Difference between revisions of "2014 AMC 10B Problems/Problem 23"

Revision as of 21:02, 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?

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

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

(Diagram edited from copeland's diagram)

Solution

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 $r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E$

@@ Line 88: / Line 88: @@
 which is equivalent to <cmath>r^2-3r+1=0</cmath>
 <math> r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}</math>
-so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow D</cmath>
+so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E</cmath>
 ==See Also==
 {{AMC10 box|year=2014|ab=B|num-b=22|num-a=24}}
 {{MAA Notice}}

2014 AMC 10B (Problems • Answer Key • Resources)
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Difference between revisions of "2014 AMC 10B Problems/Problem 23"

Revision as of 21:02, 20 February 2014

Problem

Solution

See Also