Question

考虑所有位于一个长、宽、高分别为 $3$ 、 $4$ 、 $5$ 个单位的长方体（盒子）内部，或与它的距离不超过一个单位的点。已知这些点所组成的集合体积为 $\frac{m + n\pi}{p}$ ，其中 $m$ 、 $n$ 和 $p$ 是正整数，且 $n$ 与 $p$ 互质。求 $m + n + p$ 。

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m,$ $n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$

答案

答案：505

知识点：体积长方体圆柱球

难度评级：2210

解答：

该区域由长方体本身、从六个面向外伸出的厚度为 $1$ 的薄层、沿十二条棱的半径为 $1$ 的四分之一圆柱，以及八个顶点处半径为 $1$ 的球八分体组成。长方体体积为 $3 \cdot 4 \cdot 5 = 60$ ，各薄层总计 $2(3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5) = 94$ 。

与每个维度平行的四条棱上的四分之一圆柱合并成一个完整圆柱，所以这些圆柱总体积为 $\pi \cdot 1^2 (3 + 4 + 5) = 12\pi$ 。八个球八分体合并成一个单位球，体积为 $\frac{4\pi}{3}$ 。

总体积为因此 $m + n + p$ $= 462 + 40 + 3$ $= 505$ 。 $\begin{aligned} &60 + 94 + 12\pi \\ &\quad {}+ \frac{4\pi}{3} = 154 + \frac{40\pi}{3} \\ &= \frac{462 + 40\pi}{3}, \end{aligned}$

The region consists of the box itself, six slabs of thickness $1$ projecting outward from the faces, quarter-cylinders of radius $1$ along the twelve edges, and eighth-spheres of radius $1$ at the eight corners. The box has volume $3 \cdot 4 \cdot 5 = 60,$ and the slabs total $2(3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5) = 94.$

The four quarter-cylinders along edges parallel to each dimension combine into a full cylinder, so the cylinders total $\pi \cdot 1^2 (3 + 4 + 5) = 12\pi.$ The eight octants combine into one unit sphere of volume $\frac{4\pi}{3}.$

The total volume is $\begin{aligned} &60 + 94 + 12\pi \\ &\quad {}+ \frac{4\pi}{3} = 154 + \frac{40\pi}{3} \\ &= \frac{462 + 40\pi}{3}, \end{aligned}$ so $m + n + p$ $= 462 + 40 + 3$ $= 505.$

2003 AIME I 第 5 题

答案：505

难度评级：2210

解答：

其他年份的第 5 题