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Perimeter Equals Area (Posted on 2003-09-12) Difficulty: 3 of 5
The perimeter of a rectangle in units equals its area in units squared. (4,4), (3,6), and (6,3) are three possible pairs of lengths for this rectangle.

Give another pair of positive integral sides for this rectangle or prove why there isn't another pair.

  Submitted by Gamer    
Rating: 4.0000 (4 votes)
Solution: (Hide)
John Reid had another solution here. Here's my solution:

Other than the three examples given, there isn't another pair.

Creating a formula for one side in terms of the other and calling the sides a and b, you know that 2a+2b=ab. Adding -2b to both sides gives 2a=ab-2b. Factoring this gives 2a=(a-2)b, and dividing by (a-2) gives 2a/(a-2)=b.

Looking at this, you can rule out any values below 2 for a, as you will end up with a negative result for either a or b. 2 can be eliminated for a as it gives an undefined answer.

The only integers not greater than 4 but greater than 2 are accounted for. (3,6) and (4,4), and 6 is accounted for in (6,3). No odd numbers above 3 work for a, because you are dividing an even (2a) by an odd (a-2).

No even numbers over 6 work either for a. Because as a gets larger, b gets smaller, b must be less than 3 if a is greater than 6. But b can't ever be less than or equal to 2, because (a/(a-2)) is never going to quite equal 1-- it will always be more than 1. This means that 2a/(a-2) must be over 2.

Since b has to be less than 3 and greater than 2, it can't be an integer, because there are no integers less than 3 but greater than 2. This proves that both the sides of a rectangle can't be integers if they aren't (4,4), (3,6), or (6,3).

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutionproblem solutionK Sengupta2007-04-30 13:08:35
re: additional solution - questionableGamer2003-09-16 19:06:15
additional solution - questionableLorne Hrynkiw2003-09-14 20:16:44
solution without proofabc2003-09-12 19:57:51
re: Another way of thinking about thisGamer2003-09-12 19:17:37
re: Another way of thinking about thisSilverKnight2003-09-12 18:10:04
SolutionAnother way of thinking about thisJohn Reid2003-09-12 18:01:11
Solution (and spoiler) I thinkSilverKnight2003-09-12 14:07:39
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