Pythagorean Golden Means
Age 16 to 18
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 $a=b\varphi^3.$