Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2014 AMC 10B Problems/Problem 23

Revision as of 16:40, 16 February 2016 by TheMaskedMagician (talk | contribs) (→‎Solution 2)

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 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. [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: \[V_{frustum}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)\] Using $V_{sphere}=\dfrac{4r^{3}\pi}{3}$, we get \[V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\] so we have: \[\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.\] Dividing by $\frac{2\pi\sqrt{r}}{3}$, we get \[r^2+r+1=4r\] which is equivalent to \[r^2-3r+1=0\] $r=\dfrac{3\pm\sqrt{(-3)^2-4\cdot1\cdot1}}{2\cdot1}$ , so \[r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}\]

Solution 2

Similar to above, draw a smaller cone top with the base of the smaller circle with radius $r_1$ and height $h$. The smaller right triangle is similar to the blue highlighted one in Solution 1. Then $\frac{r_1}{r_2-r_1}=\frac{h}{2R}$ where $R$ is the radius of the sphere. Then $h=\frac{2Rr_1}{r_2-r_1}$.

From the Pythagorean theorem on the blue triangle in Solution 1, we get similarly that $R=\sqrt{r_2r_1}$.

From the volume requirements, we get that $\frac{8}{3}\pi R^3=\frac{\pi r_2^2(h+2R)-\pi r_1^2h}{3}$ which yields $8R^3=r_2^2(h+2R)-r_1^2h$.

The small right triangle on top is similar to the big right triangle of the entire big cone. So $\frac{r_1}{r_2}=\frac{h}{h+2R}\implies h+2R=\frac{r_2h}{r_1}$.

Substituting yields $8R^3=\frac{h(r_2^3-r_1^3)}{r_1}$.

Substituting $R=\sqrt{r_2r_1}$ and $h=\frac{2Rr_1}{r_2-r_1}$ yields $8(r_2r_1)^{\frac{3}{2}}=2(r_2r_1)^{\frac{1}{2}}(r_2^2+r_2r_1+r_1^2)$ which yields $r_2^2-3r_2r_1+r_1^2=0$.

Solving in $r_2$ yields $r_2=\frac{3r_1\pm r_1\sqrt{5}}{2}$ so $\frac{r_2}{r_1}=\frac{3+\sqrt{5}}{2}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_23&oldid=76336"