PR 4 # 18
 

Given five points in or on a unit square, prove that at least two points are no farther than [tex]sqrt2 / 2[/tex] apart.

[spoiler]Pigeonhole principle[/spoiler]

I think that you are on the right track. However, you need to supply more specifics in order to have a proof.

[QUOTE=Wacky]Given five points in or on a unit square, prove that at least two points are no farther than [tex]sqrt2 / 2[/tex] apart.[/QUOTE]
[spoiler]Note that 1 > sqrt2 / 2, so four points can be placed at the corners of the
square, to give a mimimum separation of 1. These points are as far as possible
from the centre. The centre is sqrt2 /2 from each of the corners, so place the
fifth point there. This proves the existence of a solution with the property required.

Proving it is the unique and maximal solution is another matter entirely![/spoiler]

Paul

The "official" answer is very similar to what has been decribed here.
[spoiler]Draw two lines through the center of the square, perpendicular to each other, such that each is parallel to a side of the unit square. These two lines partition the unit square into four 1/2 unit squares. At least two of the 5 points must be in (or on the perimeter of) one of these smaller squares. This pair of points cannot be farther apart than the length of the small square
s diagonal, [tex] sqrt2 /2 [/tex] units.[/spoiler]