Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Interesting Problem Again
Replies: 1   Last Post: Nov 21, 2002 8:38 PM

 Messages: [ Previous | Next ]
 Leroy Quet Posts: 422 Registered: 12/13/04
Re: Interesting Problem Again
Posted: Nov 21, 2002 8:38 PM

On 21 Nov 2002, Manimaran wrote:
>Hi I did post a message before but it is not appearing properly.
>
>I am repeating the problem here.
>
>Let us take a rectangle having sides l, b
>
>l > b
>
>let us cut off a perfect square out of this rectangle. so the
>remaining will be a rectangle of sides b, (l-b)
>
>lets us cut off a perfect square out of this rectangle. we will be
>left with a rectangle of sides (l-b), (2b -l)
>
>iteration 1 : sides (l,b)
>iteration 2 : sides (b, l-b)
>iteration 3 : sides (l-b, 2b-l)
>.....
>....
>iteration n : sides (function(l,b,n)???, function(l,b,n)
>
>At what point the left out rectangle becomes a square and no more
>iterations is possible?
>
>iteration end : sides (a,a) [a = ?] and iteration number of end = ?
>
>Is there a series for this problem? Anyone tried to solve this?
>This seems to be interesting to me...
>Since the process we do looks repetitive to the rectangle,
>there may be a way to solve these two problems
>as functions of the sides of the given triangle...
>
>Any takers?

First, this has not appeared on Google yet. If you posted to Math Forum, and they haven't fixed their problems yet, then your post (and my reply) will NOT leave the Math Forum system.

I will assume that you mean that one side is 1, not the lower-case L.
In any case, this does not really matter.
I'll just call one side "a" and the other "b" (b as before),
where a and b are both positive integers, and
a > b.
If the sides are in irrational proportion (ie. no a and b = integers exist) then your subdividing can be done forever without reaching a square dead-end.

And also, your assumption is that a and b are in a range of
proportions. It is possible that more than one square can be taken
from a given rectangle: example
(2 squares removable)
--------------------
| | | |
| 1 | 2 | |
| | | |
--------------------

But, as far as I know, this problem comes down to continued fractions.

Take the continued fraction of (a/b).

a/b = c(1) +
1/(c(2) + 1/(c(3) +....+1/(c(m-1) +1/c(m))...))

where the c()'s are positive integers.

Each c(k) is the number of squares that can be taken from the k_th rectangle.

So, the number of iterations is m, the number of terms in the continued fraction.

And the sum of the c()'s is the total number of squares to be taken out.

As an example of an irrational ratio of side to adjacent side.
Let a side be 1, and the other be the golden ratio
(sqrt(5)+1)/2.
Then the continued fraction is made up of an infinite number
of 1's.
So the rectangle can be subdivided by removing one square at a time,
an infinite number of times.
This rectangle subdivision is well-known.

Thanks,
Leroy Quet