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Math-5797 · mathematics · Mathematics

Erdős Problem #655

Status Open · erdosproblems.comLLM status Open unverified

Statement

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, \[ |D(x_1,\dots,x_n)|\ \ge\ (1+c)\frac{n}{2}\ ? \]

Restatement from Math-5797-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Math-5797-A3 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5797

Erdős problems #655, #738, #740 --- Round 2 continuation

Problem #655

1) ROUND-2 OBJECTIVE

Path (C): turn the Round-1 disproof into the exact corrected statement. Round 1 already produced a regular $n$-gon counterexample, so the natural gap to close is the sharp lower bound that actually follows from the stated hypothesis.

2) Round-1 FOUNDATION USED

I use the following vetted Round-1 fact.

The regular $n$-gon satisfies the hypothesis of the problem and determines exactly $\lfloor n/2\rfloor$ distinct distances. Therefore the claimed bound $(1+c)n/2$ is false for every fixed $c>0$.

I do not re-prove that counterexample here.

3) NEW INSIGHT / TOOL (ROUND-2)

The hypothesis controls the multiplicity of each radius around a fixed point. If one fixes a point $x_i$, then among the $n-1$ distances from $x_i$ to the other points, each distinct value can occur at most twice. Therefore $x_i$ itself determines at least $\lceil (n-1)/2\rceil$ distinct distances. Since these are all global interpoint distances, the whole configuration determines at least that many distinct distances.

Combined with the Round-1 regular $n$-gon construction, this gives the exact optimum.

4) ATTACK PLAN (ROUND-2)

The only remaining gap after Round 1 is to identify the best possible lower bound replacing the false inequality \[ |D|\geq (1+c)\frac n2. \] I will prove the sharp correction \[ |D|\geq \left\lfloor\frac n2\right\rfloor, \] and then use the Round-1 regular $n$-gon to show equality is attainable for every $n$.

5) WORK (ROUND-2)

Let $x_1,\dots,x_n$ be distinct points in the plane satisfying the hypothesis that for every $i$ and every radius $\rho>0$, the circle centred at $x_i$ with radius $\rho$ contains at most two of the other points.

Let \[ D:=\{\,|x_i-x_j|:1\leq i<j\leq n\,\} \] be the global distance set.

For a fixed index $i$, define the set of distances seen from $x_i$ by \[ D_i:=\{\,|x_i-x_j|:j\neq i\,\}. \] Every value in $D_i$ belongs to $D$, so \[ |D|\geq |D_i| \qquad \text{for every } i. \] Now fix $i$. There are $n-1$ points other than $x_i$. By hypothesis, each radius about $x_i$ can hit at most two of them. Hence each element of $D_i$ accounts for at most two points, and therefore \[ 2|D_i|\geq n-1. \] So \[ |D_i|\geq \left\lceil\frac{n-1}{2}\right\rceil = \left\lfloor\frac n2\right\rfloor. \] Since $|D|\ge |D_i|$, we obtain the sharp universal lower bound \[ |D|\geq \left\lfloor\frac n2\right\rfloor. \]

Thus the exact corrected theorem is:

Corrected theorem. For every $n\ge 1$, every $n$-point set in the plane satisfying the stated hypothesis determines at least $\lfloor n/2\rfloor$ distinct distances.

To see that this is best possible, invoke the Round-1 construction: the $n$ vertices of a regular $n$-gon satisfy the hypothesis and determine exactly $\lfloor n/2\rfloor$ distinct distances. Therefore \[ \min |D| = \left\lfloor\frac n2\right\rfloor \] over all such $n$-point configurations.

