divine proportion Archives - Math and Multimedia

The Golden Ratio

August 19, 2011 GB High School Geometry, High School Mathematics

Problem: Find point $P$ on segment $AB$ such that AP:PB = AB:AP.

Solution:

If we let $r$ be the length of $AP$ and $s$ be the length of $PB$ , then the ratio above can be written as

$\displaystyle\frac{r}{s} =\frac{r+s}{r}$

Simplifying the equation we have $r^2 = rs + s^2$ which is equivalent to $r^2-rs - s^2$ , a quadratic equation in $r$ . So, we solve for r using the quadratic formula with $a = 1$ , $b = s$ and $c = -s^2$ :

$r = \displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-s \pm \sqrt{s^2-4(1)(-s^2)}}{2(10)}$ .

Simplifying this gives us $r = \frac{s + s\sqrt{5}}{2}$ . Dividing both sides by s, we have

$\displaystyle\frac{r}{s} = \frac{1 + \sqrt{5}}{2} \approx 1.61803398874989.$ » Read more

8 comments divine proportion, the golden ratio, the golden rectangle