### Gold Again

Without using a calculator, computer or tables find the exact values of cos36cos72 and also cos36 - cos72.

### Golden Triangle

Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.

### Golden Powers

You add 1 to the golden ratio to get its square. How do you find higher powers?

# Pythagorean Golden Means

##### Age 16 to 18Challenge 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 $a=b\varphi^3.$