What ordinary motion incidences do not detect

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The attempted motion-space approach gives the following exact obstructions, not a proof of either assertion of Problem 132.

A rich radius can have only ordinary intersections

Let $J(x,y)=(-y,x)$. Fix an occurring radius $r$ and artificial anchors $a,b$ with $|a-b|=r$, choosing $b-a$ unequal to every oriented length-$r$ difference of $A$. Only finitely many directions are excluded. For $p\in A$, define the line in $\mathbb R^2\times\mathbb R$

$$L_{a,p}=\left\{\left(\frac{a+p}{2} +\frac t2J(p-a),t\right):t\in\mathbb R\right\},$$

and similarly define $L_{b,q}$ for $q\in A$.

Within either family the lines are pairwise disjoint. An intersection for distinct $p,p'$ would require

$$p-p'+tJ(p-p')=0,$$

which is impossible over $\mathbb R$. Put $d=b-a$ and $e=q-p$. A cross-family intersection is equivalent to

$$d+e=tJ(d-e). \tag{1}$$

When $d\ne e$, a real $t$ exists uniquely precisely when $(d+e)\cdot(d-e)=0$, or $|d|=|e|$. The anchor choice excludes $d=e$ for all pairs at distance $r$. Thus

$$L_{a,p}\cap L_{b,q}\ne\varnothing \quad\Longleftrightarrow\quad |p-q|=r.$$

There are exactly $2m_A(r)$ intersection points. Each belongs to exactly two lines, since within each family the lines are disjoint. Every intersection is therefore ordinary, even for $m_A(r)>n$.

The presence or abundance of ordinary intersections in this representation does not identify a rare radius. A useful argument would have to compare the arrangements for different radii, or use more metric information. No such implication was established.

Global motion energy loses the leading-order distinction

For the global arrangement $\{L_{a,p}:a,p\in A\}$, an intersection represents a nontranslation direct isometry mapping at least two points of $A$ into $A$: the first two coordinates are its rotation center, and $t=\cot(\theta/2)$ for its nonzero rotation angle $\theta$. Put $\nu(g)=|A\cap gA|$. Let $\mathcal I$ be the nontranslation direct isometries with $\nu(g)\ge2$, and $\mathcal T$ the translations with that property. Counting ordered pairs of congruent unordered segments and their two endpoint matchings gives

$$\sum_{g\in\mathcal I} \binom{\nu(g)}2 =2\sum_r m_A(r)^2 -\sum_{t\in\mathcal T} \binom{\nu(t)}2. \tag{2}$$

The translation sum includes the identity. All sums are finite: matching two distinct source points to two distinct target points determines a direct isometry uniquely. Formula (2) counts motion overlaps, not individual distance-class sizes.

If at most $B$ classes are rare, their total multiplicity is at most $Bn$, and every rich class has multiplicity at least $n+1$. Hence

$$\sum_r m_A(r)^2 \ge(n+1)\left(\binom n2-Bn\right) =\tfrac12n^3-O_B(n^2).$$

But a regular odd $n$-gon has every class rare and energy exactly $n^2(n-1)/2$. Thus an energy estimate at this leading order cannot distinguish the hypothesized bounded-rarity spectrum from a spectrum in which all classes are rare. This does not rule out finer, distance-labeled incidence information.

A checked inverse criterion and the missing moment step

The unrestricted attempt tested whether the large convex layer forced by bounded rarity could be treated using an inverse theorem for rigid motions. The following conditional implication is valid; its needed hypothesis has not been deduced from bounded rarity.

Criterion. Fix $\alpha,\beta>0$. Let $C$ be a strictly convex $m$-point set. If at least $\alpha m$ distinct direct rigid motions $g$ satisfy $|C\cap gC|\ge\beta m$, then, for sufficiently large $m$, a circle contains $\Omega_{\alpha,\beta}(m)$ points of $C$.

The external input is Mansfield--Passant, A Structural Theorem for Sets with Few Triangles, Combinatorica 44 (2024), Theorem 1.6: these motion hypotheses force either cubic additive energy or a positive circular fraction. The full statement and its use of at-least-rich actions in Lemma 3.1 and Corollary 3.2 were checked. Their Theorem 1.4 also supplies the circular conclusion for a convex set with only $O(m^2)$ triangle congruence classes. Primary source.

Here is the elementary reason the additive-energy alternative is impossible in strict convex position. For a nonzero vector $v$, write $r(v)=\#\{(p,q)\in C^2:p-q=v\}$. Then $r(v)\le2$. Rotate so $v=(a,0)$ with $a>0$. Write the horizontal sections of $\operatorname{conv}C$ as $[\ell(y),u(y)]$. A representing pair must be the two section endpoints, so $u(y)-\ell(y)=a$. Three representing pairs have three different heights $y_1<y_2<y_3$; otherwise there would be three collinear vertices. The width function is concave. Equality with the same value at these three heights forces it to be constant throughout $[y_1,y_3]$. Since $u$ and $-\ell$ are concave, they must then both be affine on that interval. The middle pair is therefore on two polygon sides rather than at vertices, a contradiction.

Consequently the additive energy satisfies

$$E_+(C)=\sum_v r(v)^2 \le m^2+2\sum_{v\ne0}r(v)=3m^2-2m,$$

excluding a lower bound of order $m^3$ and proving the criterion. The source theorem is imported; this convex exclusion was checked by the author and is author-recorded.

Why the available distance count does not meet the criterion

Put $E_D(C)=\sum_r m_C(r)^2$. If $q(C)\le C_0m$, then

$$E_D(C)\ge\frac{\binom m2^2}{C_0m}=\Omega_{C_0}(m^3).$$

In (2), the translation contribution is at most $m(m-1)$: the identity contributes $\binom m2$, while $\sum_{v\ne0}\binom{r(v)}2\le\tfrac12\sum_{v\ne0}r(v)=\binom m2$. Thus the hypothesis yields a cubic second moment of nontranslation motion overlaps. It does not yet yield linearly many motions with linearly many overlapping points.

The usual rich-motion estimate $\#\{g:|C\cap gC|\ge k\}=O(m^3/k^2)$ is also recalled in the checked source, equation (1). Those numerical bounds alone allow a cubic second moment to be supported on motions of bounded richness. This is a failure of that numerical inference, not a constructed convex configuration. A lower bound of order $m^4$ for triangle energy, or an $O(m^2)$ bound for triangle congruence classes, would supply the stronger input. The palette estimate gives only $O(q(C)^3)=O(m^3)$ possible triples of side lengths.

For a hypothetical bounded-rarity family, the layer theorem supplies $|C|\ge cn$ and hence $q(C)\le n/2+O(1)=O_c(|C|)$. The criterion would finish this route if the missing moment step were proved: a positive circular fraction of $C$ would be one of $A$, contradicting the circle-incidence theorem. Neither that step nor the general convex few-distance inverse statement was proved or refuted. These are provisional directions, not reductions whose remaining hypotheses can be omitted.

A labeled short-edge hypothesis does supply the missing motions

The subsequent short-edge incidence inverse proves the needed rich-motion conclusion when a fixed smallest-distance rank has linearly many edges touching the convex subset. Its overloaded-bin deletion gives linearly many anchored triangle types with quadratically many realizations, rather than relying on the unlabeled second moment. This proves rarity divergence for convex sets and fixed convex covers. The general $q(C)=O(|C|)$ inverse and the arbitrary-layer application remain unproved: a large convex layer need not meet the short rich graph.