Question

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 $m + n.$

Answer

Solution:

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 $6:$ $r_i = \frac{11}{8} \cdot 6 = \frac{33}{4}, \qquad r_o = \frac{11}{14} \cdot 6 = \frac{33}{7}.$

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 Problem 8

Solution:

Problem 8 in Other Years