Question

一个直圆锥形状的实心体高 $4$ 英寸，底面半径为 $3$ 英寸。整个圆锥表面，包括底面，都被涂上了油漆。一个平行于圆锥底面的平面将圆锥分成两个实心体：一个较小的圆锥形实心体 $\mathcal{C}$ 和一个圆台形实心体 $\mathcal{F}$ ，使得 $\mathcal{C}$ 与 $\mathcal{F}$ 的已涂表面积之比，和 $\mathcal{C}$ 与 $\mathcal{F}$ 的体积之比，都等于 $k$ 。已知 $k = m/n$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

A solid in the shape of a right circular cone is $4$ inches tall and its base has a $3$ -inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $\mathcal{C}$ and a frustum-shaped solid $\mathcal{F},$ in such a way that the ratio between the areas of the painted surfaces of $\mathcal{C}$ and $\mathcal{F}$ and the ratio between the volumes of $\mathcal{C}$ and $\mathcal{F}$ are both equal to $k.$ Given that $k = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：512

知识点：圆锥表面积长度、面积与体积的缩放关系

难度评级：2710

解答：

该圆锥半径为 $3$ ，高为 $4$ ，斜高为 $5$ ，所以已涂表面包括侧面积 $\pi \cdot 3 \cdot 5 = 15\pi$ 和底面积 $9\pi$ ，总共 $24\pi$ 。设切割得到的相似比为 $t$ ，即 $\mathcal{C}$ 是半径 $3t$ 、斜高 $5t$ 的圆锥。那么 $\mathcal{C}$ 的已涂表面只有其侧面积 $15\pi t^2$ ，而 $\mathcal{F}$ 的已涂表面是剩余部分 $24\pi - 15\pi t^2$ 。体积之比为 $t^3$ 比 $1 - t^3$ 。

令两个比值相等， $\frac{15 t^2}{24 - 15 t^2} = \frac{t^3}{1 - t^3},$ 所以 $15 t^2 (1 - t^3) = t^3 (24 - 15 t^2)$ ，化简得 $15 t^2 = 24 t^3$ ，因而 $t = \frac{5}{8}$ 。

于是 $k = \frac{t^3}{1 - t^3} = \frac{125/512}{387/512} = \frac{125}{387}$ ，这已经是最简分数，因为 $387 = 3^2 \cdot 43$ ，所以 $m + n = 125 + 387 = 512$ 。

The cone has radius $3,$ height $4,$ and slant height $5,$ so its painted surface consists of lateral area $\pi \cdot 3 \cdot 5 = 15\pi$ and base area $9\pi,$ totaling $24\pi.$ Suppose the cut is at similarity ratio $t,$ so $\mathcal{C}$ is a cone with radius $3t$ and slant height $5t.$ Then $\mathcal{C}$ 's painted surface is only its lateral area $15\pi t^2,$ and $\mathcal{F}$ 's painted surface is the rest, $24\pi - 15\pi t^2.$ The volumes are in ratio $t^3$ to $1 - t^3.$

Setting the two ratios equal, $\frac{15 t^2}{24 - 15 t^2} = \frac{t^3}{1 - t^3},$ so $15 t^2 (1 - t^3) = t^3 (24 - 15 t^2),$ which simplifies to $15 t^2 = 24 t^3,$ giving $t = \frac{5}{8}.$

Then $k = \frac{t^3}{1 - t^3} = \frac{125/512}{387/512} = \frac{125}{387},$ which is in lowest terms since $387 = 3^2 \cdot 43,$ so $m + n = 125 + 387 = 512.$

2004 AIME I 第 11 题

答案：512

难度评级：2710

解答：

其他年份的第 11 题