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This is the canonical argument for the symmetric polygon census. The deductions below and the completed census are author-recorded by Plasma AI. No independent proof review is asserted. The census exhausts exactly the finite domains specified here and finds no counterexample.

Target and specified domains

Write $D_v$ for the number of distinct positive distances from vertex $v$ to the other vertices. A witness against Problem 982 must be strictly convex and satisfy $\max_v D_v<\lfloor n/2\rfloor$.

The first domain consists of every integer $3\le m\le32$ and all real $r>0$, with vertices

$$A_k=e^{2\pi i k/m},\qquad B_k=r e^{(2k+1)\pi i/m}, \qquad 0\le k<m.$$

All positive critical radii are enumerated, including nonconvex ones, then filtered to $\cos(\pi/m)<r<\sec(\pi/m)$. There are $n=2m$ vertices and two rotation orbits, so a witness needs $D_A<m$ and $D_B<m$ simultaneously.

The second domain consists of every integer $8\le n\le24$ and all real $s>0$, with vertices

$$p_k=(\cos(2\pi k/n),\sqrt{s}\sin(2\pi k/n)), \qquad 0\le k<n.$$

All positive critical values of $s$ are enumerated and every vertex is checked. The generic complement is covered symbolically. Every positive $s$ preserves strict convexity by an invertible affine map of a regular polygon. The stretching axes are exactly those in the displayed coordinates: arbitrary orientations of those axes relative to the polygon are not included. Neither finite order range exhausts unbounded orders or other families.

For $r=1$ or $s=1$ all vertices are on one circle. Any circle centered at a vertex intersects that common circumcircle in at most two points. Therefore $D_v\ge\lceil(n-1)/2\rceil=\lfloor n/2\rfloor$ at every vertex. This argument disposes of the common-circle case without searching that family.

Concentric distance and convexity reduction

Put $\theta=\pi/m$, $h=\lfloor m/2\rfloor$, and $q=\lceil m/2\rceil$. The distinct within-ring squared distances at $A_0$ are

$$a_i=2-2\cos(2i\theta)\quad(1\le i\le h),$$

and at $B_0$ they are $r^2a_i$. The cross-ring squared distances at both orbits are

$$b_j(r)=1+r^2-2c_jr,\qquad c_j=\cos((2j+1)\theta)\quad(0\le j<q).$$

Each displayed list is injective for $r>0$: its angles run strictly through $(0,\pi]$ and cosine is strictly decreasing there. Reflection accounts for all omitted vertices. Thus the generic count at either orbit is $h+q=m$. Every exceptional radius is a positive real root of

$$P_{ij}(r)=r^2-2c_jr+1-a_i,\qquad Q_{ij}(r)=(1-a_i)r^2-2c_jr+1=r^2P_{ij}(1/r).$$

There are no within-list exceptional equalities. Let $u$ count the pairs $(i,j)$ with $P_{ij}(r)=0$, and $v$ those with $Q_{ij}(r)=0$. Injectivity makes each matching a disjoint identification of two lengths, so

$$(D_A,D_B)=(m-u,m-v).$$

In particular, a witness requires at least one event of each type at the same radius. Radius reciprocity exchanges the two counts. Repeated roots, linear equations, zero roots and repeated descriptions of one radius must be handled exactly.

In increasing angular order the polygon is $A_0,B_0,A_1,B_1,\ldots,A_{m-1},B_{m-1}$. Each edge stays inside its angular sector, so nonadjacent edges do not meet. Its successive edge determinants at the two vertex types are

$$2r\sin\theta(1-r\cos\theta),\qquad 2\sin\theta(r-\cos\theta).$$

The polygon is therefore strictly convex exactly when $\cos\theta<r<1/\cos\theta$. Both endpoints are degenerate and excluded. For any possible counterexample, reconstruct its full exact vertex list, verify all supporting-edge inequalities, and retain the distance partition at every vertex, as an additional check of the reduction.

The eight-point example in [[research/distance_multiplicity_interfaces/exact_optimizers|the local obstruction account]] has $m=4$, $r=(1+\sqrt3)/\sqrt2$ after similarity, and counts $(3,3)$. Since $r>\sqrt2$, it fails the strict-convexity test. It is a negative control, not a witness against Problem 982.

