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Difference between revisions of "2014 AMC 12B Problems/Problem 19"

(Created page with "==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);...")
 
(Added templates)
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==Problem==
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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?
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<asy>
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real r=(3+sqrt(5))/2;
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real s=sqrt(r);
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real Brad=r;
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real brad=1;
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real Fht = 2*s;
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import graph3;
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import solids;
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currentprojection=orthographic(1,0,.2);
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currentlight=(10,10,5);
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revolution sph=sphere((0,0,Fht/2),Fht/2);
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//draw(surface(sph),green+white+opacity(0.5));
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//triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));}
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triple f(pair t) {
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triple v0 = Brad*(cos(t.x),sin(t.x),0);
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triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht);
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return (v0 + t.y*(v1-v0));
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}
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triple g(pair t) {
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return (t.y*cos(t.x),t.y*sin(t.x),0);
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}
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surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2);
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surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2);
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surface base = surface(g,(0,0),(2pi,Brad),80,2);
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draw(sback,gray(0.9));
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draw(sfront,gray(0.5));
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draw(base,gray(0.9));
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draw(surface(sph),gray(0.4));</asy>
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<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>
 +
 
==Solution==
 
==Solution==
−
 
 
First, we draw the vertical cross-section passing through the middle of the frustum.
 
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
 
let the top base equal 2 and the bottom base to be equal to 2r
Line 31: Line 62:
 
label("$r$",(B+P)/2,NE);
 
label("$r$",(B+P)/2,NE);
 
</asy>
 
</asy>
−
(diagram by DivideBy0?)
 
  
 
then using the Pythagorean theorem we have:
 
then using the Pythagorean theorem we have:
Line 55: Line 85:
 
<math> r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}</math>
 
<math> r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}</math>
 
so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E</cmath>
 
so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E</cmath>
 +
 +
{{AMC12 box|year=2014|ab=B|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Revision as of 12:30, 21 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,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4));[/asy] $\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$

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]

then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$ subtracting $r^2-2r+1$ from both sides $4r=4s^2$ solving for s we get: \[s=\sqrt{r}\] next we can find the area of the frustum and of the sphere and we know $V_{frustum}=2V_{sphere}$ so we can solve for $s$ using $V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)$ we get: \[V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)\] using $V_{sphere}=\dfrac{4r^{3}\pi}{3}$ we get \[V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\] so we have: \[\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}\] dividing by $\frac{2\pi*\sqrt{r}}{3}$ we get \[r^2+r+1=4r\] which is equivalent to \[r^2-3r+1=0\] $r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}$ so \[r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E\]

2014 AMC 12B (Problems • Answer Key • Resources)
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Problem 18 		Followed by
Problem 20
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All AMC 12 Problems and Solutions

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