Difference between revisions of "2014 AMC 12B Problems/Problem 19"
m (→Solution 1) |
|||
(11 intermediate revisions by 7 users not shown) | |||
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 71: | Line 75: | ||
solving for s we get: | solving for s we get: | ||
<cmath>s=\sqrt{r}</cmath> | <cmath>s=\sqrt{r}</cmath> | ||
− | next we can find the | + | 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> |
− | 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*2\sqrt{r}}{3}(r^2+r+1)</cmath> | + | <cmath>V_{\text{frustum}}=\frac{\pi*2\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 | 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*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> | ||
Line 85: | Line 89: | ||
<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> | ||
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) === | ||
+ | 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 | ||
+ | |||
+ | == Video Solution by icematrix == | ||
+ | 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}} | ||
+ | {{AMC10 box|year=2014|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:38, 15 June 2024
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 volume 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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.