Difference between revisions of "2014 AMC 12B Problems/Problem 19"
(Added templates) |
m |
||
Line 86: | Line 86: | ||
so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E</cmath> | so <cmath>r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E</cmath> | ||
+ | == 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}} |
Revision as of 13:25, 22 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?
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.