Difference between revisions of "2003 AIME II Problems/Problem 5"
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== Solution == | == Solution == | ||
The volume of the wedge is half the volume of a cylinder with height <math>12</math> and radius <math>6</math>. (Imagine taking another identical wedge and sticking it to the existing one). Thus, <math>V=\dfrac{6^2\cdot 12\pi}{2}=\boxed{216}\pi</math>. | The volume of the wedge is half the volume of a cylinder with height <math>12</math> and radius <math>6</math>. (Imagine taking another identical wedge and sticking it to the existing one). Thus, <math>V=\dfrac{6^2\cdot 12\pi}{2}=\boxed{216}\pi</math>. | ||
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| + | I'm Cool! | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=4|num-a=6}} | {{AIME box|year=2003|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:43, 12 May 2017
Problem
A cylindrical log has diameter 
inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a 
angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as 
, where n is a positive integer. Find 
.
Solution
The volume of the wedge is half the volume of a cylinder with height and radius 
. (Imagine taking another identical wedge and sticking it to the existing one). Thus, 
.
I'm Cool!
See also
| 2003 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
