Difference between revisions of "2014 AMC 12B Problems/Problem 19"
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solving for s we get: | solving for s we get: | ||
<cmath>s=\sqrt{r}</cmath> | <cmath>s=\sqrt{r}</cmath> | ||
| − | next we can find the area of the frustum and of the sphere and we know <math>V_{frustum}=2V_{sphere}</math> so we can solve for <math>s</math> | + | next we can find the area of the frustum and of the sphere and we know <math>V_{\text{frustum}}=2V_{\text{sphere}}</math> so we can solve for <math>s</math> |
| − | using <math>V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math> | + | using <math>V_{\text{frustum}}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math> |
we get: | we get: | ||
| − | <cmath>V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)</cmath> | + | <cmath>V_{\text{frustum}}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)</cmath> |
| − | using <math>V_{sphere}=\dfrac{4r^{3}\pi}{3}</math> | + | using <math>V_{\text{sphere}}=\dfrac{4r^{3}\pi}{3}</math> |
we get | we get | ||
| − | <cmath>V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath> | + | <cmath>V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath> |
so we have: | so we have: | ||
<cmath>\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath> | <cmath>\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath> | ||
Revision as of 17:16, 10 February 2018
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,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4));[/asy]](http://latex.artofproblemsolving.com/d/0/7/d07f08e4aea2e999536cab8632cc10fb0282d034.png)
Solution
First, we draw the vertical cross-section passing through the middle of the frustum.
let the top base equal 2 and the bottom base to be equal to 2r
![[asy] size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label("$r-1$",(X+B)/2,S); label("$1$",(Y+C)/2,N); label("$s$",(O+Y)/2,W); label("$s$",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label("$1$",(C+P)/2,NE); label("$r$",(B+P)/2,NE); [/asy]](http://latex.artofproblemsolving.com/0/1/6/016c1f2f2e25e86be1364ec138a943e0955b8e6d.png)
then using the Pythagorean theorem we have:
which is equivalent to:
subtracting 
from both sides
solving for s we get:
![\[s=\sqrt{r}\]](http://latex.artofproblemsolving.com/7/6/1/761f734737f3d130b9c1f3714b65c6062426b51c.png)
next we can find the area of the frustum and of the sphere and we know 
so we can solve for 
using 
we get:
![\[V_{\text{frustum}}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)\]](http://latex.artofproblemsolving.com/d/7/e/d7eef8d7362d698016a192e684324b236ecda676.png)
using 
we get
![\[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\]](http://latex.artofproblemsolving.com/8/2/d/82da68e37e5b3401b8de5a8588b92760176358cd.png)
so we have:
![\[\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}\]](http://latex.artofproblemsolving.com/7/3/1/731ee8c7ba48dd60c6b5cba6156967cd2d371542.png)
dividing by 
we get ![\[r^2+r+1=4r\]](http://latex.artofproblemsolving.com/e/f/f/eff48a7e2a483ec769c53595e6616aae042ab70e.png)
which is equivalent to ![\[r^2-3r+1=0\]](http://latex.artofproblemsolving.com/d/b/2/db29bc3b20d4947107253c8633c1e77dec9c4443.png)
so ![\[r=\dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}\]](http://latex.artofproblemsolving.com/d/4/3/d4392ed60ab25c1ad1394398225bf6abdf959a50.png)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
