BMAD 63 Report post Posted February 27, 2014 There are 6 points in a rectangle with the sides, 3 and 4. Prove that the distance between at least two of these points is smaller than the square root of 5. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted February 28, 2014 Square root of five screams out for an hypotenuse, and in this puzzle, that's what it is. For two points inside a rectangle their greatest separation is the diagonal. For a 1 x 2 rectangle the diagonal is sqrt(5). And in a 3 x 4 rectangle, there are exactly six such rectangles. So a weaker, immediate result applies to seven points: By the pigeon hole principle, at least two of the seven would be in the same 1 x 2 rectangle and thus be closer than sqrt(5). But the result holds also for six points, as shown in the figure. Optimally placing six points, all on ends of diagonals, the vertical pairs are only 2 units apart. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted February 27, 2014 (edited) I place two points on the diagonal end points. I place two more a distance x from them along the long edges. I place the final two a distance x from them along the short edges. Requiring nearest points of type 2 and 3 to have a spacing of x givesx^{2} - 14x + 25 = 0 or x = 2.10102, somewhat less than 5^{1/2}. Edit: I suspect there is a configuration where the minimum distance is 5^{1/2}. Edited February 27, 2014 by bonanova Add comment Share this post Link to post Share on other sites

There are 6 points in a rectangle with the sides, 3 and 4. Prove that the distance between at least two of these points is smaller than the square root of 5.

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