Difference between revisions of "2014 AMC 12B Problems/Problem 19"
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− | ==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? | ||
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<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) === | |
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− | ==Solution 2(ADD DIAGRAM)== | ||
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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> | ||
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~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}} |
Latest revision as of 13:16, 19 January 2021
Contents
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?
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
then using the Pythagorean theorem we have: which is equivalent to: subtracting from both sides solving for s we get: next we can find the area of the frustum and of the sphere and we know so we can solve for using we get: using we get so we have: dividing by we get which is equivalent to so
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 This implies that the angle from the center of the sphere to a point on the circumference of the top circle is Hence the bottom radius is and the top radius is This means that the radio between the bottom radius and top radius is Using the frustum volume formula, we find that the are of this figure is We can equate this to Simplifying, we are left with a quadratic conveniently in The quadratic is This gives us
~NeeNeeMath
Video Solution by icematrix
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 |
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