2014 AMC 10B Problems/Problem 23

Revision as of 19:45, 20 February 2014 by Bobthesmartypants (talk | contribs) (Problem)

Problem

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$

[asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); //draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4));[/asy]

(Diagram edited from copeland's diagram)

Solution

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png