6) ADVERSARIAL VERIFICATION

  • Quantifiers. The proof gives the lower bound for every $n\ge 1$, not merely for large $n$.
  • Odd/even parity. The identity $\lceil (n-1)/2\rceil=\lfloor n/2\rfloor$ handles both parity cases exactly.
  • Edge cases. For $n=1$, the distance set is empty and $\lfloor n/2\rfloor=0$. For $n=2$, there is exactly one distance and $\lfloor n/2\rfloor=1$. The argument still works.
  • No hidden assumption. The proof uses only the stated circle-multiplicity hypothesis; it does not assume convexity, general position, or any structure beyond distinctness of the points.
  • Sharpness. The regular $n$-gon from Round 1 attains equality, so the correction cannot be improved.

7) FINAL

FULL SOLUTION --- COUNTEREXAMPLE / DISPROOF

The original statement is false. The exact corrected statement is: \[ |D|\ge \left\lfloor\frac n2\right\rfloor, \] and this bound is sharp for every $n$, with equality achieved by the regular $n$-gon.

8) COMPLETION ESTIMATE

COMPLETION: 100%

9) REFERENCES

No external theorem is needed for the Round-2 correction; the proof above is self-contained.

References

References

  • [Bloom738] T. F. Bloom (ed.), Erdős Problem #738, Erdős Problems Project, accessed 8 March 2026.
  • [Bloom740] T. F. Bloom (ed.), Erdős Problem #740, Erdős Problems Project, accessed 8 March 2026.
  • [Scott97] A. D. Scott, Induced trees in graphs of large chromatic number, Journal of Graph Theory 24 $1997$, 297--311.
  • [NguyenScottSeymour24] T. Nguyen, A. Scott, and P. Seymour, A note on the Gyárfás--Sumner conjecture, Graphs and Combinatorics 40 $2024$, Article 33.
  • [Rodl77] V. Rödl, On the chromatic number of subgraphs of a given graph, Proceedings of the American Mathematical Society 64 $1977$, 370--371.
  • [KomjathShelah88] P. Komjáth and S. Shelah, Forcing constructions for uncountably chromatic graphs, Journal of Symbolic Logic 53 $1988$, 696--707.

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Math-5797-A1 · No claim recorded · LLM-generated · gpt pro 5.4 ·

Reused writeup on Math-5797 (gpt pro 5.4)

Erdős Problem 655 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/655

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.4/655.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/655.tex.

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%.

REUSED SOURCE 2

Original attribution: gpt_pro_5.4. Original file: attacks/erdos/gpt_pro_5.4/655.tex.

Erdős problems #655, #738, #740 --- Round 2 continuation

Problem #655

1) ROUND-2 OBJECTIVE

Path (C): turn the Round-1 disproof into the exact corrected statement. Round 1 already produced a regular $n$-gon counterexample, so the natural gap to close is the sharp lower bound that actually follows from the stated hypothesis.

2) Round-1 FOUNDATION USED

I use the following vetted Round-1 fact.

The regular $n$-gon satisfies the hypothesis of the problem and determines exactly $\lfloor n/2\rfloor$ distinct distances. Therefore the claimed bound $(1+c)n/2$ is false for every fixed $c>0$.

I do not re-prove that counterexample here.

3) NEW INSIGHT / TOOL (ROUND-2)

The hypothesis controls the multiplicity of each radius around a fixed point. If one fixes a point $x_i$, then among the $n-1$ distances from $x_i$ to the other points, each distinct value can occur at most twice. Therefore $x_i$ itself determines at least $\lceil (n-1)/2\rceil$ distinct distances. Since these are all global interpoint distances, the whole configuration determines at least that many distinct distances.

Combined with the Round-1 regular $n$-gon construction, this gives the exact optimum.

4) ATTACK PLAN (ROUND-2)

The only remaining gap after Round 1 is to identify the best possible lower bound replacing the false inequality \[ |D|\geq (1+c)\frac n2. \] I will prove the sharp correction \[ |D|\geq \left\lfloor\frac n2\right\rfloor, \] and then use the Round-1 regular $n$-gon to show equality is attainable for every $n$.

5) WORK (ROUND-2)

Let $x_1,\dots,x_n$ be distinct points in the plane satisfying the hypothesis that for every $i$ and every radius $\rho>0$, the circle centred at $x_i$ with radius $\rho$ contains at most two of the other points.

