2014 AMC 10B Problems/Problem 23
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]](https://latex.artofproblemsolving.com/b/c/a/bcae66453a62181e1866ee687a7aed449a6594da.png)
Solution 1
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.
(If you don't get why the blue side is
, reference this (https://www.ck12.org/geometry/Tangent-Lines/lesson/Tangent-Lines-GEOM/)
![[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("$r-1$",(r/2+1/2,0),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]](https://latex.artofproblemsolving.com/3/4/b/34b3cd62c08642105b975a53c232c20e13bc53c4.png)

Then using the Pythagorean theorem we have:




Solving
for
, we end up with
Next, we can find the volume of the frustum (truncated cone)
and of the sphere. Since we know
, we can solve for
using
we get:
Using
, we get
so we have:
Dividing by
, we get
which is equivalent to
by the Quadratic Formula,
, so![\[r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}\]](https://latex.artofproblemsolving.com/4/6/3/463b0f30cd0d1f995e8334d04cc9497d32cf8e76.png)

![\[s=\sqrt{r}.\]](https://latex.artofproblemsolving.com/6/d/f/6dfde7da597d8e7793a419e59f3bd49a686e64c5.png)



![\[V_{\text{frustum}}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)\]](https://latex.artofproblemsolving.com/c/2/4/c24cd9f71d863e02bc76f9a299e0ebc537573a20.png)

![\[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\]](https://latex.artofproblemsolving.com/8/2/d/82da68e37e5b3401b8de5a8588b92760176358cd.png)
![\[\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.\]](https://latex.artofproblemsolving.com/2/3/d/23da0dd79a5b1c17a4012126e13a3723da1778f6.png)

![\[r^2+r+1=4r\]](https://latex.artofproblemsolving.com/e/f/f/eff48a7e2a483ec769c53595e6616aae042ab70e.png)
![\[r^2-3r+1=0\]](https://latex.artofproblemsolving.com/d/b/2/db29bc3b20d4947107253c8633c1e77dec9c4443.png)

![\[r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}\]](https://latex.artofproblemsolving.com/4/6/3/463b0f30cd0d1f995e8334d04cc9497d32cf8e76.png)
Solution 2
Similar
to above, draw a smaller cone top with the base of the smaller circle with
radius
and height
. The smaller right triangle is similar to the blue
highlighted one in Solution 1. Then
where
is the radius of the sphere. Then
.





From
the Pythagorean theorem on the blue triangle in Solution 1, we get similarly
that
.

From
the volume requirements, we get that
which yields
.


The
small right triangle on top is similar to the big right triangle of the entire
big cone. So
.

Substituting
yields
.

Substituting
and
yields
which yields
.




Solving
in
yields
so
.



AoPS
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