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It’s been an awfully long time since I’ve posted anything; time to finally break the silence.

This problem appeared elsewhere on the internet some months ago; some of you may have already seen it. I don’t want to say right away where I saw it, because there was some commentary which included some rough hints which I don’t want to give, but I’ll be sure to give credit when the solution is published. I’ll bet some of you will be able to find a solution, and will agree it’s quite cute. Here it is:

Given integers $m, n \geq 1$, show that it is possible to construct a set of $m + n$ points in the plane, let’s say $x_1, \ldots, x_m, y_1, \ldots, y_n$, so that no three points of the set are collinear, and for which there exist points $z_1, z_2, \ldots, z_{m+n-1}$, all lying on a straight line, and arranged so that on the line between any $x_i$ and any $y_j$, some $z_k$ lies between them.

So no $x_i$ can “see” any $y_j$, because there’s always some $z_k$ blocking the view. As the proposer observed, the problem would be easy if we had $m n$ $z$‘s to play with, one for each pair $(x_i, y_j)$. But here there are only $m+ n-1$ $z$‘s, so some of them will have to do more than their fair share, blocking the view between quite a few $(x, y)$-pairs simultaneously. Thus, you have to arrange the $x$‘s and $y$‘s so that a lot of the lines between them will be coincident at a point $z$, and subject to the constraint I italicized above.

Please submit solutions to topological[dot]musings[At]gmail[dot]com by Friday, July 17, 11:59 pm (UTC); do not submit solutions in Comments. Everyone with a correct solution will be inducted into our Hall of Fame! We look forward to your response.

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