Let $ABCD$ be a square with side length $1.$ A laser is located at vertex $A,$ which fires a laser beam at point $X$ on side $\overline{BC},$ such that $BX = \frac{1}{2}.$ The beam reflects off the sides of the square, until it ends up at another vertex; at this point, the beam will stop. Find the length of the total path of the laser beam.
Once the laser hits line BC, it will immediately be reflected to the upper right corner. This will make two right triangles, each with legs of 1 and 0.5. We can calculte the hypotenuse of each of these triangles.
\(1^2 + \frac{1}{2}^2 = h^2\)
\(4/4+1/4 = h^2\)
\(5/4 = h^2\)
\(h = \sqrt5/2\)
Since the laser's path is two of these identical hypotenuses, we double this about to get the length of sqrt(5).