Equal Area and Perimeter: Rectangles

Date: 09/09/2001 at 13:56:35
From: Jessica
Subject: Area and perimeter

Hi!

There are only two rectangles whose area is exactly the same as their 
perimeter if the dimensions of each are whole numbers. What are the 
dimensions?

Thanks,
Jessica

Date: 09/10/2001 at 12:46:10
From: Doctor Ian
Subject: Re: Area and perimeter

Hi Jessica,

Let's look at a rectangle for a moment:

    a
  +------------+
  |            |
  |            | b
  +------------+

The area will be a*b; the perimeter will be 2a + 2b.  So we want to 
find pairs of numbers (a,b) such that

  a*b = 2a + 2b

More specifically, we want to find pairs of whole numbers for which 
this equation is true.  One simple way to attack a problem like this 
is to start making a table:

   a    b    a*b   2a+2b
  ---  ---   ---   -----
   0    0     0      0
   0    1     0      2
   1    0     0      2

So we've found one solution already.  A 0 by 0 rectangle has an area 
equal to its perimeter.  

(This seems kind of strange, I know, but it fits the definition. And
note that the problem explicitly says that the numbers must be 'whole
numbers', not 'counting numbers'.  The only difference between
counting numbers and whole numbers is that the whole numbers include
zero. So whenever you see a problem that talks about 'whole numbers',
that's a tip that zero might turn out to be important.)
   
Now, the tough part about constructing a table like this is to make 
sure that you cover all the possibilities. How can we guarantee that 
we get all the possible (a,b) pairs?  What if we miss one, and that 
turns out to be the answer? We could keep looking forever.

Well, here is a trick for ordering these pairs:

   |
   10  11   12   13
   |
   5    6    7   14
   | 
   2    3    8   15
   |
   1----4----9---16---

The coordinates of the points, in order, are:

   1:  (0,0)

   2;  (0,1)
   3:  (1,1)
   4:  (1,0)

   5:  (0,2) 
   6:  (1,2)
   7:  (2,2)
   8:  (2,1)
   9:  (2,0)
   
  10:  (0,3)
  11:  (1,3)

and so on. Do you see why this will get _every_ pair of whole numbers? 

Of course, it will get some of them twice (for example, (2,1) and 
(1,2) actually describe the same rectangle), which creates some extra 
work. (I included both (0,1) and (1,0) in the table above before I 
realized this. I didn't need to work out both of them.)

If we want to be lazy (which in math is often a good thing), we can 
just look at the points above the diagonal:

   |
   7    8    9   10
   |
   4    5    6   
   | 
   2    3    
   |
   1------------------

(Do you see why this works?)

If we want to be even lazier, we can note that since 

  a*b = 2a + 2b

      = 2(a + b)

the product a*b must be an even number. This is a break, because it 
means that we can ignore any (a,b) pair in which both a and b are odd.  
So we can forget about possibilities like (1,3), or (3,5), since there 
is no way to multiply two odd numbers to get an even number. 

Anyway, now you can make a table, which will let you find the other 
answer to the problem. 

Now, when making a table like this, there are two possibilities:

  1. You'll run across the answer very quickly. If that happens, 
     then using a table will turn out to have been a good idea.

  2.  You won't run across the answer very quickly.  If that happens,
      there are two more possibilities. 

      a.  You'll notice some pattern in the table that will 
          tell you where the answer has to be, without your 
          having to construct all the table entries in between. 
          If that happens, then using a table will turn out to 
          have been a good idea.

      b. You won't notice any pattern like that. In that 
         case, you'll want to start looking around for another 
         way to solve the problem. 

Fortunately, in this case, you should come across the answer pretty 
quickly, if you use both of the shortcuts that I mentioned. 

Does this help?  Write back if you'd like to talk about this some 
more, or if you have any other questions. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/