A wire is cut into two pieces, one of length $a$ and the other of length $b$. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a square. If the area of the triangle is twice the area of the square, what is $\frac ab$?

1. The equilateral triangle has side $\frac a3$ and area $\frac{\sqrt3}{36}a^2$. The square has side $\frac b4$ and area $\frac{b^2}{16}$.
2. Setting the triangle's area to twice the square's gives $\frac{\sqrt3}{36}a^2=\frac{b^2}{8}$, so $\frac{a^2}{b^2}=\frac{3\sqrt3}{2}$.
3. Therefore $\dfrac ab=\boxed{\sqrt{\frac{3\sqrt3}{2}}}$.