gpt pro 5.4 attempt on Math-5796

Problem #654

1. ROUND-2 OBJECTIVE

Path (B): explicit counterexample / disproof.

Round 1 left the main gap completely open: no counterexample family was known there. In Round 2 I give an explicit infinite family showing that the strongest form of the conjecture is false.

2. Round-1 FOUNDATION USED

I use only the Round-1 definition of \[ f(n):=\min_{|X|=n}\max_{x\in X} R_X(x) \] under the constraint that $X\subset \mathbf{R}^2$ has no four points on a circle, together with the trivial lower bound $f(n)\ge (n-1)/3$.

No nontrivial Round-1 lemma is needed for the disproof.

3. NEW INSIGHT / TOOL (ROUND 2)

The new construction has two ingredients.

• Put symmetric pairs of points on each coordinate axis. Then from any point on one axis, the distances to the opposite axis come in equal pairs, so the cross-axis contribution is halved.
• Choose the $x$-coordinates to be powers of $2$ and the $y$-coordinates to be powers of $3$. Then any product of two signed $x$-coordinates is a signed power of $2$, while any product of two signed $y$-coordinates is a signed power of $3$. This forbids the only possible $4$-point concyclicity pattern.

4. ATTACK PLAN (ROUND 2)

For each $t\ge 1$, define \[ X_t:=\{(\pm 2^i,0):1\le i\le t\}\cup\{(0,\pm 3^j):1\le j\le t\}. \] Then $|X_t|=4t$.

I prove two claims.

1. No four points of $X_t$ lie on a common circle.
2. Every point of $X_t$ determines at most $3t-1$ distinct distances to the other points.

These two claims imply \[ f(4t)\le 3t-1=\frac{3}{4}(4t)-1, \] which disproves the asymptotic statement $f(n)>(1-o(1))n$.

5. WORK (ROUND 2)

Fix $t\ge 1$, and define \[ A_t:=\{\pm 2^i:1\le i\le t\},\qquad B_t:=\{\pm 3^j:1\le j\le t\}, \] \[ X_t:=\{(a,0):a\in A_t\}\cup \{(0,b):b\in B_t\}. \] Then $|X_t|=4t$.

Claim 1: no four points of $X_t$ are concyclic.

A circle meets any fixed line in at most two points. Since all points of $X_t$ lie on the two coordinate axes, any set of four concyclic points from $X_t$ would have to consist of exactly two points on the $x$-axis and exactly two points on the $y$-axis. So it would have the form \[ (a,0),(a',0),(0,b),(0,b') \] with $a\ne a'$ and $b\ne b'$.

Suppose such four points lie on a circle. Write the equation of the circle as \[ x^2+y^2+ux+vy+w=0. \] Plugging in $(a,0)$ and $(a',0)$ shows that $a$ and $a'$ are roots of \[ x^2+ux+w=0, \] so by Viète's formula, \[ aa'=w. \] Similarly, plugging in $(0,b)$ and $(0,b')$ shows that $b$ and $b'$ are roots of \[ y^2+vy+w=0, \] so \[ bb'=w. \] Hence \[ aa'=bb'. \] But $aa'$ is a signed power of $2$, while $bb'$ is a signed power of $3$. These can never be equal. Contradiction.

Therefore no four points of $X_t$ lie on a common circle.

Claim 2: every point of $X_t$ sees at most $3t-1$ distinct distances.

Take first a point $p=(a,0)$ on the $x$-axis.

• There are $2t-1$ other points of $X_t$ on the $x$-axis, so their contribution to $R_{X_t}(p)$ is at most $2t-1$.
• For each $j\in\{1,\dots,t\}$, the two points $(0,3^j)$ and $(0,-3^j)$ are both at distance \[ \sqrt{a^2+3^{2j}} \] from $p$. So all $2t$ points on the $y$-axis contribute at most $t$ distinct distances.

Hence \[ R_{X_t}(p)\le (2t-1)+t=3t-1. \]

The same argument applies to a point on the $y$-axis by symmetry. Thus every point of $X_t$ determines at most $3t-1$ distinct distances.

So for $n=4t$ we have constructed an $n$-point set with no four points on a circle and with \[ \max_{x\in X_t}R_{X_t}(x)\le 3t-1=\frac{3n}{4}-1. \] Therefore \[ f(4t)\le \frac{3}{4}(4t)-1. \] In particular, \[ \limsup_{t\to\infty}\frac{f(4t)}{4t}\le \frac{3}{4}<1. \] So the statement $f(n)>(1-o(1))n$ is false.

6. ADVERSARIAL VERIFICATION

• The concyclicity proof covers all sign patterns: any product of two signed $x$-coordinates is $\pm 2^m$, and any product of two signed $y$-coordinates is $\pm 3^n$.
• A circle cannot contain three points from one axis and one from the other, because each axis meets a circle in at most two points.
• In the distance count I used only upper bounds, so possible accidental overlaps between same-axis and cross-axis distances can only help.
• The family is given for $n=4t$. That is enough to disprove an asymptotic assertion of the form $(1-o(1))n$, since such a statement must hold along every infinite subsequence.

7. FINAL

Partial known SOLUTION -- COUNTEREXAMPLE / DISPROOF

The strongest form of Problem #654 is false. The explicit family \[ X_t=\{(\pm 2^i,0):1\le i\le t\}\cup\{(0,\pm 3^j):1\le j\le t\} \] has no four points on a circle and satisfies \[ \max_{x\in X_t}R_{X_t}(x)\le \frac{3}{4}|X_t|-1. \] Thus $f(n)$ is not $(1-o(1))n$ in general. This agrees with the current status note on the Erdős Problems site [EP654].

8. COMPLETION ESTIMATE

COMPLETION: 30%

The conjecture that $f(n)>(1-o(1))n$ is false. Aletheia $2026$ constructed (n)-point sets in the plane with no four points on a circle such that every point determines at most $(3/4)n+O(1)$ distinct distances. However, this construction places all points on two lines and therefore does not resolve the stronger version of the problem for point sets in general position. Determining whether $f(n)>(1/3+c)n$ for some $c>0$ in general position remains open.