Question

一块平板上有一个半径为 $1$ 的圆孔和一个半径为 $2$ 的圆孔，两个圆孔圆心之间的距离为 $7$ 。两个半径相等的球分别放在这两个圆孔中，并且这两个球互相相切。球半径的平方为 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：173

知识点：球勾股定理根式

难度评级：2450

解答：

半径为 $r$ 的球放在半径为 $a$ 的圆孔中时，球心位于圆孔的轴线上；因为球心到孔边缘每一点的距离都是 $r$ ，所以它到平板平面的距离为 $\sqrt{r^2 - a^2}$ 。因此两个球心位于平板同一侧，深度分别为 $\sqrt{r^2 - 1}$ 和 $\sqrt{r^2 - 4}$ ，水平距离为 $7$ 。

两球相切意味着球心距离为 $2r$ ： $\begin{aligned} &49 + \left(\sqrt{r^2 - 1} - \sqrt{r^2 - 4}\right)^2 \\ &= 4r^2. \end{aligned}$ 展开得 $49 + 2r^2 - 5$ $- 2\sqrt{(r^2 - 1)(r^2 - 4)}$ $= 4r^2$ ，所以 $\sqrt{(r^2 - 1)(r^2 - 4)} = 22 - r^2$ 。两边平方，得 $r^4 - 5r^2 + 4 = 484 - 44r^2 + r^4$ ，因此 $39r^2 = 480$ ， $r^2 = \frac{160}{13}$ 。

所以 $m + n = 160 + 13 = 173$ 。

A sphere of radius $r$ resting in a circular hole of radius $a$ has its center on the axis of the hole; since the center is at distance $r$ from every point of the hole's rim, it sits at distance $\sqrt{r^2 - a^2}$ from the plane of the board. So the two centers lie at depths $\sqrt{r^2 - 1}$ and $\sqrt{r^2 - 4}$ on the same side of the board, with horizontal separation $7.$

Tangency of the spheres means the centers are $2r$ apart: $\begin{aligned} &49 + \left(\sqrt{r^2 - 1} - \sqrt{r^2 - 4}\right)^2 \\ &= 4r^2. \end{aligned}$ Expanding gives $49 + 2r^2 - 5$ $- 2\sqrt{(r^2 - 1)(r^2 - 4)}$ $= 4r^2,$ so $\sqrt{(r^2 - 1)(r^2 - 4)} = 22 - r^2.$ Squaring, $r^4 - 5r^2 + 4 = 484 - 44r^2 + r^4,$ hence $39r^2 = 480$ and $r^2 = \frac{160}{13}.$

Thus $m + n = 160 + 13 = 173.$

2020 AIME I 第 6 题

答案：173

难度评级：2450

解答：

其他年份的第 6 题