Question

$ABCD$ 和 $BCFG$ 是一个立方体的两个面，且 $AB = 12$ 。一束光从顶点 $A$ 发出，在面 $BCFG$ 上的点 $P$ 处反射，该点到 $\overline{BG}$ 的距离为 $7$ ，到 $\overline{BC}$ 的距离为 $5$ 。光束继续在立方体的各个面上反射。从光束离开点 $A$ 到它下一次到达立方体顶点为止，光路长度为 $m\sqrt{n}$ ，其中 $m$ 和 $n$ 是整数，且 $n$ 不被任何质数的平方整除。求 $m + n$ 。

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

答案：230

知识点：正方体反射（几何）整除性

难度评级：2840

解答：

设 $A = (0, 0, 0)$ ，立方体为 $[0, 12]^3$ ，且 $P = (12, 7, 5)$ 在平面 $x = 12$ 上。每次反射时把立方体关于相关面翻折，可把反射光路拉直成从 $A$ 经过 $P$ 的一条射线：每穿过一个平面 $x = 12k$ 、 $y = 12k$ 或 $z = 12k$ 就对应一次反射，而光束恰好在三个坐标同时为 $12$ 的倍数时到达立方体顶点。

射线由点 $(12t, 7t, 5t)$ 组成。因为 $7$ 和 $5$ 都与 $12$ 互质，坐标 $7t$ 与 $5t$ 第一次被 $12$ 整除发生在 $t = 12$ ，对应点为 $(144, 84, 60)$ 。路径长度等于直线距离 $\begin{aligned} &\sqrt{144^2 + 84^2 + 60^2} \\ &= 12\sqrt{12^2 + 7^2 + 5^2} \\ &= 12\sqrt{218}. \end{aligned}$

由于 $218 = 2 \cdot 109$ 是平方因子自由的， $m + n = 12 + 218 = 230$ 。

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 $\begin{aligned} &\sqrt{144^2 + 84^2 + 60^2} \\ &= 12\sqrt{12^2 + 7^2 + 5^2} \\ &= 12\sqrt{218}. \end{aligned}$

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

2002 AIME I 第 11 题

答案：230

难度评级：2840

解答：

其他年份的第 11 题