2230.22 – The Golden Ratio

Tags:Problem Set 10Algebra/TrigonometryArithmetic: Q & R

You see here a rectangle with a special shape. The shape has the property that if you make a square at the bottom and cut it off, the remaining rectangle is similar to the original rectangle. That is,

lx=xx–l.\frac{l}{x} = \frac{x}{x – l}.

Find the ratio of xx to ll. This is a famous number, known as the golden ratio. Rectangles of this shape are frequently used in architecture.

Solution

First,

lx=xl–x⇝l(l–x)=x2⇝l2–lx–x2=0\frac{l}{x} = \frac{x}{l – x} \leadsto l(l – x) = {x^2} \leadsto l^2 – lx – x^2 = 0

We want xl\frac{x}{l}, so we divide by l2l^2:

1–xl–x2l2=0⇝(xl)2+xl–1=0.1 – \frac{x}{l} – \frac{x^2}{l^2} = 0 \leadsto \left(\frac{x}{l}\right)^2 + \frac{x}{l} – 1 = 0.

We then invite the quadratic formula to the show:

xl=−1±(1−(−4)2=−1±√52.\frac{x}{l}=\frac{-1±\sqrt{(1-(-4)}}{2}=\frac{-1±√5}{2}.

We take the positive root:

xl=−1+√52≈.06180.\frac{x}{l} = \frac{-1 + √5}{2} ≈ .06180.

Coincidentally, the ratio of 1km1mile≈.06214\frac{1km}{1 mile} ≈ .06214.