27. [13] $A B C D E$ is a regular pentagon inscribed in a circle of radius 1 . What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex?
Answer: $\frac{\pi}{5}$ Draw the perpendicular bisectors of all the sides and diagonals of the pentagon with one endpoint at $A$. These lines all intersect in the center of the circle, because they are the set of points equidistant from two points on the circle. Now, a given point is farther from $A$ than from point $X$ if it is on the $X$ side of the perpendicular bisector of segment $A X$. So, we want to find the area of the set of all points which are separated from $A$ by all of these perpendicular bisectors, which turns out to be a single $72^{\circ}$ sector of the circle, which has area $\frac{\pi}{5}$.