A fun Fibonacci number surprise with a 1, 2, Sqrt[5] right triangle

My son had a little trouble with this problem from the 2009 AMC 10 B yesterday:

I should him how to solve the problem a few different ways – including by folding!

One of the ways I talked about was finding a rough answer by approximating by the fraction 9/4. We found this number by looking at a calculator and seeing that

As I thought about the problem more last night, I realized that 9/4 is part of the continued fraction approximation for . The first couple of approximations that you find using continued fractions are:

2, 9/4, 38/17, and 161/72.

If I approximate the hypotenuse of the original right triangle with those numbers, I get the following approximations for the length of BD, which are all ratios of consecutive Fibonacci numbers:

If you are familiar with continued fractions and especially the continued fraction approximations for the golden ratio, the emergence of the Fibonacci numbers probably isn’t a huge surprise. I missed it the first time, though, and think that students might really enjoy seeing this little Fibonacci surprise.