Exact second-distance extremality without five concyclic points

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Proposition. For every $m\ge8$ divisible by four, there is a set of $N=2m$ planar points such that its second-largest distance occurs exactly $3m=3N/2$ times, yet no five of its points are concyclic.

Moreover, its strictly convex outer $m$-gon has uniform lengths in both of its largest cyclic separation classes. Thus the parity condition in long_chord_rigidity is essential. Exact or near extremality in Vesztergombi's second-distance bound cannot by itself force a circular fraction. This construction does not establish a small total distance palette or bounded rarity, and is not a counterexample to either assertion of Problem 132.

1. A regular initial configuration

Put $k=m/2$, $\theta=\pi/m$, $c=\cos\theta$, $s=\sin\theta$, and

$$p_j=e^{2ij\theta},\qquad r=2c,\qquad D_0=2,$$

where indices are modulo $m$. The regular outer polygon has $k$ diameter pairs, $m$ pairs of length $r$ at cyclic separation $k-1$, and all other distances strictly below $r$. Define

$$\rho=\sqrt{5c^2-1}-c,\qquad H=\rho+c,\qquad u_j=-\rho e^{i(2j+1)\theta}.$$

We have $0<\rho<c$, the upper inequality following from $\sqrt{5c^2-1}<2c$. Thus the inner points lie strictly inside the outer polygon's incircle. The identity

$$1+\rho^2+2\rho c=4c^2=r^2$$

gives $|u_j-p_j|=|u_j-p_{j+1}|=r$. More generally,

$$|u_j-p_l|^2=1+\rho^2+2\rho\cos((2l-2j-1)\theta).$$

Its maximum occurs only for $l=j,j+1$. Also all inner--inner distances are at most $2\rho<r$. Hence there are exactly $3m$ pairs at $r$, and all undesignated inequalities are strict.

2. A full-rank analytic deformation

Set $a=k-1$. Since $k$ is even, $\gcd(a,m)=1$. In the variables $v_0,\ldots,v_{m-1}\in\mathbb R^2$ and one additional real variable $D$, impose

$$|v_j-v_{j+a}|^2=r^2\quad(0\le j<m),\qquad |v_j-v_{j+k}|^2=D^2\quad(0\le j<k). \tag{1}$$

The differential of these $m+k$ equations has full row rank at $(p_0,\ldots,p_{m-1},2)$. To see this, consider first a dependence on the coordinate columns alone. Give the first family coefficients $\alpha_j$, and the diameter edges coefficients $\beta_j=\beta_{j+k}$. Write $\phi=\pi-2\theta$. The equilibrium equation at $p_j$, after rotation by $p_j^{-1}$ and removal of the common factor two, is

$$\alpha_j(1-e^{i\phi})+ \alpha_{j-a}(1-e^{-i\phi})+2\beta_j=0.$$

Its imaginary part gives $\alpha_j=\alpha_{j-a}$, so every $\alpha_j$ is one common $\alpha$. Its real part then gives $\beta_j=-2c^2\alpha$. This is the entire coordinate stress space. The additional $D$ column has pairing $-4\sum_{j=0}^{k-1}\beta_j=-4k\beta_j$, which vanishes only when all coefficients vanish. This proves full row rank.

The analytic implicit-function theorem gives a local solution manifold of dimension $2m+1-(m+k)=k+1$. For every nearby outer solution, construct $u_j(v)$ as the intersection near $u_j$ of the two radius-$r$ circles about $v_j,v_{j+1}$. The two circle gradients there are independent, so this construction is analytic. The strict inequalities from Section 1 persist. Consequently every sufficiently nearby solution still has exactly $k$ diameter pairs and $3m$ second-distance pairs.

3. Free anti-periodic radial velocities

Write an outer velocity as

$$\dot v_j=(f_j+i g_j)p_j,\qquad f_j,g_j\in\mathbb R.$$

The coordinate stress above, applied to the differentiated equations, gives $\dot D=0$. The diameter equations give

$$f_{j+k}=-f_j. \tag{2}$$

The other equations give

$$g_{j+a}-g_j=-\cot\theta(f_j+f_{j+a}). \tag{3}$$

Conversely, any $k$ real values $f_0,\ldots,f_{k-1}$, extended by (2), admit a solution of (3). Its only compatibility condition around the single step-$a$ cycle is $\sum_j f_j=0$, already satisfied. The $g_j$ are unique up to a common rotation; choose this constant so that

$$g_{j+k}=-g_j.$$

Indeed, (3) makes $g_{j+k}+g_j$ constant on the step-$a$ cycle. Thus the anti-periodic radial data are completely free and every such tangent is realized by an analytic curve in the solution manifold.

