Thai Mathematical Olympiad

Let $F$ denote this figure. Consider the translations of $F$ under the following vectors in the plane: ${\vec{u}}_1=0.001\vec{i}$, and $${\vec{u}}_{k+1}=R_{\pi /3}({\vec{u}}_k),k=1,2,3,4,5,$$ where $R_{\pi /3}$ denotes the rotation by $\pi /3$ counterclockwise. Now let $F_0=F$ and, for $k=1,\dots ,6$, $F_k$ = the translation of $F$ in the direction ${\vec{u}}_k$.

We will show that $F_k\cap F_l=\varnothing$ for all $k,l\in \{0,\dots ,6\},k\ne l$. Since the distance between any two points of $F$ is not equal to $0.001$, $F_0$ and $F_k$ have no common points for $k=1,\dots ,6$. By the same reasoning, $F_k$ and $F_{k+1}$ have no common points for $k=1,\dots ,6$ ($F_7=F_1$). Next since the distance between any two points in $F$ is not equal to $0.001\times \sqrt{3}$, it follows that $F_k$ and $F_{k+2}$ have no common points. Finally, that the distance between any two points in $F$ is not equal to $0.002$ implies $F_k$ and $F_{k+2}$ have no points in common. Thus $F_k\cap F_l=\varnothing$ for all $k\ne l$ as claimed.

The figures $F_k$ are all lying in the square of side $5.002$, and they have pairwise empty intersection. Thus $$7\times \text{area}(F)=\sum _{i=0}^6\text{area}(F_i)\le 5.002^2;$$ therefore $\text{area}(F)\le 5.002^2/7<3.575$. $\square$