Japan 2013 Initial Round

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

First note that we have $\angle A_2{A}_3{A}_4=\angle A_2{A}_3{A}_1+\angle A_1{A}_3{A}_5+\angle A_5{A}_3{A}_4$. By the theorem on subtended angles by arcs on the circumference on a circle, we have $\angle A_2{A}_3{A}_1=\angle A_2{A}_4{A}_1=30^{\circ }$, $\angle A_5{A}_3{A}_4=\angle A_5{A}_2{A}_4=30^{\circ }$. We also have $\angle A_1{A}_3{A}_5=30^{\circ }$, and therefore, we conclude that $\angle A_2{A}_3{A}_4=90^{\circ }$, which implies that the line segment $A_2{A}_4$ is a diameter of the given circle. Similarly, we get that the line segment $A_3{A}_5$ is also a diameter of the given circle, and therefore, the point $B_1$ of the intersection of $A_2{A}_4$ and $A_3{A}_5$ is the center of the circle.

Let us denote by $H$ the foot of the perpendicular line drawn from $B_1$ to the line $A_2{A}_5$. For the triangle $A_2{B}_{1}H$, we have $\angle H{A}_2{B}_1=30^{\circ }$ and $\angle B_{1}H{A}_2=90^{\circ }$, from which we get $\angle A_2{B}_{1}H=60^{\circ }$. Using the fact that $A_2{B}_1=1$, we get $B_{1}H=\frac{1}{2}$ and $H{A}_2=\frac{\sqrt{3}}{2}$, and then we obtain that the area of the triangle $A_2{B}_{1}H=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}$.

For the triangle $A_3{B}_1{B}_5$, we have $\angle B_5{A}_3{B}_1=30^{\circ }$ and $\angle A_3{B}_1{B}_5=\angle B_1{A}_5{A}_2+\angle B_1{A}_2{A}_5=30^{\circ }+30^{\circ }=60^{\circ }$, from which it follows that $\angle B_1{B}_5{A}_3=90^{\circ }$. We therefore, conclude that $B_1{B}_5=\frac{1}{2}$ and $B_5{A}_2=B_1{A}_2-B_1{B}_5=1-\frac{1}{2}=\frac{1}{2}$.

For the triangle $A_2{B}_4{B}_5$, we have $\angle B_5{A}_2{B}_4=30^{\circ }$, $\angle B_4{B}_5{A}_2=\angle B_1{B}_5{A}_3=90^{\circ }$, from which we get $B_4{B}_5=B_5{A}_2\tan 30^{\circ }=\frac{1}{2}\cdot \frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{6}$. Therefore, we conclude that the area of the triangle $A_2{B}_4{B}_5$ 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 $B_{1}H{B}_4{B}_5$ 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 $B_{1}H{B}_3{B}_2$ also equals $\frac{\sqrt{3}}{12}$, we conclude that the area of the pentagon $B_1{B}_2{B}_3{B}_4{B}_5$ equals $2\cdot \frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{6}$.