A third of the area

Answer: $\dfrac{1}{\sqrt3-1} = \dfrac{1+\sqrt3}2 =1.366$ to 3 d.p.

Using the relationship between the areas

The area of the small square is $\frac13$ of the area of the large square, so $$\frac{x^2}{(x+y{)}^2}=\frac{1}{3}\Rightarrow {\left(\frac{x}{x+y}\right)}^2=\frac{1}{3}\Rightarrow \frac{x}{x+y}={\displaystyle \frac{1}{\sqrt{3}}}$$

We can make $\dfrac x y$ the subject of this equation:

$$\begin{array}{rl}\frac{x}{x+y}& =\frac{1}{\sqrt{3}}\\ \Rightarrow \sqrt{3}x& =x+y\\ \Rightarrow \sqrt{3}x-x& =y\\ \Rightarrow (\sqrt{3}-1)x& =y\\ \Rightarrow \frac{x}{y}& =\frac{1}{\sqrt{3}-1}\end{array}$$

To see how to express this as a fraction with a whole number denominator, see the bottom of the solution.

Splitting the larger square into smaller squares and rectangles

Image

Green = $\frac13$ of total

$\therefore$ red + blue = green $\times$ 2

$\Rightarrow y^2+2xy=2x^2$

$\Rightarrow 1+2\dfrac xy=2\dfrac{x^2}{y^2}$

Let $\dfrac xy=w$:

$$\begin{array}{rl}1+2w=2w^2& \Rightarrow 2w^2-2w-1=0\\ & \Rightarrow w^2-w-\frac{1}{2}=0\\ & \Rightarrow {(w-\frac{1}{2})}^2-\frac{1}{4}-\frac{1}{2}=0\\ & \Rightarrow w-\frac{1}{2}=\pm \frac{\sqrt{3}}{2}\\ & \Rightarrow w=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\text{ take the positive root since }w>0\\ & \Rightarrow w=\frac{1+\sqrt{3}}{2}\end{array}$$

Splitting the larger square into trapeziums

Image

The area of each trapezium is $\frac12\left(\left(x+y\right)+x\right)y$

Green square occupies a third of the area

$\therefore$ the three shapes have equal area

$\frac12\left(\left(x+y\right)+x\right)y=x^2$

Multiplying both sides of this equation by $2$ and expanding gives $2xy+y^2=2x^2$.

$\dfrac xy$ can be found from this equation as shown above.