Given a finite collection of lines $L_1,\dots,L_m$ in ${\bf{R}}^2$, let $R_1,\dots,R_n$ be the connected components of ${\bf{R}}^2 \setminus (L_1 \cup \dots \cup L_m)$, and say that $\{L_1,\dots,L_m\}$ is *triangulating away from infinity* iff every $R_j$ that is bounded is triangular.

Can every finite line-arrangement be augmented (by adding more lines) to yield one that is triangulating away from infinity?

(The old and new arrangements need not be generic; they may have parallel lines and/or multiple intersections.)

Some background: A slightly different (and, in my view, less natural) question was raised by me about twenty years ago; it has never been answered either. That problem first appeared in Richard K. Guy and Richard J. Nowakowski’s American Mathematical Monthly column on open problems (see “Bite-Sized Combinatorial Geometry Problems”, AMM, Vol. 103, No. 4, Apr. 1996, p. 342 and “Monthly Unsolved Problems”, AMM, Vol. 104, No. 10, Dec. 1997, pp. 969-970). Stan Wagon invited the readers of his on-line column to completely solve a one-parameter class of problems of this kind in which three lines symmetrically divide an equilateral triangle; see Link .

A related problem that might be simpler to solve in the negative is a version in which we not only require that all bounded components have 3 sides but also require that all unbounded components have 2 or 3 sides. If nobody can solve my original problem affirmatively or negatively, a negative solution to this problem would deserve the bounty.

(Historical aside: Many years ago I proposed the original version of this problem to John Conway. He said “Let me think about that,” and as the curiosity-bug bit into his brain, he began to draw pictures, make observations, and formulate conjectures. But he was no fool; he could see that I was deliberately trying to entice him into working on the problem, and he was too proud to want to be seen as one who is so easily seduced. He shuddered as if shaking off an unpleasant memory and said “You know, I don’t have to work on just ANY damned problem!”)

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