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

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

(→‎Solution)
(→‎Solution)
Line 38: Line 38:
  
 
First, we draw the vertical cross-section passing through the middle of the frustum.
 
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>
 
<asy>
 
size(7cm);
 
size(7cm);
Line 56: Line 56:
 
Y = (0,2*s);
 
Y = (0,2*s);
 
draw(X--Y);
 
draw(X--Y);
−
label("$r$",(X+B)/2,S);
+
label("$r-1$",(X+B)/2,S);
 
label("$1$",(Y+C)/2,N);
 
label("$1$",(Y+C)/2,N);
 
label("$s$",(O+Y)/2,W);
 
label("$s$",(O+Y)/2,W);
Line 67: Line 67:
 
label("$r$",(B+P)/2,NE);
 
label("$r$",(B+P)/2,NE);
 
</asy>
 
</asy>
 +
then using the Pythagorean theorem we have:
 +
<math>(r+1)^2=(2s)^2+(r-1)^2</math>
 +
which is equivalent to:
 +
<math>r^2+2r+1=4s^2+r^2-2r+1</math>
 +
subtracting <math>r^2-2r+1</math> from both sides
 +
<math>4r=4s^2</math>
 +
solving for s we get:
 +
<cmath>s=\sqrt{r}</cmath>
 +
next we can find the area of the frustum and of the sphere and we know <math>V_{frustum}=2V_{sphere}</math> so we can solve for <math>s</math>
 +
using <math>V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math>
 +
we get:
 +
<cmath>V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)</cmath>
 +
using  <math>V_{sphere}=\dfrac{4r^{3}\pi}{3}</math>
 +
we get
 +
<cmath>V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
 +
so we have:
 +
<cmath>\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
 +
dividing by <math>\frac{2\pi*\sqrt{r}}{3}</math>
 +
we get <cmath>r^2+r+1=4r</cmath>
 +
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>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2014|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:01, 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 \[r=\dfrac{3+\sqrt{5}}{2}\longrightarrow D\]

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=60236"