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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 , show that it is possible to construct a set of points in the plane, let’s say , so that

no three points of the set are collinear, and for which there exist points , all lying on a straight line, and arranged so that on the line between any and any , some lies between them.

So no can “see” any , because there’s always some blocking the view. As the proposer observed, the problem would be easy if we had ‘s to play with, one for each pair . But here there are only ‘s, so some of them will have to do more than their fair share, blocking the view between quite a few -pairs simultaneously. Thus, you have to arrange the ‘s and ‘s so that a lot of the lines between them will be coincident at a point , 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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