Three congruent isosceles triangles are constructed outward with their bases on the sides of an equilateral triangle of side length $1$. The sum of their areas is twice the area of the equilateral triangle. What is the length of one of the two congruent sides of an isosceles triangle?

1. Let $h$ be the altitude of each isosceles triangle. The area condition gives $3(\frac12h)=2(\frac{\sqrt3}{4})$, so $h=\frac{\sqrt3}{3}$.
2. Each congruent side is $\sqrt{(\frac12)^2+h^2}=\sqrt{\frac14+\frac13}=\boxed{\frac{\sqrt{21}}6}$.