Two rare distances when all diameters share an endpoint

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Theorem. Let $A\subset\mathbb R^2$, $|A|=n\ge5$. If every diameter pair contains one common point, then $R(A)\ge2$. If there are at least three diameter pairs, the second-largest distance itself has multiplicity at most $n$.

This extends the unique-diameter theorem. It does not settle configurations whose diameter graph has vertex-cover number at least two, or the unrestricted growth question.

The sharp bound and its equality case

Normalize the second-largest distance to one. Let $a$ be the common diameter endpoint, let $B$ be its diameter neighbors, and write $D>1$ for the diameter. Every pair in $A\setminus\{a\}$ has length at most one.

Let $d$ be the unit degree of $a$. If $d\le1$, deleting $a$ and using the diameter bound gives $m_A(1)\le n-1+d\le n$. Otherwise, fix $b\in B$. After putting $a=0$ and $b=(D,0)$, each unit neighbor $(\cos\theta,\sin\theta)$ of $a$ satisfies $\cos\theta\ge D/2$. The neighbors therefore lie in an arc of width less than $2\pi/3$. Their extreme members $u,v$ are at distance at most one, so their angular separation is at most $\pi/3$.

Let $L$ be the $d-2$ neighbors strictly between $u,v$. The unit-circle arc lemma shows that each $w\in L$ has unit degree exactly one: for every $x\ne a$, both $|x-u|,|x-v|\le1$, and consequently $|x-w|<1$. Deleting $a\cup L$ removes exactly $d$ unit edges and leaves $n-d+1$ points of diameter at most one. Hence

$$m_A(1)\le n+1. \tag{1}$$

Equality implies that the remaining set has as many diameter-one edges as vertices. No remaining vertex is isolated in that graph, since deleting an isolate would violate the diameter bound. The deleted points are also unit-active. Thus equality in (1) makes every point of $A$ unit-active.

Three diameter neighbors make the second distance rare

All mutual distances in $B$ are at most one. Put $\alpha=2\arcsin(1/(2D))<\pi/3$. Relative to one member of $B$, all arguments lie in $[-\alpha,\alpha]$. Their extreme difference is less than $\pi$; the distance condition on these extremes then makes that difference at most $\alpha$. Thus $B$ lies on a minor arc of its radius-$D$ circle.

If $|B|\ge3$, choose $b$ strictly between its extremes $b_-,b_+$. With $a=0$, write

$$b=\lambda b_-+\mu b_+,\qquad \lambda,\mu>0,\quad s=\lambda+\mu>1.$$

For every $y\ne a$, the bounds $|y-b_-|,|y-b_+|\le1$ give

$$|y-b|^2\le1-(s-1)(|y|^2+D^2-1)<1.$$

The remaining point $a$ is at distance $D$ from $b$. Hence $b$ is unit-inactive. Equality in (1) is impossible, proving $m_A(1)\le n$ whenever $|B|\ge3$.

Exactly two diameter neighbors

The case $|B|=1$ is the checked unique-diameter theorem. Suppose $B=\{b,c\}$ and, for contradiction, $R(A)=1$. Then (1) forces $m_A(1)=n+1$, and all points are unit-active. The diameter has multiplicity two, so the pair budget gives

$$\binom n2\ge2+(q(A)-1)(n+1),$$

and therefore

$$q(A)\le \begin{cases} (n-1)/2,&n\text{ odd},\\ n/2-1,&n\text{ even}. \end{cases} \tag{2}$$

Put $C=A\setminus\{a,b,c\}$ and $m=n-3$. For $x\in C$, choose a unit neighbor $y\in A$. The unit disk centered at $y$ contains all of $C$, because both endpoints of every exceptional pair belong to $\{a,b,c\}$. Its tangent at $x$ supports $C$ uniquely. Thus $C$ is strictly convex. Altman's lower bound, deletion of the diameter, and (2) imply

$$\lfloor m/2\rfloor\le q(C)\le q(A)-1 \le\lfloor m/2\rfloor. \tag{3}$$

All inequalities are equalities, so the palette of $A$ is exactly the palette of $C$ together with $D$.

Orders $5\le n\le10$ are already proved. For $n\ge11$ we have $m\ge8$, and the checked Altman--Fishburn classifications in the supplied full paper give three possibilities: $C$ is regular odd, regular even, or a regular odd polygon with one vertex deleted.

A regular odd polygon admits no new point whose distances to its vertices all lie in its chord palette. A noncentral point has at most two neighbors at each prescribed distance, which is insufficient; its center would create the circumradius, which is not an odd regular polygon's chord. A regular odd polygon with one vertex deleted admits only that missing vertex. The reflection-and-cosine proof of this latter assertion is in the unique-diameter argument. Either alternative contradicts the three distinct added points $a,b,c$ and palette equality in (3).

If $C$ is regular even, its polygon diameter is one by (3). It supplies $m/2$ unit pairs. Each added point contributes at most two unit pairs to $C$; a circumcenter contributes none. Of the three added--added pairs, $ab,ac$ have length $D$, so only $bc$ can be unit. Consequently

$$m_A(1)\le\frac{n-3}{2}+6+1 =\frac{n+11}{2}\le n,$$

contradicting $m_A(1)=n+1$. This proves the theorem.

The author checked the short-arc deletion, the strict radius-$D$ inequality, the parity in (2)--(3), and all classification and multiplicity steps; the argument is author-recorded. In particular the even-order bound in (2) follows directly from counting; no circular use of the unique-diameter theorem is involved.

A checked bound for two disjoint diameter pairs

Suppose instead that the only diameter pairs are disjoint $ab,cd$. For a diameter endpoint $v$, let $k_v\in\{0,1,2\}$ count its unit neighbors among the other diameter pair. Then

$$m_A(1)\le n+2+\min_v k_v\le n+4. \tag{4}$$

Here are the details beyond the star proof. For an endpoint $a$ with unit degree $d_a\ge2$, its unit neighbors again lie in an arc of width less than $120$ degrees. Neither $c$ nor $d$ can be an interior member of this arc: if both are neighbors, their separation exceeds $60$ degrees and they must be the two extremes; if only one is a neighbor, the unit-circle arc lemma applied to its diameter partner rules out an interior position.

Let $L$ be the interior unit neighbors. Each can have unit neighbors only among $a,c,d$. An extra edge to $c$ requires $d$ to be an extreme neighbor, and conversely. Each of $c,d$ supplies at most one extra edge: the two intersections of its unit circle with the circle about $a$ have angular separation at least $120$ degrees, since its distance from $a$ is at most one. Thus at most $k_a$ extra edges are incident with $L$.

Deleting $a\cup L$ leaves $n-d_a+1$ points with unique diameter $cd$. Applying the sharp unique-diameter bound gives

$$m_A(1)\le d_a+k_a+(n-d_a+1)+1=n+2+k_a.$$

If $d_a\le1$, simply deleting $a$ gives the stronger bound $m_A(1)\le n+1$. This proves (4). Equality in an endpoint-specific bound makes every point unit-active. In particular, $m_A(1)=n+4$ forces all four cross-sides to be unit, so the four diameter endpoints form a unit-side square.

The bound alone does not give a second rare distance. It is sharp: take a unit square and the four inward intersections of unit circles centered at its adjacent vertices. The only diameter pairs are the square diagonals and there are twelve unit pairs on eight points. This is not an Erdős-132 counterexample.