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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July 14, 2009 at 9:14 am

basil mohammedThanks for the problem

I could only find one possible reason for the relevance of the number m+n-1 in the problem. That is let z_k be the point which blocks x_i from y_k-i. There only m+n-1 possible values for k. But the surprise is that when I started constructing such a sequence of points starting from x_1 (i used geogebra). we find that some z_i s automatically block some pairs of points which we wish to block.

July 14, 2009 at 10:10 am

Todd TrimbleBasil,

If you believe you have a solution, then by all means send it in. [I ask all readers not to give away any hints in comments; requests for clarification are of course fine (I don’t think any hints were given away here).]

I should probably think about this more carefully, but I’m guessing the true significance of is that you can’t do any better than that. That is, if you have points in the plane with no three collinear, then you will need at least points to block every from every .

July 14, 2009 at 10:45 am

basil mohammedNO it works well, I tried one hour on geogebra. Also used reflection on the line containing z_i to get X_i from Y_i and vice versa. it should work . now the only job i have to do is to find a sequence of initial points from which i can construct the remaining.

how to use latex here?

July 14, 2009 at 12:11 pm

Todd TrimbleDamn it, do

NOTdiscuss your attempts at solution here!! I just finished saying that!!Send your solution to our gmail address, as indicated in the problem statement. (Replace [dot] by a period sign, and replace [at] by the ‘a’ with a circle around it.) Your solution doesn’t have to be in latex. But we will NOT discuss it here in comments until I post the solution, and any attempts to do so will be ruthlessly removed.

July 14, 2009 at 12:24 pm

basil mohammedsorry

July 20, 2009 at 7:34 am

basil mohammedplease post the solution

July 20, 2009 at 2:10 pm

Todd Trimblebasil, it’s coming; I’m waiting for permission to quote from someone. I hope to post the solution later today.

July 22, 2009 at 11:08 pm

Solution to POW-13: Highly coincidental! « Todd and Vishal’s blog[…] | Tags: elliptic curves, pascal's mystic hexagon | by Todd Trimble Huh — no solutions to POW-13 came in! I guess I was surprised by […]