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"

m (→‎Solution 1)
(fixed diagram)
 
(8 intermediate revisions by 7 users not shown)
Line 15: Line 15:
 
currentlight=(10,10,5);
 
currentlight=(10,10,5);
 
revolution sph=sphere((0,0,Fht/2),Fht/2);
 
revolution sph=sphere((0,0,Fht/2),Fht/2);
−
draw(surface(sph),green+white+opacity(0.5));
+
//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) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));}
 
triple f(pair t) {
 
triple f(pair t) {
Line 28: Line 28:
 
surface sfront=surface(f,(7pi/4,0),(11pi/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);
 
surface base = surface(g,(0,0),(2pi,Brad),80,2);
−
draw(sback,gray(0.9));
+
draw(sback,gray(0.3));
 
draw(sfront,gray(0.5));
 
draw(sfront,gray(0.5));
 
draw(base,gray(0.9));
 
draw(base,gray(0.9));
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 have a diameter of <math>2</math>, and the bottom base have a diameter of <math>2r</math>.
+
Let the top base have a diameter of <math>2</math> and the bottom base has a diameter of <math>2r</math>.
 +
 
 
<asy>
 
<asy>
 
size(7cm);
 
size(7cm);
Line 67: Line 68:
 
label("$r$",(B+P)/2,NE);
 
label("$r$",(B+P)/2,NE);
 
</asy>
 
</asy>
−
(If you don't get why the blue hypotenuse is <math>1+r</math>, reference this link: https://www.ck12.org/geometry/Tangent-Lines/lesson/Tangent-Lines-GEOM/)
 
−
 
  
 
Then using the Pythagorean theorem we have:
 
Then using the Pythagorean theorem we have:
Line 75: Line 74:
 
<math>r^2+2r+1=4s^2+r^2-2r+1</math>.
 
<math>r^2+2r+1=4s^2+r^2-2r+1</math>.
 
Subtracting <math>r^2-2r+1</math> from both sides,
 
Subtracting <math>r^2-2r+1</math> from both sides,
−
<math>4r=4s^2</math>
+
<math>4r=4s^2</math>, and solving for <math>s</math>, we end up with
−
 
 
−
Solving for <math>s</math>, we end up with
 
 
<cmath>s=\sqrt{r}.</cmath>
 
<cmath>s=\sqrt{r}.</cmath>
−
Next, we can find the volume of the frustum (truncated cone) and of the sphere. Since we know <math>V_{\text{frustum}}=2V_{\text{sphere}}</math>, we can solve for <math>s</math>
+
Next, we can find the volume of the frustum (truncated cone) and the sphere. Since we know <math>V_{\text{frustum}}=2V_{\text{sphere}}</math>, we can solve for <math>s</math>
 
using <math>V_{\text{frustum}}=\frac{\pi h}{3}(R^2+r^2+Rr)</math>
 
using <math>V_{\text{frustum}}=\frac{\pi h}{3}(R^2+r^2+Rr)</math>
 
we get:
 
we get:
 
<cmath>V_{\text{frustum}}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)</cmath>
 
<cmath>V_{\text{frustum}}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)</cmath>
−
Using  <math>V_{\text{sphere}}=\dfrac{4r^{3}\pi}{3}</math>, we get  
+
Using  <math>V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}</math>, we get  
 
<cmath>V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
 
<cmath>V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath>
 
so we have:
 
so we have:
Line 113: Line 110:
  
 
Similar to the previous solutions, we let <math>R</math>, <math>r</math>, and <math>s</math> be the radii of the larger base, smaller base, and the sphere, respectively. By the Pythagorean Theorem, we have <math>s=\sqrt{Rr}</math>. Let <math>H</math>, <math>h</math> be the heights of the large and small cones, respectively, then we have <math>H=h+2s</math> and
 
Similar to the previous solutions, we let <math>R</math>, <math>r</math>, and <math>s</math> be the radii of the larger base, smaller base, and the sphere, respectively. By the Pythagorean Theorem, we have <math>s=\sqrt{Rr}</math>. Let <math>H</math>, <math>h</math> be the heights of the large and small cones, respectively, then we have <math>H=h+2s</math> and
−
<cmath>h=2s(\frac{r}{R-r})==\frac{2sr}{R-r}=\frac{2(\sqrt{Rr})r}{R-r}</cmath>  
+
<cmath>h=2s\left(\frac{r}{R-r}\right)==\frac{2sr}{R-r}=\frac{2\left(\sqrt{Rr}\right)r}{R-r}</cmath>  
 
Thus we have the equation  
 
Thus we have the equation  
 
<cmath>V_{large cone}-V_{small cone}=2V_{sphere}</cmath>
 
<cmath>V_{large cone}-V_{small cone}=2V_{sphere}</cmath>
−
<cmath>\frac{\pi R^2H}{3}-\frac{\pi r^2h}{3}=2(\frac{4\pi s^3}{3})</cmath>
+
<cmath>\frac{\pi R^2H}{3}-\frac{\pi r^2h}{3}=2\left(\frac{4\pi s^3}{3}\right)</cmath>
 
<cmath>\pi R^2H-\pi r^2h=8\pi s^3</cmath>
 
<cmath>\pi R^2H-\pi r^2h=8\pi s^3</cmath>
 
<cmath>R^2H-r^2h=8s^3</cmath>
 
<cmath>R^2H-r^2h=8s^3</cmath>
 
Substituting the expressions for <math>H</math>, <math>h</math>, <math>s</math> in order into the equation yields
 
