Rare distances from an arbitrary positive-fraction circular subset

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Statement and scope

For a finite planar set $A$, write $n=|A|$, let $q(A)$ be the number of distinct positive distances, and let

$$R(A)=\#\{r>0:1\le m_A(r)\le n\},$$

where $m_A(r)$ counts unordered pairs. The argument below proves:

Theorem. Fix $0<c\le1$. There is a constant $C_c$ such that, whenever $A\subset\mathbb R^2$ has $n$ points and contains $m\ge cn$ points on one circle,

$$R(A)\ge \frac{m-1}{2}-C_c n^{2/3}. \tag{1}$$

More precisely, if $q(A)\le n$, then

$$R(A)\ge q(A)-C_c n^{2/3}. \tag{2}$$

No regularity of the circular subset and no restriction on the remaining points is assumed. Thus any fixed positive circular fraction gives linearly many rare distances for sufficiently large $n$.

This does not solve the unrestricted problem: a hypothetical bounded-rarity family is not yet known to contain a positive-fraction circular subset. The theorem removes the previous circular-majority or regular-polygon assumptions, not this remaining structural gap.

1. An incidence bound for translated cosine curves

Write $\mathbb T=\mathbb R/(2\pi\mathbb Z)$. Fix $a>0$, $b\in\mathbb R$, with

$$(a,b)\ne(1,0),$$

and consider the translates

$$\Gamma_{u,v}=\{(x,y)\in\mathbb T^2: \cos(y-v)=a\cos(x-u)+b\}. \tag{3}$$

All constants in this section are absolute and uniform in $a,b$. If the base real curve is empty or consists of a single point, the incidence and overlap bounds below hold directly. In the remaining case it has a nontrivial real arc, so coincident real curves have coincident complex algebraic closures.

Algebraic irreducibility and coincidence

Put $X=e^{ix}$, $Y=e^{iy}$. Multiplication of the unshifted equation by $2XY$ gives

$$F(X,Y)=X(Y^2+1)-aY(X^2+1)-2bXY. \tag{4}$$

As a quadratic in $Y$, its discriminant is

$$[aX^2+2(b-1)X+a]\,[aX^2+2(b+1)X+a]. \tag{5}$$

The two quadratic factors are coprime: their difference is $4X$, and their constant terms are the nonzero number $a$. Consequently their product is a square in $\mathbb C(X)$ only if both quadratics are squares. This requires

$$(b-1)^2=a^2=(b+1)^2,$$

which forces $b=0$, $a^2=1$, and hence $a=1$. Thus (4) is irreducible over $\mathbb C(X)$ as a polynomial in $Y$; it is primitive over $\mathbb C[X]$, so it is irreducible in $\mathbb C[X,Y]$. Translating angles rescales $X,Y$ by nonzero constants and preserves irreducibility.

Two distinct translated complex curves therefore have at most nine intersections, by the ordinary degree-three Bézout bound. Only a fixed constant matters here.

To identify coincident translates, compare Laurent coefficients in

$$\frac12e^{-iv}Y+\frac12e^{iv}Y^{-1} -\frac a2e^{-iu}X-\frac a2e^{iu}X^{-1}-b.$$

Proportionality of the coefficients of $X,X^{-1}$ forces the proportionality constant to be $1$ or $-1$; the same holds from $Y,Y^{-1}$. If $b\ne0$, the constant must be $1$, so the parameters agree. If $b=0$, there is exactly the further possibility

$$(u',v')=(u+\pi,v+\pi). \tag{6}$$

Thus any family indexed by distinct parameter pairs has curve-coincidence multiplicity at most two.

The same calculation supplies the two-degrees-of-freedom condition. For a fixed point $(x,y)$, the parameters of curves through it satisfy

$$\cos(v-y)=a\cos(u-x)+b,$$

which is another translate of the same curve, now in the parameter torus. Two different points give distinct parameter curves except when $b=0$ and the points differ by $(\pi,\pi)$. In the absence of this antipodal exception, at most nine distinct curves pass through two points.

