Two rare distances with all but two points on a circle

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Theorem. Let $A$ be an $n$-point planar set, $n\ge5$. If $n-2$ points of $A$ are concyclic, then $R(A)\ge2$.

The subsequent three-off-circle theorem extends this conclusion to $n-3$ concyclic points. The palette lemmas below remain inputs to its proof.

The two remaining points may be anywhere. This strengthens the cocircular-hull extension, which required one noncentral added point to lie inside the circular subset's hull. It uses the checked small-order and unique-diameter theorems, but no classification of general near-minimal convex polygons.

1. A small difference set on a circle

Let $S$ be a finite nonempty subset of an abelian group, $|S|=h$, and put $D=S-S$. If $|D|<3h/2$, then $D$ is a subgroup. Indeed, for $x=a-b\in D$, the sets $S-a,S-b\subset D$ intersect in at least $2h-|D|>h/2$ points. Equivalently, $|S\cap(S+x)|>h/2$. For $x,y\in D$, the two subsets $S\cap(S+x)$ and $S\cap(S+y)$ of $S$ intersect. A point $z$ in their intersection has $z-x,z-y\in S$, so $x-y\in D$. Thus the finite nonempty set $D$ is closed under subtraction.

For the angles $S\subset\mathbb R/(2\pi\mathbb Z)$ of $h$ circular points $C$, every positive chord length accounts for at most two elements of $D$, and zero accounts for one. Therefore

$$h\le |D|\le2q(C)+1.$$

When $2q(C)+1<3h/2$, the lemma makes $D$ a finite subgroup of the circle group. It has the form $(2\pi/k)\mathbb Z/(2\pi\mathbb Z)$, where $k=|D|$. Since $S$ lies in one coset of $D$, $C$ is a subset of a regular $k$-gon. No additive-combinatorial theorem is imported here.

2. A regular-polygon palette extension lemma

Let $V$ be a regular $k$-gon, $k\ge5$, with chord-length palette $P(V)$. In each of the following cases, there is no point off the circumcircle and off its center whose distances to $C$ all lie in $P(V)$:

• $k$ is even and $C$ is $V$ or $V$ minus one vertex;
• $k$ is odd and $C$ is $V$ minus one or two vertices.

Write $M=\lfloor k/2\rfloor$ and normalize the circle to radius one. A noncentral point has at most two circle points at each prescribed distance. Here $|C|$ equals $2M$ or $2M-1$, so every one of the $M$ palette values must occur. The pinned list consists of pairs, or of pairs and exactly one singleton. There is at least one pair. Reflection in the line joining the center to the pin exchanges that pair; as both vertices belong to $V$, this line is a reflection axis of $V$.

Even order

For $k=2M$, an edge-midpoint axis has no fixed vertex of $V$. Deleting at most one vertex does not remove any pinned value. Writing the pin's radius as $\rho>0$, its squared palette is

$$\left\{1+\rho^2-2\rho\cos\frac{(2j-1)\pi}{2M}:1\le j\le M\right\}.$$

Equality with the squared chord palette $\{2-2\cos(j\pi/M):1\le j\le M\}$ gives, from means and ranges,

$$\rho^2=1+2/M,\qquad \rho=\cos(\pi/(2M)),$$

a contradiction. Here the cosine sums are respectively zero and $-1$; the ranges are $4\rho c$ and $4c^2$, where $c=\cos(\pi/(2M))$.

A vertex axis has two fixed vertices. It is impossible when $C=V$, since all pinned values must be paired. With one vertex deleted, that vertex must be fixed; otherwise the two fixed vertices and the reflected mate of the deleted vertex give three singletons. Normalize the deleted vertex to $1$ and write the pin as signed $\rho\ne0$ on the real axis. The pinned and chord squared palettes are affine images of the same nonconstant list $\{\cos(j\pi/M):1\le j\le M\}$. Equality of their ranges forces $|\rho|=1$, contrary to being off the circle.

Odd order with one or two deletions

For $k=2M+1$, a reflection has one fixed vertex. With one deletion, all surviving vertices must be paired, so the fixed vertex is the deleted one. The same-list range argument again gives $|\rho|=1$.

With two deletions, exactly one singleton is allowed. Either one deleted vertex is fixed, or the reflection exchanges the two deleted vertices. In the first case the other deleted vertex has a surviving mate, so the pinned palette is unchanged from deleting only the fixed vertex. The preceding argument applies.

