GPT 6 Astra Ultra attempt on Math-5500

FORMAL RESTATEMENT

Erdős #209; explicit counterexample, 2026-09-23. Given $d\geq4$ pairwise nonparallel lines in the real plane, with no four concurrent, must three of the lines form a triangle whose three vertices are each incident with exactly two of the given lines?

QUICK LITERATURE AND CONTEXT CHECK

The current source records a negative answer, with broader families due to Füredi--Palásti and Escudero. A single admissible value of $d$ suffices to disprove the universal statement. Here $d=4$ already gives a completely explicit counterexample. This does not assert a new result or a construction for every possible $d$.

OBJECTIVE AND ATTACK PLAN

Put three lines through one point and use a fourth transversal. Check all hypotheses and all four possible choices of three lines.

WORK

Take \[ L_1:\ x=0,\qquad L_2:\ y=0,\qquad L_3:\ y=x,\qquad L_4:\ x+y=1. \tag{1} \] Their directions are vertical, horizontal, slope $1$, and slope $-1$, so no two are parallel. The first three meet at $O=(0,0)$, and $O\notin L_4$. Thus no four lines are concurrent.

The intersections with the fourth line are \[ L_1\cap L_4=(0,1),\quad L_2\cap L_4=(1,0),\quad L_3\cap L_4=(1/2,1/2). \tag{2} \] These are distinct, and direct substitution shows that each lies on exactly the displayed two lines. They are the only ordinary intersection points, since the remaining three pairs all meet at $O$, which has multiplicity three.

There are only two types of three-line selections. The selection $\{L_1,L_2,L_3\}$ is concurrent and forms no triangle. Every other selection contains $L_4$ and two of $L_1,L_2,L_3$. Those three lines form a nondegenerate triangle, but its vertex at the intersection of the latter two lines is $O$, incident with three lines of the arrangement. Such a triangle is therefore not ordinary in the required sense. This exhausts all four selections.

VERIFICATION AND SCOPE

The concurrency of three lines is permitted by the hypothesis, which excludes only concurrency of four. The three ordinary points in $2$ lie on one line; connecting ordinary points is not the same as selecting three arrangement lines forming an ordinary triangle. The example has no parallel lines or hidden fourth line through $O$. It settles the universal yes/no question negatively, while leaving stronger requirements such as counterexamples at every $d$ outside this file's scope.

SOURCES

• Exact statement, known counterexample attributions, and discussion: https://www.erdosproblems.com/209.

CONCLUSION

Attempt status: solved by an explicit counterexample. The four lines in $1$ satisfy every hypothesis and admit no required triangle. No novelty is claimed. Completion rate estimate: 100%.