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Points in a rectangle

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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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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).

post-1048-0-03811800-1393564413_thumb.gi

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.

Edit:

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

Edited by bonanova
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