GPT 6 Astra Ultra attempt on Math-5428
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Contents
FORMAL RESTATEMENT
Erdős #105; exact audit of a known counterexample, 2026-09-23. If $A,B\subset\mathbb R^2$ are disjoint, $|A|=n$, $|B|=n-3$, and $A$ is not collinear, must some line contain at least two points of $A$ and no point of $B$? The answer is no. We give ten points $A$ whose every determined line meets a six-point set $B_0$; adding one further point gives the required $|B|=7$.QUICK LITERATURE AND CONTEXT CHECK
The source records that the assertion was disproved in 2025. The explicit stronger ten/six configuration used here was posted in the problem discussion by Core_65536 in July 2026. This file independently checks that posted configuration using integer arithmetic and gives a complete line certificate. It does not claim to have discovered the example.OBJECTIVE AND ATTACK PLAN
List all the distinct lines determined by pairs from $A$, and exhibit a point of $B_0$ on every one. An exact incidence table avoids relying on a diagram, rounded coordinates, or a numerical collinearity tolerance.WORK
In the order used for the indices below, let \[ \begin{aligned} A=\{&(168,0),(280,560),(280,140),(420,630),(360,240),\\ &(504,672),(280,-280),(420,0),(840,840),(0,-840)\},\\ B_0=\{&(0,420),(840,3360),(336,336),(280,0),(420,315),(-840,-2520)\}. \end{aligned} \tag{1} \] All ten points of $A$ are distinct, all six of $B_0$ are distinct, and inspection of their coordinate pairs gives $A\cap B_0=\varnothing$. The first three points of $A$ are not collinear: the determinant of their two difference vectors from $A_1$ is \[ \det\begin{pmatrix}112&560\\112&140\end{pmatrix}=-47040\ne0. \]In the table, $(a,b,c)$ denotes the line $ax+by+c=0$. The next column lists its points of $A$, and the final column gives an index of a point of $B_0$ on it. Every incidence in the table is verified by substitution in that integer equation. \[ \begin{array}{r|r|r|l|c} a&b&c&A\text{ indices}&B_0\text{ index}\\\hline 0&1&0&1,8&4\\ 1&-2&840&2,4,6,9&1\\ 1&0&-420&4,8&5\\ 1&0&-280&2,3,7&4\\ 1&1&-420&3,8&1\\ 2&-1&-840&7,8,9,10&6\\ 2&-1&-336&1,6&3\\ 3&-1&-840&5,6,10&4\\ 4&1&-1680&2,5,8&3\\ 5&-4&-840&1,3,5,9&5\\ 5&-2&-840&1,4&6\\ 5&-1&-840&1,2,10&2\\ 5&2&-840&1,7&1\\ 7&-2&-1680&3,4,10&3\\ 8&-1&-3360&6,8&2\\ 13&-2&-4200&4,5,7&2\\ 17&-4&-5880&6,7&5\\ 19&-8&-4200&3,6&6 \end{array} \tag{2} \] These are eighteen distinct lines: their coefficient triples are primitive and their first nonzero entries are positive, so distinct triples represent distinct lines. They contain three sets of four $A$-points, six sets of three, and nine sets of two. Consequently the table accounts for \[ 3\binom42+6\binom32+9\binom22=18+18+9=45=\binom{10}2 \] pairs. Two distinct lines cannot account for the same pair of distinct points. Thus every pair of points of $A$ appears on a line in the table, proving that every $A$-determined line meets $B_0$.
Finally take $B=B_0\cup\{(10000,10000)\}$. This new point is different from every listed point, so $|B|=7=|A|-3$ and $A\cap B=\varnothing$. Enlarging $B_0$ cannot remove its incidence with any line. Therefore the required line does not exist, and all hypotheses of the original question hold.
VERIFICATION AND SCOPE
An independent offline exact-arithmetic audit normalized the line equation for every one of the $45$ pairs, obtained exactly the eighteen rows in $2$, checked disjointness and noncollinearity, and checked a blocking point on each row. The table and pair count above already supply the complete finite proof, rather than an inference from a search that failed to find a line.SOURCES
- Current statement: https://www.erdosproblems.com/105.
- Core_65536's explicit ten/six configuration and preceding counterexamples: https://www.erdosproblems.com/forum/thread/105.