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2014 AMC 12B Problems/Problem 19

Revision as of 22:09, 16 January 2019 by Neeneemath (talk | contribs) (→‎Solution 2(ADD DIAGRAM))

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 $V_{\text{frustum}}=2V_{\text{sphere}}$ so we can solve for $s$ using $V_{\text{frustum}}=\frac{\pi*h}{3}(R^2+r^2+Rr)$ we get: \[V_{\text{frustum}}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)\] using $V_{\text{sphere}}=\dfrac{4r^{3}\pi}{3}$ we get \[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\] 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 \[r=\dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}\]



Solution 2(ADD DIAGRAM)

Let's once again look at the cross of the frustum. Let the angle from the center of the sphere to a point on the circumference of the bottom circle be $\theta.$ This implies that the angle from the center of the sphere to a point on the circumference of the top circle is $90 - \theta.$ Hence the bottom radius is $r\tan{\theta}$ and the top radius is $\frac {r}{\tan {\theta}}.$ This means that the radio between the bottom radius and top radius is $(\tan {\theta})^2.$ Using the frustum volume formula, we find that the are of this figure is $\frac{2\pi r}{3}(r^2(\tan {\theta})^2 + r^2 + \frac {r^2} {(\tan {\theta})^2}).$ We can equate this to $\frac {8\pi*r^3} 3.$ Simplifying, we are left with a quadratic conveniently in $(\tan {\theta})^2.$ The quadratic is $(\tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.$ This gives us $(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}$


~NeeNeeMath

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

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