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Difference between revisions of "2024 AIME II Problems/Problem 8"
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In both cases, we know <math>\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}</math>.
 
In both cases, we know <math>\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}</math>.
  
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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>.
+
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>
 
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
 
~Prof_Joker
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==See also==
 
==See also==

Revision as of 11:30, 9 February 2024

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 outside of $S$, it is externally 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 $m+n$.

Solution 1

First, let's consider a section $\mathcal{P}$ 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 $\mathcal{P}$ we took crosses one of the equator of the sphere.

Here I drew two graphs, the first one is the case when $T$ is internally tangent to $S$,

[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 $T$ is externally tangent to $S$.

[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("$E$", 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 $O$ is the center of sphere $S$, and points $A$ and $B$ are the intersections of the sphere and the axis. Point $E$ (ignoring the subscripts) is one of the circle centers of the intersection of torus $T$ with section $\mathcal{P}$. Point $G$ (again, ignoring the subscripts) is one of the tangents between the torus $T$ and sphere $S$ on section $\mathcal{P}$. $EF\bot CD$, $HG\bot CD$.

And then, we can start our calculation.

In both cases, we know $\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}$.

Hence, in the case of internal tangent, $\frac{E_iF_i}{OE_i} =\frac{G_iH_i}{OG_i}\Longrightarrow \frac{6}{11-3} =\frac{r_i}{11}\Longrightarrow r_i=\frac{33}{4}$. In the case of external tangent, $\frac{E_oF_o}{OE_o} =\frac{G_oH_o}{OG_o}\Longrightarrow \frac{6}{11+3} =\frac{r_o}{11}\Longrightarrow r_o=\frac{33}{7}$.

Thereby, $r_i-r_o=\frac{33}{4}-\frac{33}{7}=\frac{99}{28}$. And there goes the answer, $99+28=\boxed{\mathbf{127} }$

~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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