# 2024 AIME II Problems/Problem 8

## 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 $m+n$.

$[asy] unitsize(0.3 inch); draw(ellipse((0,0), 3, 1.75)); draw((-1.2,0.1)..(-0.8,-0.03)..(-0.4,-0.11)..(0,-0.15)..(0.4,-0.11)..(0.8,-0.03)..(1.2,0.1)); draw((-1,0.04)..(-0.5,0.12)..(0,0.16)..(0.5,0.12)..(1,0.04)); draw((0,2.4)--(0,-0.15)); draw((0,-0.15)--(0,-1.75), dashed); draw((0,-1.75)--(0,-2.25)); draw(ellipse((2,0), 1, 0.9)); draw((2.03,-0.02)--(2.9,-0.4)); [/asy]$

## 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 $S$,

$[asy] unitsize(0.35cm); pair O = (0, 0); real r1 = 11; real r2 = 3; draw(circle(O, r1)); pair A = O + (0, -r1); pair B = O + (0, r1); draw(A--B); pair C = O + (0, -1.25*r1); pair D = O + (0, 1.25*r1); draw(C--D, dashed); dot(O); pair E = (2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)); pair F = (0, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)); pair G = (-r2 * O + r1 * E) / (r1 - r2); pair H = (-r2 * O + r1 * F) / (r1 - r2); draw(circle(E, r2)); draw(circle((-2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)), r2)); draw(O--G, dashed); draw(F--E, dashed); draw(G--H, dashed); label("O", O, SW); label("A", A, SW); label("B", B, NW); label("C", C, NW); label("D", D, SW); label("E_i", E, NE); label("F_i", F, W); label("G_i", G, SE); label("H_i", H, W); label("r_i", 0.5 * H + 0.5 * G, NE); label("3", 0.5 * E + 0.5 * G, NE); label("11", 0.5 * O + 0.5 * G, NE); [/asy]$

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

$[asy] unitsize(0.35cm); pair O = (0, 0); real r1 = 11; real r2 = 3; draw(circle(O, r1)); pair A = O + (0, -r1); pair B = O + (0, r1); draw(A--B); pair C = O + (0, -1.25*(r1 + r2)); pair D = O + (0, 1.25*r1); draw(C--D, dashed); dot(O); pair E = (2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)); pair F = (0, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)); pair G = (r2 * O + r1 * E) / (r1 + r2); pair H = (r2 * O + r1 * F) / (r1 + r2); draw(circle(E, r2)); draw(circle((-2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)), r2)); draw(O--E, dashed); draw(F--E, dashed); draw(G--H, dashed); label("O", O, SW); label("A", A, SW); label("B", B, NW); label("C", C, NW); label("D", D, SW); label("E_o", E, NE); label("F_o", F, SW); label("G_o", G, S); label("H_o", H, W); label("r_o", 0.5 * H + 0.5 * G, NE); label("3", 0.5 * E + 0.5 * G, NE); label("11", 0.5 * O + 0.5 * G, NE); [/asy]$

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

## Video Solution

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