3. Let $A B$ be a line segment with length 2 , and $S$ be the set of points $P$ on the plane such that there exists point $X$ on segment $A B$ with $A X=2 P X$. Find the area of $S$.
Answer: $\sqrt{3}+\frac{2 \pi}{3}$
Observe that for any $X$ on segment $A B$, the locus of all points $P$ such that $A X=2 P X$ is a circle centered at $X$ with radius $\frac{1}{2} A X$. Note that the point $P$ on this circle where $P A$ forms the largest angle with $A B$ is where $P A$ is tangent to the circle at $P$, such that $\angle P A B=\arcsin (1 / 2)=30^{\circ}$. Therefore, if we let $Q$ and $Q^{\prime}$ be the tangent points of the tangents from $A$ to the circle centered at $B$ (call it $\omega$ ) with radius $\frac{1}{2} A B$, we have that $S$ comprises the two $30-60-90$ triangles $A Q B$ and $A Q^{\prime} B$, each with area $\frac{1}{2} \sqrt{3}$ and the $240^{\circ}$ sector of $\omega$ bounded by $B Q$ and $B Q^{\prime}$ with area $\frac{2}{3} \pi$. Therefore the total area is $\sqrt{3}+\frac{2 \pi}{3}$.