2021 AMC 10A Fall 第 3 题

Question

半径为

2

的黏土球最多有多少个可以完全放进边长为

6

的立方体中？假设这些球在装入立方体之前可以重新塑形，但不能被压缩。

What is the maximum number of balls of clay with radius

2

that can completely fit inside a cube of side length

6

assuming that the balls can be reshaped but not compressed before they are packed in the cube?

3

4

5

6

7

知识点：体积球估算

因为黏土可以重新塑形，所以比较体积。

Compare volumes because the clay can be reshaped

个数为

\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor

。

The count is

\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor

立方体体积为

6^3=216

。一个黏土球的体积为

\frac{4}{3}\pi\cdot 2^3=\frac{32\pi}{3}

。

因为黏土可以重新塑形但不能压缩，最多的球数为

\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor\text{。}

由于

12\lt 4\pi\lt 13

，可知

6\lt \frac{81}{4\pi}\lt \frac{81}{12}\lt 7

，因此其整数部分为

6

。

所以正确答案是 D。

The cube has volume

6^3=216.

One ball of clay has volume

\frac{4}{3}\pi\cdot 2^3=\frac{32\pi}{3}.

Because the clay may be reshaped but not compressed, the maximum number of balls is

\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.

Since

12\lt 4\pi\lt 13,

we have

6\lt \frac{81}{4\pi}\lt \frac{81}{12}\lt 7.

Therefore the floor is

6.

Thus, D is the correct answer.

其他年份的第 3 题