $I$ be the points of $A$ which are not extreme vertices of its convex hull. No point of $I$ can be an endpoint of a diameter of $X$. Indeed, if $p=\sum_i\lambda_i a_i$ is a nontrivial convex combination of distinct points of $A$, strict convexity gives, for every $q\in X$, \[ \|q-p\|^2<\sum_i\lambda_i\|q-a_i\|^2\le D^2. \] This includes points in the relative interiors of hull sides. \subsection*{The forty-nine six-distance lattice cores} Here $|S|=6$, and the sole additional squared distance is $D^2>\max S$. Every added point $q\in X\setminus A$ must therefore satisfy $\|q-p\|^2\in S$ for every $p\in I$. For each of the forty-nine cores the verifier finds three noncollinear points $A_0,B_0,C_0$ in $I$. In triangular coordinates put $Q(x,y)=x^2+xy+y^2$. An assignment \[ Q(q-A_0)=a,\qquad Q(q-B_0)=b,\qquad Q(q-C_0)=c, \qquad (a,b,c)\in S^3, \] determines $q$ uniquely. To see this, put $w=B_0-A_0$ and $v=C_0-A_0$ and subtract the first equation from the other two. The resulting system is \[ \begin{pmatrix}2w_x+w_y&w_x+2w_y\\ 2v_x+v_y&v_x+2v_y\end{pmatrix} \begin{pmatrix}q_x\\q_y\end{pmatrix} =\begin{pmatrix}Q(B_0)-Q(A_0)+a-b\\ Q(C_0)-Q(A_0)+a-c\end{pmatrix}. \] Its determinant is nonzero by noncollinearity. Solve it over $\mathbb Q$ and retain the solution only if the first original equation holds. The other two then hold as well. Exhausting $S^3$ thus produces every possible real added point; rationality follows from the equations and is not imposed on $X$. Filter the candidates against all of $I$ and $A$. A retained point either has only old distances to $A$, or has just one additional squared distance $T>\max S$. The new diameter must occur between a candidate and a core point, or between two candidates. Enumerate these possible values of $T$, and, for each, keep only candidates compatible with the palette $S\cup\{T\}$. Join two candidates when their mutual squared distance belongs to that palette. The largest admissible extension is a largest clique in this finite graph. Checking the old palette alone also covers extensions with no additional distance. \subsection*{The three five-distance lattice cores} These are $L_0\setminus\{v_3\}$, $L_4\setminus\{v_0\}$, and $L_4\setminus\{v_4\}$, with the zero-based vertex indices of Table~\ref{tab:eleven-deletions}. Each has exactly five nonextreme points. Now the palette of $X$ has the form \[ S\cup\{T,D^2\},\qquad D^2>\max(S\cup\{T\}). \] For an added point $q$, its squared distances to $I$ lie in $S\cup\{T\}$. If at most two of these distances equal $T$, at least three are old, so there are two distinct old-distance anchors. If at least three equal $T$, the corresponding three points lie on a circle centered at $q$. They are noncollinear, and $q$ is their circumcenter. Both alternatives give finite lists without first specifying $T$. For the first alternative choose distinct $A_0,B_0\in I$, assign $a,b\in S$, and put \[ w=B_0-A_0,\qquad L=Q(w),\qquad t=\frac{a-b+L}{2L},\qquad k=\frac{a-t^2L}{3L}. \] If $k<0$ there is no real solution. Otherwise the two circle intersections are exactly \begin{equation}\label{eq:quadratic-anchors} q=A_0+t w\ \mathord\pm\sqrt{k}\,(-w_x-2w_y,\,2w_x+w_y). \end{equation} The last vector is perpendicular to $w$ in the triangular-coordinate metric and has squared norm $3L$, which proves the formula. Enumerate~\eqref{eq:quadratic-anchors} for every anchor pair and every assignment in $S^2$, and add the circumcenters of every noncollinear triple of $I$. The calculation at this step permits irrational points. Write $\sqrt{k}=f\sqrt d$, with $f\in\mathbb Q$ and $d$ a positive squarefree integer, treating $k=0$ separately and absorbing $d=1$ into the rational coordinates. Candidate coordinates then have the form \[ (a_x+b_x\sqrt d,\ a_y+b_y\sqrt d),\qquad a_x,b_x,a_y,b_y\in\mathbb Q. \] Every squared distance to a core point is represented exactly as $u+v\sqrt d$. If $d>1$, this belongs to $\mathbb Q$ precisely when $v=0$. This representation permits exact comparisons, including equality of two unknown new distances. Remove points of $A$ and retain a candidate only when it has at most one distance outside $S$ toward $I$ and at most two outside $S$ toward $A$. The exhaustive calculation verifies that all survivors are rational. Their numbers, in the order above, are $10,12,11$. Take the pool of squared distances outside $S$ occurring from survivors to $A$ or between survivors. For every pair $a\max S$, use palette $S\cup\{a,b\}$ and discard candidates at squared distance $b$ from any point of $I$. The remaining compatibility graph again yields the extension maximum by a