# Pythagorean Golden Means

## Problem

This question involves the sides of a right-angled triangle, the Golden Ratio, and the arithmetic, geometric and harmonic means of two numbers. Take any two numbers $a$ and $b$, where $ 0 < b < a $.

The arithmetic mean (AM) is $ (a+b)/2 $;

the geometric mean (GM) is $ \sqrt{ab} $;

the harmonic mean (HM) is $$ {1 \over {{1 \over 2}\left( {1 \over a} + {1\over b } \right)}}; $$

and the arithmetic mean is always the largest.

Show that the AM, GM and HM of $a$ and $b$ can be the lengths of the sides of a right-angled triangle if and only if $$ a = b\varphi^3, $$ where $ \varphi = {1\over 2}(1+\sqrt{5}) $, the Golden Ratio.

## Getting Started

<p>

See the article on <a href="http://nrich.maths.org/public/viewer.php?obj_id=1402&part=index"> Pythagorean Golden Means.</a></p>

## Student Solutions

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 $a=b\varphi^3.$