Well,

It has been a long break on this blog, but I am back now and I hope to be regular posting here now on. Anyways, this problem has a long history but recently it was brought to me by sagarmoy .

**Problem : **Let there be a rectangle of type-A and type-B. type-A rectangle have at least one integral side ( that is length of the side is integer ) and type-B rectangle have no side of integral length.

1) Prove that from a given instance of a type-A rectangle, it is impossible to tile up these rectangle and make a big type-B rectangle

2) Prove that there is no way of tiling up (any) type-A rectangles to make a big type-B rectangle.

**Solution : **

It is easy to see tha version 1 of the question is a special case of version 2.

1) we have a specific type-A rectangle. so let it have two sides i ( integral ) and r (rational )

Suppose we have made a type-B rectangle using this type-A rectangle. Then the sides of type-B rectangle is

m1i + n1r, m2i + n2r , where m1,m2,n1 and n2 are positive integers. Area of the big rectangle should be equal to the area of the sum of the smaller rectangles.

where, is the total number of tiles used to form bigger rectangle.

r being a rational can be written in a form p/q where p and q are integers which are relatively prime. So

Multiplying by q

Right hand side of the equation is an integer. Also, first and second term of the left hand side is also clearly an integer. To make the equation correct q has to divide n1 or n2 because it can not divide p ( remember that p and q are relatively prime ). Without loss of generality, let us say q divides n1 then the side with length

m1i + n1(p/q ) becomes an integer ( as q divides n1 ).

2) Frankly speaking, none of us ( me, deepanjan and sagarmoy ) could come up with a proof for this second case. However, an interesting compilation of 14 different proof of this version is available at :

http://www.jstor.org/stable/2322213?seq=1

Those of you who are not subscribed to this journal, I am writing down one proof mentioned in that link which I found as most elegant.

Let our big rectangle R be aligned in a way such that bottom-left corner co-incide with (0,0), and the sides are aligned with x and y axis. Let T be the set of constituent tiles ( of type A ).

Let S = { (x,y) | x and y are integers, (x,y) is a corner of some tile in T }

We can draw a bipartite graph now with S and T forming two partitions of this graph. There is an edge from to if and only if tile t is having s as one of its corner. As all tiles in T are of type-A ( one integer side ), all tiles should have either 0,2 or 4 neighbours in S. Total number of edges between S and T has to be even. Each point in S which is not a corner of R will have 2 or 4 neighbours in T.

Note that (0,0) has only one neighbour in T, to make total number of edges even, there has to yet another point in s which lies on odd number of tiles. Clearly, this point s has to be another corner of R ( so as to have only one neighbour in T ). If R has any other corner apart from (0,0) in S then R must have width or height of integer size.

There is one proof mentioned in the linked article which works like a magic, comes out of thin air. I won’t call that proof very elegant, but it really requires imagination to come up with such a proof. But I guess, I will mentioned that proof in some alter post.

Tags: math, rectangle tiling theorem, theorem

July 31, 2008 at 9:21 am |

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