Substituting the expressions for <math>H</math>, <math>h</math>, <math>s</math> in order into the equation yields
−
<cmath>R^2(h+2s)+r^2h=8s^3</cmath>
+
<cmath>R^2\left(h+2s\right)+r^2h=8s^3</cmath>
−
<cmath>R^2((\frac{2sr}{R-r})+2s)+r^2(\frac{2sr}{R-r})=8s^3</cmath>
+
<cmath>R^2\left(\left(\frac{2sr}{R-r}\right)+2s\right)+r^2\left(\frac{2sr}{R-r}\right)=8s^3</cmath>
−
<cmath>R^2(\frac{2sR}{R-r})+r^2(\frac{2sr}{R-r})=8s^3</cmath>
+
<cmath>R^2\left(\frac{2sR}{R-r}\right)+r^2\left(\frac{2sr}{R-r}\right)=8s^3</cmath>
 
<cmath>\frac{2sR^3-2sr^3}{R-r}=8s^3</cmath>
 
<cmath>\frac{2sR^3-2sr^3}{R-r}=8s^3</cmath>
 
<cmath>\frac{R^3-r^3}{R-r}=4s^2</cmath>
 
<cmath>\frac{R^3-r^3}{R-r}=4s^2</cmath>
 
The numerator <math>R^3-r^3</math> can be factored to yield
 
The numerator <math>R^3-r^3</math> can be factored to yield
−
<cmath>\frac{(R-r)(R^2+Rr+r^2)}{R-r}=4s^2</cmath>
+
<cmath>\frac{\left(R-r\right)\left(R^2+Rr+r^2\right)}{R-r}=4s^2</cmath>
 
<cmath>R^2+Rr+r^2{R-r}=4s^2</cmath>
 
<cmath>R^2+Rr+r^2{R-r}=4s^2</cmath>
−
<cmath>R^2+Rr+r^2=4(\sqrt{Rr})^2</cmath>
+
<cmath>R^2+Rr+r^2=4\left(\sqrt{Rr}\right)^2</cmath>
 
<cmath>R^2+Rr+r^2=4Rr</cmath>
 
<cmath>R^2+Rr+r^2=4Rr</cmath>
 
<cmath>R^2-3Rr+r^2=0</cmath>
 
<cmath>R^2-3Rr+r^2=0</cmath>
Line 134: Line 131:
  
 
~ Nafer
 
~ Nafer
 +
 +
==Video Solution==
 +
https://youtu.be/3C5AYs7GoF4
  
 
==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}}

Latest revision as of 18:15, 11 August 2023

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(0.3)); 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 has 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("$r-1$",(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]

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, $4r=4s^2$, and solving for $s$, we end up with \[s=\sqrt{r}.\] Next, we can find the volume of the frustum (truncated cone) and the sphere. Since we know $V_{\text{frustum}}=2V_{\text{sphere}}$, we can solve for $s$ using $V_{\text{frustum}}=\frac{\pi h}{3}(R^2+r^2+Rr)$ we get: \[V_{\text{frustum}}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)\] Using $V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}$, we get \[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\] so we have: \[\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.\] Dividing by $\frac{2\pi\sqrt{r}}{3}$, we get \[r^2+r+1=4r\] which is equivalent to \[r^2-3r+1=0\] by the Quadratic Formula, $r=\dfrac{3\pm\sqrt{(-3)^2-4\cdot1\cdot1}}{2\cdot1}$ , 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 $\frac{r_2}{r_1}=\frac{3+\sqrt{5}}{2}$.

Solution 3: Another Way of Simplifying the Equation in Solution 2

Similar to the previous solutions, we let $R$, $r$, and $s$ be the radii of the larger base, smaller base, and the sphere, respectively. By the Pythagorean Theorem, we have $s=\sqrt{Rr}$. Let $H$, $h$ be the heights of the large and small cones, respectively, then we have $H=h+2s$ and \[h=2s\left(\frac{r}{R-r}\right)==\frac{2sr}{R-r}=\frac{2\left(\sqrt{Rr}\right)r}{R-r}\] Thus we have the equation \[V_{large cone}-V_{small cone}=2V_{sphere}\] \[\frac{\pi R^2H}{3}-\frac{\pi r^2h}{3}=2\left(\frac{4\pi s^3}{3}\right)\] \[\pi R^2H-\pi r^2h=8\pi s^3\] \[R^2H-r^2h=8s^3\] Substituting the expressions for $H$, $h$, $s$ in order into the equation yields \[R^2\left(h+2s\right)+r^2h=8s^3\] \[R^2\left(\left(\frac{2sr}{R-r}\right)+2s\right)+r^2\left(\frac{2sr}{R-r}\right)=8s^3\] \[R^2\left(\frac{2sR}{R-r}\right)+r^2\left(\frac{2sr}{R-r}\right)=8s^3\] \[\frac{2sR^3-2sr^3}{R-r}=8s^3\] \[\frac{R^3-r^3}{R-r}=4s^2\] The numerator $R^3-r^3$ can be factored to yield \[\frac{\left(R-r\right)\left(R^2+Rr+r^2\right)}{R-r}=4s^2\] \[R^2+Rr+r^2{R-r}=4s^2\] \[R^2+Rr+r^2=4\left(\sqrt{Rr}\right)^2\] \[R^2+Rr+r^2=4Rr\] \[R^2-3Rr+r^2=0\] Solving the quadratic to obtain the ratio $\frac{R}{r}=\frac{3+\sqrt{5}}{2}$ $\boxed{\textbf{(E)}}$

~ Nafer

Video Solution

https://youtu.be/3C5AYs7GoF4

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_23&oldid=196812"