Don't be greedy

$\begin{aligned} area(XBY) & = area(AXYC) \\ 2\cdot area(XBY) & = area(ABC) \\ \frac{area(ABC)}{area(XBY)} &=2 \\ \therefore \frac{AB}{BX} & =\sqrt{2} \\ \frac{AX}{AB} & =\frac{AB-BX}{AB}=1-\frac{1}{\sqrt{2}} \\ \Rightarrow 1-0.7071 & =0.2928\sim \boxed{0.293}\end{aligned}$

First, let's find the area of a general equilateral triangle given the base length $b$ . We know that all angles in an equilateral triangle are $\frac{\pi}{3}$ , so, assuming we have the triangle centered on the origin, the height must be $b\sin(\frac{\pi}{3})=\frac{b\sqrt{3}}{2}$ . As we are only interested in the ratio between $AX$ and $AB$ , we can let $b=1$ WLOG. Applying the formula for area of a triangle, we then have $a(\Delta ABC) = \frac{(\frac{b\sqrt(3)}{2})(b)}{2}=\frac{b^2\sqrt{3}}{4}$ Since $b=1, a(\Delta ABC) = \frac{\sqrt{3}}{4}$ . We are given that $a(\Delta AXY) = \frac{a(\Delta ABC)}{2} = \frac{\sqrt{3}}{8}$ . Using the equation above, $\frac{\sqrt{3}}{8} = \frac{b^2\sqrt{3}}{4}$ $\frac{\sqrt{3}}{2} = b^2\sqrt{3}$ $\frac{1}{2} = b^2$ $b = \frac{\sqrt{2}}{2}$

Therefore, $BX=BY=\frac{\sqrt{2}}{2}$ . It can be seen that $AX=1-BX=1-\frac{\sqrt{2}}{2}$ . The desired ratio is then $\frac{1-\frac{\sqrt{2}}{2}}{1}=1-\frac{\sqrt{2}}{2}$