jmc

Let the side length of $△ABC$ be $s$. Then the areas of $△APB$, $△BPC$, and $△CPA$ are, respectively, $s/2$, $s$, and $3s/2$. The area of $△ABC$ is the sum of these, which is $3s$. The area of $△ABC$ may also be expressed as $(\sqrt{3}/4)s^2$, so $3s=(\sqrt{3}/4)s^2$. The unique positive solution for $s$ is $\boxed{4\sqrt{3}}$.