, 2019

Solution:

Observe that for any $X$ on segment $AB$, the locus of all points $P$ such that $AX = 2 PX$ is a circle centered at $X$ with radius $\frac{1}{2} AX$. Note that the point $P$ on this circle where $PA$ forms the largest angle with $AB$ is where $PA$ is tangent to the circle at $P$, such that $\angle PAB = \arcsin(1/2) = 30^\circ$.

Therefore, if we let $Q$ and $Q'$ be the tangent points of the tangents from $A$ to the circle centered at $B$ (call it $\omega$) with radius $\frac{1}{2} AB$, we have that $S$ comprises the two $30$-$60$-$90$ triangles $AQB$ and $AQ'B$, each with area $\frac{1}{2} \sqrt{3}$, and the $240^\circ$ sector of $\omega$ bounded by $BQ$ and $BQ'$ with area $\frac{2}{3} \pi$.

Therefore the total area is $\sqrt{3} + \frac{2\pi}{3}$.