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2014 AMC 10B Problems/Problem 23

Revision as of 17:41, 26 January 2015 by Equationcrunchor (talk | contribs) (→‎Solution)

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]

Solution

First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of 2, and the bottom base have a diameter of 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 end up with \[s=\sqrt{r}\]. Next, we can find the volume of the frustum and of the sphere. Since we know $V_{frustum}=2V_{sphere}$, 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 \[r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}\]

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

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