Question

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 $m + n.$

答案

Solution:

Place $A = (0, 0, 0)$ with the cube $[0, 12]^3$ and $P = (12, 7, 5)$ on the face $x = 12.$ Reflecting the cube across the relevant face at each bounce straightens the reflected path into the straight ray from $A$ through $P:$ each crossing of a plane $x = 12k,$ $y = 12k,$ or $z = 12k$ corresponds to a bounce, and the beam reaches a vertex of the cube exactly when all three coordinates are simultaneously multiples of $12.$

The ray consists of the points $(12t, 7t, 5t).$ Since $7$ and $5$ are relatively prime to $12,$ the coordinates $7t$ and $5t$ are first divisible by $12$ when $t = 12,$ at the point $(144, 84, 60).$ The path length equals the straight-line distance $\sqrt{144^2 + 84^2 + 60^2} = 12\sqrt{12^2 + 7^2 + 5^2} = 12\sqrt{218}.$

Since $218 = 2 \cdot 109$ is squarefree, $m + n = 12 + 218 = 230.$

2002 AIME I Problem 11

Solution:

Problem 11 in Other Years