2000 AIME I 第 8 题

Question

一个右圆锥形容器高

12

英寸，底面半径为

5

英寸。容器中密封有液体，当圆锥尖端朝下、底面水平放置时，液体深

9

英寸。当圆锥尖端朝上、底面水平放置时，液体深度为

m - n\sqrt[3]{p}

英寸，其中

m

、

n

、

p

是正整数，且

p

不被任何质数的立方整除。求

m + n + p

。

A container in the shape of a right circular cone is

12

inches tall and its base has a

5

-inch radius. The liquid that is sealed inside is

9

inches deep when the cone is held with its point down and its base horizontal. When the cone is held with its point up and its base horizontal, the liquid is

m - n\sqrt[3]{p}

inches deep, where

m,

n,

and

p

are positive integers and

p

is not divisible by the cube of any prime number. Find

m + n + p.

答案

答案：52

知识点：圆锥相似长度、面积与体积的缩放关系体积

难度评级：2390

小提示：

尖端朝下时，液体形成一个与容器相似、比例为

\frac{9}{12}

的圆锥，所以占总体积的

\frac{27}{64}

。

Point down, the liquid is a cone similar to the container with ratio

\frac{9}{12},

so it fills

\frac{27}{64}

of the volume

大提示：

尖端朝上时，上方空隙是一个相似圆锥，占总体积的

\frac{37}{64}

；取立方根求它的高度。

Point up, the empty space is a similar cone at the top; its share of the volume is

\frac{37}{64},

so take a cube root to get its height

解答：

尖端朝下时，液体形成一个与容器相似的圆锥，线性比例为

\frac{9}{12} = \frac{3}{4}

，所以体积占比为

\left(\frac{3}{4}\right)^3 = \frac{27}{64}

。

尖端朝上时，顶端空隙是相似圆锥，体积占比为

1 - \frac{27}{64} = \frac{37}{64}

，所以其高度为

12 \sqrt[3]{\frac{37}{64}} = \frac{12 \sqrt[3]{37}}{4} = 3\sqrt[3]{37}\text{。}

因此液体深度为

12 - 3\sqrt[3]{37}

英寸。

37

不含质数的立方因子，所以

m + n + p = 12 + 3 + 37 = 52

。

Held point down, the liquid forms a cone similar to the container with ratio

\frac{9}{12} = \frac{3}{4},

so its volume is

\left(\frac{3}{4}\right)^3 = \frac{27}{64}

of the container’s volume.

Held point up, the empty space is a similar cone at the apex with

1 - \frac{27}{64} = \frac{37}{64}

of the volume, so its height is

12 \sqrt[3]{\frac{37}{64}} = \frac{12 \sqrt[3]{37}}{4} = 3\sqrt[3]{37}

inches. The liquid is therefore

12 - 3\sqrt[3]{37}

inches deep.

Since

37

is cube-free,

m + n + p = 12 + 3 + 37 = 52.

2000 AIME I 第 8 题

答案：52

难度评级：2390

小提示：

大提示：

解答：

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