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Difference between revisions of "2014 AMC 12B Problems/Problem 19"

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==Problem==
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== 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;
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draw(sfront,gray(0.5));
 
draw(sfront,gray(0.5));
 
draw(base,gray(0.9));
 
draw(base,gray(0.9));
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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>
  
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==Solution==
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== 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
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solving for s we get:
 
solving for s we get:
 
<cmath>s=\sqrt{r}</cmath>
 
<cmath>s=\sqrt{r}</cmath>
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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>
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next we can find the volume of the frustum and of the sphere and we know <math>V_{\text{frustum}}=2V_{\text{sphere}}</math> so we can solve for <math>s</math>
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using <math>V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math>
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using <math>V_{\text{frustum}}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math>
 
we get:
 
we get:
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<cmath>V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)</cmath>
+
<cmath>V_{\text{frustum}}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)</cmath>
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using  <math>V_{sphere}=\dfrac{4r^{3}\pi}{3}</math>
+
using  <math>V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}</math>
 
we get  
 
we get  
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<cmath>V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
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<cmath>V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
 
so we have:
 
so we have:
 
<cmath>\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
 
<cmath>\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
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which is equivalent to <cmath>r^2-3r+1=0</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>
 
<math> r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}</math>
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so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E</cmath>
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so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</cmath>
 +
 
 +
=== 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>
 +
 
 +
~NeeNeeMath
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 +
=== Remark ===
 +
For problems that involve a circle inscribed into an isosceles trapezoid, the following fact is very useful. If we let the bases be <math>x</math> and <math>y</math>, and the height be <math>h</math>, then <math>h = \sqrt{xy}</math>. Then, we could solve from solution 1 directly knowing the radius in terms of the base. By letting the upper base be <math>1</math> and the lower base be <math>x</math>, we can find <math>r</math> in terms of <math>x</math>, and solve like in solution one.
 +
~Puck_0
 +
 
 +
== Video Solution by icematrix ==
 +
https://youtu.be/3C5AYs7GoF4
  
 +
== 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}}
 +
{{AMC10 box|year=2014|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:08, 20 October 2024

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 volume 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 $V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}$ 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

Remark

For problems that involve a circle inscribed into an isosceles trapezoid, the following fact is very useful. If we let the bases be $x$ and $y$, and the height be $h$, then $h = \sqrt{xy}$. Then, we could solve from solution 1 directly knowing the radius in terms of the base. By letting the upper base be $1$ and the lower base be $x$, we can find $r$ in terms of $x$, and solve like in solution one. ~Puck_0

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
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

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