Local Mathematical Competitions

In the sequel $[XYZ]$ will represent the area of a triangle $XYZ$. Denote by $S$ the area of the polygon $A_1{A}_2...A_{2010}$. Notice that $[O{A}_1{A}_2]=[O{A}_2{A}_3]=\cdots =[O{A}_{2010}{A}_1]=\frac{1}{2010}S$, since the polygon $A_1{A}_2...A_{2010}$ is given as being regular. Then for each $k=1,2,...,2009$ we have $$\frac{[O{B}_k{B}_{k+1}]}{[O{A}_k{A}_{k+1}]}=\frac{O{B}_k\cdot O{B}_{k+1}}{O{A}_k\cdot O{A}_{k+1}}=\frac{1}{k(k+1)},$$ as well as $$\frac{[O{B}_{2010}{B}_1]}{[O{A}_{2010}{A}_1]}=\frac{O{B}_{2010}\cdot O{B}_1}{O{A}_{2010}\cdot O{A}_1}=\frac{1}{2010},$$ since the ratio of the areas of two triangles sharing a same angle is equal to the ratio of the products of the corresponding sides. Therefore denoting by $T$ the area of the polygon $B_1{B}_2...B_{2010}$, we have $$\begin{array}{rl}T& =\sum _{k=1}^{2009}[O{B}_k{B}_{k+1}]+[O{B}_{2010}{B}_1]\\ & =\frac{S}{2010}\left(\sum _{k=1}^{2009}\frac{1}{k(k+1)}+\frac{1}{2010}\right)\\ & =\frac{S}{2010}\left(\sum _{k=1}^{2009}\left(\frac{1}{k}-\frac{1}{k+1}\right)+\frac{1}{2010}\right)\\ & =\frac{S}{2010}\left(\left(1-\frac{1}{2010}\right)+\frac{1}{2010}\right)=\frac{S}{2010},\end{array}$$ hence $\frac{T}{S}=\frac{1}{2010}$.