Math-5502 · mathematics · Mathematics

Math-5502

Erdős Problem #211

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Statement

Erdős #211; reconstruction using the incidence theorem, 2026-09-23. Let $P$ be a set of $n$ real planar points, and suppose no line contains more than $n-k$ of them, where $1\leq k<n$. Prove that the number $L(P)$ of lines determined by pairs of points satisfies $L(P)\geq ckn$ for an absolute $c>0$.

Restatement from Math-5502-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-5502-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5502

FORMAL RESTATEMENT

Erdős #211; reconstruction using the incidence theorem, 2026-09-23. Let $P$ be a set of $n$ real planar points, and suppose no line contains more than $n-k$ of them, where $1\leq k<n$. Prove that the number $L(P)$ of lines determined by pairs of points satisfies $L(P)\geq ckn$ for an absolute $c>0$.

QUICK LITERATURE AND CONTEXT CHECK

This is the quantitative form of Beck's theorem, proved by Beck and also by Szemerédi--Trotter. The external input here is the Szemerédi--Trotter rich-line estimate, not the desired $ckn$ conclusion. The conversion from that estimate to the precise dependence on $k$ is given below.

OBJECTIVE AND ATTACK PLAN

Separate configurations having no very rich line from configurations with one. In the first case, control how many pairs lie on rich lines by a dyadic sum. In the second, count lines between a richest line and a suitably sized subset outside it.

WORK

Declared external input. There is an absolute constant $C\geq1$ such that for every real planar $n$-point set and every integer $r\geq2$, \[ |\{\ell:|P\cap\ell|\geq r\}|\leq C(n^2/r^3+n/r). \tag{1} \] This standard consequence of the Szemerédi--Trotter incidence theorem is the sole substantial geometric theorem used without proof.

Write $M=\max_\ell|P\cap\ell|$. Choose an integer $r_0\geq64C$ and put $\alpha=1/(128C)$. Suppose first that $M\leq\alpha n$. A line containing between $r$ and $2r$ points accounts for at most $2r^2$ pairs. Summing $1$ over $r=r_0,2r_0,4r_0,\ldots\leq M$, the number of pairs on lines containing at least $r_0$ points is at most \[ \sum_r2C(n^2/r+nr) \leq4Cn^2/r_0+4CnM \leq 3n^2/32. \tag{2} \] If there are no such $r$, the sum is zero. For $n\geq2$, the total number of pairs is $\binom n2\geq n^2/4$. Thus at least $n^2/8$ pairs lie on lines with fewer than $r_0$ points. Each such line accounts for at most $r_0^2/2$ pairs, so \[ L(P)\geq n^2/(4r_0^2)\geq kn/(4r_0^2). \tag{3} \]

Now suppose $M>\alpha n$. Fix a richest line $\ell_0$ and write $h=n-M\geq k$. Choose any $q=\min(h,M)$ points off $\ell_0$. For every line $\ell\ne\ell_0$ joining one of the $M$ points on $\ell_0$ to at least one chosen outside point, let $b_\ell$ be the number of chosen outside points on it. Such a line meets $\ell_0$ in exactly one point, so \[ \sum_\ell b_\ell=Mq,\qquad \sum_\ell\binom{b_\ell}{2}\leq\binom q2. \tag{4} \] The second inequality holds because each outside pair determines only one line; that line may or may not meet $\ell_0$ at a point of $P$. Since $b-1\leq\binom b2$ for every positive integer $b$, the number of these lines is at least \[ Mq-\binom q2\geq Mq/2. \tag{5} \] If $h\leq M$, this is at least $\alpha nh/2\geq\alpha nk/2$. If $h>M$, it is at least $M^2/2>\alpha^2n^2/2\geq\alpha^2nk/2$. Combining $3$--$5$, an admissible absolute constant is \[ c=\min\{1/(4r_0^2),\alpha^2/2\}>0. \]

VERIFICATION AND SCOPE

The dyadic calculation includes the $n/r$ term in $1$, which cannot be dropped for rich lines. In the second case only $q\leq M$ outside points are used, preventing the subtraction in $5$ from overwhelming the main term. All constants are independent of $n$, $k$, and the configuration. The proof is complete relative to the explicitly stated incidence theorem; it is not a new proof of that theorem.

SOURCES

CONCLUSION

Attempt status: solved using a named external theorem. The argument proves the required uniform lower bound $L(P)\gg kn$. No novelty is claimed. Completion rate estimate: 100%.

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