2024 AIME II Problems/Problem 8
Torus 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 be a sphere with a radius 11. When rests on the outside of , it is externally 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 .
<asy> unitsize(0.5cm); 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, SW); label("", A, SW); label("", B, NW); label("", C, NW); label("", D, SW); label("", E, NE); label("", F, W); label("", G, SE); label("", H, W); label("", 0.5 * H + 0.5 * G, NE); label("", 0.5 * E + 0.5 * G, NE); label("", 0.5 * O + 0.5 * G, NE); <asy>
<asy> unitsize(0.5cm); 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, SW); label("", A, SW); label("", B, NW); label("", C, NW); label("", D, SW); label("", E, NE); label("", F, SW); label("", G, N); label("", H, W); label("", 0.5 * H + 0.5 * G, NE); label("", 0.5 * E + 0.5 * G, NE); label("", 0.5 * O + 0.5 * G, NE); <asy>
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
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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