Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO

Solution:

The answer is (E). The intersection of the two circles is divided by $XY$ into two congruent figures, bounded by the arcs $XY$, so to get the result it suffices to compute the area of one of these two figures and multiply by two. Let $O$ be the center of the circle on which $P$ lies; by the properties of central and inscribed angles, $X\widehat{O}Y = X\widehat{P}Y = 120^\circ$. Hence the triangle $XOY$ is isosceles with base angles of $30^\circ$ and therefore $OX = OY = \frac{2}{\sqrt{3}} \cdot \frac{XY}{2} = \sqrt{3}$. The area bounded by the segment $XY$ and the arc on which $P$ lies can be computed by difference

as the area of the circular sector $XOY$ minus the area of the triangle $XOY$.
Since the angle at $O$ is $120^\circ$, the area of the circular sector is $\frac{120^\circ}{360^\circ} \pi OX^2 = \pi$; in the triangle $XOY$ the height with respect to $XY$ is $\frac{XO}{2}$ and hence its area is $\frac{3 \sqrt{3}}{4}$. Thus the required area is $2\left(\pi-\frac{3 \sqrt{3}}{4}\right)$.