Suppose 6 points are placed in a 3x4 rectangle. Prove that there are 2 of them at distance of =<sqrt(5) from each other

Any help is appreciated

i suspect this has to do with the pigeon hole principle. maybe starting at a corner, and construct some shape whose distance from the corner is sqrt(5). repeat for each corner. and ultimately show that there are 5 such shapes. so a 6th would have to share one of the holes.

something to that effect. might get you started.

Alright thanks for the help, ill try and see if i can figure this out

Yes this will be an easy problem to solve using the Dirichlet's box principle or as carteblanche stated, the pigeon hole principle.

i believe if you construct a pentagon of width 2 and height 2. something like
/ \
|_ _|

then you can fit 3 of these and 2 other pentagons of half the size as the one above.

It is easy to prove that any length inside of the pentagon is less or equal to sqrt5

edit: wrote bullshit


This post was edited by HbSoe on Oct 7 2014 04:20am

try the following:



the rectangle is divided into 5 areas, since there are 6 points at least 2 are in the same area.
2 points inside the same area can't be at a distance > sqrt(5) from each other

That picture is incorrect because within your large pentagon there is a length of sqrt8, therefore the Dirichlet's box principle cannot be used.
However something like

would work

This post was edited by cdexswzaq on Oct 7 2014 02:50pm

winner