The Golden Ratio

Consider a square with side length $x$ and a rectangle, call it $R_1$, with side lengths $x$ and $y$ with $y < x$. Suppose that we push this square and rectangle together so that they form a new rectangle, call it $R_2$, with side lengths $x + y$ and $x$ as illustrated below:

Consider the following problem: For what values of $x$ and $y$ make the ratio of the larger side length to the shorter side lengths of rectangles $R_1$ and $R_2$ equal? This ratio for $R_1$ is given by $\displaystyle{\frac{x}{y}}$, and this ratio for $R_2$ is given by $\displaystyle{\frac{x+y}{x}}$.

Let $\displaystyle{\phi = \frac{x}{y}}$, that is, $\phi$ will denote the ratio of $R_1$. If $R_2$ is to have this same ratio then $\displaystyle{\phi = \frac{x+y}{x}}$. Hence:

(1)

\begin{align} \quad \phi = \frac{x +y}{x} = 1 + \frac{y}{x} = 1 + \frac{1}{\phi} \end{align}

Therefore $\displaystyle{\phi = 1 + \frac{1}{\phi}}$ which is equivalent to the quadratic equation $\phi^2 - \phi - 1 = 0$. We can now solve for the ratio $\phi$ by using the quadratic formula with $a = 1$, $b = -1$, and $c = -1$:

(2)

\begin{align} \quad \phi &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{1 \pm \sqrt{1 + 4}}{2} \\ &= \frac{1 \pm \sqrt{5}}{2} \end{align}

So there are two values for the ratio $\phi$. The positive solution is given the special name known as the Golden Ratio.

Definition: The Golden Ratio is the number $\displaystyle{\phi = \frac{1 + \sqrt{5}}{2}}$ and is one of the solutions to the quadratic equation $x^2 - x - 1 = 0$. The other solution is denoted $\displaystyle{\Phi = \frac{1 - \sqrt{5}}{2}}$.

The golden ratio may seem somewhat arbitrary but in fact, it pops up frequently in nature, art, infrastructure, human anatomy, etc…

The golden ratio can be derived in many ways. One way involves the Fibonacci sequence and is given as an example on this page. We will now examine some nice properties of the golden ratio.

Theorem 1: $\phi - 1 = -\Phi$.

• Proof: We have that:

(3)

\begin{align} \quad \phi - 1 &= \frac{1 + \sqrt{5}}{2} - 1 \\ &= \frac{1 + \sqrt{5}}{2} - \frac{2}{2} \\ &= \frac{-1 + \sqrt{5}}{2} \\ &= - \left ( \frac{1 - \sqrt{5}}{2} \right ) \\ &= - \Phi \quad \blacksquare \end{align}

Theorem 2: $\displaystyle{\frac{1}{\phi} = -\Phi}$.

• Proof: We have that:

(4)

\begin{align} \quad \frac{1}{\phi} &= \frac{2}{1 + \sqrt{5}} \\ &= \frac{2(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})} \\ &= \frac{2 - 2\sqrt{5}}{-4} \\ &= -\frac{1 - \sqrt{5}}{2} \\ &= -\Phi \quad \blacksquare \end{align}

By combining theorems 1 and 2 above we see that $\phi^2 - \phi - 1 = 0$ which verifies that indeed, $\phi$ is a solution to $x^2 - x - 1 = 0$.