largest-clique calculation. This covers the actual pair $\{T,D^2\}$ even when one or both values occur only between added points. \subsection*{Exact calculation and lattice results} The lattice verifier uses rational arithmetic, squarefree extraction of positive integers, and an exact include/exclude clique search. Its search invariant is that every retained candidate is compatible with all points already selected. At each recursive call, branching on one remaining candidate covers both its inclusion and its exclusion; a branch is pruned only when the selected cardinality plus the number of remaining candidates cannot improve the best clique already found. The entire computation can be summarized as follows. \begin{enumerate} \item Reconstruct each coordinate core from the deletion table and verify that its squared-distance equality pattern is the specified colored matrix. \item Compute its convex hull and nonextreme set $I$ with integer orientation tests. Form the complete candidate list by the appropriate anchor procedure above, using exact arithmetic. \item Form the complete pool of possible additional distance values, filter for the largest-distance restriction, and exhaust the resulting compatibility graphs. \end{enumerate} The results are \begin{center} \begin{tabular}{lrr} Core type & Number of cores & Upper bound for $|X|$\\\hline Six distances, first group & 32 & 12\\ Six distances, second group & 17 & 16\\ Five distances & 3 & 16 \end{tabular} \end{center} For each five-distance core, cardinality sixteen is attained only in the computed extra-value pair $\{12,13\}$. The certificate records the palette, nonextreme points, candidate list, extension palettes, and largest-clique witnesses for each core. It is reproduced by \begin{verbatim} python3 core_extensions/eleven_lattice_extensions.py \end{verbatim} Its final check is that all fifty-two bounds are at most sixteen. \subsection*{The ten circular cores} The circular configurations in Table~\ref{tab:circular-eleven} are covered by four containment assertions: \begin{enumerate}[label=(\alph*)] \item A seven-distance set containing eleven vertices of $R_{13}$ is contained in $R_{13}$ together with its center. \item A seven-distance set containing at least ten vertices of $R_{12}$ is contained in $R_{12}$ together with its center. \item A seven-distance set containing at least ten vertices of $R_{11}$ is contained in $R_{11}$ together with its center. \item A seven-distance set containing $R_{10}$ and its center has no additional points. \end{enumerate} Here the regular polygons have circumradius one. Deleting the stated numbers of vertices leaves every chord length of the corresponding polygon: each nondiameter chord length occurs on $n$ pairs and each vertex belongs to two of them, whereas a diameter occurs on $n/2$ disjoint pairs. The removals cannot eliminate all pairs of any length. We verify the assertions by one common exact procedure. For a noncentral point $q$ compatible with at most seven distances, the eight or more retained polygon vertices cannot all have different distances from $q$. An equal-distance pair places $q$ on a reflection axis of the full regular polygon. After rotation, write $q=t e$, where $e$ is a unit vector along such an axis and $t\in\mathbb R\setminus\{0\}$. Odd polygons have one axis type; even polygons have vertex and edge-midpoint axes. If $C_j$ is twice the projection of a polygon vertex onto $e$, the distinct squared radii from $q$ to the full polygon are \[ f_j(t)=t^2+1-tC_j. \] For $t\ne0$ distinct projections give distinct radii. For $R_{13}$ there are seven radii, one occurring once and six occurring twice. Removing two vertices leaves at least six radii. As the old palette has six values, at least five retained radii must be old. For $R_{12}$, vertex axes have seven radii, of which two occur once; edge axes have six radii, all occurring twice. Removing at most two vertices leaves at least five radii, and hence at least four old matches. For $R_{11}$ there are six radii, only one occurring once. Removing at most one vertex leaves at least five, and the five old chord lengths force at least three old matches. Finally, $R_{10}$ has six radii on a vertex axis and five on an edge axis. The core $R_{10}$ together with its center has six old squared distances, including $1$, so the respective requirements are five and four old matches. In every case at least two distinct radii must match two distinct old values. If $f_j(t)=a$ and $f_k(t)=b$, subtraction gives the finite candidate set \begin{equation}\label{eq:axis-candidates} t=\frac{b-a}{C_j-C_k}\qquad(j