Computing a Geometrical ratio $\frac{a}{b}$.

Question

$XYZ$ is an equilateral triangle as shown on the image below.

The aim is to find the ratio $\frac{a}{b}$.

So far from the picture, it is easy to see that $b= \frac{YZ}{2}$.

Does anyone have an idea on how to find $b$ in term of $a$?

It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.

Answer 1

Another way to find the ratio is to solve the $\triangle ROS$, for which we know two sides and an angle:

\begin{align} \triangle ROS:\quad |OR|&=r=\tfrac13\cdot2b\cdot\tfrac{\sqrt3}2 =\tfrac{\sqrt3}3\,b ,\\ |OS|&=2r=\tfrac{2\sqrt3}3\,b ,\\ \angle SRO&=\angle ZRO+\angle SRZ=90^\circ+60^\circ =150^\circ , \end{align}
so we can apply the cosine rule to get

\begin{align} |OS|^2&=|OR|^2+|RS|^2-2\cdot|OR|\cdot|RS|\cos\angle SRO ,\\ \tfrac43\,b^2&= \tfrac13\,b^2+a^2-2\cdot\tfrac{\sqrt3}3\,b\cdot a\cdot\cos150^\circ ,\\ \tfrac43\,b^2&= \tfrac13\,b^2+a^2+2\cdot\tfrac{\sqrt3}3\,b\cdot a\cdot\cos30^\circ , \end{align}

\begin{align} b^2 -b\cdot a -a^2 &=0 ,\\ (\tfrac{b}a)^2 - \tfrac{b}a -1 &=0 , \end{align} which gives the answer (positive root) as

\begin{align} \frac{b}a &=\frac{1+\sqrt{5}}{2} , \end{align}

which is indeed the Golden ratio.

Answer 2

Use the power of the point $R$ (or $Q$) with respect to the big circle.:

$$ (a+b)\cdot a = s^2$$

where $s= YZ/2$.

Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0\implies q={1\pm \sqrt{5}\over2}$$