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Points in a rectangle



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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.

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  1. I place two points on the diagonal end points.
  2. I place two more a distance x from them along the long edges.
  3. 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 gives

x2 - 14x + 25 = 0 or x = 2.10102, somewhat less than 51/2.


I suspect there is a configuration where the minimum distance is 51/2.

Edited by bonanova
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