Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12.$ A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P,$ which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}.$ The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n},$ where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n.$
Here's the following solution:
It moves $5$ to the side and $7$ up, and we must end up at another vertex. Every time it reflects, it will move $5$ units to the side and $7$ units either up or down. If it hits a face partway through a reflection, it will continue the remaining number of units in the opposite direction. We first have that both $5x$ and $7x$ are multiples of $12$ for $x=12$. The length $AP$ is $\sqrt{5^2+7^2+12^2}=\sqrt{218}$, so we multiply that by $12$, and we have $12\sqrt{218}\Rightarrow 12+218=\boxed{230}$.
However, I can't se that every time it reflects, it will move $5$ units to the side and $7$ units either up or down.