# Solution

In the arts, the golden rectangle, often considered to be the rectangle with the most aesthetically pleasing proportions, has the property that if you remove a square whose edge length matches that of the smallest side of the rectangle (shown in blue below), you are left with a smaller rectangle (shown in red) with the same proportions as the original one. That is to say $(a+b)/a = a/b$.

Find the value of $a/b$, dubbed the golden ratio and its reciprocal -- how are they related?

We know that

$$\frac{a+b}{a} = \frac{a}{b}$$

If we wish to solve for $a/b$, it would be helpful if this was the only unknown in the equation. That is to say, if $x=a/b$, can we rewrite our equation so that it is in terms of only $x$?

We can turn the $a$ in the denominator into an $a/b$ by dividing it by $b$. So that we don't alter the value of the expression on the left, let us do this to the numerator as well, yielding

$$\frac{a/b + 1}{a/b} = \frac{a}{b}$$

Now, we can replace each $a/b$ with $x$ as suggested above, and then simply solve for $x$

$$\frac{x+1}{x} = x$$

Clearing the fractions by multiplying by $x$, we have

$$x+1 = x^2$$

This of course is quadratic, so its solution is routine.

$$x^2-x-1=0$$

which yields

$$x = \frac{1 \pm \sqrt{5}}{2}$$

Knowing that $x = a/b$ is a ratio of side lengths, it must be true that $x>0$, so we can eliminate the negative solution found. Thus, the remaining solution must be the value of the golden ratio, $\varphi$, that we seek:

$$\varphi = \frac{1 +\sqrt{5}}{2}$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}