AMC10 2020 B

AMC10 2020 B · Q20

AMC10 2020 B · Q20. It mainly tests 3D geometry (volume), 3D geometry (surface area).

Let $B$ be a right rectangular prism (box) with edge lengths 1, 3, and 4, together with its interior. For real $r \ge 0$, let $S(r)$ be the set of points in 3-dimensional space that lie within a distance $r$ of some point in $B$. The volume of $S(r)$ can be expressed as $ar^3 + br^2 + cr + d$, where $a$, $b$, $c$, and $d$ are positive real numbers. What is $\frac{bc}{ad}$?

设 $B$ 是一个边长为 1、3 和 4 的直角长方体（盒子）连同其内部。对于实数 $r \ge 0$，设 $S(r)$ 为三维空间中距离 $B$ 中某点的距离不超过 $r$ 的点的集合。$S(r)$ 的体积可以表示为 $ar^3 + br^2 + cr + d$，其中 $a$、$b$、$c$ 和 $d$ 是正实数。$\frac{bc}{ad}$ 是多少？

(A) 6 6

(B) 19 19

(D) 26 26

(E) 38 38

Answer

Correct choice: (B)

正确答案：（B）

Solution

The solid $S(r)$ consists of the original prism, a prism with height $r$ on each face, a quarter cylinder of radius $r$ on each edge, and $\frac{1}{8}$ of a sphere of radius $r$ at each vertex, together with their interiors. Therefore the volume is equal to $$4 \cdot 3 \cdot 1 + 2(4 \cdot 3 \cdot r + 4 \cdot 1 \cdot r + 3 \cdot 1 \cdot r) + 4 \left( \frac{1}{4} \pi r^2 (4 + 3 + 1) \right) + 8 \left( \frac{1}{8} \cdot \frac{4}{3} \pi r^3 \right),$$ which simplifies to $\frac{4}{3}\pi r^3 + 8\pi r^2 + 38r + 12$. Thus the requested fraction is $$\frac{bc}{ad} = \frac{8\pi \cdot 38}{\frac{4}{3}\pi \cdot 12} = 19.$$

固体 $S(r)$ 由原长方体、各面上高度为 $r$ 的长方体、各棱上的四分之一圆柱体半径 $r$、各顶点的八分之一球体半径 $r$ 连同其内部组成。因此体积等于 $$4 \cdot 3 \cdot 1 + 2(4 \cdot 3 \cdot r + 4 \cdot 1 \cdot r + 3 \cdot 1 \cdot r) + 4 \left( \frac{1}{4} \pi r^2 (4 + 3 + 1) \right) + 8 \left( \frac{1}{8} \cdot \frac{4}{3} \pi r^3 \right)，$$ 简化为 $\frac{4}{3}\pi r^3 + 8\pi r^2 + 38r + 12$。因此请求的分数为 $$\frac{bc}{ad} = \frac{8\pi \cdot 38}{\frac{4}{3}\pi \cdot 12} = 19。$$

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