Two rare distances in nearly convex sets

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Proposition. Let $n\ge5$. If $A$ contains an $(n-1)$-point subset in strictly convex position, then $R(A)\ge2$.

The convex subset need not consist of all hull vertices of $A$. This is a restricted result, not the unrestricted first assertion of Problem 132.

Use the checked small-order theorem for $5\le n\le10$, and assume $n\ge11$. If $R(A)=1$, its sole rare class is the diameter. Writing $q=q(A)$ gives

$$\binom n2\ge1+(q-1)(n+1),\qquad q\le\lfloor n/2\rfloor. \tag{1}$$

For even $n$ and $q=n/2$, equality forces a unique diameter pair and multiplicity exactly $n+1$ for every other class.

Let $H=A\setminus\{p\}$ be strictly convex. We use precisely Theorems 1--2 of the supplied Fishburn paper: an odd convex $v$-gon with $(v-1)/2$ distances is regular; for even $v\ge8$, a convex $v$-gon with $v/2$ distances is regular or a regular $(v+1)$-gon minus one vertex. The digest and these full-source statements were checked; no near-minimum classification is assumed.

If $n$ is odd, Altman's bound and (1) force $q(H)=(n-1)/2$. The even-order classification makes $H$ concyclic. This contradicts the all-but-one-circle theorem, which gives two rare distances in the full set $A$.

If $n$ is even and $q\le n/2-1$, then $q(H)=(n-2)/2$ by Altman's bound. Its odd equality case makes $H$ regular, giving the same contradiction.

Finally suppose $n$ is even and $q=n/2$, with unique diameter pair $ab$. If $p$ is one of $a,b$, deletion removes that distance, so $q(H)\le q-1=(n-2)/2$ and the odd equality case again makes $H$ regular. Otherwise $a,b\in H$. Both convex sets $H\setminus\{a\}$ and $H\setminus\{b\}$ have $n-2$ points and at most $(n-2)/2$ distances. The even classification makes both concyclic. They share $n-3\ge3$ noncollinear points, so their circles agree, and their union $H$ is concyclic. Apply the all-but-one-circle theorem once more. This proves the proposition.

The author checked the deletion counts, parity cases, common-circle argument, and exact orders required by the source classification; the argument is author-recorded. This gives no bound when the number of points outside the chosen convex subset is unbounded, nor an asymptotic rarity theorem even for every convex set.

Comparison with the second independent input

Section 5.2 of the second independent input recovers the hull-vertex case of the proposition by the same two endpoint deletions. The proposition above is stronger: its convex $(n-1)$-point subset need not be the full hull. Section 5.1 is also covered by the stronger all-but-one-circle theorem, which allows the extra point anywhere and supplies a linear lower bound.

The input's two-nonvertex result is valid under its additional circular hull hypothesis; a shorter proof and a modest extension are recorded in circular configurations. It does not establish the arbitrary two-nonvertex case. The lack of an asymptotic conclusion from the classification argument itself is not a current gap for this family: the later short-edge inverse proves divergence for fixed convex covers with $o(n)$ exceptions, including all-but-one convex sets.

Two points outside convex position at even order

Corollary. If $n\ge6$ is even and $A$ contains an $(n-2)$-point subset $C$ in strictly convex position, then $R(A)\ge2$.

Orders through ten are already covered by small orders. For even $n\ge12$, suppose $R(A)=1$. The unique-diameter theorem excludes the even counting boundary, so $q(A)\le n/2-1$. Altman's bound gives $q(C)\ge(n-2)/2$, hence equality. Fishburn's even equality classification quoted above applies to $|C|=n-2\ge10$ and makes $C$ cocircular. The all-but-two-circle theorem now contradicts $R(A)=1$.

At odd order this argument leaves a convex $(n-2)$-gon with one more than its minimum number of distances. No general classification of that case is assumed, and the arbitrary odd-order two-exception case remains untreated.

Sections 7.2--7.3 of the third independent report were also checked. Its one-nonvertex classification argument and its two-nonvertex reflection/affine-permutation argument are correct under the stated hull hypotheses. They do not strengthen the results above: the one-exception convex subset need not be the hull, and the two-off-circle theorem allows the two added points anywhere. The report's unhandled two-nonvertex case is therefore superseded at even order, but not at arbitrary odd order.