Difference between revisions of "2024 AIME II Problems/Problem 8"

Revision as of 19:41, 9 February 2024

Problem

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 $\tfrac{m}{n}$ , 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_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 $O$ 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

Video Solution

https://youtu.be/-1HLRjtLCSM

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 1: / Line 1: @@
-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 outside of <math>S</math>, it is externally 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>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+==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 outside of <math>S</math>, it is externally 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>.
 ==Solution 1==
@@ Line 104: / Line 105: @@
 {{AIME box|year=2024|num-b=7|num-a=9|n=II}}
-[[Category:]]
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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Difference between revisions of "2024 AIME II Problems/Problem 8"

Revision as of 19:41, 9 February 2024

Contents

Problem

Solution 1

Video Solution

See also