Affine distance reduction

With $\alpha_k=2\pi k/n$, every squared distance from pin $i$ has the form $X_{ij}+sY_{ij}$, where

$$\begin{aligned} X_{ij}&=1+\tfrac12(\cos(2\alpha_i)+\cos(2\alpha_j)) -\cos(\alpha_i-\alpha_j)-\cos(\alpha_i+\alpha_j),\\ Y_{ij}&=1-\tfrac12(\cos(2\alpha_i)+\cos(2\alpha_j)) -\cos(\alpha_i-\alpha_j)+\cos(\alpha_i+\alpha_j). \end{aligned}$$

Equivalently these are the squares of the horizontal and vertical coordinate differences before scaling. Two such lines coincide for every $s$ exactly when both coefficients agree. Otherwise their only possible equality is

$$s=\frac{X_{ik}-X_{ij}}{Y_{ij}-Y_{ik}},$$

provided the denominator is nonzero. Equal slopes and unequal intercepts never meet. Enumerating these positive quotients for every pin and every unordered pair of its other vertices covers every exceptional parameter. Away from them the symbolic pairs $(X,Y)$ give the complete partition. Specialization can only merge generic classes, so every generic pin count is at least its count at $s=1$, which meets the target by the circle argument. Only positive critical parameters can be counterexamples.

Exact arithmetic and completeness

Use rational polynomials modulo $\Phi_{2m}$ for concentric coefficients and modulo $\Phi_n$ for affine coefficients, with the embedding $\zeta_N=e^{2\pi i/N}$. Polynomial remainders decide equality; certified real enclosures select the intended embedding and decide strict signs. Any unresolved enclosure must refine or fail, never count as equality. Concentric radii are roots of degree-at-most-two equations over this field; common-root tests and their selected real branches identify duplicates. Affine parameters are exact field quotients.

For completeness of the common-root test, normalize two quadratics to $f=x^2+bx+c$ and $g=x^2+dx+e$. If $b=d$, they have a common root only when $c=e$, and their selected branches must agree. If $b\ne d$, their only possible common root is $x=(e-c)/(b-d)$. Substitute this exact field value in both polynomials. For a positive discriminant, its branch is determined by the sign of $2x+b$. Linear and double-root cases are field values. This covers reducible quadratics too; no irreducibility assumption on an event polynomial is used. Disjoint certified intervals prove inequality; overlap alone never identifies roots. The precision cap causes failure if distinct values cannot be separated.

The checker enumerates every equation or quotient above, retains every positive real root, merges exact duplicates, and checks the distance counts. Concentric counts use the complete matching lists; the full $2m-1$ vertex labels are additionally partitioned directly at every strictly convex event, every event lowering both orbits, and all events for $m\le5$. For the affine family every one of the $n-1$ distances at every pin is evaluated exactly at every critical parameter. The resulting new equalities must agree exactly with the generated events. These checks and the generic-complement arguments compose into the stated bounded exclusion.

Outcome and remaining gap

The census contains no surviving counterexample. At each tested concentric order the only admissible events are $P_{h,q-1}=0$ and its reciprocal $Q_{h,q-1}=0$, at distinct radii. They give $(D_A,D_B)=(m-1,m)$ and $(m,m-1)$. The first radius has the explicit form

$$\rho_m= \begin{cases} \sqrt{3+\cos^2(\pi/m)}-\cos(\pi/m),&m\text{ even},\\ 2\cos(\pi/(2m))-1,&m\text{ odd}. \end{cases}$$

These identities follow by solving the displayed farthest-distance equation. The assertion that no other events are admissible is certified here only for $3\le m\le32$. The two formulas do not extend that finite coverage.

For the affine family, the table gives the minimum of $\max_i D_i$ among noncircular critical parameters. Together with the generic counts and the circle argument this excludes every $s>0$ for each $8\le n\le24$.

The failed step in the intended disproof is constructing simultaneous distance reductions at every vertex while preserving strict convexity. That existence input is false in these two bounded domains according to the author-recorded exact census. It remains unresolved outside them. No external classification or provisional literature claim is a premise. There is no remaining computation inside the specified domains and no independent review of this account.

Reopening requires a different domain with a plausible source of more simultaneous equalities: an order outside the bounds, a different angular offset, or additional independently placed vertex orbits, for example. The next discriminating test is whether such a construction can lower every orbit's count at one exact parameter while passing all strict convexity inequalities. Merely enlarging a floating grid would not answer it. This bounded exclusion supplies no claim about all convex polygons.