Let C be a point on Cassidy's path. Because C is twice as far from B as it is from A, it must lie on the circle with center B and radius 3 inches. Similarly, it must lie on the circle with center A and radius 1.5 inches. Thus, Cassidy's path traces out the region between two circles.
To find the area of this region, we first consider the intersection of the two circles. Let D and E be the points of intersection. From the similarity of triangles ABE and ACB, we have [\frac{BE}{AB} = \frac{BC}{AC}.] From the Pythagorean theorem, we get BC=3−1.5=1.5, so BE=2AB=3. From the Pythagorean theorem again, we get AC=4.52−1.52=3, so [\frac{BE}{AB} = \frac{BC}{AC} = \frac{3}{6} = \frac12.] This shows that △BDE is a 30-60-90 triangle, so BD=21BE=1.5 inches and DE=3⋅BD=1.53 inches.
Let the area of the region between the two circles be X. The area of the region inside Cassidy's path is then
[X - \frac12 \pi (1.5)^2 - \frac12 \pi (3)^2.]
We can divide the region between the two circles into △ABE and four identical regions like △BDE. The area of △ABE is 41⋅6⋅3=4.5 square inches. The area of one of the four regions like △BDE is 21⋅1.5⋅1.53=433 square inches. Therefore,
[X = 4.5 + 4 \cdot \frac{3\sqrt{3}}{4} = 6 + 3\sqrt{3}.]
Finally, the area of the region inside Cassidy's path is
[6 + 3\sqrt{3} - \frac12 \pi (1.5)^2 - \frac12 \pi (3)^2 = 6 + 3\sqrt{3} - \frac{9}{2}\pi = \boxed{\frac{1}{2} \pi + 3 \sqrt{3}}.]
The final answer is 1/2*pi + 3*sqrt(3).
