jmc

Let the coordinates of $A$ be $(a_1,a_2)$. Then since $A$ is on the graph of $y=-\frac{1}{2}{x}^2$, we know that $a_2=-\frac{1}{2}{a}_1^2$. We can also use our knowledge of special right triangles to write $a_2$ in terms of $a_1$. Let $C$ be the midpoint of $A$ and $B$ and let $O$ be the origin. Then $OCA$ is a 30-60-90 right triangle, so the ratio of the length of $OC$ to the length of $CA$ is $\sqrt{3}:1$. Now the coordinates of C are $(0,a_2)$, so the length of $OC$ is just $-a_2$ (since $a_2$ is negative) and the length of $CA$ is $a_1$. This means $\frac{-a_2}{a_1}=\sqrt{3}⟹a_2=-\sqrt{3}{a}_1$.

We can now set our two equations for $a_2$ equal to each other and get $-\sqrt{3}{a}_1=-\frac{1}{2}{a}_1^2$. Multiplying both sides by $-\frac{2}{a_1}$ immediately gives $a_1=2\sqrt{3}$. From here we could solve for $a_2$ using one of our equations and then use the Pythagorean Theorem to solve for the side length of the equilateral triangle, but there's a better way. We remember that the hypotenuse of our special triangle is twice as long as the shortest side of it, which has length $a_1=2\sqrt{3}$. Therefore our answer is $\boxed{4\sqrt{3}}$.