2024 AIME II Problems/Problem 8
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[hide]Problem
Torus is the surface produced by revolving a circle with radius around an axis in the plane of the circle that is a distance from the center of the circle (so like a donut). Let be a sphere with a radius . When rests on the inside of , it is internally tangent to along a circle with radius , and when rests on the outside of , it is externally tangent to along a circle with radius . The difference can be written as , where and are relatively prime positive integers. Find .
Solution 1
First, let's consider a section 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 we took crosses one of the equator of the sphere.
Here I drew two graphs, the first one is the case when is internally tangent to ,
and the second one is when is externally tangent to .
For both graphs, point is the center of sphere , and points and are the intersections of the sphere and the axis. Point (ignoring the subscripts) is one of the circle centers of the intersection of torus with section . Point (again, ignoring the subscripts) is one of the tangents between the torus and sphere on section . , .
And then, we can start our calculation.
In both cases, we know .
Hence, in the case of internal tangent, .
In the case of external tangent, .
Thereby, . And there goes the answer,
~Prof_Joker
Solution 2
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution(中文)subtitle in English
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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