Difference between revisions of "2003 AIME II Problems/Problem 5"

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(Video Solution by Sal Khan)
 
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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, $V=\dfrac{6^2\cdot 12\pi}{2}=\boxed{216\pi}\$.
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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}=216\pi</math>, so <math>n=\boxed{216}</math>.
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== Video Solution by Sal Khan ==
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Part 1: https://www.youtube.com/watch?v=EzE53aPGbrQ&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=5
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Part 2: https://www.youtube.com/watch?v=yXg5CYuCcU4&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=25
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- AMBRIGGS
  
 
== 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}}
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[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:41, 30 July 2022

Problem

A cylindrical log has diameter $12$ 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 $45^\circ$ 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 $n\pi$, where n is a positive integer. Find $n$.

Solution

The volume of the wedge is half the volume of a cylinder with height $12$ and radius $6$. (Imagine taking another identical wedge and sticking it to the existing one). Thus, $V=\dfrac{6^2\cdot 12\pi}{2}=216\pi$, so $n=\boxed{216}$.

Video Solution by Sal Khan

Part 1: https://www.youtube.com/watch?v=EzE53aPGbrQ&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=5

Part 2: https://www.youtube.com/watch?v=yXg5CYuCcU4&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=25

- AMBRIGGS

See also

2003 AIME II (ProblemsAnswer KeyResources)
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

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