nmtc 2013

Interesting answer!

Let the side length of square $ABCD$ be $a=2013 \space cm$ . It can be seen that $\triangle AFD$ is similar to $\triangle DCE$ . Therefore,

$\dfrac {AF}{AD} = \dfrac {DC}{DE} = \dfrac {a}{\sqrt{(\frac{a}{2})^2 + a^2}} = \dfrac {2}{\sqrt{5}}\quad \Rightarrow AF = \dfrac {2}{\sqrt{5}} AD = \dfrac {2a} {\sqrt{5}}$

Now draw $FG$ perpendicular to $AB$ . Again $\triangle FGA$ is similar to $\triangle DCE$ . Therefore<

$\Rightarrow FG = \dfrac {2}{\sqrt{5}} AF = \dfrac {2}{\sqrt{5}} \times \dfrac {2a}{\sqrt{5}} = \dfrac {4a}{5 }\quad \Rightarrow AG = \dfrac {1}{2} \times \dfrac {4a}{5 } = \dfrac {2a}{5}$

$\Rightarrow BG = AB - AG = a - \dfrac {2a}{5} = \dfrac {3a}{5}$

$\Rightarrow FB = \sqrt{BG^2+FG^2} = \sqrt{ (\frac {3a}{5})^2 + (\frac {4a}{5})^2} = \dfrac {5a}{5} = a = \boxed {2013} \space cm$

Notice that $\triangle FGB$ is a $3$ - $4$ - $5$ Pythagoras triangle.