Question

一个直圆锥的底面半径为 $r$ ，高为 $h$ 。圆锥侧放在平桌上。当圆锥在桌面上无滑动滚动时，圆锥底面与桌面接触的点会描出一段圆弧，其圆心是顶点接触桌面的点。圆锥完成 $17$ 整圈滚动后第一次回到桌面上的原始位置。 $h/r$ 的值可写成 $m\sqrt{n}$ 的形式，其中 $m$ 和 $n$ 为正整数，且 $n$ 不被任何质数的平方整除。求 $m + n$ 。

A right circular cone has base radius $r$ and height $h.$ The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making $17$ complete rotations. The value of $h/r$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$

答案

答案：14

知识点：圆锥圆周长勾股定理

难度评级：2300

解答：

底面的接触点与固定顶点的距离始终为 $\ell = \sqrt{r^2 + h^2}$ （斜高），所以它描出半径为 $\ell$ 的圆。无滑动滚动时，圆锥每滚过一个底面周长的弧长就转一圈，因此恰好 $17$ 圈后回到原位意味着 $2\pi\sqrt{r^2 + h^2} = 17 \cdot 2\pi r,$ 即 $\sqrt{r^2 + h^2} = 17r.$

平方得 $h^2 = 288r^2$ ，所以 $h/r = \sqrt{288} = 12\sqrt{2}$ ，于是 $m + n = 12 + 2 = 14$ 。

The contact point of the base stays at distance $\ell = \sqrt{r^2 + h^2}$ (the slant height) from the fixed vertex, so it traces a circle of radius $\ell.$ Rolling without slipping, the cone makes one rotation for each base circumference of arc, so returning after exactly $17$ rotations means $2\pi\sqrt{r^2 + h^2} = 17 \cdot 2\pi r,$ i.e. $\sqrt{r^2 + h^2} = 17r.$

Squaring gives $h^2 = 288r^2,$ so $h/r = \sqrt{288} = 12\sqrt{2},$ and $m + n = 12 + 2 = 14.$

2008 AIME I 第 5 题

答案：14

难度评级：2300

解答：

其他年份的第 5 题