Difference between revisions of "2014 AMC 10B Problems/Problem 23"
(→See Also) |
(fixed diagram) |
||
(43 intermediate revisions by 13 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. | + | 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 2 | + | 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 73: | 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 |
− | |||
− | |||
<cmath>s=\sqrt{r}.</cmath> | <cmath>s=\sqrt{r}.</cmath> | ||
− | Next, we can find the volume of the frustum and | + | 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_{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_{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_{sphere}=\dfrac{ | + | Using <math>V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}</math>, we get |
− | <cmath>V_{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: | ||
<cmath>\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.</cmath> | <cmath>\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.</cmath> | ||
Line 88: | Line 87: | ||
<cmath>r^2+r+1=4r</cmath> | <cmath>r^2+r+1=4r</cmath> | ||
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\cdot1\cdot1}}{2\cdot1}</math> | + | by the Quadratic Formula, <math> r=\dfrac{3\pm\sqrt{(-3)^2-4\cdot1\cdot1}}{2\cdot1}</math> |
, so | , so | ||
<cmath>r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}</cmath> | <cmath>r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}</cmath> | ||
Line 94: | Line 93: | ||
==Solution 2== | ==Solution 2== | ||
− | Similar to above, draw a smaller cone top with the base of the smaller circle with radius <math>r_1</math> and height <math>h</math>. The smaller right triangle is similar to the blue highlighted one in Solution 1. Then <math>\frac{r_1}{r_2-r_1}=\frac{h}{2R}</math> where <math>R</math> is the radius of the sphere. Then <math>h=\frac{2Rr_1}{r_2-r_1}</math>. | + | Similar to above, draw a smaller cone top with the base of the smaller circle with radius <math>r_1</math> and height <math>h</math>. The smaller right triangle is similar to the blue highlighted one in Solution <math>1</math>. Then <math>\frac{r_1}{r_2-r_1}=\frac{h}{2R}</math> where <math>R</math> is the radius of the sphere. Then <math>h=\frac{2Rr_1}{r_2-r_1}</math>. |
− | From the Pythagorean theorem on the blue triangle in Solution 1, we get similarly that <math>R=\sqrt{r_2r_1}</math>. | + | From the Pythagorean theorem on the blue triangle in Solution <math>1</math>, we get similarly that <math>R=\sqrt{r_2r_1}</math>. |
From the volume requirements, we get that <math>\frac{8}{3}\pi R^3=\frac{\pi r_2^2(h+2R)-\pi r_1^2h}{3}</math> which yields <math>8R^3=r_2^2(h+2R)-r_1^2h</math>. | From the volume requirements, we get that <math>\frac{8}{3}\pi R^3=\frac{\pi r_2^2(h+2R)-\pi r_1^2h}{3}</math> which yields <math>8R^3=r_2^2(h+2R)-r_1^2h</math>. | ||
Line 107: | Line 106: | ||
Solving in <math>r_2</math> yields <math>r_2=\frac{3r_1\pm r_1\sqrt{5}}{2}</math> so <math>\frac{r_2}{r_1}=\frac{3+\sqrt{5}}{2}</math>. | Solving in <math>r_2</math> yields <math>r_2=\frac{3r_1\pm r_1\sqrt{5}}{2}</math> so <math>\frac{r_2}{r_1}=\frac{3+\sqrt{5}}{2}</math>. | ||
+ | |||
+ | == Solution 3: Another Way of Simplifying the Equation in Solution 2 == | ||
+ | |||
+ | 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\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 | ||
+ | <cmath>V_{large cone}-V_{small cone}=2V_{sphere}</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>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 | ||
+ | <cmath>R^2\left(h+2s\right)+r^2h=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\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{R^3-r^3}{R-r}=4s^2</cmath> | ||
+ | The numerator <math>R^3-r^3</math> can be factored to yield | ||
+ | <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=4\left(\sqrt{Rr}\right)^2</cmath> | ||
+ | <cmath>R^2+Rr+r^2=4Rr</cmath> | ||
+ | <cmath>R^2-3Rr+r^2=0</cmath> | ||
+ | Solving the quadratic to obtain the ratio <math>\frac{R}{r}=\frac{3+\sqrt{5}}{2}</math> <math>\boxed{\textbf{(E)}}</math> | ||
+ | |||
+ | ~ 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
Contents
[hide]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?
Solution 1
First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of and the bottom base has a diameter of .
Then using the Pythagorean theorem we have: , which is equivalent to: . Subtracting from both sides, , and solving for , we end up with Next, we can find the volume of the frustum (truncated cone) and the sphere. Since we know , we can solve for using we get: Using , we get so we have: Dividing by , we get which is equivalent to by the Quadratic Formula, , so
Solution 2
Similar to above, draw a smaller cone top with the base of the smaller circle with radius and height . The smaller right triangle is similar to the blue highlighted one in Solution . Then where is the radius of the sphere. Then .
From the Pythagorean theorem on the blue triangle in Solution , we get similarly that .
From the volume requirements, we get that which yields .
The small right triangle on top is similar to the big right triangle of the entire big cone. So .
Substituting yields .
Substituting and yields which yields .
Solving in yields so .
Solution 3: Another Way of Simplifying the Equation in Solution 2
Similar to the previous solutions, we let , , and be the radii of the larger base, smaller base, and the sphere, respectively. By the Pythagorean Theorem, we have . Let , be the heights of the large and small cones, respectively, then we have and Thus we have the equation Substituting the expressions for , , in order into the equation yields The numerator can be factored to yield Solving the quadratic to obtain the ratio
~ Nafer
Video Solution
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 |