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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