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What is the relationship between the arithmetic, geometric and harmonic means of two numbers, the sides of a right angled triangle and the Golden Ratio?

### Classical Means

Use the diagram to investigate the classical Pythagorean means.

# Pythagorean Golden Means

##### Stage: 5 Challenge Level:

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 $\varphi = {1\over 2}(1+\sqrt{5})$.

$$\begin{eqnarray} \varphi^2 &=& {1\over 4}(5+2\sqrt{5}+1) \\ \varphi^3 &=& {1\over 8}(1 + \sqrt {5})(6 + 2\sqrt{5}) \\ &=& {1\over 8}(16 + 8\sqrt{5}) \\ &=& 2 + \sqrt{5}. \end{eqnarray}$$

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

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

Let the ratio $a/b = x$ then

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

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

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