Difference between revisions of "2024 AIME II Problems/Problem 8"
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− | Torus <math>T</math> 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 <math>S</math> be a sphere with a radius 11. When <math>T</math> rests on the | + | ==Problem== |
+ | Torus <math>T</math> is the surface produced by revolving a circle with radius <math>3</math> around an axis in the plane of the circle that is a distance <math>6</math> from the center of the circle (so like a donut). Let <math>S</math> be a sphere with a radius <math>11</math>. When <math>T</math> rests on the inside of <math>S</math>, it is internally tangent to <math>S</math> along a circle with radius <math>r_i</math>, and when <math>T</math> rests on the outside of <math>S</math>, it is externally tangent to <math>S</math> along a circle with radius <math>r_o</math>. The difference <math>r_i-r_o</math> can be written as <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | <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 <math>\mathcal{P} </math> 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 <math>\mathcal{P} </math> we took crosses one of the equator of the sphere. | ||
+ | |||
+ | Here I drew two graphs, the first one is the case when <math>T</math> is internally tangent to <math>S</math>, | ||
+ | |||
+ | <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 <math>T</math> is externally tangent to <math>S</math>. | ||
+ | |||
+ | <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 <math>O</math> is the center of sphere <math>S</math>, and points <math>A</math> and <math>B</math> are the intersections of the sphere and the axis. Point <math>E</math> (ignoring the subscripts) is one of the circle centers of the intersection of torus <math>T</math> with section <math>\mathcal{P} </math>. Point <math>G</math> (again, ignoring the subscripts) is one of the tangents between the torus <math>T</math> and sphere <math>S</math> on section <math>\mathcal{P} </math>. <math>EF\bot CD</math>, <math>HG\bot CD</math>. | ||
+ | |||
+ | And then, we can start our calculation. | ||
+ | |||
+ | In both cases, we know <math>\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}</math>. | ||
+ | |||
+ | Hence, in the case of internal tangent, <math>\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} </math>. | ||
+ | |||
+ | In the case of external tangent, <math>\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} </math>. | ||
+ | |||
+ | Thereby, <math>r_i-r_o=\frac{33}{4}-\frac{33}{7}=\frac{99}{28}</math>. And there goes the answer, <math>99+28=\boxed{\mathbf{127} }</math> | ||
+ | |||
+ | ~Prof_Joker | ||
+ | ==Solution 2== | ||
+ | [[File:2024 AIME II 8.png|230px|right]] | ||
+ | <cmath>OC = OD = 11, AC = BD = 3, EC' = FD' = 6.</cmath> | ||
+ | <cmath>\frac {CC'}{C'E} = \frac{AC}{OA} \implies CC' = \frac {3 \cdot 6}{11-3}</cmath> | ||
+ | <cmath>\frac {DD'}{DB} = \frac{FD'}{OB} \implies DD' = \frac {3 \cdot 6}{11+3}</cmath> | ||
+ | <cmath>CC' + DD' = \frac {9}{4}+\frac {9}{7} = \frac {99}{28}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/-1HLRjtLCSM | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution(中文)subtitle in English == | ||
+ | https://youtu.be/YdQdDBROG8U | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2024|num-b=7|num-a=9|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:31, 22 October 2024
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.