Incidence estimate

Let $P\subset\mathbb T^2$ be finite and let $\mathcal C$ be a set of distinct curves (3). In the case $b=0$, suppose initially that $P$ contains no pair differing by $(\pi,\pi)$. Then

$$I(P,\mathcal C) \le C\bigl(|P|^{2/3}|\mathcal C|^{2/3}+|P|+|\mathcal C|\bigr). \tag{7}$$

Here are sufficient details to reduce (7) to the elementary planar crossing lemma. Cut each angular coordinate at a value avoided by all points of $P$, and work in the resulting open rectangle. Each curve meets each cut at at most two points. It has only $O(1)$ singularities or vertical turning points: away from them it is one of the two graphs

$$y=v\pm\arccos(a\cos(x-u)+b),$$

and the endpoints of their domains occur among the at most four solutions of $a\cos(x-u)+b=\pm1$. Hence a curve splits into $O(1)$ simple arcs and isolated points; discarded cut, singular, or isolated-point incidences contribute $O(|\mathcal C|)$.

Join consecutive points of $P$ along every such arc. This draws a multigraph with at least $I(P,\mathcal C)-O(|\mathcal C|)$ edges. Edge multiplicity is bounded by an absolute constant because two points lie on at most nine curves, and a curve has only $O(1)$ arcs. Different curves have at most nine intersections, so the drawing has $O(|\mathcal C|^2)$ crossings. Passing to a simple subgraph loses only a constant factor. The crossing lemma then gives (7).

For completeness, the required form of the crossing lemma follows from $e\le3v+\operatorname{cr}$ and random vertex sampling. If a simple drawn graph has $e\ge4v$, sample every vertex with probability $p=4v/e$. Taking expectations gives $p^2e\le3pv+p^4\operatorname{cr}$, hence $\operatorname{cr}\ge e^3/(64v^2)$. Adjacent-edge crossings can first be removed, as in the standard proof. This proves the asserted estimate. Equivalently, (7) is the $k=2$ case of the standard Pach--Sharir bounded-degree curve-incidence theorem.

When $b=0$ and antipodal pairs are present, retain one point from each equivalence class modulo $(\pi,\pi)$. Every curve is invariant under this shift, so the retained point set has at least half the original incidences. Applying (7) to it proves the same bound for all $P$, with a different absolute constant. Alternatively partition $P$ into two half-open $x$-semicircles and apply the non-antipodal case to each.

A primary source for the general incidence theorem is the original Pach--Sharir paper, On the Number of Incidences Between Points and Curves, Combinatorics, Probability and Computing 7 (1998), 121--127. A checked precise restatement is Theorem 2.1 of Pach--de Zeeuw, Distinct distances on algebraic curves in the plane, full paper. The proof above uses only the elementary two-degrees-of-freedom case.

2. Small-difference-set cosine overlap

Lemma. Let $U,V\subset\mathbb T$ be finite nonempty sets, with

$$u=|U|,\quad v=|V|,\quad d_U=|U-U|,\quad d_V=|V-V|.$$

For $a>0$, $(a,b)\ne(1,0)$, put

$$E=\#\{(x,y)\in U\times V:\cos y=a\cos x+b\}.$$

Then

$$E\le C\left( \frac{(d_Ud_V)^{2/3}}{(uv)^{1/3}} +\frac{d_Ud_V}{uv}+1\right). \tag{8}$$

Proof. Take the point set $P=(U-U)\times(V-V)$. For every $(s,t)\in U\times V$, use the curve

$$\cos(y+t)=a\cos(x+s)+b.$$

Every solution $(x_0,y_0)$ counted by $E$, together with each $(s,t)$, yields the incidence of $(x_0-s,y_0-t)\in P$ with that curve. A fixed incidence with a distinct curve comes from at most two parameter pairs, by (6); for each parameter pair it determines the original solution uniquely. Thus the number of incidences with distinct curves is at least $Euv/2$. There are at most $uv$ distinct curves. Apply (7), including its antipodal extension, and divide by $uv/2$. This proves (8). $\square$

In particular, if $U,V$ are translates of one $m$-element set $B$ with

$$|B-B|\le Km,$$

then

$$E\le C\bigl(K^{4/3}m^{2/3}+K^2+1\bigr). \tag{9}$$

The same bound holds for phase-shifted cosines, because translating $U,V$ does not change their sizes or difference-set sizes.

3. Few distances and one large circle force finitely many concentric circles

We may take all points of $A$ on the chosen circle as the anchor, increasing $m$ if necessary; proving (1) for this larger $m$ also proves the stated bound. Assume henceforth that $q=q(A)\le n$, and take the circle containing $m\ge cn$ points as the circle of radius $R>0$ centered at the origin. Write its angular set as $B\subset\mathbb T$, $|B|=m$, and let $D_0$ be the set of occurring squared distances with zero adjoined, so $|D_0|=q+1\le n+1$.

