The hero of an animated cartoon once again faces his arch-nemesis and needs to deliver his famous Reflective Ray move. In the square of the figure below, the ray, starting from $F$, must bounce exactly once off the sides $C D, A D$, and $A B$, in that order, before hitting the enemy at position $E$. Whenever the ray bounces off one of the sides of the square, the angle of incidence is equal to the angle of reflection as shown in the right figure. Given that $B E=E F=F C=2 \mathrm{~m}$ and that the ray travels at $1 \mathrm{~m} / \mathrm{s}$, determine the time elapsed between the ray's discharge at $F$ and its arrival at point $E$.
![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-16.jpg?height=352&width=650&top_left_y=744&top_left_x=607)
Solution

If the wall where the ray reflects were a mirror, its trajectory would also "appear" on the other side of the mirror as a straight-line continuation of the initial trajectory. Thus, after reflecting the square three times along the sides where the ray hits, we can trace an imaginary trajectory with the same length as the real trajectory. In the drawing below, the square $A B C D$ was initially reflected with respect to side $D C$ and then followed by reflections on sides $D J$ and $J O$. The sum of the four segments that make up the real trajectory coincides with the length of segment $F W$. Since $O W=B E=E F$, it follows that $O E=F W$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-16.jpg?height=545&width=782&top_left_y=1738&top_left_x=543)

By the Pythagorean Theorem in triangle $O M E$, we have:

$$
O E^{2}=O M^{2}+E M^{2}=12^{2}+10^{2}=244
$$

Therefore, $O E=2 \sqrt{61} \mathrm{~m}$. Consequently, the ray will take $2 \sqrt{61}$ seconds to reach the target.