SAT1000 - P888

Extend reflections of the mirror cuboid so that the path of the ray is represented as a straight line. Since the ray goes through $(0, 0, 0)$ and $(4, 3, 12)$ , it has parametric equations $x = 4t$ , $y = 3t$ , and $z = 12t$ and will intersect a mirror plane of one of the reflected cuboids at $x = 11n$ , $y = 7n$ , and $z = 12n$ for integers $n \geq 0$ .

Combining equations above gives $t = \frac{11}{4}n$ , $t = \frac{7}{3}n$ , and $t = n$ , and testing the first few integer values of $n$ and sorting by $t$ give $I_n = (t, x, y, z)$ values of $I_0 = (0, 0, 0, 0)$ , $I_1 = (1, 4, 3, 12)$ , $I_2 = (2, 8, 6, 24)$ , $I_3 = (\frac{7}{3}, \frac{28}{3}, 7, 28)$ , and $I_4 = (\frac{11}{4}, 11, \frac{33}{4}, 33)$ .

Using the distance equation on the values above, $L_1 = 13$ , $L_2 = 13$ , $L_3 = \frac{13}{3}$ , and $L_4 = \frac{65}{12}$ , so $L_1 = L_2 > L_3 < L_4$ , which is choice C .