, 2013

$\frac{\sqrt{3}}{6}$

First note that we have $\angle A_2A_3A_4 = \angle A_2A_3A_1 + \angle A_1A_3A_5 + \angle A_5A_3A_4$. By the theorem on subtended angles by arcs on the circumference on a circle, we have $\angle A_2A_3A_1 = \angle A_2A_4A_1 = 30^{\circ}$, $\angle A_5A_3A_4 = \angle A_5A_2A_4 = 30^{\circ}$. We also have $\angle A_1A_3A_5 = 30^{\circ}$, and therefore, we conclude that $\angle A_2A_3A_4 = 90^{\circ}$, which implies that the line segment $A_2A_4$ is a diameter of the given circle. Similarly, we get that the line segment $A_3A_5$ is also a diameter of the given circle, and therefore, the point $B_1$ of the intersection of $A_2A_4$ and $A_3A_5$ is the center of the circle.

Let us denote by $H$ the foot of the perpendicular line drawn from $B_1$ to the line $A_2A_5$. For the triangle $A_2B_1H$, we have $\angle HA_2B_1 = 30^{\circ}$ and $\angle B_1HA_2 = 90^{\circ}$, from which we get
$\angle A_2B_1H = 60^{\circ}$. Using the fact that $A_2B_1 = 1$, we get $B_1H = \frac{1}{2}$ and $HA_2 = \frac{\sqrt{3}}{2}$, and then we obtain that the area of the triangle $A_2B_1H = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.

For the triangle $A_3B_1B_5$, we have $\angle B_5A_3B_1 = 30^{\circ}$ and $\angle A_3B_1B_5 = \angle B_1A_5A_2 + \angle B_1A_2A_5 = 30^{\circ} + 30^{\circ} = 60^{\circ}$, from which it follows that $\angle B_1B_5A_3 = 90^{\circ}$. We therefore, conclude that $B_1B_5 = \frac{1}{2}$ and $B_5A_2 = B_1A_2 - B_1B_5 = 1 - \frac{1}{2} = \frac{1}{2}$.

For the triangle $A_2B_4B_5$, we have $\angle B_5A_2B_4 = 30^{\circ}$, $\angle B_4B_5A_2 = \angle B_1B_5A_3 = 90^{\circ}$, from which we get $B_4B_5 = B_5A_2 \tan 30^{\circ} = \frac{1}{2} \cdot \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{6}$. Therefore, we conclude that the area of the triangle $A_2B_4B_5$ equals $\frac{1}{2} \cdot \frac{\sqrt{3}}{6} \cdot \frac{1}{2} = \frac{\sqrt{3}}{24}$.

Summarizing what we obtained so far, we now get that the area of the quadrilateral $B_1HB_4B_5$ equals $\frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{24} = \frac{\sqrt{3}}{12}$. Since we get in the same way that the area of the quadrilateral $B_1HB_3B_2$ also equals $\frac{\sqrt{3}}{12}$, we conclude that the area of the pentagon $B_1B_2B_3B_4B_5$ equals $2 \cdot \frac{\sqrt{3}}{12} = \frac{\sqrt{3}}{6}$.