4. Breaking outer five-point circles

If a fixed five-element index set $S$ remains concyclic to first order, its radial velocities necessarily have the form

$$f_j=A+B\cos(2j\theta)+C\sin(2j\theta),\qquad j\in S. \tag{4}$$

These are the variations of the radius and the two center coordinates. Condition (4) defines a proper linear subspace of the anti-periodic data. If $S$ contains no full antipodal pair, its five values are independent, whereas (4) has only three parameters. If it contains such a pair, (2) forces $A=0$; five indices represent at least three different antipodal pairs, giving at least three independent values but only two remaining parameters.

There are finitely many five-element subsets. A tangent chosen outside all their proper subspaces breaks every outer five-point circle along its analytic curve, for all sufficiently small nonzero parameters. Formally, the circle through any three of the selected points varies analytically near the initial configuration. Failure of (4) makes one of the other two circle equations have a nonzero first derivative.

5. Simultaneously breaking inner five-point circles

Put $w_j=e^{i(2j+1)\theta}$ and write $\dot u_j=-w_j(\widehat f_j+i\widehat g_j)$, so $\widehat f_j$ is the physical radial velocity on the inner circle. Differentiating the two circle-intersection equations and adding them gives

$$\widehat f_j=-\frac{(1+\rho c)(f_j+f_{j+1}) -\rho s(g_{j+1}-g_j)}{2H}. \tag{5}$$

This is an anti-periodic sequence. The linear map from $f$ to $\widehat f$ is invertible on the anti-periodic space. To check this, complexify and take its Fourier basis $f_j=z^j$, where $z^k=-1$. Since $k$ is even, $z\ne-1$. Equations (2)--(3) give

$$g_j=\cot\theta\frac{z-1}{z+1}z^j.$$

Substitution into (5) gives the multiplier

$$\widehat f_j=-\frac{(1+z)^2+4\rho c z}{2H(1+z)}z^j. \tag{6}$$

It is nonzero: for $z=e^{i\omega}$, the numerator divided by $z$ is $4\cos^2(\omega/2)+4\rho c>0$.

The argument of Section 4 applies to the inner circle as well, with its angular offset. Since the radial map is invertible, each inner five-point concyclicity condition pulls back to a proper linear subspace of the free outer data. Choose a tangent outside the finitely many inner and outer subspaces simultaneously.

Every mixed five-tuple is already nonconcyclic at the initial configuration: a circle other than the two concentric circles meets each of them in at most two points, and either original circle contains no point of the other. These finitely many nonconcyclicity conditions persist by continuity. Therefore a sufficiently small nonzero point of the chosen analytic curve has no five concyclic points, while retaining all the distance counts. This proves the proposition.

6. The entire local family has linearly many rare distances

The preceding construction has a stronger limitation than merely failing to control the palette. Its regular initial configuration has exactly

$$q=3k-1=3N/4-1,\qquad R=3k-2=3N/4-2. \tag{7}$$

Moreover every sufficiently nearby $N$-point configuration, whether or not it satisfies (1), has at least $3N/4-2$ rare distances. Thus none of the sufficiently small deformations used above has bounded rarity.

The palette computation applies to every even $m=2k\ge6$. In the notation of Section 1, the outer, inner, and cross palettes are respectively

$$O_j=2\sin(j\theta),\qquad I_j=\rho O_j,\qquad C_j=\sqrt{(1-\rho)^2+4\rho\sin^2((j-1/2)\theta)} \quad(1\le j\le k).$$

