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Rosebush Puzzle 2*: Solution More solutions from Sean Peters Here are a two more solutions to the ten points in 5 rows of 4 problem. And here's an observation: no point can lie on three (or more) of the rows. Consider that if it does, then you have used all ten points to construct those lines, call them A, B, and C: Assume there is another row of 4 points, call it D. By the pigeonhole principle, it must contain at least 2 points from one of the 3 lines A, B, or C. Without loss of generality, assume that D contains at least 2 points that lie on A; then A and D are collinear, a contradiction. Note that the illustration above isn't the only way that 3 lines can all share a point, but the logic still holds for all cases. I have strong heuristic evidence that these 6 are the only solutions but I haven't yet written a rigorous proof. If I do complete a proof, you will be among the first to know. - Sean Peters Back to the puzzle

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