2011 Fermat Problem 18

A 4 \times 4 square piece of paper is cut into two identical pieces along its diagonal. The resulting triangular pieces of paper are each cut into two identical pieces.

Each of the four resulting pieces is cut into two identical pieces. Each of the eight new resulting pieces is finally cut into two identical pieces. The length of the longest edge of one of these final sixteen pieces of paper is

\textbf{(A)}\ 1\quad \textbf{(B)}\ 2\quad \textbf{(C)}\ \frac{1}{2}\quad \textbf{(D)}\ \frac{1}{\sqrt{2}}\quad \textbf{(E)}\ 2\sqrt{2}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.