Difference between revisions of "2014 AMC 10B Problems/Problem 23"
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==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? | ||
| + | |||
| + | <math>\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</math> | ||
| + | |||
| + | <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,rgb(0,1,0)); | ||
| + | draw(sfront,rgb(.3,1,.3)); | ||
| + | draw(base,rgb(.4,1,.4)); | ||
| + | draw(surface(sph),rgb(.3,1,.3)); | ||
| + | </asy> | ||
| + | |||
| + | (Credits to the Math Jam for the diagram) | ||
==Solution== | ==Solution== | ||
Revision as of 19:19, 20 February 2014
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?
![[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,rgb(0,1,0)); draw(sfront,rgb(.3,1,.3)); draw(base,rgb(.4,1,.4)); draw(surface(sph),rgb(.3,1,.3)); [/asy]](http://latex.artofproblemsolving.com/8/0/0/80052dc5a6e11e63d494595cc82a6ee4ca08bfb3.png)
(Credits to the Math Jam for the diagram)
Solution
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
