AIME 2024 II · 第 8 题

AIME 2024 II — Problem 8

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

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 $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

$AIME diagram$

解析

Solution 1

First, let's consider a section $\mathcal{P}$ of the solids, along the axis. By some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\mathcal{P}$ we took crosses one of the equator of the sphere.

Here I drew two graphs, the first one is the case when $T$ is internally tangent to $S$ ,

$AIME diagram$

and the second one is when $T$ is externally tangent to $S$ .

$AIME diagram$

For both graphs, point $O$ is the center of sphere $S$ , and points $A$ and $B$ are the intersections of the sphere and the axis. Point $E$ (ignoring the subscripts) is one of the circle centers of the intersection of torus $T$ with section $\mathcal{P}$ . Point $G$ (again, ignoring the subscripts) is one of the tangents between the torus $T$ and sphere $S$ on section $\mathcal{P}$ . $EF\bot CD$ , $HG\bot CD$ .

And then, we can start our calculation.

In both cases, we know $\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}$ .

Hence, in the case of internal tangent, $\frac{E_iF_i}{OE_i} =\frac{G_iH_i}{OG_i}\Longrightarrow \frac{6}{11-3} =\frac{r_i}{11}\Longrightarrow r_i=\frac{33}{4}$ .

In the case of external tangent, $\frac{E_oF_o}{OE_o} =\frac{G_oH_o}{OG_o}\Longrightarrow \frac{6}{11+3} =\frac{r_o}{11}\Longrightarrow r_o=\frac{33}{7}$ .

Thereby, $r_i-r_o=\frac{33}{4}-\frac{33}{7}=\frac{99}{28}$ . And there goes the answer, $99+28=\boxed{\mathbf{127} }$

~Prof_Joker

Solution 2

$OC = OD = 11, AC = BD = 3, EC' = FD' = 6.$ $\frac {CC'}{C'E} = \frac{AC}{OA} \implies CC' = \frac {3 \cdot 6}{11-3}$ $\frac {DD'}{DB} = \frac{FD'}{OB} \implies DD' = \frac {3 \cdot 6}{11+3}$ $CC' + DD' = \frac {9}{4}+\frac {9}{7} = \frac {99}{28}.$ vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/-1HLRjtLCSM

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution（spoken in Chinese）subtitle in English

https://youtu.be/YdQdDBROG8U