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]