Question

环面 $T$ 是由一个半径为 $3$ 的圆绕一条轴旋转得到的曲面，这条轴在该圆所在平面内，并且到圆心的距离为 $6$ （形状像甜甜圈）。

设 $S$ 是半径为 $11$ 的球。当 $T$ 靠在 $S$ 的内部时，它沿半径为 $r_i$ 的圆与 $S$ 内切；当 $T$ 靠在 $S$ 的外部时，它沿半径为 $r_o$ 的圆与 $S$ 外切。差 $r_i - r_o$ 可写成 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Torus $T$ is the surface produced by revolving a circle with radius $3$ around an axis in the plane of the circle that is a distance $6$ from the center of the circle (so like a donut).

Let $S$ be a sphere with a radius $11.$ When $T$ rests on the inside of $S,$ it is internally tangent to $S$ along a circle with radius $r_i,$ and when $T$ rests on the outside of $S,$ it is externally tangent to $S$ along a circle with radius $r_o.$ The difference $r_i - r_o$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：127

知识点：立体几何相切圆相似

难度评级：2650

解答：

由对称性，环面的旋转轴经过球心 $O$ 。在经过旋转轴的平面中观察：环面显示为一个半径为 $3$ 的圆（管道截面），其圆心到轴的距离为 $6$ ，球显示为以 $O$ 为圆心、半径为 $11$ 的圆。两个曲面沿截面相切点旋转出的圆相切；该相切点位于从 $O$ 经过管道截面圆心的射线上。内切时，管道截面圆心到 $O$ 的距离为 $11 - 3 = 8$ ；外切时为 $11 + 3 = 14$ 。

相切点在这条射线上且到 $O$ 的距离为 $11$ ，所以它是管道截面圆心相对于 $O$ 按 $\frac{11}{8}$ （或 $\frac{11}{14}$ ）缩放得到的点；它到轴的距离也是管道截面圆心到轴距离 $6$ 的同一倍数： $r_i = \frac{11}{8} \cdot 6 = \frac{33}{4},$ $r_o = \frac{11}{14} \cdot 6 = \frac{33}{7}.$

因此 $r_i - r_o = \frac{33 \cdot 3}{28} = \frac{99}{28}$ ，已经是最简分数，所以 $m + n = 99 + 28 = 127$ 。

By symmetry the axis of the torus passes through the center $O$ of the sphere. Work in a plane through the axis: there the torus appears as a circle of radius $3$ (the tube) whose center sits at distance $6$ from the axis, and the sphere appears as a circle of radius $11$ centered at $O.$ The two surfaces are tangent along the circle swept by the tangency point of these cross-sections, which lies on the ray from $O$ through the tube's center. For internal tangency the tube's center is at distance $11 - 3 = 8$ from $O;$ for external tangency, $11 + 3 = 14.$

The tangency point lies at distance $11$ from $O$ along that ray, so it is the tube center scaled by $\frac{11}{8}$ (resp. $\frac{11}{14}$ ) from $O,$ and its distance from the axis is the same multiple of the tube center's distance $6:$ $r_i = \frac{11}{8} \cdot 6 = \frac{33}{4},$ $r_o = \frac{11}{14} \cdot 6 = \frac{33}{7}.$

Then $r_i - r_o = \frac{33 \cdot 3}{28} = \frac{99}{28},$ which is in lowest terms, so $m + n = 99 + 28 = 127.$

2024 AIME II 第 8 题

答案：127

难度评级：2650

解答：

其他年份的第 8 题