PR 4 # 18

Wacky · 2006-06-18, 12:55

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

axn · 2006-06-18, 13:57

Pigeonhole principle

Wacky · 2006-06-18, 14:24

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

xilman · 2006-06-18, 15:14

Quote:

Originally Posted by Wacky

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

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!

Paul

Wacky · 2006-06-18, 15:29

The "official" answer is very similar to what has been decribed here.
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.

2006-06-18, 12:55	#1
Wacky Jun 2003 The Texas Hill Country 3²×11² Posts	PR 4 # 18 Given five points in or on a unit square, prove that at least two points are no farther than $sqrt2 / 2$ apart.

2006-06-18, 13:57	#2
axn Jun 2003 3²×5×113 Posts	Pigeonhole principle

2006-06-18, 14:24	#3
Wacky Jun 2003 The Texas Hill Country 3²·11² Posts	I think that you are on the right track. However, you need to supply more specifics in order to have a proof.

2006-06-18, 15:29	#5
Wacky Jun 2003 The Texas Hill Country 3²×11² Posts	The "official" answer is very similar to what has been decribed here. 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.