They satisfy

$$O_{j-1}<C_j<I_j<O_j\quad(1\le j\le k-1), \tag{8}$$

where $O_0=0$, and

$$I_{k-1}<I_k<O_{k-1}=C_k<O_k. \tag{9}$$

Here is an algebraic verification of the strict comparisons. We have $\rho^2+2c\rho+1=4c^2$ and $0<\rho<c$. With $x=2j\theta$,

$$I_j^2-C_j^2 =\rho^2-1+2\rho\bigl((c-\rho)\cos x+s\sin x\bigr).$$

A function $a\cos x+b\sin x$, $a,b>0$, has no interior minimum on $[0,\pi]$. On $[2\theta,\pi-2\theta]$ its smaller endpoint is $\pi-2\theta$. Substitution into the preceding expression gives

$$8c(2c^2-1)(c-\rho)>0.$$

Similarly, with $y=2(j-1)\theta$,

$$C_j^2-O_{j-1}^2 =\rho^2-1+2\bigl((1-\rho c)\cos y+\rho s\sin y\bigr).$$

On $[0,\pi-4\theta]$ its minimum is at an endpoint. The endpoint values are $C_1^2>0$ and $4\rho^2s^2>0$, respectively. This proves (8); (9) follows from $I_k=2\rho<2c=O_{k-1}=C_k$ and monotonicity. Thus the three $k$-element palettes have precisely one coincidence, namely $C_k=O_{k-1}$.

Each outer or inner class has multiplicity $m$, except its antipodal class, which has multiplicity $k$. Each cross class has $2m=N$ pairs. Consequently only $C_k=O_{k-1}$ is rich, with multiplicity $3m$; all the other $3k-2$ classes are rare. This proves (7).

Finally, for any fixed finite configuration, $R$ cannot decrease under sufficiently small perturbations. Choose disjoint intervals around its distinct positive distances, and perturb so little that each labeled pair stays in its original interval. An initially rare class can split, but each resulting class still contains at most its original $N$ pairs, and cannot merge with a different original class. It therefore supplies at least one rare distance in the perturbed set. Applied to (7), this proves the neighborhood assertion.

The interlacing calculation, the exact multiplicities, and the neighborhood argument were checked by the author and are author-recorded. No floating-point calculation is used as a certificate.

7. An odd-polygon caution for a possible palette bound

The even initial configuration satisfies $m(r_2)=2q+2$. That equality does not extend to a universal upper bound. Put

$$\zeta=e^{2\pi i/5},\qquad \rho=2\sin(\pi/30),\qquad A=\{\zeta^j,\rho\zeta^j:0\le j<5\}.$$

Let $r=2\sin(\pi/5)$, $D=2\sin(2\pi/5)$, and $b=\sqrt{1+\rho^2-2\rho\cos(2\pi/5)}$. The exact spectrum is

$$\begin{array}{c|rrrrrr} \text{length}&\rho r&\rho D&1-\rho&b&r&D\\\hline \text{multiplicity}&5&5&5&10&15&5. \end{array}$$

The displayed lengths are strictly increasing. For example $0<\rho<1/4$, $\rho D<1/2<1-\rho$, and $1-\rho<b<1<r<D$. For $b<1$ use $\rho<1/4<2\cos(2\pi/5)$. The longest cross distance equals $r$: writing $c=\cos(\pi/5)$, the identity $\rho=\sqrt3\sin(\pi/5)-c$ gives $1+\rho^2+2c\rho=4\sin^2(\pi/5)=r^2$.

Thus $q=6$ and $m_A(r_2)=15>2q+2$. This is not a counterexample to Problem 132: it has five rare distances. Whether any bound $m_A(r_2)\le2q(A)+C$ with an absolute constant $C$ holds is not established here. Its relevance to a hypothetical $R=1$ sequence is explained in first_exceptional_distance.

Relation to the remaining gap

The one-rare asymptotic reduction forces near-extremality of the second distance on the first two layers of a hypothetical unbounded $R=1$ family. The proposition rules out deriving a circular fraction from that extremality alone, even if both uniform long-chord families can be extracted. Any successful inverse statement must also use additional hypotheses, such as the sharp global distance palette. Section 6 excludes the entire sufficiently small neighborhood of this regular extremizer, but does not characterize remote portions of the long-edge framework variety. The strengthened unrestricted inverse statement is not proved here.

The author checked the row-rank calculation, all strict inequalities, the anti-periodic tangent equations, the inner radial multiplier, and the five-point transversality argument; the argument is author-recorded. No numerical approximation is used as a certificate.