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 12B Problems/Problem 19"
m (added solutions header)
Line 1: Line 1:
−
==Problem==
+
== 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?
 
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>
 
<asy>
 
real r=(3+sqrt(5))/2;
 
real r=(3+sqrt(5))/2;
Line 28: Line 29:
 
draw(sfront,gray(0.5));
 
draw(sfront,gray(0.5));
 
draw(base,gray(0.9));
 
draw(base,gray(0.9));
−
draw(surface(sph),gray(0.4));</asy>
+
draw(surface(sph),gray(0.4));
 +
</asy>
 +
 
 
<math>\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</math>
 
<math>\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</math>
  
−
==Solution==
+
== Solutions ==
 +
=== Solution 1 ===
 
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
 
let the top base equal 2 and the bottom base to be equal to 2r
Line 86: Line 90:
 
so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</cmath>
 
so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</cmath>
  
−
 
+
=== Solution 2(ADD DIAGRAM) ===
−
 
 
−
 
 
−
 
 
−
==Solution 2(ADD DIAGRAM)==
 
−
 
 
 
Let's once again look at the cross section of the frustum. Let the angle from the center of the sphere to a point on the circumference of the bottom circle be <math>\theta.</math> This implies that the angle from the center of the sphere to a point on the circumference of the top circle is <math>90 - \theta.</math> Hence the bottom radius is <math>r\tan{\theta}</math> and the top radius is <math>\frac {r}{\tan {\theta}}.</math> This means that the radio between the bottom radius and top radius is <math>(\tan {\theta})^2.</math> Using the frustum volume formula, we find that the are of this figure is <math>\frac{2\pi r}{3}(r^2(\tan {\theta})^2 + r^2 + \frac {r^2} {(\tan {\theta})^2}).</math> We can equate this to <math>\frac {8\pi*r^3} 3.</math> Simplifying, we are left with a quadratic conveniently in <math>(\tan {\theta})^2.</math> The quadratic is <math>(\tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.</math> This gives us <math>(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</math>
 
Let's once again look at the cross section of the frustum. Let the angle from the center of the sphere to a point on the circumference of the bottom circle be <math>\theta.</math> This implies that the angle from the center of the sphere to a point on the circumference of the top circle is <math>90 - \theta.</math> Hence the bottom radius is <math>r\tan{\theta}</math> and the top radius is <math>\frac {r}{\tan {\theta}}.</math> This means that the radio between the bottom radius and top radius is <math>(\tan {\theta})^2.</math> Using the frustum volume formula, we find that the are of this figure is <math>\frac{2\pi r}{3}(r^2(\tan {\theta})^2 + r^2 + \frac {r^2} {(\tan {\theta})^2}).</math> We can equate this to <math>\frac {8\pi*r^3} 3.</math> Simplifying, we are left with a quadratic conveniently in <math>(\tan {\theta})^2.</math> The quadratic is <math>(\tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.</math> This gives us <math>(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</math>
−
 
  
 
~NeeNeeMath
 
~NeeNeeMath
  
−
==Video Solution by icematrix==
+
== Video Solution by icematrix ==
 
https://youtu.be/3C5AYs7GoF4
 
https://youtu.be/3C5AYs7GoF4
 +
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2014|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:16, 19 January 2021

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]

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

Solutions

Solution 1

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 section 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

Video Solution by icematrix

https://youtu.be/3C5AYs7GoF4

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

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_12B_Problems/Problem_19&oldid=142767"