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Monday, March 23, 2015

Problem of the Week

Happy Monday and Happy Spring! Unfortunately, it still doesn't feel like Spring in the Boston/Cambridge area. We at the Center still need to bundle up to our necks to make it across the street for lunch. Fortunately, I've got a Problem of the Week to share to distract us from the cold:

This problem is pretty easy. But you need to use mathematics to make sure you've exhausted all possible solutions- don't simply guess and check. Like usual, I've shared my solution below...

The solution is pretty easy to follow. We're simply looking for a rectangle where the sides have integer number lengths- and because you can't have a negative length, we can restrict our search to natural numbers greater than zero. The trickiest part of this solution was where I marked "creative step" with an arrow- factoring (2x) into 2(x - 2) + 4 was necessary to make the x exist only in the denominator below 4.

We were able to determine the three possible choices for the x-value because of the restriction that we are looking for natural numbers. (x - 2) was required to divide evenly into 4, so (x - 2) had to be equal to one of the divisors of 4. I then used each x value that I found to determine its y pair.

In conclusion, I found three answers, but two unique rectangles which have integer side lengths where the perimeter is equal to the area. The first is a rectangle with sides 3 and 6, and the second is a square with side length 4.

I'll be back next week with a calculus based Problem of the Week! Remember to comment if you have any questions about my method, and please share any alternate solutions!

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