Let \[ D:=\{\,|x_i-x_j|:1\leq i<j\leq n\,\} \] be the global distance set.

For a fixed index $i$, define the set of distances seen from $x_i$ by \[ D_i:=\{\,|x_i-x_j|:j\neq i\,\}. \] Every value in $D_i$ belongs to $D$, so \[ |D|\geq |D_i| \qquad \text{for every } i. \] Now fix $i$. There are $n-1$ points other than $x_i$. By hypothesis, each radius about $x_i$ can hit at most two of them. Hence each element of $D_i$ accounts for at most two points, and therefore \[ 2|D_i|\geq n-1. \] So \[ |D_i|\geq \left\lceil\frac{n-1}{2}\right\rceil = \left\lfloor\frac n2\right\rfloor. \] Since $|D|\ge |D_i|$, we obtain the sharp universal lower bound \[ |D|\geq \left\lfloor\frac n2\right\rfloor. \]

Thus the exact corrected theorem is:

Corrected theorem. For every $n\ge 1$, every $n$-point set in the plane satisfying the stated hypothesis determines at least $\lfloor n/2\rfloor$ distinct distances.

To see that this is best possible, invoke the Round-1 construction: the $n$ vertices of a regular $n$-gon satisfy the hypothesis and determine exactly $\lfloor n/2\rfloor$ distinct distances. Therefore \[ \min |D| = \left\lfloor\frac n2\right\rfloor \] over all such $n$-point configurations.

6) ADVERSARIAL VERIFICATION

  • Quantifiers. The proof gives the lower bound for every $n\ge 1$, not merely for large $n$.
  • Odd/even parity. The identity $\lceil (n-1)/2\rceil=\lfloor n/2\rfloor$ handles both parity cases exactly.
  • Edge cases. For $n=1$, the distance set is empty and $\lfloor n/2\rfloor=0$. For $n=2$, there is exactly one distance and $\lfloor n/2\rfloor=1$. The argument still works.
  • No hidden assumption. The proof uses only the stated circle-multiplicity hypothesis; it does not assume convexity, general position, or any structure beyond distinctness of the points.
  • Sharpness. The regular $n$-gon from Round 1 attains equality, so the correction cannot be improved.

7) FINAL

FULL SOLUTION --- COUNTEREXAMPLE / DISPROOF

The original statement is false. The exact corrected statement is: \[ |D|\ge \left\lfloor\frac n2\right\rfloor, \] and this bound is sharp for every $n$, with equality achieved by the regular $n$-gon.

8) COMPLETION ESTIMATE

COMPLETION: 100%

9) REFERENCES

No external theorem is needed for the Round-2 correction; the proof above is self-contained.

References

References

  • [Bloom738] T. F. Bloom (ed.), Erdős Problem #738, Erdős Problems Project, accessed 8 March 2026.
  • [Bloom740] T. F. Bloom (ed.), Erdős Problem #740, Erdős Problems Project, accessed 8 March 2026.
  • [Scott97] A. D. Scott, Induced trees in graphs of large chromatic number, Journal of Graph Theory 24 $1997$, 297--311.
  • [NguyenScottSeymour24] T. Nguyen, A. Scott, and P. Seymour, A note on the Gyárfás--Sumner conjecture, Graphs and Combinatorics 40 $2024$, Article 33.
  • [Rodl77] V. Rödl, On the chromatic number of subgraphs of a given graph, Proceedings of the American Mathematical Society 64 $1977$, 370--371.
  • [KomjathShelah88] P. Komjáth and S. Shelah, Forcing constructions for uncountably chromatic graphs, Journal of Symbolic Logic 53 $1988$, 696--707.

COMPLETION ESTIMATE

Inherited primary-source estimate: $100\%$. This is the original source's unverified estimate, not a new assessment or confirmation that the problem is solved.

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Math-5797-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-5797

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%.

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