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/
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