Difference between revisions of "2014 AMC 10B Problems/Problem 23"
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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 | |
<asy> | <asy> | ||
size(7cm); | size(7cm); | ||
Line 56: | Line 56: | ||
Y = (0,2*s); | Y = (0,2*s); | ||
draw(X--Y); | draw(X--Y); | ||
− | label("$r$",(X+B)/2,S); | + | label("$r-1$",(X+B)/2,S); |
label("$1$",(Y+C)/2,N); | label("$1$",(Y+C)/2,N); | ||
label("$s$",(O+Y)/2,W); | label("$s$",(O+Y)/2,W); | ||
Line 67: | Line 67: | ||
label("$r$",(B+P)/2,NE); | label("$r$",(B+P)/2,NE); | ||
</asy> | </asy> | ||
+ | then using the Pythagorean theorem we have: | ||
+ | <math>(r+1)^2=(2s)^2+(r-1)^2</math> | ||
+ | which is equivalent to: | ||
+ | <math>r^2+2r+1=4s^2+r^2-2r+1</math> | ||
+ | subtracting <math>r^2-2r+1</math> from both sides | ||
+ | <math>4r=4s^2</math> | ||
+ | solving for s we get: | ||
+ | <cmath>s=\sqrt{r}</cmath> | ||
+ | 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> | ||
+ | using <math>V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math> | ||
+ | we get: | ||
+ | <cmath>V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)</cmath> | ||
+ | using <math>V_{sphere}=\dfrac{4r^{3}\pi}{3}</math> | ||
+ | we get | ||
+ | <cmath>V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath> | ||
+ | so we have: | ||
+ | <cmath>\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}</cmath> | ||
+ | dividing by <math>\frac{2\pi*\sqrt{r}}{3}</math> | ||
+ | we get <cmath>r^2+r+1=4r</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> | ||
+ | so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow D</cmath> | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:01, 20 February 2014
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?
(Diagram edited from copeland's diagram)
Solution
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
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 |
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