Newsgroups: sci.math
From: kirill@topaz.kiev.ua  (Kirill E. Semenoff)
Subject: A Cute Task! (Re: Your favourite proofs)
Date: Mon, 18 Jul 1994 14:47:00 GMT

I propose you a little task, in which I was impressed by both 
the problem and the solution. Here you are:
    There are square regular grid on the infinite plane. The
    length of each cell is 1 unit (accordingly, it's square 
    is 1 sq.unit). Consider a blot on the plane that has 
    arbitrary shape and has ssquare less than 1 sq.unit.
    ---------------------------------------------------------
    Prove that the blot can be moved (with no rotation) on 
    the plane in such a way that it would not overlay any 
    node of the grid.
I know a beautiful solution for this task, which is, probably,
my favourite proof. I'll post it soon, but first try to solve
the problem. Good luck !!!

____________________________________________________

       Everything You've Done Is A Mistake         

   Kirill Semenoff, Topaz-Inform, Kiev, Ukraine    
____________________________________________________

==============================================================================

Newsgroups: sci.math
From: kfoster@rmii.com (Kurt Foster)
Subject: A Cute Task! (Re: Your favourite proofs)
Date: 18 Jul 1994 16:50:04 GMT

Kirill E. Semenoff (kirill@topaz.kiev.ua) wrote:
: I propose you a little task, in which I was impressed by both 
: the problem and the solution. Here you are:
:     There are square regular grid on the infinite plane. The
:     length of each cell is 1 unit (accordingly, it's square 
:     is 1 sq.unit). Consider a blot on the plane that has 
:     arbitrary shape and has ssquare less than 1 sq.unit.
:     ---------------------------------------------------------
:     Prove that the blot can be moved (with no rotation) on 
:     the plane in such a way that it would not overlay any 
:     node of the grid.

     Form a set T(S) in the region {(x,y) | each of x and y are in the 
interval [0,1) } by translating each point of S into that region.  
Clearly T(S) has area less than or equal to that of S (different points 
of S could translate to the same point in T(S) ).  Also, if a translate 
of S includes a node, the same translate of T(S) also will include a 
node.
     Now, since the area of S is less than 1, so is that of T(S).  
There is therefore a point (a,b) with 0 < a < 1 and 0 < b < 1 which is 
NOT in T(S). The translation of T(S) by (-a, -b) does not include the 
node (0,0) since (a,b) is not in T(S).  All the points (x-a, y-b) where 
(x,y) is in T(S) satisfy x-a in [-a, 1-a) and y-b in [-b, 1-b), so this 
translate of T(S) does not include any node points other than (0,0).  The 
translation of S by (-a, -b) will therefore not include any nodes either.
     Standard stuff for the "geometry of numbers".

==============================================================================

Newsgroups: sci.math
From: kfoster@rmii.com (Kurt Foster)
Subject: A Cute Task! (Re: Your favourite proofs)
Date: 19 Jul 1994 01:54:54 GMT

Kurt Foster (kfoster@rmii.com) wrote:
:      Form a set T(S) in the region {(x,y) | each of x and y are in the 
: interval [0,1) } by translating each point of S into that region.  
: Clearly T(S) has area less than or equal to that of S (different points 
: of S could translate to the same point in T(S) ).  Also, if a translate 
: of S includes a node, the same translate of T(S) also will include a 
: node.

     I forgot to specify that the points of S are translated into T(S) by 
a node - in fact, you subtract the greatest integer from each coordinate.
  ^^^^
[deletia -- djr]