Since the chord-squared map $\theta\mapsto2R^2-2R^2\cos\theta$ is at most two-to-one,

$$|B-B|\le2(q+1)\le K m, \tag{10}$$

where one may fix $K=4/c$.

For any noncentral point $p=\rho e^{i\phi}\in A$, define its pinned squared-distance spectrum on the anchor circle by

$$S_p=\{R^2+\rho^2-2R\rho\cos(\theta-\phi):\theta\in B\}.$$

It is contained in $D_0$, and $|S_p|\ge m/2$. This remains valid if $p$ itself is on the circle, because zero was adjoined.

If $p'=\sigma e^{i\psi}$ has $\rho\ne\sigma$, equality of one value in $S_p,S_{p'}$ gives

$$\cos(\eta-\psi)=\frac\rho\sigma\cos(\theta-\phi) +\frac{\sigma^2-\rho^2}{2R\sigma}. \tag{11}$$

The coefficient $a=\rho/\sigma$ is positive and different from one. Lemma (9), with phase-shifted copies of $B$, therefore gives

$$|S_p\cap S_{p'}|\le C_c(m^{2/3}+1). \tag{12}$$

This bounds intersection cardinality because each common squared-distance value gives at least one pair counted by the lemma.

Choose $L=\lceil4/c\rceil$. If $A$ occupied $L$ distinct positive radii about the origin, choose one point at each radius. By the elementary union lower bound and (12),

$$q+1\ge\left|\bigcup_{j=1}^L S_{p_j}\right| \ge\frac{Lm}{2}-C_c\binom L2(m^{2/3}+1) \ge2n-O_c(n^{2/3}),$$

contrary to $q+1\le n+1$, for sufficiently large $n$. Thus $A\setminus\{0\}$ lies on $k=O_c(1)$ concentric circles. Denote their pairwise distinct positive radii by

$$R_1,\dots,R_k,$$

and their nonempty angular sets by $B_1,\dots,B_k$. Take $B_1=B$, $R_1=R$.

4. One common angular set with bounded doubling

For every $i$, the distance formula between radius $R_i$ and the anchor radius $R$ gives

$$|B_i-B|\le2(q+1)\le Km. \tag{13}$$

This is valid even when $i=1$, again including zero in $D_0$.

An elementary form of Ruzsa's covering lemma supplies $X_i\subset B_i$, $|X_i|\le K$, such that

$$B_i\subset X_i+(B-B). \tag{14}$$

Indeed, choose a maximal family of disjoint translates $x-B$, with $x\in B_i$. Their union lies in $B_i-B$, so there are at most $K$ of them. Every other translate $b-B$ meets one of the chosen translates; consequently $b\in x+(B-B)$.

Let $X=\bigcup_iX_i$, so $|X|=O_c(1)$, and define the symmetric angular set

$$T=X-X+2B-2B. \tag{15}$$

Then every difference set $B_i-B_j$ is contained in $T$.

We use the standard Plünnecke--Ruzsa inequality in its two-set form:

$$|U+V|\le K|U| \quad\Longrightarrow\quad |rV-sV|\le K^{r+s}|U|\qquad(r+s>1). \tag{16}$$

Apply it to $U=B$, $V=-B$, using (10). We obtain

$$|2B-2B|\le K^4m, \qquad |4B-4B|\le K^8m.$$

Thus

$$m\le|T|\le |X|^2K^4m=O_c(m), \qquad |T-T|\le|X|^4K^8m=O_c(m). \tag{17}$$

The lower bound follows because $T$ contains $B-B$, which has at least $m$ elements. Hence $T$ has size comparable to $m$ and bounded difference-set ratio, with constants depending only on $c$.

For a primary checked source for (16), see Giorgis Petridis, New Proofs of Plünnecke-type Estimates for Product Sets in Groups, arXiv:1101.3507v3, Theorem 1.2 (PDF p.2), whose proof is on PDF p.8. The statement holds in every Abelian group, so it applies to $\mathbb T$, including torsion. Theorem 3.1 and Ruzsa's triangle inequality give the proof. Full paper.

