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Math-5797-A2 · revision r1 · on Math-5797

gpt pro 5.2 attempt on Math-5797

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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Contents
  1. PROBLEM 1 $\#655$: Distinct distances with bounded concentric multiplicity
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE/CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. VERIFICATION
  7. FINAL
  8. COMPLETION ESTIMATE

PROBLEM 1 $\#655$: Distinct distances with bounded concentric multiplicity

FORMAL RESTATEMENT

Fix $n\in\mathbb{N}$ and distinct points $x_1,\dots,x_n\in\mathbb{R}^2$.
  • Convention: a circle with center $x_i$ means the set $\{y\in\mathbb{R}^2: \|y-x_i\|=\rho\}$ for some radius $\rho>0$ (not the closed disk).
  • Hypothesis ("no centered circle contains three other points"): for every $i\in\{1,\dots,n\}$ and every radius $\rho>0$, the circle centered at $x_i$ of radius $\rho$ contains at most two points of $\{x_1,\dots,x_n\}\setminus\{x_i\}$. Equivalently, \[ \forall i\ \forall \rho>0:\quad \left|\{j\neq i: \|x_i-x_j\|=\rho\}\right|\le 2. \]
  • Let the distance set \[ D(x_1,\dots,x_n):=\{\|x_i-x_j\|: 1\le i<j\le n\}, \] and write $|D|$ for its cardinality.

Question (as written): Does there exist an absolute constant $c>0$ and an $n_0\in\mathbb{N}$ such that for all $n\ge n_0$ and all point sets satisfying the hypothesis, \[ |D(x_1,\dots,x_n)|\ \ge\ (1+c)\frac{n}{2}\ ? \]

QUICK LITERATURE/CONTEXT CHECK

A quick scan of the online discussion of Erdős problem #655 reports that the conjectured bound is false, and specifically notes an observation (attributed there to Zach Hunter) that $n$ equally spaced points on a circle give a counterexample.

ATTACK PLAN

  1. Disproof strategy: take the vertex set of a regular $n$-gon on a circle. Check the hypothesis by computing, for a fixed vertex, how many other vertices occur at the same distance. Then compute the number of distinct interpoint distances.
  2. Verification strategy: handle odd/even $n$ carefully (the "opposite vertex" case when $n$ is even) and confirm that the chord lengths are all distinct for step sizes $1,\dots,\lfloor n/2\rfloor$.

WORK

Theorem (Counterexample): The statement in Problem #655 is false. In fact, for every $n\ge 3$ there exist points $x_1,\dots,x_n\in\mathbb{R}^2$ satisfying the hypothesis but with \[ |D(x_1,\dots,x_n)|=\left\lfloor\frac{n}{2}\right\rfloor. \] Consequently, for every $c>0$ and all sufficiently large $n$, one has $|D(x_1,\dots,x_n)|<(1+c)\frac{n}{2}$.
Proof: Fix $n\ge 3$ and set $R=1$. Define points on the unit circle by \[ x_k := \left(\cos(2\pi k/n),\,\sin(2\pi k/n)\right)\in\mathbb{R}^2,\qquad k=0,1,\dots,n-1. \] Relabel them as $x_1,\dots,x_n$ if desired.

Step 1: Verify the hypothesis. Fix a vertex $x_i$. For each integer $t$ with $1\le t\le n-1$, the two vertices $x_{i+t}$ and $x_{i-t}$ (indices modulo $n$) have the same distance to $x_i$, and that distance depends only on $t$. Indeed, the central angle between $x_i$ and $x_{i+t}$ is $2\pi t/n$, so the chord length is \[ d_t := \|x_i-x_{i+t}\| = 2\sin\left(\frac{\pi t}{n}\right). \] For a fixed $t$ with $1\le t< n/2$, there are exactly two vertices at distance $d_t$ from $x_i$ (namely $x_{i\pm t}$). If $n$ is even and $t=n/2$, there is exactly one vertex at that distance (the opposite vertex). Hence for every radius $\rho>0$, the circle centered at $x_i$ of radius $\rho$ contains at most two other points of the configuration. This is exactly the hypothesis.

Step 2: Compute the number of distinct distances. Every interpoint distance in this configuration is some $d_t$ with $1\le t\le \lfloor n/2\rfloor$ (because $d_t=d_{n-t}$). We claim these values are all distinct: if $1\le s<t\le\lfloor n/2\rfloor$, then \[ 0<\frac{\pi s}{n}<\frac{\pi t}{n}\le\frac{\pi}{2}, \] and $\sin$ is strictly increasing on $(0,\pi/2]$, so $\sin(\pi s/n)<\sin(\pi t/n)$ and therefore $d_s<d_t$. Thus \[ D(x_1,\dots,x_n)=\{d_t: 1\le t\le \lfloor n/2\rfloor\} \quad\text{and}\quad |D(x_1,\dots,x_n)|=\left\lfloor\frac{n}{2}\right\rfloor. \]

Step 3: Contradict the claimed lower bound. Let $c>0$ be arbitrary. For all $n>\frac{2}{c}$, \[ \left\lfloor\frac{n}{2}\right\rfloor\le \frac{n}{2} < (1+c)\frac{n}{2}, \] so the required inequality fails for this $n$ and this configuration. Since this works for arbitrarily large $n$, no such absolute constant $c>0$ can exist. ∎

(Tiny-case computation check.) For the regular $n$-gon construction, one gets $|D|=\lfloor n/2\rfloor$ for small $n$ as well: \[ \begin{array}{c|cccccccc} n & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline |D| & 1 & 2 & 2 & 3 & 3 & 4 & 4 & 5 \end{array} \]

VERIFICATION

  • Quantifiers: the counterexample is constructed for every $n\ge 3$, so it defeats any universal $c>0$ and any $n_0$.
  • Hypothesis check: for each center $x_i$ and each radius $\rho$, the points at distance $\rho$ correspond to a step size $t$; there are at most two such points (or one if $n$ is even and $t=n/2$).
  • Distinctness of distances: proved via strict monotonicity of $\sin$ on $(0,\pi/2]$.
  • Edge cases: for $n=1,2$ the hypothesis is vacuous and the conclusion is irrelevant; the statement asks for $n$ sufficiently large anyway.

FINAL

LABEL: FULL SOLUTION
SUBLABEL: COUNTEREXAMPLE/DISPROOF
A regular $n$-gon (equally spaced points on a circle) satisfies the hypothesis but determines exactly $\lfloor n/2\rfloor$ distinct distances, contradicting any lower bound of the form $(1+c)\,n/2$ with fixed $c>0$.

COMPLETION ESTIMATE

COMPLETION: 100%.