Perimeter Equals Area (Posted on 2003-09-12)

	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).

	Subject	Author	Date
	Better late than never	Larry	2020-10-31 13:07:29
	problem solution	K Sengupta	2007-04-30 13:08:35
	re: additional solution - questionable	Gamer	2003-09-16 19:06:15
	additional solution - questionable	Lorne Hrynkiw	2003-09-14 20:16:44
	solution without proof	abc	2003-09-12 19:57:51
	re: Another way of thinking about this	Gamer	2003-09-12 19:17:37
	re: Another way of thinking about this	SilverKnight	2003-09-12 18:10:04
	Another way of thinking about this	John Reid	2003-09-12 18:01:11
	Solution (and spoiler) I think	SilverKnight	2003-09-12 14:07:39