Two consecutive long-diagonal families force regularity

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Lemma. Let $n=2m+1\ge5$, and let $p_0,\ldots,p_{n-1}$ be the vertices of a strictly convex polygon in counterclockwise order. If, for every $i$,

$$|p_{i+m}-p_i|=a,\qquad |p_{i+m-1}-p_i|=b,$$

then the polygon is regular. Indices are modulo $n$.

These are the two largest cyclic separations, not an assumption that they are already the two largest numerical distances.

Proof. Let $\varepsilon_i\in(0,\pi)$ be the angle at $p_i$ between the rays to $p_{i+m}$ and $p_{i+m+1}$. Consider the $n$ triangles $(p_i,p_{i+m},p_{i+m+1})$. At a given vertex, its two appearances as a base vertex contribute its polygon interior angle plus $\varepsilon_i$: their angular intervals cover the interior angle and overlap precisely in $\varepsilon_i$. Adding their angles, including the apex appearances, gives

$$n\pi=(n-2)\pi+2\sum_i\varepsilon_i, \qquad\sum_i\varepsilon_i=\pi. \tag{1}$$

Reindex by $q_j=p_{mj}$. This visits every vertex because $\gcd(m,n)=1$. Its consecutive edges all have length $a$, and its three-step chords all have length $b$, since $3m\equiv m-1\pmod n$. Its positive turns $\delta_j$ are the numbers $\pi-\varepsilon_i$ in permuted order. Thus

$$0<\delta_j<\pi,\qquad \sum_j(\pi-\delta_j)=\pi. \tag{2}$$

Expanding the squared length of three successive edges gives

$$F(\delta_j,\delta_{j+1})=b^2/a^2, \quad F(x,y)=3+2\cos x+2\cos y+2\cos(x+y).$$

Subtracting consecutive equations and dividing by $4\cos(\delta_{j+1}/2)>0$ yields

$$\cos(\delta_j+\delta_{j+1}/2) =\cos(\delta_{j+2}+\delta_{j+1}/2).$$

The angle ranges imply either $\delta_j=\delta_{j+2}$ or $\delta_j+\delta_{j+1}+\delta_{j+2}=2\pi$. The latter would make three of the positive quantities $\pi-\delta_j$ sum to $\pi$, contradicting (2), since $n\ge5$ leaves other positive terms. The turns are therefore two-periodic. Since $n$ is odd, all turns are equal, necessarily $2\pi m/n$. Equal edge lengths and equal turns make $q$ a regular star polygon; its vertex set, and hence the original polygon, is regular. $\square$

Scope and remaining gap

The author checked the angle-sum identity, star orientation, the cosine alternatives, and the odd-period argument; the argument is author-recorded. No premise here follows from merely having many rich distance classes. In particular, richness has not supplied spanning cycles with these prescribed cyclic separations. This lemma therefore does not resolve either unrestricted question in Problem 132.

The constant-width inverse theorem uses a related local isometry-propagation mechanism, but obtains its compatible rows only under a bounded distance-count excess on a constant-width boundary.

The parity restriction cannot be removed. The even-order deformation construction has both largest cyclic separation classes uniform, attains the exact $3N/2$ second-distance bound after adding its inner points, and has no five concyclic points. Thus extracting those two families at even order would not prove circularity.