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Claim. The answer to Problem 105 is no. There are disjoint finite sets $A,B\subset\mathbb{R}^2$ with $|A|=n$ and $|B|=n-3$, $A$ not contained in a line, such that every line through two points of $A$ passes through a point of $B$. The claimant, credited by the site as Xichuan and posting under the forum account eigensolver, found three such configurations by a computer search that treats the lines spanned by $A$ as sets to be covered by the points of $B$, and posted them in the problem's discussion on 2025-10-24; projective coordinates for one configuration are given in the thread.

Acceptance. Thomas Bloom, the site's curator, marked the problem disproved on 2025-10-25 and credits the three counterexamples on the problem page. Boris Alexeev reported in the same thread that they had checked all three configurations by computer and one of them by hand (2025-10-25), and on 2025-11-17 posted a Lean proof of the disproof in Alexeev's repository of formalized Erdős problems, which the site's label notes. The Lean file names Wu Xichuan as its informal author and ChatGPT Pro (Thinking), Aristotle and Boris Alexeev as its formal authors. That Lean development is linked above at its pinned commit; this corpus has not built or audited it, so it is not listed as evidence. No refereed publication exists; the claim is accepted on the curator's documented acceptance and Alexeev's independent check.

What remains. The site's remark (last edited 2025-10-25) leaves open whether the statement holds with $n-4$ blocking points. A comment in the problem's thread on 2026-07-24, by the forum account Core_65536, gives a counterexample to that version with $n=10$: its ten points of $A$, not all on a line, span eighteen lines, and each of those lines contains exactly one of its six points of $B$. The comment says the construction arose while testing AI-assisted search in Cursor with the Grok model; it is a forum comment, not a refereed result. Adding to its $B$ any new point outside $A$ gives a further counterexample to the problem itself, with $n=10$. Whether the statement holds with $n-O(1)$ or $(1-o(1))n$ blocking points is open. The result of Beck and of Szemerédi and Trotter gives it with $n-3$ replaced by $cn$ for some constant $c>0$.