5. Distinct radius-pair spectra almost never coincide

For an unordered radius pair $\tau=(i,j)$, $1\le i\le j\le k$, write

$$u_\tau=R_i^2+R_j^2, \qquad v_\tau=2R_iR_j>0,$$

and define

$$S_\tau=\{u_\tau-v_\tau\cos\theta:\theta\in B_i-B_j\}. \tag{18}$$

These are the squared-distance spectra for each unordered pair of circles; the same-circle spectra include zero.

Two distinct unordered radius pairs have distinct parameter pairs $(u_\tau,v_\tau)$. Indeed, their values determine

$$(R_i+R_j)^2=u_\tau+v_\tau, \qquad (R_i-R_j)^2=u_\tau-v_\tau,$$

which determine the unordered pair of positive radii.

For $\tau\ne\upsilon$, equality of a value in $S_\tau,S_\upsilon$ gives, for some $x,y\in T$,

$$\cos y=\frac{v_\tau}{v_\upsilon}\cos x +\frac{u_\upsilon-u_\tau}{v_\upsilon}. \tag{19}$$

The coefficient $a=v_\tau/v_\upsilon$ is positive, and $(a,b)\ne(1,0)$ by the preceding uniqueness observation. Apply (8) with $U=V=T$, using (17), to obtain

$$|S_\tau\cap S_\upsilon|=O_c(m^{2/3}+1)=O_c(n^{2/3}). \tag{20}$$

There are only $k(k+1)/2=O_c(1)$ pair types. Thus only $O_c(n^{2/3})$ positive squared-distance values occur for two or more pair types.

If the origin belongs to $A$, distances incident with it have squared values among $R_1^2,\dots,R_k^2$; discard these at most $k$ values as well. Every remaining occurring positive distance is represented entirely within exactly one unordered pair of concentric circles.

For a same-circle type $i=i$, each point has at most two neighbors at a fixed positive distance, so the unordered multiplicity is at most $|B_i|\le n$. For a cross-circle type $i\ne j$, each point on either circle has at most two neighbors on the other, so the multiplicity is at most

$$2\min\{|B_i|,|B_j|\}\le |B_i|+|B_j|\le n.$$

Consequently every non-discarded distance is rare, proving (2):

$$R(A)\ge q(A)-O_c(n^{2/3}).$$

The anchor circle itself determines at least $(m-1)/2$ positive distances, by fixing one of its points and using the same at-most-two-neighbors observation. Therefore (2) implies (1) when $q\le n$.

Finally, when $q>n$, the number of rich classes is less than $n/2$, because each consumes at least $n+1$ of the $\binom n2$ unordered pairs. Hence

$$R(A)>q-n/2>n/2\ge(m-1)/2,$$

which is stronger than (1). The proof above assumed sufficiently large $n$ only when bounding the number of occupied radii; enlarging $C_c$ absorbs the finitely many smaller values of $n$. This completes the proof.

Checks of potential degeneracies

• All distances are counted as unordered pairs. The factor two in the cross-circle bound is an actual degree bound, and $2\min(m_i,m_j)\le n$.
• The anchor circle has positive radius. The only zero-radius circle is the possible singleton at its center, whose at most $k$ distance values are explicitly removed.
• Coincident positive-radius circles are grouped as a single radius before defining pair types.
• The only reducible cosine relation for $a>0$ is $(a,b)=(1,0)$. It is excluded in the pinned-spectrum step by different radii, and in the pair-spectrum step by uniqueness of the unordered radius pair from its sum of squares and product.
• The $b=0$ simultaneous-antipodal degeneracy is only order two. It is handled in both point and curve incidence counts; it never produces unbounded multiplicity.
• All additive-combinatorial arguments take place in the compact Abelian group $\mathbb T$; no torsion-free or rational-angle assumption occurs.
• Every implied constant depends only on the fixed fraction $c$, not on the radius ratios, angular shifts, or the particular configuration.

Status and checks

The author checked the argument, including the uniform incidence bound, the two uses of additive covering, the algebraic degeneracies, and the multiplicity count; the argument is author-recorded. The cited Plünnecke–Ruzsa statement and its application were checked in the full primary paper. No novelty claim is made.

Together with the existing layer reduction, this leaves a precise necessary condition for a hypothetical bounded-rarity sequence: it has a positive fraction of points in finitely many outer convex layers, but every circle contains only $o(n)$ points. There is no proved implication from that layer concentration to a circular subset. Neither unrestricted assertion has been settled.