## Challenge #5 Solution

Okay, time for a solution to Challenge #5

To recap: Suppose you have a rectangle which is one unit tall and has this special property: if you cut off a square piece from the end of the rectangle, you’re left with a smaller rectangle that has the same proportions as the original rectangle. How long is the original rectangle? Well, name and conquer — let’s start by calling the length of the rectangle x (since, um, x is short for “length”). This corresponds to the question mark in the picture above. The height of the rectangle, we are told, is 1. Now, how about the small blue rectangle? Its “width” (which is vertical, since the rectangle has been turned sideways relative to the big blue rectangle) is obviously 1; and its “height”? It’s not too hard to see that the height is (x – 1). OK, we’re almost there: what else do we know? Well, we know that the small blue rectangle is just a scaled-down copy of the big blue rectangle; that is, the two rectangles are proportional. So we can set up a proportion as follows: $\displaystyle \frac{1}{x} = \frac{x - 1}{1}$

Can you solve for x from here without reading the rest of the solution?

To solve this equation we can first multiply both sides by x, which gives us the equation $1 = x^2 - x$. Moving the 1 to the other side yields $x^2 - x - 1 = 0$. Hmm… haven’t we seen this equation somewhere before!? That’s right, the width of the rectangle is the golden ratio, $\phi = \frac{1 + \sqrt{5}}{2}$! (We want the positive solution of the equation, although it is interesting to think about what could be meant by a rectangle with negative width). Astounding. I’ll leave it to you to think about what the connection is between infinite continued fractions and recursively defined rectangles. 