Description: Circles contain an equilateral triangle, a square and a pentagon. Find the ration of XZ to XY in each diagram.
golden solutions

Taking the radius of the circle as 1 unit, in the equilateral triangle

OY = \frac{1}{2}

XP = PY = \frac{\sqrt{3}}{4}

and

OP = \frac{1}{4}

. Then, by Pythagoras Theorem,

PZ = \frac{\sqrt{1} 5}{4} .

XZ = \frac{\sqrt{3}}{4} (1 + \sqrt{5}) .

Hence

\frac{XZ}{XY} = \frac{1}{2} (1 + \sqrt{5})

which is the golden ratio.

Taking the side of the square as one unit and using Pythagoras Theorem, the raduis of the circle is given by $OS = \frac{\sqrt{5}}{2}$ . Hence $XZ = \frac{1}{2} (1 + \sqrt{5})$ and

$\frac{XZ}{XY} = \frac{1}{2} (1 + \sqrt{5})$

which is the golden ratio again.

Take the side of the pentagon as one unit. Then $XY = 1$ because it is in the isosceles triangle with angles $72^{o}, 72^{o}$ and $36^{o}$ . Let the chord length $XZ = x$ , then the ratio of the long to the short side in this triangle is $\frac{1}{x - 1}$ . The triangle $XZW$ is similar and the ratio of the long to the short side is $\frac{x}{1}$ . Hence we have

$\frac{x}{1} = \frac{1}{x - 1}$

which gives the quadratic equation

$x^{2} - x - 1 = 0$

and, as $x$ must be the positive root of this equation, this gives $x = \frac{1}{2} (1 + \sqrt{5})$ . Hence

$\frac{XZ}{XY} = \frac{1}{2} (1 + \sqrt{5})$

which is the golden ratio yet again. T