Pythagorean golden means

Freddie Manners, age 11 from Packwood Haugh School, Shropshire sent in the following beautiful solution. Freddie asks ``Is this relationship to the Golden Ratio coincidental?'' The answer is probably not. Mathematics if full of connections which at first seem surprising. The question involves the sides of a right-angled triangle, the cube of the Golden Ratio $\varphi = {1\over 2}(1+\sqrt{5})$, and the arithmetic, geometric and harmonic means of two number (AM, GM and HM respectively). Firstly Freddie found the cube of $\varphi = {1\over 2}(1+\sqrt{5})$.

$$\begin{array}{rcl}{\phi }^2& =& \frac{1}{4}(5+2\sqrt{5}+1)\\ {\phi }^3& =& \frac{1}{8}(1+\sqrt{5})(6+2\sqrt{5})\\ & =& \frac{1}{8}(16+8\sqrt{5})\\ & =& 2+\sqrt{5}.\end{array}$$

Take any two numbers $a$ and $b$, where $0< b< a$. Because the AM is the largest we have

$$\begin{array}{rcl}{\left(\frac{(a+b)}{2}\right)}^2& =& ab+\frac{1}{{(\frac{1}{2}(\frac{1}{a}+\frac{1}{b}))}^2}\\ & =& ab+\frac{(2ab{)}^2}{(a+b{)}^2}\\ \frac{(2ab{)}^2}{(a+b{)}^2}& =& {\left(\frac{(a+b)}{2}\right)}^2-ab\\ & =& {\left(\frac{(a-b)}{2}\right)}^2\\ \frac{2ab}{(a+b)}& =& \frac{(a-b)}{2}\\ 4ab& =& a^2-b^2\\ \frac{4a}{b}& =& {\left(\frac{a}{b}\right)}^2-1.\end{array}$$

Let the ratio $a/b = x$ then

$$\begin{array}{rcl}4x& =& x^2-1\\ x^2-4x-1& =& 0\\ x& =& 2\pm \sqrt{5}\end{array}$$

As $\sqrt 5 > 2$ the solution $2-\sqrt5$ would give a minus number.

So $a/b = 2 + \sqrt5 = \varphi^3$ and $a=b\varphi^3.$