# More On The Rectangle Puzzle

Last week, I posted a solution to a puzzle – how can you find a rectangle whose area equals its perimeter?

This week, I’ll post a simpler solution.

Again, we need a bit of algebra, but the algebra will be simpler this time. I’ll call the sides of the rectangle A and B, but instead of inventing a new variable S to equal the area and perimeter, I’ll make T equal to the ratio A/B.

Now, this is a bit nonintuitive. After all, this puzzle doesn’t seem to have anything to do with the ratio of the sides. However, let’s carry on.

• If T is A/B, then A is BT.
• The area AB, that is, B2T.
• The perimeter is 2A + 2B, that is, 2BT+2B, that is, B times 2T+2.
• Since the area equals the perimeter, this means B2T equals B(2T+2).
• If I cancel out B, I get BT = 2T+2

But BT is A, so I’ve instantly got a formula for A. Specifically, A = 2T + 2. Dividing this by T gives me B = 2 + 2/T.

So, I have another puzzle solution factory – much simpler, you’ll agree, than last week’s. The puzzle solution factory works like this:

• Choose a value of T,
• Work out 2 + 2T and 2 + 2/T
• Those are the sides of the rectangle!

Let’s try it.

• Choose T=1
• 2+2T = 4 and 2+2/T is also 4.
• So the 4×4 square has perimeter equal to its area!

And again…

• Choose T=4
• 2+2T is 10, and 2+2/T is 2.5.
• So, a 2.5 x 10 rectangle has area equal to its perimeter!

An obvious moral of this little story is that there’s often more than one way to solve a math problem, and some ways are harder than others. But we kind of knew that already. Next week, I’ll talk about a much more powerful thing this little puzzle teaches us about math.

Stay tuned!