For exchanged deletions, normalize the fixed vertex to $1$ and put

$$a=\pi/k,\quad c=\cos a,\quad u=2c-1,\quad c_j=\cos(2ja).$$

The holes are $e^{\pm2ita}$ for some $1\le t\le M$. Palette equality for the signed real pin $\rho\ne0$ is precisely

$$F=\rho E+(1-\rho^2)/2,\quad F=\{c_1,\ldots,c_M\},\quad E=\{c_0,\ldots,c_M\}\setminus\{c_t\}. \tag{1}$$

Since $c_M=-c$ and $c_{M-1}=-\cos3a$, the ranges give

$$|\rho|=u\quad(t<M),\qquad |\rho|=1/u\quad(t=M). \tag{2}$$

Explicitly, $\operatorname{range}F=(1+c)u$, whereas the two possible ranges of $E$ are $1+c$ and $(1+c)u^2$. If $\rho>0$, matching maxima in (1) gives $|\rho-1|=2\sin a$. But

$$1-u=2(1-c)<2\sin a,\qquad 1/u-1<2\sin a. \tag{3}$$

For the second inequality, squaring and canceling $1-c>0$ reduces to $1-c<(1+c)(2c-1)^2$. The difference is $2c(2c^2-1)>0$, since $k\ge5$ gives $c>1/\sqrt2$. If $\rho<0$, matching minima instead gives $|\rho|=2\cos(a/2)-1$, strictly between $u$ and $1$ and hence equal to neither value in (2). This proves the extension lemma.

3. Proof of the theorem

The small-order theorem handles $5\le n\le10$. Assume $n\ge11$ and $R(A)=1$, and let $C$ be the $h=n-2$ points on the chosen circle. If another point lies on that circle, the all-but-one-circle theorem already gives a contradiction. Thus both remaining points are off the circle.

Pair counting and the unique-diameter theorem give

$$q(A)\le \begin{cases} n/2-1,&n\text{ even},\\ (n-1)/2,&n\text{ odd}. \end{cases} \tag{4}$$

Indeed, equality $q(A)=n/2$ at even order would force a uniquely attained diameter. This use of the unique-diameter theorem is not circular: its small-order base cases were proved independently.

If $h=2m$ is even, then $|S-S|\le h+1<3h/2$. Section 1 puts $C$ in a regular polygon of order $h$ or $h+1$. All $m$ chord lengths of that polygon survive the possible single deletion, exhausting the palette in (4). At least one of the two added points is noncentral, contradicting the extension lemma.

If $h=2m+1$ is odd, then $|S-S|\le h+2<3h/2$. Thus $C$ is a regular $h$-gon, a regular $(h+1)$-gon minus one vertex, or a regular $(h+2)$-gon minus two vertices. In the latter two cases all $m+1$ chord values survive and exhaust the palette, again giving a contradiction. For these survival assertions, a deleted vertex removes at most two pairs of any nondiametral chord class; such a class starts with $k$ pairs. At even order the diameter has $k/2$ pairs and loses at most one per deleted vertex.

It remains that $C$ is a regular odd $h=2m+1$ polygon. Normalize its circumradius to one. If one added point is the center, its unit distance has at most $h+2=n$ pairs: there are no unit chords in an odd regular polygon, the other pin adds at most two, and the pin-pin distance is not one because the other pin is off the circle. This rare distance is smaller than the polygon diameter, a contradiction.

Both added points are therefore noncentral. Each sees at least $m+1$ lengths on $C$, so there is exactly one new value beyond its $m$ chord values. It occurs at most $2+2+1=5\le n$ times. Hence it is rare and, under $R(A)=1$, is the full diameter $D$.

Each pin uses every value of this $(m+1)$-element palette, with exactly one singleton. Its center line is a reflection axis of $C$. Normalize its fixed vertex to $1$, write the pin as signed $\rho$, and use $a=\pi/h$, $c_j=\cos(2ja)$, and $u=2\cos a-1$. If $\rho<0$, its largest pinned value $D$ comes from the fixed vertex. Removing it leaves the same cosine list $c_1,\ldots,c_m$ as the chord palette. Equality of ranges forces $\rho=-1$, on the circle. If $\rho>0$, its largest value comes from $c_m$. The remaining pinned list uses $c_0,\ldots,c_{m-1}$; the range and minimum comparisons in (2)--(3) give $\rho=1/u$ and $\rho-1=2\sin a$, contradicting (3). This excludes the last case and proves the theorem.

Scope

The author checked the small-difference argument, reflection singleton counts, palette survival, signed-radius comparisons, and parity in (4); the argument is author-recorded. The positive-circle theorem already gives linear rarity growth for this family; the new conclusion is the exact finite bound $R\ge2$ at every order. The resulting even-order convex corollary is also proved. Arbitrary odd-order sets with two points outside convex position, and the unrestricted planar problem, remain untreated.